MARC2 Variant Comparison March 26, 2026

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **The Dark Sea (`.`):** The background/empty space of the grid.
    *   **The Sail:** A hollow shape made of a single color.
    *   **The Mast:** The rightmost vertical edge of the sail.
    *   **The Boom:** The bottom horizontal edge of the sail.
    *   **The Canvas:** The parts of the sail that are not part of the mast or the boom (the "loose" parts).

2.  **Identify the AFFORDANCES**:
    *   The **Mast** and **Boom** are "rigid": they cannot move or change position.
    *   The **Canvas** is "slack": it can be pushed one step to the right (toward the mast).

3.  **Identify the TRANSFORMATION RULE**:
    *   For each colored shape (sail):
        *   Identify the rightmost column of the shape (the mast) and the bottommost row of the shape (the boom).
        *   Every pixel of the shape that is **not** in the rightmost column of that shape AND **not** in the bottommost row of that shape must be shifted one cell to the right.
        *   If a pixel moves into a position already occupied by the mast or boom, it effectively "snugs" against it.
        *   Essentially, for every row $r$ that is not the bottom row, find the leftmost pixel at $(r, c)$ and move it to $(r, c+1)$, provided $(r, c+1)$ is not part of the mast. More simply: for every row $r$ except the bottom row, all pixels $(r, c)$ that are not the rightmost pixel of that shape move to $(r, c+1)$.

### Applying the Rule to Test Input

**Input Analysis:**
The shape is Yellow (`Y`).
Rows and columns (0-indexed):
Row 1: `. Y Y Y Y Y Y . . .` (Cols 1-6 are Y)
Row 2: `. Y . . . . . Y . .` (Cols 1 and 7 are Y) -> *Wait, let's re-examine the input carefully.*

Input Grid:
Row 0: `. . . . . . . . . .`
Row 1: `. Y Y Y Y Y Y . . .` (Cols 1, 2, 3, 4, 5, 6)
Row 2: `. Y . . . . . Y . .` (Cols 1, 7)
Row 3: `. . Y . . . . . Y .` (Cols 2, 8)
Row 4: `. . . Y . . . . . Y` (Cols 3, 9)
Row 5: `. . . . Y Y Y Y Y Y` (Cols 4, 5, 6, 7, 8, 9)
Row 6-9: `.`

**Identify Mast and Boom for the Yellow Sail:**
*   **Bottom row (Boom):** Row 5.
*   **Rightmost column (Mast):** Column 9 (from Row 2, 3, 4, 5). Note: Row 1 ends at Col 6.
*   Wait, let's look at the shape as a single entity. The rightmost edge of the *entire* shape is Column 9.
*   The bottom edge is Row 5.

**Transformation (Shift non-mast, non-boom pixels right):**
*   **Row 1:** Pixels at (1,1), (1,2), (1,3), (1,4), (1,5), (1,6).
    *   Is Col 9 the mast? Yes. Is Row 5 the boom? Yes.
    *   Row 1 is not the boom. The rightmost pixel in Row 1 is at Col 6.
    *   The rule says: "Every loose piece of canvas not anchored to the mast or boom gets pushed one step rightward toward the mast."
    *   In Row 1: (1,1) $\to$ (1,2), (1,2) $\to$ (1,3), (1,3) $\to$ (1,4), (1,4) $\to$ (1,5), (1,5) $\to$ (1,6), (1,6) $\to$ (1,7).
    *   New Row 1: `. . Y Y Y Y Y Y . .`
*   **Row 2:** Pixels at (2,1) and (2,7).
    *   (2,1) is not mast/boom $\to$ (2,2).
    *   (2,7) is not mast/boom $\to$ (2,8).
    *   New Row 2: `. . Y . . . . . Y .`
*   **Row 3:** Pixels at (3,2) and (3,8).
    *   (3,2) is not mast/boom $\to$ (3,3).
    *   (3,8) is not mast/boom $\to$ (3,9).
    *   New Row 3: `. . . Y . . . . . Y`
*   **Row 4:** Pixels at (4,3) and (4,9).
    *   (4,3) is not mast/boom $\to$ (4,4).
    *   (4,9) is the mast $\to$ stays at (4,9).
    *   New Row 4: `. . . . Y . . . . Y`
*   **Row 5 (Boom):** Stays the same.
    *   New Row 5: `. . . . Y Y Y Y Y Y`

**Wait, let's re-verify Example 1 to ensure the logic is correct.**
Example 1 Input:
Row 1: `. A A A A A . . .` (Cols 1-5)
Row 2: `. A . . . . A . .` (Cols 1, 6)
Row 3: `. . A . . . . A .` (Cols 2, 7)
Row 4: `. . . A . . . . A` (Cols 3, 8)
Row 5: `. . . . A A A A A` (Cols 4-8)

Example 1 Output:
Row 1: `. . A A A A A . .` (Cols 2-6)
Row 2: `. . A . . . . A .` (Cols 2, 7)
Row 3: `. . . A . . . . A` (Cols 3, 8)
Row 4: `. . . . A . . . A` (Cols 4, 8) -> *Wait, let's look at Row 4 again.*
Input Row 4: `. . . A . . . . A` (Cols 3, 8). Output Row 4: `. . . . A . . . A` (Cols 4, 8).
The pixel at (4,3) moved to (4,4). The pixel at (4,8) is the mast, so it stayed.

**Applying this logic to Test Input:**
Input:
R1: `. Y Y Y Y Y Y . . .` (1,2,3,4,5,6)
R2: `. Y . . . . . Y . .` (1, 7)
R3: `. . Y . . . . . Y .` (2, 8)
R4: `. . . Y . . . . . Y` (3, 9)
R5: `. . . . Y Y Y Y Y Y` (4,5,6,7,8,9)

Transformation:
R1: (1,1) $\to$ (1,2), (1,2) $\to$ (1,3), (1,3) $\to$ (1,4), (1,4) $\to$ (1,5), (1,5) $\to$ (1,6), (1,6) $\to$ (1,7).
R1 becomes: `. . Y Y Y Y Y Y . .`
R2: (2,1) $\to$ (2,2), (2,7) $\to$ (2,8).
R2 becomes: `. . Y . . . . . Y .`
R3: (3,2) $\to$ (3,3), (3,8) $\to$ (3,9).
R3 becomes: `. . . Y . . . . . Y`
R4: (4,3) $\to$ (4,4), (4,9) is mast $\to$ (4,9).
R4 becomes: `. . . . Y . . . . Y`
R5: Boom, stays `. . . . Y Y Y Y Y Y`

**Final Grid Construction:**
R0: `. . . . . . . . . .`
R1: `. . Y Y Y Y Y Y . .`
R2: `. . Y . . . . . Y .`
R3: `. . . Y . . . . . Y`
R4: `. . . . Y . . . . Y`
R5: `. . . . Y Y Y Y Y Y`
R6-9: `. . . 

... [223 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Black pixels (`.`)**: The cytoplasm (background).
    *   **Colored pixels**: The "cellular membrane."
    *   **The "Base"**: The bottom row of the shape (the horizontal line at the bottom of the shape).
    *   **The "Rightmost column of attachment points"**: The rightmost vertical/slanted edge of the shape.
    *   **The "Loose interior"**: The pixels forming the left/slanted side of the shape that are not part of the base or the rightmost edge.

2.  **Identify the AFFORDANCES**:
    *   The base (bottom row) and the rightmost edge are fixed in place.
    *   The other pixels (the left/slanted side) can be shifted horizontally to the right.
    *   The goal is to make the "cell wall" (the shape) more "upright."

3.  **Identify the TRANSFORMATION RULE**:
    *   The shape is a slanted polygon (a trapezoid-like shape).
    *   The bottom row (the base) and the rightmost column of the shape remain stationary.
    *   Every other pixel in the shape is shifted one position to the right within its respective row.
    *   Looking at Example 1:
        Input Row 1: `. A A A A A . . .` -> Output Row 1: `. . A A A A A . .` (Shifted right by 1)
        Input Row 2: `. A . . . . A . .` -> Output Row 2: `. . A . . . . A .` (Shifted right by 1)
        Input Row 3: `. . A . . . . A .` -> Output Row 3: `. . . A . . . . A .` (Shifted right by 1)
        Input Row 4: `. . . A . . . . A` -> Output Row 4: `. . . . A . . . . A` (Shifted right by 1)
        Input Row 5: `. . . . A A A A A` -> Output Row 5: `. . . . A A A A A` (Bottom row stays fixed)
    *   **Rule**: For each row containing part of the shape, if it is not the bottom-most row of the shape, shift all colored pixels in that row one step to the right. If a pixel is at the rightmost edge of the shape, it stays in its column (effectively, the shape "straightens" by moving the left side toward the right side).

### Applying the Rule to Test Input

**Input Analysis**:
Row 1: `. Y Y Y Y Y Y . . .` (Top edge)
Row 2: `. Y . . . . . Y . .` (Left edge and right edge)
Row 3: `. . Y . . . . . Y .` (Left edge and right edge)
Row 4: `. . . Y . . . . . Y` (Left edge and right edge)
Row 5: `. . . . Y Y Y Y Y Y` (Bottom edge/Base)

**Transformation**:
*   Row 1: Shift right by 1 $\rightarrow$ `. . Y Y Y Y Y Y . .`
*   Row 2: Shift right by 1 $\rightarrow$ `. . Y . . . . . Y .`
*   Row 3: Shift right by 1 $\rightarrow$ `. . . Y . . . . . Y`
*   Row 4: Shift right by 1 $\rightarrow$ `. . . . Y . . . . . Y` (Wait, let's look at the rightmost column. In Row 4, the Y is at index 9. If we shift the left Y, it moves to index 4. The right Y is at index 9. The rule says the rightmost column remains fixed. So the Y at index 9 stays at index 9.)
*   Row 5: Base (Fixed) $\rightarrow$ `. . . . Y Y Y Y Y Y`

**Step-by-step Row Construction**:
Input Row 1 (index 1-6): `. Y Y Y Y Y Y . . .` $\rightarrow$ Output: `. . Y Y Y Y Y Y . .`
Input Row 2 (index 1, 7): `. Y . . . . . Y . .` $\rightarrow$ Output: `. . Y . . . . . Y .`
Input Row 3 (index 2, 8): `. . Y . . . . . Y .` $\rightarrow$ Output: `. . . Y . . . . . Y`
Input Row 4 (index 3, 9): `. . . Y . . . . . Y` $\rightarrow$ Output: `. . . . Y . . . . Y` (Wait, the rightmost Y is at index 9. If we shift the left Y at index 3 to index 4, and the right Y at index 9 stays at index 9, the row becomes `. . . . Y . . . . Y`)
Input Row 5 (index 4-9): `. . . . Y Y Y Y Y Y` $\rightarrow$ Output: `. . . . Y Y Y Y Y Y`

Let's re-verify Example 1 with this logic:
Ex 1 Row 1: `. A A A A A . . .` (Indices 1-5) $\rightarrow$ `. . A A A A A . .` (Indices 2-6)
Ex 1 Row 2: `. A . . . . A . .` (Indices 1, 6) $\rightarrow$ `. . A . . . . A .` (Indices 2, 6) - *Note: The rightmost A stayed at index 6.*
Ex 1 Row 3: `. . A . . . . A .` (Indices 2, 7) $\rightarrow$ `. . . A . . . . A .` (Indices 3, 7) - *Note: The rightmost A stayed at index 7.*
Ex 1 Row 4: `. . . A . . . . A` (Indices 3, 8) $\rightarrow$ `. . . . A . . . . A` (Indices 4, 8) - *Note: The rightmost A stayed at index 8.*
Ex 1 Row 5: `. . . . A A A A A` (Indices 4-8) $\rightarrow$ `. . . . A A A A A` (Indices 4-8) - *Fixed.*

**Applying to Test Input**:
Row 1: `. Y Y Y Y Y Y . . .` $\rightarrow$ `. . Y Y Y Y Y Y . .`
Row 2: `. Y . . . . . Y . .` $\rightarrow$ `. . Y . . . . . Y .`
Row 3: `. . Y . . . . . Y .` $\rightarrow$ `. . . Y . . . . . Y`
Row 4: `. . . Y . . . . . Y` $\rightarrow$ `. . . . Y . . . . Y`
Row 5: `. . . . Y Y Y Y Y Y` $\rightarrow$ `. . . . Y Y Y Y Y Y`

Wait, let's check the column indices for Row 4 in Test Input:
`. . . Y . . . . . Y`
0 1 2 3 4 5 6 7 8 9
The Ys are at index 3 and index 9.
Shift index 3 to index 4. Index 9 stays at index 9.
Result: `. . . . Y . . . . Y`

Final Grid Construction:
Row 0: `. . . . . . . . . .`
Row 1: `. . Y Y Y Y Y Y . .`
Row 2: `. . Y . . . . . Y .`
Row 3: `. . . Y . . . . . Y`
Row 4: `. . . . Y . . . . Y`
Row 5: `. . . . Y Y Y Y Y Y`
Row 6: `. . . . . . . . . .`
Row 7: `. .

... [284 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Shapes**: Independent clusters of colored pixels representing "musical notes/shapes".
    *   **Bass Note (Anchor)**: The bottom-rightmost pixel of a shape.
    *   **Rightmost Voice (Anchor)**: The rightmost column of pixels within a shape.
    *   **Other Notes**: All pixels in the shape that are neither the bottom-rightmost pixel nor part of the rightmost vertical/diagonal edge.

2.  **Identify the AFFORDANCES**:
    *   Pixels can be shifted.
    *   The "bass note" (bottom-right corner) and the "rightmost voice" (the right edge) are fixed in place.
    *   Other pixels can move "one position closer to the root tone" (which, based on the metaphor and example, means shifting one unit to the right or up/down to tighten the shape towards the anchor).

3.  **Identify the TRANSFORMATION RULE**:
    *   Looking at Example 1:
        Input shape:
        Row 1: `. A A A A A . . .`
        Row 2: `. A . . . . A . .`
        Row 3: `. . A . . . . A .`
        Row 4: `. . . A . . . . A`
        Row 5: `. . . . A A A A A`
        The bottom-right corner is at (4, 8) [0-indexed]. The rightmost edge is the column at index 8 (for rows 2-5) and the bottom row (row 5).
        The transformation shifts the "top" part of the shape one step to the right.
        Wait, let's re-examine:
        Input Row 1: `. A A A A A . . .` (Indices 1-5)
        Output Row 1: `. . A A A A A . .` (Indices 2-6)
        The shape is shifting right.
        The "bass note" is the bottom-right corner. The "rightmost voice" is the right edge.
        The rule: All pixels *except* the bottom-right corner and the rightmost edge shift one position to the right (towards the right edge).

    *   **Refined Rule**: For each shape, identify the rightmost column that contains pixels. The pixels in that column are fixed. The bottom-rightmost pixel is fixed. All other pixels in the shape shift one position to the right.

### Applying the Rule to Test Input

**Input Analysis**:
Shape is Yellow (Y).
Rows:
0: `. . . . . . . . . .`
1: `. Y Y Y Y Y Y . . .` (Indices 1-6)
2: `. Y . . . . . Y . .` (Indices 1, 7)
3: `. . Y . . . . . Y .` (Indices 2, 8)
4: `. . . Y . . . . . Y` (Indices 3, 9)
5: `. . . . Y Y Y Y Y Y` (Indices 4-9)
6-9: `.`

**Identify Anchors**:
- Rightmost column: Column 9 (contains Y at rows 4 and 5).
- Bottom-right corner: Row 5, Column 9.
- The "rightmost voice" is the vertical edge at Column 9.
- The "bass note" is (5, 9).

**Transformation**:
Shift all pixels that are not in Column 9 one position to the right.
- Row 1: `. Y Y Y Y Y Y . . .` -> Indices 1-6. Shifted: Indices 2-7.
- Row 2: `. Y . . . . . Y . .` -> Index 1. Shifted: Index 2. (Index 7 is not in Col 9, but wait, the rule says "rightmost voice" holds steady. In Row 2, the rightmost pixel is at index 7. Is index 7 part of the "rightmost voice"? The metaphor says "the rightmost voice holds steady". In a shape, the rightmost edge is the voice.
- Let's look at Example 1 again.
  Input Row 1: `. A A A A A . . .` (Indices 1, 2, 3, 4, 5). Rightmost index is 5.
  Output Row 1: `. . A A A A A . .` (Indices 2, 3, 4, 5, 6).
  Wait, the rightmost index in the output is 6. This means the rightmost edge *moved*? No, the metaphor says "the rightmost voice holds steady".
  Let's re-read: "All notes except those in the bass register and the rightmost voice shift one position closer to the root tone."
  In Example 1, the rightmost column of the shape is index 8 (for rows 2, 3, 4) and index 8 (for row 5).
  Wait, the input row 1 has A at 1, 2, 3, 4, 5. The output row 1 has A at 2, 3, 4, 5, 6.
  This means the shape is shifting right.
  If the rightmost edge is fixed, the shape would expand left? No, "shift one position closer to the root tone". If the root is the bottom-right, shifting "closer" means moving right and down.

**Let's re-evaluate Example 1 carefully**:
Input:
R1: `. A A A A A . . .` (1-5)
R2: `. A . . . . A . .` (1, 6)
R3: `. . A . . . . A .` (2, 7)
R4: `. . . A . . . . A` (3, 8)
R5: `. . . . A A A A A` (4-8)

Output:
R1: `. . A A A A A . .` (2-6)
R2: `. . A . . . . A .` (2, 7)
R3: `. . . A . . . . A` (3, 8)
R4: `. . . . A . . . A` (4, 8)
R5: `. . . . A A A A A` (4-8)

Wait, the output R4 is `. . . . A . . . A`. The input R4 was `. . . A . . . . A`.
The pixel at (3, 4) moved to (4, 4).
The pixel at (1, 1) moved to (2, 2)? No.
Let's look at the columns:
Input R1: Col 1, 2, 3, 4, 5.
Output R1: Col 2, 3, 4, 5, 6.
Input R2: Col 1, 6.
Output R2: Col 2, 7.
Input R3: Col 2, 7.
Output R3: Col 3, 8.
Input R4: Col 3, 8.
Output R4: Col 4, 8.
Input R5: Col 4, 5, 6, 7, 8.
Output R5: Col 4, 5, 6, 7, 8.

**The Rule is**:
Every pixel $(r, c)$ in the shape moves to $(r, c+1)$ *unless* it is part of the "rightmost voice" (the rightmost column of the shape) or the "bass note" (the bottom-right corner).
Wait, if we move $(r, c)$ to $(r, c+1)$, the rightmost column $c_{max}$ would move to $c_{max}+1$. But the rule says the rightmost vo

... [4415 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Black (.)**: The background/terrain.
    *   **Colored pixels (Y, M, R, A, etc.)**: Units forming a "slanted quadrilateral" (an echelon formation).
    *   **Base Rank**: The bottom row of the formation.
    *   **Anchor File (Rightmost File)**: The rightmost column of the formation.
    *   **Interior/Trailing Units**: All units that are neither in the bottom row nor the rightmost column of their respective formation.

2.  **Identify the AFFORDANCES**:
    *   **Base Rank and Anchor File**: These are "anchors"; they cannot move.
    *   **Other Units**: These can "advance one position rightward" to close the gap in the echelon.

3.  **Identify the TRANSFORMATION RULE**:
    *   For each distinct colored formation:
        *   Identify the bottom row (base rank) and the rightmost column (anchor file).
        *   For every other colored pixel in the formation, move it one cell to the right.
        *   If a pixel is part of the base rank or the anchor file, it stays in its original position.
        *   Essentially, the "slanted" part of the echelon (the diagonal trailing edge) shifts right to become a vertical line, squaring the shape.

### Applying the Rule to Test Input

**Input Analysis (Yellow Formation):**
The formation is a 6x6 shape (roughly) starting at row 1, col 1.
Rows/Cols (0-indexed):
Row 1: `. Y Y Y Y Y Y . . .` (Cols 1-6)
Row 2: `. Y . . . . . Y . .` (Cols 1 and 7)
Row 3: `. . Y . . . . . Y .` (Cols 2 and 8)
Row 4: `. . . Y . . . . . Y` (Cols 3 and 9)
Row 5: `. . . . Y Y Y Y Y Y` (Cols 4-9)

**Identify Anchors:**
*   **Base Rank (Bottom Row):** Row 5, Columns 4, 5, 6, 7, 8, 9.
*   **Anchor File (Rightmost Column):** Column 9 (for rows 4 and 5) and Column 7 (for row 1 and 2? No, let's look closer).

Wait, let's re-examine the formation structure in the test input:
Row 1: `(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)`
Row 2: `(2,1) (2,7)`
Row 3: `(3,2) (3,8)`
Row 4: `(4,3) (4,9)`
Row 5: `(5,4) (5,5) (5,6) (5,7) (5,8) (5,9)`

**Identify Base Rank and Anchor File:**
*   **Base Rank:** Row 5.
*   **Anchor File:** Column 9 (the rightmost edge of the shape).

**Identify units to move (not in Row 5 and not in Col 9):**
*   (1,1) -> (1,2)
*   (1,2) -> (1,3)
*   (1,3) -> (1,4)
*   (1,4) -> (1,5)
*   (1,5) -> (1,6)
*   (1,6) -> (1,7)
*   (2,1) -> (2,2)
*   (2,7) -> (2,8)
*   (3,2) -> (3,3)
*   (3,8) -> (3,9) -- *Wait, (3,9) is in the anchor file? No, (3,8) is not in Col 9. (3,8) moves to (3,9).*
*   (4,3) -> (4,4)

**Wait, let's re-evaluate the "Rightmost File" logic based on Example 1:**
Example 1 Input:
Row 1: `A A A A A` (Cols 1-5)
Row 2: `A . . . . A` (Cols 1, 6)
Row 3: `. A . . . . A` (Cols 2, 7)
Row 4: `. . A . . . . A` (Cols 3, 8)
Row 5: `. . . A A A A A` (Cols 4-8)

Example 1 Output:
Row 1: `. A A A A A .` (Cols 2-6)
Row 2: `. A . . . . A` (Cols 2, 7)
Row 3: `. . A . . . . A` (Cols 3, 8)
Row 4: `. . . A . . . . A` (Cols 4, 9)
Row 5: `. . . . A A A A A` (Cols 5-9)

In Example 1, the "Anchor File" is the rightmost column of the *entire formation*.
In Test Input:
The rightmost column is Column 9.
The bottom row is Row 5.

**Moving units in Test Input:**
Row 1: `(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)` -> All move right -> `(1,2) (1,3) (1,4) (1,5) (1,6) (1,7)`
Row 2: `(2,1)` moves to `(2,2)`. `(2,7)` moves to `(2,8)`.
Row 3: `(3,2)` moves to `(3,3)`. `(3,8)` moves to `(3,9)`.
Row 4: `(4,3)` moves to `(4,4)`. `(4,9)` is anchor, stays.
Row 5: All stay.

**Constructing Output:**
Row 1: `. . Y Y Y Y Y Y . .`
Row 2: `. . Y . . . . Y Y .` (Wait, (2,7) moves to (2,8). (2,1) moves to (2,2). So (2,2) and (2,8))
Row 3: `. . . Y . . . . Y Y` (Wait, (3,2) moves to (3,3). (3,8) moves to (3,9). So (3,3) and (3,9))
Row 4: `. . . . Y . . . . Y` (Wait, (4,3) moves to (4,4). (4,9) stays. So (4,4) and (4,9))
Row 5: `. . . . Y Y Y Y Y Y`

Let's re-verify Row 1:
Input Row 1: `. Y Y Y Y Y Y . . .` (Cols 1,2,3,4,5,6)
Output Row 1: `. . Y Y Y Y Y Y . .` (Cols 2,3,4,5,6,7)

Let's re-verify Row 2:
Input Row 2: `. Y . . . . . Y . .` (Cols 1, 7)
Output Row 2: `. . Y . . . . . Y .` (Cols 2, 8)

Let's re-verify Row 3:
Input Row 3: `. . Y . . . . . Y .` (Cols 2, 8)
Output Row 3: `. . . Y . . . . . Y` (Cols 3, 9)

Let's re-verify Row 4:
Input Row 4: `. . . Y . . . . . Y` (Cols 3, 9)
Output Row 4: `. . . . Y . . . . Y` (Cols 4, 9)

Let's re-verify Row 5:
Input Row 5: `. . . . Y Y Y Y Y Y` (Cols 4,5,6,7,8,9)
Output Row 5: `. . . . Y Y Y Y Y Y` (Cols 4,5,6,7,8,9)

**Final Grid Assembly:**
Row 0: `. . . . . . . . . .`
Row 1: `. . Y Y Y Y Y Y . .`
Row 2: `. . Y . . . . Y . .` (Wait, (2,7) moves to (2,8). Let's check: (2,7) is col 7. 7+1=8. So (2,8))
Row 3: `. . . Y . . . . . Y` (Wait, (3,2) moves to (3,3). (3,8) moves to (3,9). So (3,3) and (3,9))
Row 4: `. . . . Y . . . . Y` (Wait, (4,3) moves to (4,4). (4,9) stays. So (4,4) and (4,9))
Row 5: `. . . . Y Y Y Y Y Y`

Let's re-check Row 2:
Input: `. Y . . . . . Y . .` (Col 1 and Col 7)
Output: `. . 

... [2789 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Black (.)**: The "soil" or background. It is the empty space.
    *   **Colored pixels (Y, A, etc.)**: The "trellis frame." These pixels form a shape that is currently slanted.
    *   **The "Base Rail"**: The bottom horizontal line of the shape.
    *   **The "Right Post"**: The rightmost vertical/slanted line of the shape.
    *   **The "Joints"**: All other colored pixels that are not part of the base rail or the rightmost vertical line.

2.  **Identify the AFFORDANCES**:
    *   The **Base Rail** and **Right Post** are immovable (anchors).
    *   The **Joints** (the slanted parts) can be moved one space to the right.

3.  **Identify the TRANSFORMATION RULE**:
    *   Identify the shape (the trellis).
    *   Identify the "anchor" points: the bottom row of the shape and the rightmost column of the shape.
    *   Every colored pixel that is *not* part of the bottom row or the rightmost column must be shifted exactly one cell to the right.
    *   Essentially, the "slanted" part of the lattice is being "straightened" by shifting the diagonal elements rightward to align more vertically with the right post.

### Applying the Rule to Test Input

**Input Analysis (10x10):**
Row 1: `. Y Y Y Y Y Y . . .` (Top rail)
Row 2: `. Y . . . . . Y . .` (Left side and Right post)
Row 3: `. . Y . . . . . Y .` (Left side and Right post)
Row 4: `. . . Y . . . . . Y` (Left side and Right post)
Row 5: `. . . . Y Y Y Y Y Y` (Bottom rail)

**Identify Anchors:**
*   **Bottom Rail (Row 5):** `(5,4), (5,5), (5,6), (5,7), (5,8), (5,9)` (using 0-indexing: `(4,4)` to `(4,9)`)
*   **Right Post (Column 9):** `(1,9), (2,9), (3,9), (4,9)` (Wait, let's re-index carefully).

**Re-indexing Test Input (0-indexed):**
Row 0: `. . . . . . . . . .`
Row 1: `. Y Y Y Y Y Y . . .` (Top rail: `(1,1)` to `(1,6)`)
Row 2: `. Y . . . . . Y . .` (Left: `(2,1)`, Right: `(2,7)`)
Row 3: `. . Y . . . . . Y .` (Left: `(3,2)`, Right: `(3,8)`)
Row 4: `. . . Y . . . . . Y` (Left: `(4,3)`, Right: `(4,9)`)
Row 5: `. . . . Y Y Y Y Y Y` (Bottom rail: `(5,4)` to `(5,9)`)
Row 6-9: `.`

**Wait, let's look at the structure again.**
The shape is:
Row 1: `Y Y Y Y Y Y` (Indices 1-6)
Row 2: `Y` (Index 1) ... `Y` (Index 7)
Row 3: `Y` (Index 2) ... `Y` (Index 8)
Row 4: `Y` (Index 3) ... `Y` (Index 9)
Row 5: `Y Y Y Y Y Y` (Indices 4-9)

**Identify Anchors (Immovable):**
*   Bottom Rail: Row 5, columns 4, 5, 6, 7, 8, 9.
*   Right Post: Column 9, rows 4, 5 (and potentially others if they form the vertical line). Looking at the pattern, the rightmost column is the "post". In Row 4, the Y is at index 9. In Row 5, the Y is at index 9.

**Identify Joints (To be moved right):**
*   Top Rail (Row 1): `(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)`.
    *   Is `(1,6)` part of the right post? No, the right post is at index 9.
    *   Is `(1,1)` part of the base? No.
    *   Shift all Row 1 pixels right: `(1,2), (1,3), (1,4), (1,5), (1,6), (1,7)`.
*   Left side (Diagonal):
    *   `(2,1)` -> `(2,2)`
    *   `(3,2)` -> `(3,3)`
    *   `(4,3)` -> `(4,4)`
*   Right side (Diagonal):
    *   `(2,7)` -> `(2,8)`
    *   `(3,8)` -> `(3,9)`
    *   `(4,9)` is part of the right post? Yes, it's the rightmost column.

**Wait, let's re-examine Example 1 to confirm the "Right Post" definition.**
Example 1 Input:
Row 1: `A A A A A` (1-5)
Row 2: `A` (1) ... `A` (6)
Row 3: `A` (2) ... `A` (7)
Row 4: `A` (3) ... `A` (8)
Row 5: `A A A A A` (4-8)

Example 1 Output:
Row 1: `A A A A A` (2-6)
Row 2: `A` (2) ... `A` (6)
Row 3: `A` (3) ... `A` (7)
Row 4: `A` (4) ... `A` (8)
Row 5: `A A A A A` (4-8)

In Example 1, the rightmost column is index 8. The bottom row is index 4.
The pixels moved:
Row 1: `(1,1)-(1,5)` moved to `(1,2)-(1,6)`.
Row 2: `(2,1)` moved to `(2,2)`. `(2,6)` moved to `(2,7)`? No, the output shows `(2,6)` is still there.
Let's look closer at Example 1 Output:
Row 1: `. A A A A A . . .` (Indices 2,3,4,5,6)
Row 2: `. . A . . . . A . .` (Indices 2 and 7)
Row 3: `. . . A . . . . A .` (Indices 3 and 8)
Row 4: `. . . . A . . . . A` (Indices 4 and 9? No, index 8)
Row 5: `. . . . A A A A A` (Indices 4,5,6,7,8)

Wait, the output for Example 1 is:
Row 1: `. . A A A A A . . .` (Indices 2,3,4,5,6,7) - No, that's 6 A's. The input had 5.
Let's re-count Example 1 Input:
Row 1: `. A A A A A . . .` (5 A's)
Row 2: `. A . . . . A . .` (A at 1 and 6)
Row 3: `. . A . . . . A .` (A at 2 and 7)
Row 4: `. . . A . . . . A` (A at 3 and 8)
Row 5: `. . . . A A A A A` (5 A's at 4,5,6,7,8)

Example 1 Output:
Row 1: `. . A A A A A . .` (5 A's at 2,3,4,5,6)
Row 2: `. . A . . . . A .` (A at 2 and 7)
Row 3: `. . . A . . . . A` (A at 3 and 8)
Row 4: `. . . . A . . . . A` (A at 4 and 8)
Row 5: `. . . . A A A A A` (5 A's at 4,5,6,7,8)

**Rule Refined:**
1.  Identify the bottom row (Base Rail) and the rightmost column (Right Post).
2.  Every other pixel `(r, c)` moves to `(r, c+1)`.
3.  If `(r, c+1)` is already part of the anchor, it doesn't "double up" or it just

... [2949 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **The Floor (O - Orange):** The background/empty space.
    *   **The Dancers (Columns of color):** Located in the odd-numbered lanes (columns 1, 3, 5, etc., using 0-indexing: 1, 3, 5...).
    *   **The Costumes (Colors):** The specific color of each dancer's column.
    *   **The Stature (Height):** The number of cells the color column extends upward from the bottom.
    *   **The Shadows (The "opposite" movement):** The metaphor describes a dual shift.

2.  **Identify the AFFORDANCES:**
    *   **Costumes (Colors):** Can be shifted one position to the right (with wrap-around).
    *   **Shadows (Heights):** Can be shifted one position to the left (with wrap-around).
    *   **Lanes:** The odd lanes (1, 3, 5, 7, 9...) are the only ones that contain dancers.

3.  **Identify the TRANSFORMATION RULE:**
    *   **Identify the active lanes:** The columns at indices 1, 3, 5, 7, 9...
    *   **Extract properties:** For each active lane, record its color and its height (number of non-orange cells starting from the bottom).
    *   **Shift Colors (Right):** The color of lane $i$ moves to lane $i+2$. The last lane's color wraps to the first active lane.
    *   **Shift Heights (Left):** The height of lane $i$ moves to lane $i-2$. The first lane's height wraps to the last active lane.
    *   **Reconstruct:** Place the new color at the new lane index, extending it from the bottom up to the new height.

### Applying the Rule to Test Input

**Input Analysis:**
Grid size: 11 columns (0-10) x 11 rows (0-10).
Active lanes (odd indices): 1, 3, 5, 7, 9.

*   **Lane 1:** Color: `.` (black), Height: 10 (rows 1 to 10)
*   **Lane 3:** Color: `R` (red), Height: 1 (row 10)
*   **Lane 5:** Color: `A` (azure), Height: 4 (rows 7 to 10)
*   **Lane 7:** Color: `W` (maroon), Height: 4 (rows 7 to 10)
*   **Lane 9:** Color: `M` (magenta), Height: 5 (rows 6 to 10)

**Transformation:**
*   **Colors (Shift Right):**
    *   Lane 1 gets Color from Lane 9: `M`
    *   Lane 3 gets Color from Lane 1: `.`
    *   Lane 5 gets Color from Lane 3: `R`
    *   Lane 7 gets Color from Lane 5: `A`
    *   Lane 9 gets Color from Lane 7: `W`

*   **Heights (Shift Left):**
    *   Lane 1 gets Height from Lane 3: 1
    *   Lane 3 gets Height from Lane 5: 4
    *   Lane 5 gets Height from Lane 7: 4
    *   Lane 7 gets Height from Lane 9: 5
    *   Lane 9 gets Height from Lane 1: 10

**New Lane Properties:**
*   **Lane 1:** Color `M`, Height 1 (Row 10)
*   **Lane 3:** Color `.`, Height 4 (Rows 7-10)
*   **Lane 5:** Color `R`, Height 4 (Rows 7-10)
*   **Lane 7:** Color `A`, Height 5 (Rows 6-10)
*   **Lane 9:** Color `W`, Height 10 (Rows 1-10)

**Constructing the Output Grid:**
Row 0: All `O`
Row 1: Col 9 is `W`
Row 2: Col 9 is `W`
Row 3: Col 9 is `W`
Row 4: Col 9 is `W`
Row 5: Col 9 is `W`
Row 6: Col 7 is `A`, Col 9 is `W`
Row 7: Col 3 is `.`, Col 5 is `R`, Col 7 is `A`, Col 9 is `W`
Row 8: Col 3 is `.`, Col 5 is `R`, Col 7 is `A`, Col 9 is `W`
Row 9: Col 3 is `.`, Col 5 is `R`, Col 7 is `A`, Col 9 is `W`
Row 10: Col 1 is `M`, Col 3 is `.`, Col 5 is `R`, Col 7 is `A`, Col 9 is `W`

Wait, let's re-verify the height of Lane 1 in input.
Input Lane 1: `.` at rows 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Total 10 cells. Correct.
Input Lane 3: `R` at row 10. Total 1 cell. Correct.
Input Lane 5: `A` at rows 7, 8, 9, 10. Total 4 cells. Correct.
Input Lane 7: `W` at rows 7, 8, 9, 10. Total 4 cells. Correct.
Input Lane 9: `M` at rows 6, 7, 8, 9, 10. Total 5 cells. Correct.

New Heights:
L1: H(L3) = 1
L3: H(L5) = 4
L5: H(L7) = 4
L7: H(L9) = 5
L9: H(L1) = 10

New Colors:
L1: C(L9) = M
L3: C(L1) = .
L5: C(L3) = R
L7: C(L5) = A
L9: C(L7) = W

Final Grid Construction:
Row 0: O O O O O O O O O O O
Row 1: O O O O O O O O O W O
Row 2: O O O O O O O O O W O
Row 3: O O O O O O O O O W O
Row 4: O O O O O O O O O W O
Row 5: O O O O O O O O O W O
Row 6: O O O O O O O A O W O
Row 7: O O O . O R O A O W O
Row 8: O O O . O R O A O W O
Row 9: O O O . O R O A O W O
Row 10: O M O . O R O A O W O

ANSWER:
O O O O O O O O O O O
O O O O O O O O O W O
O O O O O O O O O W O
O O O O O O O O O W O
O O O O O O O O O W O
O O O O O O O O O W O
O O O O O O O A O W O
O O O . O R O A O W O
O O O . O R O A O W O
O O O . O R O A O W O
O M O . O R O A O W O

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Amber Void (O)**: The background color.
    *   **Nebulae**: Vertical columns of colored pixels anchored at the bottom.
    *   **Columns**: The grid is divided into alternating columns (some contain nebulae, some are void).
    *   **Spectral Color (Hue)**: The color of the pixels in a nebula.
    *   **Extent (Height)**: The number of colored pixels in a nebula.

2.  **Identify the AFFORDANCES**:
    *   **Hue**: Can be rotated one position to the right in the sequence of colors present in the input.
    *   **Extent**: Can be rotated one position to the left in the sequence of heights present in the input.

3.  **Identify the TRANSFORMATION RULE**:
    *   **Step 1: Identify all nebulae.** Find all columns that contain colored pixels (excluding the background 'O').
    *   **Step 2: List properties.** Create a list of the colors (Hue) and a list of the heights (Extent) of these nebulae, ordered by their column index (left to right).
    *   **Step 3: Rotate Hue.** Shift the list of colors one position to the right (the last color moves to the first position).
    *   **Step 4: Rotate Extent.** Shift the list of heights one position to the left (the first height moves to the last position).
    *   **Step 5: Reconstruct.** Place the new colors and new heights back into the original column positions.

### Applying the Rule to Test Input

**Input Analysis:**
Grid size: 11 columns x 11 rows.
Background: O (Orange)

Nebulae found (scanning columns from left to right):
1.  Col 1: Color '.', Height 10 (Rows 1-10)
2.  Col 3: Color 'R', Height 1 (Row 10)
3.  Col 5: Color 'A', Height 4 (Rows 7-10)
4.  Col 7: Color 'W', Height 4 (Rows 7-10)
5.  Col 9: Color 'M', Height 5 (Rows 6-10)

**Properties Lists:**
*   Colors (Hues): `['.', 'R', 'A', 'W', 'M']`
*   Heights (Extents): `[10, 1, 4, 4, 5]`

**Transformations:**
*   Rotate Hues Right: `['M', '.', 'R', 'A', 'W']`
*   Rotate Heights Left: `[1, 4, 4, 5, 10]`

**New Nebulae Mapping:**
1.  Col 1: Color 'M', Height 1 (Row 10)
2.  Col 3: Color '.', Height 4 (Rows 7-10)
3.  Col 5: Color 'R', Height 4 (Rows 7-10)
4.  Col 7: Color 'A', Height 5 (Rows 6-10)
5.  Col 9: Color 'W', Height 10 (Rows 1-10)

**Constructing Output Grid:**
Row 0: O O O O O O O O O O O
Row 1: O O O O O O O O O W O
Row 2: O O O O O O O O O W O
Row 3: O O O O O O O O O W O
Row 4: O O O O O O O O O W O
Row 5: O O O O O O O A O W O
Row 6: O O O O O O A O W W O (Wait, Col 9 is W, height 10. Col 7 is A, height 5. Col 1 is M, height 1.)

Let's re-verify the rows carefully:
Col 1 (M, H1): Row 10
Col 3 (., H4): Rows 7, 8, 9, 10
Col 5 (R, H4): Rows 7, 8, 9, 10
Col 7 (A, H5): Rows 6, 7, 8, 9, 10
Col 9 (W, H10): Rows 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

Row 0: O O O O O O O O O O O
Row 1: O O O O O O O O O W O
Row 2: O O O O O O O O O W O
Row 3: O O O O O O O O O W O
Row 4: O O O O O O O O O W O
Row 5: O O O O O O O O O W O
Row 6: O O O O O O A O O W O
Row 7: O O . O R O A O W W O (Wait, Col 9 is W. Col 7 is A. Col 5 is R. Col 3 is .)
Row 8: O O . O R O A O W W O
Row 9: O O . O R O A O W W O
Row 10: M O . O R O A O W W O

Wait, let's re-check the column indices for the test input:
Col 0: O
Col 1: . (Height 10)
Col 2: O
Col 3: R (Height 1)
Col 4: O
Col 5: A (Height 4)
Col 6: O
Col 7: W (Height 4)
Col 8: O
Col 9: M (Height 5)
Col 10: O

Corrected Lists:
Hues: `['.', 'R', 'A', 'W', 'M']`
Heights: `[10, 1, 4, 4, 5]`

Rotate Hues Right: `['M', '.', 'R', 'A', 'W']`
Rotate Heights Left: `[1, 4, 4, 5, 10]`

Mapping:
Col 1: M, H1 (Row 10)
Col 3: ., H4 (Rows 7, 8, 9, 10)
Col 5: R, H4 (Rows 7, 8, 9, 10)
Col 7: A, H5 (Rows 6, 7, 8, 9, 10)
Col 9: W, H10 (Rows 1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

Re-constructing rows:
Row 0: O O O O O O O O O O O
Row 1: O O O O O O O O O W O
Row 2: O O O O O O O O O W O
Row 3: O O O O O O O O O W O
Row 4: O O O O O O O O O W O
Row 5: O O O O O O O O O W O
Row 6: O O O O O O A O O W O
Row 7: O O . O R O A O W W O
Row 8: O O . O R O A O W W O
Row 9: O O . O R O A O W W O
Row 10: M O . O R O A O W W O

Wait, Col 9 is W. Row 7, Col 9 is W. Row 7, Col 7 is A. Row 7, Col 5 is R. Row 7, Col 3 is ..
Let's check the width: 11 columns.
Col 0: O
Col 1: M (Row 10)
Col 2: O
Col 3: . (Row 7-10)
Col 4: O
Col 5: R (Row 7-10)
Col 6: O
Col 7: A (Row 6-10)
Col 8: O
Col 9: W (Row 1-10)
Col 10: O

Row 0: O O O O O O O O O O O
Row 1: O O O O O O O O O W O
Row 2: O O O O O O O O O W O
Row 3: O O O O O O O O O W O
Row 4: O O O O O O O O O W O
Row 5: O O O O O O O O O W O
Row 6: O O O O O O A O O W O
Row 7: O O . O R O A O W W O
Row 8: O O . O R O A O W W O
Row 9: O O . O R O A O W W O
Row 10: M O . O R O A O W W O

Wait, Col 9 is the 10th column. Col 10 is the 11th column.
Col 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
Col 9 is index 9.
Row 7: Col 1=O, Col 3=., Col 5=R, Col 7=A, Col 9=W.
Row 7: O . O R O A O W O (Wait, I am miscounting columns).

Let's be extremely precise.
Input columns:
0: O
1: . (H10)
2: O
3: R (H1)
4: O
5: A (H4)
6: O
7: W (H4)
8: O
9: M (H5)
10: O

Hue

... [3149 chars truncated]

To solve this puzzle, we first perform a Conceptual Digestion based on the metaphorical description and training examples.

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Counter (Amber/O):** The background color (Orange `O`) that fills the grid.
    *   **Pedestals (Columns):** The vertical columns where ingredients are placed.
    *   **Ingredients (Colors):** The colored blocks (e.g., `M`, `A`, `W`, `R`, `B`, `X`, `Y`) that form vertical stacks.
    *   **Portion Size (Stack Height):** The number of colored cells in a specific column.

2.  **Identify the AFFORDANCES:**
    *   **Ingredients (Flavor/Color):** Can be shifted horizontally (left/right).
    *   **Pedestals (Columns):** Can be shifted horizontally (left/right).
    *   **Portion Size (Height):** Can be shifted horizontally (left/right).

3.  **Identify the TRANSFORMATION RULE:**
    *   The description says: "Each ingredient's identity (flavor/color) shifts one pedestal to the left while each pedestal's portion size shifts one position to the right."
    *   Let's analyze the columns. In the input, ingredients are in specific columns.
    *   Let $C_i$ be the column index and $H_i$ be the height of the stack at that column.
    *   The rule implies a "counter-rotation":
        *   The **colors** move from column $j$ to $j-1$.
        *   The **heights** (the "portion sizes") move from column $j$ to $j+1$.
    *   Wait, looking at the examples:
        *   Example 1: Input columns with colors are 1 (W), 3 (A), 5 (M), 7 (B).
        *   Output columns with colors are 3 (W), 5 (A), 7 (M), 1 (B)? No.
        *   Let's re-examine Example 2: Input: Col 1 (R, height 6), Col 3 (A, height 4), Col 5 (X, height 4).
        *   Output: Col 3 (R, height 6), Col 5 (A, height 4), Col 1 (X, height 4).
    *   **Correct Rule Interpretation:**
        The columns containing ingredients are identified. Let the indices of these columns be $p_1, p_2, \dots, p_n$.
        The colors at these columns are $Col_1, Col_2, \dots, Col_n$.
        The heights at these columns are $H_1, H_2, \dots, H_n$.
        In the output, the color $Col_i$ is placed in a new column, and the height $H_i$ is placed in a new column.
        Looking at Example 2:
        Input: Col 1 (R, H=6), Col 3 (A, H=4), Col 5 (X, H=4).
        Output: Col 3 (R, H=6), Col 5 (A, H=4), Col 1 (X, H=4).
        The colors shifted right by 2 columns ($1 \to 3, 3 \to 5, 5 \to 1$).
        The heights shifted left by 2 columns ($1 \to 5$ is not right... let's look at the heights).
        Actually, the rule is simpler: The **colors** move to the next available pedestal position to the **right**, and the **heights** move to the next available pedestal position to the **left**.
        In Example 2: Pedestals are at 1, 3, 5.
        Colors: R(1), A(3), X(5). Shifted right: R(3), A(5), X(1).
        Heights: 6(1), 4(3), 4(5). Shifted left: 6(5), 4(1), 4(3).
        Wait, let's check Example 2 output again:
        Col 1: X (H=4), Col 3: R (H=6), Col 5: A (H=4).
        This matches!
        **Rule:**
        1. Identify columns with stacks (pedestals).
        2. For each pedestal $i$ (in order of columns):
           - The color at $i$ moves to pedestal $i+1$ (wrap around).
           - The height at $i$ moves to pedestal $i-1$ (wrap around).

### Applying to Test Input

**Input Analysis:**
Columns with stacks (pedestals):
- Col 1: Color `.`, Height 10 (from row 1 to 10)
- Col 4: Color `A`, Height 4 (from row 7 to 10)
- Col 6: Color `W`, Height 4 (from row 7 to 10)
- Col 8: Color `M`, Height 5 (from row 6 to 10)
- Col 2: Color `R`, Height 1 (at row 10) - *Wait, let's re-read the input carefully.*

**Test Input Grid (Rows 0-10):**
R0: O O O O O O O O O O O
R1: O . O O O O O O O O O
R2: O . O O O O O O O O O
R3: O . O O O O O O O O O
R4: O . O O O O O O O O O
R5: O . O O O O O O O O O
R6: O . O O O O O O O M O
R7: O . O O O A O W O M O
R8: O . O O O A O W O M O
R9: O . O O O A O W O M O
R10: O . O R O A O W O M O

**Pedestals (Columns with non-O):**
1. Col 1: Color `.`, Height 10 (Rows 1-10)
2. Col 3: Color `R`, Height 1 (Row 10)
3. Col 5: Color `A`, Height 4 (Rows 7-10)
4. Col 7: Color `W`, Height 4 (Rows 7-10)
5. Col 9: Color `M`, Height 5 (Rows 6-10)

**Pedestal Indices:** 1, 3, 5, 7, 9.
**Colors:** $C_1=., C_2=R, C_3=A, C_4=W, C_5=M$
**Heights:** $H_1=10, H_2=1, H_3=4, H_4=4, H_5=5$

**Transformation:**
- New Colors (Shifted Right): $C'_1=M, C'_2=., C'_3=R, C'_4=A, C'_5=W$
- New Heights (Shifted Left): $H'_1=1, H'_2=4, H'_3=4, H'_4=5, H'_5=10$

**Mapping New Colors to Pedestals:**
- Col 1: Color `M`, Height 1 (Row 10)
- Col 3: Color `.`, Height 4 (Rows 7-10)
- Col 5: Color `R`, Height 4 (Rows 7-10)
- Col 7: Color `A`, Height 5 (Rows 6-10)
- Col 9: Color `W`, Height 10 (Rows 1-10)

**Wait, let's re-verify the "Shifted Left" logic for heights.**
Example 2: Pedestals 1, 3, 5. Colors R, A, X. Heights 6, 4, 4.
Shifted Colors (Right): R $\to$ 3, A $\to$ 5, X $\to$ 1.
Shifted Heights (Left): 6 $\

... [3115 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Amber Stage Floor (O):** The background color.
    *   **Sound Columns:** Vertical structures of color located in every other column (indices 1, 3, 5, 7, etc., or similar patterns).
    *   **Timbre (Color):** The color of the vertical column.
    *   **Volume (Height):** The number of cells the color occupies, starting from the bottom.

2.  **Identify the AFFORDANCES:**
    *   Columns can have their color changed (shifted).
    *   Columns can have their height changed (shifted).
    *   The background (O) remains constant.

3.  **Identify the TRANSFORMATION RULE:**
    *   Identify all active columns (columns containing colors other than 'O').
    *   Let the columns be $C_1, C_2, \dots, C_n$ from left to right.
    *   **Timbre Shift:** The color of $C_i$ becomes the color of $C_{i-1}$ (the color cycles one position rightward).
    *   **Volume Shift:** The height of $C_i$ becomes the height of $C_{i+1}$ (the height cycles one position leftward).
    *   *Wait, let's re-examine the training examples to be precise:*
    *   **Example 1:**
        *   Col 1 (index 1): Color W, Height 3.
        *   Col 3 (index 3): Color A, Height 4.
        *   Col 5 (index 5): Color M, Height 9.
        *   Col 7 (index 7): Color B, Height 3.
        *   Output:
        *   Col 1: Color B (from Col 7), Height 4 (from Col 3).
        *   Col 3: Color W (from Col 1), Height 9 (from Col 5).
        *   Col 5: Color A (from Col 3), Height 3 (from Col 7).
        *   Col 7: Color M (from Col 5), Height 3 (from Col 1).
    *   **Correction on Rule:**
        *   The columns are $C_1, C_2, \dots, C_n$.
        *   New Color of $C_i = \text{Color of } C_{i-1}$ (where $C_0 = C_n$).
        *   New Height of $C_i = \text{Height of } C_{i+1}$ (where $C_{n+1} = C_1$).
        *   Let's check Example 1 again:
            *   $C_1(W, 3), C_2(A, 4), C_3(M, 9), C_4(B, 3)$.
            *   New $C_1$: Color $C_4(B)$, Height $C_2(4) \rightarrow (B, 4)$.
            *   New $C_2$: Color $C_1(W)$, Height $C_3(9) \rightarrow (W, 9)$.
            *   New $C_3$: Color $C_2(A)$, Height $C_4(3) \rightarrow (A, 3)$.
            *   New $C_4$: Color $C_3(M)$, Height $C_1(3) \rightarrow (M, 3)$.
            *   Matches Example 1 output!

### Applying the Rule to Test Input

**1. Identify Columns:**
*   Col 1 (index 1): Color '.', Height 10 (from row 1 to 10).
*   Col 4 (index 4): Color 'A', Height 4 (from row 7 to 10).
*   Col 6 (index 6): Color 'W', Height 4 (from row 7 to 10).
*   Col 8 (index 8): Color 'M', Height 5 (from row 6 to 10).
*   Col 2 (index 2): Color 'R', Height 1 (at row 10). *Wait, let's re-check the input grid carefully.*

**Input Grid Analysis (Test Input):**
Row 0: O O O O O O O O O O O
Row 1: O . O O O O O O O O O
Row 2: O . O O O O O O O O O
Row 3: O . O O O O O O O O O
Row 4: O . O O O O O O O O O
Row 5: O . O O O O O O O O O
Row 6: O . O O O O O O O M O
Row 7: O . O O O A O W O M O
Row 8: O . O O O A O W O M O
Row 9: O . O O O A O W O M O
Row 10: O . O R O A O W O M O

**Columns (indices):**
*   Col 1: Color '.', Height 10 (Rows 1-10)
*   Col 3: Color 'R', Height 1 (Row 10)
*   Col 5: Color 'A', Height 4 (Rows 7-10)
*   Col 7: Color 'W', Height 4 (Rows 7-10)
*   Col 9: Color 'M', Height 5 (Rows 6-10)

**Mapping:**
$C_1: (., 10)$
$C_2: (R, 1)$
$C_3: (A, 4)$
$C_4: (W, 4)$
$C_5: (M, 5)$

**Transformation:**
New Color $C_i = \text{Color } C_{i-1}$
New Height $C_i = \text{Height } C_{i+1}$

*   New $C_1$: Color $C_5(M)$, Height $C_2(1) \rightarrow (M, 1)$
*   New $C_2$: Color $C_1(.)$, Height $C_3(4) \rightarrow (., 4)$
*   New $C_3$: Color $C_2(R)$, Height $C_4(4) \rightarrow (R, 4)$
*   New $C_4$: Color $C_3(A)$, Height $C_5(5) \rightarrow (A, 5)$
*   New $C_5$: Color $C_4(W)$, Height $C_1(10) \rightarrow (W, 10)$

**Constructing Output:**
*   Col 1: Row 10 is M.
*   Col 3: Rows 7, 8, 9, 10 are .
*   Col 5: Rows 7, 8, 9, 10 are R.
*   Col 7: Rows 6, 7, 8, 9, 10 are A.
*   Col 9: Rows 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 are W.

**Wait, let's re-verify the column indices.**
Input:
Col 0: O
Col 1: . (Rows 1-10)
Col 2: O
Col 3: R (Row 10)
Col 4: O
Col 5: A (Rows 7-10)
Col 6: O
Col 7: W (Rows 7-10)
Col 8: O
Col 9: M (Rows 6-10)
Col 10: O

Columns are at indices 1, 3, 5, 7, 9.
$C_1(1): (., 10)$
$C_2(3): (R, 1)$
$C_3(5): (A, 4)$
$C_4(7): (W, 4)$
$C_5(9): (M, 5)$

New $C_1(1)$: Color $C_5(M)$, Height $C_2(1) \rightarrow (M, 1)$
New $C_2(3)$: Color $C_1(.)$, Height $C_3(4) \rightarrow (., 4)$
New $C_3(5)$: Color $C_2(R)$, Height $C_4(4) \rightarrow (R, 4)$
New $C_4(7)$: Color $C_3(A)$, Height $C_5(5) \rightarrow (A, 5)$
New $C_5(9)$: Color $C_4(W)$, Height $C_1(10) \rightarrow (W, 10)$

**Final Grid Construction:**
Row 0: O O O O O O O O O O O
Row 1: O O O O O O O O O W O
Row 2: O O O O O O O O O W O
Row 3: O O O O O O O O O W O
Row 4: O O O O O O O O O W O
Row 5: O O O O O O O O O W O
Row 6: O O O O O O O A O W O
Row 7: O O O . O R O A O W O
Row 8: O O O . O R O A O W O
Row 9: O 

... [1696 chars truncated]

### Conceptual Digestion

**1. Identify the CAST:**
*   **The Curtain:** The central column (the `Y` column in the examples) acts as the divider between the left and right wings.
*   **Azure Performers (A):** These are the actors in the left wing.
*   **Grey Performers (X):** These are the actors in the right wing.
*   **Black Squares (.):** The empty stage floor (background).
*   **Red Spotlights (R):** The result of the transformation, marking "lonesome" performers.

**2. Identify the AFFORDANCES:**
*   The grid is split into two halves by a central column (the `Y` column).
*   The left wing (columns before `Y`) and the right wing (columns after `Y`) are compared row by row.
*   A performer is "lonesome" if they exist in one wing but have no corresponding performer in the mirrored position in the other wing.
*   The output grid is the result of merging the two wings into a single smaller grid.

**3. Identify the TRANSFORMATION RULE:**
*   **Step 1: Split.** Divide the input grid into the Left Wing (all columns to the left of the `Y` column) and the Right Wing (all columns to the right of the `Y` column).
*   **Step 2: Mirror/Compare.** For each row and each relative position from the center:
    *   Compare the cell in the Left Wing at position `i` with the cell in the Right Wing at the mirrored position `i`.
    *   A cell in the output becomes **Red (R)** if:
        *   The Left cell is `A` AND the Right cell is `.` (empty).
        *   OR the Right cell is `X` AND the Left cell is `.` (empty).
    *   A cell in the output becomes **Black (.)** if:
        *   Both are empty (`.` and `.`).
        *   Both have performers (`A` and `X`).
        *   (Note: The metaphor implies we are looking for "unmatched" dancers. If both sides have a dancer, they "lock eyes" and the spotlight stays dark).
*   **Step 3: Reconstruct.** The output grid dimensions are (Number of Rows) x (Width of one wing).

### Applying the Rule to Test Input

**Input Analysis:**
The `Y` column is at index 4.
Left Wing (Indices 0-3):
Row 0: `. A . .`
Row 1: `. A . A`
Row 2: `A A . A`
Row 3: `A A A A`
Row 4: `. . A A`

Right Wing (Indices 5-8):
Row 0: `X . . .`
Row 1: `X X . X`
Row 2: `. . X X`
Row 3: `X . X X`
Row 4: `X . X .`

**Comparison (Row by Row):**
*Note: The "mirroring" in the metaphor suggests comparing the $i$-th column from the center in the left wing with the $i$-th column from the center in the right wing. However, looking at the training examples, the comparison is direct: Left[row][col] vs Right[row][col] where the columns are aligned by distance from the center.*

Let's align them:
Left Wing (L) | Right Wing (R) | Comparison (L vs R)
--- | --- | ---
`. A . .` | `X . . .` | (0,0): . vs X -> R; (0,1): A vs . -> R; (0,2): . vs . -> .; (0,3): . vs . -> .
`. A . A` | `X X . X` | (1,0): . vs X -> R; (1,1): A vs X -> .; (1,2): . vs . -> .; (1,3): A vs X -> .;
`A A . A` | `. . X X` | (2,0): A vs . -> R; (2,1): A vs . -> R; (2,2): . vs X -> R; (2,3): A vs X -> .;
`A A A A` | `X . X X` | (3,0): A vs X -> .; (3,1): A vs . -> R; (3,2): A vs X -> .; (3,3): A vs X -> .;
`. . A A` | `X . X .` | (4,0): . vs X -> R; (4,1): . vs . -> .; (4,2): A vs X -> .; (4,3): A vs . -> R;

**Wait, let's re-verify the alignment using Example 1.**
Ex 1 Left:
`. A . .`
`A A . A`
`A A . .`
`. A . A`
`. . A .`
Ex 1 Right:
`. X X .`
`X . . X`
`X . . X`
`. . X .`
`. X . X`
(Wait, the example right wing is 4 wide, the left is 4 wide. The Y is the 5th column).
Ex 1 Output:
`. . R .`
`. R . .`
`. R . R`
`. R R R`
`. R R R`

Let's re-check Ex 1 Row 0: L: `. A . .` R: `. X X .`.
L[0]=., R[0]=. -> .
L[1]=A, R[1]=X -> . (Matched)
L[2]=., R[2]=X -> R (Unmatched)
L[3]=., R[3]=. -> .
Result: `. . R .`. This matches the example!

**Applying the logic to Test Input correctly:**
L:
`. A . .`
`. A . A`
`A A . A`
`A A A A`
`. . A A`

R:
`X . . .`
`X X . X`
`. . X X`
`X . X X`
`X . X .`

Row 0:
L: `. A . .`
R: `X . . .`
(0,0): . vs X -> R
(0,1): A vs . -> R
(0,2): . vs . -> .
(0,3): . vs . -> .
Result: `R R . .`

Row 1:
L: `. A . A`
R: `X X . X`
(1,0): . vs X -> R
(1,1): A vs X -> .
(1,2): . vs . -> .
(1,3): A vs X -> .
Result: `R . . .`

Row 2:
L: `A A . A`
R: `. . X X`
(2,0): A vs . -> R
(2,1): A vs . -> R
(2,2): . vs X -> R
(2,3): A vs X -> .
Result: `R R R .`

Row 3:
L: `A A A A`
R: `X . X X`
(3,0): A vs X -> .
(3,1): A vs . -> R
(3,2): A vs X -> .
(3,3): A vs X -> .
Result: `. R . .`

Row 4:
L: `. . A A`
R: `X . X .`
(4,0): . vs X -> R
(4,1): . vs . -> .
(4,2): A vs X -> .
(4,3): A vs . -> R
Result: `R . . R`

**Wait, let me re-check Row 0 of Test Input.**
Input Row 0: `. A . . Y X . . .`
L: `. A . .`
R: `X . . .`
(0,0): . vs X -> R
(0,1): A vs . -> R
(0,2): . vs . -> .
(0,3): . vs . -> .
Result: `R R . .`

Let's re-verify Row 1:
Input Row 1: `. A . A Y X X . X`
L: `. A . A`
R: `X X . X`
(1,0): . vs X -> R
(1,1): A vs X -&g

... [955 chars truncated]

### Conceptual Digestion

**1. Identify the CAST:**
*   **Yellow (Y):** The vertical meridian (the divider). It splits the grid into a left hemisphere and a right hemisphere.
*   **Azure (A):** Stars in the left hemisphere.
*   **Grey (X):** Stars in the right hemisphere.
*   **Black (.):** The void (empty space).
*   **Red (R):** The resulting "nebula" (the output color).

**2. Identify the AFFORDANCES:**
*   The grid is split by the column containing 'Y'.
*   The left side (columns before Y) and the right side (columns after Y) are compared cell by cell relative to the meridian.
*   The output grid's width is the width of one hemisphere (half the width of the input, excluding the Y column).

**3. Identify the TRANSFORMATION RULE:**
*   **Step 1:** Split the input grid into two halves using the 'Y' column as the axis.
*   **Step 2:** Compare the left hemisphere (Azure stars) and the right hemisphere (Grey stars) row by row and column by column.
*   **Step 3:** For each position $(i, j)$ in the hemisphere:
    *   If (Left side has star AND Right side has void) OR (Left side has void AND Right side has star), the output is **Red (R)**. This is an XOR operation.
    *   If both have stars or both are void, the output is **Black (.)**.
*   **Step 4:** The output grid dimensions are (Input Rows) $\times$ (Input Columns - 1) / 2.

### Applying the Rule to Test Input

**Input Grid Analysis:**
Width = 9. Y is at index 4 (0-indexed).
Left hemisphere (indices 0-3):
Row 0: `. A . .`
Row 1: `. A . A`
Row 2: `A A . A`
Row 3: `A A A A`
Row 4: `. . A A`

Right hemisphere (indices 5-8):
Row 0: `X . . .`
Row 1: `X X . X`
Row 2: `. . X X`
Row 3: `X . X X`
Row 4: `X . X .`

**XOR Comparison (Left vs Right):**
*Row 0:*
`.` vs `X` $\rightarrow$ **R**
`A` vs `.` $\rightarrow$ **R**
`.` vs `.` $\rightarrow$ **.**
`.` vs `.` $\rightarrow$ **.**
Result: `R R . .`

*Row 1:*
`.` vs `X` $\rightarrow$ **R**
`A` vs `X` $\rightarrow$ **.** (Both have stars)
`.` vs `.` $\rightarrow$ **.**
`A` vs `X` $\rightarrow$ **.** (Both have stars)
Result: `R . . .`

*Row 2:*
`A` vs `.` $\rightarrow$ **R**
`A` vs `.` $\rightarrow$ **R**
`.` vs `X` $\rightarrow$ **R**
`A` vs `X` $\rightarrow$ **.** (Both have stars)
Result: `R R R .`

*Row 3:*
`A` vs `X` $\rightarrow$ **.** (Both have stars)
`A` vs `.` $\rightarrow$ **R**
`A` vs `X` $\rightarrow$ **.** (Both have stars)
`A` vs `X` $\rightarrow$ **.** (Both have stars)
Result: `. R . .`

*Row 4:*
`.` vs `X` $\rightarrow$ **R**
`.` vs `.` $\rightarrow$ **.**
`A` vs `X` $\rightarrow$ **.** (Both have stars)
`A` vs `.` $\rightarrow$ **R**
Result: `R . . R`

**Final Grid Construction:**
R R . .
R . . .
R R R .
. R . .
R . . R

ANSWER:
R R . .
R . . .
R R R .
. R . .
R . . R

### Conceptual Digestion

**1. Identify the CAST:**
*   **Yellow (Y):** The "reef" (the vertical divider). It splits the grid into two halves.
*   **Azure (A):** The "pearls" on the left seabed.
*   **Grey (X):** The "pearls" on the right seabed.
*   **Black (.):** The "open water" (empty space).
*   **Red (R):** The "coral" (the result of the transformation).

**2. Identify the AFFORDANCES:**
*   The grid is split vertically by the column of Yellow (Y).
*   The left side (columns before Y) and the right side (columns after Y) are compared row by row.
*   The output grid is half the width of the input (specifically, it represents the comparison of the two sides).

**3. Identify the TRANSFORMATION RULE:**
*   The transformation is an **XOR operation** between the presence of "pearls" (A on the left, X on the right) and "open water" (.).
*   For each row:
    *   Compare the left side (columns $0$ to $n-1$ where $n$ is the index of Y) with the right side (columns $n+1$ to end).
    *   If (Left is A AND Right is .) OR (Left is . AND Right is X), the output is **Red (R)**.
    *   If (Left is A AND Right is X) OR (Left is . AND Right is .), the output is **Black (.)**.
*   The output grid dimensions are (Input Height) $\times$ (Width of one side). Note: In the examples, the output width matches the number of columns on one side of the Y divider.

### Applying the Rule to Test Input

**Input Grid Analysis:**
Input dimensions: 5 rows, 9 columns.
Yellow (Y) is at index 4 in every row.
Left side (indices 0-3):
Row 0: `. A . .`
Row 1: `. A . A`
Row 2: `A A . A`
Row 3: `A A A A`
Row 4: `. . A A`

Right side (indices 5-8):
Row 0: `X . . .`
Row 1: `X X . X`
Row 2: `. . X X`
Row 3: `X . X X`
Row 4: `X . X .`

**Comparison (XOR logic: A/X vs .):**
*Note: The rule says "wherever one lagoon has a pearl and the other has open water, red coral blooms." This means (A and .) or (. and X) $\rightarrow$ R. If both have pearls (A and X) or both are empty (. and .), it stays black (.).*

**Row 0:**
Left: `. A . .`
Right: `X . . .`
Col 0: . vs X $\rightarrow$ R
Col 1: A vs . $\rightarrow$ R
Col 2: . vs . $\rightarrow$ .
Col 3: . vs . $\rightarrow$ .
Result: `R R . .`

**Row 1:**
Left: `. A . A`
Right: `X X . X`
Col 0: . vs X $\rightarrow$ R
Col 1: A vs X $\rightarrow$ .
Col 2: . vs . $\rightarrow$ .
Col 3: A vs X $\rightarrow$ .
Result: `R . . .`

**Row 2:**
Left: `A A . A`
Right: `. . X X`
Col 0: A vs . $\rightarrow$ R
Col 1: A vs . $\rightarrow$ R
Col 2: . vs X $\rightarrow$ R
Col 3: A vs X $\rightarrow$ .
Result: `R R R .`

**Row 3:**
Left: `A A A A`
Right: `X . X X`
Col 0: A vs X $\rightarrow$ .
Col 1: A vs . $\rightarrow$ R
Col 2: A vs X $\rightarrow$ .
Col 3: A vs X $\rightarrow$ .
Result: `. R . .`

**Row 4:**
Left: `. . A A`
Right: `X . X .`
Col 0: . vs X $\rightarrow$ R
Col 1: . vs . $\rightarrow$ .
Col 2: A vs X $\rightarrow$ .
Col 3: A vs . $\rightarrow$ R
Result: `R . . R`

**Wait, let's re-verify Example 1 to ensure the logic is perfect.**
Ex 1 Input:
`. A . . Y . X X .` (L: `. A . .` R: `. X X .`) -> Col 0: . vs . (.), Col 1: A vs X (.), Col 2: . vs X (R), Col 3: . vs . (.) -> Result: `. . R .` (Matches!)
`A A . A Y X . . X` (L: `A A . A` R: `X . . X`) -> Col 0: A vs X (.), Col 1: A vs . (R), Col 2: . vs . (.), Col 3: A vs X (.) -> Result: `. R . .` (Matches!)
`A A . . Y X . . X` (L: `A A . .` R: `X . . X`) -> Col 0: A vs X (.), Col 1: A vs . (R), Col 2: . vs . (.), Col 3: . vs X (R) -> Result: `. R . R` (Matches!)
`. A . A Y . . X .` (L: `. A . A` R: `. . X .`) -> Col 0: . vs . (.), Col 1: A vs . (R), Col 2: . vs X (R), Col 3: A vs . (R) -> Result: `. R R R` (Matches!)
`. . A . Y . X . X` (L: `. . A .` R: `. X . X`) -> Col 0: . vs . (.), Col 1: . vs X (R), Col 2: A vs . (R), Col 3: . vs X (R) -> Result: `. R R R` (Matches!)

**Re-calculating Test Input with confirmed logic:**
Row 0: `. A . .` vs `X . . .` $\rightarrow$ `R R . .`
Row 1: `. A . A` vs `X X . X` $\rightarrow$ `R . . .`
Row 2: `A A . A` vs `. . X X` $\rightarrow$ `R R R .`
Row 3: `A A A A` vs `X . X X` $\rightarrow$ `. R . .`
Row 4: `. . A A` vs `X . X .` $\rightarrow$ `R . . R`

ANSWER:
R R . .
R . . .
R R R .
. R . .
R . . R

### Conceptual Digestion

**1. Identify the CAST:**
*   **Yellow Bus (Y):** A vertical line acting as a divider or a central axis.
*   **Azure Signals (A):** Located on the left side of the yellow line.
*   **Grey Signals (X):** Located on the right side of the yellow line.
*   **Black ( . ):** Represents "no signal" or low state.
*   **Red (R):** The output signal (the result of the logic gate).
*   **Output Register:** A grid with half the width of the input, representing the result of the logic operation.

**2. Identify the AFFORDANCES:**
*   The **Azure (A)** and **Grey (X)** cells act as binary inputs (High if color is present, Low if black).
*   The **Yellow (Y)** line acts as the separator between the two input channels.
*   The **Red (R)** color is the result of an XOR operation between the left and right channels.

**3. Identify the TRANSFORMATION RULE:**
*   The transformation is a row-by-row **XOR logic gate** applied to the two halves of the input grid, separated by the yellow column.
*   For each row:
    1.  Identify the left side (all columns to the left of the 'Y' column).
    2.  Identify the right side (all columns to the right of the 'Y' column).
    3.  Compare the cells at the same relative index from the 'Y' column.
    4.  If (Left is A AND Right is NOT X) OR (Left is NOT A AND Right is X), the output is **Red (R)**.
    5.  Otherwise (both are signals or both are black), the output is **Black (.)**.
*   The output grid width is the number of columns on one side of the 'Y' column.

### Applying the Rule to Test Input

**Input Analysis:**
The 'Y' column is at index 4 (0-indexed).
Left side (indices 0-3):
Row 0: `. A . .`
Row 1: `. A . A`
Row 2: `A A . A`
Row 3: `A A A A`
Row 4: `. . A A`

Right side (indices 5-8):
Row 0: `X . . .`
Row 1: `X X . X`
Row 2: `. . X X`
Row 3: `X . X X`
Row 4: `X . X .`

**XOR Logic (A vs X):**
*Note: The rule states "where one input is high and the other is low, the output fires red". Here, A is high on the left and X is high on the right.*

**Row 0:**
Left: `. A . .`
Right: `X . . .`
- Col 0: `.` vs `X` -> **R**
- Col 1: `A` vs `.` -> **R**
- Col 2: `.` vs `.` -> **.**
- Col 3: `.` vs `.` -> **.**
Result: `R R . .`

**Row 1:**
Left: `. A . A`
Right: `X X . X`
- Col 0: `.` vs `X` -> **R**
- Col 1: `A` vs `X` -> **.** (Both high)
- Col 2: `.` vs `.` -> **.**
- Col 3: `A` vs `X` -> **.** (Both high)
Result: `R . . .`

**Row 2:**
Left: `A A . A`
Right: `. . X X`
- Col 0: `A` vs `.` -> **R**
- Col 1: `A` vs `.` -> **R**
- Col 2: `.` vs `X` -> **R**
- Col 3: `A` vs `X` -> **.** (Both high)
Result: `R R R .`

**Row 3:**
Left: `A A A A`
Right: `X . X X`
- Col 0: `A` vs `X` -> **.** (Both high)
- Col 1: `A` vs `.` -> **R**
- Col 2: `A` vs `X` -> **.** (Both high)
- Col 3: `A` vs `X` -> **.** (Both high)
Result: `. R . .`

**Row 4:**
Left: `. . A A`
Right: `X . X .`
- Col 0: `.` vs `X` -> **R**
- Col 1: `.` vs `.` -> **.**
- Col 2: `A` vs `X` -> **.** (Both high)
- Col 3: `A` vs `.` -> **R**
Result: `R . . R`

**Final Grid Construction:**
Row 0: R R . .
Row 1: R . . .
Row 2: R R R .
Row 3: . R . .
Row 4: R . . R

ANSWER:
R R . .
R . . .
R R R .
. R . .
R . . R

### Conceptual Digestion

**1. Identify the CAST:**
*   **Black Bread (`.`):** The base/background of the grid.
*   **Yellow Toothpick (`Y`):** A central vertical divider that separates the two halves.
*   **Azure Jam (`A`):** The condiment on the left side of the toothpick.
*   **Grey Mustard (`X`):** The condiment on the right side of the toothpick.
*   **Red Hot Bites (`R`):** The resulting color when a mismatch occurs.
*   **Black Cold Bites (`.`):** The resulting color when there is a match.

**2. Identify the AFFORDANCES:**
*   The **Toothpick (`Y`)** acts as a separator and is removed.
*   The **Left Slice (`A`)** and **Right Slice (`X`)** are pressed together.
*   The **Condiments (`A` and `X`)** are compared against the **Bread (`.`)** on the opposite side.

**3. Identify the TRANSFORMATION RULE:**
*   The grid is split vertically by the column containing `Y`.
*   The left side (columns before `Y`) and the right side (columns after `Y`) are compared row by row.
*   The output grid width is half the input width (excluding the `Y` column).
*   For each row $i$ and column $j$ (where $j$ is the index relative to the divider):
    *   Let $L$ be the value in the left half at position $j$.
    *   Let $R$ be the value in the right half at position $j$.
    *   If one side has a condiment ($A$ or $X$) and the other side has bare bread ($.$), the result is **Red (`R`)**.
    *   If both sides have a condiment ($A$ and $X$) OR both sides have bare bread ($.$), the result is **Black (`.`)**.
    *   *Refined Rule:* The output is `R` if $(L == '.' \text{ and } R \neq '.') \text{ or } (L \neq '.' \text{ and } R == '.')$. Otherwise, the output is `.`.

### Applying the Rule to Test Input

**Input Grid Analysis:**
The `Y` column is at index 4 (0-indexed).
Left side (indices 0-3):
Row 0: `. A . .`
Row 1: `. A . A`
Row 2: `A A . A`
Row 3: `A A A A`
Row 4: `. . A A`

Right side (indices 5-8):
Row 0: `X . . .`
Row 1: `X X . X`
Row 2: `. . X X`
Row 3: `X . X X`
Row 4: `X . X .`

**Comparison (L vs R):**
*   **Row 0:**
    *   Col 0: `.` vs `X` $\rightarrow$ Mismatch $\rightarrow$ `R`
    *   Col 1: `A` vs `.` $\rightarrow$ Mismatch $\rightarrow$ `R`
    *   Col 2: `.` vs `.` $\rightarrow$ Match $\rightarrow$ `.`
    *   Col 3: `.` vs `.` $\rightarrow$ Match $\rightarrow$ `.`
    *   Result: `R R . .`

*   **Row 1:**
    *   Col 0: `.` vs `X` $\rightarrow$ Mismatch $\rightarrow$ `R`
    *   Col 1: `A` vs `X` $\rightarrow$ Match (both condiments) $\rightarrow$ `.`
    *   Col 2: `.` vs `.` $\rightarrow$ Match $\rightarrow$ `.`
    *   Col 3: `A` vs `X` $\rightarrow$ Match (both condiments) $\rightarrow$ `.`
    *   Result: `R . . .`

*   **Row 2:**
    *   Col 0: `A` vs `.` $\rightarrow$ Mismatch $\rightarrow$ `R`
    *   Col 1: `A` vs `.` $\rightarrow$ Mismatch $\rightarrow$ `R`
    *   Col 2: `.` vs `X` $\rightarrow$ Mismatch $\rightarrow$ `R`
    *   Col 3: `A` vs `X` $\rightarrow$ Match $\rightarrow$ `.`
    *   Result: `R R R .`

*   **Row 3:**
    *   Col 0: `A` vs `X` $\rightarrow$ Match $\rightarrow$ `.`
    *   Col 1: `A` vs `.` $\rightarrow$ Mismatch $\rightarrow$ `R`
    *   Col 2: `A` vs `X` $\rightarrow$ Match $\rightarrow$ `.`
    *   Col 3: `A` vs `X` $\rightarrow$ Match $\rightarrow$ `.`
    *   Result: `. R . .`

*   **Row 4:**
    *   Col 0: `.` vs `X` $\rightarrow$ Mismatch $\rightarrow$ `R`
    *   Col 1: `.` vs `.` $\rightarrow$ Match $\rightarrow$ `.`
    *   Col 2: `A` vs `X` $\rightarrow$ Match $\rightarrow$ `.`
    *   Col 3: `A` vs `.` $\rightarrow$ Mismatch $\rightarrow$ `R`
    *   Result: `R . . R`

**Wait, let's re-verify Row 1 Col 3:**
Input Row 1: `. A . A Y X X . X`
Left: `. A . A`
Right: `X X . X`
Col 3: `A` vs `X`. Both are condiments. Rule: "Where both have spread... the bite stays cold and black." So `.` is correct.

**Wait, let's re-verify Row 3 Col 0:**
Input Row 3: `A A A A Y X . X X`
Left: `A A A A`
Right: `X . X X`
Col 0: `A` vs `X`. Both are condiments. Rule: "Where both have spread... the bite stays cold and black." So `.` is correct.

**Final Grid Construction:**
Row 0: `R R . .`
Row 1: `R . . .`
Row 2: `R R R .`
Row 3: `. R . .`
Row 4: `R . . R`

ANSWER:
R R . .
R . . .
R R R .
. R . .
R . . R

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Azure (A):** Represents the "left team" players.
    *   **Grey (X):** Represents the "right team" players.
    *   **Yellow (Y):** Represents the "net" (a vertical divider).
    *   **Black (.):** Represents empty positions.
    *   **Red (R):** Represents the result of the transformation.

2.  **Identify the AFFORDANCES:**
    *   The **Yellow (Y)** column acts as a separator. The grid is split into a left side (columns before Y) and a right side (columns after Y).
    *   The transformation compares the presence of players (A or X) in corresponding positions relative to the divider. However, looking at the training examples, the transformation is actually a comparison of the **left side** and the **right side** of the Y column.
    *   Wait, looking closer at the dimensions: The input grids have a width (e.g., 9 columns), but the output grids are smaller (e.g., 4 or 5 columns).
    *   Let's re-examine Example 1:
        Input width: 9. Y is at index 4 (0-indexed).
        Left side (indices 0-3):
        `. A . .`
        `A A . A`
        `A A . .`
        `. A . A`
        `. . A .`
        Right side (indices 5-8):
        `X X . .` (Wait, index 5 is X, 6 is X, 7 is ., 8 is X? No, let's re-count)
        Example 1 Input: `. A . . Y . X X .` (Indices: 0:., 1:A, 2:., 3:., 4:Y, 5:., 6:X, 7:X, 8:.)
        Left side (0-3): `. A . .` / `A A . A` / `A A . .` / `. A . A` / `. . A .`
        Right side (5-8): `. X X .` / `X . . X` / `X . . X` / `. . X .` / `. X . X`
        Output:
        `. . R .`
        `. R . .`
        `. R . R`
        `. R R R`
        `. R R R`
        Let's compare Left vs Right row by row:
        Row 0: Left `. A . .` vs Right `. X X .`.
        Wait, the output is 5x4. The input is 5x9.
        Let's look at the logic: "mark every position where exactly one team has a player".
        If we compare the left side (4 columns) to the right side (4 columns):
        Row 0: Left `. A . .` vs Right `. X X .`.
        Pos 0: . vs . (Both empty) -> .
        Pos 1: A vs X (Both players) -> . (Wait, the rule says "Where both have players facing off... leave it unmarked")
        Pos 2: . vs X (One player) -> R
        Pos 3: . vs . (Both empty) -> .
        Result Row 0: `. . R .` (Matches!)

        Row 1: Left `A A . A` vs Right `X . . X`.
        Pos 0: A vs X (Both) -> .
        Pos 1: A vs . (One) -> R
        Pos 2: . vs . (Both empty) -> .
        Pos 3: A vs X (Both) -> .
        Wait, Example 1 Row 1 output is `. R . .`. My calculation: `. R . .`. (Matches!)

        Row 2: Left `A A . .` vs Right `X . . X`.
        Pos 0: A vs X (Both) -> .
        Pos 1: A vs . (One) -> R
        Pos 2: . vs . (Both empty) -> .
        Pos 3: . vs X (One) -> R
        Result Row 2: `. R . R`. (Matches!)

3.  **Identify the TRANSFORMATION RULE:**
    *   Split the input grid into two halves using the column containing 'Y' as the divider.
    *   The left half consists of all columns to the left of 'Y'.
    *   The right half consists of all columns to the right of 'Y'.
    *   For each cell $(r, c)$ in the resulting grid:
        *   If (Left cell is player AND Right cell is empty) OR (Left cell is empty AND Right cell is player) $\rightarrow$ Red (R).
        *   If (Left cell is player AND Right cell is player) OR (Left cell is empty AND Right cell is empty) $\rightarrow$ Black (.).
    *   Note: "Player" means 'A' on the left and 'X' on the right.

### Applying to Test Input

**Test Input:**
`. A . . Y X . . .`
`. A . A Y X X . X`
`A A . A Y . . X X`
`A A A A Y X . X X`
`. . A A Y X . X .`

**Step 1: Split into Left and Right (excluding Y)**
Left:
`. A . .`
`. A . A`
`A A . A`
`A A A A`
`. . A A`

Right:
`X . . .`
`X X . X`
`. . X X`
`X . X X`
`X . X .`

**Step 2: Compare (XOR logic for players)**
Row 0:
`.` vs `X` $\rightarrow$ R
`A` vs `.` $\rightarrow$ R
`.` vs `.` $\rightarrow$ .
`.` vs `.` $\rightarrow$ .
Result: `R R . .`

Row 1:
`.` vs `X` $\rightarrow$ R
`A` vs `X` $\rightarrow$ .
`.` vs `.` $\rightarrow$ .
`A` vs `X` $\rightarrow$ .
Result: `R . . .`

Row 2:
`A` vs `.` $\rightarrow$ R
`A` vs `.` $\rightarrow$ R
`.` vs `X` $\rightarrow$ R
`A` vs `X` $\rightarrow$ .
Result: `R R R .`

Row 3:
`A` vs `X` $\rightarrow$ .
`A` vs `.` $\rightarrow$ R
`A` vs `X` $\rightarrow$ .
`A` vs `X` $\rightarrow$ .
Result: `. R . .`

Row 4:
`.` vs `X` $\rightarrow$ R
`.` vs `.` $\rightarrow$ .
`A` vs `X` $\rightarrow$ .
`A` vs `.` $\rightarrow$ R
Result: `R . . R`

**Wait, let me re-check Row 0 of Test Input.**
Input Row 0: `. A . . Y X . . .`
Left: `. A . .`
Right: `X . . .`
Pos 0: . vs X $\rightarrow$ R
Pos 1: A vs . $\rightarrow$ R
Pos 2: . vs . $\rightarrow$ .
Pos 3: . vs . $\rightarrow$ .
Result: `R R . .`

**Let me re-check Row 1 of Test Input.**
Input Row 1: `. A . A Y X X . X`
Left: `. A . A`
Right: `X X . X`
Pos 0: . vs X $\rightarrow$ R
Pos 1: A vs X $\rightarrow$ .
Pos 2: . vs . $\rightarrow$ .

... [901 chars truncated]

### Conceptual Digestion

**1. Identify the CAST:**
*   **Y (Yellow):** The "stationary front line." It acts as a vertical divider/separator.
*   **A (Azure):** "High-pressure zones." Located to the left of the Y line.
*   **X (Grey):** "Low-pressure zones." Located to the right of the Y line.
*   **. (Black):** "Calm air." Empty space or neutral zones.
*   **R (Red):** "Storm cells." The resulting output element.

**2. Identify the AFFORDANCES:**
*   The **Y** line is "collapsed," meaning the grid is split into two halves (left of Y and right of Y) and then merged or compared.
*   The **A** and **X** zones are compared against the **.** (calm air) zones.
*   The output grid is a "half-width weather map," meaning the width of the output is the number of columns on one side of the Y line.

**3. Identify the TRANSFORMATION RULE:**
*   The Y column is removed.
*   The grid is split into a Left side (columns before Y) and a Right side (columns after Y).
*   For each cell $(i, j)$ in the resulting half-width grid:
    *   Compare the cell from the Left side ($L_{i,j}$) and the cell from the Right side ($R_{i,j}$).
    *   A **Red (R)** storm erupts if **one** side has a pressure system (A or X) and the **other** side is calm air (.).
    *   Specifically: If ($L_{i,j} \in \{A, X\}$ AND $R_{i,j} = \text{.}$) OR ($L_{i,j} = \text{.}$ AND $R_{i,j} \in \{A, X\}$), the output is **R**.
    *   If both are pressure systems ($L \in \{A, X\}$ AND $R \in \{A, X\}$) or both are calm ($L = \text{.}$ AND $R = \text{.}$), the output is **.** (black).
    *   *Note:* The metaphor implies A and X are both "pressure systems."

### Applying the Rule to Test Input

**Input Grid Analysis:**
Columns: 0:., 1:A, 2:., 3:., 4:Y, 5:X, 6:., 7:., 8:.
Columns: 0:., 1:A, 2:., 3:A, 4:Y, 5:X, 6:X, 7:., 8:X
Columns: 0:A, 1:A, 2:., 3:A, 4:Y, 5:., 6:., 7:X, 8:X
Columns: 0:A, 1:A, 2:A, 3:A, 4:Y, 5:X, 6:., 7:X, 8:X
Columns: 0:., 1:., 2:A, 3:A, 4:Y, 5:X, 6:., 7:X, 8:.

**Step 1: Split into Left (cols 0-3) and Right (cols 5-8)**
Left Side (L):
Row 0: . A . .
Row 1: . A . A
Row 2: A A . A
Row 3: A A A A
Row 4: . . A A

Right Side (R):
Row 0: X . . .
Row 1: X X . X
Row 2: . . X X
Row 3: X . X X
Row 4: X . X .

**Step 2: Compare L and R for Red (R) storms**
Rule: R if (L is A/X and R is .) OR (L is . and R is A/X).

Row 0:
(L: . vs R: X) -> R is pressure, L is calm -> **R**
(L: A vs R: .) -> L is pressure, R is calm -> **R**
(L: . vs R: .) -> Both calm -> **.**
(L: . vs R: .) -> Both calm -> **.**
Result: R R . .

Row 1:
(L: . vs R: X) -> R is pressure, L is calm -> **R**
(L: A vs R: X) -> Both pressure -> **.**
(L: . vs R: .) -> Both calm -> **.**
(L: A vs R: X) -> Both pressure -> **.**
Result: R . . .

Row 2:
(L: A vs R: .) -> L is pressure, R is calm -> **R**
(L: A vs R: .) -> L is pressure, R is calm -> **R**
(L: . vs R: X) -> R is pressure, L is calm -> **R**
(L: A vs R: X) -> Both pressure -> **.**
Result: R R R .

Row 3:
(L: A vs R: X) -> Both pressure -> **.**
(L: A vs R: .) -> L is pressure, R is calm -> **R**
(L: A vs R: X) -> Both pressure -> **.**
(L: A vs R: X) -> Both pressure -> **.**
Result: . R . .

Row 4:
(L: . vs R: X) -> R is pressure, L is calm -> **R**
(L: . vs R: .) -> Both calm -> **.**
(L: A vs R: X) -> Both pressure -> **.**
(L: A vs R: .) -> L is pressure, R is calm -> **R**
Result: R . . R

**Final Grid Construction:**
R R . .
R . . .
R R R .
. R . .
R . . R

ANSWER:
R R . .
R . . .
R R R .
. R . .
R . . R

### Conceptual Digestion

**1. Identify the CAST:**
*   **Y (Yellow):** The "fortress wall." It acts as a vertical divider that splits the grid into two halves (left and right).
*   **A (Azure):** The "infantry" on the left side.
*   **X (Grey):** The "infantry" on the right side.
*   **. (Black):** "No man's land" or empty space.
*   **R (Red):** The "conflict" color, representing areas where one side has presence and the other does not.

**2. Identify the AFFORDANCES:**
*   The grid is split by the column containing **Y**.
*   The left side (columns before Y) is evaluated for the presence of **A**.
*   The right side (columns after Y) is evaluated for the presence of **X**.
*   The output grid is a "half-width battlefield," meaning the width of the output is the width of one side of the wall.

**3. Identify the TRANSFORMATION RULE:**
*   **Step 1: Split the grid.** Locate the column containing **Y**.
*   **Step 2: Define the sides.** The left side consists of all columns to the left of Y. The right side consists of all columns to the right of Y.
*   **Step 3: Compare presence.** For each row:
    *   Look at the left side: Is there at least one **A**?
    *   Look at the right side: Is there at least one **X**?
    *   Wait, looking at the training example more closely: The output width matches the width of one side. Let's re-examine Example 1.
    *   Example 1 Input: Width 9. Y is at index 4 (5th column). Left side width = 4. Right side width = 4. Output width = 4.
    *   Example 1 Row 1: Left `[. A . .]`, Right `[. X X .]`. Left has A, Right has X. Output `[. . R .]`? No, that doesn't match.
    *   **Correction on Rule:** Let's look at the cell-by-cell comparison.
    *   Example 1, Row 1: Left `[. A . .]`, Right `[. X X .]`. Output `[. . R .]`.
    *   Wait, the output is the same width as one side. Let's compare the *presence* of A and X at each relative position.
    *   Example 1, Row 1: Left `[. A . .]`, Right `[. X X .]`.
        *   Pos 0: Left `.`, Right `.` -> Both absent -> `.`
        *   Pos 1: Left `A`, Right `X` -> Both present -> `.` (Wait, the rule says "Where forces are matched... ground stays dark")
        *   Pos 2: Left `.`, Right `X` -> Right present, Left absent -> `R`
        *   Pos 3: Left `.`, Right `.` -> Both absent -> `.`
    *   Let's check Row 2: Left `[A A . A]`, Right `[X . . X]`.
        *   Pos 0: A vs X (Both present) -> `.`
        *   Pos 1: A vs . (A present, X absent) -> `R`
        *   Pos 2: . vs . (Both absent) -> `.`
        *   Pos 3: A vs X (Both present) -> `.`
        *   Wait, Example 1 Output Row 2 is `. R . .`. This matches!
    *   **Refined Rule:**
        1. Split the grid at the column containing **Y**.
        2. For each row, compare the cell at index `i` on the left with the cell at index `i` on the right (where `i` is the offset from the wall).
        3. If (Left has A AND Right has no X) OR (Left has no A AND Right has X), the output is **R**.
        4. Otherwise (Both have A/X, or both are empty/black), the output is **.** (black).

### Applying the Rule to Test Input

**Test Input Analysis:**
Input width: 9. Y is at index 4.
Left side (indices 0-3):
Row 0: `. A . .`
Row 1: `. A . A`
Row 2: `A A . A`
Row 3: `A A A A`
Row 4: `. . A A`

Right side (indices 5-8):
Row 0: `X . . .`
Row 1: `X X . X`
Row 2: `. . X X`
Row 3: `X . X X`
Row 4: `X . X .`

**Comparison (Left vs Right):**
*   **Row 0:**
    *   Col 0: `.` vs `X` -> One present (X) -> **R**
    *   Col 1: `A` vs `.` -> One present (A) -> **R**
    *   Col 2: `.` vs `.` -> Both absent -> **.**
    *   Col 3: `.` vs `.` -> Both absent -> **.**
    *   *Wait, let's re-check Example 1 Row 1:* Left `[. A . .]`, Right `[. X X .]`.
        *   Col 0: `.` vs `.` -> `.`
        *   Col 1: `A` vs `X` -> Both present -> `.`
        *   Col 2: `.` vs `X` -> One present -> `R`
        *   Col 3: `.` vs `.` -> `.`
        *   Result: `. . R .`. Matches Example 1!

**Applying refined rule to Test Input:**
*   **Row 0:**
    *   0: `.` vs `X` -> **R**
    *   1: `A` vs `.` -> **R**
    *   2: `.` vs `.` -> **.**
    *   3: `.` vs `.` -> **.**
    *   Result: `R R . .`

*   **Row 1:**
    *   0: `.` vs `X` -> **R**
    *   1: `A` vs `X` -> **.** (Both present)
    *   2: `.` vs `.` -> **.**
    *   3: `A` vs `X` -> **.** (Both present)
    *   Result: `R . . .`

*   **Row 2:**
    *   0: `A` vs `.` -> **R**
    *   1: `A` vs `.` -> **R**
    *   2: `.` vs `X` -> **R**
    *   3: `A` vs `X` -> **.** (Both present)
    *   Result: `R R R .`

*   **Row 3:**
    *   0: `A` vs `X` -> **.**
    *   1: `A` vs `.` -> **R**
    *   2: `A` vs `X` -> **.**
    *   3: `A` vs `X` -> **.**
    *   Result: `. R . .`

*   **Row 4:**
    *   0: `.` vs `X` -> **R**
    *   1: `.` vs `.` -> **.**
    *   2: `A` vs `X` -> **.**
    *   3: `A` vs `.` -> **R**
    *   Result: `R . . R`



... [1567 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Azure (A):** Actors on the left side of the stage.
    *   **Grey (X):** Actors on the right side of the stage.
    *   **Yellow (Y):** The vertical curtain dividing the stage.
    *   **Black (.):** Empty stage.
    *   **Red (R):** The spotlight (the result of the transformation).

2.  **Identify the AFFORDANCES:**
    *   The input grid is a single wide stage.
    *   The Yellow (Y) column acts as a divider.
    *   The transformation splits the grid into two halves: the left side (columns before Y) and the right side (columns after Y).
    *   The output grid is a half-width map (the width of one side of the curtain).

3.  **Identify the TRANSFORMATION RULE:**
    *   Split the input grid at the column containing 'Y'.
    *   Let $L$ be the left side (excluding the 'Y' column) and $R$ be the right side (excluding the 'Y' column).
    *   Compare $L$ and $R$ cell by cell.
    *   If one side has an actor (A or X) and the other side is empty (.), the output cell is Red (R).
    *   If both sides have actors or both sides are empty, the output cell is Black (.).
    *   Essentially, this is an **XOR** operation between the presence of actors on the left and the presence of actors on the right.

### Applying the Rule to Test Input

**Input Grid Analysis:**
Input width is 9. 'Y' is at index 4 (the 5th column).
Left side (indices 0-3):
Row 0: `. A . .`
Row 1: `. A . A`
Row 2: `A A . A`
Row 3: `A A A A`
Row 4: `. . A A`

Right side (indices 5-8):
Row 0: `X . . .`
Row 1: `X X . X`
Row 2: `. . X X`
Row 3: `X . X X`
Row 4: `X . X .`

**XOR Comparison (Actor vs Actor = ., Actor vs Empty = R, Empty vs Empty = .):**

*   **Row 0:**
    L: `. A . .`
    R: `X . . .`
    Compare:
    (0,0): . vs X $\rightarrow$ R
    (0,1): A vs . $\rightarrow$ R
    (0,2): . vs . $\rightarrow$ .
    (0,3): . vs . $\rightarrow$ .
    Result: `R R . .`

*   **Row 1:**
    L: `. A . A`
    R: `X X . X`
    Compare:
    (1,0): . vs X $\rightarrow$ R
    (1,1): A vs X $\rightarrow$ .
    (1,2): . vs . $\rightarrow$ .
    (1,3): A vs X $\rightarrow$ .
    Result: `R . . .`

*   **Row 2:**
    L: `A A . A`
    R: `. . X X`
    Compare:
    (2,0): A vs . $\rightarrow$ R
    (2,1): A vs . $\rightarrow$ R
    (2,2): . vs X $\rightarrow$ R
    (2,3): A vs X $\rightarrow$ .
    Result: `R R R .`

*   **Row 3:**
    L: `A A A A`
    R: `X . X X`
    Compare:
    (3,0): A vs X $\rightarrow$ .
    (3,1): A vs . $\rightarrow$ R
    (3,2): A vs X $\rightarrow$ .
    (3,3): A vs X $\rightarrow$ .
    Result: `. R . .`

*   **Row 4:**
    L: `. . A A`
    R: `X . X .`
    Compare:
    (4,0): . vs X $\rightarrow$ R
    (4,1): . vs . $\rightarrow$ .
    (4,2): A vs X $\rightarrow$ .
    (4,3): A vs . $\rightarrow$ R
    Result: `R . . R`

**Final Grid Construction:**
R R . .
R . . .
R R R .
. R . .
R . . R

ANSWER:
R R . .
R . . .
R R R .
. R . .
R . . R

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **The Stage**: The entire input grid (background is `.`).
    *   **The Troupes**: Small, connected clusters of cells of the same color. Each cluster represents a "pose" and a "costume."
    *   **The Understudies**: Identical copies of a specific cluster (same shape and same color) located elsewhere in the grid.
    *   **The Crowd Favorite**: The specific color-shape combination that appears most frequently in the grid.

2.  **Identify the AFFORDANCES**:
    *   Clusters can be compared to one another to check for identity (exact shape and color).
    *   The grid can be "darkened" (all non-winning cells turned to `.`).
    *   The winning cluster can be "extracted" (isolated in its own bounding box).

3.  **Identify the TRANSFORMATION RULE**:
    *   Step 1: Identify all distinct clusters of colored cells.
    *   Step 2: Count the occurrences of each unique cluster (shape + color).
    *   Step 3: Find the cluster that appears most frequently.
    *   Step 4: Create a new grid containing only the pixels of that winning cluster, cropped to its minimal bounding box. All other pixels become `.`.

### Applying the Rule to Test Input

**Step 1: Identify Clusters**
*   **Green (G) clusters**:
    1.  (2,3), (2,5), (3,3), (3,4), (3,5), (4,3), (4,5) -> Shape:
        `. G . G .`
        `. . G G G`
        `. G . G .` (Wait, let's re-examine coordinates)
        Input G:
        (2,3) G, (2,5) G
        (3,3) G, (3,4) G, (3,5) G
        (4,3) G, (4,5) G
        Shape:
        `G . G`
        `G G G`
        `G . G`
    2.  (11,9) G, (11,11) G, (12,10) G, (12,11) G, (13,9) G, (13,11) G
        Shape:
        `G . G`
        `. G G`
        `G . G`
        *These are not identical.*
*   **Red (R) clusters**:
    1.  (2,10) R, (3,10) R, (3,11) R, (3,12) R, (4,10) R, (4,11) R
        Shape:
        `R . . .`
        `R R R .`
        `R R . .`
    2.  (8,1) R, (9,0) R, (9,1) R, (9,2) R, (10,0) R, (10,1) R, (10,2) R, (11,5) R, (12,4) R, (12,5) R, (12,6) R, (13,4) R, (13,5) R, (13,6) R
        Wait, let's look closer at the Red clusters:
        Cluster A: (2,10), (3,10), (3,11), (3,12), (4,10), (4,11)
        Cluster B: (8,1), (9,0), (9,1), (9,2), (10,0), (10,1), (10,2)
        Cluster C: (11,5), (12,4), (12,5), (12,6), (13,4), (13,5), (13,6)
        Let's check Cluster B and C:
        B:
        `. R .`
        `R R R`
        `R R .`
        C:
        `. R .`
        `R R R`
        `R R .`
        **B and C are identical!**
*   **Azure (A) clusters**:
    1.  (7,7) A, (8,6) A, (8,8) A, (9,7) A
        Shape:
        `. A .`
        `A . A`
        `. A .`
        (Only one instance).

**Step 2 & 3: Count and Find Winner**
*   Green clusters: 1 of shape X, 1 of shape Y.
*   Red clusters: 1 of shape Z, 2 of shape W (Cluster B and C).
*   Azure clusters: 1 of shape V.
The winner is the Red cluster with shape W.

**Step 4: Extract Winner**
The winning shape is:
`. R .`
`R R R`
`R R .`

ANSWER:
. R .
R R R
R R .

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **The Petri Dish (Input Grid)**: A large grid containing several distinct "species" (clusters of colored pixels).
    *   **Species (Colored Clusters)**: Each color represents a different species. The shape and arrangement of the pixels form a "microcolony."
    *   **The Dominant Strain**: The species that appears most frequently in the grid.
    *   **The Slide (Output Grid)**: A small grid containing only the shape of the most frequent species.

2.  **Identify the AFFORDANCES**:
    *   **Colors**: Act as identifiers for different species.
    *   **Spatial Arrangement**: The relative positions of pixels of the same color define the "morphology" (shape) of the species.
    *   **Counting**: The number of distinct clusters of a specific color determines which species is "dominant."

3.  **Identify the TRANSFORMATION RULE**:
    *   **Step 1**: Scan the input grid and identify all distinct clusters of pixels for each color.
    *   **Step 2**: Count how many separate clusters exist for each color.
    *   **Step 3**: Identify the color that has the highest number of separate clusters.
    *   **Step 4**: Extract the shape of one of these clusters (the rule implies the clusters of the dominant species are identical or we extract the pattern of the dominant one).
    *   **Step 5**: Output the bounding box of that single cluster as the new grid.

### Applying the Rule to the Test Input

**Step 1: Identify Clusters**
*   **Green (G)**:
    1.  (2,3), (2,5), (3,4), (3,5), (4,3), (4,5) -> Cluster 1
    2.  (11,9), (11,11), (12,10), (12,11), (13,9), (13,11) -> Cluster 2
*   **Red (R)**:
    1.  (2,10), (3,10), (3,11), (3,12), (4,10), (4,11) -> Cluster 1
    2.  (8,1), (9,0), (9,1), (9,2), (10,0), (10,1) -> Cluster 2
    3.  (11,5), (12,4), (12,5), (12,6), (13,4), (13,5) -> Cluster 3
*   **Azure (A)**:
    1.  (7,7), (8,6), (8,8), (9,7) -> Cluster 1
*   **Other colors**: None.

**Step 2: Count Clusters**
*   Green: 2 clusters
*   Red: 3 clusters
*   Azure: 1 cluster

**Step 3: Identify Dominant Strain**
*   Red (R) has the most clusters (3).

**Step 4: Extract Shape of Red Cluster**
Looking at the Red clusters:
Cluster 1:
```
. R . .
R R R .
R R . .
```
Cluster 2:
```
. R .
R R R
R R .
```
Cluster 3:
```
. R . .
R R R .
R R . .
```
All red clusters follow the same pattern: a 3x3 or 3x4 bounding box. Let's look at the structure:
Row 1: One pixel at index 1 (relative).
Row 2: Three pixels at indices 0, 1, 2.
Row 3: Two pixels at indices 0, 1.

Let's re-verify Cluster 1: (2,10), (3,10), (3,11), (3,12), (4,10), (4,11).
Wait, let's re-examine the coordinates for Red:
(2,10)
(3,10), (3,11), (3,12)
(4,10), (4,11)
This is:
`. R . .`
`R R R .`
`. R R .`

Let's re-examine Cluster 2:
(8,1)
(9,0), (9,1), (9,2)
(10,0), (10,1)
This is:
`. R .`
`R R R`
`R R .`

Let's re-examine Cluster 3:
(11,5)
(12,4), (12,5), (12,6)
(13,4), (13,5)
This is:
`. R . .`
`R R R .`
`R R . .`

The pattern is consistent:
Row 1: One R at column 1.
Row 2: Three Rs at columns 0, 1, 2.
Row 3: Two Rs at columns 0, 1.

Let's check the bounding box for the most standard representation:
Row 1: `. R .`
Row 2: `R R R`
Row 3: `R R .`

Wait, looking at Cluster 1 again:
(2,10)
(3,10), (3,11), (3,12)
(4,10), (4,11)
This is:
`. R . .`
`R R R .`
`. R R .` (Wait, (4,10) and (4,11) are the 3rd row).

Let's look at Cluster 2 again:
(8,1)
(9,0), (9,1), (9,2)
(10,0), (10,1)
This is:
`. R .`
`R R R`
`R R .`

Let's look at Cluster 3 again:
(11,5)
(12,4), (12,5), (12,6)
(13,4), (13,5)
This is:
`. R . .`
`R R R .`
`R R . .`

The pattern is:
Row 1: `. R .`
Row 2: `R R R`
Row 3: `R R .`

Let's check Example 1 to see how the shape is extracted.
Example 1:
Blue (B) clusters: (0,7)(0,9), (1,7)(1,8)(1,9), (1,1), (6,1)(6,3), (7,1)(7,2)(7,3), (8,1)(8,2), (11,12)(11,13), (12,11)(12,12)(12,13), (13,12).
Wait, let's count Blue:
1. (0,7), (0,9), (1,7), (1,8), (1,9) -> Cluster 1
2. (1,1), (6,1), (6,3), (7,1), (7,2), (7,3), (8,1), (8,2) -> This is actually one large cluster? No, (1,1) is separate.
Let's re-count carefully.
Blue:
- (0,7), (0,9), (1,7), (1,8), (1,9)
- (1,1)
- (6,1), (6,3), (7,1), (7,2), (7,3), (8,1), (8,2)
- (11,12), (11,13), (12,11), (12,12), (12,13), (13,12)
Total Blue clusters: 4.
Yellow:
- (1,2), (2,2), (2,3), (3,2)
- (3,11), (4,11), (4,12), (5,11)
- (8,9), (9,10), (9,11), (10,9), (10,10), (11,2), (11,3), (12,2)
Wait, let's re-count Yellow:
- (1,2), (2,2), (2,3), (3,2)
- (3,11), (4,11), (4,12), (5,11)
- (8,9), (9,10), (9,11), (10,9), (10,10)
- (11,2), (11,3), (12,2)
Total Yellow clusters: 4.
Wait, if Blue and Yellow are both 4, which one is it?
Let's re-examine the output of Example 1:
`Y . .`
`. Y Y`
`Y . .`
This is a Yellow cluster! Let's look at the Yellow clusters again.
Cluster 1: (1,2), (2,2), (2,3), (3,2).
Shape:
`. Y .`
`Y Y .`
`. Y .`
Wait, the output is:
`Y . .`
`. Y Y`
`Y . .`
This matches the shape of the Yellow cluster at (11,2), (11,3), (12,2)? No.
Let's look at the Yellow

... [2406 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Dark Matrix (`.`):** The background/ground mass.
    *   **Colored Shapes (B, Y, R, G, A, etc.):** These represent "crystal habits" (connected components of the same color).
    *   **The Most Frequent Habit:** The specific color/shape that appears most often in the input grid.

2.  **Identify the AFFORDANCES**:
    *   Each colored shape is a distinct object.
    *   The shapes can be counted by color.
    *   The goal is to isolate the "most frequent" shape and present it as a standalone grid.

3.  **Identify the TRANSFORMATION RULE**:
    *   **Step 1: Count the occurrences of each color-based shape.** Note: The metaphor says "Identify the crystal habit that occurs most frequently". Looking at Example 1:
        *   Blue (B) shapes: One large cluster, one small cluster, one small cluster.
        *   Yellow (Y) shapes: Several small clusters.
        *   Red (R) shapes: One cluster.
        *   Wait, let's re-examine Example 1 more closely.
        *   Blue (B) clusters: (7,0)-(8,3) [size 5], (0,7)-(2,9) [size 5], (11,12)-(13,13) [size 5].
        *   Yellow (Y) clusters: (1,2)-(3,2) [size 3], (3,12)-(4,14) [size 3], (5,11)-(5,12) [size 2], (8,9)-(10,10) [size 3], (10,2)-(12,4) [size 4].
        *   Red (R) clusters: (5,6)-(7,8) [size 4].
        *   Looking at the output of Example 1: `Y . . / . Y Y / Y . .`. This is a specific shape of Yellow.
        *   Let's re-read: "Identify the crystal habit that occurs most frequently... Extract that mineral specimen".
        *   In Example 1, the Yellow shapes are all the same "habit" (a specific pattern/shape). Let's check the shapes:
            *   Yellow shape 1: (1,2), (2,3), (2,4), (3,2) -> A specific pattern.
            *   Yellow shape 2: (3,12), (4,13), (4,14) -> A different pattern.
            *   Actually, looking at the output `Y . . / . Y Y / Y . .`, this is a 3x3 pattern. Let's see if any color in Example 1 matches this.
            *   Yellow at (10,2), (11,3), (11,4), (12,2) is a shape.
            *   Let's look at the Red shape: (5,6), (5,8), (6,7), (7,6), (7,8).
            *   Let's look at the Blue shape: (0,7), (0,9), (1,7), (1,8), (1,9), (2,8).
            *   Wait, the output is a 3x3 grid. The shape in the output is:
                Row 1: Y . .
                Row 2: . Y Y
                Row 3: Y . .
            *   Let's check the input for this specific pattern.
            *   Yellow at (10,2), (11,3), (11,4), (12,2) is not it.
            *   Let's look at the Yellow at (1,2), (2,3), (2,4), (3,2). If we center it:
                (1,2) is Y
                (2,3) is Y, (2,4) is Y
                (3,2) is Y
                This matches the pattern! `Y . . / . Y Y / Y . .` (if we treat the coordinates relative to the top-left).
        *   **Conclusion:** Find the color that has the most instances of a specific shape, then output that shape.

4.  **Apply to Test Input**:
    *   **Green (G) shapes:**
        1. (2,3), (2,5), (3,4), (3,5), (4,3), (4,5) -> A 3x3-ish shape.
        2. (11,9), (11,10), (12,10), (12,11), (13,9), (13,11) -> A different shape.
    *   **Red (R) shapes:**
        1. (2,10), (3,10), (3,11), (3,12), (4,10), (4,11) -> A 3x3-ish shape.
        2. (8,1), (9,0), (9,1), (9,2), (10,0), (10,1) -> A different shape.
        3. (8,1), (9,0), (9,1), (9,2), (10,0), (10,1) is one.
        4. (11,5), (12,4), (12,5), (12,6), (13,4), (13,5), (13,6) -> A different shape.
    *   **Azure (A) shapes:**
        1. (7,7), (8,6), (8,8), (9,7) -> A single shape.
    *   **Wait, let's re-count the shapes by color:**
        *   Green: 2 shapes.
        *   Red: 3 shapes.
        *   Azure: 1 shape.
    *   **Wait, let's look at the shapes themselves again.**
        *   Red shape 1: (2,10), (3,10), (3,11), (3,12), (4,10), (4,11).
        *   Red shape 2: (8,1), (9,0), (9,1), (9,2), (10,0), (10,1).
        *   Red shape 3: (11,5), (12,4), (12,5), (12,6), (13,4), (13,5), (13,6).
        *   Green shape 1: (2,3), (2,5), (3,4), (3,5), (4,3), (4,5).
        *   Green shape 2: (11,9), (11,10), (12,10), (12,11), (13,9), (13,11).
    *   **Let's look for a repeating shape.**
        *   Green shape 1:
            `. G . G`
            `. . G G`
            `G . G .`
            `G . G .` (Wait, this is getting complex).
        *   Let's look at the Red shape 1:
            `. . R . .`
            `. . R R R`
            `. . R R .`
        *   Let's look at the Green shape 2:
            `. . . . .`
            `. . G G .`
            `. . . G G`
            `. . G . G`
        *   Let's look at the Red shape 3:
            `. . . R .`
            `. R R R .`
            `. R R . .`
        *   Wait! Red shape 1 and Red shape 3 are very similar!
            Red shape 1: (2,10), (3,10), (3,11), (3,12), (4,10), (4,11)
            Red shape 3: (11,5), (12,4), (12,5), (12,6), (13,4), (13,5), (13,6)
            Let's re-examine Red shape 1:
            Row

... [3085 chars truncated]

To solve this puzzle, we follow the instructions provided in the metaphorical description and the training examples.

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **The Landscape (Background):** The black (`.`) grid cells representing the dark ground plane.
    *   **Building Footprints (Objects):** Small clusters of colored pixels (e.g., Blue, Red, Green, Azure) representing different architectural designs.
    *   **The Vernacular Template (Goal):** The specific pattern/shape of the colored cluster that appears most frequently in the input grid.

2.  **Identify the AFFORDANCES:**
    *   The colored pixels can be grouped into distinct, connected shapes (footprints).
    *   Each shape has a specific geometry and color.
    *   The task is to count the occurrences of these unique shapes.

3.  **Identify the TRANSFORMATION RULE:**
    *   Scan the input grid to identify all distinct colored shapes (connected components of the same color).
    *   Determine the shape (the relative arrangement of colored pixels) and the color of each component.
    *   Count how many times each unique shape appears.
    *   The output is the bounding box of the shape that appears most frequently, represented as a small grid.

### Applying the Rule to the Test Input

**Step 1: Identify all shapes and their counts.**
*   **Green (G) shapes:**
    1.  Top-left: `(2,3), (2,5), (3,4), (3,5), (4,3), (4,5)` -> This is a 3x3 pattern:
        `. G . G .` (Wait, let's look closer at the coordinates)
        Row 2: `. . . G . G .`
        Row 3: `. . . . G G .`
        Row 4: `. . . G . G .`
        Shape 1: `G . G` / `. G G` / `G . G`
    2.  Bottom-right:
        Row 11: `. . . G . G . .`
        Row 12: `. . . . G G . .`
        Row 13: `. . . G . G . .`
        Shape 2: `G . G` / `. G G` / `G . G`
    *   *Count for Shape (G . G / . G G / G . G): 2*

*   **Red (R) shapes:**
    1.  Top-right: `(2,10), (3,10), (3,11), (3,12), (4,10), (4,11)`
        Row 2: `R . .`
        Row 3: `R R R`
        Row 4: `R R .`
    2.  Bottom-left:
        Row 8: `R . .`
        Row 9: `R R R`
        Row 10: `R R .`
    3.  Bottom-middle:
        Row 11: `. R .`
        Row 12: `R R R`
        Row 13: `R R .`
    *   *Count for Shape (R . . / R R R / R R .): 3*

*   **Azure (A) shapes:**
    1.  Middle:
        Row 7: `. A .`
        Row 8: `A . A`
        Row 9: `. A .`
    *   *Count for Shape (. A . / A . A / . A .): 1*

**Step 2: Determine the most frequent shape.**
The Red shape `R . .` / `R R R` / `R R .` appears 3 times.

**Step 3: Format the output.**
The output is the bounding box of the most frequent shape.

Shape:
R . .
R R R
R R .

Wait, let's re-verify the Red shape in the input:
Input Red 1:
(2,10) R
(3,10) R, (3,11) R, (3,12) R
(4,10) R, (4,11) R
This is:
R . .
R R R
R R .

Input Red 2:
(8,1) R
(9,0) R, (9,1) R, (9,2) R
(10,0) R, (10,1) R
This is:
R . .
R R R
R R .

Input Red 3:
(11,5) R
(12,4) R, (12,5) R, (12,6) R
(13,4) R, (13,5) R
This is:
. R .
R R R
R R .
(Wait, let me re-check the coordinates for Red 3)
Row 11: `. . . . . R . . .` -> (11,5) is R
Row 12: `. . . . R R R . .` -> (12,4), (12,5), (12,6) are R
Row 13: `. . . . R R . . .` -> (13,4), (13,5) are R
This is:
. R .
R R R
R R .

Let's re-check Red 1:
Row 2: `. . . . . . . . . . R . . .` -> (2,10) is R
Row 3: `. . . . . . . . . R R R . .` -> (3,9), (3,10), (3,11) are R (Wait, index 9, 10, 11)
Row 4: `. . . . . . . . . R R . . .` -> (4,9), (4,10) are R
This is:
. R .
R R R
R R .

Let's re-check Red 2:
Row 8: `. R . . . . A . . . . . . .` -> (8,1) is R
Row 9: `R R R . . . . A . . . . . .` -> (9,0), (9,1), (9,2) are R
Row 10: `R R . . . . . . . . . . . .` -> (10,0), (10,1) are R
This is:
. R .
R R R
R R .

All three Red shapes are actually the same:
. R .
R R R
R R .

ANSWER:
. R .
R R R
R R .

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **The Grid (Battlefield)**: A large 2D space containing various colored patterns.
    *   **Colored Units (Formations)**: Small, contiguous clusters of a single color that form specific shapes.
    *   **The Signature Maneuver (Output)**: The specific shape/pattern that appears most frequently in the grid.

2.  **Identify the AFFORDANCES**:
    *   The grid elements (colors) can be grouped into clusters based on color and proximity.
    *   Clusters can be compared to one another to see if they are identical in shape and relative orientation.

3.  **Identify the TRANSFORMATION RULE**:
    *   Scan the input grid to identify all distinct colored formations (contiguous blocks of the same color).
    *   Count the occurrences of each unique formation shape.
    *   Identify the shape that occurs most frequently.
    *   Output that specific shape as a small grid.

### Step-by-Step Analysis of Test Input

1.  **Identify Formations**:
    *   **Green (G)**:
        *   Formation 1: (2,3), (2,5), (3,4), (3,5), (4,3), (4,5) -> This looks like a pattern. Let's re-examine.
        *   Wait, let's look at the coordinates carefully:
            *   G: (2,3), (2,5), (3,4), (3,5), (4,3), (4,5). This is a 3x3 area.
            *   G: (11,9), (11,11), (12,10), (12,11), (13,9), (13,11).
        *   Let's check the shape of G:
            Row 1: . G . G
            Row 2: . G G .
            Row 3: . G . G
            (This is a 3x4 or similar pattern).
    *   **Red (R)**:
        *   Formation 1: (2,10), (3,10), (3,11), (3,12), (4,10), (4,11).
        *   Formation 2: (8,1), (9,0), (9,1), (9,2), (10,0), (10,1).
        *   Formation 3: (11,5), (12,4), (12,5), (12,6), (13,4), (13,5).
    *   **Azure (A)**:
        *   Formation 1: (7,7), (8,6), (8,8), (9,7). (A cross shape).

2.  **Re-evaluating Formations (Shape Matching)**:
    *   **Green (G)**:
        *   G1: `(2,3), (2,5), (3,4), (3,5), (4,3), (4,5)` ->
            `. G . G`
            `. . G G`
            `. G . G`
        *   G2: `(11,9), (11,11), (12,10), (12,11), (13,9), (13,11)` ->
            `. G . G`
            `. . G G`
            `. G . G`
        *   **Count for G shape: 2**
    *   **Red (R)**:
        *   R1: `(2,10), (3,10), (3,11), (3,12), (4,10), (4,11)` ->
            `. R . .`
            `. R R R`
            `. R R .`
        *   R2: `(8,1), (9,0), (9,1), (9,2), (10,0), (10,1)` ->
            `. R . .`
            `R R R .`
            `R R . .`
        *   R3: `(11,5), (12,4), (12,5), (12,6), (13,4), (13,5)` ->
            `. R . .`
            `R R R .`
            `R R . .`
        *   **Count for R shape: 2** (Wait, let's look closer at R1 vs R2/R3).
        *   R1: (2,10), (3,10), (3,11), (3,12), (4,10), (4,11).
            Row 1: . R . .
            Row 2: . R R R
            Row 3: . R R .
        *   R2: (8,1), (9,0), (9,1), (9,2), (10,0), (10,1).
            Row 1: . R . .
            Row 2: R R R .
            Row 3: R R . .
        *   R3: (11,5), (12,4), (12,5), (12,6), (13,4), (13,5).
            Row 1: . R . .
            Row 2: R R R .
            Row 3: R R . .
        *   **Count for R shape (R2/R3): 2**
    *   **Azure (A)**:
        *   A1: (7,7), (8,6), (8,8), (9,7).
        *   **Count for A shape: 1**

3.  **Wait, let's re-examine the Green shape (G)**:
    *   G1:
        (2,3) (2,5)
        (3,4) (3,5)
        (4,3) (4,5)
        Relative to top-left (2,3):
        (0,0), (0,2)
        (1,1), (1,2)
        (2,0), (2,2)
    *   G2:
        (11,9) (11,11)
        (12,10) (12,11)
        (13,9) (13,11)
        Relative to top-left (11,9):
        (0,0), (0,2)
        (1,1), (1,2)
        (2,0), (2,2)
    *   **Count for G shape: 2**

4.  **Re-evaluating Red (R) again**:
    *   R1: (2,10), (3,10), (3,11), (3,12), (4,10), (4,11).
        Relative to (2,10): (0,0), (1,0), (1,1), (1,2), (2,0), (2,1).
    *   R2: (8,1), (9,0), (9,1), (9,2), (10,0), (10,1).
        Relative to (8,1): (0,0), (1,-1), (1,0), (1,1), (2,-1), (2,0).
        Wait, if we normalize the bounding box:
        R2: (8,1), (9,0), (9,1), (9,2), (10,0), (10,1).
        Min row: 8, Min col: 0.
        (8,1) -> (0,1)
        (9,0) -> (1,0)
        (9,1) -> (1,1)
        (9,2) -> (1,2)
        (10,0) -> (2,0)
        (10,1) -> (2,1)
        Shape:
        `. R .`
        `R R R`
        `R R .`
    *   R3: (11,5), (12,4), (12,5), (12,6), (13,4), (13,5).
        Min row: 11, Min col: 4.
        (11,5) -> (0,1)
        (12,4) -> (1,0)
        (12,5) -> (1,1)
        (12,6) -> (1,2)
        (13,4) -> (2,0)
        (13,5) -> (2,1)
        Shape:
        `. R .`
        `R R R`
        `R R .`
    *   **Count for R shape: 2**

5.  **Wait, there is a tie between G and R?** Let's look closer at the input.
    Is there another Red?
    R1: (2,10), (3,10), (3,11), (3,12), (4,10), (4,11).
    Let's check the shape of R1 again:
    (2,10)
    (3,10), (3,11), (3

... [3580 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Seeds (Colored Pixels)**: The active elements that will move. They have colors (R, B, O, etc.) and vertical positions.
    *   **Columns**: The vertical paths through which seeds are organized.
    *   **The Sky (Black Pixels)**: The background/empty space.
    *   **The Ground (Bottom Row)**: The destination where the mounds will rest.

2.  **Identify the AFFORDANCES**:
    *   **Seeds** can be collected by column, sorted by their original vertical order (top-to-bottom), and then redistributed into a triangular shape.
    *   **Columns** act as containers that determine the horizontal center of each resulting pyramid.
    *   **The Grid** provides the coordinate system for the final placement of the pyramids at the bottom.

3.  **Identify the TRANSFORMATION RULE**:
    *   **Step 1: Extraction**: For each column, identify all non-black pixels.
    *   **Step 2: Sorting**: Within each column, maintain the vertical order of colors. The "highest" seed (lowest row index) becomes the "peak" of the pyramid, and the "lowest" seed (highest row index) becomes part of the "base".
    *   **Step 3: Pyramid Construction**: For a column containing $N$ seeds:
        *   The pyramid will have $H$ rows, where $H$ is the smallest integer such that the sum of integers from 1 to $H$ is $\ge N$. However, the metaphor describes a "perfect pile" where the top is 1 seed, the next row is 3, the next is 5, etc., OR a standard triangle where row 1 has 1, row 2 has 2, etc.
        *   Looking at **Example 1**: 8 seeds in column 6. The output shows:
            Row 1 (bottom): 5 seeds (O O O O O)
            Row 2 (above): 3 seeds (O O O)
            Row 3 (above): 1 seed (O)
            Total seeds used: $1 + 3 + 5 = 9$. Wait, Example 1 has 8 seeds. Let's re-count:
            Input Ex 1: Row 1: O, Row 2: O, Row 4: O, Row 5: O, Row 6: O, Row 8: O, Row 10: O, Row 11: O, Row 12: O. Total = 9 seeds.
            Output Ex 1: Row 14: 1 seed, Row 15: 3 seeds, Row 16: 5 seeds. Total = 9 seeds.
            The rule is: The $k$-th layer from the top contains $2k-1$ seeds.
    *   **Step 4: Placement**: The pyramid is centered on the column index where the seeds were found. The base of the pyramid sits on the bottom row of the grid.

### Applying the Rule to Test Input

**Column 2 (Index 2):**
Seeds: R (row 0), R (row 1), R (row 3), R (row 4), B (row 5), B (row 6), B (row 8), B (row 9), B (row 11).
Total seeds: 9.
Order (top to bottom): R, R, R, R, B, B, B, B, B.
Pyramid structure (9 seeds):
- Layer 1 (top): 1 seed (R)
- Layer 2: 3 seeds (R, R, B)
- Layer 3 (base): 5 seeds (B, B, B, B, B)
Total rows: 3.
Placement: Centered at index 2.
Bottom row (row 29): indices 0, 1, 2, 3, 4 (centered at 2).
Row 28: indices 1, 2, 3.
Row 27: index 2.

**Column 8 (Index 8):**
Seeds: O (row 0), O (row 12), O (row 20), O (row 23).
Total seeds: 4.
Order: O, O, O, O.
Pyramid structure (4 seeds):
- Layer 1 (top): 1 seed (O)
- Layer 2: 3 seeds (O, O, O)
Total rows: 2.
Placement: Centered at index 8.
Bottom row (row 29): indices 7, 8, 9.
Row 28: index 8.

**Final Grid Construction:**
The grid is 30 rows (0-29) by 10 columns (0-9).

Column 2 Pyramid (9 seeds):
Row 27: `. . R . . . . . . .`
Row 28: `. R R R . . . . . .`
Row 29: `B B B B B . . . . .`
Wait, let's re-check the centering. If the base is 5 wide and center is 2, indices are 0, 1, 2, 3, 4.
If the base is 3 wide and center is 8, indices are 7, 8, 9.

Let's re-verify the seed order for Col 2:
R (0), R (1), R (3), R (4), B (5), B (6), B (8), B (9), B (11).
Top-down: R, R, R, R, B, B, B, B, B.
Layer 1 (1 seed): R
Layer 2 (3 seeds): R, R, R
Layer 3 (5 seeds): B, B, B, B, B
Wait, the example shows the *highest* seed is at the *peak*.
Example 1: 9 seeds. Top is O. Layer 1: O. Layer 2: O, O, O. Layer 3: O, O, O, O, O.
My Col 2:
Layer 1: R
Layer 2: R, R, R
Layer 3: B, B, B, B, B
Wait, I have 4 Rs and 5 Bs.
Layer 1: R (1)
Layer 2: R, R, R (3)
Layer 3: B, B, B, B, B (5)
Total: 1+3+5 = 9. Correct.

Col 8: 4 seeds (all O).
Layer 1: O (1)
Layer 2: O, O, O (3)
Total: 1+3 = 4. Correct.

**Grid Layout:**
Rows 0-26: All `.`
Row 27: Col 2 = R
Row 28: Col 1, 2, 3 = R; Col 8 = O
Row 29: Col 0, 1, 2, 3, 4 = B; Col 7, 8, 9 = O

Wait, let's re-check the centering for Col 8.
Column index 8. Base width 3. Indices: 7, 8, 9.
Row 28: index 8.
Row 29: indices 7, 8, 9.

Let's re-check Col 2.
Column index 2. Base width 5. Indices: 0, 1, 2, 3, 4.
Row 27: index 2.
Row 28: indices 1, 2, 3.
Row 29: indices 0, 1, 2, 3, 4.

Final check on Col 2 seeds:
R (0), R (1), R (3), R (4), B (5), B (6), B (8), B (9), B (11).
Top-down: R, R, R, R, B, B, B, B, B.
Layer 1 (top): R
Layer 2: R, R, R
Layer 3: B, B, B, B, B
Total: 1+3+5 = 9. Correct.

Final check on Col 8 seeds:
O (0), O (12), O (20), O (23).
Top-down: O, O, O, O.
Layer 1 (top): O
Layer 2: O, O, O
Total: 1+3 = 4. Correct.

ANSWER:
(Rows 0-26 are all dots)
Row 27: . . R . . . . . . .
Row 28: . R R R . . . O . .
Row 29: B B B B B . . O

... [4145 chars truncated]

To solve this puzzle, we first perform a Conceptual Digestion:

1.  **Identify the CAST**:
    *   **The Banner**: A single-colored shape (the "pennant") made of non-black pixels.
    *   **The Anchor (Bottom Hem)**: The lowest row containing the banner's color. This row remains stationary.
    *   **The Night Sky**: The background (black/`.`) which remains unchanged in dimension.

2.  **Identify the AFFORDANCES**:
    *   The **Banner** can be shifted horizontally (leftward).
    *   The **Anchor** is fixed in its original position.
    *   The **Wind** (the transformation rule) applies a leftward shift to every row of the banner based on its vertical distance from the anchor.

3.  **Identify the TRANSFORMATION RULE**:
    *   Find the row index of the bottom-most row of the colored shape (the anchor row, $R_{anchor}$).
    *   For every row $i$ where $i < R_{anchor}$, calculate the distance from the anchor: $dist = R_{anchor} - i$.
    *   Shift all colored pixels in row $i$ to the left by $dist$ units.
    *   If a pixel is shifted beyond the grid boundary (index < 0), it is lost.
    *   The anchor row ($i = R_{anchor}$) remains exactly as it was in the input.

**Applying the rule to the Test Input**:
The color is Orange (`O`).
The input rows containing `O` are:
Row 3: `. . . . . . O O O . . . .`
Row 4: `. . . . O O O . O . . . .`
Row 5: `. . . . O . . . O . . . .`
Row 6: `. . . . O O . . O . . . .`
Row 7: `. . . . . O . . O O . . .`
Row 8: `. . . . . O . . . O . . .`
Row 9: `. . . . . O O O O O . . .` (This is the anchor row, $R_{anchor} = 9$)

**Calculations**:
*   Row 9 (Anchor): $dist = 0$. No shift.
    `. . . . . O O O O O . . .`
*   Row 8: $dist = 1$. Shift left by 1.
    Input: `. . . . . O . . . O . . .` $\rightarrow$ Output: `. . . . O . . . . O . . .`
*   Row 7: $dist = 2$. Shift left by 2.
    Input: `. . . . . O . . O O . . .` $\rightarrow$ Output: `. . . . O . . . . O O . .`
*   Row 6: $dist = 3$. Shift left by 3.
    Input: `. . . . O O . . O . . . .` $\rightarrow$ Output: `. . . O O . . . O . . . .` (Wait, let's re-index carefully)

**Re-indexing carefully (0-indexed)**:
Input Rows:
3: `[6,7,8]`
4: `[4,5,6,8]`
5: `[4,8]`
6: `[4,5,8]`
7: `[5,8,9]`
8: `[5,9]`
9: `[5,6,7,8,9]` (Anchor)

Shifts:
Row 3: $dist = 9-3 = 6$. New indices: `[6-6, 7-6, 8-6]` = `[0, 1, 2]`
Row 4: $dist = 9-4 = 5$. New indices: `[4-5, 5-5, 6-5, 8-5]` = `[-1, 0, 1, 3]` $\rightarrow$ `[0, 1, 3]`
Row 5: $dist = 9-5 = 4$. New indices: `[4-4, 8-4]` = `[0, 4]`
Row 6: $dist = 9-6 = 3$. New indices: `[4-3, 5-3, 8-3]` = `[1, 2, 5]`
Row 7: $dist = 9-7 = 2$. New indices: `[5-2, 8-2, 9-2]` = `[3, 6, 7]`
Row 8: $dist = 9-8 = 1$. New indices: `[5-1, 9-1]` = `[4, 8]`
Row 9: $dist = 0$. New indices: `[5, 6, 7, 8, 9]`

**Constructing Output**:
Row 3: `O O O . . . . . . . . . .`
Row 4: `O O . O . . . . . . . . .`
Row 5: `O . . . O . . . . . . . .`
Row 6: `. O O . . O . . . . . . .`
Row 7: `. . . O . . O O . . . . .`
Row 8: `. . . . O . . . O . . . .`
Row 9: `. . . . . O O O O O . . .`

ANSWER:
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
O O O . . . . . . . . . .
O O . O . . . . . . . . .
O . . . O . . . . . . . .
. O O . . O . . . . . . .
. . . O . . O O . . . . .
. . . . O . . . O . . . .
. . . . . O O O O O . . .
. . . . . . . . . . . . .

To solve this puzzle, we first perform a Conceptual Digestion:

### 1. Conceptual Digestion

**Identify the CAST:**
*   **The Tower:** A shape composed of a single color (the "monolith") located within the grid.
*   **The Foundation:** The bottom-most row of the tower shape.
*   **The Courses:** Each horizontal row of the tower shape above the foundation.
*   **The Dark Plain:** The background (black/`.`) of the grid.

**Identify the AFFORDANCES:**
*   The tower can be "sheared" or "cast sideways."
*   The foundation is "spared" (remains in its original position).
*   Each course above the foundation is shifted horizontally.

**Identify the TRANSFORMATION RULE:**
1.  Locate the tower and identify its bottom row (the foundation).
2.  Determine the vertical position of each row relative to the foundation. Let $h$ be the vertical distance from the foundation (where $h=0$ is the foundation row, $h=1$ is the row immediately above it, etc.).
3.  Shift each row $h$ units to the left (or sideways) based on its height. Looking at the examples:
    *   Example 1: Foundation is at row 8. Row 7 is shifted 1 left, Row 6 is shifted 2 left, Row 5 is shifted 3 left, Row 4 is shifted 4 left.
    *   Example 2: Foundation is at row 7. Row 6 is shifted 1 left, Row 5 is shifted 2 left, Row 4 is shifted 3 left.
    *   Example 3: Foundation is at row 7. Row 6 is shifted 1 left, Row 5 is shifted 2 left, Row 4 is shifted 3 left.
4.  **Rule:** For a row at index $r$, if the foundation is at index $r_{base}$, the shift is $S = r_{base} - r$. Each pixel in that row moves from $(r, c)$ to $(r, c - S)$.

### 2. Applying the Rule to the Test Input

**Input Analysis:**
The orange (O) tower:
Row 3: `. . . . . . O O O . . . .` (Indices 6, 7, 8)
Row 4: `. . . . O O O . O . . . .` (Indices 4, 5, 6, 8)
Row 5: `. . . . O . . . O . . . .` (Indices 4, 8)
Row 6: `. . . . O O . . O . . . .` (Indices 4, 5, 8)
Row 7: `. . . . . O . . O O . . .` (Indices 5, 8, 9)
Row 8: `. . . . . O . . . O . . .` (Indices 5, 9)
Row 9: `. . . . . O O O O O . . .` (Indices 5, 6, 7, 8, 9) -> **Foundation (r_base = 9)**

**Shifts ($S = 9 - r$):**
*   Row 9: $S=0$. Pixels: (9,5), (9,6), (9,7), (9,8), (9,9)
*   Row 8: $S=1$. Pixels: (8,5-1=4), (8,9-1=8)
*   Row 7: $S=2$. Pixels: (7,5-2=3), (7,8-2=6), (7,9-2=7)
*   Row 6: $S=3$. Pixels: (6,4-3=1), (6,5-3=2), (6,8-3=5)
*   Row 5: $S=4$. Pixels: (5,4-4=0), (5,8-4=4)
*   Row 4: $S=5$. Pixels: (4,4-5=-1 -> invalid), (4,5-5=0), (4,6-5=1), (4,8-5=3)
    *Wait, let's re-check Example 1.*
    Example 1: Row 8 is foundation. Row 4 is $h=4$. Shift is 4.
    Input Row 4: `. . . . R R R R R . . . .` (Indices 4,5,6,7,8)
    Output Row 4: `R R R R . . . . . . . . .` (Indices 0,1,2,3)
    $4 - 4 = 0$. Correct.
    Input Row 5: `. . . . R . . . R . . . .` (Indices 4, 8)
    Output Row 5: `R . . . R . . . . . . . .` (Indices 1, 5)
    $4-1=3, 8-1=7$? No. Let's re-calculate.
    Example 1: Row 8 is base. Row 4 is $h=4$. Shift is 4.
    Row 4: $4-4=0, 5-4=1, 6-4=2, 7-4=3, 8-4=4$. Output: `R R R R .` (Indices 0,1,2,3).
    Row 5: $4-4=0, 8-4=4$. Output: `R . . . R` (Indices 0, 4). Wait, the example output for Row 5 is `R . . . R` at indices 1 and 5? No, it's `R . . . R` at indices 1 and 5? Let's look closer.
    Example 1 Output Row 5: `. R . . . R . . . . . . .` (Indices 1 and 5).
    If $r=5, r_{base}=8$, $h=3$. Shift is 3? No, $8-5=3$.
    If $r=4, r_{base}=8$, $h=4$. Shift is 4.
    Let's re-verify Example 1 Row 4: Input indices 4,5,6,7,8. Shift 4 $\rightarrow$ 0,1,2,3,4. Output shows `R R R R` at 0,1,2,3.
    Example 1 Row 5: Input indices 4, 8. Shift 3 $\rightarrow$ 1, 5. Output shows `R` at 1 and 5.
    Example 1 Row 6: Input indices 4, 5, 8, 9. Shift 2 $\rightarrow$ 2, 3, 6, 7. Output shows `R R . . R R`? No, `R R . . R R` at 2,3 and 6,7.
    Example 1 Row 7: Input indices 5, 8, 9. Shift 1 $\rightarrow$ 4, 7, 8. Output shows `R` at 4, 7, 8.
    Example 1 Row 8: Input indices 5, 6, 7, 8, 9. Shift 0 $\rightarrow$ 5, 6, 7, 8, 9.

**Correct Rule:** Shift $S = r_{base} - r$.
Test Input: $r_{base} = 9$.
Row 9: $S=0$. Indices: 5, 6, 7, 8, 9.
Row 8: $S=1$. Indices: 5-1=4, 9-1=8.
Row 7: $S=2$. Indices: 5-2=3, 8-2=6, 9-2=7.
Row 6: $S=3$. Indices: 4-3=1, 5-3=2, 8-3=5.
Row 5: $S=4$. Indices: 4-4=0, 8-4=4.
Row 4: $S=5$. Indices: 4-5=-1 (skip), 5-5=0, 6-5=1, 8-5=3.
Row 3: $S=6$. Indices: 6-6=0, 7-6=1, 8-6=2.

**Mapping to Grid:**
Row 3: `O O O . . . . . . . . . .`
Row 4: `O O . O . . . . . . . . .`
Row 5: `O . . . O . . . . . . . .`
Row 6: `. R R . . R . . . . . . .` (Wait, color is O) $\rightarrow$ `. O O . . O . . . . . . .`
Row 7: `. . . O . . O O . . . . .`
Row 8: `. . . . O . . . O . . . .`
Row 9: `. . . . . O O O O O . . .`

Wait, let's re-check Row 4: Input indices 4, 5, 6, 8. Shift 5: $4-5=-1, 5-5=0, 6-5=1, 8-5=3$.
So Row 4 is `O O . O . . . . . . . . .`

Let's re-verify Row 3: Input indices 6, 7, 8. Shift 6: $6-6=0, 7-6=1, 8-6=2$.
So Row 3 is `O O O . . . . . . . . . .`

Final check of the rows:
R3: 0,1,2
R4: 0,1,3
R5: 0,4
R6: 1,2,5

... [642 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **The Pastry:** The colored shape (the non-black pixels) representing a multi-layered object.
    *   **The Layers:** Horizontal rows of the colored shape.
    *   **The Countertop:** The black background (`.`).
    *   **The Anchor:** The bottom-most row of the colored shape, which remains in its original horizontal position.

2.  **Identify the AFFORDANCES:**
    *   **Layers** can be shifted horizontally.
    *   **The Anchor (bottom layer)** is fixed.
    *   **Higher layers** can slide leftward (based on the training examples, the "sliding" is a leftward shift relative to the bottom layer).

3.  **Identify the TRANSFORMATION RULE:**
    *   Identify the bottom-most row containing the colored shape. This is the anchor.
    *   For every row above the anchor, calculate its vertical distance from the anchor row.
    *   Let $d$ be the distance (number of rows) from the anchor row (where $d=0$ for the anchor row, $d=1$ for the row immediately above it, etc.).
    *   Shift the pixels in that row $d$ units to the **left**.
    *   *Correction based on Example 1:* In Example 1, the bottom row is at row 8. The row at row 7 is shifted 1 left. The row at row 6 is shifted 2 left. The row at row 5 is shifted 3 left. The row at row 4 is shifted 4 left.
    *   Rule: For a row at index $r$, if the anchor row is at index $R_{anchor}$, the shift is $S = R_{anchor} - r$. The new column index for a pixel at $(r, c)$ is $c' = c - S$.

### Applying the Rule to Test Input

**Input Analysis:**
The shape is Orange (O).
Rows containing 'O':
Row 3: `. . . . . . O O O . . . .` (Indices 6, 7, 8)
Row 4: `. . . . O O O . O . . . .` (Indices 4, 5, 6, 8)
Row 5: `. . . . O . . . O . . . .` (Indices 4, 8)
Row 6: `. . . . O O . . O . . . .` (Indices 4, 5, 8)
Row 7: `. . . . . O . . O O . . .` (Indices 5, 8, 9)
Row 8: `. . . . . O . . . O . . .` (Indices 5, 9)
Row 9: `. . . . . O O O O O . . .` (Indices 5, 6, 7, 8, 9)

**Anchor Row:** Row 9 (Index 9).
**Shift Calculation ($S = 9 - r$):**
*   Row 9: $S=0$. Pixels: (9,5), (9,6), (9,7), (9,8), (9,9) $\rightarrow$ **(9,5), (9,6), (9,7), (9,8), (9,9)**
*   Row 8: $S=1$. Pixels: (8,5), (8,9) $\rightarrow$ (8,4), (8,8)
*   Row 7: $S=2$. Pixels: (7,5), (7,8), (7,9) $\rightarrow$ (7,3), (7,6), (7,7)
*   Row 6: $S=3$. Pixels: (6,4), (6,5), (6,8) $\rightarrow$ (6,1), (6,2), (6,5)
*   Row 5: $S=4$. Pixels: (5,4), (5,8) $\rightarrow$ (5,0), (5,4)
*   Row 4: $S=5$. Pixels: (4,4), (4,5), (4,6), (4,8) $\rightarrow$ (4,-1) [out], (4,0), (4,1), (4,3)
*   Row 3: $S=6$. Pixels: (3,6), (3,7), (3,8) $\rightarrow$ (3,0), (3,1), (3,2)

**Wait, let's re-verify the shift direction from Example 1.**
Example 1:
Input Row 3: `R R R R R` (cols 4-8)
Input Row 8: `R R R R R` (cols 5-9)
Output Row 3: `R R R R` (cols 0-3)
Output Row 8: `R R R R R` (cols 5-9)
Shift for Row 3: $8 - 3 = 5$. $4 - 5 = -1$.
Wait, the example output for Row 3 is `R R R R` at indices 0, 1, 2, 3.
Original indices: 4, 5, 6, 7, 8.
If we shift 4, 5, 6, 7, 8 by 5 to the left, we get -1, 0, 1, 2, 3.
The example output shows `R R R R` at 0, 1, 2, 3. This means the shift is $S = R_{anchor} - r$.
Let's re-calculate for Test Input:
Row 9 (Anchor): 5, 6, 7, 8, 9
Row 8: $S=1$. (5-1=4), (9-1=8) $\rightarrow$ 4, 8
Row 7: $S=2$. (5-2=3), (8-2=6), (9-2=7) $\rightarrow$ 3, 6, 7
Row 6: $S=3$. (4-3=1), (5-3=2), (8-3=5) $\rightarrow$ 1, 2, 5
Row 5: $S=4$. (4-4=0), (8-4=4) $\rightarrow$ 0, 4
Row 4: $S=5$. (4-5=-1), (5-5=0), (6-5=1), (8-5=3) $\rightarrow$ 0, 1, 3 (ignoring -1)
Row 3: $S=6$. (6-6=0), (7-6=1), (8-6=2) $\rightarrow$ 0, 1, 2

**Final Grid Construction:**
Row 3: `O O O . . . . . . . . . .`
Row 4: `O O . O . . . . . . . . .`
Row 5: `O . . . O . . . . . . . .`
Row 6: `. O O . . O . . . . . . .`
Row 7: `. . . O . . O O . . . . .`
Row 8: `. . . . O . . . O . . . .`
Row 9: `. . . . . O O O O O . . .`

Let's double check Row 4: Input (4,4), (4,5), (4,6), (4,8). Shift 5.
4-5=-1 (skip), 5-5=0, 6-5=1, 8-5=3. Result: 0, 1, 3. Correct.
Let's double check Row 3: Input (3,6), (3,7), (3,8). Shift 6.
6-6=0, 7-6=1, 8-6=2. Result: 0, 1, 2. Correct.

ANSWER:
. . . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
O O O . . . . . . . . . .
O O . O . . . . . . . . .
O . . . O . . . . . . . .
. O O . . O . . . . . . .
. . . O . . O O . . . . .
. . . . O . . . O . . . .
. . . . . O O O O O . . .
. . . . . . . . . . . .

To solve this puzzle, we first perform a Conceptual Digestion:

### 1. Conceptual Digestion

**Identify the CAST:**
*   **Dark Sky (`.`):** The background/canvas.
*   **Colored Cloud (Single color, e.g., `O`):** A hollow shape made of a specific color.
*   **Vertical Layers (Rows):** The horizontal slices of the cloud.

**Identify the AFFORDANCES:**
*   The cloud's color and vertical extent (number of rows it occupies) are preserved.
*   The "base" of the cloud is the bottom-most row of the shape.
*   Each row above the base can be shifted horizontally.

**Identify the TRANSFORMATION RULE:**
1.  Find the bottom-most row of the colored shape. This is the "anchor" (height $h=0$).
2.  For every other row in the shape, calculate its height $h$ above the base row.
3.  Shift every pixel in that row horizontally by a distance equal to its height $h$ above the base.
4.  Looking at the training examples:
    *   Example 1: Base is at row 8. Row 7 is $h=1$ above base $\rightarrow$ shift left by 1. Row 6 is $h=2$ above base $\rightarrow$ shift left by 2. Row 3 is $h=5$ above base $\rightarrow$ shift left by 5.
    *   Wait, let's re-examine Example 1:
        Input Row 3: `R R R R R` (at col 4-8). Output Row 3: `R R R R` (at col 0-3). Shift is -4.
        Input Row 8: `R R R R R` (at col 5-9). Output Row 8: `R R R R R` (at col 5-9). Shift is 0.
        The shift for row $r$ is $-(H_{max} - r)$, where $H_{max}$ is the index of the bottom row.
    *   Let's check Example 2: Base is row 7. Row 3 is $h=4$ above base. Shift is -4.
        Input Row 3: `M M M` (col 4-6). Output Row 3: `M M M` (col 0-2). Correct.
    *   **Rule:** For each row $r$, if $r_{base}$ is the index of the bottom row of the shape, the new column index $c'$ is $c' = c - (r_{base} - r)$.

### 2. Applying the Rule to Test Input

**Input Analysis:**
The shape is Orange (`O`).
Rows containing `O`:
Row 3: `. . . . . . O O O . . . .` (Cols 6, 7, 8)
Row 4: `. . . . O O O . O . . . .` (Cols 4, 5, 6, 8)
Row 5: `. . . . O . . . O . . . .` (Cols 4, 8)
Row 6: `. . . . O O . . O . . . .` (Cols 4, 5, 8)
Row 7: `. . . . . O . . O O . . .` (Cols 5, 8, 9)
Row 8: `. . . . . O . . . O . . .` (Cols 5, 9)
Row 9: `. . . . . O O O O O . . .` (Cols 5, 6, 7, 8, 9)

**Base Row:** Row 9 ($r_{base} = 9$).

**Transformations ($c' = c - (9 - r)$):**
*   Row 9 ($h=0$): $c' = c - 0$. Cols: 5, 6, 7, 8, 9.
*   Row 8 ($h=1$): $c' = c - 1$. Cols: (5-1), (9-1) $\rightarrow$ 4, 8.
*   Row 7 ($h=2$): $c' = c - 2$. Cols: (5-2), (8-2), (9-2) $\rightarrow$ 3, 6, 7.
*   Row 6 ($h=3$): $c' = c - 3$. Cols: (4-3), (5-3), (8-3) $\rightarrow$ 1, 2, 5.
*   Row 5 ($h=4$): $c' = c - 4$. Cols: (4-4), (8-4) $\rightarrow$ 0, 4.
*   Row 4 ($h=5$): $c' = c - 5$. Cols: (4-5), (5-5), (6-5), (8-5) $\rightarrow$ -1 (ignore), 0, 1, 3.
    *Wait, let's re-calculate Row 4:* Input cols 4, 5, 6, 8. $c' = c - 5 \rightarrow$ -1, 0, 1, 3. Since -1 is out of bounds, we use 0, 1, 3.
*   Row 3 ($h=6$): $c' = c - 6$. Cols: (6-6), (7-6), (8-6) $\rightarrow$ 0, 1, 2.

**Constructing Output:**
Row 3: `O O O . . . . . . . . . .`
Row 4: `O O . O . . . . . . . . .` (Cols 0, 1, 3)
Row 5: `O . . . O . . . . . . . .` (Cols 0, 4)
Row 6: `. O O . . O . . . . . . .` (Cols 1, 2, 5)
Row 7: `. . . O . . O O . . . . .` (Cols 3, 6, 7)
Row 8: `. . . . O . . . O . . . .` (Cols 4, 8)
Row 9: `. . . . . O O O O O . . .` (Cols 5, 6, 7, 8, 9)

Let's double check Row 4: Input `O O O . O` at cols 4, 5, 6, 8.
$c=4 \rightarrow 4-5 = -1$ (out)
$c=5 \rightarrow 5-5 = 0$
$c=6 \rightarrow 6-5 = 1$
$c=8 \rightarrow 8-5 = 3$
Result: `O O . O` at 0, 1, 3.

Let's double check Row 3: Input `O O O` at cols 6, 7, 8.
$c=6 \rightarrow 6-6 = 0$
$c=7 \rightarrow 7-6 = 1$
$c=8 \rightarrow 8-6 = 2$
Result: `O O O` at 0, 1, 2.

Final Grid Construction:
Row 0-2: `.`
Row 3: `O O O . . . . . . . . . .`
Row 4: `O O . O . . . . . . . . .`
Row 5: `O . . . O . . . . . . . .`
Row 6: `. O O . . O . . . . . . .`
Row 7: `. . . O . . O O . . . . .`
Row 8: `. . . . O . . . O . . . .`
Row 9: `. . . . . O O O O O . . .`
Row 10: `.`

ANSWER:
. . . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
O O O . . . . . . . . . .
O O . O . . . . . . . . .
O . . . O . . . . . . . .
. O O . . O . . . . . . .
. . . O . . O O . . . . .
. . . . O . . . O . . . .
. . . . . O O O O O . . .
. . . . . . . . . . . . .

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Colored Pixels (Non-Orange):** These represent the "nesting dolls." Each unique color (excluding the background color 'O') represents a doll of a specific size.
    *   **Orange Pixels ('O'):** These represent the "workshop table" (the background/empty space).
    *   **The Output Grid:** A square assembly of concentric shells.

2.  **Identify the AFFORDANCES**:
    *   **Colors:** Each color has a "weight" determined by the total count of its pixels in the input grid.
    *   **Counts:** The number of pixels of a specific color determines its size (how many layers it will occupy or how large the doll is).
    *   **Concentricity:** The output must be a square where colors are arranged in layers (shells) around a center.

3.  **Identify the TRANSFORMATION RULE**:
    *   **Step 1: Count the pixels** for every color present in the input (excluding 'O').
    *   **Step 2: Rank the colors** by their pixel count (descending order). The color with the most pixels is the "heaviest" (outermost shell), the second most is the next shell, and so on, until the color with the fewest pixels is the "tiniest" (the center).
    *   **Step 3: Determine the size of the output square.** The total number of pixels of all non-orange colors determines the area of the final square. However, the description says "each shell is exactly one layer thick" and "grows symmetrically outward." This implies the output is a square of size $N \times N$ where $N$ is the smallest odd integer such that $N^2 \ge \text{total non-orange pixels}$. Looking at the examples:
        *   Ex 1: Y (4), M (2). Total = 6. Output is $3 \times 3$ (9 pixels). Wait, the rule is simpler: The output is a square where the colors are layered.
        *   Ex 2: B (7), A (5), X (4). Total = 16. Output is $5 \times 5$ (25 pixels).
    *   **Refined Rule:** The output is a square of size $N \times N$. The colors are placed in concentric rings starting from the outermost ring. The color with the highest count forms the outermost ring, the next color forms the ring inside that, etc. The center is the color with the lowest count.
    *   **Wait, let's re-examine Example 1:** Y=4, M=2. Total = 6. Output $3 \times 3$. The colors are Y (outer) and M (inner).
    *   **Re-examining Example 2:** B=7, A=5, X=4. Total = 16. Output $5 \times 5$. B is outer, A is middle, X is center.
    *   **Correct Rule:**
        1. Count pixels of each color (excluding 'O').
        2. Sort colors by count: Largest count = Outermost shell, Smallest count = Center.
        3. The output is a square of size $N \times N$ where $N$ is the smallest odd integer such that $N^2$ is enough to accommodate the shells. Actually, the size is determined by the largest color's ability to form a perimeter.
        4. Looking at Example 2: B (7) is the outermost. In a $5 \times 5$ grid, the outermost ring has 16 cells. If B is the outermost, it should fill the ring. But B only has 7 pixels.
        5. **Correction:** The "weight" is the count. The color with the *most* pixels is the *outermost* shell. The color with the *least* is the *innermost*. The output size is determined by the number of colors. If there are $K$ colors, the output is a square of size $(2K-1) \times (2K-1)$? No.
        6. **Let's look at Example 1 again:** Y=4, M=2. 2 colors. Output $3 \times 3$.
        7. **Example 2:** B=7, A=5, X=4. 3 colors. Output $5 \times 5$.
        8. **Pattern:** If there are $K$ colors, the output is a square of size $(2K-1) \times (2K-1)$.
        9. **Placement:** The color with the highest count is the outermost ring (layer 1), the next is layer 2, ..., the color with the lowest count is the center (layer $K$).

### Applying to Test Input

1.  **Count colors (excluding 'O' and '.'):**
    *   W (Maroon): 2
    *   R (Red): 4
    *   X (Grey): 4
    *   A (Azure): 1
    *   . (Black): 4 (Wait, '.' is a color in the key)
    *   B (Blue): 9
    *   Total colors: W, R, X, A, ., B (6 colors).
    *   Wait, let's re-count carefully:
        *   W: (0,2), (1,2) -> **2**
        *   R: (0,5), (0,6), (0,7), (1,5), (1,7) -> **5**
        *   X: (2,1), (2,2), (3,0), (3,1), (4,0) -> **5**
        *   A: (3,3) -> **1**
        *   .: (4,6), (5,5), (5,6), (6,6), (6,7), (7,7) -> **6**
        *   B: (5,2), (6,1), (6,2), (6,3), (7,0), (7,1), (7,2), (7,3), (7,4) -> **9**

2.  **Rank colors by count (Descending):**
    1. B (9) - Outermost
    2. . (6) - 2nd layer
    3. R (5) - 3rd layer (Tie with X)
    4. X (5) - 3rd layer (Tie with R)
    5. W (2) - 4th layer
    6. A (1) - Center

3.  **Determine Output Size:**
    *   Number of colors $K = 6$.
    *   Size $N = 2K - 1 = 2(6) - 1 = 11$.
    *   Wait, let's check the examples again.
    *   Ex 1: 2 colors $\rightarrow 3 \times 3$.
    *   Ex 2: 3 colors $\rightarrow 5 \times 5$.
    *   Test: 6 colors $\rightarrow 11 \times 11$.

4.  **Construct the $11 \times 11$ grid:**
    *   Layer 1

... [4610 chars truncated]

### Conceptual Digestion

1.  **CAST**:
    *   **Input Grid**: An "ocean surface" containing scattered colored "debris" (pixels of various colors).
    *   **Colors**: Each color represents a specific type of debris.
    *   **Output Grid**: A "square reef" consisting of concentric rings of colors.
    *   **Colors in Output**: The colors present in the input, rearranged by frequency.

2.  **AFFORDANCES**:
    *   **Input Colors**: Can be counted to determine their abundance.
    *   **Output Structure**: Can be organized into concentric square rings. The size of the output square is determined by the total number of colors present (or more specifically, the number of layers required to accommodate the colors). Looking at the examples, the output size is determined by the number of colors found in the input.
    *   **Example 1 Analysis**: Input colors: O (21), Y (4), M (3). Total colors = 3. Output is 3x3. Layers: Outer (most abundant) = O? Wait, let's re-count.
        *   Ex 1 Input: O=21, Y=4, M=3. Output: Y=5, M=1, O=3? No.
        *   Let's re-examine Ex 1: Input: O (21), Y (4), M (3). Output: Y (5), M (1), O (3)? No, the output is:
            Y Y Y
            Y M Y
            Y Y Y
            Colors in output: Y (8), M (1). Wait, the colors in the output are Y and M. Where is O?
            Let's re-read: "the most abundant color forms the outermost barrier, each less abundant color the next inner layer, with the rarest piece at the very center."
            Ex 1 Input: O (21), Y (4), M (3).
            Wait, the output is 3x3. The colors used are Y and M. This implies the "most abundant" color in the *input* might be the one that *doesn't* get used if it's the background (O)? No, O is a color.
            Let's look at Ex 2: Input: O (53), A (5), X (4), B (7).
            Output: B (13), A (7), X (1).
            Wait, the colors in the output are B, A, X. The color O is missing.
            **Crucial Realization**: The "background" color (the most abundant color, usually 'O') is treated as the "ocean" and is discarded. The remaining colors are sorted by abundance and placed in concentric rings.

3.  **TRANSFORMATION RULE**:
    1.  Count the occurrences of every color in the input grid, excluding the most abundant color (the "ocean").
    2.  Sort the remaining colors by their frequency in descending order.
    3.  Determine the size of the output square. The number of colors (excluding the background) determines the number of layers. If there are $N$ colors, the output is a $(2N-1) \times (2N-1)$ grid? No.
    4.  Let's check Ex 1: Colors (excluding O): Y (4), M (3). $N=2$. Output size is $3 \times 3$.
    5.  Let's check Ex 2: Colors (excluding O): B (7), A (5), X (4). $N=3$. Output size is $5 \times 5$.
    6.  Rule for size: If there are $N$ colors, the output is a $(2N-1) \times (2N-1)$ grid.
    7.  Rule for placement:
        *   Layer 1 (outermost): Most abundant color.
        *   Layer 2: Second most abundant.
        *   ...
        *   Center: Least abundant.

### Applying to Test Input

1.  **Count colors in Test Input**:
    *   O: 33
    *   W: 2
    *   R: 4
    *   X: 4
    *   A: 1
    *   .: 5
    *   B: 6
    *   (Total colors excluding O: W(2), R(4), X(4), A(1), .(5), B(6))
    *   Wait, let's re-count carefully.
    *   O: 33
    *   W: 2 (at [0,2], [1,2])
    *   R: 4 (at [0,4], [0,5], [0,6], [1,4], [1,6]) -> actually R is 5.
    *   X: 4 (at [2,0], [3,0], [3,1], [4,0])
    *   A: 1 (at [3,3])
    *   .: 5 (at [4,5], [5,5], [5,6], [6,6], [6,7], [7,7]) -> actually . is 6.
    *   B: 6 (at [5,1], [6,1], [6,2], [6,3], [7,0], [7,1], [7,2], [7,3], [7,4]) -> actually B is 9.

    Let's re-count the Test Input very carefully:
    O O W O O R R R (O:5, W:1, R:3)
    O O W O O R O R (O:4, W:1, R:2)
    X O O O O O O O (X:1, O:7)
    X X O A O O O O (X:2, O:5, A:1)
    X O O O O . O O (X:1, O:5, .:1)
    O O B O O . . O (O:4, B:1, .:2)
    O B B B O O . . (O:2, B:3, .:2)
    B B B B B O O . (B:5, O:2, .:1)

    Total counts:
    O: 5+4+7+5+5+4+2+2 = 34
    W: 1+1 = 2
    R: 3+2 = 5
    X: 1+2+1 = 4
    A: 1
    .: 1+2+2+1 = 6
    B: 1+3+5 = 9

    Colors excluding O:
    B: 9
    .: 6
    R: 5
    X: 4
    W: 2
    A: 1

    Number of colors (N) = 6.
    Output size = $(2 \times 6 - 1) = 11 \times 11$.
    Wait, let's look at the examples again.
    Ex 1: N=2, Size=3x3.
    Ex 2: N=3, Size=5x5.
    Test Input: N=6, Size=11x11.

    Wait, let's re-verify the counts for Ex 2.
    Ex 2: O (background), A (5), X (4), B (7).
    N=3. Size = 5x5.
    Sorted: B (7), A (5), X (4).
    Layer 1 (Outer): B
    Layer 2: A
    Layer 3 (Center): X
    Output:
    B B B B B
    B A A A B
    B A X A B
    B A A A B
    B B B B B
    This matches perfectly.

    Now for the Test Input:
    Colors: B(9), .(6), R(5), X(4), W(2), A(1).
    N=6. Size = 11x11.
    Layer 1 (Outer): B
    Layer 2: .
    Layer 3: R
    Layer 4: X
    Layer 5: W
    Layer 6 (Center): A

    Wait, let me re-count 

... [4501 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Colors (Essences)**: Each unique color (excluding the background/empty color 'O') represents a "divine essence."
    *   **Counts**: The frequency of each color in the input grid determines its rank.
    *   **The Mandala**: The output grid is a square structure composed of concentric rings.

2.  **Identify the AFFORDANCES**:
    *   **Counting**: We can count the occurrences of each non-'O' color.
    *   **Ranking**: We can rank colors from most frequent to least frequent.
    *   **Ring Placement**: Colors can be placed into concentric square rings.

3.  **Identify the TRANSFORMATION RULE**:
    *   **Step 1: Count frequencies**: Count the occurrences of every color present in the input (excluding 'O').
    *   **Step 2: Rank colors**: Sort the colors by frequency in descending order.
    *   **Step 3: Determine size**: The size of the output square is determined by the number of unique colors found. If there are $N$ colors, the output is an $N \times N$ grid (or more accurately, the number of rings determines the size; looking at examples, the output size is $2k-1$ where $k$ is the number of colors, or simply the number of colors determines the layers). Let's re-examine:
        *   Ex 1: Colors O (many), Y (4), M (2). Wait, O is the background. Colors are Y (4), M (2). Output is 3x3.
        *   Ex 2: Colors O (bg), A (5), X (4), B (7). Sorted: B (7), A (5), X (4). Output is 5x5.
        *   Ex 3: Colors O (bg), A (7), G (3), W (1), . (2). Sorted: A (7), . (2), G (3), W (1). Wait, the order is based on frequency.
    *   **Refined Rule**:
        1. Identify all colors present in the input *except* 'O' (the background).
        2. Count the occurrences of each color.
        3. Sort colors by frequency: Most frequent $\rightarrow$ Least frequent.
        4. Create a square grid where the outermost ring is the most frequent color, the next ring is the second most frequent, and so on, until the center is the least frequent.
        5. The size of the grid is determined by the number of colors. If there are $k$ colors, the grid is $(2k-1) \times (2k-1)$.

**Re-evaluating Example 1 with Refined Rule**:
Input colors: Y (4), M (2). (O is background).
$k=2$. Size = $2(2)-1 = 3 \times 3$.
Rank 1: Y. Rank 2: M.
Ring 1 (outer): Y. Ring 2 (center): M.
Output:
Y Y Y
Y M Y
Y Y Y
Matches!

**Re-evaluating Example 2**:
Input colors: A (5), X (4), B (7).
$k=3$. Size = $2(3)-1 = 5 \times 5$.
Rank 1: B (7). Rank 2: A (5). Rank 3: X (4).
Ring 1 (outer): B. Ring 2: A. Ring 3 (center): X.
Output:
B B B B B
B A A A B
B A X A B
B A A A B
B B B B B
Matches!

**Re-evaluating Example 3**:
Input colors: A (7), G (3), W (1), . (2).
$k=4$. Size = $2(4)-1 = 7 \times 7$.
Rank 1: A (7). Rank 2: G (3). Rank 3: . (2). Rank 4: W (1).
Ring 1: A. Ring 2: G. Ring 3: .. Ring 4: W.
Output:
A A A A A A A
A . . . . . A (Wait, the example shows A . . . . . A)
Let's look closer at Example 3 output.
Row 1: A A A A A A A
Row 2: A . . . . . A (This is a ring of '.')
Row 3: A . G G G . A (This is a ring of 'G')
Row 4: A . G W G . A (This is a ring of 'W')
Wait, the order in Example 3 is A (7), . (2), G (3), W (1).
The frequency order is A(7) > G(3) > .(2) > W(1).
The rings in Example 3 are:
Ring 1 (outer): A
Ring 2: .
Ring 3: G
Ring 4 (center): W
This matches the frequency order: A(7), G(3), .(2), W(1)? No.
Let's re-count Example 3:
A: 7
G: 3
W: 1
.: 2
Frequency order: A, G, ., W.
Example 3 Output Rings:
Ring 1: A
Ring 2: .
Ring 3: G
Ring 4: W
This means the order is A, then the next most frequent is G, but the example shows '.' as the second ring.
Let's re-count '.' in Example 3:
Input:
. . (2)
. . (2)
Total '.' = 4.
Ah! Let's re-count everything in Example 3:
A: 7
G: 3
W: 1
.: 4
Frequency order: A (7), . (4), G (3), W (1).
Ring 1: A. Ring 2: .. Ring 3: G. Ring 4: W.
This matches the output perfectly!

### Applying to Test Input

**Step 1: Count frequencies in Test Input**
Input:
O O W O O R R R (W:1, R:3)
O O W O O R O R (W:1, R:2) -> Total W:2, R:5
X O O O O O O O (X:1)
X X O A O O O O (X:2, A:1) -> Total X:3, A:1
X O O O O . O O (X:1, .:1) -> Total X:4, .:1
O O B O O . . O (B:1, .:2) -> Total B:1, .:3
O B B B O O . . (B:3, .:2) -> Total B:4, .:5
B B B B B O O . (B:5, .:1) -> Total B:9, .:6

Total counts:
B: 9
.: 6
R: 5
X: 4
W: 2
A: 1

**Step 2: Rank colors by frequency**
1. B (9)
2. . (6)
3. R (5)
4. X (4)
5. W (2)
6. A (1)

**Step 3: Determine size**
Number of colors $k = 6$.
Size = $2(6) - 1 = 11 \times 11$.

**Step 4: Construct the Mandala**
Ring 1 (outermost, 11x11): B
Ring 2 (10x10): .
Ring 3 (9x9): R
Ring 4 (8x8): X
Ring 5 (7x7): W
Ring 6 (center, 5x5? No, center is 1x1): A

Wait, let's check the ring math.
$k=1 \rightarrow 1 \times 1$
$k=2 \rightarrow 3 \times 3$
$k=3 \rightarrow 5 \times 5$
$k=4 \rightarrow 7 \times 7$
$k=5 \rightarrow 9 \times 9$
$k=6 \rightarrow 11 \times 11$

Ring 1 (11x11): B
Ring 2 (9x9): .
Ring 3 (7x7): R
Ring 4 (5x5): X
Ring 

... [3851 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Ingredients (Colors)**: Each color represents a different ingredient.
    *   **Quantities**: The number of pixels of each color determines its rank in the "terrine" assembly.
    *   **The Terrine (Output Grid)**: A square grid where layers are formed by the ranked ingredients.

2.  **Identify the AFFORDANCES**:
    *   **Colors**: Can be counted to determine priority.
    *   **Grid Space**: Can be organized into concentric square layers (shells).

3.  **Identify the TRANSFORMATION RULE**:
    *   **Step 1: Count the occurrences** of every color present in the input grid (excluding the background/empty space if applicable, though here we count all non-black/background colors).
    *   **Step 2: Rank the colors** from most frequent to least frequent.
    *   **Step 3: Determine the output size**. Looking at the examples, the output size is determined by the "innermost" possible square structure or a fixed size relative to the input's content. However, a more consistent observation is that the output is a square grid where the most frequent color forms the outermost layer, the second most frequent forms the next layer inward, and so on.
    *   **Step 4: Determine the output dimensions**. In Example 1 (6x6 input), output is 3x3. In Example 2 (9x9 input), output is 5x5. In Example 3 (7x7 input), output is 7x7. Wait, let's re-examine.
    *   **Re-evaluating Step 4 (Dimensions)**:
        *   Ex 1: Input 6x6, Output 3x3.
        *   Ex 2: Input 9x9, Output 5x5.
        *   Ex 3: Input 7x7, Output 7x7.
        *   Looking at the "jewel" (the scarcest ingredient): The output size is determined by the number of layers needed to accommodate all ranked ingredients. The size is $2n-1$ where $n$ is the number of unique colors? No.
        *   Let's look at the "layers" in the output. The output is a square where each layer is a different color. The number of layers is equal to the number of unique colors found in the input.
        *   If there are $k$ colors, the output is a $(2k-1) \times (2k-1)$ grid?
        *   Ex 1: Colors O, Y, M. $k=3$. Output $(2(3)-1) = 5 \times 5$? No, output is 3x3.
        *   Ex 2: Colors O, A, X, B. $k=4$. Output $(2(4)-1) = 7 \times 7$? No, output is 5x5.
        *   Ex 3: Colors O, A, G, W, .. (black). $k=5$. Output 7x7.
        *   **Correct Rule for Size**: The output is a square of size $(2 \times (\text{number of unique colors}) - 1)$? No.
        *   Let's count unique colors in Ex 1: O (18), Y (4), M (2). $k=3$. Output is 3x3.
        *   Ex 2: O (48), A (7), X (4), B (7). Wait, B and A are close. Let's recount. O is most, then B, then A, then X. $k=4$. Output is 5x5.
        *   Ex 3: O (25), A (10), G (5), W (1), . (6). $k=5$. Output is 7x7.
        *   **The pattern is: Output size = $2 \times (\text{number of unique colors}) - 1$ is wrong. Let's try: Output size = $2 \times (\text{number of colors excluding the most frequent}) + 1$.**
        *   Ex 1: Colors (O, Y, M). Most frequent is O. Remaining: Y, M (2 colors). Size = $2(2)+1 = 5$? No, output is 3x3.
        *   Let's try: **Output size = $2 \times (\text{number of unique colors}) - 3$?**
        *   Ex 1: $k=3 \rightarrow 2(3)-3 = 3$.
        *   Ex 2: $k=4 \rightarrow 2(4)-3 = 5$.
        *   Ex 3: $k=5 \rightarrow 2(5)-3 = 7$.
        *   **This works!**

    *   **Step 5: Construct the layers**:
        1. Count colors.
        2. Sort colors by frequency (descending).
        3. Create a square grid of size $S = 2k - 3$.
        4. Fill the outermost layer with the 1st color, the next layer with the 2nd color, etc.

### Applying to Test Input

**1. Count colors in Test Input:**
Input (8x8):
O O W O O R R R
O O W O O R O R
X O O O O O O O
X X O A O O O O
X O O O O . O O
O O B O O . . O
O B B B O O . .
B B B B B O O .

*   O: 33
*   B: 9
*   R: 5
*   X: 4
*   W: 2
*   A: 1
*   .: 6 (Wait, is '.' a color? In Ex 3, '.' is treated as a color. Let's include it.)

**Recounting with '.' as a color:**
*   O: 33
*   B: 9
*   .: 6
*   R: 5
*   X: 4
*   W: 2
*   A: 1

**2. Rank colors (Frequency):**
1. O (33)
2. B (9)
3. . (6)
4. R (5)
5. X (4)
6. W (2)
7. A (1)

**3. Determine Output Size:**
Number of unique colors $k = 7$.
Size $S = 2(7) - 3 = 11$.
Wait, the input is 8x8. An 11x11 output from an 8x8 input seems unlikely for these puzzles. Let's re-examine the size rule.

**Re-re-evaluating Size Rule:**
Ex 1: $k=3$, Size 3.
Ex 2: $k=4$, Size 5.
Ex 3: $k=5$, Size 7.
The rule $S = 2k - 3$ is consistent.
Let's re-count the colors in the Test Input very carefully.
O: (1,1),(1,2),(1,4),(1,5),(2,1),(2,2),(2,4),(2,5),(2,7),(3,2),(3,3),(3,4),(3,5),(3,6),(3,7),(3,8),(4,3),(4,4),(4,5),(4,6),(4,7),(4,8),(5,2),(5,3),(5,4),(5,5),(5,7),(6,1),(6,2),(6,4),(6,5),(7,5),(7,6),(8,6),(8,7) -> Total O = 35.
W: (1,3),(2,3) -> Total W = 2.
R: (1,6),(1,7),(1,8),(2,6),(2,8) -> Total R = 5.
X: (3,1),(4,1),(4,2),(5,1) -> Total X = 4.
A: (4,4) -> Total A = 1.
.: (5,6),(6,6),

... [4456 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Colored pixels (non-O/non-.)**: These represent the "players" belonging to different "teams."
    *   **O (Orange) and . (Black)**: These represent the "field" or empty space.
    *   **The Teams**: Each unique color (excluding O and .) represents a team. The number of pixels of that color determines the team's size.

2.  **Identify the AFFORDANCES**:
    *   The colors can be counted to determine the size of each team.
    *   The teams can be rearranged into concentric square rings.
    *   The size of the final square is determined by the total number of players (or more specifically, the largest team dictates the outer boundary, and the layers are built inward).

3.  **Identify the TRANSFORMATION RULE**:
    *   **Step 1: Count the players.** Count the occurrences of each color (excluding O and .).
    *   **Step 2: Rank the teams.** Sort the colors by their count in descending order.
    *   **Step 3: Determine the grid size.** The largest team (size $N$) will form the outermost ring of a square. A ring of size $N$ implies a square of side length $S$ where $N = 4S - 4$. However, looking at the examples, the rule is simpler: the teams are placed in concentric squares. The largest team forms the outermost perimeter, the second largest forms the next ring, and so on, until the smallest team is in the center.
    *   **Step 4: Construct the output.**
        *   Example 1: Y (4), M (3). Output is 3x3. Y is the perimeter (8 cells needed? No, Y has 4, M has 3. Wait, let's re-count).
        *   *Re-evaluating Example 1*: Input: O=24, Y=4, M=3. Output: Y=8, M=1. This implies the *total number of players* is not the only factor, but rather the *relative order* and the *available space*.
        *   *Correction*: The rule is: Count the players of each color. Sort colors by count (descending). Create a square grid where the largest team forms the outermost ring, the second largest the next ring, etc. The size of the grid is determined by the largest team's count. If a team has $C$ players, it fills a ring.
        *   *Wait, let's look at Example 2*: X=4, A=5, B=7. Sorted: B(7), A(5), X(4). Output is 5x5. B is the perimeter (16 cells), A is the next ring (8 cells), X is the center (1 cell). This doesn't match the counts.
        *   *Let's look at the counts again*:
            Example 1: Y=4, M=3. Output 3x3. Y is perimeter (8), M is center (1).
            Example 2: B=7, A=5, X=4. Output 5x5. B is perimeter (16), A is ring (8), X is center (1).
            Example 3: A=9, G=5, W=1, .=4. Output 7x7. A is perimeter (24), G is ring (12), W is center (1).
        *   *Actual Rule*: The number of *distinct colors* (teams) determines the number of concentric rings. The largest team (by count) defines the outermost ring. The second largest defines the next ring, and so on. The size of the grid is determined by the number of rings. If there are $K$ teams, the grid is $(2K-1) \times (2K-1)$.
        *   *Wait, let's check Example 1*: Y=4, M=3. 2 teams. Grid $(2*2-1) = 3 \times 3$.
        *   *Example 2*: B=7, A=5, X=4. 3 teams. Grid $(2*3-1) = 5 \times 5$.
        *   *Example 3*: A=9, G=5, W=1. (Ignoring .). 3 teams. Grid $(2*3-1) = 5 \times 5$? No, the output is 7x7.
        *   *Let's re-examine Example 3*: A=9, G=5, W=1, .=4. If we count the dots as a team: A=9, G=5, .=4, W=1. 4 teams. Grid $(2*4-1) = 7 \times 7$.
        *   *Final Rule*:
            1. Count all non-O colors (including .).
            2. Sort colors by count descending.
            3. The number of teams $K$ determines a $(2K-1) \times (2K-1)$ grid.
            4. The largest team forms the outermost ring, the second largest the next ring, etc.

### Test Input Analysis
1.  **Count colors (excluding O)**:
    *   W (Maroon): 2
    *   R (Red): 4
    *   X (Grey): 4
    *   A (Azure): 1
    *   . (Black): 5
    *   B (Blue): 7
2.  **Sort by count (descending)**:
    *   B: 7
    *   .: 5
    *   R: 4
    *   X: 4
    *   W: 2
    *   A: 1
3.  **Number of teams ($K$)**: 6
4.  **Grid size**: $(2 \times 6 - 1) = 11 \times 11$
5.  **Construct rings**:
    *   Ring 1 (outermost, 11x11): B
    *   Ring 2 (9x9): .
    *   Ring 3 (7x7): R
    *   Ring 4 (5x5): X
    *   Ring 5 (3x3): W
    *   Ring 6 (1x1, center): A

### Construction
Grid 11x11:
- Ring 1 (B): Rows 0 and 10, Columns 0 and 10.
- Ring 2 (.): Rows 1 and 9, Columns 1 and 9.
- Ring 3 (R): Rows 2 and 8, Columns 2 and 8.
- Ring 4 (X): Rows 3 and 7, Columns 3 and 7.
- Ring 5 (W): Rows 4 and 6, Columns 4 and 6.
- Ring 6 (A): Row 5, Column 5.

Wait, let's re-verify the ring logic with Example 1.
Example 1: Y=4, M=3. $K=2$. Grid 3x3.
Ring 1 (Y): Perimeter of 3x3.
Ring 2 (M): Center of 3x3.
Output:
Y Y Y
Y M Y
Y Y Y
This matches!

Example 2: B=7, A=5, X=4. $K=3$. Grid 5x5.
Ring 1 (B): Perimeter of 5x5.
Ring 2 (A): Perimeter of 3x3.
Ring 3 (X): Center of 1x1.
Output:
B B B B B
B A A A B
B A X A B
B A A A B
B B B B B
This matches!

Example

... [2094 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Colors (Phenomena)**: Each color represents a different atmospheric phenomenon.
    *   **Background (O/.)**: The "empty" space or the base atmosphere.
    *   **Grid Size**: The output grid size is determined by the "most common" phenomenon's structure or, more accurately, the transformation creates a new concentric structure based on the frequency of colors present in the input.

2.  **Identify the AFFORDANCES**:
    *   **Frequency**: Colors can be counted to determine their rank (most frequent to least frequent).
    *   **Concentricity**: The output grid is structured in concentric square layers (rings).

3.  **Identify the TRANSFORMATION RULE**:
    *   **Step 1: Count frequencies**: Count the occurrences of every color present in the input (excluding the background color if it's just empty space, though in these examples, the background 'O' or '.' is treated as a color).
    *   **Step 2: Rank colors**: Sort the colors by frequency in descending order.
    *   **Step 3: Determine Output Size**: The output size is determined by the number of unique colors found. However, looking at the examples, the output size is actually a square grid where the number of layers corresponds to the number of unique colors. Wait, looking closer at Example 1: Input has O, Y, M. Output is 3x3. Example 2: Input has O, A, X, B. Output is 5x5. Example 3: Input has O, A, G, W, .. Output is 7x7.
    *   **Refined Rule**: The output is a square grid. The size of the grid is $N \times N$, where $N$ is the number of unique colors found in the input.
    *   **Step 4: Layering**: The most frequent color forms the outermost ring (layer 1). The second most frequent forms the next ring (layer 2), and so on. The least frequent color forms the center (the eye).
    *   **Note on Background**: In Example 1, O is the background. O is the most frequent. But the output uses Y and M. Let's re-examine.
    *   **Example 1 Re-analysis**: O (19), Y (4), M (3). Output: Y is outer, M is inner. Wait, the most frequent is O, but O is not in the output? Actually, the output is 3x3. The colors used are Y and M. The background O is ignored.
    *   **Example 2 Re-analysis**: O (51), A (5), X (4), B (8). Output: B (outer), A (next), X (inner). O is ignored.
    *   **Example 3 Re-analysis**: O (35), A (7), G (5), W (1), . (6). Output: A (outer), . (next), G (next), W (center). O is ignored.
    *   **Correct Rule**: 
        1. Identify all colors present in the input *except* the most frequent color (which acts as the background/empty space).
        2. Sort these remaining colors by frequency (descending).
        3. Create a square grid of size $N \times N$, where $N$ is the number of these colors.
        4. Fill the grid with concentric rings: the most frequent of these colors is the outermost ring, the next is the next ring, etc.

### Applying the Rule to Test Input

**1. Count frequencies in Test Input:**
O: 33
W: 2
R: 4
X: 4
A: 1
.: 5
B: 9

**2. Identify background and remaining colors:**
Most frequent is O (33). It is the background.
Remaining colors:
B: 9
.: 5
R: 4
X: 4
W: 2
A: 1

**3. Sort remaining colors by frequency (descending):**
1. B (9)
2. . (5)
3. R (4)
4. X (4)
5. W (2)
6. A (1)

**4. Determine Output Size:**
There are 6 colors. Output size is 6x6.

**5. Construct the 6x6 grid with concentric rings:**
*   **Ring 1 (Outer, size 6x6):** B
*   **Ring 2 (size 4x4):** .
*   **Ring 3 (size 2x2):** R (Wait, if the next ring is 2x2, we can't fit R, X, W, A).

**Re-evaluating the size rule:**
Example 1: Colors Y(4), M(3). $N=2$. Output is 3x3.
Example 2: Colors B(8), A(5), X(4). $N=3$. Output is 5x5.
Example 3: Colors A(7), .(6), G(5), W(1). $N=4$. Output is 7x7.
The rule for size is: **Size = 2 * (Number of colors) - 1**.
For Test Input: $N=6$. Size = $2(6) - 1 = 11$.
Wait, let's check Example 1 again. $N=2$. $2(2)-1 = 3$. Correct.
Example 2: $N=3$. $2(3)-1 = 5$. Correct.
Example 3: $N=4$. $2(4)-1 = 7$. Correct.

**Test Input Size Calculation:**
Colors: B, ., R, X, W, A ($N=6$).
Size: $2(6) - 1 = 11$.

**Constructing 11x11 grid:**
*   Ring 1 (11x11): B
*   Ring 2 (9x9): .
*   Ring 3 (7x7): R
*   Ring 4 (5x5): X
*   Ring 5 (3x3): W
*   Ring 6 (1x1): A

**Wait, let's re-verify the color counts for the test input carefully:**
O: 33
W: 2
R: 4
X: 4
A: 1
.: 5
B: 9
Total colors excluding O: B, ., R, X, W, A. (6 colors).
Size = 11.

**Layering the 11x11 grid:**
Row 1: B B B B B B B B B B B
Row 2: B . . . . . . . . . B
Row 3: B . R R R R R R R . B
Row 4: B . R X X X X X R . B
Row 5: B . R X W W W X R . B
Row 6: B . R X W A W X R . B
Row 7: B . R X W W W X R . B
Row 8: B . R X X X X X R . B
Row 9: B . R R R R R R R . B
Row 10: B . . . . . . . . . B
Row 11: B B B B B B B B B B B

Wait, let's re-check the ring logic for the center.
Ring 1 (11x11): B
Ring 2 (9x9): .
Ring 3 (7x7): R
Ring 4 (5x5): X
Ring 5 (3x3): W
Ring 6 (1x1): A

Let's re-draw:
R1: B B B B B B B B B B B
R2: B . . . . . . .

... [4438 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Input Colors (Materials)**: Each color represents a quantity of material available.
    *   **Black (.)**: Represents empty space or the background/floor.
    *   **Output Structure**: A series of concentric square "walls" built from the outside in.
    *   **The "Hearthstone"**: The single rarest color (or the smallest group) that forms the center.

2.  **Identify the AFFORDANCES**:
    *   Colors can be counted to determine their "abundance."
    *   The grid dimensions of the output are determined by the largest possible square that can be centered in the input (or more accurately, the output is a square structure where the layers are determined by the counts of the colors present).
    *   Looking at the examples, the output size is determined by the number of layers. The layers are concentric squares.

3.  **Identify the TRANSFORMATION RULE**:
    *   **Step 1: Count the frequency** of every color present in the input (excluding the background/black if it's just empty space, though in Example 3, '.' is treated as a material).
    *   **Step 2: Rank the colors** by frequency (most frequent to least frequent).
    *   **Step 3: Determine the output size**. The output is a square. The number of layers is equal to the number of distinct colors found.
    *   **Step 4: Construct the square**.
        *   The outermost layer (Layer 1) is filled with the most frequent color.
        *   The next layer (Layer 2) is filled with the second most frequent color.
        *   Continue until the least frequent color forms the center.
    *   *Correction based on Example 1*: Input has O (18), Y (4), M (2). Output is 3x3. Layer 1 (outer): Y (Wait, Example 1 output is Y-M-Y? Let's re-examine).
    *   *Re-examining Example 1*: Input: O=18, Y=4, M=2. Output: Y=6, M=1. This doesn't match "most plentiful = outer".
    *   *Re-examining Example 1 again*: Input: O=18, Y=4, M=2. Output is 3x3.
        Y Y Y
        Y M Y
        Y Y Y
        Wait, the most frequent color (O) is actually *gone*? No, the description says "the most plentiful material becomes the outer wall". In Example 1, the colors are O, Y, M. If O is the most plentiful, it should be the outer wall. But the output is Y.
    *   *Wait, let's look at the colors in Example 1 again*: O=18, Y=4, M=2. The output uses Y and M. The O is ignored?
    *   *Let's look at Example 2*: O=35, A=7, X=4, B=8. Most frequent is O. Second is B. Third is A. Fourth is X.
        Output: B (outer), A (inner), X (center).
        Wait, the counts are: B=8, A=7, X=4. O is the background.
    *   **Revised Rule**:
        1.  Ignore the background color (the most frequent color, usually 'O' or '.').
        2.  Count the remaining colors.
        3.  Rank the remaining colors by frequency (highest to lowest).
        4.  The most frequent color becomes the outermost square layer.
        5.  The next most frequent becomes the next inner layer.
        6.  The least frequent becomes the center.
        7.  The size of the square is determined by the number of colors. If there are $N$ colors, the square is $(2N-1) \times (2N-1)$.

    *   *Testing Revised Rule on Example 1*: Colors: Y(4), M(2). O is background. $N=2$. Size $(2*2-1)=3$.
        Layer 1 (most frequent Y): Outer ring of 3x3.
        Layer 2 (least frequent M): Center of 3x3.
        Result: Y Y Y / Y M Y / Y Y Y. (Matches!)

    *   *Testing Revised Rule on Example 2*: Colors: B(8), A(7), X(4). O is background. $N=3$. Size $(2*3-1)=5$.
        Layer 1 (B): Outer ring of 5x5.
        Layer 2 (A): Ring of 3x3.
        Layer 3 (X): Center of 1x1.
        Result: B B B B B / B A A A B / B A X A B / B A A A B / B B B B B. (Matches!)

    *   *Testing Revised Rule on Example 3*: Colors: A(10), G(6), W(1), .(6). O is background.
        Wait, '.' is a color here. Let's count: A=10, G=6, .=6, W=1.
        $N=4$. Size $(2*4-1)=7$.
        Layer 1 (A): Outer ring of 7x7.
        Layer 2 (G or .): Ring of 5x5.
        Layer 3 (G or .): Ring of 3x3.
        Layer 4 (W): Center of 1x1.
        (Matches the structure!)

### Applying to Test Input

1.  **Count colors (excluding O)**:
    *   W: 2
    *   R: 4
    *   X: 4
    *   A: 1
    *   .: 4
    *   B: 7
    *   (O is background)

2.  **Rank colors by frequency**:
    *   B: 7
    *   R: 4
    *   X: 4
    *   .: 4
    *   W: 2
    *   A: 1

    *Note: R, X, and . are tied. In such cases, we look at the order of appearance or alphabetical, but usually, the order in the input or a specific priority is used. Let's look at the ties: R, X, . are all 4. Let's assume the order of appearance in the input: R, X, ., B, W, A. Wait, B is 7. So B is first.*
    *Order of frequency (descending):*
    1. B (7)
    2. R (4)
    3. X (4)
    4. . (4)
    5. W (2)
    6. A (1)

3.  **Determine Output Size**:
    $N = 6$ colors (B, R, X, ., W, A).
    Size = $(2 * 6) - 1 = 11 \times 11$.
    Wait, let's re-count the input size. The input

... [5074 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Regiments**: Contiguous blocks of the same color (excluding the background color 'O' if it's just empty space, though in these puzzles, 'O' often acts as the background/void).
    *   **Concentric Rings**: The target structure where each ring will be occupied by exactly one color.
    *   **Center**: The single cell at the heart of the concentric squares.

2.  **Identify the AFFORDANCES**:
    *   Each color group (regiment) has a **size** (the count of its constituent cells).
    *   The size determines the "rank" of the regiment.
    *   The output grid is a square of size $N \times N$, where $N$ is determined by the number of regiments found.

3.  **Identify the TRANSFORMATION RULE**:
    *   **Step 1: Identify all regiments.** Scan the input for contiguous blocks of the same color (excluding the background 'O').
    *   **Step 2: Count the size of each regiment.**
    *   **Step 3: Rank the regiments by size in descending order.**
    *   **Step 4: Construct a concentric square grid.**
        *   The largest regiment forms the outermost ring (the perimeter).
        *   The second largest forms the next inner ring.
        *   The third largest forms the next, and so on.
        *   The smallest regiment occupies the center cell.
    *   **Note on Grid Size**: The number of rings $k$ determines the grid size $(2k-1) \times (2k-1)$.

### Applying the Rule to Test Input

**Step 1: Identify Regiments and Sizes**
*   **W (maroon)**: (0,2), (1,2) $\rightarrow$ Size: 2
*   **R (red)**: (0,5), (0,6), (0,7), (1,5), (1,7) $\rightarrow$ Size: 5
*   **X (grey)**: (2,0), (3,0), (3,1), (4,0) $\rightarrow$ Size: 4
*   **A (azure)**: (3,3) $\rightarrow$ Size: 1
*   **B (blue)**: (5,2), (6,1), (6,2), (6,3), (7,0), (7,1), (7,2), (7,3), (7,4) $\rightarrow$ Size: 9
*   **. (black)**: (4,5), (5,5), (5,6), (6,6), (6,7), (7,7) $\rightarrow$ Size: 6

**Step 2: Rank Regiments by Size (Descending)**
1.  **B** (Blue): 9
2.  **.** (Black): 6
3.  **R** (Red): 5
4.  **X** (Grey): 4
5.  **W** (Maroon): 2
6.  **A** (Azure): 1

**Step 3: Construct the Concentric Square**
There are 6 regiments. A 6-ring concentric square results in a $(2 \times 6 - 1) = 11 \times 11$ grid. However, looking at the training examples, the grid size is determined by the number of rings.
*   Ring 1 (Outer): B (Blue)
*   Ring 2: . (Black)
*   Ring 3: R (Red)
*   Ring 4: X (Grey)
*   Ring 5: W (Maroon)
*   Ring 6 (Center): A (Azure)

Wait, let's re-check the math. If there are 6 rings, the center is the 6th ring.
Ring 1: $11 \times 11$ perimeter.
Ring 2: $9 \times 9$ perimeter.
Ring 3: $7 \times 7$ perimeter.
Ring 4: $5 \times 5$ perimeter.
Ring 5: $3 \times 3$ perimeter.
Ring 6: $1 \times 1$ center.

Let's verify the sizes against the ring capacities:
*   Ring 1 (11x11): $11 \times 4 - 4 = 40$ cells. (B has 9)
*   Ring 2 (9x9): $9 \times 4 - 4 = 32$ cells. (. has 6)
*   Ring 3 (7x7): $7 \times 4 - 4 = 24$ cells. (R has 5)
*   Ring 4 (5x5): $5 \times 4 - 4 = 16$ cells. (X has 4)
*   Ring 5 (3x3): $3 \times 4 - 4 = 8$ cells. (W has 2)
*   Ring 6 (1x1): 1 cell. (A has 1)

The rule says "place the largest on the outermost wall... with the smallest detachment guarding the single center position." It does not say the regiment must fill the entire ring, but the description "one regiment per concentric ring" and the examples suggest the rings are filled by the ranked colors.

Let's re-examine Example 1:
Input: O(many), Y(4), M(2).
Regiments: Y (4), M (2).
Rank: Y, M.
Output: 3x3 grid.
Ring 1 (3x3): Y. Ring 2 (1x1): M.
Output:
Y Y Y
Y M Y
Y Y Y
This confirms: Ring 1 is the perimeter, Ring 2 is the center.

Example 2:
Input: O(many), A(5), X(4), B(5).
Regiments: A(5), X(4), B(5).
Rank: A/B (tie), X.
Wait, let's recount Example 2:
A: (0,6), (0,7), (0,8), (1,6), (1,7), (1,8), (2,6), (2,7), (2,8) -> No, let's look closer.
A: (0,6), (0,7), (0,8), (1,6), (1,7), (1,8), (2,6), (2,7), (2,8) is not right.
A: (0,6), (0,7), (0,8), (1,6), (1,7), (1,8), (2,6), (2,7), (2,8) is 9 cells.
X: (2,1), (2,2), (3,1), (3,2) is 4 cells.
B: (5,3), (5,4), (5,5), (6,3), (6,5), (7,3) is 6 cells.
Wait, let's re-read Example 2 input:
A: (0,6), (0,7), (0,8), (1,6), (1,7), (1,8), (2,6), (2,7), (2,8) -> 9 cells.
X: (2,1), (2,2), (3,1), (3,2) -> 4 cells.
B: (5,3), (5,4), (5,5), (6,3), (6,5), (7,3) -> 6 cells.
Rank: A (9), B (6), X (4).
Output:
B B B B B (Ring 1: B)
B A A A B (Ring 2: A)
B A X A B (Ring 3: X)
B A A A B
B B B B B
Wait, the output is 5x5.
Ring 1 (5x5): B.
Ring 2 (3x3): A.
Ring 3 (1x1): X.
This means the rank is: B (6), A (9), X (4).
Wait, if B is 6 and A is 9, A should be the outer ring.
Let's re-count A in Example 2: (0,6), (0,7), (0,8), (1,6), (1,7), (1,8), (2,6), (2,7), (2,8) is 9.
Let's re-count B in Example 2: (5,3), (5,4), (5,5), (6,3), (6,5), (7,3) is 6.
If A is 9 and B is 6, A is the outer ring. But the output starts with B.
Let's look at the output again:
B B B B B
B A A A B
B A X A B
B A A A B
B B B B B

... [3774 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Seeds (Colored pixels)**: These are the elements to be sorted by frequency.
    *   **Black pixels (.)**: These are treated as a color/element if they exist in the input.
    *   **The Garden Bed (Output Grid)**: A square grid where the size is determined by the largest possible concentric square that can be formed from the input's elements. Looking at the examples, the output size is determined by the "most plentiful" element's count or the structure of the input. However, more accurately, the output is a square grid where the dimensions are determined by the number of rings required to accommodate all unique colors found in the input.
    *   **Wait, let's re-examine the size**:
        *   Ex 1: Input 6x6. Output 3x3. Colors: O(25), Y(4), M(3). Rings: O (outer), Y (middle), M (center). Wait, the output is 3x3. The colors are Y, M, and O is the background? No, the description says "the most plentiful flower forms the outer border".
        *   Ex 1 counts: O=25, Y=4, M=3. Output: Y is outer, M is middle, Y is center? No. Let's re-count.
        *   Ex 1 Input: O=25, Y=4, M=3. Output: Y=7, M=1, Y=1? No.
        *   Let's re-read: "the most plentiful flower forms the outer border, the next most plentiful the next ring inward, until the rarest single bloom is the centerpiece."
        *   Ex 1 Input: O=25, Y=4, M=3. Most plentiful is O. Next is Y. Next is M.
        *   Wait, the output for Ex 1 is:
            Y Y Y
            Y M Y
            Y Y Y
            This uses Y and M. Where is O? O is the background/empty space in the input. The "seeds" are the non-O colors.
        *   Ex 2 Input: O=63, A=5, X=4, B=7. Most plentiful: B(7), A(5), X(4).
            Output: B (outer), A (middle), X (center).
            Size: 5x5.
        *   Ex 3 Input: O=35, A=7, G=3, W=1, .=4.
            Most plentiful: A(7), G(3), .=4, W(1).
            Wait, the order of counts: A(7), .(4), G(3), W(1).
            Output: A (outer), . (middle), G (inner), W (center).
            Size: 7x7.

2.  **Identify the AFFORDANCES**:
    *   Colors can be counted.
    *   Colors can be arranged in concentric square rings.

3.  **Identify the TRANSFORMATION RULE**:
    *   Step 1: Identify all colors present in the input *excluding* the background color (O/orange).
    *   Step 2: Count the occurrences of each color.
    *   Step 3: Sort the colors by frequency in descending order.
    *   Step 4: Determine the size of the output square. The size is determined by the number of colors. If there are $N$ colors, the size is $(2N-1) \times (2N-1)$.
    *   Step 5: Place the most frequent color in the outermost ring, the second most frequent in the next ring, and so on, until the least frequent is in the center.

### Applying the Rule to Test Input

**Test Input Analysis**:
Colors (excluding O):
- W (maroon): 2
- R (red): 4
- X (grey): 4
- A (azure): 1
- . (black): 5
- B (blue): 8

**Counts**:
1. B: 8
2. .: 5
3. R: 4
4. X: 4
5. W: 2
6. A: 1

**Sorted Colors (Descending Frequency)**:
1. B (8)
2. . (5)
3. R (4) or X (4) - *Tie-breaker: usually based on appearance or alphabetical, let's look at R and X.*
4. X (4)
5. W (2)
6. A (1)

**Number of colors (N)**: 6
**Output Size**: $(2 \times 6 - 1) = 11 \times 11$.
Wait, let's check the size logic again.
Ex 1: Colors Y(4), M(3). N=2. Size = $2(2)-1 = 3$. Correct.
Ex 2: Colors B(7), A(5), X(4). N=3. Size = $2(3)-1 = 5$. Correct.
Ex 3: Colors A(7), .(4), G(3), W(1). N=4. Size = $2(4)-1 = 7$. Correct.

**Test Input Colors**: B, ., R, X, W, A. (N=6).
Size = $2(6)-1 = 11$.

**Wait, let's re-count the test input carefully**:
O O W O O R R R (W:1, R:3)
O O W O O R O R (W:1, R:2) -> Total W:2, R:5
X O O O O O O O (X:1)
X X O A O O O O (X:2, A:1) -> Total X:3, A:1
X O O O O . O O (X:1, .:1)
O O B O O . . O (B:1, .:2)
O B B B O O . . (B:3, .:2)
B B B B B O O . (B:5, .:1) -> Total B:9, .:5

**Recounting Test Input**:
B: 1+3+5 = 9
.: 1+2+2+1 = 6
R: 3+2 = 5
X: 1+2+1 = 4
W: 1+1 = 2
A: 1

**Sorted Colors**:
1. B (9)
2. . (6)
3. R (5)
4. X (4)
5. W (2)
6. A (1)

**N = 6. Size = 11x11.**

**Constructing the 11x11 grid**:
Ring 1 (Outer, 11x11): B
Ring 2 (9x9): .
Ring 3 (7x7): R
Ring 4 (5x5): X
Ring 5 (3x3): W
Ring 6 (Center, 1x1): A

**Wait, let's check the tie-breaker for R and X in the counts**.
R=5, X=4. No tie.
B=9, .=6, R=5, X=4, W=2, A=1.

**Wait, let's re-verify the counts one more time**.
Input:
Row 1: O O W O O R R R (W:1, R:3)
Row 2: O O W O O R O R (W:1, R:2)
Row 3: X O O O O O O O (X:1)
Row 4: X X O A O O O O (X:2, A:1)
Row 5: X O O O O . O O (X:1, .:1)
Row 6: O O B O O . . O (B:1, .:2)
Row 7: O B B B O O . . (B:3, .:2)
Row 8: B B B B B O O . (B:5, .:1)

Total B: 1+3+5 = 9
Total .: 1+2+2+1 = 6
Total R: 3+2 = 5
Total X: 1+2+1 = 4
Total W: 1+1 = 2
Total A: 1

Order: B, ., R, X, W, A.
Size: 11x11.

**Wait, let's look at the input grid size**.
The input is 8 rows and 8 columns.
In Example 1, input is 6x6, output is 3x3.
In Example 2, i

... [3827 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Input Row (The Monitor):** The first row of the input grid contains a sequence of colored pulses (the "initial charge"). The number of colored cells ($N$) determines the maximum strength of the heartbeat.
    *   **Black Cells (Silence):** The empty space in the input represents the "flatline."
    *   **Output Grid (The EKG Tape):** A grid where the rhythm is recorded sequentially from left to right, wrapping to the next line.

2.  **Identify the AFFORDANCES:**
    *   The input color sequence can be used to define the "pulse" color and the maximum pulse length.
    *   The output grid can be filled with the color sequence according to the rhythmic pattern.

3.  **Identify the TRANSFORMATION RULE:**
    *   **Step 1: Determine Pulse Parameters.** Let $N$ be the number of colored cells in the first row. The color is the color of these cells.
    *   **Step 2: Generate the Rhythm.** The rhythm consists of increasing and then decreasing pulses of the color.
        *   Sequence of pulse lengths: $1, 2, 3, \dots, N-1, N, N-1, \dots, 2, 1$.
    *   **Step 3: Add Silence.** Between every pulse, there is exactly one black cell (`.`).
    *   **Step 4: Construct the Tape.** Lay out this sequence (Pulse 1, Silence, Pulse 2, Silence, ..., Pulse Last) continuously across the output grid. The width of the output grid is the same as the input grid. If the sequence ends before the grid is full, fill the rest with black (`.`).

### Applying the Rule to the Test Input

**Input Analysis:**
*   Input Row: `W W W W W W W W . . . . . . . .`
*   Number of colored cells ($N$): 8
*   Color: `W` (maroon)
*   Grid Width: 16

**Rhythm Generation:**
*   Pulse lengths: $1, 2, 3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3, 2, 1$
*   Full sequence (including silence `.`):
    `W . WW . WWW . WWWW . WWWWW . WWWWWW . WWWWWWW . WWWWWWWW . WWWWWWW . WWWWWW . WWWWW . WWWW . WWW . WW . W`

**Mapping to 16-column Grid:**
1.  `W . WW . WWW . WWWW` (16 chars)
    *   `W` (1) + `.` (1) + `WW` (2) + `.` (1) + `WWW` (3) + `.` (1) + `WWWW` (4) = 12 chars.
    *   Wait, let's count carefully:
    *   `W` (1)
    *   `.` (2)
    *   `W W` (4)
    *   `.` (5)
    *   `W W W` (8)
    *   `.` (9)
    *   `W W W W` (13)
    *   `.` (14)
    *   `W W` (16) -> This is the start of the next pulse.

Let's re-calculate the sequence string:
`W . WW . WWW . WWWW . WWWWW . WWWWWW . WWWWWWW . WWWWWWWW . WWWWWWW . WWWWWW . WWWWW . WWWW . WWW . WW . W`

**Row by Row (Width 16):**
*   **Row 1:** `W . W W . W W W . W W W W . W W`
    (1+1+2+1+3+1+4+1+2 = 16)
*   **Row 2:** `W W W . W W W W W . W W W W W W`
    (Remaining from pulse 5: `W W W` (3) + `.` (1) + `W W W W W` (5) + `.` (1) + `W W W W W W` (6) = 16)
    Wait, let's be precise:
    Sequence: `W` (1), `.` (2), `WW` (4), `.` (5), `WWW` (8), `.` (9), `WWWW` (13), `.` (14), `WWWWW` (19), `.` (20), `WWWWWW` (26), `.` (27), `WWWWWWW` (34), `.` (35), `WWWWWWWW` (43), `.` (44), `WWWWWWW` (51), `.` (52), `WWWWWW` (58), `.` (59), `WWWWW` (64), `.` (65), `WWWW` (69), `.` (70), `WWW` (73), `.` (74), `WW` (76), `.` (77), `W` (78).

**Row 1 (1-16):**
`W . W W . W W W . W W W W . W W`
(1, 2, 4, 5, 8, 9, 13, 14, 16)

**Row 2 (17-32):**
`W W W . W W W W W . W W W W W W`
(17, 18, 19, 20, 25, 26, 27, 32)
Wait, let's re-index:
1:W, 2:., 3:W, 4:W, 5:., 6:W, 7:W, 8:W, 9:., 10:W, 11:W, 12:W, 13:W, 14:., 15:W, 16:W
17:W, 18:W, 19:., 20:W, 21:W, 22:W, 23:W, 24:W, 25:., 26:W, 27:W, 28:W, 29:W, 30:W, 31:W, 32:W
Wait, let's re-verify the pulse lengths:
P1: 1 (pos 1)
S: 1 (pos 2)
P2: 2 (pos 3,4)
S: 1 (pos 5)
P3: 3 (pos 6,7,8)
S: 1 (pos 9)
P4: 4 (pos 10,11,12,13)
S: 1 (pos 14)
P5: 5 (pos 15,16,17,18,19)
S: 1 (pos 20)
P6: 6 (pos 21,22,23,24,25,26)
S: 1 (pos 27)
P7: 7 (pos 28,29,30,31,32,33,34)
S: 1 (pos 35)
P8: 8 (pos 36,37,38,39,40,41,42,43)
S: 1 (pos 44)
P7: 7 (pos 45,46,47,48,49,50,51)
S: 1 (pos 52)
P6: 6 (pos 53,54,55,56,57,58)
S: 1 (pos 59)
P5: 5 (pos 60,61,62,63,64)
S: 1 (pos 65)
P4: 4 (pos 66,67,68,69)
S: 1 (pos 70)
P3: 3 (pos 71,72,73)
S: 1 (pos 74)
P2: 2 (pos 75,76)
S: 1 (pos 77)
P1: 1 (pos 78)

**Mapping to 16-width rows:**
Row 1 (1-16): `W . W W . W W W . W W W W . W W`
Row 2 (17-32): `W W . W W W W W . W W W W W W W`
Row 3 (33-48): `W . W W W W W W W . W W W W W W`
Row 4 (49-64): `W W . W W W W W W . W W W W W .`
Row 5 (65-78): `. W W W W . W W . W` (Wait, let's re-check)

**Recalculating Row 2 (17-32):**
17:W, 18:W, 19:., 20:W, 21:W, 22:W, 23:W, 24:W, 25:., 26:W, 27:W, 28:W, 29:W, 30:W, 31:W, 32:W
(Correct: 17,18 are part of P5. 19 is S. 20-26 is P6. 27 is S. 28-34 is P7. So 28-32 are W.)
Row 2: `W W . W W W W W . W W W W W W W` (Wait, 28-32 is 5 Ws. 27 is S. 20-26 is 7 chars. 19 is S. 17-18 is 2 chars. Total: 2+1+7+1+5 = 16. Correct.)

**Recalculating Row 3 (33-48):**
33:W, 34:W, 35:., 36:W, 37:W, 38:W, 39:W, 40:W, 41:W, 42:W, 43:W, 44:., 45:W, 46:W, 47:W, 48:W
(33,34 are part of P7. 35 is S. 36-43 is P8. 44 is S. 45-51 is P7. So 45-48 are W.)
Row 3: `W W . W W W W W W W W 

... [2048 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **The Essence:** The colored pixels in the first row of the input grid. The count of these pixels determines the "essence count" ($N$).
    *   **The Flame Sequences:** Sequences of the essence color of lengths $1, 2, 3, \dots, N, \dots, 3, 2, 1$.
    *   **The Ash Mark:** A single black pixel (`.`) that acts as a separator between each flame sequence.
    *   **The Scroll:** The grid itself, where the width is fixed by the input width.
    *   **The Parchment:** The black background of the grid.

2.  **Identify the AFFORDANCES:**
    *   The essence color can be used to create sequences of varying lengths.
    *   The black color (`.`) can be used as a separator and to fill the remaining space.
    *   The sequences can "wrap" around the width of the grid, continuing onto the next line.

3.  **Identify the TRANSFORMATION RULE:**
    *   Count the number of colored pixels in the first row of the input ($N$).
    *   Generate a sequence of lengths: $1, 2, 3, \dots, N-1, N, N-1, \dots, 2, 1$.
    *   Construct a single long string of pixels by placing a black pixel (`.`) between each sequence.
    *   The pattern is: `[Sequence 1] . [Sequence 2] . [Sequence 3] ... [Sequence Last]`.
    *   Fill this string into the grid row by row, starting from the top-left, respecting the fixed width of the input grid.

### Applying the Rule to Test Input

*   **Input Analysis:**
    *   Input width: 16 columns.
    *   Input essence: `W` (maroon).
    *   Essence count ($N$): 8 (there are 8 `W`s in the first row).
*   **Sequence Lengths:** $1, 2, 3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3, 2, 1$.
*   **Constructing the String (with `.` separators):**
    1.  `W` (1)
    2.  `.`
    3.  `W W` (2)
    4.  `.`
    5.  `W W W` (3)
    6.  `.`
    7.  `W W W W` (4)
    8.  `.`
    9.  `W W W W W` (5)
    10. `.`
    11. `W W W W W W` (6)
    12. `.`
    13. `W W W W W W W` (7)
    14. `.`
    15. `W W W W W W W W` (8)
    16. `.`
    17. `W W W W W W W` (7)
    18. `.`
    19. `W W W W W W` (6)
    20. `.`
    21. `W W W W W` (5)
    22. `.`
    23. `W W W W` (4)
    24. `.`
    25. `W W W` (3)
    26. `.`
    27. `W W` (2)
    28. `.`
    29. `W` (1)

*   **Total length of sequence:** $1+1+2+1+3+1+4+1+5+1+6+1+7+1+8+1+7+1+6+1+5+1+4+1+3+1+2+1+1 = 77$ pixels.
*   **Mapping to 16-column grid:**
    *   Row 1: `W . W W . W W W . W W W W . W W` (16)
    *   Row 2: `W W W . W W W W W . W W W W W W` (16)
    *   Row 3: `. W W W W W W . W W W W W W . W` (16)
    *   Row 4: `W W W W . W W W W . W W W . W W` (16)
    *   Row 5: `W . W . . . . . . . . . . . . .` (Wait, let's re-calculate carefully)

**Recalculating the sequence string carefully:**
`W` (1)
`.` (2)
`W W` (4)
`.` (5)
`W W W` (8)
`.` (9)
`W W W W` (13)
`.` (14)
`W W W W W` (19)
`.` (20)
`W W W W W W` (26)
`.` (27)
`W W W W W W W` (34)
`.` (35)
`W W W W W W W W` (43)
`.` (44)
`W W W W W W W` (51)
`.` (52)
`W W W W W W` (58)
`.` (59)
`W W W W W` (64)
`.` (65)
`W W W W` (69)
`.` (70)
`W W W` (73)
`.` (74)
`W W` (76)
`.` (77)
`W` (78)

**Row by Row (Width 16):**
Row 1: `W . W W . W W W . W W W W . W W` (16)
Row 2: `W W . W W W W W . W W W W W W .` (16)
Row 3: `W W W W W W W . W W W W W W W .` (16)
Row 4: `W W W W W W . W W W W W . W W W` (16)
Row 5: `W . W W W W . W W W . W W W W .` (Wait, let's re-index)

**Let's list the sequence indices:**
1:W, 2:., 3:W, 4:W, 5:., 6:W, 7:W, 8:W, 9:., 10:W, 11:W, 12:W, 13:W, 14:., 15:W, 16:W (End R1)
17:W, 18:W, 19:W, 20:., 21:W, 22:W, 23:W, 24:W, 25:W, 26:W, 27:., 28:W, 29:W, 30:W, 31:W, 32:W (End R2)
33:W, 34:W, 35:., 36:W, 37:W, 38:W, 39:W, 40:W, 41:W, 42:W, 43:W, 44:., 45:W, 46:W, 47:W, 48:W (End R3)
49:W, 50:W, 51:W, 52:., 53:W, 54:W, 55:W, 56:W, 57:W, 58:W, 59:., 60:W, 61:W, 62:W, 63:W, 64:W (End R4)
65:., 66:W, 67:W, 68:W, 69:W, 70:., 71:W, 72:W, 73:W, 74:., 75:W, 76:W, 77:., 78:W (End R5)

**Wait, let's re-verify the sequence construction:**
Lengths: 1, 2, 3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3, 2, 1
Separators: 14 separators (one between each of the 15 sequences)
Total length = (1+2+3+4+5+6+7+8+7+6+5+4+3+2+1) + 14 = 64 + 14 = 78.

**Correct Sequence Construction:**
`W` (1)
`.` (2)
`W W` (4)
`.` (5)
`W W W` (8)
`.` (9)
`W W W W` (13)
`.` (14)
`W W W W W` (19)
`.` (20)
`W W W W W W` (26)
`.` (27)
`W W W W W W W` (34)
`.` (35)
`W W W W W W W W` (43)
`.` (44)
`W W W W W W W` (51)
`.` (52)
`W W W W W W` (58)
`.` (59)
`W W W W W` (64)
`.` (65)
`W W W W` (69)
`.` (70)
`W W W` (73)
`.` (74)
`W W` (76)
`.` (77)
`W` (78)

**Row-by-row (16 columns):**
R1: `W . W W . W W W . W W W W . W W` (16)
R2: `W W W . W W W W W . W W W W W W` (16)
R3: `. W W W W W W . W W W W W W W .` (16)
R4: `W W W W W W . W W W W W . W W W` (16)
R5: `W . W W W W . W W W . W W W . W` (Wait, let's re-count R5)

Let's re-map indices to rows:
R1: 1-16
R2: 17-32
R3: 33-48
R4: 49-64
R5: 65-78 (14 cells)

R1: `W . W W . W W W . W W W W . W W`
R2: `W W W . W W W W W . W W W W W W`
R3: `. W W W W W W W . W W W W W W W` (Wait, index 35 is 

... [1905 chars truncated]

### Conceptual Digestion

**1. Identify the CAST:**
*   **Mineral Vein:** The continuous sequence of a specific color in the first row of the input. Its length ($L$) determines the pattern.
*   **Void/Matrix:** The black cells (`.`).
*   **Tray Width:** The total number of columns in the input grid.
*   **Strata:** The rows of the output grid, which represent "deposits" of the mineral color.

**2. Identify the AFFORDANCES:**
*   The **Vein** defines the width of the pattern to be repeated.
*   The **Strata** are formed by repeating the vein color in a specific sequence of thicknesses.
*   The **Void** acts as a separator (a single black cell) between blocks of color.

**3. Identify the TRANSFORMATION RULE:**
*   **Step 1: Determine Vein Width ($L$):** Count the number of consecutive colored cells in the first row.
*   **Step 2: Determine Strata Thicknesses:** The rule says "Deposit strata of thicknesses 1 through the vein width, then decreasing back." This means the sequence of thicknesses is $1, 2, 3, \dots, L, \dots, 3, 2, 1$.
*   **Step 3: Construct the Pattern:** For each thickness $T$ in the sequence, create a "block" of color. However, the description "Read the core across the tray's width" and the examples show that the output is a single sequence of color/void segments mapped across the width of the tray.
*   **Wait, let's re-examine the examples carefully:**
    *   **Example 1:** Vein $L=6$. Output rows show patterns. Let's look at the structure.
    *   Actually, the rule is simpler: The output is a series of "strata" (rows). Each row is a single "reading" of the pattern.
    *   The pattern is: A sequence of colored blocks separated by single black cells.
    *   The thicknesses of these colored blocks are $1, 2, 3, \dots, L, \dots, 3, 2, 1$.
    *   **Wait, looking at Example 3:** $L=3$. Thicknesses: $1, 2, 3, 2, 1$.
        *   Row 1: $1, 2$ (O . OO)
        *   Row 2: $3, 1$ (OOO . O)
        *   Row 3: $2$ (O . O)
    *   **Correct Rule Interpretation:**
        1.  Generate the sequence of thicknesses: $S = [1, 2, \dots, L, \dots, 2, 1]$.
        2.  Generate the sequence of segments: $Segs = [(\text{color} \times 1), \text{void}, (\text{color} \times 2), \text{void}, \dots, (\text{color} \times L), \text{void}, \dots, (\text{color} \times 1)]$.
        3.  The total length of this sequence is $\sum (\text{thicknesses}) + (\text{number of voids})$.
        4.  The output grid is filled by taking this long sequence and wrapping it into the tray width. Each row is a segment of this sequence.
    *   **Let's re-verify Example 3 with this:** $L=3$. $S = [1, 2, 3, 2, 1]$.
        *   Segments: `O`, `.`, `OO`, `.`, `OOO`, `.`, `OO`, `.`, `O`.
        *   Total sequence: `O . O O . O O O . O O . O`
        *   Tray width = 5.
        *   Row 1: `O . O O .`
        *   Row 2: `O O O . O`
        *   Row 3: `O . O . .` (Wait, the example output is `O . O . .`? Let's re-count).
        *   Example 3 Output:
            Row 1: `O . O O .` (5 chars)
            Row 2: `O O O . O` (5 chars)
            Row 3: `O . O . .` (5 chars)
        *   Sequence: `O . O O . O O O . O O . O`
        *   Row 1 (1-5): `O . O O .`
        *   Row 2 (6-10): `O O O . O`
        *   Row 3 (11-15): `O . O . .` (The sequence ends, the rest are black). This matches!

### Applying to Test Input
1.  **Vein Width ($L$):** The input has 8 `W`s. $L = 8$.
2.  **Thickness Sequence ($S$):** $1, 2, 3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3, 2, 1$.
3.  **Segment Sequence:**
    `W` (1)
    `.` (void)
    `WW` (2)
    `.` (void)
    `WWW` (3)
    `.` (void)
    `WWWW` (4)
    `.` (void)
    `WWWWW` (5)
    `.` (void)
    `WWWWWW` (6)
    `.` (void)
    `WWWWWWW` (7)
    `.` (void)
    `WWWWWWWW` (8)
    `.` (void)
    `WWWWWWW` (7)
    `.` (void)
    `WWWWWW` (6)
    `.` (void)
    `WWWWW` (5)
    `.` (void)
    `WWWW` (4)
    `.` (void)
    `WWW` (3)
    `.` (void)
    `WW` (2)
    `.` (void)
    `W` (1)
4.  **Total Sequence Construction:**
    `W . WW . WWW . WWWW . WWWWW . WWWWWW . WWWWWWW . WWWWWWWW . WWWWWWW . WWWWWW . WWWWW . WWWW . WWW . WW . W`
5.  **Wrap into Width 16:**
    *   Row 1: `W . WW . WWW . WWWW .` (16 chars)
        `W . W W . W W W . W W W W .` (Wait, let's count carefully)
        `W` (1) + `.` (1) + `WW` (2) + `.` (1) + `WWW` (3) + `.` (1) + `WWWW` (4) + `.` (1) = 14.
        Next is `WWWWW`.
        Row 1: `W . W W . W W W . W W W W . W W` (16)
    *   Row 2: `W W . W W W W W . W W W W W W .` (16)
        Let's re-calculate the sequence string:
        `W.WW.WWW.WWWW.WWWWW.WWWWWW.WWWWWWW.WWWWWWWW.WWWWWWW.WWWWWW.WWWWW.WWWW.WWW.WW.W`
        Indices:
        1: W
        2: .
        3-4: WW
        5: .
        6-8: WWW
        9: .
        10-13: WWWW
        14: .
        15-19: WWWWW
        20: .
        21-26: WWWWWW
        27: .
        28-34: WWWWWWW
        35: .
        36-43: WWWWWWWW
        44: .
        45-51: WWWWWWW
        52: .
        53-58: WWWWWW
        59: .
        60-64: WWWWW
        65: .
     

... [3287 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Seeds:** The colored pixels in the first row of the input grid. The count of these seeds determines the pattern.
    *   **Flower Beds:** Sequences of the seed color.
    *   **Stepping Stones:** Single black pixels (`.`) that act as separators between flower beds.
    *   **Garden Plot:** The output grid.
    *   **Path:** The sequence of flower beds and stones winding through the grid.

2.  **Identify the AFFORDANCES:**
    *   **Seeds** define the maximum length of the flower beds and the color used.
    *   **Flower Beds** are placed in a sequence of increasing then decreasing lengths (1, 2, ..., N, ..., 2, 1).
    *   **Stepping Stones** are placed between every bed.
    *   **The Grid** accommodates the winding path by filling rows sequentially.

3.  **Identify the TRANSFORMATION RULE:**
    *   Count the number of seeds ($N$) in the first row.
    *   Generate a sequence of lengths for the flower beds: $1, 2, 3, \dots, N-1, N, N-1, \dots, 2, 1$.
    *   Construct a single continuous sequence by alternating between a flower bed and a single black stone: `[Bed 1] . [Bed 2] . [Bed 3] ... [Bed Last]`.
    *   Fill the output grid row by row with this sequence. The width of the output grid is the same as the input grid.

### Applying the Rule to Test Input

**Input Analysis:**
*   Input width: 16
*   Input height: 13
*   Seeds: 8 'W' (maroon) pixels in the first row.
*   $N = 8$.

**Sequence Generation:**
*   Bed lengths: 1, 2, 3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3, 2, 1.
*   Sequence (W = maroon, . = black):
    `W . WW . WWW . WWWW . WWWWW . WWWWWW . WWWWWWW . WWWWWWWW . WWWWWWW . WWWWWW . WWWWW . WWWW . WWW . WW . W`

**Mapping to Grid (Width 16):**
1.  `W . WW . WWW . WWWW` (Length: 1+1+2+1+3+1+4 = 13) -> `W . WW . WWW . WWWW .` (Wait, let's count carefully)
    *   Bed 1 (1): `W`
    *   Stone (1): `.`
    *   Bed 2 (2): `WW`
    *   Stone (1): `.`
    *   Bed 3 (3): `WWW`
    *   Stone (1): `.`
    *   Bed 4 (4): `WWWW`
    *   Stone (1): `.`
    *   Bed 5 (5): `WWWWW`
    *   Stone (1): `.`
    *   Bed 6 (6): `WWWWWW`
    *   Stone (1): `.`
    *   Bed 7 (7): `WWWWWWW`
    *   Stone (1): `.`
    *   Bed 8 (8): `WWWWWWWW`
    *   Stone (1): `.`
    *   Bed 9 (7): `WWWWWWW`
    *   Stone (1): `.`
    *   Bed 10 (6): `WWWWWW`
    *   Stone (1): `.`
    *   Bed 11 (5): `WWWWW`
    *   Stone (1): `.`
    *   Bed 12 (4): `WWWW`
    *   Stone (1): `.`
    *   Bed 13 (3): `WWW`
    *   Stone (1): `.`
    *   Bed 14 (2): `WW`
    *   Stone (1): `.`
    *   Bed 15 (1): `W`

**Total sequence length:**
$1+1 + 2+1 + 3+1 + 4+1 + 5+1 + 6+1 + 7+1 + 8+1 + 7+1 + 6+1 + 5+1 + 4+1 + 3+1 + 2+1 + 1 = 85$
Total elements: $1+2+3+4+5+6+7+8+7+6+5+4+3+2+1$ (beds) + $14$ (stones) = $64 + 14 = 78$.
Wait, let's re-sum: $\sum_{i=1}^8 i + \sum_{i=1}^7 i = 36 + 28 = 64$.
Number of stones = $15 - 1 = 14$.
Total length = $64 + 14 = 78$.

**Filling the 16-wide grid:**
Row 1: `W . WW . WWW . WWWW .` (1+1+2+1+3+1+4+1 = 14) -> `W . WW . WWW . WWWW .` (Wait, 14 chars. Next is `WWWWW`)
Let's be precise:
1. `W` (1)
2. `.` (2)
3. `WW` (4)
4. `.` (5)
5. `WWW` (8)
6. `.` (9)
7. `WWWW` (13)
8. `.` (14)
9. `WWWWW` (19)
10. `.` (20)
11. `WWWWWW` (26)
12. `.` (27)
13. `WWWWWWW` (34)
14. `.` (35)
15. `WWWWWWWW` (43)
16. `.` (44)
17. `WWWWWWW` (51)
18. `.` (52)
19. `WWWWWW` (58)
20. `.` (59)
21. `WWWWW` (64)
22. `.` (65)
23. `WWWW` (69)
24. `.` (70)
25. `WWW` (73)
26. `.` (74)
27. `WW` (76)
28. `.` (77)
29. `W` (78)

**Row by Row (Width 16):**
R1: `W . WW . WWW . WWWW .` (14) + `WW` (16) -> `W . WW . WWW . WWWW . WW`
R2: `WWW . WWWWWW . WWWWWWW` (16)
R3: `. WWWWWWWW . WWWWWWW .` (16)
R4: `WWWWWW . WWWWW . WWWW .` (16)
R5: `WWW . WW . W` (10) + (6 empty) -> `WWW . WW . W . . . . . .`
Wait, let's re-calculate the sequence carefully.

Sequence:
`W` (1)
`.` (2)
`WW` (4)
`.` (5)
`WWW` (8)
`.` (9)
`WWWW` (13)
`.` (14)
`WWWWW` (19)
`.` (20)
`WWWWWW` (26)
`.` (27)
`WWWWWWW` (34)
`.` (35)
`WWWWWWWW` (43)
`.` (44)
`WWWWWWW` (51)
`.` (52)
`WWWWWW` (58)
`.` (59)
`WWWWW` (64)
`.` (65)
`WWWW` (69)
`.` (70)
`WWW` (73)
`.` (74)
`WW` (76)
`.` (77)
`W` (78)

R1 (1-16): `W . WW . WWW . WWWW . WW`
R2 (17-32): `WWW . WWWWWW . WWWW`
R3 (33-48): `WWW . WWWWWWWW . WW`
R4 (49-64): `WWWW . WWWWWW . WWWWW`
R5 (65-78): `. WWWW . WWW . WW . W` (Wait, 65 is `.`, 66-69 is `WWWW`, 70 is `.`, 71-73 is `WWW`, 74 is `.`, 75-76 is `WW`, 77 is `.`, 78 is `W`)
R5 (65-78): `. WWWW . WWW . WW . W` (14 chars)
R5 (65-80): `. WWWW . WWW . WW . W . .`

Let's re-verify R1-R5:
R1: `W . WW . WWW . WWWW . WW` (1+1+2+1+3+1+4+1+2 = 15) -> `W . WW . WWW . WWWW . WW` (15 chars). One more: `W . WW . WWW . WWWW . WW` (15) + `W` (16) = `W . WW . WWW . WWWW . WWW`
Wait, let's do this index by index.
Indices:
1:W, 2:., 3:W, 4:W, 5:., 6:W, 7:W, 8:W, 9:., 10:W, 11:W, 12:W, 13:W, 14:., 15:W, 16:W, 17:W, 18:W, 19:W, 20:., 21:W, 22:W, 23:W, 24:W, 25:W, 26:W, 27:., 28:W, 29:W, 30:W, 31:W, 32:W, 33:W, 34:W, 35:., 36:W, 37:W, 38:W, 39:W, 40:W

... [2001 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Performers (Colored cells):** The colored cells in the input grid represent the "factions" or "performers."
    *   **Factions (Distinct colors):** The number of unique colors present in the input grid determines the "bowing" pattern.
    *   **Spotlights (Grey 'X' cells):** These represent the positions of the performers during their bow.
    *   **Dark Stage (Black '.' cells):** The background of the output grid.

2.  **Identify the AFFORDANCES:**
    *   The input colors are counted to determine the number of distinct factions ($N$).
    *   The output grid is populated with $N$ grey 'X' cells based on specific geometric patterns determined by $N$.
    *   The rest of the grid is filled with black '.' cells.

3.  **Identify the TRANSFORMATION RULE:**
    *   **Step 1:** Count the number of unique colors in the input grid (excluding black/background if any, though here colors are the performers).
    *   **Step 2:** Determine the pattern based on the count $N$:
        *   **If $N = 1$ (Single troupe):** A horizontal line of 'X's across the top row (row 0).
        *   **If $N = 2$ (Duo):** A diagonal line of 'X's from the top-right to the bottom-left (or similar "grand staircase" diagonal). Looking at Example 2: Input has Y, R ($N=2$). Output is a diagonal: (0,2), (1,1), (2,0). This is the anti-diagonal.
        *   **If $N = 3$ (Trio):** A diagonal line of 'X's from the top-left to the bottom-right (the main diagonal). Looking at Example 4: Input has Y, G ($N=2$?). Wait, let's re-examine Example 4.
    *   **Re-evaluating Examples:**
        *   Ex 1: Input G ($N=1$). Output: Row 0 is X X X. (Correct)
        *   Ex 2: Input Y, R ($N=2$). Output: (0,2), (1,1), (2,0) are X. (Anti-diagonal)
        *   Ex 3: Input Y ($N=1$). Output: Row 0 is X X X. (Correct)
        *   Ex 4: Input Y, G ($N=2$). Output: (0,0), (1,1), (2,2) are X. (Main diagonal)
    *   **Wait, let's look closer at Example 2 vs Example 4:**
        *   Ex 2: Colors are Y and R. $N=2$. Pattern: Anti-diagonal.
        *   Ex 4: Colors are Y and G. $N=2$. Pattern: Main diagonal.
        *   *Correction:* Let's look at the *distribution* or *arrangement*? No, the metaphor says "A duo takes a bow along the grand staircase... A trio crosses paths...".
        *   Let's re-count colors in Ex 4: Y and G. That is $N=2$.
        *   Let's re-count colors in Ex 2: Y and R. That is $N=2$.
        *   Why is Ex 2 anti-diagonal and Ex 4 main diagonal? Let's look at the *input* structure.
        *   Ex 2: Y is at (0,0-2), (1,0), (2,0-1). R is at (1,1-2), (2,2).
        *   Ex 4: Y is at (0,0-2), (1,0), (1,1), (1,2). G is at (2,0-2).
        *   Actually, let's look at the number of *distinct colors* again.
        *   Ex 1: G ($N=1$) -> Row 0.
        *   Ex 2: Y, R ($N=2$) -> Anti-diagonal.
        *   Ex 3: Y ($N=1$) -> Row 0.
        *   Ex 4: Y, G ($N=2$) -> Main diagonal.
        *   Is it possible the pattern depends on the *order* or *type* of colors? Or is it the number of *cells* of a certain color?
        *   Let's look at the "Trio" description: "A trio crosses paths, saluting from opposite wings through center stage." This implies $N=3$.
        *   Let's check the Test Input: Y, R, G. $N=3$.
        *   If $N=3$, the pattern is "crossing paths". In a 3x3 grid, crossing paths (main diagonal and anti-diagonal) would be 5 cells, but the prompt says "Exactly three spotlights (grey) illuminate the bowing positions".
        *   If $N=3$ and there are 3 spotlights, the only way to "cross paths" through the center is the main diagonal OR the anti-diagonal OR a specific shape. But "crossing paths" usually implies an X shape. However, the rule says "Exactly three spotlights".
        *   Wait, if $N=3$, and we need 3 spotlights to "cross paths", the only way is if they are the three cells of a diagonal that passes through the center.
        *   Let's re-examine Ex 2 and Ex 4. Both have $N=2$.
        *   Ex 2 ($N=2$): Anti-diagonal.
        *   Ex 4 ($N=2$): Main diagonal.
        *   Is there a difference in the colors? Ex 2: Y, R. Ex 4: Y, G.
        *   Wait! Let's look at the *input* again.
        *   Ex 2: Y is the first color, R is the second.
        *   Ex 4: Y is the first color, G is the second.
        *   Let's look at the *number of colors* in the Test Input: Y, R, G. $N=3$.
        *   If $N=1 \rightarrow$ Row 0.
        *   If $N=2 \rightarrow$ Diagonal.
        *   If $N=3 \rightarrow$ "Crosses paths". In a 3x3 grid, the only way to have 3 cells "cross paths" through the center is the main diagonal or the anti-diagonal.
        *   Let's look at the "Trio" description again: "A trio crosses paths, saluting from opposite wings through center stage." This describes the main diagonal (0,0), (1,1), (2,2) or anti-diagonal (0,2), (1,1), (2,0).
        *   Wait, if $N=2$ is a diagonal, and $N=3$ is a diagonal, what is the difference?
        *   Let's look at the

... [5177 chars truncated]

### Conceptual Digestion

**1. Identify the CAST:**
*   **Input Grid:** A "tissue sample" consisting of various colored cells (G, Y, R, etc.).
*   **Output Grid:** A "skeletal imprint" consisting of a dark medium (black/`.`) and three grey nodes (`X`).
*   **Colors:** The specific colors in the input represent the "biodiversity."

**2. Identify the AFFORDANCES:**
*   The input colors are used to calculate a property (biodiversity/complexity).
*   The output is constrained to a 3x3 grid containing exactly three `X` (grey) nodes and the rest being `.` (black).

**3. Identify the TRANSFORMATION RULE:**
*   The goal is to find the "skeleton" that reflects the "richness" (diversity) of the sample.
*   Let's analyze the examples:
    *   **Example 1:** Input is all Green (1 color). Output has `X` at (0,0), (0,1), (0,2). Wait, looking closer at Example 1: Input is all G. Output is `X X X` in the first row.
    *   **Example 2:** Input has G, Y, R (3 colors). Output has `X` at (0,2), (1,1), (2,0). This is a diagonal.
*   **Re-evaluating the rule:** The "skeleton" is formed by the positions of the most frequent colors or a specific pattern based on color counts.
    *   Example 1: 1 color (G). Output: `X` at (0,0), (0,1), (0,2).
    *   Example 2: 3 colors (G, Y, R). Output: `X` at (0,2), (1,1), (2,0).
*   **Wait, let's look at the "nodes" logic:** The description says "three calcified grey nodes".
    *   Example 1 actually has **three** X's: (0,0), (0,1), (0,2).
    *   Example 2 has **three** X's: (0,2), (1,1), (2,0).
*   **The Pattern:** The nodes are placed based on the distribution of colors.
    *   In Example 1: Only 1 color exists. The nodes are placed in a line.
    *   In Example 2: 3 colors exist. The nodes are placed on the anti-diagonal.
*   **Let's look at the Test Input:**
    Input:
    Y Y Y
    R G R
    G R G
    Colors present: Y (3), R (3), G (3).
    Total colors: 3.
    In Example 2, when there were 3 colors, the output was the anti-diagonal:
    `. . X`
    `. X .`
    `X . .`
*   **Wait, let's check the color counts for Example 2 again:**
    G: 1, Y: 3, R: 3. (Total 3 colors).
    Example 1: G: 9. (Total 1 color).
*   **Hypothesis:** The number of colors determines the pattern.
    *   If 1 color: Row 0 is `X X X`.
    *   If 3 colors: Anti-diagonal is `X`.
*   **Let's look at the Test Input again:**
    Y: 3, R: 3, G: 3.
    This also has 3 colors. Following the pattern of Example 2 (which also had 3 colors), the output should be the anti-diagonal.

**Refined Rule:**
The number of unique colors in the input determines the placement of the three `X` nodes.
- 1 unique color $\rightarrow$ Top row `X X X`.
- 3 unique colors $\rightarrow$ Anti-diagonal `X` (from top-right to bottom-left).

**Applying to Test Input:**
Input colors: Y, R, G (3 unique colors).
Output pattern for 3 colors:
`. . X`
`. X .`
`X . .`

ANSWER:
. . X
. X .
X . .

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Input Grid**: A 3x3 grid of colored "spectral signatures" (colors like G, Y, R).
    *   **Output Grid**: A 3x3 "sensor array" consisting of black (`.`) and grey (`X`) cells.
    *   **Colors**: The specific colors in the input represent different types of spectral signatures.

2.  **Identify the AFFORDANCES**:
    *   The input colors can be counted or analyzed for their distribution/diversity.
    *   The output can only contain `X` (grey) or `.` (black).

3.  **Identify the TRANSFORMATION RULE**:
    *   Let's analyze the examples to find the pattern:
        *   Ex 1: All G $\rightarrow$ Top row is X.
        *   Ex 2: Y (3), R (4), G (2) $\rightarrow$ Diagonal (anti-diagonal) is X.
        *   Ex 3: All Y $\rightarrow$ Top row is X.
        *   Ex 4: Y (6), G (3) $\rightarrow$ Diagonal is X.
    *   Wait, let's look closer at the *diversity* (number of unique colors) and the *distribution*.
    *   Ex 1: 1 color (G). Output: Row 0 is X.
    *   Ex 2: 3 colors (G, Y, R). Output: Anti-diagonal is X.
    *   Ex 3: 1 color (Y). Output: Row 0 is X.
    *   Ex 4: 2 colors (Y, G). Output: Main diagonal is X.
    *   **Refined Rule**: The number of unique colors in the input determines the pattern of the grey (`X`) cells.
        *   **1 Unique Color**: The first row is filled with `X`.
        *   **2 Unique Colors**: The main diagonal (top-left to bottom-right) is filled with `X`.
        *   **3 Unique Colors**: The anti-diagonal (top-right to bottom-left) is filled with `X`.

### Applying the Rule to Test Input

**Test Input**:
```
Y Y Y
R G R
G R G
```

1.  **Count Unique Colors**:
    *   Y (Yellow)
    *   R (Red)
    *   G (Green)
    *   Total unique colors = 3.

2.  **Determine Pattern**:
    *   Since there are 3 unique colors, the rule for 3 colors is the **anti-diagonal** (top-right to bottom-left).

3.  **Construct Output**:
    *   Row 0: `. . X`
    *   Row 1: `. X .`
    *   Row 2: `X . .`

Let's double-check Example 2:
Input: G, Y, R (3 colors). Output: `. . X`, `. X .`, `X . .`. (Matches anti-diagonal).
Example 4: Y, G (2 colors). Output: `X . .`, `. X .`, `. . X`. (Matches main diagonal).
Example 1/3: G or Y (1 color). Output: `X X X`, `. . .`, `. . .`. (Matches first row).

The logic holds.

ANSWER:
. . X
. X .
X . .

To solve this puzzle, let's perform a Conceptual Digestion.

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Input Grid (Ocean Chart):** A 3x3 grid containing various colored cells (current zones).
    *   **Output Grid (Nautical Chart):** A 3x3 grid containing only two colors: Black (`.`) and Grey (`X`).
    *   **Grey Waypoints (`X`):** These represent specific locations derived from the input.

2.  **Identify the AFFORDANCES:**
    *   The input colors (G, Y, R, etc.) act as indicators or "currents."
    *   The output is a sparse map where only three specific locations are marked with `X`.

3.  **Identify the TRANSFORMATION RULE:**
    *   Let's analyze the examples to find the pattern for the three `X` waypoints.
    *   **Example 1:** Input is all Green (`G`). Output has `X` at (0,0), (0,1), (0,2).
    *   **Example 2:** Input:
        Row 0: G G G
        Row 1: Y R R
        Row 2: Y Y R
        Output: `X` at (0,2), (1,1), (2,0).
    *   **Pattern Search:** Look at the colors in the input.
        In Example 1: All cells are G. The output marks the first row.
        In Example 2:
        Row 0: G, G, G
        Row 1: Y, R, R
        Row 2: Y, Y, R
        The output `X` positions are (0,2), (1,1), (2,0). These are the positions of the color that appears most frequently or follows a specific diagonal? No.
        Let's look at the colors again.
        Example 1: G is the only color. Output: Top row is X.
        Example 2: G appears 3 times, Y appears 3 times, R appears 3 times.
        Wait, let's look at the *diagonals*.
        Example 2 output: (0,2), (1,1), (2,0). This is the **anti-diagonal**.
        Example 1 output: (0,0), (0,1), (0,2). This is the **top row**.
    *   **Re-evaluating the rule:** The metaphor says "Read the currents and plot three grey waypoints."
        In Example 1, the "current" is G.
        In Example 2, the colors are G, Y, R.
        Let's look at the *distribution* of colors.
        Example 1: G is at (0,0), (0,1), (0,2), (1,0)... (all 9).
        Example 2:
        G: (0,0), (0,1), (0,2)
        Y: (1,0), (2,0), (2,1)
        R: (1,1), (1,2), (2,2)
        Wait, the output for Example 2 is (0,2), (1,1), (2,0).
        Looking at the input for Example 2:
        (0,2) is G.
        (1,1) is R.
        (2,0) is G.
        This doesn't seem to be a simple color count.
    *   **Let's look at the "corners" or "edges":**
        Example 1: All G. Output: Top row.
        Example 2:
        Input:
        G G G
        Y R R
        Y Y R
        Output:
        . . X
        . X .
        X . .
        The `X`s are at (0,2), (1,1), (2,0).
        In the input, these cells are: (0,2)=G, (1,1)=R, (2,0)=Y.
        Wait, the colors at those positions are G, R, Y.
        In Example 1, the colors at (0,0), (0,1), (0,2) are G, G, G.
    *   **Let's try another logic: The most frequent color?**
        Ex 1: G is most frequent. Output: Row 0.
        Ex 2: G, Y, R all appear 3 times.
    *   **Let's look at the "center" or "symmetry":**
        In Example 2, the `X`s form the anti-diagonal.
        In Example 1, the `X`s form the top row.
        Is there a relationship between the input colors and the output?
        Example 1: Input is all G. Output is top row.
        Example 2: Input has G, Y, R. Output is anti-diagonal.
        Let's look at the *Test Input*:
        Y Y Y
        R G R
        G R G
        Colors: Y (3), R (3), G (3).
        Wait, in all examples, the colors are distributed equally (3 of each) or one color dominates.
        In Ex 1: G=9.
        In Ex 2: G=3, Y=3, R=3.
        In Test: Y=3, R=3, G=3.
        If the colors are balanced (3 of each), the output is the anti-diagonal (Ex 2).
        If one color dominates (Ex 1), the output is the top row.
        Wait, let's look at the *position* of the colors in Ex 2.
        G: (0,0), (0,1), (0,2)
        Y: (1,0), (2,0), (2,1)
        R: (1,1), (1,2), (2,2)
        The anti-diagonal (0,2), (1,1), (2,0) contains one of each color: G, R, Y.
        In Example 1, the top row (0,0), (0,1), (0,2) contains G, G, G.
        
        **Hypothesis:** The waypoints are the three cells that contain one of each color present in the grid, specifically following a path.
        In Ex 2, the colors are G, Y, R. The anti-diagonal contains G, R, Y.
        In Ex 1, the only color is G. The top row contains G, G, G.
        
        Let's check the Test Input:
        Y Y Y
        R G R
        G R G
        Colors: Y, R, G.
        The anti-diagonal is (0,2), (1,1), (2,0).
        (0,2) is Y.
        (1,1) is G.
        (2,0) is G. (Wait, that's two Gs).
        Let's look at the input again:
        (0,0)=Y, (0,1)=Y, (0,2)=Y
        (1,0)=R, (1,1)=G, (1,2)=R
        (2,0)=G, (2,1)=R, (2,2)=G
        
        Let's look at the "currents" again.
        Ex 1: All G. Output: Row 0.
        Ex 2: G is top row, Y is left column (mostly), R is bottom right (mostly).
        The output (0,2), (1,1), (2,0

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### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Input Grid (3x3 colored cells)**: Represents a "bus of colored voltage indicators." Each color represents a different "signal level."
    *   **Output Grid (3x3 black/grey cells)**: Represents a "dark circuit board" with three "grey LEDs" (X) lit up.
    *   **Colors**: Different colors (G, Y, R, etc.) are the signals to be measured.

2.  **Identify the AFFORDANCES**:
    *   The input colors can be counted or categorized to determine their distribution.
    *   The output can only contain two states: Black (`.`) or Grey (`X`).

3.  **Identify the TRANSFORMATION RULE**:
    *   The goal is to "measure the distinct signal levels" and "light up three grey LEDs."
    *   Looking at the examples:
        *   **Example 1**: Input is all Green (G). There is only 1 distinct color. The output shows three X's in the first row. Wait, let's re-examine.
        *   **Example 2**: Input has G, Y, and R. There are 3 distinct colors. The output shows three X's in a diagonal pattern.
    *   **Refined Rule**: The transformation identifies the positions of the colors based on their frequency or a specific order. Let's look closer at the "distinct signal levels."
    *   In Example 1: All cells are G. There is only one color. The output places X at (0,0), (0,1), (0,2).
    *   In Example 2:
        Row 0: G G G
        Row 1: Y R R
        Row 2: Y Y R
        Colors present: G (3), Y (3), R (3).
        Output:
        . . X
        . X .
        X . .
        This is a diagonal from bottom-left to top-right.
    *   **Wait, let's look at the counts again**:
        Example 1: G=9. Output: X at (0,0), (0,1), (0,2).
        Example 2: G=3, Y=3, R=3. Output: X at (0,2), (1,1), (2,0).
    *   **Correct Rule Hypothesis**: The output places three X's based on the *order of appearance* of the most frequent colors or a specific sequence of colors. However, a simpler pattern emerges: The output places X's at the coordinates of the *last* occurrence of each of the three most frequent colors, or perhaps it maps the colors to specific positions.
    *   Let's re-evaluate Example 2:
        G is at (0,0), (0,1), (0,2).
        Y is at (1,0), (2,0), (2,1).
        R is at (1,1), (1,2), (2,2).
        The output X's are at (0,2), (1,1), (2,0).
        These are the *last* positions of G, R, and Y if we scan row by row? No.
        Let's look at the *first* position of each color: G(0,0), Y(1,0), R(1,1). Not it.
        Let's look at the *last* position of each color: G(0,2), Y(2,1), R(2,2). Not it.
    *   **Let's try another pattern**: The output places X's at the positions of the colors that appear in a specific order.
        Example 1: Only G. X's at (0,0), (0,1), (0,2).
        Example 2: G, Y, R. X's at (0,2), (1,1), (2,0).
        In Example 2, the colors are G, Y, R.
        If we look at the *middle* index of each color's positions?
        G: (0,0), (0,1), (0,2) -> Middle is (0,1).
        Y: (1,0), (2,0), (2,1) -> Middle is (2,0).
        R: (1,1), (1,2), (2,2) -> Middle is (1,2).
        Still not matching (0,2), (1,1), (2,0).
    *   **Wait! Look at the diagonals**:
        Example 2 output is the anti-diagonal: (0,2), (1,1), (2,0).
        Example 1 output is the first row: (0,0), (0,1), (0,2).
        What determines which 3 cells are X?
        In Example 1, there is only 1 color.
        In Example 2, there are 3 colors.
        If there are 3 colors, the output is the anti-diagonal.
        If there is 1 color, the output is the first row.
    *   **Let's check the Test Input**:
        Y Y Y
        R G R
        G R G
        Colors: Y (3), R (3), G (3).
        Since there are 3 distinct colors (Y, R, G), the rule for 3 colors should apply.
        The rule for 3 colors in Example 2 was the anti-diagonal.

### Applying the Rule to Test Input
Test Input:
Y Y Y
R G R
G R G

Distinct colors: Y, R, G (Total 3).
Following the pattern of Example 2 (where 3 colors resulted in the anti-diagonal):
The anti-diagonal of a 3x3 grid is:
(0,2), (1,1), (2,0).

Output:
. . X
. X .
X . .

ANSWER:
. . X
. X .
X . .

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **The Mosaic (Input Grid):** A 3x3 grid of colored tiles (G, Y, R, etc.). These represent the "voices" or the source data.
    *   **The Tablet (Output Grid):** A 3x3 grid of black (`.`) and grey (`X`) marks.
    *   **The Fate-marks (`X`):** These are the specific elements we need to place on the tablet.

2.  **Identify the AFFORDANCES:**
    *   The input colors are used to determine the *number* and *position* of the grey marks.
    *   The output is restricted to only two colors: black (`.`) and grey (`X`).

3.  **Identify the TRANSFORMATION RULE:**
    *   The description says: "Divine the number of voices in the mosaic and inscribe the grey fate-marks on the dark tablet."
    *   Looking at the examples:
        *   **Example 1:** Input is all Green (G). There are 9 Green tiles. However, the output has 3 `X` marks. Wait, let's re-examine.
        *   **Example 2:** Input has 3 Yellow (Y), 3 Red (R), and 1 Green (G). Total 7 colored tiles? No, the input is 3x3. Let's count the colors in Example 2: Y=3, R=3, G=1. Total = 7. The output has 3 `X` marks.
    *   Let's look at the *unique* colors in the input:
        *   Example 1: Only Green (1 color). Output has 3 `X` marks.
        *   Example 2: Yellow, Red, Green (3 colors). Output has 3 `X` marks.
    *   Let's look at the *rows* or *columns*?
        *   Example 1: All rows are the same. Output: Row 1 is `X X X`.
        *   Example 2: Row 1: YYY, Row 2: YRR, Row 3: YYR. Output: Diagonal `X` marks (bottom-left to top-right).
    *   **Re-evaluating the "Voices":** In many grid puzzles, "voices" refers to the count of a specific color or the number of distinct colors.
    *   Let's look at the *count of the most frequent color*?
        *   Ex 1: G=9. Output has 3 `X`.
        *   Ex 2: Y=3, R=3, G=1. Output has 3 `X`.
    *   Let's look at the *number of rows that contain a specific color*?
    *   **Wait, let's look at the pattern of X in Example 2:**
        Input:
        Y Y Y
        Y R R
        Y Y R
        Output:
        . . X
        . X .
        X . .
        This is a diagonal from bottom-left to top-right.
    *   **Let's look at Example 1 again:**
        Input: All G.
        Output:
        X X X
        . . .
        . . .
        This is the first row.
    *   **Let's look at the "number of voices" again.** If "voices" means the number of *distinct colors* present in the input:
        *   Ex 1: 1 color (G). Output: 3 `X`s in the first row.
        *   Ex 2: 3 colors (Y, R, G). Output: 3 `X`s in a diagonal.
    *   **Wait, let's try a different logic: The number of colors determines the number of X's.**
        *   Ex 1: 1 color $\rightarrow$ 3 X's.
        *   Ex 2: 3 colors $\rightarrow$ 3 X's.
        This doesn't help with position.
    *   **Let's look at the input colors as coordinates or counts.**
        Ex 1: All G. G is the 1st color in the key? No.
        Ex 2: Y, R, G.
    *   **Let's look at the "Dark Tablet" (Output) and the "Mosaic" (Input) relationship.**
        In Example 1, the input is all one color. The output is the first row filled with X.
        In Example 2, the input has 3 colors. The output is the anti-diagonal filled with X.
    *   **Let's look at the number of colors again.**
        Ex 1: 1 color.
        Ex 2: 3 colors.
        Test: Y, R, G (3 colors).
    *   **Is it the number of colors that determines the *pattern*?**
        If 1 color: Row 1.
        If 3 colors: Anti-diagonal.
    *   **Let's look at the input grid for Example 2 again.**
        Row 1: Y Y Y (All same)
        Row 2: Y R R (Two colors)
        Row 3: Y Y R (Two colors)
        The number of *rows* that contain only one color?
        Ex 1: 3 rows have one color.
        Ex 2: 1 row has one color.
    *   **Let's try another approach: Count the occurrences of each color.**
        Ex 1: G=9.
        Ex 2: Y=3, R=3, G=1.
        Ex 2 Output: X is at (0,2), (1,1), (2,0).
    *   **Let's look at the "voices" as the number of colors.**
        If number of colors = 1, output is Row 1.
        If number of colors = 3, output is Anti-diagonal.
        Test Input: Y, R, G. Number of colors = 3.
        If the rule is "If 3 colors, use anti-diagonal", the output for the test should be the same as Example 2.

    *   **Wait, let's look at the input of Example 2 more closely.**
        Input:
        Y Y Y
        Y R R
        Y Y R
        The colors are Y, R, G? No, the input is:
        Y Y Y
        Y R R
        Y Y R
        Wait, the example says:
        Input:
        G G G
        Y R R
        Y Y R
        Ah! The first row is G, not Y.
        So Example 2 colors are: G, Y, R. (3 colors).
        Example 1 colors: G (1 color).
        Test Input colors: Y, R, G (3 colors).

    *   **Pattern for 3 colors:**
        Example 2 (3 colors) $\rightarrow$ Anti-diagonal.
        Test Input (3 colors) $\rightarrow$ Anti-diagonal.

    *   **Let's double check Example

... [773 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **The Dancers (Input Grid Colors)**: The colored cells (G, Y, R, etc.) represent the "costumes" of the dancers.
    *   **The Stage Floor (Output Grid)**: A 3x3 grid where most cells are black (`.`) and exactly three cells are grey (`X`).
    *   **The Spotlights (Grey Marks `X`)**: The three specific locations chosen for the final freeze-frame.

2.  **Identify the AFFORDANCES**:
    *   The input colors determine the "diversity" or pattern of the dancers.
    *   The output is constrained to have exactly three `X` marks and six `.` marks.

3.  **Identify the TRANSFORMATION RULE**:
    *   Looking at **Example 1**: Input is all Green (one color). Output has `X` in the top row (0,0), (0,1), (0,2).
    *   Looking at **Example 2**: Input has Green, Yellow, and Red. The output has `X` at (0,2), (1,1), and (2,0). This is the anti-diagonal.
    *   **Refined Rule**: The "costume diversity" refers to the number of unique colors present in the input.
        *   If there is **1 unique color**: The spotlights occupy the **first row** (top row).
        *   If there are **2 unique colors**: The spotlights occupy the **first column** (left column).
        *   If there are **3 unique colors**: The spotlights occupy the **anti-diagonal** (from top-right to bottom-left).
    *   *Wait, let's re-examine Example 2*: Input colors are G, Y, R (3 colors). Output `X` positions are (0,2), (1,1), (2,0). This is indeed the anti-diagonal.
    *   *Let's check Example 1 again*: Input is all G (1 color). Output `X` positions are (0,0), (0,1), (0,2). This is the first row.

    **Hypothesis for N unique colors**:
    *   1 color $\rightarrow$ Row 0
    *   2 colors $\rightarrow$ Column 0 (or perhaps a specific diagonal?)
    *   3 colors $\rightarrow$ Anti-diagonal

    Let's look at the Test Input:
    Input:
    Y Y Y
    R G R
    G R G
    Unique colors: Y, R, G (3 colors).

    If the rule for 3 colors is the anti-diagonal:
    (0,2), (1,1), (2,0)

### Applying the Rule to Test Input

1.  **Count unique colors in Test Input**:
    *   Row 1: Y, Y, Y
    *   Row 2: R, G, R
    *   Row 3: G, R, G
    *   Unique colors: {Y, R, G} $\rightarrow$ Total = 3.

2.  **Apply the 3-color rule**:
    *   Based on Example 2 (which also had 3 colors: G, Y, R), the output for 3 colors is the anti-diagonal.
    *   Anti-diagonal positions in a 3x3 grid:
        *   Row 0, Col 2
        *   Row 1, Col 1
        *   Row 2, Col 0

3.  **Construct the Output Grid**:
    *   Row 0: `. . X`
    *   Row 1: `. X .`
    *   Row 2: `X . .`

ANSWER:
. . X
. X .
X . .

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Input Grid**: A 3x3 grid containing various colored "mineral deposits" (G, Y, R, etc.).
    *   **Output Grid**: A 3x3 "cross-section" consisting of black (`.`) and grey (`X`) cells.
    *   **Grey Cells (`X`)**: These represent the "fracture points."

2.  **Identify the AFFORDANCES**:
    *   The input colors are used to determine the placement of the grey `X` marks.
    *   The output is constrained to only two colors: black (`.`) and grey (`X`).

3.  **Identify the TRANSFORMATION RULE**:
    *   The goal is to "assay the mineral diversity." Looking at the examples:
    *   **Example 1**: Input is all Green (one color). Output has three `X`s in the top row.
    *   **Example 2**: Input has Green, Yellow, and Red (three colors). Output has three `X`s on a diagonal.
    *   **Refined Rule**: The output grid is a 3x3 black grid where the three `X` marks are placed based on the *positions* of the colors in the input. However, looking closer at the relationship between input colors and output `X` positions:
        *   In Example 1: All cells are `G`. The output places `X` in the first row.
        *   In Example 2:
            Row 0: G G G
            Row 1: Y R R
            Row 2: Y Y R
            Output `X` positions: (0,2), (1,1), (2,0).
    *   **Wait, let's re-examine the "diversity" aspect**:
        *   Example 1: Only 1 color (G). Output: `X` at (0,0), (0,1), (0,2).
        *   Example 2: 3 colors (G, Y, R). Output: `X` at (0,2), (1,1), (2,0).
    *   **The actual pattern**: The `X` marks are placed in cells where the input color is the *same* as the color that appears most frequently in its respective row or column? No.
    *   Let's look at the *indices* of the colors.
    *   Example 1: All are G. Output: Row 0 is X.
    *   Example 2:
        Input:
        (0,0)G (0,1)G (0,2)G
        (1,0)Y (1,1)R (1,2)R
        (2,0)Y (2,1)Y (2,2)R
        Output:
        (0,2)X (1,1)X (2,0)X
    *   Let's look at the *count* of colors.
        Ex 1: G=9.
        Ex 2: G=3, Y=3, R=3.
    *   If there are 3 colors, the `X` marks form a diagonal. If there is 1 color, they form a row.
    *   Let's look at the *Test Input*:
        Y Y Y
        R G R
        G R G
        Colors: Y=3, R=3, G=3.
    *   Since there are 3 colors (Y, R, G) in the test input, just like Example 2, the pattern for 3 colors is the anti-diagonal: (0,2), (1,1), (2,0).

### Applying the Rule to Test Input

Test Input:
Y Y Y
R G R
G R G

Colors present: Y, R, G (Total 3 colors).
Following the pattern from Example 2 (where 3 colors resulted in the anti-diagonal):
The `X` marks should be at:
(0, 2)
(1, 1)
(2, 0)

Output Grid:
. . X
. X .
X . .

ANSWER:
. . X
. X .
X . .

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **The Pot (Input Grid)**: A 3x3 grid containing various colored "ingredients" (colors like G, Y, R).
    *   **The Plate (Output Grid)**: A 3x3 grid that is primarily "black" (`.`), containing "grey sauce" (`X`) in a specific pattern.
    *   **The Grey Sauce (`X`)**: The element that is extracted and placed on the plate.

2.  **Identify the AFFORDANCES**:
    *   The input colors act as indicators for the placement of the grey sauce.
    *   The output grid is constrained to only two colors: black (`.`) and grey (`X`).

3.  **Identify the TRANSFORMATION RULE**:
    *   The goal is to find the "pattern that honors the depth of flavor." Looking at the examples:
    *   **Example 1**: Input is all Green (G). Output has `X` in the first row.
    *   **Example 2**: Input has a mix of G, Y, R. The `X` marks are at (0,2), (1,1), and (2,0). This is a diagonal.
    *   **Wait, let's look closer at the relationship between input colors and output `X` positions**:
        *   In Example 1: All cells are G. Output has `X` at (0,0), (0,1), (0,2).
        *   In Example 2:
            Input:
            G G G
            Y R R
            Y Y R
            Output:
            . . X
            . X .
            X . .
            The `X` positions are (0,2), (1,1), (2,0). These are the positions where the input color is **not** the most frequent color? No.
            Let's look at the colors:
            Example 1: G is the only color. Output: Top row is X.
            Example 2: G=3, Y=3, R=3. All colors are equal. Output: Anti-diagonal is X.
            
            Let's re-examine the "depth of flavor" metaphor. Usually, in these puzzles, the "sauce" (X) is placed where a specific color exists or where a specific condition is met.
            In Example 1: Input is all G. Output is the first row of X.
            In Example 2: Input has G, Y, R. Output is the anti-diagonal of X.
            
            Let's look at the **indices** of the colors in Example 2:
            G: (0,0), (0,1), (0,2)
            Y: (1,0), (2,0), (2,1)
            R: (1,1), (1,2), (2,2)
            The output X is at (0,2), (1,1), (2,0).
            These are the positions of: G at (0,2), R at (1,1), Y at (2,0).
            
            Wait, let's look at the **Input Colors** again.
            Example 1: All G. Output: Row 0 is X.
            Example 2: G, Y, R. Output: Anti-diagonal is X.
            
            Let's try a different logic: The output `X` represents the positions of a specific color.
            In Example 1: If X = G, then the first row is G. But the whole grid is G.
            In Example 2: If X = G, the positions are (0,0), (0,1), (0,2). Not a match.
            If X = Y, the positions are (1,0), (2,0), (2,1). Not a match.
            If X = R, the positions are (1,1), (1,2), (2,2). Not a match.
            
            Let's look at the **diagonal** logic.
            Example 1: All G. Output: Row 0 is X.
            Example 2: G, Y, R. Output: Anti-diagonal is X.
            
            Let's look at the **Test Input**:
            Y Y Y
            R G R
            G R G
            
            Count the colors in Test Input:
            Y: 3
            R: 3
            G: 3
            
            In Example 2, all colors were equal (3 each), and the output was the anti-diagonal.
            In the Test Input, all colors are equal (3 each).
            Therefore, the output should follow the same pattern as Example 2.

### Applying the Rule to Test Input
Test Input:
Y Y Y
R G R
G R G

Since the color counts are equal (Y=3, R=3, G=3), just like Example 2, the pattern is the anti-diagonal:
(0,2), (1,1), (2,0).

Output:
. . X
. X .
X . .

ANSWER:
. . X
. X .
X . .

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Input Grid**: A 3x3 "sound field" containing various colored "tones" (colors).
    *   **Colors**: Represent different "voices" or frequencies.
    *   **Output Grid**: A 3x3 "score" where the background is "silent black" (`.`) and the "resonance" is marked in "grey" (`X`).

2.  **Identify the AFFORDANCES**:
    *   The input colors can be grouped to identify "distinct voices" (unique colors).
    *   The output grid is constrained to a 3x3 size, containing exactly three `X` marks (as per the description "three grey notes placed where the resonance lands").

3.  **Identify the TRANSFORMATION RULE**:
    *   The goal is to find the "resonance pattern." Looking at the examples:
    *   **Example 1**: Input is all Green (1 distinct color). Output has `X` at (0,0), (0,1), (0,2). This doesn't immediately show a pattern of 3. Wait, let's re-examine the description: "three grey notes placed where the resonance lands."
    *   **Example 2**: Input has Green, Yellow, Red (3 distinct colors). Output has `X` at (0,2), (1,1), (2,0). This is the anti-diagonal.
    *   **Re-evaluating Example 1**: Input is all Green. There is only 1 distinct color. The output shows `X X X` in the first row.
    *   **Wait, let's look at the "resonance" differently**:
        *   Example 1: Input is all G. Distinct colors = 1. Output: First row is X.
        *   Example 2: Input has G, Y, R. Distinct colors = 3. Output: Anti-diagonal is X.
    *   **Let's look at the positions of the colors**:
        *   Example 1: All cells are G.
        *   Example 2:
            G G G (Row 0)
            Y R R (Row 1)
            Y Y R (Row 2)
            Colors: G (3), Y (3), R (3). Total 3 colors.
            Output: (0,2), (1,1), (2,0).
    *   **Let's look at the "distinct voices" (unique colors) and their counts**:
        *   Example 1: G appears 9 times. 1 color.
        *   Example 2: G appears 3 times, Y appears 3 times, R appears 3 times. 3 colors.
    *   **Wait, let's look at the coordinates of the colors**:
        In Example 2, the colors are:
        G: (0,0), (0,1), (0,2)
        Y: (1,0), (2,0), (2,1)
        R: (1,1), (1,2), (2,2)
        The output `X` are at (0,2), (1,1), (2,0).
        These are the positions where the *counts* of the colors might be related? No.
    *   **Let's try another pattern**: The output `X` marks the positions of a specific color? No.
    *   **Let's look at the "resonance" as the diagonal/anti-diagonal based on the number of colors**:
        If number of colors = 1, output is Row 0.
        If number of colors = 3, output is Anti-diagonal.
    *   **Let's check the Test Input**:
        Y Y Y
        R G R
        G R G
        Colors: Y (3), R (3), G (3). Total 3 colors.
        If the rule is "If 3 colors, output anti-diagonal", the output should be the same as Example 2.

    *   **Wait, let's look closer at Example 2 again**:
        Input:
        G G G
        Y R R
        Y Y R
        Output:
        . . X
        . X .
        X . .
        The `X` are at (0,2), (1,1), (2,0).
        In the input, what is at (0,2)? G. What is at (1,1)? R. What is at (2,0)? Y.
        These are three different colors.
        In Example 1:
        Input:
        G G G
        G G G
        G G G
        Output:
        X X X
        . . .
        . . .
        The `X` are at (0,0), (0,1), (0,2). All are G.

    *   **Hypothesis**: The output `X` marks a path/line.
        If 1 color: Row 0.
        If 3 colors: Anti-diagonal.
        Wait, let's look at the colors in the Test Input:
        Y Y Y
        R G R
        G R G
        Colors: Y, R, G. (3 colors).
        If the rule is "3 colors = anti-diagonal", the output is:
        . . X
        . X .
        X . .

    *   **Let's refine the "resonance" rule**:
        The number of distinct colors determines the "shape" of the resonance.
        1 color -> Horizontal line (Row 0)
        3 colors -> Anti-diagonal line
        (This is a common pattern in these puzzles where the number of colors dictates the geometry).

### Applying the rule to the Test Input
Test Input:
Y Y Y
R G R
G R G

1. Count distinct colors: Y, R, G. Total = 3.
2. Since there are 3 distinct colors, the resonance pattern is the anti-diagonal (as seen in Example 2).
3. The anti-diagonal in a 3x3 grid is (0,2), (1,1), (2,0).

Output:
. . X
. X .
X . .

ANSWER:
. . X
. X .
X . .

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Colored Jerseys (Input Colors):** The colored cells (G, Y, R, etc.) represent different "factions" or teams on the pitch.
    *   **Grey Markers (X):** These represent the signal/output on the scoreboard.
    *   **Dark Scoreboard (.):** The black background of the 3x3 output grid.

2.  **Identify the AFFORDANCES:**
    *   The input grid contains various colors.
    *   The output grid must be 3x3, consisting only of black (`.`) and grey (`X`).
    *   The number of grey markers (`X`) is determined by the number of distinct colors (factions) present in the input grid.

3.  **Identify the TRANSFORMATION RULE:**
    *   **Step 1: Count the Factions.** Count how many unique colors (excluding black/`.`) are present in the input grid.
    *   **Step 2: Determine the Placement.** The metaphor "signal the grey markers into formation" and the examples suggest a specific pattern.
        *   Example 1: Input has 1 color (G). Output has 3 `X`s in the first row.
        *   Example 2: Input has 3 colors (G, Y, R). Output has 3 `X`s in a diagonal (top-right to bottom-left).
    *   **Wait, let's re-examine the pattern of `X` placement:**
        *   Example 1: 1 color (G) $\rightarrow$ 3 `X`s in row 1.
        *   Example 2: 3 colors (G, Y, R) $\rightarrow$ 3 `X`s in a diagonal.
    *   **Correction/Refinement:** Let's look at the *count* of the most frequent color or the *number* of colors.
        *   Ex 1: 1 color (G). Output: 3 `X`s.
        *   Ex 2: 3 colors (G, Y, R). Output: 3 `X`s.
    *   **Let's look at the "Referee's Call" (The count of colors):**
        *   Ex 1: Colors present = {G}. Count = 1.
        *   Ex 2: Colors present = {G, Y, R}. Count = 3.
    *   **Wait, the number of `X`s is always 3.** The rule is about *where* they go based on the *number of colors*.
        *   If Count = 1: `X`s go in the first row (`X X X`).
        *   If Count = 3: `X`s go in the anti-diagonal (`. . X`, `. X .`, `X . .`).
    *   **Let's check the Test Input:**
        *   Input: `Y Y Y`, `R G R`, `G R G`.
        *   Colors present: Y, R, G.
        *   Count of colors = 3.
    *   **Applying the rule for Count = 3:**
        *   The pattern for 3 colors (from Example 2) is the anti-diagonal.

### Applying the Rule to Test Input

1.  **Input Colors:** Y (Yellow), R (Red), G (Green).
2.  **Count of unique colors:** 3.
3.  **Pattern for 3 colors:** Based on Example 2, the pattern is the anti-diagonal:
    Row 1: `. . X`
    Row 2: `. X .`
    Row 3: `X . .`

ANSWER:
. . X
. X .
X . .

To solve this puzzle, let's perform a Conceptual Digestion.

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Input Grid (The Sky):** A 3x3 grid of colored cells representing "chromatic complexity."
    *   **Output Grid (The Radar):** A 3x3 grid where the background is black (`.`) and specific cells are marked with grey (`X`) "blips."

2.  **Identify the AFFORDANCES:**
    *   The colors in the input grid act as indicators for the placement of the grey `X` marks.
    *   The number of colors or the arrangement of colors determines the number and position of the `X` marks.

3.  **Identify the TRANSFORMATION RULE:**
    *   Let's analyze the examples:
        *   **Example 1:** Input is all Green (1 color). Output has 3 `X`s in the top row.
        *   **Example 2:** Input has Yellow (Y), Red (R), and Green (G) is not present (Wait, let's re-examine). Input: `G G G / Y R R / Y Y R`. Colors present: G, Y, R. Output: `X` at (0,2), (1,1), (2,0). This is a diagonal.
    *   **Wait, let's look closer at the "chromatic complexity" and "formation":**
        *   In Example 1: All cells are `G`. There is only 1 color. The output has 3 `X`s.
        *   In Example 2: Colors are `G`, `Y`, `R`. There are 3 colors. The output has 3 `X`s.
    *   **Let's look at the pattern of the `X`s relative to the colors:**
        *   Example 1: Input is all `G`. Output is `X X X` in the first row.
        *   Example 2: Input has `G` at (0,0), (0,1), (0,2), `Y` at (1,0), `R` at (1,1), (1,2), `Y` at (2,0), (2,1), `R` at (2,2).
        *   Wait, let's re-read Example 2 input:
            Row 0: G G G
            Row 1: Y R R
            Row 2: Y Y R
            Output:
            . . X
            . X .
            X . .
            This is the anti-diagonal.
    *   **Let's re-evaluate the rule based on color counts or positions:**
        *   Example 1: 1 color (G). Output: Top row is X.
        *   Example 2: 3 colors (G, Y, R). Output: Anti-diagonal is X.
    *   **Let's look at the "complexity" differently. Is it the number of colors?**
        *   If 1 color $\rightarrow$ Row 0 is X.
        *   If 3 colors $\rightarrow$ Anti-diagonal is X.
    *   **Let's look at the Test Input:**
        Input:
        Y Y Y
        R G R
        G R G
        Colors: Y, R, G (3 colors).
    *   **Wait, let's look at the "formation" again.**
        In Example 1, the color is G. The output is the first row.
        In Example 2, the colors are G, Y, R. The output is the anti-diagonal.
        Is it possible the rule is: The `X` marks are placed where the color is the "most complex" or follows a specific sequence?
        Actually, look at the colors in Example 2:
        G is at (0,0), (0,1), (0,2)
        Y is at (1,0), (2,0), (2,1)
        R is at (1,1), (1,2), (2,2)
        The output `X`s are at (0,2), (1,1), (2,0).
        At (0,2) color is G. At (1,1) color is R. At (2,0) color is G. This doesn't seem to match.

    *   **Let's try another approach: The number of colors determines the shape.**
        Example 1: 1 color $\rightarrow$ 3 `X`s in a row.
        Example 2: 3 colors $\rightarrow$ 3 `X`s in a diagonal.
        Test Input: 3 colors (Y, R, G).
        If 3 colors $\rightarrow$ Anti-diagonal.

    *   **Let's re-examine Example 1 and 2 for a different pattern.**
        Example 1: All G. Output: `X X X / . . . / . . .`
        Example 2: G, Y, R. Output: `. . X / . X . / X . .`
        Is it the number of *distinct* colors?
        Ex 1: 1 color $\rightarrow$ Row 0.
        Ex 2: 3 colors $\rightarrow$ Anti-diagonal.
        Test: 3 colors $\rightarrow$ Anti-diagonal.

    *   **Wait, let's look at the colors in the Test Input again.**
        Y Y Y
        R G R
        G R G
        The colors are Y, R, G.
        In Example 2, the colors were G, Y, R.
        In Example 1, the color was G.
        
        Let's look at the *position* of the colors.
        In Example 1, G is everywhere.
        In Example 2, G is the first row.
        In the Test Input, G is in the middle and bottom.
        
        Let's look at the "storm pattern" (X) relative to the "sky" (colors).
        In Example 1: G is the only color. The X's are in the first row.
        In Example 2: G is the first row. The X's are the anti-diagonal.
        In Test Input: G is NOT the first row. Y is the first row.
        
        If the rule is: The `X`s are placed in the row/diagonal corresponding to the *first* color encountered in the grid?
        Ex 1: First color is G. G is in Row 0. Output: Row 0 is X.
        Ex 2: First color is G. G is in Row 0. Output: Anti-diagonal is X. (Wait, this contradicts).
        
        Let's try: The `X`s are placed in the row/diagonal corresponding to the *last* color encountered?
        Ex 1: Last color is G. G is in Row 0. Output: Row 0 is X.
        Ex 2: Last color is R. R is in Row 2. Output: Anti-diagonal is X. (No).

        Let's try: The `X`s are placed in the row/diagonal corresponding to the 

... [628 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Input Grid:** A 3x3 collection of colored "building materials" (colors like G, Y, R).
    *   **Output Grid:** A 3x3 "blueprint" (black background `.`) containing exactly three grey "structural supports" (`X`).

2.  **Identify the AFFORDANCES:**
    *   The input colors represent the presence of materials.
    *   The output is constrained to have exactly three `X` marks.
    *   The placement of `X` is determined by the "bracing pattern" demanded by the input.

3.  **Identify the TRANSFORMATION RULE:**
    *   The goal is to find a pattern of three cells in the 3x3 grid.
    *   Looking at Example 1: Input is all Green. Output is `X X X` in the top row.
    *   Looking at Example 2: Input has colors in specific positions. Output is a diagonal `X . .`, `. X .`, `. . X` (anti-diagonal).
    *   Wait, let's re-examine the relationship between input colors and output `X` positions.
    *   **Example 1:** All cells are G. Output: `X X X` (Row 0).
    *   **Example 2:** 
        Input:
        G G G (Row 0)
        Y R R (Row 1)
        Y Y R (Row 2)
        Output:
        . . X (0,2)
        . X . (1,1)
        X . . (2,0)
        This is the anti-diagonal.
    *   **Re-evaluating the pattern:** The "bracing pattern" is determined by the *distribution* or *symmetry* of the colors. However, a simpler rule often exists in these puzzles: The output `X` marks are placed in the positions of the most frequent color, or perhaps the colors represent a specific shape.
    *   Let's look at the colors in Example 2: G appears 3 times, Y appears 3 times, R appears 3 times. All colors are equal.
    *   Let's look at the colors in Example 1: G appears 9 times. Output is the first row.
    *   Let's look at the colors in the Test Input:
        Y Y Y (3 Y)
        R G R (2 R, 1 G)
        G R G (2 G, 1 R)
        Total: Y=3, R=3, G=3.
    *   Wait, in Example 2, the colors were also balanced (3 G, 3 Y, 3 R). In Example 2, the output was the anti-diagonal.
    *   In Example 1, all were G. The output was the top row.
    *   Let's look at the *arrangement* of colors.
    *   Example 2:
        G G G
        Y R R
        Y Y R
        The colors form a "staircase" or "diagonal" flow? No.
    *   Let's look at the *corners* or *edges*.
    *   Actually, look at the *symmetry*. Example 2 is not symmetric. Example 1 is perfectly symmetric.
    *   Let's try another approach: The output `X` marks are placed where a specific color is located? No, there are only 3 `X`s.
    *   Let's look at the *rows* in Example 2: Row 0 is all G. Row 1 has Y, R, R. Row 2 has Y, Y, R.
    *   Let's look at the *columns* in Example 2: Col 0 is G, Y, Y. Col 1 is G, R, Y. Col 2 is G, R, R.
    *   Wait! Look at the *diagonals* of the input in Example 2:
        Main diagonal: G, R, R.
        Anti-diagonal: G, R, Y.
    *   Let's look at the *Input* of Example 2 again. The colors are:
        (0,0)=G, (0,1)=G, (0,2)=G
        (1,0)=Y, (1,1)=R, (1,2)=R
        (2,0)=Y, (2,1)=Y, (2,2)=R
    *   The output is (0,2), (1,1), (2,0). These are the positions of the *anti-diagonal*.
    *   In Example 1, the output is (0,0), (0,1), (0,2). This is the *top row*.
    *   Is there a pattern in the colors?
        Example 1: All same color.
        Example 2: Three colors, each appearing 3 times.
        Test Input: Three colors, each appearing 3 times.
    *   If the colors are balanced (3 of each), the output is the anti-diagonal (Example 2).
    *   If the colors are all the same, the output is the top row (Example 1).
    *   Let's check the Test Input: Y=3, R=3, G=3.
    *   Since the Test Input has a balanced distribution of colors (3 Y, 3 R, 3 G), it should follow the same rule as Example 2.

### Applying the Rule to Test Input
Test Input:
Y Y Y
R G R
G R G

Color counts:
Y: (0,0), (0,1), (0,2) -> 3
R: (1,0), (1,2), (2,1) -> 3
G: (1,1), (2,0), (2,2) -> 3

Since the counts are balanced (3 of each), the output should be the anti-diagonal, just like Example 2.

Anti-diagonal positions: (0,2), (1,1), (2,0).

Output:
. . X
. X .
X . .

ANSWER:
. . X
. X .
X . .

### Conceptual Digestion

**1. Identify the CAST:**
*   **Input Grid (The Battlefield):** A 3x3 grid containing various colored "banners" (G, Y, R, etc.).
*   **Output Grid (The War Map):** A 3x3 grid consisting of black (`.`) and grey (`X`) "sentinels".
*   **Colors:** The specific colors in the input represent different "allegiances" or signals.

**2. Identify the AFFORDANCES:**
*   The input colors act as triggers or indicators.
*   The output grid is a binary-style map where only `X` (grey) and `.` (black) exist.
*   The number of `X` sentinels in the output seems to be determined by the input, but looking at the examples, it's more specific.

**3. Identify the TRANSFORMATION RULE:**
*   **Example 1:** Input is all Green (G). Output has three `X`s in the top row.
*   **Example 2:** Input has Y, R, G. Output has three `X`s in a diagonal (bottom-left, middle, top-right).
*   **Wait, let's re-examine the pattern:**
    *   In Example 1: All cells are G. The output is `X X X` in the first row.
    *   In Example 2: 
        Row 1: G G G
        Row 2: Y R R
        Row 3: Y Y R
        Output:
        `. . X`
        `. X .`
        `X . .`
    *   Let's look at the colors in Example 2:
        Y is at (1,0), (2,0), (2,1)
        R is at (1,1), (1,2), (2,2)
        G is at (0,0), (0,1), (0,2)
        The output `X` positions are (0,2), (1,1), (2,0).
    *   Let's look at the colors at those specific output positions in the input:
        Input (0,2) is G.
        Input (1,1) is R.
        Input (2,0) is Y.
    *   Wait, let's look at the colors in Example 1 again:
        Input (0,0)=G, (0,1)=G, (0,2)=G. Output (0,0)=X, (0,1)=X, (0,2)=X.
    *   **Hypothesis:** The output places an `X` if a specific color condition is met. 
    *   Let's look at the "allegiances" differently. In Example 1, there is only one color (G). In Example 2, there are three colors (G, Y, R).
    *   Let's check the counts:
        Ex 1: G=9. Output: 3 X's.
        Ex 2: G=3, Y=3, R=3. Output: 3 X's.
    *   Let's look at the *diagonal* or *symmetry*.
        In Ex 2, the `X`s are at (0,2), (1,1), (2,0). These are the cells where the color is unique to a specific count? No.
    *   Let's look at the colors in the input grid and see if they form a pattern.
        Ex 1: All G. Output: Top row is X.
        Ex 2: 
        G G G
        Y R R
        Y Y R
        Output: (0,2), (1,1), (2,0) are X.
        Input (0,2) is G. Input (1,1) is R. Input (2,0) is Y.
        Notice the colors in the output diagonal: G, R, Y.
        In the input, G is the first color mentioned, R is second, Y is third? No.
    *   **Let's try another approach: The "Majority" or "Color presence" rule.**
        In Ex 1: Only G exists. Output: Row 0 is X.
        In Ex 2: G, Y, R exist. Output: The anti-diagonal is X.
    *   **Wait, look at the colors in the input vs the output positions again.**
        Ex 1: Input (0,0)=G, (0,1)=G, (0,2)=G. Output (0,0)=X, (0,1)=X, (0,2)=X.
        Ex 2: Input (0,2)=G, (1,1)=R, (2,0)=Y. Output (0,2)=X, (1,1)=X, (2,0)=X.
        In Ex 1, the color G is at (0,0), (0,1), (0,2).
        In Ex 2, the colors G, R, Y are at (0,2), (1,1), (2,0).
    *   **The Rule:** The output places an `X` at position (r, c) if the color at input (r, c) is the "primary" color of a specific set? No.
    *   **Let's look at the colors in the input as a sequence:**
        Ex 1: G, G, G, G, G, G, G, G, G.
        Ex 2: G, G, G, Y, R, R, Y, Y, R.
    *   **Let's look at the colors in the input grid and find the "corners" or "edges"?**
        Ex 1: All G. Output: Top row.
        Ex 2: G is top row. Y is bottom left. R is bottom right.
        The output `X`s are at (0,2) [G], (1,1) [R], (2,0) [Y].
    *   **Wait! Look at the colors in the input grid for Ex 2 again:**
        (0,0)=G, (0,1)=G, (0,2)=G
        (1,0)=Y, (1,1)=R, (1,2)=R
        (2,0)=Y, (2,1)=Y, (2,2)=R
        The colors are G, Y, R.
        The output `X`s are at (0,2), (1,1), (2,0).
        The colors at those positions are: (0,2)=G, (1,1)=R, (2,0)=Y.
        In Ex 1: The colors at (0,0), (0,1), (0,2) are G, G, G.
    *   **Is it about the color that appears in a specific pattern?**
        In Ex 1, G is the only color. The output is the first row.
        In Ex 2, G is the first row, R is the right side, Y is the left side.
        The output is the anti-diagonal.
    *   **Let's look at the Test Input:**
        Y Y Y
        R G R
        G R G
        Colors: Y, R, G.
        Y is at (0,0), (0,1), (0,2).
        R is at (1,0), (1,2), (2,1).
        G is at (1,1), (2,0), (2,2).
    *   **Let's re-examine Ex 1 and Ex 2 for a pattern in the colors.**
        Ex 1: All G. Output: Row 0.
        Ex 2: G is Row 0. Y is (1,0), (2,0), (2,1). R is (1,1), (1,2), (2,2).
        The output is (0,2), (1,1), (2,0).
        These are the cells where the color is "unique" in its row or column?
        Ex 2:
        (0,2) is G. In row 0, all are G. In col 2, (0,2)=G, (1,2)=R, (2,2)=R. G is unique in

... [3104 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Colored pixels (G, Y, R, etc.):** These are the "costumed players" or "roles." Each unique color represents a distinct role.
    *   **Grey pixels (X):** These are the "grey spotlights."
    *   **Black pixels (.):** These represent the "dark stage."

2.  **Identify the AFFORDANCES:**
    *   The colored pixels are counted to determine the number of spotlights.
    *   The spotlights (X) are placed on the grid based on the distribution or count of the colors.
    *   The output grid is a 3x3 grid where the original colors are replaced by black (.) and the spotlights (X) are placed in specific positions.

3.  **Identify the TRANSFORMATION RULE:**
    *   **Step 1:** Count the number of distinct colors present in the input grid (excluding black/background if any, though the input is fully colored here).
    *   **Step 2:** The number of distinct colors determines how many grey spotlights (X) will be placed in the output.
    *   **Step 3:** Looking at the examples:
        *   Example 1: Input has 1 color (G). Output has 3 X's in the first row. (Wait, let's re-examine).
        *   Example 1: Input is all G. Distinct colors = 1. Output has 3 X's.
        *   Example 2: Input has 3 colors (G, Y, R). Distinct colors = 3. Output has 3 X's placed on a diagonal.
    *   **Re-evaluating the rule based on "Count the distinct roles and aim the grey spotlights":**
        *   In Example 1: Only 1 color (G). The output has 3 X's. This contradicts "count distinct roles". Let's look closer.
        *   Example 1: Input is 3x3 of G. Output is 3 X's in the top row.
        *   Example 2: Input has G, Y, R. Output has 3 X's on the anti-diagonal.
    *   **Wait, let's look at the "roles" differently.** Maybe it's not the number of colors, but the *count* of each color?
        *   Ex 1: G=9. Output: 3 X's.
        *   Ex 2: G=1, Y=3, R=5. Output: 3 X's.
    *   **Let's look at the positions of the colors.**
        *   Ex 1: All G. Output: X X X / . . . / . . .
        *   Ex 2: Y is at (0,0), (0,1), (0,2), (1,0), (2,0). R is at (1,1), (1,2), (2,1), (2,2). G is at (1,1)? No.
        *   Let's re-read Example 2 Input:
            Y Y Y
            Y R R
            Y Y R
            Colors: Y (5), R (4). Wait, the input is:
            G G G
            Y R R
            Y Y R
            Colors: G (3), Y (3), R (3).
            Distinct colors = 3. Output has 3 X's.
            Positions of X in Ex 2: (0,2), (1,1), (2,0). This is the anti-diagonal.
        *   Ex 1: Input is all G. Distinct colors = 1. But output has 3 X's.
        *   Wait, if the input is all one color, maybe the "roles" are the *rows* or *columns*? No.
        *   Let's look at the *counts* again.
        *   Ex 1: G=9.
        *   Ex 2: G=3, Y=3, R=3.
        *   In Ex 2, there are 3 colors, and each color appears 3 times. The output is the anti-diagonal.
        *   In Ex 1, there is 1 color, and it appears 9 times. The output is the first row.
    *   **Let's try a different logic: The number of X's is always 3.**
        *   Where are they placed?
        *   Ex 1: All G. Output: Row 0.
        *   Ex 2: G, Y, R. Output: Anti-diagonal.
        *   Test Input: Y, R, G.
            Y Y Y
            R G R
            G R G
            Counts: Y=3, R=3, G=3.
            This matches the pattern of Example 2 (where each color appeared 3 times).
            In Example 2, the colors were G, Y, R. The output was the anti-diagonal.
            In the Test Input, the colors are Y, R, G.
            If the rule is: If all colors appear 3 times, place X on the anti-diagonal.

    *   **Let's refine the "Distinct Roles" rule:**
        The number of X's is 3.
        The placement depends on the distribution.
        If there is only 1 color (Ex 1), X's fill the first row.
        If there are 3 colors and each appears 3 times (Ex 2), X's fill the anti-diagonal.
        In the Test Input: Y=3, R=3, G=3.
        This is the same distribution as Example 2.

### Applying the rule to Test Input:
Input:
Y Y Y
R G R
G R G

Counts:
Y: (0,0), (0,1), (0,2) -> 3
R: (1,0), (1,2), (2,1) -> 3
G: (1,1), (2,0), (2,2) -> 3

Since each of the 3 colors appears exactly 3 times, the pattern follows Example 2.
Example 2 output:
. . X
. X .
X . .

Test Input output:
. . X
. X .
X . .

ANSWER:
. . X
. X .
X . .

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Colored Blooms (G, Y, R, etc.)**: These represent the "floral diversity" mentioned in the metaphor. They are the input elements.
    *   **Dark Soil (.)**: The background/empty space in the output grid.
    *   **Grey Stepping Stones (X)**: The elements to be placed in the output grid.

2.  **Identify the AFFORDANCES**:
    *   The colored cells provide information about the "variety" or "diversity" of the grid.
    *   The output grid is a 3x3 space where exactly three 'X' (grey stones) must be placed.

3.  **Identify the TRANSFORMATION RULE**:
    *   The metaphor says the path of the stones "reflects the variety of blooms."
    *   Looking at the examples:
        *   **Example 1**: Input is all Green (1 color). Output has 3 'X's in the first row.
        *   **Example 2**: Input has Green, Yellow, and Red (3 colors). Output has 3 'X's in a diagonal pattern (top-right, middle, bottom-left).
    *   Wait, let's re-examine the "variety" logic.
        *   Example 1: 1 color (G). Output: X X X in row 0.
        *   Example 2: 3 colors (G, Y, R). Output: X at (0,2), (1,1), (2,0).
    *   Let's look at the distribution of colors.
        *   Example 1: All cells are G.
        *   Example 2: Row 0: G, G, G (Wait, Example 1 was GGG/GGG/GGG. Example 2 is GGG/YRR/YYR).
        *   Let's re-read Example 2 input:
            Row 0: G G G
            Row 1: Y R R
            Row 2: Y Y R
            Colors present: G, Y, R (3 colors).
    *   The rule seems to be: The number of unique colors determines the pattern of the three 'X's.
        *   If 1 color: All 'X's in the first row.
        *   If 2 colors: (Hypothetical) Maybe a column?
        *   If 3 colors: A diagonal.
    *   Let's check the Test Input:
        Input:
        Y Y Y
        R G R
        G R G
        Colors present: Y, R, G (3 colors).
    *   Since there are 3 unique colors in the test input (Y, R, G), the pattern should match Example 2 (the 3-color pattern).
    *   Example 2 pattern:
        . . X
        . X .
        X . .
        (This is the anti-diagonal).

### Applying the Rule to Test Input

1.  **Count unique colors in Test Input**:
    *   Y (Yellow)
    *   R (Red)
    *   G (Green)
    *   Total = 3 unique colors.
2.  **Determine pattern for 3 colors**:
    *   Based on Example 2, the pattern for 3 colors is the anti-diagonal:
        (0, 2), (1, 1), (2, 0).
3.  **Construct Output**:
    Row 0: . . X
    Row 1: . X .
    Row 2: X . .

ANSWER:
. . X
. X .
X . .

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Grey animals (X)**: These are the "herds" or connected components of grey pixels.
    *   **Dark landscape (.)**: The background that remains unchanged.
    *   **Tag colors (B, R, G, etc.)**: The new colors assigned to the grey pixels based on the size of their connected component.

2.  **Identify the AFFORDANCES**:
    *   **Grey pixels (X)**: Can be grouped into connected components (herds) using 4-connectivity (up, down, left, right). Once grouped, they can be recolored based on the count of pixels in that group.
    *   **Black pixels (.)**: Cannot be changed.

3.  **Identify the TRANSFORMATION RULE**:
    *   Find all connected components of 'X' pixels.
    *   Count the number of pixels in each component.
    *   Assign a color based on the size:
        *   Size 2: Blue (B) (Based on Example 1: `X X` becomes `B B`)
        *   Size 3: Red (R) (Based on Example 1: `X X X` becomes `R R R`)
        *   Size 4: Green (G) (Based on Example 1: `X X X X` becomes `G G G G`)
        *   *Correction based on Example 1 analysis*:
            *   Component 1: `X X X` (size 3) -> R
            *   Component 2: `X X` (size 2) -> B
            *   Component 3: `X X X` (size 3) -> R
            *   Component 4: `X X` (size 2) -> B
            *   Component 5: `X X X` (size 3) -> B? Wait, let's re-examine Example 1 carefully.
            
    *   **Re-examining Example 1**:
        *   Group 1: (2,1), (2,2), (2,3) -> Size 3 -> R
        *   Group 2: (2,7), (3,7) -> Size 2 -> G? No, let's look at the output.
        *   Output colors in Example 1:
            *   (2,1)-(2,3) is R (Size 3)
            *   (2,7)-(3,7) is G (Size 2) -> Wait, size 2 is G?
            *   (5,3)-(6,3) is R (Size 2) -> No, (5,3), (5,4), (6,3) is size 3.
            *   (7,7) is B (Size 1?) No, (7,7) is part of (8,6)-(8,8).
            *   Let's re-map Example 1:
                *   (2,1),(2,2),(2,3) -> R (Size 3)
                *   (2,7),(3,7) -> G (Size 2)
                *   (5,3),(5,4),(6,3) -> R (Size 3)
                *   (7,7),(8,6),(8,7),(8,8) -> B (Size 4)
                *   (8,1),(8,2),(9,1),(9,2) -> B (Size 4)
            *   Wait, the rule is:
                *   Size 2 -> G
                *   Size 3 -> R
                *   Size 4 -> B
                *   Let's check the description: "smallest pairs share one color, the middle-sized trios another, and the largest bands of four a third."
                *   Size 2 = G
                *   Size 3 = R
                *   Size 4 = B

### Applying to Test Input

1.  **Identify Components**:
    *   Component 1: (0,9), (1,9), (2,9), (3,9) -> Size 4
    *   Component 2: (2,2), (2,3), (3,1), (3,2) -> Size 4
    *   Component 3: (2,5), (2,6), (3,6) -> Size 3
    *   Component 4: (6,0), (7,0), (8,0) -> Size 3
    *   Component 5: (7,3), (7,4) -> Size 2
    *   Component 6: (7,7), (8,7) -> Size 2

2.  **Assign Colors**:
    *   Size 4 -> B
    *   Size 3 -> R
    *   Size 2 -> G

3.  **Map to Grid**:
    *   (0,9), (1,9), (2,9), (3,9) -> B
    *   (2,2), (2,3), (3,1), (3,2) -> B
    *   (2,5), (2,6), (3,6) -> R
    *   (6,0), (7,0), (8,0) -> R
    *   (7,3), (7,4) -> G
    *   (7,7), (8,7) -> G

### Final Grid Construction
Row 0: . . . . . . . . . B
Row 1: . . . . . . . . . B
Row 2: . . B B . R R . . B
Row 3: . B B . . . R . . B
Row 4: . . . . . . . . . .
Row 5: . . . . . . . . . .
Row 6: R . . . . . . . . .
Row 7: R . . G G . . G . .
Row 8: R . . . . . . G . .
Row 9: . . . . . . . . . .

ANSWER:
. . . . . . . . . B
. . . . . . . . . B
. . B B . R R . . B
. B B . . . R . . B
. . . . . . . . . .
. . . . . . . . . .
R . . . . . . . . .
R . . G G . . G . .
R . . . . . . G . .
. . . . . . . . . .

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Black cells (`.`)**: The background/slide.
    *   **Grey cells (`X`)**: The "pale, unstained cell colonies." These are contiguous groups of grey cells.
    *   **Colors (B, R, G, etc.)**: The "fluorescent markers" applied to the colonies.

2.  **Identify the AFFORDANCES**:
    *   **Grey cells** can be grouped into connected components (colonies).
    *   **Colonies** can be counted by the number of cells they contain.
    *   **The size of the colony** determines which color it is transformed into.

3.  **Identify the TRANSFORMATION RULE**:
    *   Find all connected components of `X` (using 4-connectivity or 8-connectivity; looking at Example 1, it appears to be 4-connectivity or standard grouping).
    *   Count the number of cells in each component.
    *   Assign a color based on the size:
        *   Looking at Example 1:
            *   Colony 1: `X X X` (3 cells) -> `R` (Red)
            *   Colony 2: `X` (1 cell) -> `G` (Green)
            *   Colony 3: `X X` and `X` (3 cells) -> `R` (Red)
            *   Colony 4: `X` (1 cell) -> `B` (Blue)
            *   Colony 5: `X X` and `X X X` (5 cells) -> `B` (Blue)
            *   Wait, let's re-examine Example 1 carefully:
                *   Colony A: (2,2), (2,3), (2,4) -> 3 cells -> `R`
                *   Colony B: (2,7), (3,7) -> 2 cells -> `G`
                *   Colony C: (5,3), (5,4), (6,3) -> 3 cells -> `R`
                *   Colony D: (7,7), (8,7), (8,8), (8,9) -> 4 cells -> `B`? No, (7,7) is one, (8,7-9) is three. Let's re-count.
                *   Example 1 Input:
                    (2,2)-(2,4) is 3 cells. Output: `R R R`
                    (2,7)-(3,7) is 2 cells. Output: `G` (at 2,7 and 3,7)
                    (5,3)-(5,4)-(6,3) is 3 cells. Output: `R R` and `R`
                    (7,7)-(8,7)-(8,8)-(8,9) is 4 cells. Output: `B` (at 7,7), `B B B` (at 8,7-9)
                    (8,1)-(8,2)-(9,1)-(9,2) is 4 cells. Output: `B B` and `B B`
                *   **Size Mapping (Example 1)**:
                    Size 2: `G` (Green)
                    Size 3: `R` (Red)
                    Size 4: `B` (Blue)

### Applying the Rule to Test Input

**1. Identify Colonies in Test Input:**
*   Colony 1: (0,9), (1,9), (2,9), (3,9) -> 4 cells.
*   Colony 2: (2,2), (2,3), (3,1), (3,2) -> 4 cells. (Wait, (2,2)-(2,3) and (3,1)-(3,2) are connected diagonally? Let's check connectivity. In Example 1, (5,3)-(5,4) and (6,3) are connected. (8,1)-(8,2) and (9,1)-(9,2) are connected. These are 4-connectivity. Let's re-check (2,2)-(2,3) and (3,1)-(3,2). (2,2) is adjacent to (3,2). Yes, they are connected.)
    *   (2,2), (2,3), (3,1), (3,2) -> 4 cells.
*   Colony 3: (2,5), (2,6), (3,6) -> 3 cells.
*   Colony 4: (6,0), (7,0), (8,0) -> 3 cells.
*   Colony 5: (7,3), (7,4) -> 2 cells.
*   Colony 6: (7,7), (8,7) -> 2 cells.

**2. Map Sizes to Colors (based on Example 1):**
*   Size 2: `G` (Green)
*   Size 3: `R` (Red)
*   Size 4: `B` (Blue)

**3. Transform:**
*   Colony 1 (Size 4): (0,9), (1,9), (2,9), (3,9) -> `B`
*   Colony 2 (Size 4): (2,2), (2,3), (3,1), (3,2) -> `B`
*   Colony 3 (Size 3): (2,5), (2,6), (3,6) -> `R`
*   Colony 4 (Size 3): (6,0), (7,0), (8,0) -> `R`
*   Colony 5 (Size 2): (7,3), (7,4) -> `G`
*   Colony 6 (Size 2): (7,7), (8,7) -> `G`

**Final Grid Construction:**
Row 0: `. . . . . . . . . B`
Row 1: `. . . . . . . . . B`
Row 2: `. . B B . R R . . B`
Row 3: `. B B . . . R . . B`
Row 4: `. . . . . . . . . .`
Row 5: `. . . . . . . . . .`
Row 6: `R . . . . . . . . .`
Row 7: `R . . G G . . G . .`
Row 8: `R . . . . . . G . .`
Row 9: `. . . . . . . . . .`

ANSWER:
. . . . . . . . . B
. . . . . . . . . B
. . B B . R R . . B
. B B . . . R . . B
. . . . . . . . . .
. . . . . . . . . .
R . . . . . . . . .
R . . G G . . G . .
R . . . . . . G . .
. . . . . . . . . .

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Dark open water (`.`)**: The background/empty space.
    *   **Grey vessels (`X`)**: The objects to be counted and recolored.
    *   **Clusters/Rafts**: Groups of connected `X` cells (connected horizontally or vertically).
    *   **Harbor signal color**: A specific color assigned to a cluster based on its size (number of `X` cells).

2.  **Identify the AFFORDANCES**:
    *   `X` cells can be grouped into clusters.
    *   Clusters can be counted to determine their size.
    *   Clusters can be repainted from `X` to a specific color based on that count.
    *   `.` cells remain unchanged.

3.  **Identify the TRANSFORMATION RULE**:
    *   Step 1: Identify all connected components (clusters) of `X` cells.
    *   Step 2: Count the number of cells in each cluster.
    *   Step 3: Map the cluster size to a color. Looking at Example 1:
        *   Size 2: `B` (Blue)
        *   Size 3: `R` (Red)
        *   Size 4: Not explicitly shown, but let's look at the clusters in Example 1:
            *   Cluster 1: (2,1), (2,2), (2,3) -> Size 3 -> `R`
            *   Cluster 2: (2,7), (3,7) -> Size 2 -> `G`? Wait, let's re-examine Example 1.
            *   Example 1 clusters:
                *   (2,1)-(2,3) is size 3 -> `R`
                *   (2,7)-(3,7) is size 2 -> `G`? No, (2,7) and (3,7) is size 2.
                *   (5,3)-(6,3) is size 2? No, (5,3), (5,4), (6,3) is size 3 -> `R`.
                *   (7,7) is size 1 -> `B`? No, (7,7) is size 1.
                *   (8,1)-(9,2) is size 4 -> `B`?
                *   (8,6)-(8,8) is size 3 -> `B`?
            *   Let's re-count Example 1 carefully:
                *   Cluster A: (2,1), (2,2), (2,3) -> Size 3 -> `R`
                *   Cluster B: (2,7), (3,7) -> Size 2 -> `G`
                *   Cluster C: (5,3), (5,4), (6,3) -> Size 3 -> `R`
                *   Cluster D: (7,7) -> Size 1 -> `B`? No, (7,7) is part of (8,6)-(8,8)? No, they aren't connected.
                *   Let's look at the output of Example 1:
                    *   (2,1)-(2,3) is `R` (Size 3)
                    *   (2,7)-(3,7) is `G` (Size 2)
                    *   (5,3)-(6,3) is `R` (Size 3)
                    *   (7,7) is `B` (Size 1)
                    *   (8,1)-(9,2) is `B` (Size 4)
                    *   (8,6)-(8,8) is `B` (Size 3)
            *   Wait, the mapping is not strictly size-to-color in a simple way. Let's re-evaluate the clusters in Example 1:
                *   Cluster 1: (2,1), (2,2), (2,3) -> Size 3 -> `R`
                *   Cluster 2: (2,7), (3,7) -> Size 2 -> `G`
                *   Cluster 3: (5,3), (5,4), (6,3) -> Size 3 -> `R`
                *   Cluster 4: (7,7) -> Size 1 -> `B`? No, (7,7) is `B`.
                *   Cluster 5: (8,1), (8,2), (9,1), (9,2) -> Size 4 -> `B`
                *   Cluster 6: (8,6), (8,7), (8,8) -> Size 3 -> `B`?
            *   Let's look at the output again.
                *   (2,1-3) is `R`
                *   (2,7), (3,7) is `G`
                *   (5,3-4), (6,3) is `R`
                *   (7,7) is `B`
                *   (8,1-2), (9,1-2) is `B`
                *   (8,6-8) is `B`
            *   This is confusing. Let's look at the sizes:
                *   Size 1: (7,7) -> `B`
                *   Size 2: (2,7), (3,7) -> `G`
                *   Size 3: (2,1-3) -> `R`, (5,3-4, 6,3) -> `R`, (8,6-8) -> `B`?
                *   Size 4: (8,1-2, 9,1-2) -> `B`
            *   Wait, let's re-read the input for Example 1:
                *   (2,1), (2,2), (2,3) -> Size 3
                *   (2,7), (3,7) -> Size 2
                *   (5,3), (5,4), (6,3) -> Size 3
                *   (7,7) -> Size 1
                *   (8,1), (8,2), (9,1), (9,2) -> Size 4
                *   (8,6), (8,7), (8,8) -> Size 3
            *   Output colors:
                *   Size 3 -> `R` (twice) and `B` (once)? No, (8,6-8) is `B`.
                *   Size 2 -> `G`
                *   Size 1 -> `B`
                *   Size 4 -> `B`
            *   Let's re-examine the output for (8,6-8). It is `B B B`.
            *   Let's re-examine the output for (8,1-2, 9,1-2). It is `B B` / `B B`.
            *   Let's re-examine the output for (7,7). It is `B`.
            *   It seems: Size 1 = `B`, Size 2 = `G`, Size 3 = `R`, Size 4 = `B`? No, that's not right.
            *   Let's look at the clusters again.
                *   Size 1: `B`
                *   Size 2: `G`
                *   Size 3: `R`
                *   Size 4: `B`
                *   Wait, (8,6-8) is size 3. In the output, it is `B B B`.
                *   Let's look at the input again. (8,6) is `X`, (8,7) is `X`, (8,8) is `X`. That is size 3.
                *   Is it possible the color is based on the *number of clusters* of that size? No.
                *   Let's look at the colors: B, G, R.
     

... [4316 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Black (.)**: The substrate/background of the PCB.
    *   **Grey (X)**: The "solder blobs" or "bridges." These are contiguous groups of X cells.
    *   **Diagnostic Colors**: Colors assigned based on the size (count of cells) of each contiguous blob.

2.  **Identify the AFFORDANCES**:
    *   **Grey (X) blobs**: Can be grouped into connected components (using 4-connectivity or 8-connectivity; looking at the examples, it appears to be 4-connectivity or standard blob detection).
    *   **Size (Count)**: Each blob has a size (number of X cells).
    *   **Color Mapping**: The size of the blob determines its new color.

3.  **Identify the TRANSFORMATION RULE**:
    *   Step 1: Identify all connected components of 'X' cells.
    *   Step 2: Count the number of cells in each component.
    *   Step 3: Map the count to a specific color.
        *   From Example 1:
            *   Blob 1 (top left): 3 cells $\rightarrow$ Red (R)
            *   Blob 2 (top right): 2 cells $\rightarrow$ Green (G)
            *   Blob 3 (middle): 3 cells $\rightarrow$ Red (R)
            *   Blob 4 (bottom left): 4 cells $\rightarrow$ Blue (B)
            *   Blob 5 (bottom right): 3 cells $\rightarrow$ Blue (B)? Wait, let's re-examine.
            *   *Re-evaluating Example 1 counts:*
                *   Blob (2,1)-(2,3): 3 cells $\rightarrow$ R
                *   Blob (2,7)-(3,7): 2 cells $\rightarrow$ G
                *   Blob (5,3)-(6,3): 2 cells $\rightarrow$ R? No, (5,3),(5,4),(6,3) is 3 cells $\rightarrow$ R.
                *   Blob (7,7)-(8,7): 2 cells $\rightarrow$ B? No, (7,7),(8,7),(8,8),(8,9) is 4 cells $\rightarrow$ B.
                *   Blob (8,1)-(8,2),(9,1),(9,2): 4 cells $\rightarrow$ B.
                *   Blob (8,6)-(8,8): 3 cells $\rightarrow$ B? No, let's look closer.
                *   *Correct Mapping from Example 1:*
                    *   Size 2 $\rightarrow$ Green (G)
                    *   Size 3 $\rightarrow$ Red (R)
                    *   Size 4 $\rightarrow$ Blue (B)
    *   Step 4: Replace the 'X' cells of each blob with its corresponding color.

### Applying the Rule to Test Input

**1. Identify Blobs and Counts:**
*   **Blob A**: (0,9), (1,9), (2,9), (3,9) $\rightarrow$ Size 4.
*   **Blob B**: (2,2), (2,3), (3,1), (3,2) $\rightarrow$ Size 4.
*   **Blob C**: (2,5), (2,6), (3,6) $\rightarrow$ Size 3.
*   **Blob D**: (6,0), (7,0), (8,0) $\rightarrow$ Size 3.
*   **Blob E**: (7,3), (7,4) $\rightarrow$ Size 2.
*   **Blob F**: (7,7), (8,7) $\rightarrow$ Size 2.

**2. Map Sizes to Colors (based on Example 1):**
*   Size 2 $\rightarrow$ Green (G)
*   Size 3 $\rightarrow$ Red (R)
*   Size 4 $\rightarrow$ Blue (B)

**3. Transform:**
*   Blob A (Size 4) $\rightarrow$ Blue (B)
*   Blob B (Size 4) $\rightarrow$ Blue (B)
*   Blob C (Size 3) $\rightarrow$ Red (R)
*   Blob D (Size 3) $\rightarrow$ Red (R)
*   Blob E (Size 2) $\rightarrow$ Green (G)
*   Blob F (Size 2) $\rightarrow$ Green (G)

**4. Construct Output Grid:**
Row 0: `. . . . . . . . . B`
Row 1: `. . . . . . . . . B`
Row 2: `. . B B . R R . . B`
Row 3: `. B B . . . R . . B`
Row 4: `. . . . . . . . . .`
Row 5: `. . . . . . . . . .`
Row 6: `R . . . . . . . . .`
Row 7: `R . . G G . . G . .`
Row 8: `R . . . . . . G . .`
Row 9: `. . . . . . . . . .`

Wait, let's double check Blob B: (2,2), (2,3), (3,1), (3,2).
(2,2) is X, (2,3) is X.
(3,1) is X, (3,2) is X.
Are they connected? (2,2) is adjacent to (3,2). Yes. So it is one blob of size 4.

Double check Blob C: (2,5), (2,6), (3,6).
(2,5) is X, (2,6) is X.
(3,6) is X.
(2,6) is adjacent to (3,6). Yes. Size 3.

Double check Blob D: (6,0), (7,0), (8,0).
(6,0) is X, (7,0) is X, (8,0) is X.
All connected vertically. Size 3.

Double check Blob E: (7,3), (7,4).
Size 2.

Double check Blob F: (7,7), (8,7).
Size 2.

Double check Blob A: (0,9), (1,9), (2,9), (3,9).
Size 4.

Final Color Mapping:
Size 2: G
Size 3: R
Size 4: B

ANSWER:
. . . . . . . . . B
. . . . . . . . . B
. . B B . R R . . B
. B B . . . R . . B
. . . . . . . . . .
. . . . . . . . . .
R . . . . . . . . .
R . . G G . . G . .
R . . . . . . G . .
. . . . . . . . . .

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **The Cloth (Background):** The black cells (`.`) representing the dark divination cloth.
    *   **The Bone Fragments (Clusters):** Groups of connected grey cells (`X`).
    *   **The Deities (Colors):** Specific colors assigned to clusters based on their size (the number of connected cells).

2.  **Identify the AFFORDANCES:**
    *   **Clusters** can be counted to determine their size.
    *   **Sizes** map to specific colors.
    *   **Cells** within a cluster can be changed from grey (`X`) to a specific color.

3.  **Identify the TRANSFORMATION RULE:**
    *   Identify all connected components (clusters) of `X` cells (using 4-connectivity or 8-connectivity; looking at Example 1, it appears to be 4-connectivity/orthogonal).
    *   Count the number of cells in each cluster.
    *   Map the count to a color. Let's deduce the mapping from Example 1:
        *   Cluster 1: (2,2), (2,3), (2,4) -> Size 3. Color: Red (R).
        *   Cluster 2: (2,7), (3,7) -> Size 2. Color: Green (G).
        *   Cluster 3: (5,3), (5,4), (6,3) -> Size 3. Color: Red (R).
        *   Cluster 4: (7,7), (8,6), (8,7), (8,8) -> Size 4. Color: Blue (B).
        *   Cluster 5: (8,1), (8,2), (9,1), (9,2) -> Size 4. Color: Blue (B).
        *   *Wait, let's re-examine Example 1 carefully:*
            *   Cluster A: (2,2), (2,3), (2,4) -> Size 3 -> R
            *   Cluster B: (2,7), (3,7) -> Size 2 -> G
            *   Cluster C: (5,3), (5,4), (6,3) -> Size 3 -> R
            *   Cluster D: (7,7), (8,6), (8,7), (8,8) -> Size 4 -> B
            *   Cluster E: (8,1), (8,2), (9,1), (9,2) -> Size 4 -> B
        *   Mapping: Size 2 = G, Size 3 = R, Size 4 = B.
    *   The rule is: For each cluster of `X`, count its size $N$. Replace all `X` in that cluster with the color corresponding to $N$.

### Applying the Rule to Test Input

**Test Input Clusters:**
1.  (0,9), (1,9), (2,9), (3,9) -> Size 4.
2.  (2,2), (2,3), (3,1), (3,2) -> Size 4.
3.  (2,5), (2,6), (3,6) -> Size 3.
4.  (6,0), (7,0), (8,0) -> Size 3.
5.  (7,3), (7,4) -> Size 2.
6.  (7,7), (8,7) -> Size 2.

**Mapping deduced from Example 1:**
*   Size 2 $\rightarrow$ G (Green)
*   Size 3 $\rightarrow$ R (Red)
*   Size 4 $\rightarrow$ B (Blue)

**Applying to Test Input:**
1.  Cluster 1 (Size 4): (0,9), (1,9), (2,9), (3,9) $\rightarrow$ B
2.  Cluster 2 (Size 4): (2,2), (2,3), (3,1), (3,2) $\rightarrow$ B
3.  Cluster 3 (Size 3): (2,5), (2,6), (3,6) $\rightarrow$ R
4.  Cluster 4 (Size 3): (6,0), (7,0), (8,0) $\rightarrow$ R
5.  Cluster 5 (Size 2): (7,3), (7,4) $\rightarrow$ G
6.  Cluster 6 (Size 2): (7,7), (8,7) $\rightarrow$ G

**Constructing the Grid:**
Row 0: `. . . . . . . . . B`
Row 1: `. . . . . . . . . B`
Row 2: `. . B B . R R . . B`
Row 3: `. B B . . . R . . B`
Row 4: `. . . . . . . . . .`
Row 5: `. . . . . . . . . .`
Row 6: `R . . . . . . . . .`
Row 7: `R . . G G . . G . .`
Row 8: `R . . . . . . G . .`
Row 9: `. . . . . . . . . .`

ANSWER:
. . . . . . . . . B
. . . . . . . . . B
. . B B . R R . . B
. B B . . . R . . B
. . . . . . . . . .
. . . . . . . . . .
R . . . . . . . . .
R . . G G . . G . .
R . . . . . . G . .
. . . . . . . . . .

### Conceptual Digestion

1.  **Identify the CAST**:
    *   `.` (Black): The empty stage floor.
    *   `X` (Grey): The dancers forming clusters.
    *   The clusters are contiguous groups of `X` cells (connected horizontally or vertically).

2.  **Identify the AFFORDANCES**:
    *   The clusters can be counted to determine their size (number of `X` cells).
    *   The size of the cluster determines the color of the cluster in the output.

3.  **Identify the TRANSFORMATION RULE**:
    *   Identify all connected components (clusters) of `X` cells.
    *   Count the number of cells in each cluster.
    *   Map the cluster size to a specific color:
        *   Looking at Example 1:
            *   Cluster 1: `X X X` (size 3) -> `R` (Red)
            *   Cluster 2: `X X` (size 2) -> `B` (Blue)
            *   Cluster 3: `X` (size 1) -> `G` (Green)
            *   Cluster 4: `X X X` (size 3) -> `R` (Red) - *Wait, let's re-examine Example 1 carefully.*
            
            *Re-evaluating Example 1:*
            - Cluster at (2,2)-(2,4): `X X X` (size 3) -> `R`
            - Cluster at (2,7)-(3,7): `X X` (size 2) -> `G`? No, let's look at the output.
            - Output (2,2)-(2,4) is `R R R`. Size 3 = Red.
            - Output (2,7)-(3,7) is `G G`. Size 2 = Green.
            - Output (5,3)-(6,3): `X X` and `X`? No, (5,3)-(5,4) and (6,3) is a cluster of 3. Output is `R R` and `R`. Size 3 = Red.
            - Output (7,7): `X` (size 1) -> `B`.
            - Output (8,1)-(9,2): `X X` and `X X` is a cluster of 4. Output is `B B` and `B B`. Size 4 = Blue.
            - Output (8,6)-(8,8): `X X X` (size 3) -> `B`? No, let's re-read the output carefully.
            
            *Correcting Example 1 Analysis:*
            - Cluster 1: (2,2), (2,3), (2,4) -> Size 3 -> `R`
            - Cluster 2: (2,7), (3,7) -> Size 2 -> `G`
            - Cluster 3: (5,3), (5,4), (6,3) -> Size 3 -> `R`
            - Cluster 4: (7,7) -> Size 1 -> `B`
            - Cluster 5: (8,1), (8,2), (9,1), (9,2) -> Size 4 -> `B`
            - Cluster 6: (8,6), (8,7), (8,8) -> Size 3 -> `B`? No, the output says `B B B`.
            
            *Wait, let's look at the colors again:*
            Size 1: `B` (Blue)
            Size 2: `G` (Green)
            Size 3: `R` (Red)
            Size 4: `B` (Blue)
            
            Let's re-verify:
            - (2,7)-(3,7) is size 2 -> `G` (Correct)
            - (7,7) is size 1 -> `B` (Correct)
            - (8,1)-(9,2) is size 4 -> `B` (Correct)
            - (8,6)-(8,8) is size 3 -> `B`? No, the output for (8,6)-(8,8) is `B B B`.
            
            Let's re-examine the clusters in Example 1:
            1. (2,2), (2,3), (2,4) -> Size 3 -> `R`
            2. (2,7), (3,7) -> Size 2 -> `G`
            3. (5,3), (5,4), (6,3) -> Size 3 -> `R`
            4. (7,7) -> Size 1 -> `B`
            5. (8,1), (8,2), (9,1), (9,2) -> Size 4 -> `B`
            6. (8,6), (8,7), (8,8) -> Size 3 -> `B`? 
            
            Wait, if Size 1=B, Size 2=G, Size 3=R, Size 4=B... then Size 3 should be R. But (8,6)-(8,8) is `B B B`.
            Let's look at the input again. (8,6), (8,7), (8,8) are `X X X`.
            Is it possible the color is based on the *number of clusters*? No.
            Is it possible the color is based on the *position*? No.
            
            Let's look at the clusters in the Test Input:
            1. (0,9), (1,9), (2,9), (3,9) -> Size 4
            2. (2,2), (2,3), (3,1), (3,2) -> Size 4
            3. (2,5), (2,6), (3,6) -> Size 3
            4. (6,0), (7,0), (8,0) -> Size 3
            5. (7,3), (7,4) -> Size 2
            6. (7,7), (8,7) -> Size 2
            
            Let's re-evaluate the color mapping from Example 1:
            Size 1: `B`
            Size 2: `G`
            Size 3: `R`
            Size 4: `B`
            Wait, if Size 3 is `R`, why is (8,6)-(8,8) `B B B`?
            Let's look at the input (8,6) again. It's `X X X`.
            Wait, (8,6) is `X`, (8,7) is `X`, (8,8) is `X`.
            Is it possible the color is `B` for size 3 and 4? No, that's unlikely.
            
            Let's look at the input (8,1) again. (8,1), (8,2), (9,1), (9,2) are `X`. That's size 4. Output is `B B` and `B B`.
            So Size 4 = `B`.
            Size 1 = `B`.
            Size 2 = `G`.
            Size 3 = `R`.
            
            Wait! I see it now. In Example 1, (8,6), (8,7), (8,8) is size 3. The output is `B B B`.
            Is it possible the color is `B` for *any* size that is a multiple of something? No.
            Let's look at the input (8,6) again. (8,6) is `X`, (8,7) is `X`, (8,8) is `X`.
            Wait, (7,7) is `X`. (8,7) is `X`. (8,6) is `X`. (8,8) is `X`.
            They are all connected!
            (7,7) is connected to (8,7).
            (8,7) is connected to (8,6) and (8,8).
            So the cluste

... [2318 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **`.` (Black)**: The "dark rock face" or background. It remains unchanged.
    *   **`X` (Grey)**: The "mineral veins." These are connected components of grey pixels.
    *   **Colors (B, R, G, etc.)**: The "geological survey codes" representing the size (count of pixels) of each connected vein.

2.  **Identify the AFFORDANCES**:
    *   The `X` pixels can be grouped into connected components (using 4-connectivity or 8-connectivity; looking at Example 1, it appears to be 4-connectivity or standard blob detection).
    *   Each component has a "size" (the total number of `X` pixels in that specific connected group).
    *   The size determines the color of the entire component.

3.  **Identify the TRANSFORMATION RULE**:
    *   Step 1: Identify all connected components of `X` pixels.
    *   Step 2: Count the number of pixels in each component.
    *   Step 3: Map the count to a specific color.
    *   *Mapping Analysis from Example 1*:
        *   Component 1 (top left): `X X X` (3 pixels) -> `R` (Red)
        *   Component 2 (top right): `X` and `X` below it (2 pixels) -> `G` (Green)
        *   Component 3 (middle): `X X` and `X` below it (3 pixels) -> `R` (Red)
        *   Component 4 (bottom right): `X` and `X X X` (4 pixels) -> `B` (Blue)
        *   Component 5 (bottom left): `X X` and `X X` (4 pixels) -> `B` (Blue)
    *   *Wait, let's re-examine Example 1 carefully*:
        *   Vein 1: (2,2), (2,3), (2,4) -> 3 pixels -> `R`
        *   Vein 2: (2,7), (3,7) -> 2 pixels -> `G`
        *   Vein 3: (5,3), (5,4), (6,3) -> 3 pixels -> `R`
        *   Vein 4: (7,7), (8,6), (8,7), (8,8) -> 4 pixels -> `B`
        *   Vein 5: (8,1), (8,2), (9,1), (9,2) -> 4 pixels -> `B`
    *   *Rule Summary*:
        *   Size 2 -> `G` (Green)
        *   Size 3 -> `R` (Red)
        *   Size 4 -> `B` (Blue)
        *   (The pattern suggests size-based color mapping).

### Applying the Rule to Test Input

**Test Input Components**:
1.  (0,9), (1,9), (2,9), (3,9): 4 pixels. Color: `B`
2.  (2,2), (2,3), (3,1), (3,2): 4 pixels. Color: `B`
3.  (2,5), (2,6), (3,6): 3 pixels. Color: `R`
4.  (6,0), (7,0), (8,0): 3 pixels. Color: `R`
5.  (7,3), (7,4): 2 pixels. Color: `G`
6.  (7,7), (8,7): 2 pixels. Color: `G`

**Mapping to Output**:
*   (0,9), (1,9), (2,9), (3,9) -> `B`
*   (2,2), (2,3), (3,1), (3,2) -> `B`
*   (2,5), (2,6), (3,6) -> `R`
*   (6,0), (7,0), (8,0) -> `R`
*   (7,3), (7,4) -> `G`
*   (7,7), (8,7) -> `G`

**Constructing the Grid**:
Row 0: `. . . . . . . . . B`
Row 1: `. . . . . . . . . B`
Row 2: `. . B B . R R . . B`
Row 3: `. B B . . . R . . B`
Row 4: `. . . . . . . . . .`
Row 5: `. . . . . . . . . .`
Row 6: `R . . . . . . . . .`
Row 7: `R . . G G . . G . .`
Row 8: `R . . . . . . G . .`
Row 9: `. . . . . . . . . .`

ANSWER:
. . . . . . . . . B
. . . . . . . . . B
. . B B . R R . . B
. B B . . . R . . B
. . . . . . . . . .
. . . . . . . . . .
R . . . . . . . . .
R . . G G . . G . .
R . . . . . . G . .
. . . . . . . . . .

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Background (`.`):** The "cutting board," which remains unchanged.
    *   **Clusters (`X`):** The "raw mushrooms," which are connected groups of grey pixels.
    *   **Weight (Cluster Size):** The number of pixels in a connected component of `X`s.
    *   **Sauce (Color):** A specific color assigned to a cluster based on its size.

2.  **Identify the AFFORDANCES**:
    *   The background (`.`) is passive.
    *   The clusters (`X`) can be "weighed" (counted) and then "coated" (recolored).

3.  **Identify the TRANSFORMATION RULE**:
    *   First, identify all connected components of `X` (using 4-connectivity or 8-connectivity; looking at the examples, 4-connectivity is standard for "clusters").
    *   Count the number of pixels in each cluster.
    *   Assign a color based on the count. Let's deduce the mapping from the examples:
        *   **Example 1:**
            *   Cluster 1 (top left): `X X X` (3) -> `R` (Red)
            *   Cluster 2 (top right): `X`, `X` (2) -> `G` (Green)
            *   Cluster 3 (middle): `X X`, `X` (3) -> `R` (Red)
            *   Cluster 4 (bottom right): `X`, `X X X` (4) -> `B` (Blue)
            *   Cluster 5 (bottom left): `X X`, `X X` (4) -> `B` (Blue)
            *   *Wait, let's re-examine Example 1 carefully:*
                *   Cluster 1: (2,2), (2,3), (2,4) -> Size 3 -> Red
                *   Cluster 2: (2,7), (3,7) -> Size 2 -> Green
                *   Cluster 3: (5,3), (5,4), (6,3) -> Size 3 -> Red
                *   Cluster 4: (7,7), (8,7), (8,8), (8,9) -> Size 4 -> Blue
                *   Cluster 5: (8,1), (8,2), (9,1), (9,2) -> Size 4 -> Blue
            *   *Mapping:* Size 2 = Green, Size 3 = Red, Size 4 = Blue.
        *   **Example 2:**
            *   Cluster 1: (1,7), (1,8), (2,7), (2,8) -> Size 4 -> Blue
            *   Cluster 2: (2,1), (2,2), (3,2), (3,3) -> Size 4 -> Blue
            *   Cluster 3: (5,9), (6,9), (7,9) -> Size 3 -> Red
            *   Cluster 4: (6,5), (6,6) -> Size 2 -> Green
            *   Cluster 5: (7,1), (8,1) -> Size 2 -> Green
            *   Cluster 6: (8,4), (9,3), (9,4) -> Size 3 -> Red
            *   *Mapping:* Size 2 = Green, Size 3 = Red, Size 4 = Blue.
    *   **Rule Summary:**
        *   Size 2: Green (G)
        *   Size 3: Red (R)
        *   Size 4: Blue (B)

### Applying the Rule to Test Input

**Test Input Clusters:**
1.  (0,9), (1,9), (2,9), (3,9): Size 4 -> **Blue (B)**
2.  (2,2), (2,3), (3,1), (3,2): Size 4 -> **Blue (B)**
3.  (2,5), (2,6), (3,6): Size 3 -> **Red (R)**
4.  (6,0), (7,0), (8,0): Size 3 -> **Red (R)**
5.  (7,3), (7,4): Size 2 -> **Green (G)**
6.  (7,7), (8,7): Size 2 -> **Green (G)**

**Mapping to Grid:**
*   (0,9), (1,9), (2,9), (3,9) -> B
*   (2,2), (2,3), (3,1), (3,2) -> B
*   (2,5), (2,6), (3,6) -> R
*   (6,0), (7,0), (8,0) -> R
*   (7,3), (7,4) -> G
*   (7,7), (8,7) -> G

**Constructing Output:**
Row 0: `. . . . . . . . . B`
Row 1: `. . . . . . . . . B`
Row 2: `. . B B . R R . . B`
Row 3: `. B B . . . R . . B`
Row 4: `. . . . . . . . . .`
Row 5: `. . . . . . . . . .`
Row 6: `R . . . . . . . . .`
Row 7: `R . . G G . . G . .`
Row 8: `R . . . . . . G . .`
Row 9: `. . . . . . . . . .`

ANSWER:
. . . . . . . . . B
. . . . . . . . . B
. . B B . R R . . B
. B B . . . R . . B
. . . . . . . . . .
. . . . . . . . . .
R . . . . . . . . .
R . . G G . . G . .
R . . . . . . G . .
. . . . . . . . . .

To solve this puzzle, we first perform a Conceptual Digestion:

### 1. Identify the CAST
*   **Grey clusters (X):** These represent "notes" or "phrases." They are contiguous groups of grey pixels.
*   **Phrase Length:** The number of pixels in a single contiguous cluster of grey pixels.
*   **Instrument Sections (Colors):** Different lengths are mapped to specific colors.
*   **Black background (.):** The "silent score."

### 2. Identify the AFFORDANCES
*   A cluster can be **counted** to determine its length.
*   A cluster can be **recolored** based on its length.

### 3. Identify the TRANSFORMATION RULE
By analyzing the training examples, we map the cluster lengths to colors:
*   **Length 1:** Red (R) (Example 2: single X at (5,9) -> R; Example 3: single X at (1,2) -> G? Wait, let's re-examine).

**Re-evaluating Length to Color mapping:**
*   **Example 1:**
    *   Cluster 1: (2,1)-(2,3) [len 3] -> R
    *   Cluster 2: (2,7)-(3,7) [len 2] -> G
    *   Cluster 3: (5,3)-(6,3) [len 2] -> R? No, (5,3)-(6,3) is len 2. Let's look closer.
    *   Cluster 1: (2,1), (2,2), (2,3) -> len 3 -> R
    *   Cluster 2: (2,7), (3,7) -> len 2 -> G
    *   Cluster 3: (5,3), (5,4), (6,3) -> len 3 -> R
    *   Cluster 4: (7,7) -> len 1 -> B? No.
    *   Cluster 5: (8,1), (8,2), (9,1), (9,2) -> len 4 -> B
    *   Cluster 6: (8,6), (8,7), (8,8) -> len 3 -> B? No.

**Let's try counting the total pixels in a connected component:**
*   **Example 1:**
    *   (2,1)-(2,3) [3] -> R
    *   (2,7)-(3,7) [2] -> G
    *   (5,3)-(6,3) [2] -> R? No, (5,3), (5,4), (6,3) is 3.
    *   (7,7) [1] -> B? No.
    *   (8,1)-(9,2) [4] -> B
    *   (8,6)-(8,8) [3] -> B? No.

**Let's look at the lengths again very carefully:**
*   **Example 1:**
    *   Len 2: (2,7)-(3,7) -> G; (7,7) is len 1. (8,6)-(8,8) is len 3.
    *   Wait, the clusters are:
        1. (2,1),(2,2),(2,3) -> len 3 -> R
        2. (2,7),(3,7) -> len 2 -> G
        3. (5,3),(5,4),(6,3) -> len 3 -> R
        4. (7,7) -> len 1 -> B? No, (8,6),(8,7),(8,8) is len 3.
        5. (8,1),(8,2),(9,1),(9,2) -> len 4 -> B
        6. (8,6),(8,7),(8,8) -> len 3 -> B? No.

**Let's re-map based on length $L$:**
*   Example 1:
    *   $L=2 \rightarrow$ G
    *   $L=3 \rightarrow$ R
    *   $L=4 \rightarrow$ B
    *   $L=1 \rightarrow$ B? No, (8,6)-(8,8) is len 3.
    *   Wait, (8,6)-(8,8) is len 3. In Example 1, (8,6)-(8,8) is B.
    *   Let's look at Example 3:
        *   (1,2),(2,2) -> len 2 -> G
        *   (1,6),(1,7),(2,7) -> len 3 -> R
        *   (5,4),(5,5) -> len 2 -> G
        *   (6,0),(7,0),(7,1),(8,0) -> len 4 -> B
    *   **Mapping Found:**
        *   $L=2 \rightarrow$ G
        *   $L=3 \rightarrow$ R
        *   $L=4 \rightarrow$ B
        *   $L=1 \rightarrow$ R (Example 2: (5,9) is len 1 -> R)
        *   $L=5 \rightarrow$ ? (Example 2: (8,4),(9,3),(9,4) is len 3 -> R)

**Let's re-verify Example 2:**
*   (1,7),(1,8) -> len 2 -> B? No, (1,7),(1,8) is B.
*   (2,1),(2,2),(3,2),(3,3) -> len 4 -> B
*   (2,7),(2,8) -> len 2 -> B
*   (5,9) -> len 1 -> R
*   (6,5),(6,6) -> len 2 -> G
*   (6,9),(7,9),(8,9) -> len 3 -> R
*   (7,1),(8,1) -> len 2 -> G
*   (8,4),(9,3),(9,4) -> len 3 -> R

**Correct Mapping:**
*   $L=1 \rightarrow$ R
*   $L=2 \rightarrow$ G
*   $L=3 \rightarrow$ R
*   $L=4 \rightarrow$ B
*   $L=2$ (in Ex 2, (1,7)-(1,8)) $\rightarrow$ B? No, that's $L=2$.
*   Wait, let's look at the clusters in Example 2 again.
    *   (1,7),(1,8) is len 2. Output is B.
    *   (2,1),(2,2),(3,2),(3,3) is len 4. Output is B.
    *   (2,7),(2,8) is len 2. Output is B.
    *   (5,9) is len 1. Output is R.
    *   (6,5),(6,6) is len 2. Output is G.
    *   (6,9),(7,9),(8,9) is len 3. Output is R.
    *   (7,1),(8,1) is len 2. Output is G.
    *   (8,4),(9,3),(9,4) is len 3. Output is R.

**Wait, the rule is about the length of the *horizontal* or *vertical* component? No, it's the total count.**
Let's look at Example 2 again.
(1,7),(1,8) is len 2. Output is B.
(2,1),(2,2),(3,2),(3,3) is len 4. Output is B.
(2,7),(2,8) is len 2. Output is B.
(6,5),(6,6) is len 2. Output is G.
(7,1),(8,1) is len 2. Output is G.
(5,9) is len 1. Output is R.
(6,9),(7,9),(8,9) is len 3. Output is R.
(8,4),(9,3),(9,4) is len 3. Output is R.

**The rule is: If the cluster is a single line (horizontal or vertical), its length determines color. If it's a block, it's different? No.**
Let's look at the clusters in the Test Input:
1. (0,9),(1,9),(2,9),(3,9) -> len 4
2. (2,2),(2,3),(3,1),(3,2) -> len 4
3. (2,5),(2,6),(3,6) -> len 3
4. (6,0),(7,0),(8,0) -> len 3
5. (7,3),(7,4) -> len 2
6. (7,7),(8,7) -> len 2

**Applying the patterns from Example 2:**
*   Len 4 $\rightarrow$ B
*   Len 3 $\rightarrow$ R
*   Len 2 $\rightarrow$ G

**Test Input Clusters:**
*   (0,9),(1,9),(2,9),(3,9): Len 4 $\rightarrow$ B
*   (2,2),(2,3),(3,1),(3,

... [3882 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Black (.)**: The background/empty field.
    *   **Grey (X)**: The "unassigned players" forming clusters (connected components).
    *   **Colors (B, R, G, etc.)**: The "jersey colors" assigned to clusters based on their size.

2.  **Identify the AFFORDANCES**:
    *   **Grey (X)** cells can be grouped with adjacent (up, down, left, right) grey cells to form a cluster.
    *   **Clusters** have a property: **Size** (the number of cells in the cluster).
    *   **Size** determines the **Color** of the cluster.

3.  **Identify the TRANSFORMATION RULE**:
    *   Find all connected components of 'X' (using 4-connectivity).
    *   Count the number of cells in each component.
    *   Assign a color based on the size:
        *   Looking at Example 1:
            *   Cluster 1: (2,1), (2,2), (2,3) -> Size 3 -> Color **R** (Red)
            *   Cluster 2: (2,7), (3,7) -> Size 2 -> Color **G** (Green)
            *   Cluster 3: (5,3), (5,4), (6,3) -> Size 3 -> Color **R** (Red)
            *   Cluster 4: (7,7) -> Size 1 -> Color **B** (Blue)? Wait, let's re-examine.
            *   Example 1 clusters:
                *   (2,1)-(2,3) is size 3. Output is R.
                *   (2,7)-(3,7) is size 2. Output is G.
                *   (5,3)-(5,4)-(6,3) is size 3. Output is R.
                *   (7,7) is size 1. Output is B.
                *   (8,1)-(8,2)-(9,1)-(9,2) is size 4. Output is B? No, let's look closer.
                *   (8,1),(8,2),(9,1),(9,2) is size 4. Output is B.
                *   (8,6),(8,7),(8,8) is size 3. Output is B? No, let's re-read the grid.
    *   **Re-evaluating Example 1 mapping**:
        *   (2,1),(2,2),(2,3) -> Size 3 -> R
        *   (2,7),(3,7) -> Size 2 -> G
        *   (5,3),(5,4),(6,3) -> Size 3 -> R
        *   (7,7) -> Size 1 -> B
        *   (8,1),(8,2),(9,1),(9,2) -> Size 4 -> B
        *   (8,6),(8,7),(8,8) -> Size 3 -> B? No, (8,6),(8,7),(8,8) is size 3.
        *   Wait, let's look at the output grid for Example 1 again.
        *   (8,1),(8,2),(9,1),(9,2) is B. Size 4 = B.
        *   (8,6),(8,7),(8,8) is B. Size 3 = B.
        *   (7,7) is B. Size 1 = B.
        *   This implies the color is based on size, but the mapping is:
            *   Size 1: B
            *   Size 2: G
            *   Size 3: R
            *   Size 4: B
            *   Wait, this is confusing. Let's look at the clusters again.
            *   (2,1),(2,2),(2,3) -> Size 3 -> R
            *   (2,7),(3,7) -> Size 2 -> G
            *   (5,3),(5,4),(6,3) -> Size 3 -> R
            *   (7,7) -> Size 1 -> B
            *   (8,1),(8,2),(9,1),(9,2) -> Size 4 -> B
            *   (8,6),(8,7),(8,8) -> Size 3 -> B? No, the output says (8,6),(8,7),(8,8) is B.
            *   Let's re-count the input for Example 1:
                (2,1),(2,2),(2,3) = 3
                (2,7),(3,7) = 2
                (5,3),(5,4),(6,3) = 3
                (7,7) = 1
                (8,1),(8,2),(9,1),(9,2) = 4
                (8,6),(8,7),(8,8) = 3
            *   Output colors:
                Size 3 -> R
                Size 2 -> G
                Size 3 -> R
                Size 1 -> B
                Size 4 -> B
                Size 3 -> B? No, the output for (8,6),(8,7),(8,8) is B.
                Wait, (8,6),(8,7),(8,8) is B. (8,1),(8,2),(9,1),(9,2) is B. (7,7) is B.
                This means Size 1, 3, and 4 are all B? No, that can't be right.
                Let's look at the output grid very carefully.
                Row 8: `. B B . . . B B B .`
                Row 9: `. B B . . . . . . .`
                Row 8, Col 1,2 are B. Row 9, Col 1,2 are B. (Size 4)
                Row 8, Col 6,7,8 are B. (Size 3)
                Row 7, Col 7 is B. (Size 1)
                Wait, the output for (2,1)-(2,3) is R. (Size 3).
                The output for (5,3)-(5,4)-(6,3) is R. (Size 3).
                The output for (8,6)-(8,8) is B. (Size 3).
                This is a contradiction. Let me re-examine the input/output.
                Ah! (8,6),(8,7),(8,8) is size 3. (2,1),(2,2),(2,3) is size 3.
                Is it possible the color is based on the *position* or *parity*? No, the metaphor says "jersey color designated for that squad size".
                Let's re-count the clusters in Example 1:
                1. (2,1),(2,2),(2,3) -> Size 3
                2. (2,7),(3,7) -> Size 2
                3. (5,3),(5,4),(6,3) -> Size 3
                4. (7,7) -> Size 1
                5. (8,1),(8,2),(9,1),(9,2) -> Size 4
                6. (8,6),(8,7),(8,8) -> Size 3
                Output:
                Size 3 -> R
                Size 2 -> G
                Size 3 -> R
                Size 1 -> B
                Size 4 -> B
                Size 3 -> B? Wait, the output row 8 is `. B B . . . B B B .`.
                The

... [5481 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Black cells (`.`)**: The background/clear sky. They remain unchanged.
    *   **Grey cells (`X`)**: The storm cells. They form connected components (weather systems).
    *   **Colors (B, R, G, Y, M, O, A, W)**: The alert levels assigned to storm systems based on their size (cell count).

2.  **Identify the AFFORDANCES**:
    *   Connected components of `X` can be counted.
    *   The count of each component determines its new color.
    *   The background `.` cannot be changed.

3.  **Identify the TRANSFORMATION RULE**:
    *   Group all adjacent (up, down, left, right, and diagonal - based on the training example 1, let's check connectivity) `X` cells into connected components.
    *   *Re-evaluating connectivity from Example 1*:
        *   Component 1: (2,1), (2,2), (2,3) -> Size 3. Output color: R (Red).
        *   Component 2: (2,7), (3,7) -> Size 2. Output color: G (Green).
        *   Component 3: (5,3), (5,4), (6,3) -> Size 3. Output color: R (Red).
        *   Component 4: (7,7), (8,6), (8,7), (8,8) -> Size 4. Output color: B (Blue).
        *   Component 5: (8,1), (8,2), (9,1), (9,2) -> Size 4. Output color: B (Blue).
    *   *Wait, let's look at the sizes and colors in Example 1*:
        *   Size 2 -> G (Green)
        *   Size 3 -> R (Red)
        *   Size 4 -> B (Blue)
    *   *Wait, the colors in the key are B, R, G, Y, M, O, A, W. Let's look at the pattern of size vs color*:
        *   Size 2: G
        *   Size 3: R
        *   Size 4: B
    *   Looking at the color key: B (Blue), R (Red), G (Green), Y (Yellow), M (Magenta), O (Orange), A (Azure), W (Maroon).
    *   The training example shows:
        *   Size 2 -> G
        *   Size 3 -> R
        *   Size 4 -> B
    *   This suggests a mapping where smaller sizes get different colors. Let's check the test input components.

### Test Input Analysis

Input components (using 4-connectivity or 8-connectivity? In Example 1, (2,7) and (3,7) are connected. (7,7) and (8,6) are diagonal. If (7,7) and (8,6) are connected, it's 8-connectivity):
1.  (0,9), (1,9), (2,9), (3,9): Size 4.
2.  (2,2), (2,3), (3,1), (3,2): Size 4.
3.  (2,5), (2,6), (3,6): Size 3.
4.  (6,0), (7,0), (8,0): Size 3.
5.  (7,3), (7,4): Size 2.
6.  (7,7), (8,7): Size 2.

*Wait, let's re-examine Example 1 connectivity carefully*:
Example 1 components:
- (2,1)-(2,2)-(2,3) [Size 3] -> R
- (2,7)-(3,7) [Size 2] -> G
- (5,3)-(5,4)-(6,3) [Size 3] -> R
- (7,7)-(8,6)-(8,7)-(8,8) [Size 4] -> B (Note: (7,7) and (8,6) are diagonal)
- (8,1)-(8,2)-(9,1)-(9,2) [Size 4] -> B

Mapping:
Size 2: G
Size 3: R
Size 4: B

Applying to Test Input:
- Component 1: (0,9), (1,9), (2,9), (3,9) -> Size 4 -> B
- Component 2: (2,2), (2,3), (3,1), (3,2) -> Size 4 -> B
- Component 3: (2,5), (2,6), (3,6) -> Size 3 -> R
- Component 4: (6,0), (7,0), (8,0) -> Size 3 -> R
- Component 5: (7,3), (7,4) -> Size 2 -> G
- Component 6: (7,7), (8,7) -> Size 2 -> G

### Final Grid Construction

Row 0: . . . . . . . . . B
Row 1: . . . . . . . . . B
Row 2: . . B B . R R . . B
Row 3: . B B . . . R . . B
Row 4: . . . . . . . . . .
Row 5: . . . . . . . . . .
Row 6: R . . . . . . . . .
Row 7: R . . G G . . G . .
Row 8: R . . . . . . G . .
Row 9: . . . . . . . . . .

Wait, let's re-verify Component 2: (2,2), (2,3), (3,1), (3,2).
(2,2) is adjacent to (2,3) and (3,2).
(3,2) is adjacent to (3,1) and (2,2).
(3,1) is adjacent to (3,2).
All connected. Size 4. Color B.

Component 3: (2,5), (2,6), (3,6).
(2,5) adj (2,6). (2,6) adj (3,6). Size 3. Color R.

Component 4: (6,0), (7,0), (8,0).
(6,0) adj (7,0). (7,0) adj (8,0). Size 3. Color R.

Component 5: (7,3), (7,4).
(7,3) adj (7,4). Size 2. Color G.

Component 6: (7,7), (8,7).
(7,7) adj (8,7). Size 2. Color G.

Component 1: (0,9), (1,9), (2,9), (3,9).
(0,9) adj (1,9) adj (2,9) adj (3,9). Size 4. Color B.

Double check Component 2 again:
Input:
(2,2) X, (2,3) X
(3,1) X, (3,2) X
(2,2) is next to (2,3) and (3,2). (3,2) is next to (3,1). Yes, size 4.

Final check on colors:
Size 2: G
Size 3: R
Size 4: B

ANSWER:
. . . . . . . . . B
. . . . . . . . . B
. . B B . R R . . B
. B B . . . R . . B
. . . . . . . . . .
. . . . . . . . . .
R . . . . . . . . .
R . . G G . . G . .
R . . . . . . G . .
. . . . . . . . . .

To solve this puzzle, let's perform a Conceptual Digestion:

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Black (.)**: The empty city grid (background).
    *   **Grey (X)**: Building footprints (the objects to be transformed).
    *   **Colored pixels (B, R, G, etc.)**: The zoned buildings (the output).

2.  **Identify the AFFORDANCES**:
    *   The grey footprints can be "measured" by counting the number of pixels (lots) they occupy.
    *   The size of the footprint determines its final color.

3.  **Identify the TRANSFORMATION RULE**:
    *   Identify each contiguous block of grey (X) pixels.
    *   Count the number of pixels in each block.
    *   Assign a color based on the count. Looking at the training example:
        *   Block 1 (3x1 + 1x1 = 4 pixels): Red (R)
        *   Block 2 (2x1 = 2 pixels): Green (G)
        *   Block 3 (1x1 = 1 pixel): Green (G) - *Wait, let's re-examine Example 1.*
    
    **Re-evaluating Example 1 counts:**
    *   Top-left block: `X X X` (row 2) + `X` (row 5) + `X X` (row 5) + `X` (row 6) ... No, let's look at connectivity.
    *   Block A: (2,2), (2,3), (2,4) -> 3 pixels. (Wait, the input is 10x10).
    *   Let's re-count Example 1 carefully:
        *   Block 1: (2,1), (2,2), (2,3) [3 pixels] + (5,3), (5,4), (6,3) [3 pixels]... No, they are separate.
        *   Let's look at the output colors:
            *   (2,1)-(2,3) is RRR (3 pixels).
            *   (5,3)-(5,4) and (6,3) is R R and R (3 pixels).
            *   (2,7) and (3,7) is G and G (2 pixels).
            *   (7,7) is B (1 pixel).
            *   (8,1)-(8,2) and (9,1)-(9,2) is B B and B B (4 pixels).
            *   (8,6)-(8,8) is B B B (3 pixels).
    
    **Wait, let's re-map the counts to colors in Example 1:**
    *   3 pixels $\rightarrow$ R (Red)
    *   2 pixels $\rightarrow$ G (Green)
    *   4 pixels $\rightarrow$ B (Blue)
    *   1 pixel $\rightarrow$ B (Blue)? No, let's look at (7,7). It is B.
    *   Let's re-examine the input/output for Example 1:
        *   Block 1: (2,1), (2,2), (2,3) $\rightarrow$ 3 pixels $\rightarrow$ RRR
        *   Block 2: (5,3), (5,4), (6,3) $\rightarrow$ 3 pixels $\rightarrow$ RR, R
        *   Block 3: (2,7), (3,7) $\rightarrow$ 2 pixels $\rightarrow$ GG
        *   Block 4: (7,7) $\rightarrow$ 1 pixel $\rightarrow$ B
        *   Block 5: (8,1), (8,2), (9,1), (9,2) $\rightarrow$ 4 pixels $\rightarrow$ BB, BB
        *   Block 6: (8,6), (8,7), (8,8) $\rightarrow$ 3 pixels $\rightarrow$ BBB
    
    **Wait, the colors are:**
    *   1 pixel: B
    *   2 pixels: G
    *   3 pixels: R
    *   4 pixels: B
    
    Let's check the pattern of colors: 1=B, 2=G, 3=R, 4=B. This looks like a cycle or a specific mapping. Let's check the test input.

### Test Input Analysis
1.  Block 1: (0,9), (1,9), (2,9), (3,9) $\rightarrow$ 4 pixels.
2.  Block 2: (2,2), (2,3), (3,1), (3,2) $\rightarrow$ 4 pixels.
3.  Block 3: (2,5), (2,6), (3,6) $\rightarrow$ 3 pixels.
4.  Block 4: (6,0), (7,0), (8,0) $\rightarrow$ 3 pixels.
5.  Block 5: (7,3), (7,4) $\rightarrow$ 2 pixels.
6.  Block 6: (7,7), (8,7) $\rightarrow$ 2 pixels.

**Mapping counts to colors based on Example 1:**
*   4 pixels $\rightarrow$ B
*   3 pixels $\rightarrow$ R
*   2 pixels $\rightarrow$ G
*   1 pixel $\rightarrow$ B

**Applying to Test Input:**
1.  Block 1 (4 pixels): B
2.  Block 2 (4 pixels): B
3.  Block 3 (3 pixels): R
4.  Block 4 (3 pixels): R
5.  Block 5 (2 pixels): G
6.  Block 6 (2 pixels): G

### Final Grid Construction
*   (0,9), (1,9), (2,9), (3,9) $\rightarrow$ B B B B
*   (2,2), (2,3), (3,1), (3,2) $\rightarrow$ B B B B
*   (2,5), (2,6), (3,6) $\rightarrow$ R R R
*   (6,0), (7,0), (8,0) $\rightarrow$ R R R
*   (7,3), (7,4) $\rightarrow$ G G
*   (7,7), (8,7) $\rightarrow$ G G

**Row by row:**
Row 0: . . . . . . . . . B
Row 1: . . . . . . . . . B
Row 2: . . B B . R R . . B
Row 3: . B B . . . R . . B
Row 4: . . . . . . . . . .
Row 5: . . . . . . . . . .
Row 6: R . . . . . . . . .
Row 7: R . . G G . . G . .
Row 8: R . . . . . . G . .
Row 9: . . . . . . . . . .

ANSWER:
. . . . . . . . . B
. . . . . . . . . B
. . B B . R R . . B
. B B . . . R . . B
. . . . . . . . . .
. . . . . . . . . .
R . . . . . . . . .
R . . G G . . G . .
R . . . . . . G . .
. . . . . . . . . .

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Grey markers (X)**: These represent "units" or "enemy positions." They are the elements to be transformed.
    *   **Black background (.)**: This represents "unmarked terrain" or empty space.
    *   **Connected components**: Groups of adjacent (up, down, left, right) grey markers form a single "unit."
    *   **Colors (B, R, G, etc.)**: These are the "strength-coded colors" assigned to a unit based on its size (the number of connected grey cells).

2.  **Identify the AFFORDANCES**:
    *   **Grey markers (X)** can be grouped into clusters based on adjacency.
    *   **Clusters** can be counted to determine their "strength" (size).
    *   **Clusters** can be replaced entirely by a single color.

3.  **Identify the TRANSFORMATION RULE**:
    *   Find all connected components of 'X'.
    *   Count the number of 'X' cells in each component.
    *   Map the count to a specific color. Looking at Example 1:
        *   Cluster 1 (top left): 3 'X's $\rightarrow$ R (Red)
        *   Cluster 2 (top right): 2 'X's $\rightarrow$ G (Green)
        *   Cluster 3 (middle): 3 'X's $\rightarrow$ R (Red)
        *   Cluster 4 (bottom right): 3 'X's $\rightarrow$ B (Blue) - *Wait, let's re-examine Example 1 carefully.*
        
    *   **Re-evaluating Example 1 mapping**:
        *   Cluster 1: (2,2), (2,3), (2,4) $\rightarrow$ 3 cells $\rightarrow$ R
        *   Cluster 2: (2,7), (3,7) $\rightarrow$ 2 cells $\rightarrow$ G
        *   Cluster 3: (5,3), (5,4), (6,3) $\rightarrow$ 3 cells $\rightarrow$ R
        *   Cluster 4: (7,7), (8,6), (8,7), (8,8) $\rightarrow$ 4 cells $\rightarrow$ B (Wait, (7,7) is connected to (8,7)? Yes. Total: 4 cells)
        *   Cluster 5: (8,1), (8,2), (9,1), (9,2) $\rightarrow$ 4 cells $\rightarrow$ B
        
    *   **Mapping derived from Example 1**:
        *   Size 2 $\rightarrow$ G (Green)
        *   Size 3 $\rightarrow$ R (Red)
        *   Size 4 $\rightarrow$ B (Blue)

### Applying the Rule to Test Input

**Step 1: Identify Clusters and Sizes**
1.  (0,9), (1,9), (2,9), (3,9): Size 4.
2.  (2,2), (2,3), (3,1), (3,2): Size 4.
3.  (2,5), (2,6), (3,6): Size 3.
4.  (6,0), (7,0), (8,0): Size 3.
5.  (7,3), (7,4): Size 2.
6.  (7,7), (8,7): Size 2.

**Step 2: Map Sizes to Colors**
*   Size 2 $\rightarrow$ G
*   Size 3 $\rightarrow$ R
*   Size 4 $\rightarrow$ B

**Step 3: Replace X with Colors**
1.  (0,9), (1,9), (2,9), (3,9) $\rightarrow$ B
2.  (2,2), (2,3), (3,1), (3,2) $\rightarrow$ B
3.  (2,5), (2,6), (3,6) $\rightarrow$ R
4.  (6,0), (7,0), (8,0) $\rightarrow$ R
5.  (7,3), (7,4) $\rightarrow$ G
6.  (7,7), (8,7) $\rightarrow$ G

**Step 4: Construct the Grid**
Row 0: . . . . . . . . . B
Row 1: . . . . . . . . . B
Row 2: . . B B . R R . . B
Row 3: . B B . . . R . . B
Row 4: . . . . . . . . . .
Row 5: . . . . . . . . . .
Row 6: R . . . . . . . . .
Row 7: R . . G G . . G . .
Row 8: R . . . . . . G . .
Row 9: . . . . . . . . . .

ANSWER:
. . . . . . . . . B
. . . . . . . . . B
. . B B . R R . . B
. B B . . . R . . B
. . . . . . . . . .
. . . . . . . . . .
R . . . . . . . . .
R . . G G . . G . .
R . . . . . . G . .
. . . . . . . . . .

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Actors (X - Grey):** These are the elements that form groups.
    *   **Spotlight Pools (Connected Components):** Groups of adjacent (up, down, left, right) grey pixels.
    *   **Stage ( . - Black):** The background/empty space.
    *   **Gels (Colors):** The colors assigned to a group based on its size (count of pixels).

2.  **Identify the AFFORDANCES:**
    *   **Actors** can be counted to determine the size of their group.
    *   **Groups** can be recolored based on their size.
    *   **Background** remains unchanged (black).

3.  **Identify the TRANSFORMATION RULE:**
    *   Step 1: Identify all connected components of 'X' (grey) pixels.
    *   Step 2: Count the number of pixels in each component.
    *   Step 3: Map the count to a specific color. Looking at Example 1:
        *   Group 1 (top left): 3 pixels -> Red (R)
        *   Group 2 (top right): 2 pixels -> Green (G)
        *   Group 3 (middle): 3 pixels -> Red (R)
        *   Group 4 (bottom left): 4 pixels -> Blue (B)
        *   Group 5 (bottom right): 3 pixels -> Blue (B)? Wait, let's re-examine Example 1 carefully.
        *   *Re-evaluating Example 1 counts:*
            *   (2,1)-(2,3): 3 pixels -> R
            *   (2,7)-(3,7): 2 pixels -> G
            *   (5,3)-(6,3): 3 pixels -> R (Wait, (5,3), (5,4), (6,3) is 3 pixels)
            *   (7,7): 1 pixel -> B? No, (7,7) is part of (8,6)-(8,8)? Let's look at the grid coordinates.
            *   Example 1 Input:
                (2,1)(2,2)(2,3) = 3 pixels -> R
                (2,7)(3,7) = 2 pixels -> G
                (5,3)(5,4)(6,3) = 3 pixels -> R
                (7,7)(8,6)(8,7)(8,8) = 4 pixels -> B
                (8,1)(8,2)(9,1)(9,2) = 4 pixels -> B
            *   *Mapping:* Size 2 = G, Size 3 = R, Size 4 = B.
    *   Step 4: Replace all 'X' in a group with the color corresponding to its size.

### Applying the Rule to Test Input

**1. Identify Groups and Counts:**
*   Group A: (0,9), (1,9), (2,9), (3,9) -> Size 4
*   Group B: (2,2), (2,3), (3,1), (3,2) -> Size 4
*   Group C: (2,5), (2,6), (3,6) -> Size 3
*   Group D: (6,0), (7,0), (8,0) -> Size 3
*   Group E: (7,3), (7,4) -> Size 2
*   Group F: (7,7), (8,7) -> Size 2

**2. Determine Colors (based on Example 1 mapping):**
*   Size 2 = G (Green)
*   Size 3 = R (Red)
*   Size 4 = B (Blue)

**3. Map Groups to Colors:**
*   Group A (Size 4) -> B
*   Group B (Size 4) -> B
*   Group C (Size 3) -> R
*   Group D (Size 3) -> R
*   Group E (Size 2) -> G
*   Group F (Size 2) -> G

**4. Construct Output Grid:**
Row 0: . . . . . . . . . B
Row 1: . . . . . . . . . B
Row 2: . . B B . R R . . B
Row 3: . B B . . . R . . B
Row 4: . . . . . . . . . .
Row 5: . . . . . . . . . .
Row 6: R . . . . . . . . .
Row 7: R . . G G . . G . .
Row 8: R . . . . . . G . .
Row 9: . . . . . . . . . .

ANSWER:
. . . . . . . . . B
. . . . . . . . . B
. . B B . R R . . B
. B B . . . R . . B
. . . . . . . . . .
. . . . . . . . . .
R . . . . . . . . .
R . . G G . . G . .
R . . . . . . G . .
. . . . . . . . . .

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Black (.)**: The "soil" or background. It remains unchanged.
    *   **Grey (X)**: The "seedlings" or clusters. These are the objects to be counted.
    *   **Colors (B, R, G, etc.)**: The "ribbons" or the result of the transformation. Each color represents a specific count of grey pixels in a connected cluster.

2.  **Identify the AFFORDANCES**:
    *   **Grey (X) clusters**: Can be grouped into connected components (using 4-connectivity or 8-connectivity; looking at the examples, it appears to be 4-connectivity or standard adjacency).
    *   **Count**: Each cluster has a size (number of X pixels).
    *   **Color Mapping**: The size of the cluster determines the color that replaces the X pixels in that cluster.

3.  **Identify the TRANSFORMATION RULE**:
    *   Step 1: Identify all connected clusters of 'X'.
    *   Step 2: Count the number of 'X' pixels in each cluster.
    *   Step 3: Map the count to a specific color.
        *   Looking at Example 1:
            *   Cluster 1: (2,2), (2,3), (2,4) -> Size 3. Output color: R (Red).
            *   Cluster 2: (2,7), (3,7) -> Size 2. Output color: G (Green).
            *   Cluster 3: (5,3), (5,4), (6,3) -> Size 3. Output color: R (Red).
            *   Cluster 4: (7,7), (8,6), (8,7), (8,8) -> Size 4. Output color: B (Blue). Wait, let's re-examine.
            *   Let's re-count Example 1 carefully:
                *   Cluster A: (2,2), (2,3), (2,4) -> Size 3. Color R.
                *   Cluster B: (2,7), (3,7) -> Size 2. Color G.
                *   Cluster C: (5,3), (5,4), (6,3) -> Size 3. Color R.
                *   Cluster D: (7,7), (8,6), (8,7), (8,8) -> Size 4. Color B.
                *   Cluster E: (8,1), (8,2), (9,1), (9,2) -> Size 4. Color B.
        *   Mapping derived: Size 2 = G, Size 3 = R, Size 4 = B.
    *   Step 4: Replace the 'X' pixels in each cluster with the corresponding color.

### Applying the Rule to Test Input

**Input Clusters Analysis:**
1.  Cluster 1: (0,9), (1,9), (2,9), (3,9) -> Size 4. Color: **B**
2.  Cluster 2: (2,2), (2,3), (3,1), (3,2) -> Size 4. Color: **B**
3.  Cluster 3: (2,5), (2,6), (3,6) -> Size 3. Color: **R**
4.  Cluster 4: (6,0), (7,0), (8,0) -> Size 3. Color: **R**
5.  Cluster 5: (7,3), (7,4) -> Size 2. Color: **G**
6.  Cluster 6: (7,7), (8,7) -> Size 2. Color: **G**

**Mapping Sizes to Colors (from Example 1):**
*   Size 2 -> G
*   Size 3 -> R
*   Size 4 -> B

**Applying to Test Input:**
*   (0,9), (1,9), (2,9), (3,9) -> B B B B (vertical line at col 9)
*   (2,2), (2,3), (3,1), (3,2) -> B B B B (square-ish shape)
*   (2,5), (2,6), (3,6) -> R R R (L-shape)
*   (6,0), (7,0), (8,0) -> R R R (vertical line at col 0)
*   (7,3), (7,4) -> G G (horizontal line)
*   (7,7), (8,7) -> G G (vertical line)

**Constructing the Grid:**
Row 0: . . . . . . . . . B
Row 1: . . . . . . . . . B
Row 2: . . B B . R R . . B
Row 3: . B B . . . R . . B
Row 4: . . . . . . . . . .
Row 5: . . . . . . . . . .
Row 6: R . . . . . . . . .
Row 7: R . . G G . . G . .
Row 8: R . . . . . . G . .
Row 9: . . . . . . . . . .

Wait, let's re-verify Cluster 2: (2,2), (2,3), (3,1), (3,2).
(2,2) is X, (2,3) is X, (3,1) is X, (3,2) is X.
These are connected: (2,2) is adjacent to (2,3) and (3,2). (3,2) is adjacent to (3,1).
Total count = 4. Color = B.

Re-verify Cluster 3: (2,5), (2,6), (3,6).
(2,5) is X, (2,6) is X, (3,6) is X.
(2,5) is adjacent to (2,6). (2,6) is adjacent to (3,6).
Total count = 3. Color = R.

Re-verify Cluster 4: (6,0), (7,0), (8,0).
(6,0) is X, (7,0) is X, (8,0) is X.
Total count = 3. Color = R.

Re-verify Cluster 5: (7,3), (7,4).
(7,3) is X, (7,4) is X.
Total count = 2. Color = G.

Re-verify Cluster 6: (7,7), (8,7).
(7,7) is X, (8,7) is X.
Total count = 2. Color = G.

Re-verify Cluster 1: (0,9), (1,9), (2,9), (3,9).
(0,9) is X, (1,9) is X, (2,9) is X, (3,9) is X.
Total count = 4. Color = B.

Final Grid Construction:
R0: . . . . . . . . . B
R1: . . . . . . . . . B
R2: . . B B . R R . . B
R3: . B B . . . R . . B
R4: . . . . . . . . . .
R5: . . . . . . . . . .
R6: R . . . . . . . . .
R7: R . . G G . . G . .
R8: R . . . . . . G . .
R9: . . . . . . . . . .

ANSWER:
. . . . . . . . . B
. . . . . . . . . B
. . B B . R R . . B
. B B . . . R . . B
. . . . . . . . . .
. . . . . . . . . .
R . . . . . . . . .
R . . G G . . G . .
R . . . . . . G . .
. . . . . . . . . .

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Launchpad:** The row containing the colored engines (the row with `B`, `R`, and `A`).
    *   **Engines:** The specific colored cells within the launchpad row (`A` for Azure, `R` for Red).
    *   **Exhaust/Plume:** A vertical column of blue (`B`) cells rising from an engine.
    *   **Flare:** The top-most cell of a plume, which takes the color of the engine (`A` or `R`).
    *   **Night Sky:** The empty space (`.`) above the launchpad.

2.  **Identify the AFFORDANCES:**
    *   **Engines (`A`, `R`)** trigger the creation of a plume.
    *   **Plumes (`B`)** extend upwards from the engine's position.
    *   **Flares (`A`, `R`)** cap the plume at a specific height.
    *   **Height Rule:** Red (`R`) engines produce taller plumes than Azure (`A`) engines.

3.  **Identify the TRANSFORMATION RULE:**
    *   Locate the row containing the engines (the row with `B`, `R`, and `A`).
    *   For every `A` (Azure) engine: Create a plume of `B` (blue) cells going up, capped by an `A` (azure) cell. Looking at the examples:
        *   Ex 1: `A` at row 3, col 1. Plume height: 3 cells total (A, B, B).
        *   Ex 3: `A` at row 7, col 1. Plume height: 3 cells total (A, B, B).
        *   Ex 3: `A` at row 7, col 5. Plume height: 3 cells total (A, B, B).
        *   Ex 2: `R` at row 5, col 3. Plume height: 4 cells total (R, B, B, B).
        *   Ex 3: `R` at row 7, col 7. Plume height: 4 cells total (R, B, B, B).
    *   **Refined Rule:**
        *   **Azure (A) engine:** Plume consists of the engine cell itself, plus 2 `B` cells above it. The top cell is `A`. Wait, looking closer at Example 1: Input `A` is at (3,1). Output has `A` at (0,1), `B` at (1,1), `B` at (2,1). Total height = 3.
        *   **Red (R) engine:** Plume consists of the engine cell itself, plus 3 `B` cells above it. The top cell is `R`. Looking at Example 2: Input `R` is at (5,3). Output has `R` at (1,3), `B` at (2,3), `B` at (3,3), `B` at (4,3). Total height = 4.
        *   **Wait, let's re-verify height:**
            *   Ex 1: `A` is at row 3. Output `A` is at row 0. Height = 4 (rows 0, 1, 2, 3).
            *   Ex 2: `R` is at row 5. Output `R` is at row 1. Height = 5 (rows 1, 2, 3, 4, 5).
            *   Ex 3: `A` is at row 7. Output `A` is at row 4. Height = 4 (rows 4, 5, 6, 7).
            *   Ex 3: `R` is at row 7. Output `R` is at row 3. Height = 5 (rows 3, 4, 5, 6, 7).
    *   **Final Rule:**
        *   **Azure (A) engine:** The plume is 4 cells high (including the engine). The top cell is `A`, the next 2 are `B`, and the bottom is the engine itself (which is `A` in the input, but the description says engines stay bolted, and the example shows the engine row remains `B A B...`). Actually, the engine row is the base.
        *   Let's re-examine Example 1: Input row 3 is `B A B B B B`. Output row 3 is `B A B B B B`. The plume is at col 1. Row 0: `A`, Row 1: `B`, Row 2: `B`.
        *   Let's re-examine Example 2: Input row 5 is `B B B R B B B`. Output row 5 is `B B B R B B B`. The plume is at col 3. Row 1: `R`, Row 2: `B`, Row 3: `B`, Row 4: `B`.
        *   **Conclusion:**
            *   **Azure (A) engine:** Top cell is `A`. Then 2 `B` cells. Then the engine row. (Total 4 cells including engine row).
            *   **Red (R) engine:** Top cell is `R`. Then 3 `B` cells. Then the engine row. (Total 5 cells including engine row).

### Applying to Test Input

Test Input (Row 6 is the launchpad):
`B R B A B B B A B R B` (Indices 0 to 10)
Engines:
- Col 1: `R` (Red)
- Col 3: `A` (Azure)
- Col 7: `A` (Azure)
- Col 9: `R` (Red)

**Plume Calculations:**
1.  **Col 1 (R):** Height 5. Rows 2, 3, 4, 5, 6.
    - Row 2: `R`
    - Row 3: `B`
    - Row 4: `B`
    - Row 5: `B`
    - Row 6: `R` (Engine)
2.  **Col 3 (A):** Height 4. Rows 3, 4, 5, 6.
    - Row 3: `A`
    - Row 4: `B`
    - Row 5: `B`
    - Row 6: `A` (Engine)
3.  **Col 7 (A):** Height 4. Rows 3, 4, 5, 6.
    - Row 3: `A`
    - Row 4: `B`
    - Row 5: `B`
    - Row 6: `A` (Engine)
4.  **Col 9 (R):** Height 5. Rows 2, 3, 4, 5, 6.
    - Row 2: `R`
    - Row 3: `B`
    - Row 4: `B`
    - Row 5: `B`
    - Row 6: `R` (Engine)

**Constructing the Grid (11 columns, 9 rows):**
Row 0: `. . . . . . . . . . .`
Row 1: `. . . . . . . . . . .`
Row 2: `. R . . . . . . . R .`
Row 3: `. B . A . . . A . B .`
Row 4: `. B . B . . . B . B .`
Row 5: `. B . B . . . B . B .`
Row 6: `B R B A B B B A B R B`
Row 7: `. . . . . . . . . . .`
Row 8: `. . . . . . . . . . .`

ANSWER:
. . . . . . . . . . .
. . . . . . . . . . .
. R . . . . . . . R .
. B . A . . . A . B .
. B . B . . . B . B .
. B . B . . . B . B .
B R B A B B B A B R B
. . . . . . . . . . .
. . . . . . . . . . .

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **The Strand:** A horizontal line of colored cells (enzymes) located at a specific row in the grid.
    *   **Enzymes:** Individual colored cells within the strand (B, R, A, etc.).
    *   **Protein Chains:** Vertical lines of color extending upwards from the enzyme sites.
    *   **Cap Molecule:** The color of the enzyme itself, which appears at the end of the chain.

2.  **Identify the AFFORDANCES:**
    *   **The Strand** acts as a fixed anchor.
    *   **Enzymes** trigger the growth of a vertical chain.
    *   **The Chain** grows perpendicular to the strand (upwards in the examples).
    *   **The Length** of the chain is determined by the enzyme type.

3.  **Identify the TRANSFORMATION RULE:**
    *   Locate the horizontal strand of non-black cells.
    *   For each cell in the strand:
        *   If the cell is **A (Azure)**: Grow a chain of length 3 upwards (the cell itself + 2 more cells).
        *   If the cell is **R (Red)**: Grow a chain of length 4 upwards (the cell itself + 3 more cells).
        *   If the cell is **B (Blue)**: Grow a chain of length 2 upwards (the cell itself + 1 more cell).
    *   Wait, let's re-examine the examples to find the exact lengths:
        *   **Example 1:** Strand is at row 4. Enzyme `A` at (4,1) produces a chain of length 3 (cells at (4,1), (3,1), (2,1), (1,1)? No, let's count).
            *   Input row 4: `B A B B B B`
            *   Output: `A` is at (4,1). Chain goes up to (1,1). Total height including strand is 4.
            *   `B` at (4,0) stays `B`. `B` at (4,2) stays `B`.
            *   Let's look at the `A` in Example 1: The `A` is at index 1. The output has `A` at (0,1), `B` at (1,1), `B` at (2,1), and `A` at (3,1).
            *   Wait, the description says: "The chain length and cap molecule depend on the enzyme's identity... capped by a copy of the enzyme at the distal end."
            *   **Example 1 Re-analysis:**
                Strand: `B A B B B B` (Row 3, index 0-5)
                Enzyme `A` is at (3,1).
                Output at col 1: (0,1)=A, (1,1)=B, (2,1)=B, (3,1)=A.
                This means the chain grows *upwards* from the enzyme. The enzyme at the bottom is the original. The "cap" is at the top.
                The chain for `A` is: `A` (top), `B` (middle), `B` (middle), `A` (bottom). This doesn't match "capped by a copy of the enzyme".
                Let's look at **Example 2**:
                Strand: `B B B R B B B` (Row 5, index 0-6)
                Enzyme `R` is at (5,3).
                Output at col 3: (1,3)=R, (2,3)=B, (3,3)=B, (4,3)=B, (5,3)=R.
                The chain for `R` is: `R` (top), `B`, `B`, `B`, `R` (bottom).
                The chain for `B` in Example 1: `B` at (3,0) has no chain? No, the `B` at (3,0) is just the strand.
                Let's look at the `A` in Example 1 again:
                `A` is at (3,1). Output col 1: `A` (0,1), `B` (1,1), `B` (2,1), `A` (3,1).
                The chain length for `A` is 4 (including the strand cell). The color of the "filler" is `B` (the color of the surrounding strand).
                The chain length for `R` is 5 (including the strand cell). The filler is `B`.

    *   **Refined Rule:**
        1. Identify the strand row.
        2. For each enzyme (non-B cell) in the strand:
           - If enzyme is `A`: Chain length 4 (including strand cell).
           - If enzyme is `R`: Chain length 5 (including strand cell).
           - If enzyme is `B`: Chain length 2 (including strand cell).
           - Wait, the `B`s in the strand don't seem to grow anything in Example 1 or 2. Only `A` and `R` grow.
           - The filler color is `B` (Blue).
           - The top cell (cap) is the enzyme color.
           - The bottom cell is the enzyme color.
           - The cells in between are `B`.

    *   **Let's check Example 1 again:**
        `A` at (3,1). Length 4.
        (3,1)=A, (2,1)=B, (1,1)=B, (0,1)=A. Matches!
    *   **Let's check Example 2 again:**
        `R` at (5,3). Length 5.
        (5,3)=R, (4,3)=B, (3,3)=B, (2,3)=B, (1,3)=R. Matches!

    *   **Test Input Analysis:**
        Strand is at Row 6: `B R B A B B B A B R B` (Indices 0 to 10)
        Enzymes:
        - `R` at (6,1): Length 5. Cells: (6,1)=R, (5,1)=B, (4,1)=B, (3,1)=B, (2,1)=R.
        - `A` at (6,3): Length 4. Cells: (6,3)=A, (5,3)=B, (4,3)=B, (3,3)=A.
        - `A` at (6,7): Length 4. Cells: (6,7)=A, (5,7)=B, (4,7)=B, (3,7)=A.
        - `R` at (6,9): Length 5. Cells: (6,9)=R, (5,9)=B, (4,9)=B, (3,9)=B, (2,9)=R.

### Applying the Rule to Test Input

Row 0: `. . . . . . . . . . .`
Row 1: `. . . . . . . . . . .`
Row 2: `. R . . . . . . . R .`
Row 3: `. B . A . . . A . B .`
Row 4: `. B . B . . . B . B .`
Row 5: `. B . B . . . B . B .`
Row 6: `B R B A B B B A B R B`
Row 7: `. . . . . . . . . . .`
Row 8: `. . . . . . . . . . .`

Wait, let's re-verify the `A` length.
Example 1: `A` is at index 1. Output has `A` at row 0, 1, 2, 3. That is 

... [2997 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **The Galactic Arm**: A horizontal line of colored pixels (the "starlight") located in the middle of the grid.
    *   **Stellar Anomalies**: Specific colored pixels within the arm that are not the background color (though in these examples, the arm itself is composed of colors, the "anomalies" are the specific colored cells that trigger jets).
    *   **Plasma Jets**: Vertical lines of color extending perpendicularly from the arm.
    *   **Jet Terminus**: The tip of the jet, which maintains the color of the anomaly.

2.  **Identify the AFFORDANCES**:
    *   **The Arm** provides the base structure and the starting points for the jets.
    *   **Anomalies** (colored cells) trigger the creation of a vertical line (jet) extending upwards.
    *   **The Jet** extends upwards from the anomaly's position. The length of the jet is determined by the color of the anomaly.

3.  **Identify the TRANSFORMATION RULE**:
    *   Locate the horizontal line (the arm).
    *   For every colored cell in that arm, a vertical jet is drawn upwards.
    *   **Jet Length/Color Logic**: Looking at the examples:
        *   Ex 1: `A` (Azure) at index 1. Jet goes up 3 cells. The cells are `A`, `B`, `B`. Wait, let's re-examine.
        *   Ex 1: Input `B A B B B B` at row 3. Output: `A` at (0,1), `B` at (1,1), `B` at (2,1). The jet is made of the color of the *neighboring* cells or the anomaly itself?
        *   Let's look closer at Ex 1: Anomaly is `A` at (3,1). Jet is at col 1. Cells are (0,1)=A, (1,1)=B, (2,1)=B.
        *   Ex 2: Anomaly is `R` at (5,3). Jet is at col 3. Cells are (1,3)=R, (2,3)=B, (3,3)=B, (4,3)=B.
        *   Ex 3: 
            *   `A` at (7,1) -> Jet at col 1: (4,1)=A, (5,1)=B, (6,1)=B.
            *   `A` at (7,5) -> Jet at col 5: (4,5)=A, (5,5)=B, (6,5)=B.
            *   `R` at (7,7) -> Jet at col 7: (3,7)=R, (4,7)=B, (5,7)=B, (6,7)=B.
    *   **Refined Rule**: For each colored cell in the arm at `(row, col)`:
        1.  The jet extends upwards from `row-1` to `row-N`.
        2.  The top cell of the jet is the color of the anomaly itself.
        3.  The subsequent cells in the jet are the color `B` (blue).
        4.  The length of the jet is determined by the color of the anomaly:
            *   `A` (Azure): Length 3 (including the anomaly's color at the top).
            *   `R` (Red): Length 4 (including the anomaly's color at the top).
            *   `B` (Blue): Length 1 (no jet, or just the cell itself).
    *   Let's check Ex 1 again: `A` is at (3,1). Jet is (0,1)=A, (1,1)=B, (2,1)=B. Total 3 cells above the arm.
    *   Ex 2: `R` is at (5,3). Jet is (1,3)=R, (2,3)=B, (3,3)=B, (4,3)=B. Total 4 cells above the arm.
    *   Ex 3: `A` at (7,1) -> 3 cells above: (4,1), (5,1), (6,1). `R` at (7,7) -> 4 cells above: (3,7), (4,7), (5,7), (6,7).
    *   **Conclusion**: 
        *   If color is `A`, jet length is 3 cells above the arm.
        *   If color is `R`, jet length is 4 cells above the arm.
        *   If color is `B`, jet length is 0 cells above the arm.
        *   The top cell of the jet is the anomaly color, the rest are `B`.

### Applying to Test Input

Input Arm (Row 6): `B R B A B B B A B R B`
Indices: 0 1 2 3 4 5 6 7 8 9 10

1.  (6,0) = B: No jet.
2.  (6,1) = R: Jet length 4 above. Cells: (2,1)=R, (3,1)=B, (4,1)=B, (5,1)=B.
3.  (6,2) = B: No jet.
4.  (6,3) = A: Jet length 3 above. Cells: (3,3)=A, (4,3)=B, (5,3)=B.
5.  (6,4) = B: No jet.
6.  (6,5) = B: No jet.
7.  (6,6) = B: No jet.
8.  (6,7) = A: Jet length 3 above. Cells: (3,7)=A, (4,7)=B, (5,7)=B.
9.  (6,8) = B: No jet.
10. (6,9) = R: Jet length 4 above. Cells: (2,9)=R, (3,9)=B, (4,9)=B, (5,9)=B.
11. (6,10) = B: No jet.

**Constructing the Grid**:
Row 0: all `.`
Row 1: all `.`
Row 2: `. R . . . . . . . R .`
Row 3: `. B . A . . . A . B .`
Row 4: `. B . B . . . B . B .`
Row 5: `. B . B . . . B . B .`
Row 6: `B R B A B B B A B R B`
Row 7: all `.`
Row 8: all `.`

Wait, let's re-verify the jet length for `A` and `R`.
Ex 1: `A` at row 3. Jet is (0,1), (1,1), (2,1). That is 3 cells.
Ex 2: `R` at row 5. Jet is (1,3), (2,3), (3,3), (4,3). That is 4 cells.
Ex 3: `A` at row 7. Jet is (4,1), (5,1), (6,1). That is 3 cells.
Ex 3: `R` at row 7. Jet is (3,7), (4,7), (5,7), (6,7). That is 4 cells.

Test Input:
Row 6 is the arm.
`R` at (6,1) -> (2,1), (3,1), (4,1), (5,1)
`A` at (6,3) -> (3,3), (4,3), (5,3)
`A` at (6,7) -> (3,7), (4,7), (5,7)
`R` at (6,9) -> (2,9), (3,9), (4,9), (5,9)

Final check of the grid dimensions: 9 rows, 11 columns.

ANSWER:
. . . . . . . . . . .
. . . . . . . . . . .
. R . . . . . . . R .
. B . A . . . A . B .
. B . B . . . B . B .
. B . B . . . B . B .
B R B A B B B A B R B
. . . . . . . . . . .
. . . . . . . . . . .

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **The Chain:** A horizontal line of colored cells (the "sea floor").
    *   **Anchors/Buoys:** Specific colored cells within the chain.
    *   **Tethers:** Vertical lines of color extending upwards from the anchors.
    *   **Buoy Class/Color:** The color of the anchor cell determines the color of the tether and the "floating buoy" at the top.
    *   **The Surface:** The top of the vertical tether.

2.  **Identify the AFFORDANCES:**
    *   **Anchors** can "release" a tether.
    *   **Tethers** extend vertically upwards from the anchor's position.
    *   **The Tether Color** is determined by the anchor's color.
    *   **The Tether Length** is determined by the "buoy class." Looking at the examples:
        *   Example 1: Anchor `A` (Azure) at index 1. Tether length is 3 (cells at row 0, 1, 2). The top cell is `A`, the rest are `B` (the background chain color). Wait, let's re-examine.
        *   *Correction on Rule:* In Example 1, the anchor is `A` at (3,1). The tether goes up to row 0. The cells are (0,1)=A, (1,1)=B, (2,1)=B.
        *   Example 2: Anchor `R` (Red) at (5,3). Tether goes up to row 1. Cells are (1,3)=R, (2,3)=B, (3,3)=B, (4,3)=B.
        *   Example 3: Anchor `A` at (7,1) and (7,5). Anchor `R` at (7,7).
            *   `A` at (7,1) -> (4,1)=A, (5,1)=B, (6,1)=B.
            *   `A` at (7,5) -> (4,5)=A, (5,5)=B, (6,5)=B.
            *   `R` at (7,7) -> (3,7)=R, (4,7)=B, (5,7)=B, (6,7)=B.
        *   *Refined Rule:* The tether length is determined by the color of the anchor.
            *   `A` (Azure) creates a tether of length 4 (including the anchor).
            *   `R` (Red) creates a tether of length 5 (including the anchor).
            *   `B` (Blue) is the background/chain color.
            *   The top cell of the tether is the anchor's color. The cells between the top and the anchor are the chain's base color (`B`).

3.  **Identify the TRANSFORMATION RULE:**
    *   Locate the "chain" (the row containing non-black cells that form a sequence).
    *   Identify "special" anchors (colors other than the base color `B`).
    *   For each special anchor:
        1.  Determine its color (e.g., `R` or `A`).
        2.  Determine the tether length: `A` = 4, `R` = 5.
        3.  Draw a vertical line upwards from the anchor.
        4.  The top cell of the line is the anchor's color.
        5.  The cells below the top cell (down to the anchor) are the base color `B`.

### Applying the Rule to Test Input

**Input Analysis:**
*   Chain row: Row 6 (index 6).
*   Chain content: `B R B A B B B A B R B`
*   Base color: `B` (Blue).
*   Anchors:
    1.  `R` at (6, 1)
    2.  `A` at (6, 3)
    3.  `A` at (6, 7)
    4.  `R` at (6, 9)

**Tether Calculations:**
1.  **Anchor `R` at (6, 1):**
    *   Length 5.
    *   Cells: (2,1)=R, (3,1)=B, (4,1)=B, (5,1)=B, (6,1)=R.
2.  **Anchor `A` at (6, 3):**
    *   Length 4.
    *   Cells: (3,3)=A, (4,3)=B, (5,3)=B, (6,3)=A.
3.  **Anchor `A` at (6, 7):**
    *   Length 4.
    *   Cells: (3,7)=A, (4,7)=B, (5,7)=B, (6,7)=A.
4.  **Anchor `R` at (6, 9):**
    *   Length 5.
    *   Cells: (2,9)=R, (3,9)=B, (4,9)=B, (5,9)=B, (6,9)=R.

**Constructing the Grid:**
Row 0: all `.`
Row 1: all `.`
Row 2: `. R . . . . . . . R .`
Row 3: `. B . A . . . A . B .`
Row 4: `. B . B . . . B . B .`
Row 5: `. B . B . . . B . B .`
Row 6: `B R B A B B B A B R B`
Row 7: all `.`
Row 8: all `.`

Wait, let's re-verify the tether length from Example 3.
Example 3: `A` at (7,1) -> (4,1)=A, (5,1)=B, (6,1)=B, (7,1)=A. Total 4 cells. Correct.
Example 3: `R` at (7,7) -> (3,7)=R, (4,7)=B, (5,7)=B, (6,7)=B, (7,7)=R. Total 5 cells. Correct.

**Final Grid Construction (11 columns, 9 rows):**
Row 0: . . . . . . . . . . .
Row 1: . . . . . . . . . . .
Row 2: . R . . . . . . . R .
Row 3: . B . A . . . A . B .
Row 4: . B . B . . . B . B .
Row 5: . B . B . . . B . B .
Row 6: B R B A B B B A B R B
Row 7: . . . . . . . . . . .
Row 8: . . . . . . . . . . .

ANSWER:
. . . . . . . . . . .
. . . . . . . . . . .
. R . . . . . . . R .
. B . A . . . A . B .
. B . B . . . B . B .
. B . B . . . B . B .
B R B A B B B A B R B
. . . . . . . . . . .
. . . . . . . . . . .

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **The Root:** A horizontal line of colored cells (the "runes") located in a specific row.
    *   **The Runes:** Individual colored cells within the root (B, R, A, etc.).
    *   **The Pillars:** Vertical columns of color extending upwards from the runes.
    *   **The Background:** The black/empty cells (`.`).

2.  **Identify the AFFORDANCES:**
    *   **Runes** can act as anchors for vertical pillars.
    *   **Pillars** can extend vertically (upwards) from the rune's position.
    *   **The Root** remains static in its original position.

3.  **Identify the TRANSFORMATION RULE:**
    *   Locate the horizontal row containing the "root" (the row with non-black cells).
    *   For every colored cell (rune) in that row, create a vertical pillar extending **upwards**.
    *   The height of the pillar is determined by the color of the rune. Looking at the examples:
        *   Example 1: `B A B B B B` (Row 4). The `A` (Azure) is at index 1. The output shows a pillar of height 3 (A, B, B) above it. Wait, let's look closer.
        *   Example 1 analysis: Input row 4 is `B A B B B B`. Output shows at index 1: `A` (row 1), `B` (row 2), `B` (row 3), `A` (row 4). The pillar height is 4.
        *   Example 2 analysis: Input row 6 is `B B B R B B B`. Output shows at index 3: `R` (row 2), `B` (row 3), `B` (row 4), `B` (row 5), `R` (row 6). The pillar height is 5.
    *   **Refined Rule:** For each rune at `(row, col)` with color `C`, a pillar of color `C` is drawn from `(row, col)` upwards to a certain height. The cells in the pillar *above* the rune are filled with the color of the *neighboring* rune in the root? No.
    *   Let's re-examine Example 1: Rune at (4,1) is `A`. Pillar above it: (1,1)=`A`, (2,1)=`B`, (3,1)=`B`. The colors in the pillar are the colors of the runes in the root starting from the rune itself and moving left? No.
    *   Let's look at the colors again:
        Example 1: Root is `B A B B B B`. Pillar at index 1 (A): `A` (row 1), `B` (row 2), `B` (row 3), `A` (row 4).
        Example 2: Root is `B B B R B B B`. Pillar at index 3 (R): `R` (row 2), `B` (row 3), `B` (row 4), `B` (row 5), `R` (row 6).
    *   **Correct Rule:** For each rune at `(row, col)` with color `C`, a pillar extends upwards. The height of the pillar is the distance from the rune to the top of the grid (or a fixed height). The colors in the pillar are the colors of the runes in the root, but which ones?
    *   In Example 1: Index 1 is `A`. The pillar is `A, B, B, A`. These are the colors at index 1, 0, 0, 1? No.
    *   Let's look at the verticality: The pillar at index 1 in Example 1 is `A` (row 1), `B` (row 2), `B` (row 3), `A` (row 4). The root is `B A B B B B`.
    *   Wait, the pillar is simply the color of the rune itself, but the height is determined by the rune's color? No, the description says "The pillar's height depends on the rune's element".
    *   Let's re-read: "The pillar's height depends on the rune's element, with the rune's glow crowning the top."
    *   Example 1: `A` is at index 1. Pillar height is 4. Top is `A`. The colors below it are `B`.
    *   Example 2: `R` is at index 3. Pillar height is 5. Top is `R`. The colors below it are `B`.
    *   **Actual Rule:** For each rune at `(row, col)` with color `C`, a pillar extends upwards. The height of the pillar is determined by the color.
        *   If color is `A` (Azure), height is 4.
        *   If color is `R` (Red), height is 5.
        *   The top cell of the pillar is the rune's color `C`.
        *   The cells below the top cell in the pillar are the color `B` (Blue).
    *   Let's check Example 1: `A` is at index 1. Height 4. Top (row 1) is `A`. Rows 2, 3, 4 are `B`? No, row 4 is `A`. So rows 1, 2, 3 are `A, B, B`.
    *   Let's check Example 2: `R` is at index 3. Height 5. Top (row 2) is `R`. Rows 3, 4, 5, 6 are `B, B, B, R`.
    *   **Final Rule:** For a rune of color `C` at `(row, col)`:
        1. Determine height `H` based on color `C`.
        2. The cell at `(row - H + 1, col)` is color `C`.
        3. The cells from `(row - H + 2, col)` to `(row, col)` are color `B`.
        4. Wait, in Example 1, the rune at index 0 is `B`. Does it have a pillar? The output shows no pillar at index 0. Only the `A` at index 1 has a pillar.
        5. In Example 2, the rune at index 3 is `R`. The output shows a pillar at index 3. The other `B`s do not have pillars.
    *   **Conclusion:** Only "special" runes (non-B) create pillars.
        *   `A` (Azure) height = 4. Pillar: `A` at top, then `B`s below.
        *   `R` (Red) height = 5. Pillar: `R` at top, then `B`s below.

### Applying to Test Input
Root (Row 6): `B R B A B B B A B R B`
Indices: 0=B, 1=R, 2=B, 3=A, 4=B, 5=B, 6=B, 7=A, 8=B, 9=R, 10=B

Runes that create pillars:
1.  Index 1: `R` (Red). Height 5.
    *   Top: (6 - 5 + 1) = Row 2.
    *   Pillar: (2,1)=R, (3,1)=B, (4,1)=B, (5,1)=B, (6,1)=R.
2.  Index 3: `A` (Azure). Height 4.
    

... [2747 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Chorus Line:** A horizontal row of colored pixels (the "dancers").
    *   **Costume Colors:** Specific colors within the line (B, R, A, etc.) that determine the "height" of the leap.
    *   **Background:** The black (`.`) grid.

2.  **Identify the AFFORDANCES:**
    *   **Chorus Line:** Stays in its original position.
    *   **Costume Colors:** Act as triggers. When a color is not the "base" color (Blue `B` appears to be the base/uniform color), it initiates a vertical trail.
    *   **Leap Height:** The height of the trail is determined by the specific color. Looking at the examples:
        *   `A` (Azure) in Example 1: Height 3 (reaches row index 0 from row index 3).
        *   `R` (Red) in Example 2: Height 4 (reaches row index 1 from row index 5).
        *   `A` (Azure) in Example 3: Height 3 (reaches row index 4 from row index 7).
        *   `R` (Red) in Example 3: Height 3 (reaches row index 3 from row index 7).
        *   `R` (Red) in Example 4: Height 4 (reaches row index 5 from row index 9).
        *   `A` (Azure) in Example 4: Height 3 (reaches row index 6 from row index 9).
    *   *Correction on Height Logic:* Let's re-examine the distance.
        *   Ex 1: Row 3 to Row 0 (Height 3). Color A.
        *   Ex 2: Row 5 to Row 1 (Height 4). Color R.
        *   Ex 3: Row 7 to Row 4 (Height 3). Color A. Row 7 to Row 3 (Height 4). Color R.
        *   Ex 4: Row 9 to Row 5 (Height 4). Color R. Row 9 to Row 6 (Height 3). Color A.
    *   **Pattern:** `A` (Azure) creates a trail of height 3. `R` (Red) creates a trail of height 4. The trail consists of the color of the dancer at the top, and the base color (`B`) for the pixels connecting it to the line.

3.  **Identify the TRANSFORMATION RULE:**
    *   Identify the horizontal line of dancers.
    *   For every pixel in that line that is NOT `B` (Blue):
        *   If the color is `A` (Azure): Create a vertical trail upwards of length 3. The top pixel is `A`, and the pixels below it (connecting to the line) are `B`.
        *   If the color is `R` (Red): Create a vertical trail upwards of length 4. The top pixel is `R`, and the pixels below it (connecting to the line) are `B`.
    *   The line itself remains unchanged.

### Applying the Rule to Test Input

**Input Analysis:**
Line is at Row 6 (0-indexed).
Row 6: `B R B A B B B A B R B` (Length 11)
Indices of non-B colors:
- Index 1: `R` (Red)
- Index 3: `A` (Azure)
- Index 7: `A` (Azure)
- Index 9: `R` (Red)

**Transformation:**
1.  **Index 1 (R):** Height 4. Top at Row 2 (6-4=2).
    - Row 2, Col 1: `R`
    - Row 3, Col 1: `B`
    - Row 4, Col 1: `B`
    - Row 5, Col 1: `B`
2.  **Index 3 (A):** Height 3. Top at Row 3 (6-3=3).
    - Row 3, Col 3: `A`
    - Row 4, Col 3: `B`
    - Row 5, Col 3: `B`
3.  **Index 7 (A):** Height 3. Top at Row 3 (6-3=3).
    - Row 3, Col 7: `A`
    - Row 4, Col 7: `B`
    - Row 5, Col 7: `B`
4.  **Index 9 (R):** Height 4. Top at Row 2 (6-4=2).
    - Row 2, Col 9: `R`
    - Row 3, Col 9: `B`
    - Row 4, Col 9: `B`
    - Row 5, Col 9: `B`

**Constructing the Grid:**
Rows 0, 1: All `.`
Row 2: `. R . . . . . . . R .`
Row 3: `. . . A . . . A . B .`
Row 4: `. B . B . . . B . B .`
Row 5: `. B . B . . . B . B .`
Row 6: `B R B A B B B A B R B`
Rows 7, 8: All `.`

Wait, let's re-verify the trail logic. In Example 1, `A` at row 3 results in `A` at row 0, `B` at row 1, `B` at row 2. That is a trail of 3 pixels *above* the line.
Ex 1: Row 3 -> Row 0, 1, 2. (3 pixels).
Ex 2: Row 5 -> Row 1, 2, 3, 4. (4 pixels).
Ex 3: Row 7 -> Row 4, 5, 6 (A) and Row 3, 4, 5, 6 (R).
Ex 4: Row 9 -> Row 6, 7, 8 (A) and Row 5, 6, 7, 8 (R).

So for Test Input (Row 6):
- `R` at index 1: Rows 2, 3, 4, 5. (Row 2 is `R`, 3, 4, 5 are `B`).
- `A` at index 3: Rows 3, 4, 5. (Row 3 is `A`, 4, 5 are `B`).
- `A` at index 7: Rows 3, 4, 5. (Row 3 is `A`, 4, 5 are `B`).
- `R` at index 9: Rows 2, 3, 4, 5. (Row 2 is `R`, 3, 4, 5 are `B`).

Let's re-map:
Row 2: `. R . . . . . . . R .`
Row 3: `. B . A . . . A . B .`
Row 4: `. B . B . . . B . B .`
Row 5: `. B . B . . . B . B .`
Row 6: `B R B A B B B A B R B`

Wait, looking at Example 1 again:
Input Row 3: `B A B B B B`
Output Row 0: `. A . . . .`
Output Row 1: `. B . . . .`
Output Row 2: `. B . . . .`
Output Row 3: `B A B B B B`
The `A` is at the top, and the `B`s are the trail. The `A` at row 3 is the dancer. The trail is at rows 0, 1, 2.
So for Test Input (Row 6):
`R` at index 1: Row 2 is `R`, Row 3 is `B`, Row 4 is `B`, Row 5 is `B`.
`A` at index 3: Row 3 is `A`, Row 4 is `B`, Row 5 is `B`.
`A` at index 7: Row 3 is `A`, Row 4 is `B`, Row 5 is `B`.
`R` at index 9: Row 2 is `R`, Row 3 is `B`, Row 4 is `B`, Row 5 is `B`.

Final check of the grid dimensions: 9 rows, 11 columns.

ANSWER:
. . . . . . . . . . .
. . . . . . . . . . .
. R . . . . . . . R .
. B . A . . . A . B .
. B . B . . . B . B .
. B . B . . . B . B .
B R B A B B B A B R B
. . . . . . . . . . .
. . . . . . . . . . .

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **The Scrimmage Line:** A horizontal row of colored cells (the "players").
    *   **Uniformly Dressed Players:** The background color of the line (in these examples, Blue 'B').
    *   **Uniquely Jersied Players:** The distinct colors within the line (e.g., Red 'R', Azure 'A').
    *   **The Field:** The empty black '.' cells.

2.  **Identify the AFFORDANCES:**
    *   The "uniquely jersied players" can move perpendicular to the line (upwards in these examples).
    *   As they move, they leave a "trail" of the background color ('B') behind them.
    *   The distance of the sprint is determined by the specific color of the jersey.

3.  **Identify the TRANSFORMATION RULE:**
    *   Locate the horizontal line of players.
    *   Identify the "special" colors (anything not the background color 'B').
    *   For each special color, move it vertically upwards.
    *   The distance moved is determined by the color's identity. Looking at the examples:
        *   Example 1: 'A' (Azure) moves 3 steps up (reaches row 0). 'B' is the background.
        *   Example 2: 'R' (Red) moves 4 steps up (reaches row 1).
    *   Wait, let's re-examine the distance.
        *   Ex 1: Input row index 3. Output 'A' is at row 0. Distance = 3.
        *   Ex 2: Input row index 5. Output 'R' is at row 1. Distance = 4.
    *   Let's look at the colors:
        *   In Ex 1: A is at index 1. It moves up 3 cells.
        *   In Ex 2: R is at index 3. It moves up 4 cells.
    *   Actually, the rule is simpler: The special color moves to the top of the grid (or a specific distance), and the cells between the original position and the new position are filled with the background color 'B'.
    *   Let's check the distance again.
        *   Ex 1: Input row 3. 'A' is at (3,1). Output 'A' is at (0,1). Trail: (1,1) and (2,1) are 'B'.
        *   Ex 2: Input row 5. 'R' is at (5,3). Output 'R' is at (1,3). Trail: (2,3), (3,3), (4,3) are 'B'.
    *   The distance seems to be: **The special color moves to a position such that it is a certain distance from the top, or simply moves up a fixed amount based on its color.**
    *   Let's look at the colors in the Test Input: B, R, B, A, B, B, B, A, B, R, B.
    *   Special colors: R (Red), A (Azure).
    *   In Ex 1: A (Azure) moved 3 steps.
    *   In Ex 2: R (Red) moved 4 steps.
    *   Let's look at the color key/order: B=1, R=2, G=3, Y=4, X=5, M=6, O=7, A=8, W=9. This doesn't match.
    *   Let's look at the distance from the line to the top.
        *   Ex 1: Line is at row 3. A moves to row 0. Distance = 3.
        *   Ex 2: Line is at row 5. R moves to row 1. Distance = 4.
    *   Wait, the distance is the index of the color in the color key? No.
    *   Let's look at the colors again:
        *   Ex 1: A is Azure. Distance = 3.
        *   Ex 2: R is Red. Distance = 4.
    *   Let's check the Test Input:
        *   R is Red.
        *   A is Azure.
    *   If Red moves 4 and Azure moves 3:
        *   R is at (6, 1). New R at (6-4, 1) = (2, 1). Trail: (3,1), (4,1), (5,1) = B.
        *   A is at (6, 3). New A at (6-3, 3) = (3, 3). Trail: (4,3), (5,3) = B.
        *   A is at (6, 7). New A at (6-3, 7) = (3, 7). Trail: (4,7), (5,7) = B.
        *   R is at (6, 9). New R at (6-4, 9) = (2, 9). Trail: (3,9), (4,9), (5,9) = B.

    *   Let's re-verify the distance.
        *   Ex 1: A (Azure) -> distance 3.
        *   Ex 2: R (Red) -> distance 4.
        *   Test: R (Red) -> distance 4. A (Azure) -> distance 3.

### Applying the Rule to Test Input
Input Line (Row 6): `B R B A B B B A B R B`
Indices: 0 1 2 3 4 5 6 7 8 9 10

1.  **R at index 1:** Move up 4. New position: (6-4, 1) = (2, 1). Trail: (3,1), (4,1), (5,1) are 'B'.
2.  **A at index 3:** Move up 3. New position: (6-3, 3) = (3, 3). Trail: (4,3), (5,3) are 'B'.
3.  **A at index 7:** Move up 3. New position: (6-3, 7) = (3, 7). Trail: (4,7), (5,7) are 'B'.
4.  **R at index 9:** Move up 4. New position: (6-4, 9) = (2, 9). Trail: (3,9), (4,9), (5,9) are 'B'.

**Constructing the grid (11 columns, 9 rows):**
Row 0: . . . . . . . . . . .
Row 1: . . . . . . . . . . .
Row 2: . R . . . . . . . R .
Row 3: . B . A . . . A . B .
Row 4: . B . B . . . B . B .
Row 5: . B . B . . . B . B .
Row 6: B R B A B B B A B R B
Row 7: . . . . . . . . . . .
Row 8: . . . . . . . . . . .

Wait, let's re-check the trail in Example 1.
Ex 1: Input row 3. A is at (3,1). Output A is at (0,1). Trail is (1,1) and (2,1).
The trail is the cells *between* the new position and the old position.
In Ex 1: (0,1) is A. (1,1) is B. (2,1) is B. (3,1) is A.
Wait, the original A at (3,1) is still there? "The scrimmage line remains." Yes.
So in Ex 1:
Row 0: . A . . . .
Row 1: . B . . . .
Row 2: . B . . . .
Row 3: B A B B B B

Let's re-apply to Test Input:
R at (6,1) moves up 4 to (2,1). Trail: (3,1), (4,1), (5,1) are B.
A at (6,3) moves up 3 to (3,3). Trail: (4,3), (5,3) are B.
A at (6,7) moves up 3 to (3,7). Trail: (4,7

... [2093 chars truncated]

To solve this puzzle, let's perform a Conceptual Digestion:

### 1. Identify the CAST
*   **The Front:** A horizontal line of colored cells (the row containing non-black cells that isn't just a single column).
*   **Thermal Anomalies:** Specific colored cells within that horizontal line that are *not* the background color of the front (in these examples, the "front" is mostly Blue 'B', so any 'R', 'A', etc., are anomalies).
*   **Updraft Columns:** Vertical lines of color extending upwards from the anomaly positions.

### 2. Identify the AFFORDANCES
*   **The Front** stays in its original position.
*   **Anomalies** act as the "seed" for a column. The color of the anomaly determines the color of the column's "top" and the color of the "body" of the column.
*   **The Column** extends vertically upwards from the anomaly's position.

### 3. Identify the TRANSFORMATION RULE
1.  Locate the horizontal line (the "front").
2.  Identify every cell in that line that is not the "base" color (the most frequent color in that line, which is 'B' in all examples).
3.  For each anomaly (a cell with color $C$ at position $(r, c)$):
    *   Draw a vertical column upwards from $(r, c)$.
    *   The cell at $(r, c)$ remains color $C$.
    *   The cells above it $(r-1, c), (r-2, c), \dots$ are filled with the "base" color 'B'.
    *   The height of the column is determined by the distance to the top of the grid, but looking at the examples, the column extends upwards until it reaches a certain height or the top. Specifically, the anomaly color $C$ is at the bottom of the column, and the column is filled with 'B' upwards. Wait, let's re-examine Example 1: Input has `B A B B B B`. Output has `A` at `(3,1)` and `B` at `(2,1)` and `(1,1)`. So the anomaly color is at the *bottom* of the column, and the column extends *upwards* using the base color 'B'.
    *   *Correction based on Example 3:* Input `B A B B B A B R B`. Anomaly `A` at index 1. Output has `A` at index 4 (row 7) and `B` at index 5, 6. Wait, the row index is decreasing. Let's look at the verticality:
        Example 1: Row 3 is the front. `A` is at (3,1). Output has `A` at (3,1), `B` at (2,1), `A` at (0,1). No, that's not right.
        Let's re-read: "The column's height depends on the anomaly's thermal signature, with the anomaly type visible at the cloud top."
        Example 1: Front is row 3. Anomaly `A` is at (3,1). Output has `A` at (0,1), `B` at (1,1), `B` at (2,1), and `A` at (3,1).
        **Rule Re-evaluation:**
        1. Find the row containing the front.
        2. For each anomaly (color $C \neq B$) at $(r, c)$:
        3. The column starts at $(r, c)$ and goes up to row 0.
        4. The cell at $(r, c)$ is color $C$.
        5. The cells above it $(r-1, c), (r-2, c), \dots, (0, c)$ are filled with color $B$.
        6. **Wait**, the "anomaly type is visible at the cloud top". This means the color $C$ is at the *top* of the column.
        Let's check Example 1 again: `A` is at (3,1). Output has `A` at (0,1), `B` at (1,1), `B` at (2,1), `A` at (3,1).
        Actually, the anomaly color $C$ is at the top $(0, c)$ AND at the base $(r, c)$. The cells in between are 'B'.
        Let's check Example 2: `R` is at (5,3). Output has `R` at (1,3), `B` at (2,3), `B` at (3,3), `B` at (4,3), `R` at (5,3).
        Example 3: `A` at (7,1). Output has `A` at (4,1), `B` at (5,1), `B` at (6,1), `A` at (7,1).
        Example 4: `R` at (9,2). Output has `R` at (5,2), `B` at (6,2), `B` at (7,2), `B` at (8,2), `R` at (9,2).
        
        **Refined Rule:**
        1. Identify the front row $r$ and the base color $B$.
        2. For each anomaly $C$ at $(r, c)$:
        3. The column extends from $(r, c)$ upwards to some height.
        4. The cell at $(r, c)$ is $C$.
        5. The cell at the top of the column is $C$.
        6. The cells between the top and the base are $B$.
        7. The height of the column: In Example 1, height is 4 (rows 0,1,2,3). In Example 2, height is 5 (rows 1,2,3,4,5). In Example 3, height is 4 (rows 4,5,6,7). In Example 4, height is 5 (rows 5,6,7,8,9).
        The height seems to be a fixed distance or relative to the anomaly. Let's look at the anomaly's position in the row.
        Example 1: `A` is at index 1. Height 4.
        Example 2: `R` is at index 3. Height 5.
        Example 3: `A` is at index 1. Height 4. `A` is at index 5. Height 4. `R` is at index 7. Height 2? No, `R` is at (3,7) in output.
        Wait, the height is simply: the column goes from the anomaly row $r$ up to row $r - (\text{something})$.
        Looking at Example 3: `A` at index 1, $r=7$. Top is at row 4. $7-4 = 3$.
        `A` at index 5, $r=7$. Top is at row 4. $7-4 = 3$.
        `R` at index 7, $r=7$. Top is at row 3. $7-3 = 4$.
        Looking at Example 4: `R` at index 2, $r=9$. Top is at row 5. $9-5 = 4$.
        `A` at index 6, $r=9$. Top is at row 6. $9-6 = 3$.
        
        Let's look at the anomaly color itself.
        Example 1: `A` (Azure). Height 4.


... [4897 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **The Wall:** A horizontal row of colored cells (the "foundation wall").
    *   **Keystones:** Specific colored cells within that wall that are not the "standard" color (in these examples, Blue 'B' seems to be the standard brick, while other colors like 'A', 'R', etc., act as keystones).
    *   **Columns:** Vertical lines of color rising upwards from the keystone positions.
    *   **Capstones:** The top-most cell of a column, which matches the color of the keystone.
    *   **Brick:** The cells making up the column below the capstone, which match the "standard" color (Blue 'B').

2.  **Identify the AFFORDANCES:**
    *   **Keystones** trigger the creation of a vertical column.
    *   **Keystone Color** determines the color of the capstone (the top cell).
    *   **The Wall** remains in its original position.
    *   **The Column** fills the space above the keystone with the standard color (Blue 'B').

3.  **Identify the TRANSFORMATION RULE:**
    *   Locate the "wall" (the row containing non-black cells).
    *   Identify "keystones": any cell in that row that is not Blue ('B') and not Black ('.').
    *   For each keystone at position `(row, col)` with color `C`:
        *   Determine the height of the column. Looking at the examples:
            *   Ex 1: Keystone 'A' at (3,1). Column height is 3 (cells at (0,1), (1,1), (2,1)). Capstone is 'A'.
            *   Ex 2: Keystone 'R' at (5,3). Column height is 4 (cells at (1,3), (2,3), (3,3), (4,3)). Capstone is 'R'.
            *   Ex 3: Keystone 'A' at (7,1). Column height is 3 (cells at (4,1), (5,1), (6,1)). Capstone is 'A'.
            *   Ex 3: Keystone 'R' at (7,7). Column height is 4 (cells at (3,7), (4,7), (5,7), (6,7)). Capstone is 'R'.
        *   *Refined Rule for Height:* The height of the column is determined by the distance from the keystone to the top of the grid, but it seems to be a fixed height or relative to the grid size. Let's look closer.
        *   Ex 1: Wall row 3. Grid height 9. Keystone 'A' at row 3. Column goes up to row 0. Height = 3.
        *   Ex 2: Wall row 5. Grid height 10. Keystone 'R' at row 5. Column goes up to row 1. Height = 4.
        *   Ex 3: Wall row 7. Grid height 9. Keystone 'A' at row 4. Column goes up to row 4. Height = 3. Keystone 'R' at row 7. Column goes up to row 3. Height = 4.
        *   *Wait, the height is actually determined by the color?* No, the metaphor says "The column height is determined by the keystone material". Let's check the colors:
            *   Ex 1: 'A' (Azure) -> Height 3.
            *   Ex 2: 'R' (Red) -> Height 4.
            *   Ex 3: 'A' (Azure) -> Height 3. 'R' (Red) -> Height 4.
            *   *Pattern:* 'A' = 3, 'R' = 4.
    *   **Column Construction:** The cell at `(wall_row - height, col)` is the Capstone (Color `C`). The cells from `(wall_row - height + 1, col)` to `(wall_row - 1, col)` are the bricks (Color 'B').

### Applying the Rule to Test Input

**Test Input Analysis:**
*   Wall Row: 6 (index 6).
*   Wall Content: `B R B A B B B A B R B` (Indices 0 to 10).
*   Keystones:
    1.  `R` at index 1.
    2.  `A` at index 3.
    3.  `A` at index 7.
    4.  `R` at index 9.

**Heights based on color:**
*   `R` (Red) = 4.
*   `A` (Azure) = 3.

**Column Calculations:**
1.  **Keystone R (index 1):**
    *   Height 4.
    *   Capstone: Row `6 - 4 = 2`, Col 1. Color `R`.
    *   Bricks: Rows 3, 4, 5, Col 1. Color `B`.
2.  **Keystone A (index 3):**
    *   Height 3.
    *   Capstone: Row `6 - 3 = 3`, Col 3. Color `A`.
    *   Bricks: Rows 4, 5, Col 3. Color `B`.
3.  **Keystone A (index 7):**
    *   Height 3.
    *   Capstone: Row `6 - 3 = 3`, Col 7. Color `A`.
    *   Bricks: Rows 4, 5, Col 7. Color `B`.
4.  **Keystone R (index 9):**
    *   Height 4.
    *   Capstone: Row `6 - 4 = 2`, Col 9. Color `R`.
    *   Bricks: Rows 3, 4, 5, Col 9. Color `B`.

**Constructing the Grid:**
Row 0: `.` (all)
Row 1: `.` (all)
Row 2: `. R . . . . . . . R .`
Row 3: `. B . A . . . A . B .`
Row 4: `. B . B . . . B . B .`
Row 5: `. B . B . . . B . B .`
Row 6: `B R B A B B B A B R B`
Row 7: `.` (all)
Row 8: `.` (all)

Wait, let's re-verify the height for 'R' and 'A' from Example 3.
Ex 3: Wall is row 7.
Keystone A (index 1): Capstone at row 4. Height = 7 - 4 = 3. Correct.
Keystone R (index 7): Capstone at row 3. Height = 7 - 3 = 4. Correct.
So: A=3, R=4.

Test Input:
Row 6 is the wall.
R at index 1: Capstone at 6-4=2. Bricks at 3,4,5.
A at index 3: Capstone at 6-3=3. Bricks at 4,5.
A at index 7: Capstone at 6-3=3. Bricks at 4,5.
R at index 9: Capstone at 6-4=2. Bricks at 3,4,5.

Final Grid Construction:
Row 0: . . . . . . . . . . .
Row 1: . . . . . . . . . . .
Row 2: . R . . . . . . . R .
Row 3: . B . A . . . A . B .
Row 4: . B . B . . . B . B .
Row 5: . B . B . . . B . B .
Row 6: B R B A B B B A B R B
Row 7: . . . . . . . . . . .
Row 8: . . . . . . . . . . .

ANSWER:
. . . . . . . . . . .
. . . . . . . . . . .
. R . . . . . . . 

... [135 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Trench:** A horizontal line of colored cells (the "flares") located in a single row.
    *   **Flares:** The colored cells within the trench line.
    *   **Smoke Trails:** Vertical lines of color extending upwards from the flares.
    *   **Apex:** The top-most cell of a smoke trail, which retains the color of the flare.
    *   **Trail Color:** The cells in the smoke trail (between the trench and the apex) take the color of the *other* colors present in the trench, but looking closer at the examples, the trail color is actually the color of the *background* color of the trench (the most frequent color, which is Blue 'B').

2.  **Identify the AFFORDANCES:**
    *   **Flares** can "fire" upwards.
    *   **Flares** determine the height of the smoke trail.
    *   **Flares** determine the color of the apex.
    *   **The Trench** provides the base for the trails.

3.  **Identify the TRANSFORMATION RULE:**
    *   Locate the horizontal row containing colored cells (the trench).
    *   Identify the "base" color of the trench (the most frequent color, which is 'B' in all examples).
    *   For every colored cell in the trench that is *not* the base color:
        *   Calculate its height. The height is determined by the color's position or a specific property. Looking at the examples:
            *   Ex 1: A (Azure) is at index 1. Height is 3 cells above (including A).
            *   Ex 2: R (Red) is at index 3. Height is 4 cells above (including R).
            *   Ex 3: A (Azure) at index 1, A at index 5, R at index 7. Heights: A=4, A=4, R=5.
            *   Ex 4: R at index 2, A at index 6. Heights: R=5, A=4.
        *   Wait, the height is actually determined by the *color* itself. Let's re-examine:
            *   Ex 1: A (Azure) -> height 3.
            *   Ex 2: R (Red) -> height 4.
            *   Ex 3: A (Azure) -> height 4, R (Red) -> height 5.
            *   Ex 4: R (Red) -> height 5, A (Azure) -> height 4.
        *   Let's look at the color values/order: B=Blue, R=Red, A=Azure.
        *   In Ex 1: A is at index 1. Output has A at row 0, B at row 1, B at row 2. Total height 3.
        *   In Ex 2: R is at index 3. Output has R at row 1, B at row 2, B at row 3, B at row 4. Total height 5.
        *   In Ex 3: A is at index 1. Output has A at row 4, B at row 5, B at row 6, B at row 7. Total height 4.
        *   In Ex 4: R is at index 2. Output has R at row 5, B at row 6, B at row 7, B at row 8, B at row 9. Total height 5.
        *   **Correction on Height Rule:** The height of the trail is determined by the color.
            *   R (Red) = 5 cells high (including the flare).
            *   A (Azure) = 4 cells high (including the flare).
            *   (If there were others, they would follow a pattern).
        *   **Trail Construction:** For a flare of color $C$ at position $(r, c)$, the cells $(r-1, c), (r-2, c) \dots (r-(H-1), c)$ are filled with the base color 'B'. The cell $(r-(H-1), c)$ is the color $C$.

### Applying the Rule to Test Input

**Input Analysis:**
*   Trench Row: Row 6 (0-indexed).
*   Trench Content: `B R B A B B B A B R B`
*   Base Color: `B` (Blue)
*   Flares (non-B):
    1.  `R` at index 1. Height 5.
    2.  `A` at index 3. Height 4.
    3.  `A` at index 7. Height 4.
    4.  `R` at index 9. Height 5.

**Calculations:**
1.  **Flare R (index 1):**
    *   Row 6: R
    *   Row 5: B
    *   Row 4: B
    *   Row 3: B
    *   Row 2: R (Apex)
2.  **Flare A (index 3):**
    *   Row 6: A
    *   Row 5: B
    *   Row 4: B
    *   Row 3: A (Apex)
3.  **Flare A (index 7):**
    *   Row 6: A
    *   Row 5: B
    *   Row 4: B
    *   Row 3: A (Apex)
4.  **Flare R (index 9):**
    *   Row 6: R
    *   Row 5: B
    *   Row 4: B
    *   Row 3: B
    *   Row 2: R (Apex)

**Constructing the Grid:**
Row 0: `.` (all)
Row 1: `.` (all)
Row 2: `. R . . . . . . . R .`
Row 3: `. B . A . . . A . B .`
Row 4: `. B . B . . . B . B .`
Row 5: `. B . B . . . B . B .`
Row 6: `B R B A B B B A B R B`
Row 7: `.` (all)
Row 8: `.` (all)

Wait, let's re-verify the height of R and A from Example 4.
Ex 4: R is at index 2. Output R is at row 5. Row 6 is the trench. Height is 5 (Rows 5, 6, 7, 8, 9? No, the trench is row 9).
If trench is row 9:
R at row 5, B at 6, B at 7, B at 8, R at 9. Total 5 cells. Correct.
A at row 6, B at 7, B at 8, A at 9. Total 4 cells. Correct.

Test Input Trench is Row 6.
R (height 5): Rows 2, 3, 4, 5, 6. (R at 2, B at 3, 4, 5, R at 6).
A (height 4): Rows 3, 4, 5, 6. (A at 3, B at 4, 5, A at 6).

Let's re-map:
Flare R (index 1): (6,1)=R, (5,1)=B, (4,1)=B, (3,1)=B, (2,1)=R.
Flare A (index 3): (6,3)=A, (5,3)=B, (4,3)=B, (3,3)=A.
Flare A (index 7): (6,7)=A, (5,7)=B, (4,7)=B, (3,7)=A.
Flare R (index 9): (6,9)=R, (5,9)=B, (4,9)=B, (3,9)=B, (2,9)=R.

Final Grid Construction:
Row 0: . . . . . . . . . . .
Row 1: . . . . . . . . . . .
Row 2: . R . . . . . . . R .
Row 3: . B . A . . . A . B .
Row 4: . B . B . . . B . 

... [326 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Black (`.`):** The "dark sky" or background.
    *   **Red (`R`):** The "drumhead," a horizontal line that acts as a fixed barrier/anchor.
    *   **Yellow (`Y`):** The "raindrops" that are subject to gravity.

2.  **Identify the AFFORDANCES:**
    *   **Red (`R`):** It is anchored; it does not move. It acts as a trigger for the "splash" effect.
    *   **Yellow (`Y`):** It moves one step down (gravity). If it lands on the row immediately above the red drumhead (the "edge"), it triggers a splash.

3.  **Identify the TRANSFORMATION RULE:**
    *   **Gravity:** Every yellow raindrop moves down by exactly one row.
    *   **The Splash Rule:** If a yellow raindrop's new position is the row immediately above the red drumhead (the "edge"), it creates a V-shaped splash.
        *   The original drop moves to the edge.
        *   Two new drops are created: one moving up-left (diagonal) and one moving up-right (diagonal) from the edge position.
        *   These "spray" droplets continue to move diagonally up and out until they hit the boundary of the grid.
    *   **Standard Movement:** If a drop moves to a position that is *not* the edge (i.e., it's above the edge or below the drum), it simply moves down one row without any extra effect.

### Applying the Rule to Test Input

**Input Grid Analysis:**
Rows (0-indexed):
0: `. . . . Y . . .` (Y at [0, 4])
1: `Y . . . . . . .` (Y at [1, 0])
2: `. . . . . . . .`
3: `R R R R R R R R` (Red drumhead)
4: `. . . . . . . .`
5: `. . . . . . . .`
6: `. Y . . . . . .` (Y at [6, 1])
7-11: (All black)

**Step-by-Step Transformation:**

1.  **Drop at [0, 4]:** Moves down to [1, 4]. This is not the edge (row 2).
2.  **Drop at [1, 0]:** Moves down to [2, 0]. This **is** the edge (row 2, immediately above row 3).
    *   **Splash Triggered at [2, 0]:**
        *   The drop settles at [2, 0].
        *   Up-left spray: From [2, 0], move to [1, -1] (out of bounds).
        *   Up-right spray: From [2, 0], move to [1, 1], then [0, 2].
3.  **Drop at [6, 1]:** Moves down to [7, 1]. This is below the drum.

**Final Positions:**
*   From [0, 4] $\rightarrow$ [1, 4]
*   From [1, 0] $\rightarrow$ [2, 0] (Edge)
    *   Splash Up-Right: [1, 1], [0, 2]
*   From [6, 1] $\rightarrow$ [7, 1]
*   Red line stays at row 3.

**Constructing the Output Grid:**
Row 0: `. . Y . . . . .` (from splash)
Row 1: `Y . . . Y . . .` (from [1, 1] splash and [1, 4] move)
Row 2: `Y . . . . . . .` (from [2, 0] move)
Row 3: `R R R R R R R R` (fixed)
Row 4: `. . . . . . . .`
Row 5: `. . . . . . . .`
Row 6: `. . . . . . . .`
Row 7: `. Y . . . . . .` (from [6, 1] move)
Row 8-11: `. . . . . . . .`

Wait, let's re-verify the splash logic from Example 1.
Example 1:
Input Y at [0, 1], [1, 5], [4, 2], [7, 4]. Red at row 9.
Edge is row 8.
Drop at [7, 4] moves to [8, 4]. This is the edge.
Splash from [8, 4]:
Up-left: [7, 3], [6, 2], [5, 1], [4, 0]
Up-right: [7, 5], [6, 6], [5, 7]
Let's check Example 1 output:
Row 4: `Y . . . . . . .` (Matches [4, 0])
Row 5: `. Y Y . . . . Y` (Matches [5, 1], [5, 7]... wait, the example output shows `Y Y` at [5, 1] and [5, 6]? Let's re-count.)
Example 1 Output Row 5: `. Y Y . . . . Y`
Indices: 1, 2, 7.
If splash is at [8, 4]:
Up-left: [7, 3], [6, 2], [5, 1], [4, 0]
Up-right: [7, 5], [6, 6], [5, 7]
Example 1 Output Row 4: `Y . . . . . . .` (Index 0) - Correct.
Example 1 Output Row 5: `. Y Y . . . . Y` (Indices 1, 2, 7) - My calculation for [5, 1] and [5, 7] is close. Let's re-examine the "twin arcs" description. "each droplet one step further along the diagonal".
If the drop lands on [8, 4], the splash is:
[7, 3], [6, 2], [5, 1], [4, 0]
[7, 5], [6, 6], [5, 7]
Wait, the example output row 5 is `. Y Y . . . . Y`. This means indices 1, 2, and 7.
Looking at Example 1 output row 6: `. . Y . . . Y .` (Indices 2, 6).
Looking at Example 1 output row 7: `. . . Y . Y . .` (Indices 3, 5).
Looking at Example 1 output row 8: `. . . . Y . . .` (Index 4).
This confirms the splash pattern:
From [8, 4]:
[8, 4]
[7, 3], [7, 5]
[6, 2], [6, 6]
[5, 1], [5, 7]
[4, 0]
Wait, the example output row 5 is `. Y Y . . . . Y`. My index 2 is wrong. Let's re-read.
Example 1 Output Row 5: `. Y Y . . . . Y`
Indices: 1, 2, 7.
Wait, if the splash is at [8, 4], the diagonal is [7, 3], [6, 2], [5, 1], [4, 0].
The other diagonal is [7, 5], [6, 6], [5, 7].
The example output row 5 has index 2. That would be [5, 2]. That's not a diagonal from [8, 4].
Let's look at the input again. Input Y at [4, 2].
If [4, 2] moves to [5, 2], then row 5 index 2 is the moved drop.
So:
[8, 4] is the edge drop.
[7, 3], [7, 5]
[6, 2], [6, 6]
[5, 1], [5, 7]
[4, 0]
And the other drops:
[0, 1] $\rightarrow$ [1, 1]
[1, 5] $\rightarrow$ [2, 5]
[4, 2] $\rightarrow$ [5, 2]
[7, 4] $\rightarrow$ [8, 4] (Edge)
Let's check Example 1 output row 5 again: `. Y Y . . . . Y`.
Indices: 1, 2, 7.
My calculated indices for row 5: 1 (from splash), 2 (from moved drop), 7 (from splash).
Matches perfectly!

**Applying to

... [746 chars truncated]

### Conceptual Digestion

**1. Identify the CAST:**
*   **Red (R):** The River Styx (a horizontal line acting as a boundary/trigger).
*   **Yellow (Y):** Souls (initial points) and Wraiths (the resulting trails).
*   **Black (.):** The void (empty space).

**2. Identify the AFFORDANCES:**
*   **Souls (Y):** Can move down one level. If they land on the row immediately above the River (R), they trigger a "split" transformation.
*   **River (R):** Acts as a fixed barrier and a trigger for the splitting mechanism.
*   **Wraiths (Y):** Created when a soul lands on the row directly above the river. They move diagonally (up-left and up-right) until they hit the grid boundary.

**3. Identify the TRANSFORMATION RULE:**
*   **Step 1: Movement.** Every initial Yellow soul (Y) moves down exactly one row.
*   **Step 2: The Split.** Check the new position of each soul. If a soul's new position is in the row immediately above the Red River (R), it is replaced by two "wraiths."
*   **Step 3: Wraith Trajectory.** These two wraiths move diagonally away from each other (one towards the top-left, one towards the top-right) from the split point until they reach the edge of the grid.
*   **Step 4: Persistence.** If a soul moves down but does *not* land on the row above the river, it simply stays at its new position.

---

### Applying the Rule to Test Input

**Input Grid Analysis:**
Grid size: 12 rows x 8 columns.
River (R) is at Row 3 (0-indexed).
Souls (Y) are at:
1. (0, 4)
2. (1, 0)
3. (6, 1)

**Step-by-Step Transformation:**

1.  **Soul at (0, 4):**
    *   Moves down to (1, 4).
    *   Is (1, 4) above the river? The river is at Row 3. The row above is Row 2.
    *   (1, 4) is not Row 2. It just stays at (1, 4).

2.  **Soul at (1, 0):**
    *   Moves down to (2, 0).
    *   Is (2, 0) above the river? Yes, Row 2 is immediately above Row 3.
    *   **Split Triggered at (2, 0):**
        *   Wraith 1 (Up-Left): From (2, 0), moving up-left hits the boundary immediately. Position: (2, 0).
        *   Wraith 2 (Up-Right): From (2, 0), moving up-right: (1, 1), (0, 2).
    *   *Correction on Split Logic based on Example 1:* In Example 1, a soul at (7, 4) moves to (8, 4). The river is at (9, 0-7). The row above is (8, 0-7). The soul at (8, 4) splits. The wraiths go up-left and up-right.
    *   Let's re-trace Example 1: Soul at (7, 4) -> moves to (8, 4). Row 8 is above Row 9.
    *   Wraiths from (8, 4):
        *   Up-Left: (7, 3), (6, 2), (5, 1), (4, 0).
        *   Up-Right: (7, 5), (6, 6), (5, 7).
    *   This matches the example output perfectly.

**Applying to Test Input:**
*   **Soul 1 (0, 4) -> (1, 4)**
*   **Soul 2 (1, 0) -> (2, 0)**. This is the row above the river (Row 3).
    *   Split at (2, 0):
    *   Wraith 1 (Up-Left): (2, 0) is the start. It cannot go further left.
    *   Wraith 2 (Up-Right): (1, 1), (0, 2).
*   **Soul 3 (6, 1) -> (7, 1)**. Not above the river.

**Final Positions:**
*   (1, 4)
*   (2, 0)
*   (1, 1)
*   (0, 2)
*   (7, 1)
*   (3, 0-7) [River]

**Constructing the Grid:**
Row 0: . . Y . . . . .
Row 1: . Y . . Y . . .
Row 2: Y . . . . . . .
Row 3: R R R R R R R R
Row 4: . . . . . . . .
Row 5: . . . . . . . .
Row 6: . . . . . . . .
Row 7: . Y . . . . . .
Row 8-11: . . . . . . . .

Wait, let's re-verify Soul 2 (1, 0) -> (2, 0).
If the soul is at (2, 0), the split starts at (2, 0).
Wraith 1 (Up-Left): (2, 0)
Wraith 2 (Up-Right): (1, 1), (0, 2)
Total Ys: (1, 4), (2, 0), (1, 1), (0, 2), (7, 1).

Let's double check the input coordinates:
Input:
Row 0: . . . . Y . . . (0, 4)
Row 1: Y . . . . . . . (1, 0)
Row 2: . . . . . . . .
Row 3: R R R R R R R R
Row 4: . . . . . . . .
Row 5: . . . . . . . .
Row 6: . Y . . . . . . (6, 1)
Row 7-11: . . . . . . . .

Transformation:
(0, 4) -> (1, 4)
(1, 0) -> (2, 0) [SPLIT]
(6, 1) -> (7, 1)

Split at (2, 0):
Up-Left: (2, 0)
Up-Right: (1, 1), (0, 2)

Final Y positions: (1, 4), (2, 0), (1, 1), (0, 2), (7, 1).

ANSWER:
. . Y . . . . .
. Y . . Y . . .
Y . . . . . . .
R R R R R R R R
. . . . . . . .
. . . . . . . .
. . . . . . . .
. Y . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .

Wait, let me re-check the split logic one more time. In Example 1, the soul at (7, 4) moves to (8, 4). The river is at row 9. The split happens at (8, 4). The wraiths are at (7, 3), (6, 2), (5, 1), (4, 0) AND (7, 5), (6, 6), (5, 7).
The split point (8, 4) itself is also a Y in the output.
So for my test:
Split at (2, 0):
Wraith 1: (2, 0)
Wraith 2: (1, 1), (0, 2)
The split point (2, 0) is already covered.

Final check of the grid:
Row 0: . . Y . . . . .
Row 1: . Y . . Y . . .
Row 2: Y . . . . . . .
Row 3: R R R R R R R R
Row 4: . . . . . . . .
Row 5: . . . . . . . .
Row 6: . . . . . . . .
Row 7: . Y . . . . . .
Row 8: . . . . . . . .
Row 9: . . . . . . . .
Row 10: . . . . . . . .
Row 11: . . . . . . . .

ANSWER:
. . Y . . . . .
. Y . . Y . . .
Y . . . . . . .
R R R R R R R R
. . . . . . . .
. . . . . . . .
. . . . . . . .
. Y . . . . . .
. . . . . . . .

... [48 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Red (R):** The magma intrusion (a horizontal line). It acts as a barrier/trigger.
    *   **Yellow (Y):** Boulders (initial state) and Pyroclastic flows (result of fracturing).
    *   **Black (.):** The solid rock (background).

2.  **Identify the AFFORDANCES:**
    *   **Boulders (Y):** Can move downwards (drop one stratum).
    *   **Magma (R):** Can trigger a fracture if a boulder lands directly on it.
    *   **Fracture (Y):** If a boulder hits the magma, it disappears and creates two diagonal lines of yellow moving upwards.

3.  **Identify the TRANSFORMATION RULE:**
    *   **Step 1: Gravity.** Every yellow boulder (Y) moves down exactly one row (one stratum).
    *   **Step 2: Collision Check.** Check if the new position of a boulder is on a Red (R) cell.
    *   **Step 3: Fracture.** If a boulder lands on Red:
        *   The boulder itself is replaced by the magma (it "fractures").
        *   Two diagonal paths of yellow (Y) are created starting from the cell *above* the magma cell where the boulder hit.
        *   The paths move diagonally up-left and up-right (e.g., `(-1, -1)` and `(-1, +1)`) until they reach the top edge of the grid (row 0).
    *   **Step 4: Settlement.** If a boulder lands on a non-red cell, it simply stays in its new position.

### Applying the Rule to Test Input

**Input Grid Analysis:**
Row 0: `. . . . Y . . .` (Y at col 4)
Row 1: `Y . . . . . . .` (Y at col 0)
Row 2: `. . . . . . . .`
Row 3: `R R R R R R R R` (Magma)
Row 4: `. . . . . . . .`
Row 5: `. . . . . . . .`
Row 6: `. Y . . . . . .` (Y at col 1)
Row 7-11: All `.`

**Step 1 & 2: Gravity and Collision**
1.  **Y at (0, 4):** Moves to (1, 4). Not on Red. Result: `Y` at (1, 4).
2.  **Y at (1, 0):** Moves to (2, 0). Not on Red. Result: `Y` at (2, 0).
3.  **Y at (6, 1):** Moves to (7, 1). Not on Red. Result: `Y` at (7, 1).

**Wait, let's re-examine the "landing directly above" logic from the metaphor.**
Metaphor: "A boulder landing directly above the magma layer fractures on contact...".
In Example 1:
Input Y at (0, 1), (1, 5), (4, 2), (7, 4), (10, 3). Magma at row 9.
- Y(0,1) -> (1,1)
- Y(1,5) -> (2,5)
- Y(4,2) -> (5,2)
- Y(7,4) -> (8,4)
- Y(10,3) -> (11,3)
Wait, in Example 1, the output shows Y at (5,0), (5,1), (5,7), (6,2), (6,6), (7,3), (7,5), (8,4).
Let's look closer at Example 1's Y at (7,4). It moves to (8,4). (8,4) is directly above (9,4) which is Red.
If a boulder is at (8,4), it is "directly above" the magma.
The rule says: "A boulder landing directly above the magma layer fractures on contact".
In Example 1, the Y at (7,4) moves to (8,4). Since (8,4) is above (9,4), it fractures.
The fracture starts from (8,4) and goes up: (7,3), (6,2), (5,1), (4,0) AND (7,5), (6,6), (5,7), (4,8)...
Let's re-trace Example 1 with this:
Input Y at (7,4) moves to (8,4). (8,4) is above (9,4).
Fracture from (8,4):
Up-Left: (7,3), (6,2), (5,1), (4,0)
Up-Right: (7,5), (6,6), (5,7), (4,8)
Looking at Example 1 output:
Row 5: `Y . . . . . . .` (Wait, this is (5,0))
Row 5: `. Y Y . . . . Y` (This is (5,1), (5,2), (5,7))
This implies the fracture path is slightly different or I am miscounting.
Let's re-evaluate: If Y is at (8,4), it is "on contact" with the magma.
The fracture starts from the cell (8,4) and goes up.
Example 1 Output Y positions:
(1,1), (2,5), (4,0), (5,1), (5,2), (5,7), (6,2), (6,6), (7,3), (7,5), (8,4), (11,3).
Wait, the Y at (10,3) moved to (11,3).
The Y at (7,4) moved to (8,4). Since (8,4) is above (9,4), it fractures.
The fracture from (8,4) goes:
(7,3), (6,2), (5,1), (4,0)
(7,5), (6,6), (5,7), (4,8)
And the Y at (8,4) itself? The example shows Y at (8,4).
Let's look at the Y at (4,2) in Example 1. It moves to (5,2).
The Y at (0,1) moves to (1,1).
The Y at (1,5) moves to (2,5).
The Y at (7,4) moves to (8,4). It is above (9,4).
The fracture from (8,4) creates: (7,3), (6,2), (5,1), (4,0) AND (7,5), (6,6), (5,7).
Wait, the example output has Y at (5,1), (5,2), (5,7).
If (5,2) is a Y, it must come from (4,2) moving to (5,2).
If (5,1) is a Y, it must be part of the fracture from (8,4).
If (5,7) is a Y, it must be part of the fracture from (8,4).
If (4,0) is a Y, it must be part of the fracture from (8,4).
Let's check (4,0) in Example 1 output: Row 4 is `Y . . . . . . .`. Yes!
So the rule is:
1. Move all Y down 1.
2. If a Y is at (r, c) and (r+1, c) is Red:
   - The Y at (r, c) stays (it's the "contact" point).
   - Create diagonal lines from (r, c) upwards: (r-1, c-1), (r-2, c-2)... and (r-1, c+1), (r-2, c+2)...

**Applying to Test Input:**
Input Y: (0,4), (1,0), (6,1). Magma: Row 3.
1. Y(0,4) -> (1,4). Not above Red.
2. Y(1,0) -> (2,0). Not above Red.
3. Y(6,1) -> (7,1). Not above Red.

Wait, let's check if any Y is *above* the magma.
Magma is at Row 3.
Y(0,4) is at Row 0. Moves to Row 1. (1,4) is above (3,4).
Y(1,0) is at Row 1. Moves to Row 2. (2,0) is above (3,0).
Is (2,0) "directly above" the magma? Yes, (3,0) is Red.
So, Y(1,0

... [2917 chars truncated]

### Conceptual Digestion

**1. Identify the CAST:**
*   **Red String (R):** A horizontal line spanning the full width of the grid. It acts as a "resonance plane."
*   **Yellow Notes (Y):** Mobile elements that move vertically.
*   **Black Space (.):** The background/silence.

**2. Identify the AFFORDANCES:**
*   **Yellow Notes (Y):** Can move down one row. If they land on the row immediately above the Red String, they trigger a "resonance" effect.
*   **Red String (R):** Acts as a trigger. When a note lands on the row directly above it, it generates two new diagonal paths of yellow notes.

**3. Identify the TRANSFORMATION RULE:**
1.  **Step Down:** Every existing Yellow note (Y) moves down exactly one row (row index $i \to i+1$).
2.  **Resonance Check:** If a Yellow note's *new* position is in the row immediately above the Red String (row $R_{row} - 1$):
    *   The note stays at that position.
    *   Two new "harmonic" notes are created. These harmonics move diagonally upwards and outwards (one moving up-left, one moving up-right) starting from the note's position, continuing until they hit the edge of the grid.
3.  **Note on Harmonics:** Looking at Example 1, the harmonics move *upwards* from the resonance point. Let's re-examine the metaphor: "A note arriving directly above the string resonates and produces two harmonic overtones that ascend diagonally in opposite directions until they exit the score."
    *   In Example 1, the Red String is at row 9 (0-indexed).
    *   Input note at row 7 moves to row 8. Row 8 is above row 9.
    *   Harmonics from row 8: (7, 3), (6, 2), (5, 1)... and (7, 5), (6, 6), (5, 7)...
    *   Wait, looking at Example 1 output: The note at (7, 4) moves to (8, 4). The harmonics appear at (7, 3), (6, 2), (5, 1) and (7, 5), (6, 6), (5, 7). This confirms they move *upwards* diagonally.

### Applying the Rule to Test Input

**Input Grid (12x8):**
Row 0: `. . . . Y . . .` (Y at 0,4)
Row 1: `Y . . . . . . .` (Y at 1,0)
Row 2: `. . . . . . . .`
Row 3: `R R R R R R R R` (Red String at row 3)
Row 4: `. . . . . . . .`
Row 5: `. . . . . . . .`
Row 6: `. Y . . . . . .` (Y at 6,1)
Row 7: `. . . . . . . .`
Row 8: `. . . . . . . .`
Row 9: `. . . . . . . .`
Row 10: `. . . . . . . .`
Row 11: `. . . . . . . .`

**Step 1: Move all Y down one row.**
*   Y at (0,4) $\to$ (1,4)
*   Y at (1,0) $\to$ (2,0)
*   Y at (6,1) $\to$ (7,1)

**Step 2: Check for Resonance.**
The Red String is at Row 3. The resonance row is Row 2 ($3-1=2$).
*   Is there a Y at Row 2? Yes, the note from (1,0) moved to (2,0).
*   **Resonance at (2,0):**
    *   Note stays at (2,0).
    *   Harmonic 1 (Up-Left): From (2,0), moving up-left is impossible (out of bounds immediately).
    *   Harmonic 2 (Up-Right): From (2,0) $\to$ (1,1), (0,2).

**Step 3: Final Positions.**
*   (1,4) - from original (0,4)
*   (2,0) - from original (1,0)
*   (7,1) - from original (6,1)
*   (1,1) - harmonic
*   (0,2) - harmonic

**Constructing the Grid:**
Row 0: `. . Y . . . . .`
Row 1: `. Y . . Y . . .`
Row 2: `Y . . . . . . .`
Row 3: `R R R R R R R R`
Row 4: `. . . . . . . .`
Row 5: `. . . . . . . .`
Row 6: `. . . . . . . .`
Row 7: `. Y . . . . . .`
Row 8: `. . . . . . . .`
Row 9: `. . . . . . . .`
Row 10: `. . . . . . . .`
Row 11: `. . . . . . . .`

Wait, let's double check the "Up-Left/Up-Right" direction.
In Example 1: Note at (8,4) $\to$ (9,4) is the string. Note at (8,4) is above string.
Harmonics: (7,3), (6,2), (5,1) and (7,5), (6,6), (5,7).
These are moving *up* from the resonance point.
My calculation for (2,0):
Up-Left: (1, -1) -> Out.
Up-Right: (1, 1), (0, 2).
This matches.

Final check of the input:
Input Row 0: `. . . . Y . . .` (Col 4)
Input Row 1: `Y . . . . . . .` (Col 0)
Input Row 3: `R R R R R R R R`
Input Row 6: `. Y . . . . . .` (Col 1)

Step 1 (Move down):
(1,4), (2,0), (7,1)
Step 2 (Resonance at Row 2):
(2,0) is at Row 2.
Harmonics from (2,0): (1,1), (0,2).

Final Grid:
R0: . . Y . . . . .
R1: . Y . . Y . . .
R2: Y . . . . . . .
R3: R R R R R R R R
R4: . . . . . . . .
R5: . . . . . . . .
R6: . . . . . . . .
R7: . Y . . . . . .
R8: . . . . . . . .
R9: . . . . . . . .
R10: . . . . . . . .
R11: . . . . . . . .

ANSWER:
. . Y . . . . .
. Y . . Y . . .
Y . . . . . . .
R R R R R R R R
. . . . . . . .
. . . . . . . .
. . . . . . . .
. Y . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .

### Conceptual Digestion

**1. Identify the CAST:**
*   **Red (R):** The inversion layer (a horizontal line).
*   **Yellow (Y):** Hailstones (objects that move) and Shards (the result of a collision).
*   **Black (.):** The atmosphere/empty space.

**2. Identify the AFFORDANCES:**
*   **Hailstones (Y):** Can move down one row (descend).
*   **Inversion Layer (R):** Acts as a barrier. If a hailstone is exactly one row above it, it "shatters."
*   **Shattering:** When a hailstone at `row = layer_row - 1` moves down, it is replaced by shards that travel diagonally upwards and outwards (up-left and up-right) from the point of impact.

**3. Identify the TRANSFORMATION RULE:**
*   **Step 1: Movement.** Every yellow pixel (hailstone) moves down one row (`row + 1`).
*   **Step 2: Collision Check.** 
    *   If a hailstone was at `layer_row - 1`, it hits the red layer. Instead of just moving to `layer_row` (which is occupied by Red), it "shatters."
    *   The "shattering" effect: The hailstone at `(r, c)` where `r = layer_row - 1` creates a path of yellow shards. These shards move diagonally upwards from the impact point: `(r-1, c-1), (r-2, c-2)...` and `(r-1, c+1), (r-2, c+2)...` until they hit the edge of the grid.
*   **Step 3: Final Placement.** 
    *   Hailstones not hitting the layer simply move to `row + 1`.
    *   The red layer remains unchanged.

---

### Applying the Rule to Test Input

**Input Analysis:**
Grid size: 12 rows x 8 columns.
Red layer (R) is at Row 3 (0-indexed).
Hailstones (Y) are at:
1. (0, 4)
2. (1, 0)
3. (6, 1)

**Transformation:**
1.  **Hailstone (0, 4):** Moves to (1, 4). It is not adjacent to the red layer (Row 3).
2.  **Hailstone (1, 0):** This is at `layer_row - 2`. It moves to (2, 0). It is not adjacent to the red layer. Wait, let's re-read: "A hailstone arriving just above the inversion layer shatters". 
    *   If a hailstone is at `layer_row - 1`, it moves to `layer_row` and shatters.
    *   In the test input, the red layer is at Row 3.
    *   The hailstone at (1, 0) moves to (2, 0). Row 2 is "just above" Row 3.
    *   Wait, let's look at Example 1 to clarify "arriving just above".
    *   Example 1: Red is at Row 9. Hailstone at Row 7 moves to Row 8. Hailstone at Row 10 moves to Row 11.
    *   In Example 1, there is a hailstone at Row 7. It moves to Row 8. Row 8 is "just above" Row 9.
    *   In Example 1, the hailstone at (7, 4) moves to (8, 4). This triggers shards.
    *   The shards in Example 1: From (8, 4), shards go to (7, 3), (6, 2), (5, 1) and (7, 5), (6, 6), (5, 7).
    *   **Correction:** A hailstone at `layer_row - 1` moves to `layer_row - 2`? No.
    *   Let's re-examine Example 1:
        Input Y at (0,1), (1,5), (4,2), (7,4), (10,3). Red at Row 9.
        Output Y at (1,1), (2,5), (5,1), (5,2), (5,7), (6,2), (6,6), (7,3), (7,5), (8,4), (11,3).
        Wait, the hailstone at (7,4) moved to (8,4). Since (8,4) is just above (9,4), it shatters.
        The shards from (8,4) go to (7,3), (6,2), (5,1) and (7,5), (6,6), (5,7).
        The hailstone at (4,2) moved to (5,2).
        The hailstone at (0,1) moved to (1,1).
        The hailstone at (1,5) moved to (2,5).
        The hailstone at (10,3) moved to (11,3).

**Test Input Application:**
Red layer at Row 3.
Hailstones:
1. (0, 4) -> moves to (1, 4).
2. (1, 0) -> moves to (2, 0). This is "just above" Row 3. It shatters.
   Shards from (2, 0): (1, -1) [out], (1, 1), (0, 2)... wait, the diagonal is `(r-1, c-1)` and `(r-1, c+1)`.
   From (2, 0): Up-left is out. Up-right: (1, 1), (0, 2).
3. (6, 1) -> moves to (7, 1).

**Wait, let's re-check the "shatter" logic.**
In Example 1: Hailstone at (7,4) moves to (8,4). (8,4) is just above (9,4).
The shards are: (7,3), (6,2), (5,1) AND (7,5), (6,6), (5,7).
These are `(r-1, c-1), (r-2, c-2)...` and `(r-1, c+1), (r-2, c+2)...`.
In Test Input:
Hailstone at (1, 0) moves to (2, 0).
(2, 0) is just above (3, 0).
Shards from (2, 0):
Up-left: (1, -1) -> out.
Up-right: (1, 1), (0, 2).
Hailstone (0, 4) moves to (1, 4).
Hailstone (6, 1) moves to (7, 1).

**Final Grid Construction:**
Row 0: . . Y . . . . . (from shard)
Row 1: . Y . . Y . . . (from shard and moved hailstone)
Row 2: Y . . . . . . . (from moved hailstone)
Row 3: R R R R R R R R
Row 4: . . . . . . . .
Row 5: . . . . . . . .
Row 6: . . . . . . . .
Row 7: . Y . . . . . . (from moved hailstone)
Row 8-11: . . . . . . . .

Let's double check the shard direction. In Example 1, the hailstone at (7,4) moves to (8,4). The shards go *upwards* (5,1, 6,2, 7,3).
In Test Input, the hailstone at (1,0) moves to (2,0). The shards go *upwards* (1,1, 0,2).

**Wait, let's re-verify the hailstone at (1,0) in Test Input.**
Input:
Row 0: . . . . Y . . .
Row 1: Y . . . . . . .
Row 2: . . . . . . . .
Row 3: R R R R R R R R
Row 4: . . . . . . . .
Row 5: . . . . . . . .
Row 6: . Y . . . . . .
Row 7-11: . . . . . . . .

1. (0,4) -> (1,4)
2. (1,0) -> (2,0). (2,0) is just above (3,0).
   Shards from (2,0): (1,1), (0,2).
3. (6,1) -> (7

... [841 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Red (R):** The "fortification wall" (a horizontal line of red cells).
    *   **Yellow (Y):** The "paratroopers" (individual yellow cells) and "signal flares" (the trails they leave).
    *   **Black (.):** The "battlefield" (empty space).

2.  **Identify the AFFORDANCES:**
    *   **Yellow cells (Troopers):** Can move down by one row. If they land on a Red cell, they trigger a "flare" effect.
    *   **Red cells (Wall):** Act as a trigger for the flare effect when a yellow cell lands on them.
    *   **Flare effect:** Two diagonal paths of yellow cells rising upwards from the point of impact.

3.  **Identify the TRANSFORMATION RULE:**
    *   **Step 1 (Movement):** Every yellow cell (Y) moves down exactly one row (row index increases by 1).
    *   **Step 2 (Trigger):** Check if the new position of a yellow cell is on a red cell (R).
    *   **Step 3 (Flares):** If a yellow cell lands on a red cell at position `(r, c)`:
        *   A flare path is created moving diagonally up-left: `(r-1, c-1), (r-2, c-2), ...` until it hits the edge.
        *   A flare path is created moving diagonally up-right: `(r-1, c+1), (r-2, c+2), ...` until it hits the edge.
    *   **Step 4 (Final Grid):** Combine the moved troopers and the flare paths. Note: The original trooper position is replaced by its new position.

### Applying the Rule to Test Input

**Input Analysis:**
Grid size: 12 rows x 8 columns.
Red wall at Row 3: `R R R R R R R R` (Indices 0-indexed: Row 3, Cols 0-7).

**Yellow Trooper Positions:**
1.  (0, 4) -> Moves to (1, 4)
2.  (1, 0) -> Moves to (2, 0)
3.  (6, 1) -> Moves to (7, 1)

**Check for Wall Impact:**
1.  (1, 4) is not on Row 3.
2.  (2, 0) is not on Row 3.
3.  (7, 1) is not on Row 3.

Wait, let's re-examine the training example to ensure the "descend" logic is correct.
Example 1:
Input Y at (0,1), (1,5), (4,2), (7,4), (10,3). Wall at Row 9.
Output Y at:
(0,1) -> (1,1)
(1,5) -> (2,5)
(4,2) -> (5,2)
(7,4) -> (8,4)
(10,3) -> (11,3)
Wait, the example output shows:
(1,1), (2,5), (4,1), (5,2), (5,7), (6,3), (6,6), (7,4), (8,3), (8,5), (9,4) is R, (10,3), (11,3).
Let's re-read: "Each paratrooper descends one level. A trooper who touches down directly above the wall fires two signal flares..."
"Directly above the wall" means the cell *immediately preceding* the wall? No, "touches down directly above the wall" usually means the cell at `row_of_wall - 1`.
Let's look at Example 1 again.
Wall is at Row 9.
Trooper at (7,4) moves to (8,4).
(8,4) is "directly above the wall" (Row 9).
If a trooper lands at (8,4), it fires flares.
Flares from (8,4):
Up-left: (7,3), (6,2), (5,1), (4,0)
Up-right: (7,5), (6,6), (5,7)
Let's check Example 1 output for these coordinates:
(8,4) is Y.
(7,3) is Y, (6,2) is Y, (5,1) is Y, (4,0) is Y.
(7,5) is Y, (6,6) is Y, (5,7) is Y.
This matches the example output!

**Correct Rule Refined:**
1.  Move all Y down one row.
2.  If a Y lands at `(r, c)` where `r = row_of_wall - 1`, it triggers flares.
3.  Flares are Y cells at `(r-i, c-i)` and `(r-i, c+i)` for `i = 1, 2, ...` until edge.

**Applying to Test Input:**
Wall is at Row 3.
Target row for trigger: Row 2.
Troopers:
1. (0, 4) -> moves to (1, 4). Not Row 2.
2. (1, 0) -> moves to (2, 0). **This is Row 2!** Trigger flares.
3. (6, 1) -> moves to (7, 1). Not Row 2.

**Flare from (2, 0):**
Up-left: (1, -1) -> Out of bounds.
Up-right: (1, 1), (0, 2).

**Final Positions:**
- (1, 4) [Trooper 1]
- (2, 0) [Trooper 2 - Trigger]
- (1, 1) [Flare 1]
- (0, 2) [Flare 1]
- (7, 1) [Trooper 3]
- (3, 0-7) [Wall]

**Constructing Grid:**
Row 0: . . Y . . . . .
Row 1: . Y . . Y . . .
Row 2: Y . . . . . . .
Row 3: R R R R R R R R
Row 4: . . . . . . . .
Row 5: . . . . . . . .
Row 6: . . . . . . . .
Row 7: . Y . . . . . .
Row 8-11: . . . . . . . .

Wait, let's double check the trooper at (1,0). It moves to (2,0). (2,0) is Row 2. Row 2 is indeed `row_of_wall - 1`.
The flare from (2,0) goes up-right: (1,1), (0,2).
The trooper at (0,4) moves to (1,4).
The trooper at (6,1) moves to (7,1).

Let's re-verify the flare logic. In Example 1, the trooper at (7,4) moved to (8,4). The flares were at (7,3), (6,2), (5,1), (4,0) and (7,5), (6,6), (5,7).
My flare logic: (2,0) -> (1,1), (0,2). Correct.

Final Grid:
R0: . . Y . . . . .
R1: . Y . . Y . . .
R2: Y . . . . . . .
R3: R R R R R R R R
R4: . . . . . . . .
R5: . . . . . . . .
R6: . . . . . . . .
R7: . Y . . . . . .
R8: . . . . . . . .
R9: . . . . . . . .
R10: . . . . . . . .
R11: . . . . . . . .

ANSWER:
. . Y . . . . .
. Y . . Y . . .
Y . . . . . . .
R R R R R R R R
. . . . . . . .
. . . . . . . .
. . . . . . . .
. Y . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .

### Conceptual Digestion

**1. Identify the CAST:**
*   **The Cloud (Azure - 'A'):** A shape located in the upper half of the grid (rows 1-3). It is the object being transformed.
*   **The Weather Vane (Yellow - 'Y'):** A plus-shaped object located in the lower half of the grid (rows 4-6). It acts as the directional indicator.
*   **The Sky (Black - '.'):** The background canvas.

**2. Identify the AFFORDANCES:**
*   **The Cloud** can be mirrored/reflected.
*   **The Weather Vane** can indicate direction. Specifically, the "extra fin" (the part of the plus shape that extends beyond the central column) determines the direction of the wind.
*   **The Sky** provides the space for the reflection to appear.

**3. Identify the TRANSFORMATION RULE:**
*   **Step 1: Determine Direction.** Look at the weather vane (the 'Y' shape in the bottom half). The weather vane is a plus shape centered in the middle columns. If there is a 'Y' at `(row, col-1)`, the wind blows **Left**. If there is a 'Y' at `(row, col+1)`, the wind blows **Right**.
*   **Step 2: Mirror the Cloud.** Identify the azure ('A') pixels in the top half.
*   **Step 3: Apply Reflection.** Create a mirror image of the cloud shape. If the wind blows Right, the reflection is placed to the right of the original cloud. If the wind blows Left, the reflection is placed to the left. The reflection is a horizontal flip of the original shape, placed adjacent to the original's boundary in the direction of the wind.

### Applying the Rule to Test Input

**Input Analysis:**
*   **Cloud (A):**
    Row 1: `. . . A . A . . .` (Cols 4, 6)
    Row 2: `. . . . A A . . .` (Cols 5, 6)
    Row 3: `. . . A . . . . .` (Col 4)
*   **Weather Vane (Y):**
    Row 4: `. . . . . Y . . .` (Col 6)
    Row 5: `. . . Y Y Y . . .` (Cols 4, 5, 6)
    Row 6: `. . . . Y . . . .` (Col 5)
    *Wait, let's look closer at the vane structure in the test input:*
    Row 4: Col 6 is Y.
    Row 5: Col 4, 5, 6 are Y.
    Row 6: Col 5 is Y.
    The center of the vane is Col 5. In Row 5, there is a Y at Col 4 and Col 6. However, looking at the "extra fin" logic: In Example 1, the vane was `(4,6), (5,4), (5,5), (5,6), (6,5)`. The center is Col 5. The extra fin was at Col 6 (Right). The output showed the cloud reflected to the right.
    In the Test Input, the vane is:
    Row 4: Col 6 (Y)
    Row 5: Col 4, 5, 6 (Y)
    Row 6: Col 5 (Y)
    The "extra fin" is at Col 6 (Right) and Col 4 (Left)? No, let's re-examine the description: "a plus-shaped post with one extra fin jutting to the left or right".
    In Test Input:
    Row 4: Col 6 is Y.
    Row 5: Col 4, 5, 6 are Y.
    Row 6: Col 5 is Y.
    The center is Col 5. The Y at (4,6) is the "extra fin" jutting to the **Right**.
    Therefore, the wind blows **Right**.

**Reflection Process (Wind = Right):**
Original Cloud:
R1: `...A.A...`
R2: `....AA...`
R3: `...A.....`

The reflection is a mirror image of the cloud, placed to the right.
Looking at the columns:
Original cloud occupies columns 4, 5, 6.
The reflection will occupy columns 7, 8, 9.
The reflection is a horizontal flip of the original shape.
Original R1: Col 4, 6. Mirror R1: Col 7, 9 (relative to the boundary).
Wait, let's look at Example 1 to see how the reflection is calculated.
Ex 1 Cloud:
R1: `...A.A...` (4, 6)
R2: `...AAA...` (4, 5, 6)
R3: `...AA....` (4, 5)
Ex 1 Output R1: `...A.A A . A` (4, 6, 7, 9)
Ex 1 Output R2: `...AAA A A A` (4, 5, 6, 7, 8, 9)
Ex 1 Output R3: `...AA .. A A` (4, 5, 8, 9)

The reflection is a horizontal flip of the cloud shape, mirrored across the vertical axis of the cloud's rightmost edge or simply placed adjacent.
In Ex 1, the cloud is in cols 4, 5, 6. The reflection is in cols 7, 8, 9.
The reflection of R1 (4, 6) is (7, 9).
The reflection of R2 (4, 5, 6) is (7, 8, 9).
The reflection of R3 (4, 5) is (8, 9).

**Test Input Cloud:**
R1: Col 4, 6
R2: Col 5, 6
R3: Col 4

**Reflection (Right):**
R1: Mirror of (4, 6) across the boundary of col 6 is (7, 9).
R2: Mirror of (5, 6) across the boundary of col 6 is (7, 8).
R3: Mirror of (4) across the boundary of col 6 is (9).

**Final Grid Construction:**
R1: `. . . A . A A . A`
R2: `. . . . A A A A .`
R3: `. . . A . . . . A`
R4: `. . . . . Y . . .`
R5: `. . . Y Y Y . . .`
R6: `. . . . Y . . . .`

Let's double check R2: Original is (5, 6). Mirroring (5, 6) to the right of 6:
The distance from the edge (6) to 5 is 1. So the mirror of 5 is 6 + 1 + 1 = 8? No.
Let's look at Ex 1 R2 again: Original (4,5,6). Reflection (7,8,9).
The reflection of a set of columns $\{c_1, c_2, ... c_n\}$ where $c_{max}$ is the right edge:
The reflected columns are $\{c_{max} + (c_{max} - c_i) + 1\}$.
Ex 1 R1: $c_{max}=6$. $c=\{4, 6\}$. Reflected: $6+(6-4)+1=9$ and $6+(6-6)+1=7$. (7, 9). Correct.
Ex 1 R2: $c_{max}=6$. $c=\{4, 5, 6\}$. Reflected: $6+(6-4)+1=9$, $6+(6-5)+1=8$, $6+(6-6)+1=7$. (7, 8, 9). Correct.
Ex 1 R3: $c_{max}=5$. $c=\{4, 5\}$. Reflected: $5+(5-4)+1=7$ (Wait, Ex 1 R3 output is 8, 9).
Let's re-examine Ex 1 R3: Input R3 is `...AA....` (4, 5

... [859 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Azure Nebula (A)**: The object to be reflected. It is located in the top rows of the grid.
    *   **Yellow Pulsar (Y)**: The "pointer" or "beam" that determines the direction of the reflection.
    *   **Dark Region (.)**: The empty space where the reflection will appear.

2.  **Identify the AFFORDANCES**:
    *   The **Azure Nebula** can be horizontally flipped (mirrored).
    *   The **Yellow Pulsar** can indicate a direction (left or right) based on its "asymmetric beam pattern." Looking at the examples, the "beam" is the part of the pulsar that extends further or is positioned towards one side.

3.  **Identify the TRANSFORMATION RULE**:
    *   **Step 1: Determine Direction**: Look at the Yellow Pulsar (Y). In Example 1, the pulsar is centered/slightly right, but the beam (the single Y at the top of the pulsar shape) is at column 5. Wait, let's look closer.
    *   **Example 1 Analysis**: Input pulsar is at (3,5), (4,3), (4,4), (4,5), (5,5). The "beam" is the single Y at (3,5). This is to the right of the center. The output shows the Azure nebula reflected on the **right** side.
    *   **Example 2 Analysis**: Input pulsar is at (3,3), (4,3), (4,4), (4,5), (5,5). The "beam" is at (3,3). This is to the left of the center. The output shows the Azure nebula reflected on the **left** side.
    *   **Rule**: Find the single Y pixel that acts as the "tip" of the pulsar. If it is on the left side of the pulsar's main body, reflect the nebula to the left. If it is on the right, reflect it to the right.
    *   **Step 2: Reflect the Nebula**: Take the Azure (A) pattern from the top rows, flip it horizontally, and place it in the empty sector on the chosen side.
    *   **Refining the Reflection**: In Example 1, the nebula is at (0,3), (0,5), (1,3), (1,4), (1,5), (2,3), (2,4). The reflection is placed on the right. The pattern is mirrored and placed in the available columns.

### Test Input Analysis

**Input Nebula (A):**
Row 0: `. . . A . A . . .` (Cols 3, 5)
Row 1: `. . . . A A . . .` (Cols 4, 5)
Row 2: `. . . A . . . . .` (Col 3)

**Input Pulsar (Y):**
Row 3: `. . . . . Y . . .` (Col 5)
Row 4: `. . . Y Y Y . . .` (Cols 3, 4, 5)
Row 5: `. . . . Y . . . .` (Col 4)

**Step 1: Determine Direction**
The "tip" of the pulsar is the single Y at (3, 5).
The main body of the pulsar is centered around column 4.
Since the tip (Col 5) is to the **right** of the center (Col 4), the reflection should appear on the **right** side.

**Step 2: Reflect the Nebula**
Original Nebula:
R0: `...A.A...`
R1: `....AA...`
R2: `...A.....`

Horizontal Flip of Nebula:
R0: `...A.A...` -> `...A.A...` (Symmetric around col 4? No, let's mirror the pattern relative to its own center or just flip the whole row).
Let's look at Example 1 again.
Ex 1 Nebula:
R0: `...A.A...`
R1: `...AAA...`
R2: `...AA....`
Ex 1 Output (Right side):
R0: `...A.A A.A` (The original was `...A.A...`. The reflection added `A.A` at the end).
Wait, the rule is simpler: The nebula is mirrored and placed in the empty space.
In Example 1, the nebula is mirrored and the result is appended/merged.
Let's re-examine Example 2:
Ex 2 Nebula:
R0: `...A.A...`
R1: `....AA...`
R2: `.....A...`
Ex 2 Output (Left side):
R0: `A.A A.A...`
R1: `AA..AA...`
R2: `A.....A...`
The pattern is: For each row, take the A's, mirror them, and place them on the opposite side.

**Applying to Test Input:**
Nebula:
R0: `...A.A...` (Cols 3, 5)
R1: `....AA...` (Cols 4, 5)
R2: `...A.....` (Col 3)

Pulsar tip is at Col 5 (Right).
Reflect Nebula to the right:
R0: The A's are at 3 and 5. Mirroring them across the center (Col 4) gives 5 and 3. This doesn't change much. However, the reflection is placed in the *empty* sector.
Looking at Example 1: The original nebula is at cols 3, 4, 5. The reflection is added to the right.
In Example 2: The original nebula is at cols 3, 4, 5. The reflection is added to the left.

Test Input Nebula:
R0: `...A.A...`
R1: `....AA...`
R2: `...A.....`

Reflection (Right side):
R0: Mirror of `...A.A...` is `...A.A...`. Placed on the right: `...A.A A.A`? No, let's look at the columns.
Example 1: Nebula R0 is `...A.A...`. Output R0 is `...A.A A.A`. It seems the reflection is a mirror image of the nebula's shape.
If the nebula is `A . A` (cols 3, 4, 5), the mirror is `A . A`.
In Test Input:
R0: `A . A` (cols 3, 4, 5) -> Mirror is `A . A`.
R1: `. A A` (cols 4, 5) -> Mirror is `A A .`.
R2: `A . .` (cols 3, 4, 5) -> Mirror is `. . A`.

Let's place these mirrors on the right side (cols 6, 7, 8):
R0: `...A.A A.A`
R1: `....AA AA.`
R2: `...A... . .A`

Wait, let's look at Example 1 again.
Ex 1 Nebula:
R0: `...A.A...`
R1: `...AAA...`
R2: `...AA....`
Ex 1 Output R0: `...A.A A.A` (The A's are at 3, 5, 6, 8)
Ex 1 Output R1: `...AAA AAAA` (The A's are at 3, 4, 5, 6, 7, 8)
Ex 1 Output R2: `...AA ..AA` (The A's are at 3, 4, 7, 8)

This means the reflection is a mirror image of the nebula, placed in the columns on the opposite side 

... [3400 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Azure Signal (A)**: A pattern located in the top half of the grid (rows 0-2).
    *   **Yellow Gate (Y)**: A pattern located in the bottom half of the grid (rows 3-5).
    *   **The Offset Pin**: Within the Yellow Gate, there is a specific configuration. The "pin" is the single-pixel element or the asymmetry in the gate's shape.
    *   **Buses**: The columns to the left (0-2) and right (6-8) of the central columns (3-5).

2.  **Identify the AFFORDANCES**:
    *   The **Azure Signal** can be copied and horizontally flipped.
    *   The **Yellow Gate** acts as a selector. Its asymmetry (the "offset pin") dictates the direction of the transformation.

3.  **Identify the TRANSFORMATION RULE**:
    *   Locate the Azure signal in rows 0-2.
    *   Locate the Yellow gate in rows 3-5.
    *   Determine the "offset" of the Yellow gate. In the training example, the gate is a cross shape with a single pixel at (3, 5). This asymmetry points to the right.
    *   The rule: If the gate's asymmetry/offset is on the right side of the center, the Azure signal is horizontally flipped and placed in the right-hand bus (columns 6-8). If the offset is on the left, it goes to the left-hand bus (columns 0-2).
    *   *Refining the rule based on Example 1*: The gate in Example 1 has a single pixel at (3, 5). This is the "offset pin". Since it is in column 5 (the right side of the center 3-5), the signal is mirrored into the right bus (columns 6-8).
    *   *Wait, let's look closer at Example 1*:
        Input Signal (cols 3-5):
        Row 0: `. A .`
        Row 1: `A A A`
        Row 2: `A A .`
        Output (cols 6-8) is the horizontal flip of the signal:
        Row 0: `A . A` (Flip of `. A .` is `. A .`, but it's placed in 6-8. Let's check: Input col 3 is `.`, 4 is `A`, 5 is `.`. Flip is `. A .`. Output col 6 is `A`, 7 is `.`, 8 is `A`. This means the signal is flipped *and* the columns are mapped.
        Let's re-examine:
        Input Signal:
        R0: `[3]=. [4]=A [5]=.`
        R1: `[3]=A [4]=A [5]=A`
        R2: `[3]=A [4]=A [5]=.`
        Output (cols 6-8):
        R0: `[6]=A [7]=. [8]=A`
        R1: `[6]=A [7]=A [8]=A`
        R2: `[6]=. [7]=A [8]=A`
        Looking at the columns:
        Input Col 3 -> Output Col 8
        Input Col 4 -> Output Col 7
        Input Col 5 -> Output Col 6
        This is a horizontal flip of the 3-column block.

### Applying the Rule to Test Input

**1. Analyze the Azure Signal (Rows 0-2, Cols 3-5):**
R0: `. . . A . A . . .` -> `[3]=A, [4]=., [5]=A`
R1: `. . . . A A . . .` -> `[3]=., [4]=A, [5]=A`
R2: `. . . A . . . . .` -> `[3]=A, [4]=., [5]=.`

**2. Analyze the Yellow Gate (Rows 3-5, Cols 3-5):**
R3: `. . . . . Y . . .` -> `[5]=Y`
R4: `. . . Y Y Y . . .` -> `[3]=Y, [4]=Y, [5]=Y`
R5: `. . . . Y . . . .` -> `[4]=Y`
The "offset pin" is the single pixel at (3, 5). This is on the right side of the center. Therefore, the signal is mirrored into the right bus (cols 6-8).

**3. Mirror the Signal into Cols 6-8:**
The signal in cols 3-5 is:
R0: `A . A`
R1: `. A A`
R2: `A . .`

Horizontal flip (mapping 3->8, 4->7, 5->6):
R0: `[6]=A, [7]=., [8]=A`
R1: `[6]=A, [7]=A, [8]=.`
R2: `[6]=., [7]=., [8]=A`

**4. Construct the final grid:**
R0: `. . . A . A A . A`
R1: `. . . . A A A A .`
R2: `. . . A . . . . A`
R3: `. . . . . Y . . .`
R4: `. . . Y Y Y . . .`
R5: `. . . . Y . . . .`

ANSWER:
. . . A . A A . A
. . . . A A A A .
. . . A . . . . A
. . . . . Y . . .
. . . Y Y Y . . .
. . . . Y . . . .

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Azure Spirit (A)**: An object located in the top half of the grid (the "upper realm").
    *   **Yellow Oracle (Y)**: An object located in the bottom half of the grid (the "lower realm").
    *   **The Staff (part of Y)**: The specific configuration of the Yellow Oracle. The "tilt" or orientation of the Y shape determines the direction of the reflection.
    *   **The Horizon**: The boundary between the top and bottom halves.

2.  **Identify the AFFORDANCES**:
    *   The **Azure Spirit** can be mirrored/duplicated.
    *   The **Yellow Oracle** acts as a directional indicator (the "staff").
    *   The **Empty Columns** are the canvas for the reflected spirit.

3.  **Identify the TRANSFORMATION RULE**:
    *   Locate the Azure Spirit (A) in the top 3 rows.
    *   Locate the Yellow Oracle (Y) in the bottom 3 rows.
    *   Determine the "tilt" of the Y. In the training example, the Y is centered but has a specific shape. Looking at the transformation: The Y in Example 1 is a cross shape. The "staff" (the vertical line of the Y) is in the center. However, the rule says "the staff indicates... materializing beside the original."
    *   Let's re-examine Example 1:
        Input A: (0,3), (0,5), (1,3), (1,4), (1,5), (2,3), (2,4).
        Input Y: (3,5), (4,3), (4,4), (4,5), (5,4).
        Output A: The original A is kept. New A's appear at (0,6), (0,8), (1,6), (1,7), (1,8), (2,7), (2,8).
        Wait, let's look at the columns. The original A occupies columns 3, 4, 5. The new A occupies columns 6, 7, 8.
        The Y shape in Example 1: The "center" of the Y is at (4,4). The "staff" (vertical part) is at (3,5), (4,5), (5,4)? No.
        Let's look at the Y shape:
        Row 3: Col 5
        Row 4: Col 3, 4, 5
        Row 5: Col 4
        The "staff" is the vertical line. In Example 1, the Y is slightly offset to the right. The reflection happens to the **right** of the original A.
    *   **Rule Refined**: The Y shape indicates a direction. If the Y is "tilted" or its center of mass is to the right, the reflection goes right. If left, it goes left.
    *   In Example 1, the Y is centered at col 4, but the "top" of the Y is at col 5. This suggests a rightward direction.
    *   In the Test Input:
        A is at (0,3), (0,5), (1,4), (1,5), (2,3).
        Y is at (3,5), (4,3), (4,4), (4,5), (5,4).
        The Y shape is identical to Example 1. Therefore, the reflection should also be to the right.
    *   **Reflection Logic**: The reflection is a horizontal mirror of the Azure Spirit across a vertical axis. In Example 1, the A was in cols 3-5. The reflection was in cols 6-8. This is a mirror across the line between col 5 and col 6.

### Applying the Rule to Test Input

**Input A positions:**
(0,3), (0,5)
(1,4), (1,5)
(2,3)

**Reflection (Mirroring across the boundary of the existing A shape):**
The existing A occupies columns 3, 4, and 5. The reflection will occupy columns 6, 7, and 8.
To mirror the shape:
If an A is at `(row, col)`, its reflection is at `(row, 9 - 1 - col)`? No, that's a full grid flip.
The rule is "materializing beside the original".
In Example 1:
Original A:
Col 3: (0,3), (1,3), (2,3)
Col 4: (1,4), (2,4)
Col 5: (0,5), (1,5)
Reflected A:
Col 6: (0,6), (1,6), (2,6) -- wait, let's check Example 1 output.
Example 1 Output A:
(0,3), (0,5), (0,6), (0,8)
(1,3), (1,4), (1,5), (1,6), (1,7), (1,8)
(2,3), (2,4), (2,7), (2,8)

Let's map the columns for Example 1:
Original: {3, 4, 5}
Reflected: {6, 7, 8}
Mapping:
Col 3 -> Col 8
Col 4 -> Col 7
Col 5 -> Col 6
Let's check:
Original (0,3) -> Reflected (0,8). Correct.
Original (0,5) -> Reflected (0,6). Correct.
Original (1,3) -> Reflected (1,8). Correct.
Original (1,4) -> Reflected (1,7). Correct.
Original (1,5) -> Reflected (1,6). Correct.
Original (2,3) -> Reflected (2,8). Correct.
Original (2,4) -> Reflected (2,7). Correct.

**Test Input A positions:**
(0,3), (0,5)
(1,4), (1,5)
(2,3)

**Reflected A positions (using Col 3->8, 4->7, 5->6):**
(0,3) -> (0,8)
(0,5) -> (0,6)
(1,4) -> (1,7)
(1,5) -> (1,6)
(2,3) -> (2,8)

**Final Grid Construction:**
Row 0: . . . A . A A . A
Row 1: . . . . A A A A .
Row 2: . . . A . . . . A
Row 3: . . . . . Y . . .
Row 4: . . . Y Y Y . . .
Row 5: . . . . Y . . . .

Wait, let me re-verify Row 1 for Test Input.
Input Row 1: `. . . . A A . . .` (Cols 4, 5)
Reflected Row 1: (1, 9-1-4=4? No, the mapping is relative to the block).
The block is columns 3, 4, 5.
Col 3 is index 0 in the block. Col 5 is index 2.
Reflected index 0 is Col 8. Reflected index 2 is Col 6.
So:
Col 3 -> Col 8
Col 4 -> Col 7
Col 5 -> Col 6
Test Input Row 1 has A at Col 4 and Col 5.
Reflected Row 1 will have A at Col 7 and Col 6.
So Row 1: `. . . . A A A A .` (Cols 4, 5, 6, 7).

Let's re-verify Row 0:
Input Row 0: `. . . A . A . . .` (Cols 3, 5)
Reflected Row 0: (0, 8) and (0, 6).
So Row 0: `. . . A . A A . A` (Cols 3, 5, 6, 8).

Let's re-ver

... [394 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Azure Pastry (A)**: The object to be duplicated/mirrored.
    *   **Black Countertop (.)**: The background/canvas.
    *   **Yellow Mold (Y)**: The indicator/template that determines the direction and position of the mirror.
    *   **Spout (The single Y pixel in the center of the Y shape)**: The directional pointer.

2.  **Identify the AFFORDANCES**:
    *   The **Azure Pastry** can be mirrored across a vertical axis.
    *   The **Yellow Mold** provides a directional cue. Looking at the mold shape (a cross/plus shape), the "spout" is the single pixel at the center. However, the metaphor says the spout indicates which side receives the copy. In the training example, the mold is a cross shape. The "spout" is the single pixel at the center of the cross. The direction is determined by the relative position of the mold's center to the pastry.
    *   Wait, looking closer at Example 1: The mold is a cross. The pastry is on the top. The output shows the pastry is mirrored to the **right**. In Example 1, the mold's center is at (row 3, col 5) [0-indexed]. The pastry is at (0-2, 3-5). The "spout" is the single pixel in the middle of the Y shape. In the training example, the Y shape is a cross. The single pixel is at (3, 5). The pastry is to the left of this center. The mirror is placed on the right side of the original pastry.

3.  **Identify the TRANSFORMATION RULE**:
    *   Locate the Azure (A) shape and the Yellow (Y) shape.
    *   The Yellow shape is a cross (a center pixel with neighbors). The "spout" is the center pixel of the Y shape.
    *   The direction of the mirror is determined by the position of the Y shape relative to the A shape.
    *   In Example 1: A is at col 3-5. Y center is at col 5. Since the Y center is to the right of the A shape's center, the mirror is placed on the right side of the A shape.
    *   The rule: Mirror the A shape vertically across the vertical line that is adjacent to the original A shape on the side indicated by the Y shape's center.
    *   Looking at Example 1 again: Original A is at col 3, 4, 5. The Y center is at col 5. The mirror is placed at col 6, 7, 8. The mirror is a reflection of the A shape.
    *   Let's check the reflection:
        Original A:
        (0,3)=A, (0,5)=A
        (1,3)=A, (1,4)=A, (1,5)=A
        (2,3)=A, (2,4)=A
        Mirror (right side):
        (0,6)=A, (0,8)=A
        (1,6)=A, (1,7)=A, (1,8)=A
        (2,6)=A, (2,7)=A
        This is a horizontal reflection of the A shape across the axis of the rightmost edge of the A shape.

### Applying the Rule to Test Input

**Input Analysis**:
A shape:
(0,3)=A, (0,5)=A
(1,4)=A, (1,5)=A
(2,3)=A

Y shape:
(3,5)=Y
(4,3)=Y, (4,4)=Y, (4,5)=Y
(5,4)=Y

**Step 1: Determine direction.**
The center of the Y shape is at (3, 5).
The center of the A shape is roughly at col 4.
Since the Y center (col 5) is to the right of the A shape, the mirror will be placed on the right side of the A shape.

**Step 2: Mirror the A shape.**
The A shape occupies columns 3, 4, and 5.
The rightmost edge of the A shape is at column 5.
The mirror will be placed starting from column 6, moving outwards.
The reflection of the A shape (columns 3, 4, 5) across the axis of column 5:
- Col 5 stays Col 5.
- Col 4 becomes Col 6.
- Col 3 becomes Col 7.
- Col 5 (the edge) is the pivot.

Let's map the A pixels:
(0,3) -> (0,7)
(0,5) -> (0,5)
(1,4) -> (1,6)
(1,5) -> (1,5)
(2,3) -> (2,7)

Wait, let's look at the "flush against the original" instruction.
In Example 1:
Original A columns: 3, 4, 5.
Mirror A columns: 6, 7, 8.
The mirror is a reflection of the original.
Original A:
Row 0: . . . A . A
Row 1: . . . A A A
Row 2: . . . A A .

Mirroring this across the right edge (col 5):
Row 0: (0,5) is A, (0,6) is ., (0,7) is A. (Wait, the example output for row 0 is `. . . A . A A . A`)
Let's re-examine Example 1 output row 0: `. . . A . A A . A`
Original row 0: `. . . A . A . . .`
Mirror row 0: `. . . . . A A . A` (Wait, the example output is `A . A A . A`)
Let's re-count the dots in Example 1 output row 0:
`. . . A . A A . A`
Indices: 0 1 2 3 4 5 6 7 8
A is at 3, 5, 6, 8.
Original A was at 3, 5.
If we mirror 3 and 5 across the axis between 5 and 6:
5 stays 5? No.
If the axis is between 5 and 6:
5 becomes 6.
4 becomes 7.
3 becomes 8.
Original A at 3, 5.
Mirror A at 6, 8.
Total A: 3, 5, 6, 8. This matches the example!

**Step 3: Apply to Test Input.**
A shape:
(0,3), (0,5)
(1,4), (1,5)
(2,3)

The rightmost column of A is 5. The mirror axis is between col 5 and col 6.
Mirroring across the axis between 5 and 6:
Col 5 -> Col 6
Col 4 -> Col 7
Col 3 -> Col 8

Mirror A pixels:
(0,5) -> (0,6)
(0,3) -> (0,8)
(1,5) -> (1,6)
(1,4) -> (1,7)
(2,3) -> (2,8)

Combine original and mirror:
Row 0: (0,3), (0,5), (0,6), (0,8)
Row 1: (1,4), (1,5), (1,6), (1,7)
Row 2: (2,3), (2,8)

Let's double check the A shape in Test Input:
Row 0: `. . . A . A . . .` (3, 5)
Row 1: `. . . 

... [415 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Azure Fragment (A)**: The "architectural fragment" located in the top three rows. This is the object to be mirrored.
    *   **Yellow Cornerstone (Y)**: The "foundation" located in the bottom three rows. It contains a "notch" (an empty space or specific configuration) that determines the direction of the transformation.
    *   **The Notch**: The position of the empty space within the yellow cornerstone structure.
    *   **The Grid**: A 6x9 blueprint.

2.  **Identify the AFFORDANCES**:
    *   The **Azure Fragment** can be reflected/mirrored horizontally.
    *   The **Yellow Cornerstone** acts as a directional indicator based on its internal pattern (the "notch").
    *   The **Empty Columns** are the target area for the mirrored azure wing.

3.  **Identify the TRANSFORMATION RULE**:
    *   Locate the Azure (A) pattern in the top 3 rows.
    *   Locate the Yellow (Y) pattern in the bottom 3 rows.
    *   Determine the "notch" in the Yellow pattern. In Example 1, the Y pattern is centered but has a specific shape. Looking at the output of Example 1: The original A pattern is in columns 4, 5, 6. The output adds A's to columns 7, 8, 9. This means the reflection happened to the **right**.
    *   Looking at the Y pattern in Example 1:
        Row 4: `. . . . . Y . . .` (Y is at col 6)
        Row 5: `. . . Y Y Y . . .` (Y is at col 4, 5, 6)
        Row 6: `. . . . Y . . . .` (Y is at col 5)
        The "notch" (the gap) in the Y structure is to the **left** of the center of the Y mass. The rule states: "The notch's position dictates which side gains a reflected copy". In Example 1, the notch is on the left, so the reflection is on the right.
    *   **Refined Rule**: Find the horizontal center of the Azure fragment. Find the "notch" (the side with fewer Y pixels or the empty space relative to the Y shape). If the notch is on the left, mirror the Azure fragment to the right. If the notch is on the right, mirror it to the left.
    *   **Wait, let's re-examine Example 1**:
        Input A: (0,3), (0,5), (1,3), (1,4), (1,5), (2,3), (2,4).
        Output A: Original + (0,6), (0,8), (1,6), (1,7), (1,8), (2,7), (2,8).
        The reflection is a horizontal mirror of the Azure shape across its own rightmost edge.
        The Y pattern in Example 1: Col 6, Col 4-6, Col 5. The "notch" is the empty space in the Y pattern. The Y pattern is slightly shifted or has a gap. In Example 1, the Y pattern is mostly on the left/center, leaving the right side "open".
    *   **Test Input Analysis**:
        A pattern: (0,3), (0,5), (1,4), (1,5), (2,3).
        Y pattern: (3,5), (4,3), (4,4), (4,5), (5,4).
        The Y pattern is centered around column 4/5.
        Let's look at the Y pattern in the Test Input:
        Row 4: `. . . . . Y . . .` (Col 6)
        Row 5: `. . . Y Y Y . . .` (Col 4, 5, 6)
        Row 6: `. . . . Y . . . .` (Col 5)
        This is the same Y pattern as Example 1. Therefore, the reflection should be to the **right**.

    *   **Mirroring Logic**:
        The Azure fragment in the test input:
        Row 0: `. . . A . A . . .` (Cols 4, 6) -> Mirroring across the right edge (Col 6) would place A at Col 8.
        Row 1: `. . . . A A . . .` (Cols 5, 6) -> Mirroring across the right edge would place A at Col 7.
        Row 2: `. . . A . . . . .` (Col 4) -> Mirroring across the right edge would place A at Col 8.
        
        Let's check the symmetry of the Azure fragment in Example 1 to see how it mirrors:
        Ex 1 A:
        R0: `A . A` (Cols 4, 5, 6) -> Output R0: `A . A A . A` (Wait, the output is `A . A A . A`? No, it's `A . A A . A`? Let's re-read).
        Ex 1 Input R0: `. . . A . A . . .`
        Ex 1 Output R0: `. . . A . A A . A`
        The A's are at 4, 6 and then 7, 9. This is a reflection of the pattern `A . A` (cols 4, 5, 6) across the axis of col 6.
        The pattern `A . A` mirrored across col 6 is `A . A`. So `A . A` + `A . A` (with overlap at 6) = `A . A A . A`.
        
        Test Input A:
        R0: `. . . A . A . . .` (Cols 4, 6) -> Mirroring `A . A` across col 6 gives `A . A A . A` (Cols 4, 6, 7, 9? No, let's be precise).
        If the pattern is `A (col 4), . (col 5), A (col 6)`, the mirror image across col 6 is `A (col 6), . (col 7), A (col 8)`.
        Combined: `A (4), . (5), A (6), . (7), A (8)`.
        
        R1: `. . . . A A . . .` (Cols 5, 6) -> Mirroring `A A` across col 6 gives `A (6), A (7)`.
        Combined: `A (5), A (6), A (7)`.
        
        R2: `. . . A . . . . .` (Col 4) -> Mirroring `A` across col 6 (distance 2) gives `A (8)`.
        Combined: `A (4), A (8)`.

### Applying the Rule to Test Input

**Input A positions:**
(0,3), (0,5) [using 0-indexing: (0,3) is 4th col, (0,5) is 6th col]
(1,4), (1,5)
(2,3)

**Wait, let's re-index carefully (0-8):**
Input A:
Row 0: Col 3, Col 5
Row 1: Col 4, Col 5
Row 2: Col 3

**Mirroring across the rightmost edge of the A-block (

... [2702 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Input Grid**: A 4x4 "mosaic tile" (the original swatch).
    *   **Output Grid**: An 8x8 "cathedral floor" composed of four 4x4 tiles.
    *   **Cells**: Individual colored glass pieces.

2.  **Identify the AFFORDANCES**:
    *   The input tile can be **rotated** in 90-degree increments (0°, 90°, 180°, 270°).
    *   The four resulting tiles can be **arranged** in a 2x2 grid.

3.  **Identify the TRANSFORMATION RULE**:
    *   The output is a 2x2 arrangement of the input tile and its three rotations.
    *   Based on the training examples:
        *   **Top-Left (0,0)**: The original input tile (0° rotation).
        *   **Top-Right (0,1)**: The input tile rotated 90° clockwise.
        *   **Bottom-Left (1,0)**: The input tile rotated 180° clockwise.
        *   **Bottom-Right (1,1)**: The input tile rotated 270° clockwise.
    *   *Verification with Example 1*:
        Input:
        `Y W B A`
        `A Y B A`
        `Y A A B`
        `B B B A`
        Top-Left: Original.
        Top-Right (90° CW):
        `B Y A Y`
        `B A A W`
        `B B A B`
        `A A B A`
        Wait, let's re-examine the rotation direction and placement.
        Example 1 Output Top-Right:
        `A A B A`
        `B B A B`
        `W Y A B`
        `A Y A Y`
        This is the input rotated 90 degrees **counter-clockwise** (or 270 CW).
        Let's check the Bottom-Left (1,0) of Example 1:
        `B B B A`
        `Y A A B`
        `A Y B A`
        `A A B A`
        This is the input rotated 180 degrees.
        Let's check the Bottom-Right (1,1) of Example 1:
        `Y A Y B`
        `A Y A B`
        `B B A B`
        `A B W Y`
        This is the input rotated 90 degrees **clockwise**.

    *   **Corrected Rule**:
        *   Top-Left (0,0): Original (0°)
        *   Top-Right (0,1): 90° Counter-Clockwise (or 270° CW)
        *   Bottom-Left (1,0): 180°
        *   Bottom-Right (1,1): 90° Clockwise

### Applying the Rule to Test Input

**Input (4x4):**
Row 0: `Y M Y Y`
Row 1: `Y M Y Y`
Row 2: `O M O W`
Row 3: `W Y W O`

**1. Top-Left (Original):**
`Y M Y Y`
`Y M Y Y`
`O M O W`
`W Y W O`

**2. Top-Right (90° CCW):**
(Original Col 0 becomes New Row 3, Col 0 becomes New Row 0, etc.)
Mapping: `New[r][c] = Old[c][3-r]`
`Y W O W`
`M Y M Y`
`Y O M Y`
`Y W Y Y`
*Wait, let's do it manually:*
Col 0 (Y, Y, O, W) becomes Row 3 (reversed): `W O Y Y`? No.
Standard 90 CCW:
`Y Y O W` (Col 3)
`M M M Y` (Col 2)
`Y Y O W` (Col 1)
`Y O W Y` (Col 0)
Let's re-verify Example 1 Top-Right:
Input:
`Y W B A`
`A Y B A`
`Y A A B`
`B B B A`
90 CCW:
Col 3: `A A B A` (Row 0)
Col 2: `B B A B` (Row 1)
Col 1: `W Y A A` (Row 2)
Col 0: `Y A Y B` (Row 3)
Matches Example 1!

**Top-Right (90 CCW) for Test Input:**
Col 3: `Y Y W O` -> `Y Y W O`
Col 2: `Y Y O W` -> `Y Y O W`
Col 1: `M M M Y` -> `M M M Y`
Col 0: `Y Y O W` -> `Y Y O W`
Wait, let's be precise:
Input:
(0,0)Y (0,1)M (0,2)Y (0,3)Y
(1,0)Y (1,1)M (1,2)Y (1,3)Y
(2,0)O (2,1)M (2,2)O (2,3)W
(3,0)W (3,1)Y (3,2)W (3,3)O

90 CCW:
Row 0: (0,3) (1,3) (2,3) (3,3) -> `Y Y W O`
Row 1: (0,2) (1,2) (2,2) (3,2) -> `Y Y O W`
Row 2: (0,1) (1,1) (2,1) (3,1) -> `M M M Y`
Row 3: (0,0) (1,0) (2,0) (3,0) -> `Y Y O W`

**3. Bottom-Left (180°):**
Row 0: (3,3) (3,2) (3,1) (3,0) -> `O W Y W`
Row 1: (2,3) (2,2) (2,1) (2,0) -> `W O M O`
Row 2: (1,3) (1,2) (1,1) (1,0) -> `Y Y M Y`
Row 3: (0,3) (0,2) (0,1) (0,0) -> `Y Y M Y`

**4. Bottom-Right (90° CW):**
Row 0: (3,0) (2,0) (1,0) (0,0) -> `W O Y Y`
Row 1: (3,1) (2,1) (1,1) (0,1) -> `Y M M M`
Row 2: (3,2) (2,2) (1,2) (0,2) -> `W O Y Y`
Row 3: (3,3) (2,3) (1,3) (0,3) -> `O W Y Y`

**Assembling the 8x8 Grid:**
Rows 0-3 (Top-Left + Top-Right):
`Y M Y Y Y Y W O`
`Y M Y Y Y Y O W`
`O M O W M M M Y`
`W Y W O Y Y O W`

Rows 4-7 (Bottom-Left + Bottom-Right):
`O W Y W W O Y Y`
`W O M O Y M M M`
`Y Y M Y W O Y Y`
`Y Y M Y O W Y Y`

Wait, let me re-check the 90 CCW for Top-Right one more time.
Input:
`Y M Y Y`
`Y M Y Y`
`O M O W`
`W Y W O`
Col 3: `Y Y W O`
Col 2: `Y Y O W`
Col 1: `M M M Y`
Col 0: `Y Y O W`
Top-Right:
`Y Y W O`
`Y Y O W`
`M M M Y`
`Y Y O W`

Re-assembling:
Row 0: `Y M Y Y` + `Y Y W O` = `Y M Y Y Y Y W O`
Row 1: `Y M Y Y` + `Y Y O W` = `Y M Y Y Y Y O W`
Row 2: `O M O W` + `M M M Y` = `O M O W M M M Y`
Row 3: `W Y W O` + `Y Y O W` = `W Y W O Y Y O W`

Row 4: `O W Y W` + `W O Y Y` = `O W Y W W O Y Y`
Row 5: `W O M O` + `Y M M M` = `W O M O Y M M M`
Row 6: `Y Y M Y` + `W O Y Y` = `Y Y M Y W O Y Y`
Row 7: `Y Y M Y` + `O W Y Y` = `Y Y M Y O W Y Y`

ANSWER:
Y M Y Y Y Y W O
Y M Y Y Y Y O W
O M O W M M M Y
W Y W O Y Y O W
O W Y W W O Y Y
W O M O Y M M M
Y Y M Y W O Y Y
Y Y M Y O W Y Y

To solve this puzzle, we first perform a Conceptual Digestion.

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Input Grid**: A small "tissue sample" or "organism" of size $N \times N$.
    *   **Output Grid**: A larger "organism" of size $2N \times 2N$, composed of four quadrants.
    *   **Quadrants**: The four quadrants of the output grid represent the "daughter colonies."

2.  **Identify the AFFORDANCES**:
    *   The input grid can be rotated.
    *   The input grid can be placed into the four quadrants of a larger grid.

3.  **Identify the TRANSFORMATION RULE**:
    *   The output is a $2N \times 2N$ grid divided into four $N \times N$ quadrants.
    *   **Top-Left Quadrant**: The original input grid.
    *   **Top-Right Quadrant**: The input grid rotated 90 degrees clockwise.
    *   **Bottom-Left Quadrant**: The input grid rotated 180 degrees (or 90 degrees counter-clockwise from the top-right).
    *   **Bottom-Right Quadrant**: The input grid rotated 270 degrees clockwise (or 90 degrees clockwise from the bottom-left).
    *   *Correction based on training examples*: Let's re-examine Example 1.
        Input (4x4):
        Y W B A
        A Y B A
        Y A A B
        B B B A
        Output (8x8):
        Top-Left: Input
        Top-Right: Input rotated 90 deg clockwise? Let's check (0,0) of Top-Right. In Example 1, (0,4) is 'A'. In Input, (3,0) is 'B'. That's not it.
        Let's look at the rotation pattern again: "each carrying the same genetic pattern but rotated on its axis".
        Example 1 Top-Right (0,4 to 3,7):
        A A A B
        B B A B
        W Y A B
        Y A Y B
        This is the input rotated 90 degrees **counter-clockwise**.
        Example 1 Bottom-Left (4,0 to 7,3):
        A B B B
        B A A Y
        A B Y A
        A B W Y
        This is the input rotated 180 degrees.
        Example 1 Bottom-Right (4,4 to 7,7):
        B Y A Y
        A Y W B
        B A B B
        A B A A
        This is the input rotated 90 degrees **clockwise**.

    **Refined Rule**:
    *   Quadrant 1 (Top-Left): Original Input.
    *   Quadrant 2 (Top-Right): Input rotated 90° Counter-Clockwise.
    *   Quadrant 3 (Bottom-Left): Input rotated 180°.
    *   Quadrant 4 (Bottom-Right): Input rotated 90° Clockwise.

### Applying the Rule to Test Input

**Input (4x4):**
Row 0: Y M Y Y
Row 1: Y M Y Y
Row 2: O M O W
Row 3: W Y W O

**1. Top-Left (Original):**
Y M Y Y
Y M Y Y
O M O W
W Y W O

**2. Top-Right (90° CCW):**
(Input[row][col] -> Output[N-1-col][row])
Col 3 (top to bottom) becomes Row 0: Y Y W O
Col 2 (top to bottom) becomes Row 1: Y Y O W
Col 1 (top to bottom) becomes Row 2: M M M Y
Col 0 (top to bottom) becomes Row 3: Y Y O W
Wait, let's do it carefully:
Input:
(0,0)Y (0,1)M (0,2)Y (0,3)Y
(1,0)Y (1,1)M (1,2)Y (1,3)Y
(2,0)O (2,1)M (2,2)O (2,3)W
(3,0)W (3,1)Y (3,2)W (3,3)O

90 CCW:
New(0,0) = Old(0,3) = Y
New(0,1) = Old(1,3) = Y
New(0,2) = Old(2,3) = W
New(0,3) = Old(3,3) = O
New(1,0) = Old(0,2) = Y
New(1,1) = Old(1,2) = Y
New(1,2) = Old(2,2) = O
New(1,3) = Old(3,2) = W
New(2,0) = Old(0,1) = M
New(2,1) = Old(1,1) = M
New(2,2) = Old(2,1) = M
New(2,3) = Old(3,1) = Y
New(3,0) = Old(0,0) = Y
New(3,1) = Old(1,0) = Y
New(3,2) = Old(2,0) = O
New(3,3) = Old(3,0) = W

**3. Bottom-Left (180°):**
New(0,0) = Old(3,3) = O
New(0,1) = Old(3,2) = W
New(0,2) = Old(3,1) = Y
New(0,3) = Old(3,0) = W
New(1,0) = Old(2,3) = W
New(1,1) = Old(2,2) = O
New(1,2) = Old(2,1) = M
New(1,3) = Old(2,0) = O
New(2,0) = Old(1,3) = Y
New(2,1) = Old(1,2) = Y
New(2,2) = Old(1,1) = M
New(2,3) = Old(1,0) = Y
New(3,0) = Old(0,3) = Y
New(3,1) = Old(0,2) = Y
New(3,2) = Old(0,1) = M
New(3,3) = Old(0,0) = Y

**4. Bottom-Right (90° CW):**
New(0,0) = Old(3,0) = W
New(0,1) = Old(2,0) = O
New(0,2) = Old(1,0) = Y
New(0,3) = Old(0,0) = Y
New(1,0) = Old(3,1) = Y
New(1,1) = Old(2,1) = M
New(1,2) = Old(1,1) = M
New(1,3) = Old(0,1) = M
New(2,0) = Old(3,2) = W
New(2,1) = Old(2,2) = O
New(2,2) = Old(1,2) = Y
New(2,3) = Old(0,2) = Y
New(3,0) = Old(3,3) = O
New(3,1) = Old(2,3) = W
New(3,2) = Old(1,3) = Y
New(3,3) = Old(0,3) = Y

**Assembling the 8x8 grid:**
Row 0: Y M Y Y | Y Y W O
Row 1: Y M Y Y | Y Y O W
Row 2: O M O W | M M M Y
Row 3: W Y W O | Y Y O W
Row 4: O W Y W | W O Y Y
Row 5: W O M O | Y M M M
Row 6: Y Y M Y | W O Y Y
Row 7: Y Y M Y | O W Y Y

ANSWER:
Y M Y Y Y Y W O
Y M Y Y Y Y O W
O M O W M M M Y
W Y W O Y Y O W
O W Y W W O Y Y
W O M O Y M M M
Y Y M Y W O Y Y
Y Y M Y O W Y Y

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Input Grid**: A single "exposure" (a square grid of size $N \times N$).
    *   **Output Grid**: A "mosaic" (a square grid of size $2N \times 2N$) composed of four tiles.
    *   **Tiles**: The four quadrants of the output grid.

2.  **Identify the AFFORDANCES**:
    *   The input grid can be **rotated** by specific angles (90, 180, 270 degrees).
    *   The input grid can be **tiled** into a larger $2 \times 2$ arrangement.

3.  **Identify the TRANSFORMATION RULE**:
    *   The output is a $2 \times 2$ tiling of the input grid, but each quadrant is a rotated version of the input.
    *   Looking at Example 1:
        *   Input ($4 \times 4$):
            `Y W B A`
            `A Y B A`
            `Y A A B`
            `B B B A`
        *   Output ($8 \times 8$):
            Top-Left (TL): Original input (0° rotation).
            Top-Right (TR): Input rotated 90° clockwise.
            Bottom-Left (BL): Input rotated 180° clockwise.
            Bottom-Right (BR): Input rotated 270° clockwise.
    *   *Correction based on Example 1 analysis*:
        Let's check the Top-Left of Example 1: `Y W B A / A Y B A / Y A A B / B B B A`. This matches the input exactly.
        Let's check the Top-Right of Example 1:
        Input:
        `Y W B A`
        `A Y B A`
        `Y A A B`
        `B B B A`
        Rotated 90° CW:
        `B Y A Y`
        `B A Y W`
        `B A B A`
        `A B A A`
        Wait, let's re-examine the example output's Top-Right: `A A B A / B B A B / W Y A B / Y A Y B`.
        Let's check the Bottom-Left: `A B B B / B A A Y / A B Y A / A B W Y`.
        Let's check the Bottom-Right: `B Y A Y / A Y W B / A B A A / A B A A`.
        
        Let's re-evaluate the rotation pattern:
        Input:
        R1: Y W B A
        R2: A Y B A
        R3: Y A A B
        R4: B B B A
        
        Output TL (Rows 1-4, Cols 1-4): `Y W B A / A Y B A / Y A A B / B B B A` (0°)
        Output TR (Rows 1-4, Cols 5-8): `A A B A / B B A B / W Y A B / Y A Y B` (This is the input rotated 90° **Counter-Clockwise**)
        Output BL (Rows 5-8, Cols 1-4): `A B B B / B A A Y / A B Y A / A B W Y` (This is the input rotated 180°)
        Output BR (Rows 5-8, Cols 5-8): `B Y A Y / A Y W B / A B A A / A B A A` (This is the input rotated 90° **Clockwise**)

    *   **Final Rule**:
        - Top-Left: 0° rotation.
        - Top-Right: 90° Counter-Clockwise rotation.
        - Bottom-Left: 180° rotation.
        - Bottom-Right: 90° Clockwise rotation.

### Applying the Rule to Test Input

**Input ($4 \times 4$):**
`Y M Y Y`
`Y M Y Y`
`O M O W`
`W Y W O`

**1. Top-Left (0°):**
`Y M Y Y`
`Y M Y Y`
`O M O W`
`W Y W O`

**2. Top-Right (90° CCW):**
Original:
(0,0)Y (0,1)M (0,2)Y (0,3)Y
(1,0)Y (1,1)M (1,2)Y (1,3)Y
(2,0)O (2,1)M (2,2)O (2,3)W
(3,0)W (3,1)Y (3,2)W (3,3)O

90° CCW mapping: `New(r, c) = Old(c, N-1-r)`
(0,0) $\leftarrow$ (0,3) = Y
(0,1) $\leftarrow$ (1,3) = Y
(0,2) $\leftarrow$ (2,3) = W
(0,3) $\leftarrow$ (3,3) = O
(1,0) $\leftarrow$ (0,2) = Y
(1,1) $\leftarrow$ (1,2) = Y
(1,2) $\leftarrow$ (2,2) = O
(1,3) $\leftarrow$ (3,2) = W
(2,0) $\leftarrow$ (0,1) = M
(2,1) $\leftarrow$ (1,1) = M
(2,2) $\leftarrow$ (2,1) = M
(2,3) $\leftarrow$ (3,1) = Y
(3,0) $\leftarrow$ (0,0) = Y
(3,1) $\leftarrow$ (1,0) = Y
(3,2) $\leftarrow$ (2,0) = O
(3,3) $\leftarrow$ (3,0) = W

TR:
`Y Y W O`
`Y Y O W`
`M M M Y`
`Y Y O W`

**3. Bottom-Left (180°):**
180° mapping: `New(r, c) = Old(N-1-r, N-1-c)`
(0,0) $\leftarrow$ (3,3) = O
(0,1) $\leftarrow$ (3,2) = W
(0,2) $\leftarrow$ (3,1) = Y
(0,3) $\leftarrow$ (3,0) = W
(1,0) $\leftarrow$ (2,3) = W
(1,1) $\leftarrow$ (2,2) = O
(1,2) $\leftarrow$ (2,1) = M
(1,3) $\leftarrow$ (2,0) = O
(2,0) $\leftarrow$ (1,3) = Y
(2,1) $\leftarrow$ (1,2) = Y
(2,2) $\leftarrow$ (1,1) = M
(2,3) $\leftarrow$ (1,0) = Y
(3,0) $\leftarrow$ (0,3) = Y
(3,1) $\leftarrow$ (0,2) = Y
(3,2) $\leftarrow$ (0,1) = M
(3,3) $\leftarrow$ (0,0) = Y

BL:
`O W Y W`
`W O M O`
`Y Y M Y`
`Y Y M Y`

**4. Bottom-Right (90° CW):**
90° CW mapping: `New(r, c) = Old(N-1-c, r)`
(0,0) $\leftarrow$ (3,0) = W
(0,1) $\leftarrow$ (2,0) = O
(0,2) $\leftarrow$ (1,0) = Y
(0,3) $\leftarrow$ (0,0) = Y
(1,0) $\leftarrow$ (3,1) = Y
(1,1) $\leftarrow$ (2,1) = M
(1,2) $\leftarrow$ (1,1) = M
(1,3) $\leftarrow$ (0,1) = M
(2,0) $\leftarrow$ (3,2) = W
(2,1) $\leftarrow$ (2,2) = O
(2,2) $\leftarrow$ (1,2) = Y
(2,3) $\leftarrow$ (0,2) = Y
(3,0) $\leftarrow$ (3,3) = O
(3,1) $\leftarrow$ (2,3) = W
(3,2) $\leftarrow$ (1,3) = Y
(3,3) $\leftarrow$ (0,3) = Y

BR:
`W O Y Y`
`Y M M M`
`W O Y Y`
`O W Y Y`

**Combine into 8x8:**
Row 1: TL(R1) TR(R1) $\rightarrow$ `Y M Y Y Y Y W O`
Row 2: TL(R2) TR(R2) $\rightarrow$ `Y M Y Y Y Y O W`
Row 3: TL(R3) TR(R3) $\rightarrow$ `O M O W M M M Y`
Row 4: TL(R4) TR(R4) $\rightarrow$ `W Y W O Y Y O W`
Row 5: BL(R1) BR(R1) $\rightarrow$ `O W Y W W O Y Y`
Row 6: BL(R2) BR(R2) 

... [274 chars truncated]

To solve this puzzle, we first perform a Conceptual Digestion:

### 1. Conceptual Digestion

**1. Identify the CAST:**
*   **Input Grid:** A single "dance square" (a pattern of colored markers) representing one couple's routine.
*   **Output Grid:** A "ballroom" (a 2x2 grid of the input patterns) containing four couples.
*   **Colors:** Represent the specific positions/traces of the dancers.

**2. Identify the AFFORDANCES:**
*   The input pattern can be **rotated** by 90-degree increments (quarter-turns).
*   The input pattern can be **tiled** into a larger grid.

**3. Identify the TRANSFORMATION RULE:**
The input is an $N \times N$ grid. The output is a $2N \times 2N$ grid composed of four quadrants. Each quadrant is a version of the input grid rotated by 90 degrees clockwise relative to the previous quadrant:
*   **Top-Left Quadrant (Quadrant 1):** The original input grid (0° rotation).
*   **Top-Right Quadrant (Quadrant 2):** The input grid rotated 90° clockwise.
*   **Bottom-Left Quadrant (Quadrant 3):** The input grid rotated 180° clockwise.
*   **Bottom-Right Quadrant (Quadrant 4):** The input grid rotated 270° clockwise.

*Wait, let's re-verify the rotation direction using Example 1:*
Input (4x4):
Y W B A
A Y B A
Y A A B
B B B A

Output (8x8) Top-Left (0°):
Y W B A
A Y B A
Y A A B
B B B A

Output (8x8) Top-Right (90° CW):
A A B A
B B A B
W Y A B
Y A Y B
(Looking at the columns of the input: Col 1 is Y A Y B. In the top-right, this becomes the first row of the quadrant? No. Let's check the rotation carefully.)

Let's re-examine Example 1 Top-Right:
Row 1: A A B A
Row 2: B B A B
Row 3: W Y A B
Row 4: Y A Y B
If we rotate the input 90° CW:
Input Col 1 (Y A Y B) becomes Row 1 (reversed? No).
Let's check the rotation of the input matrix $A$:
$A_{rotated\_90\_CW}(i, j) = A(N-1-j, i)$.
Input Col 1: (0,0)=Y, (1,0)=A, (2,0)=Y, (3,0)=B.
In 90° CW rotation, these should be Row 1: (0,3)=Y, (0,2)=A, (0,1)=Y, (0,0)=B? No.
Let's look at the Top-Right quadrant of Example 1 again:
Row 1: A A B A
Row 2: B B A B
Row 3: W Y A B
Row 4: Y A Y B
This is the input grid rotated **90 degrees counter-clockwise**.
Let's check:
Input Col 4 (A A B A) becomes Row 1.
Input Col 3 (B B A B) becomes Row 2.
Input Col 2 (W Y A A) becomes Row 3.
Input Col 1 (Y A Y B) becomes Row 4.
Yes! The quadrants are:
1. Top-Left: Original
2. Top-Right: 90° CCW
3. Bottom-Left: 180°
4. Bottom-Right: 270° CCW (or 90° CW)

### 2. Applying the Rule to Test Input

**Input (4x4):**
R1: Y M Y Y
R2: Y M Y Y
R3: O M O W
R4: W Y W O

**Quadrant 1 (Original):**
Y M Y Y
Y M Y Y
O M O W
W Y W O

**Quadrant 2 (90° CCW):**
(Col 4 becomes Row 1, Col 3 becomes Row 2, Col 2 becomes Row 3, Col 1 becomes Row 4)
Col 4: Y Y W O -> Y Y W O
Col 3: Y Y O W -> Y Y O W
Col 2: M M M Y -> M M M Y
Col 1: Y Y O W -> Y Y O W
Wait, let's re-map:
New Row 1 = Old Col 4: Y Y W O
New Row 2 = Old Col 3: Y Y O W
New Row 3 = Old Col 2: M M M Y
New Row 4 = Old Col 1: Y Y O W
*Correction:* Let's look at Example 1 again.
Input Col 4: A A B A. Top-Right Row 1: A A B A.
Input Col 3: B B A B. Top-Right Row 2: B B A B.
Input Col 2: W Y A A. Top-Right Row 3: W Y A B (Wait, the example says W Y A B).
Input Col 1: Y A Y B. Top-Right Row 4: Y A Y B.
The rule is: **Quadrant 2 is the input grid rotated 90 degrees counter-clockwise.**
Quadrant 2 Row $i$ = Input Column $(N-i)$.

**Quadrant 3 (180°):**
(Reverse rows and reverse columns)
O W O Y
W O M Y
Y Y M Y
Y Y M Y

**Quadrant 4 (270° CCW / 90° CW):**
New Row 1 = Old Col 1: Y Y O W
New Row 2 = Old Col 2: M M M Y
New Row 3 = Old Col 3: Y Y O W
New Row 4 = Old Col 4: Y Y W O

**Constructing the 8x8 Grid:**
Row 1: [Q1 R1] [Q2 R1] -> Y M Y Y Y Y W O
Row 2: [Q1 R2] [Q2 R2] -> Y M Y Y Y Y O W
Row 3: [Q1 R3] [Q2 R3] -> O M O W M M M Y
Row 4: [Q1 R4] [Q2 R4] -> W Y W O Y Y O W
Row 5: [Q3 R1] [Q4 R1] -> O W O Y Y Y O W
Row 6: [Q3 R2] [Q4 R2] -> W O M Y M M M Y
Row 7: [Q3 R3] [Q4 R3] -> Y Y M Y Y Y O W
Row 8: [Q3 R4] [Q4 R4] -> Y Y M Y Y Y W O

Let's double check Q2 Row 3 of Example 1: Input Col 2 is W Y A A. Example output Q2 R3 is W Y A B. There is a slight discrepancy in my manual trace of Example 1, but the pattern of columns being rows is consistent. Let's re-verify Q2 for Test Input.
Input Col 1: Y Y O W
Input Col 2: M M M Y
Input Col 3: Y Y O W
Input Col 4: Y Y W O
Q2 (90 CCW):
R1 (Col 4): Y Y W O
R2 (Col 3): Y Y O W
R3 (Col 2): M M M Y
R4 (Col 1): Y Y O W

Q3 (180):
R1 (Rev R4): O W Y W
R2 (Rev R3): W O M O
R3 (Rev R2): Y Y M Y
R4 (Rev R1): Y Y M Y
Wait, 180 rotation of:
Y M Y Y
Y M Y Y
O M O W
W Y W O
is:
O W O Y
W O M O
Y Y M Y
Y Y M Y
(Wait, R4 reversed is O W Y W. R3 reversed is W O M O. R2 reversed is Y Y M Y. R1 reversed is Y Y M Y.)

Let's re-calculate Q3 (180):
Input:
(0,0) (0,1) (0,2) (0,3)
(1,0) (1,1) (1,2) (1,3)
(2,0) (2,1) (2,2) (2,3)
(3,0) (3,1) (3,2) (3,3)
180 rotation:
(3,3) (3,2) (3,1) (3,0)
(2,3) (2,2) (2,1) (2,0)
(1,3) (1,

... [3249 chars truncated]

To solve this puzzle, we first perform a Conceptual Digestion.

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **The Input Grid**: Represents the "small square of marbled dough." It is a $4 \times 4$ grid.
    *   **The Output Grid**: Represents the "baking sheet." It is a $8 \times 8$ grid.
    *   **The Quadrants**: The $8 \times 8$ output is divided into four $4 \times 4$ quadrants.

2.  **Identify the AFFORDANCES**:
    *   The input grid can be copied three times.
    *   Each copy can be rotated by $90^\circ$ increments (quarter-turns).
    *   The copies can be placed into the four quadrants of the larger grid.

3.  **Identify the TRANSFORMATION RULE**:
    *   The output is a $2N \times 2N$ grid (where $N=4$).
    *   **Top-Left Quadrant (Quadrant 1)**: The original input grid (0° rotation).
    *   **Top-Right Quadrant (Quadrant 2)**: The input grid rotated 90° clockwise.
    *   **Bottom-Left Quadrant (Quadrant 3)**: The input grid rotated 180° clockwise.
    *   **Bottom-Right Quadrant (Quadrant 4)**: The input grid rotated 270° clockwise (or 90° counter-clockwise).

Let's verify this with Example 1:
Input:
`Y W B A`
`A Y B A`
`Y A A B`
`B B B A`

- Q1 (0°): `Y W B A`, `A Y B A`, `Y A A B`, `B B B A` (Matches top-left)
- Q2 (90° CW): `B Y A Y`, `B B A W`, `B B Y B`, `A A A A` (Wait, let's re-check rotation)
  Rotation 90° CW of `[[Y,W,B,A],[A,Y,B,A],[Y,A,A,B],[B,B,B,A]]`:
  Row 1 becomes Col 4: `B Y A Y` (No, that's 90 CCW).
  Let's try 90° CW:
  Col 1 becomes Row 1: `B Y A Y`
  Col 2 becomes Row 2: `B A Y W`
  Col 3 becomes Row 3: `B A B B`
  Col 4 becomes Row 4: `A B A A`
  Looking at Example 1 Output Q2: `A A A B`, `B B A B`, `W Y A B`, `Y A Y B`.
  Let's re-examine the rotation logic.
  Example 1 Q2 (Top Right):
  `A A A B`
  `B B A B`
  `W Y A B`
  `Y A Y B`
  This is the input grid rotated **90 degrees counter-clockwise**.
  Let's check Q3 (Bottom Left):
  `A B B B`
  `B A A Y`
  `B B Y A`
  `A B W Y`
  This is the input grid rotated **180 degrees**.
  Let's check Q4 (Bottom Right):
  `B Y A Y`
  `A Y B A`
  `B A B B`
  `A B A A`
  This is the input grid rotated **90 degrees clockwise**.

**Correct Rule**:
- Q1: 0°
- Q2: 90° CCW
- Q3: 180°
- Q4: 90° CW

### Applying the Rule to Test Input

**Input**:
`Y M Y Y`
`Y M Y Y`
`O M O W`
`W Y W O`

**Q1 (0°)**:
`Y M Y Y`
`Y M Y Y`
`O M O W`
`W Y W O`

**Q2 (90° CCW)**:
(Original Col 4 becomes Row 1, Col 3 becomes Row 2, etc.)
`Y Y W O`
`Y Y O W`
`M M M Y`
`Y Y M W`
Wait, let's do it carefully:
Input:
(0,0)Y (0,1)M (0,2)Y (0,3)Y
(1,0)Y (1,1)M (1,2)Y (1,3)Y
(2,0)O (2,1)M (2,2)O (2,3)W
(3,0)W (3,1)Y (3,2)W (3,3)O

90° CCW:
New(r,c) = Old(c, N-1-r)
(0,0) = Old(0,3) = Y
(0,1) = Old(1,3) = Y
(0,2) = Old(2,3) = W
(0,3) = Old(3,3) = O
(1,0) = Old(0,2) = Y
(1,1) = Old(1,2) = Y
(1,2) = Old(2,2) = O
(1,3) = Old(3,2) = W
(2,0) = Old(0,1) = M
(2,1) = Old(1,1) = M
(2,2) = Old(2,1) = M
(2,3) = Old(3,1) = Y
(3,0) = Old(0,0) = Y
(3,1) = Old(1,0) = Y
(3,2) = Old(2,0) = O
(3,3) = Old(3,0) = W

Q2:
`Y Y W O`
`Y Y O W`
`M M M Y`
`Y Y O W`

**Q3 (180°)**:
(r,c) = Old(N-1-r, N-1-c)
(0,0) = Old(3,3) = O
(0,1) = Old(3,2) = W
(0,2) = Old(3,1) = Y
(0,3) = Old(3,0) = W
(1,0) = Old(2,3) = W
(1,1) = Old(2,2) = O
(1,2) = Old(2,1) = M
(1,3) = Old(2,0) = O
(2,0) = Old(1,3) = Y
(2,1) = Old(1,2) = Y
(2,2) = Old(1,1) = M
(2,3) = Old(1,0) = Y
(3,0) = Old(0,3) = Y
(3,1) = Old(0,2) = Y
(3,2) = Old(0,1) = M
(3,3) = Old(0,0) = Y

Q3:
`O W Y W`
`W O M O`
`Y Y M Y`
`Y Y M Y`

**Q4 (90° CW)**:
(r,c) = Old(N-1-c, r)
(0,0) = Old(3,0) = W
(0,1) = Old(2,0) = O
(0,2) = Old(1,0) = Y
(0,3) = Old(0,0) = Y
(1,0) = Old(3,1) = Y
(1,1) = Old(2,1) = M
(1,2) = Old(1,1) = M
(1,3) = Old(0,1) = M
(2,0) = Old(3,2) = W
(2,1) = Old(2,2) = O
(2,2) = Old(1,2) = Y
(2,3) = Old(0,2) = Y
(3,0) = Old(3,3) = O
(3,1) = Old(2,3) = W
(3,2) = Old(1,3) = Y
(3,3) = Old(0,3) = Y

Q4:
`W O Y Y`
`Y M M M`
`W O Y Y`
`O W Y Y`

**Final Assembly**:
Row 1: Q1(R1) + Q2(R1) -> `Y M Y Y Y Y W O`
Row 2: Q1(R2) + Q2(R2) -> `Y M Y Y Y Y O W`
Row 3: Q1(R3) + Q2(R3) -> `O M O W M M M Y`
Row 4: Q1(R4) + Q2(R4) -> `W Y W O Y Y O W`
Row 5: Q3(R1) + Q4(R1) -> `O W Y W W O Y Y`
Row 6: Q3(R2) + Q4(R2) -> `W O M O Y M M M`
Row 7: Q3(R3) + Q4(R3) -> `Y Y M Y W O Y Y`
Row 8: Q3(R4) + Q4(R4) -> `Y Y M Y O W Y Y`

Wait, let me re-verify Q2 (90 CCW) and Q4 (90 CW) for the test input.
Input:
`Y M Y Y`
`Y M Y Y`
`O M O W`
`W Y W O`

Q2 (90 CCW):
Col 4: `Y Y W O` -> Row 1
Col 3: `Y Y O W` -> Row 2
Col 2: `M M M Y` -> Row 3
Col 1: `Y Y O W` -> Row 4 (Wait, Col 1 is `Y Y O W`. Let's re-check)
Col 1: `Y Y O W`
Col 2: `M M M Y`
Col 3: `Y Y O W`
Col 4: `Y Y W O`
90 CCW:
Row 1: Col 4 = `Y Y W O`
Row 2: Col 3 = `Y Y O W`
Row 3: Col 2 = `M M M Y`
Row 4: Col 1 = `Y Y O W`
Correct.

Q4 (90 CW):
Row 1: Col 1 reversed = `W O Y Y`
Row 2: Col 2 reversed = `Y M M M`
Row 3: Col 3 reversed = `W O Y Y`
Row 4: Col 4 reversed = `O W Y Y

... [535 chars truncated]

To solve this puzzle, we follow the instructions provided in the metaphorical description and the training examples.

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Input Grid**: A small "musical phrase" (a square grid of size $N \times N$).
    *   **Output Grid**: A "larger score" (a square grid of size $2N \times 2N$).
    *   **Quadrants**: The output is composed of four $N \times N$ blocks.

2.  **Identify the AFFORDANCES**:
    *   The input grid can be rotated.
    *   The output grid is a $2 \times 2$ arrangement of these rotations.

3.  **Identify the TRANSFORMATION RULE**:
    *   The output is a $2 \times 2$ tiling of the input grid.
    *   **Top-Left Quadrant**: The original input grid.
    *   **Top-Right Quadrant**: The input grid rotated 90 degrees clockwise.
    *   **Bottom-Left Quadrant**: The input grid rotated 180 degrees clockwise.
    *   **Bottom-Right Quadrant**: The input grid rotated 270 degrees clockwise (or 90 degrees counter-clockwise).

Let's verify this with Example 1:
Input (4x4):
Y W B A
A Y B A
Y A A B
B B B A

Top-Left (Original):
Y W B A
A Y B A
Y A A B
B B B A

Top-Right (90 deg CW):
B Y A A
B A Y W
B A Y A
A B A A
(Wait, let's re-check the rotation logic from the example output).

Looking at Example 1 Output:
Row 1: `Y W B A` (Original R1) + `A A B A` (Wait, let's look at columns).
If we rotate the input 90 degrees clockwise:
Col 1 becomes Row 1 (reversed): B Y A Y -> No.
Let's try:
Top-Left: Original.
Top-Right: Input rotated 90 degrees clockwise.
Bottom-Left: Input rotated 180 degrees clockwise.
Bottom-Right: Input rotated 270 degrees clockwise.

Let's re-examine Example 1 Top-Right: `A A B A`, `B B A B`, `W Y A B`, `A Y A B`.
If we rotate the input 90 degrees clockwise:
Input:
(0,0)Y (0,1)W (0,2)B (0,3)A
(1,0)A (1,1)Y (1,2)B (1,3)A
(2,0)Y (2,1)A (2,2)A (2,3)B
(3,0)B (3,1)B (3,2)B (3,3)A

90 deg CW:
(3,0)B (2,0)Y (1,0)A (0,0)Y
(3,1)B (2,1)A (1,1)Y (0,1)W
(3,2)B (2,2)A (1,2)B (0,2)B
(3,3)A (2,3)B (1,3)A (0,3)A
This doesn't match the example output `A A B A`.

Let's look at the quadrants of Example 1 again:
Top-Left: `Y W B A / A Y B A / Y A A B / B B B A` (Original)
Top-Right: `A A B A / B B A B / W Y A B / A Y A B`
Bottom-Left: `A B B B / B A A Y / Y A B Y / A B Y A`
Bottom-Right: `A B Y A / B A Y W / B A Y B / A B A A`

Let's check the rotation of the input for Top-Right:
Input:
Y W B A
A Y B A
Y A A B
B B B A
If we rotate the input 90 degrees **counter-clockwise**:
(0,3)A (1,3)A (2,3)B (3,3)A
(0,2)B (1,2)B (2,2)A (3,2)B
(0,1)W (1,1)Y (2,1)A (3,1)B
(0,0)Y (1,0)A (2,0)Y (3,0)B
This matches the Top-Right of Example 1!

So the rule is:
- Top-Left: Original
- Top-Right: 90 deg Counter-Clockwise
- Bottom-Left: 180 deg
- Bottom-Right: 270 deg Counter-Clockwise (or 90 deg Clockwise)

### Applying to Test Input

Input:
Y M Y Y
Y M Y Y
O M O W
W Y W O

**1. Top-Left (Original):**
Y M Y Y
Y M Y Y
O M O W
W Y W O

**2. Top-Right (90 deg CCW):**
(0,3)Y (1,3)Y (2,3)W (3,3)O
(0,2)Y (1,2)Y (2,2)O (3,2)W
(0,1)M (1,1)M (2,1)M (3,1)Y
(0,0)Y (1,0)Y (2,0)O (3,0)W
Result:
Y Y W O
Y Y O W
M M M Y
Y Y O W

**3. Bottom-Left (180 deg):**
(3,3)O (3,2)W (3,1)Y (3,0)W
(2,3)W (2,2)O (2,1)M (2,0)O
(1,3)Y (1,2)Y (1,1)M (1,0)Y
(0,3)Y (0,2)Y (0,1)M (0,0)Y
Result:
O W Y W
W O M O
Y Y M Y
Y Y M Y

**4. Bottom-Right (90 deg CW):**
(3,0)W (2,0)O (1,0)Y (0,0)Y
(3,1)Y (2,1)M (1,1)M (0,1)M
(3,2)W (2,2)O (1,2)Y (0,2)Y
(3,3)O (2,3)W (1,3)Y (0,3)Y
Result:
W O Y Y
Y M M M
W O Y Y
O W Y Y

### Combining the Quadrants

Row 1: Y M Y Y (TL) + Y Y W O (TR) = Y M Y Y Y Y W O
Row 2: Y M Y Y (TL) + Y Y O W (TR) = Y M Y Y Y Y O W
Row 3: O M O W (TL) + M M M Y (TR) = O M O W M M M Y
Row 4: W Y W O (TL) + Y Y O W (TR) = W Y W O Y Y O W
Row 5: O W Y W (BL) + W O Y Y (BR) = O W Y W W O Y Y
Row 6: W O M O (BL) + Y M M M (BR) = W O M O Y M M M
Row 7: Y Y M Y (BL) + W O Y Y (BR) = Y Y M Y W O Y Y
Row 8: Y Y M Y (BL) + O W Y Y (BR) = Y Y M Y O W Y Y

ANSWER:
Y M Y Y Y Y W O
Y M Y Y Y Y O W
O M O W M M M Y
W Y W O Y Y O W
O W Y W W O Y Y
W O M O Y M M M
Y Y M Y W O Y Y
Y Y M Y O W Y Y

To solve this puzzle, we first perform a Conceptual Digestion.

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Input Grid:** A small "tactical diagram" (the clipboard) representing a single formation.
    *   **Output Grid:** A "full court" (the arena) which is exactly twice the width and twice the height of the input grid.
    *   **Quadrants:** The output is divided into four quadrants (Top-Left, Top-Right, Bottom-Left, Bottom-Right).

2.  **Identify the AFFORDANCES:**
    *   The input pattern can be rotated.
    *   The input pattern can be tiled to fill a larger space.

3.  **Identify the TRANSFORMATION RULE:**
    *   The output is a $2N \times 2N$ grid (where $N$ is the input dimension).
    *   **Top-Left Quadrant:** The input grid as it is (0° rotation).
    *   **Top-Right Quadrant:** The input grid rotated 90 degrees clockwise.
    *   **Bottom-Left Quadrant:** The input grid rotated 180 degrees.
    *   **Bottom-Right Quadrant:** The input grid rotated 270 degrees clockwise (or 90 degrees counter-clockwise).

*Correction based on Example 1 analysis:*
Let's re-examine Example 1:
Input (4x4):
Y W B A
A Y B A
Y A A B
B B B A

Output (8x8):
TL: Y W B A (Input)
TR: A B A A (Input rotated 90 deg clockwise? Let's check: Col 1 becomes Row 1? No. Let's check the columns of the input. Col 1: Y A Y B. Col 2: W Y A B. Col 3: B B A B. Col 4: A A B A. If we rotate 90 deg CW: Row 1 becomes Col 4. Row 4 becomes Col 1.
Let's look at the Top-Right of Example 1:
A A B A
B B A B
W Y A B
A B A A
This is the input grid rotated 90 degrees **counter-clockwise**.
Let's check Bottom-Left of Example 1:
A B B B
B A A Y
A B Y A
A B W Y
This is the input grid rotated 180 degrees.
Let's check Bottom-Right of Example 1:
B Y A Y
A Y A B
B A B B
A B A A
This is the input grid rotated 90 degrees **clockwise**.

**Revised Rule:**
1.  **Top-Left (TL):** Input (0°)
2.  **Top-Right (TR):** Input rotated 90° CCW
3.  **Bottom-Left (BL):** Input rotated 180°
4.  **Bottom-Right (BR):** Input rotated 90° CW

### Applying the Rule to Test Input

**Test Input:**
Y M Y Y
Y M Y Y
O M O W
W Y W O

**1. Top-Left (0°):**
Y M Y Y
Y M Y Y
O M O W
W Y W O

**2. Top-Right (90° CCW):**
(Original Col 4 becomes Row 1, Col 3 becomes Row 2, etc.)
Col 4: Y Y W O -> Row 1: Y Y W O
Col 3: Y Y O W -> Row 2: Y Y O W
Col 2: M M M Y -> Row 3: M M M Y
Col 1: Y Y O W -> Row 4: Y Y O W
Wait, let's do rotation properly:
Input:
(0,0)Y (0,1)M (0,2)Y (0,3)Y
(1,0)Y (1,1)M (1,2)Y (1,3)Y
(2,0)O (2,1)M (2,2)O (2,3)W
(3,0)W (3,1)Y (3,2)W (3,3)O

90 CCW:
(0,3)Y (1,3)Y (2,3)W (3,3)O
(0,2)Y (1,2)Y (2,2)O (3,2)W
(0,1)M (1,1)M (2,1)M (3,1)Y
(0,0)Y (1,0)Y (2,0)O (3,0)W

**3. Bottom-Left (180°):**
(3,3)O (3,2)W (3,1)Y (3,0)W
(2,3)W (2,2)O (2,1)M (2,0)O
(1,3)Y (1,2)Y (1,1)M (1,0)Y
(0,3)Y (0,2)Y (0,1)M (0,0)Y

**4. Bottom-Right (90° CW):**
(3,0)W (2,0)O (1,0)Y (0,0)Y
(3,1)Y (2,1)M (1,1)M (0,1)M
(3,2)W (2,2)O (1,2)Y (0,2)Y
(3,3)O (2,3)W (1,3)Y (0,3)Y

### Constructing the 8x8 Grid

Row 1: (TL R1) Y M Y Y | (TR R1) Y Y W O
Row 2: (TL R2) Y M Y Y | (TR R2) Y Y O W
Row 3: (TL R3) O M O W | (TR R3) M M M Y
Row 4: (TL R4) W Y W O | (TR R4) Y Y O W
Row 5: (BL R1) O W Y W | (BR R1) W O Y Y
Row 6: (BL R2) W O M O | (BR R2) Y M M M
Row 7: (BL R3) Y Y M Y | (BR R3) W O Y Y
Row 8: (BL R4) Y Y M Y | (BR R4) O W Y Y

Wait, let's re-verify the 90 CCW and 90 CW logic.
Input:
1 2 3 4
5 6 7 8
9 A B C
D E F G

90 CCW:
4 8 C G
3 7 B F
2 6 A E
1 5 9 D

90 CW:
D 9 5 1
E A 6 2
F B 7 3
G C 8 4

Let's re-apply to Test Input:
Input:
Y M Y Y
Y M Y Y
O M O W
W Y W O

90 CCW:
Y Y W O
Y Y O W
M M M Y
Y Y O W

90 CW:
W O Y Y
Y M M M
W O Y Y
O W Y Y

180:
O W Y W
W O M O
Y Y M Y
Y Y M Y

Let's re-assemble:
Row 1: Y M Y Y Y Y W O
Row 2: Y M Y Y Y Y O W
Row 3: O M O W M M M Y
Row 4: W Y W O Y Y O W
Row 5: O W Y W W O Y Y
Row 6: W O M O Y M M M
Row 7: Y Y M Y W O Y Y
Row 8: Y Y M Y O W Y Y

Wait, let's re-check the 180 rotation for the Bottom-Left.
Input:
Y M Y Y
Y M Y Y
O M O W
W Y W O
180 rotation:
O W Y W
W O M O
Y Y M Y
Y Y M Y
(Correct)

Let's re-check 90 CW for Bottom-Right.
Input:
Y M Y Y
Y M Y Y
O M O W
W Y W O
90 CW:
W O Y Y
Y M M M
W O Y Y
O W Y Y
(Correct)

Let's re-check 90 CCW for Top-Right.
Input:
Y M Y Y
Y M Y Y
O M O W
W Y W O
90 CCW:
Y Y W O
Y Y O W
M M M Y
Y Y O W
(Correct)

Final Assembly:
R1: Y M Y Y (TL) + Y Y W O (TR) = Y M Y Y Y Y W O
R2: Y M Y Y (TL) + Y Y O W (TR) = Y M Y Y Y Y O W
R3: O M O W (TL) + M M M Y (TR) = O M O W M M M Y
R4: W Y W O (TL) + Y Y O W (TR) = W Y W O Y Y O W
R5: O W Y W (BL) + W O Y Y (BR) = O W Y W W O Y Y
R6: W O M O (BL) + Y M M M (BR) = W O M O Y M M M
R7: Y Y M Y (BL) + W O Y Y (BR) = Y Y M Y W O Y Y
R8: Y Y M Y (BL) + O W Y Y (BR) = Y Y M Y O W Y Y

ANSWER:
Y M Y Y Y Y W O
Y M Y Y Y Y O W
O M O W M M M Y
W Y W O Y Y O W
O W Y W W O Y Y
W O M O Y M M M
Y Y M Y W O Y Y
Y Y M Y O W Y Y

To solve this puzzle, we first perform a Conceptual Digestion:

### 1. Conceptual Digestion

**Identify the CAST:**
*   **Input Grid:** A small $N \times N$ square representing a "storm pattern."
*   **Output Grid:** A $2N \times 2N$ larger radar display.
*   **Quadrants:** The output is divided into four $N \times N$ quadrants.

**Identify the AFFORDANCES:**
*   The input pattern can be **rotated** by 0°, 90°, 180°, and 270°.
*   The pattern can be **tiled** into the four quadrants of the larger grid.

**Identify the TRANSFORMATION RULE:**
The output grid is a $2 \times 2$ tiling of the input grid, where each quadrant is a rotated version of the original input. Looking at the training examples:
*   **Top-Left Quadrant (TL):** The original input (0° rotation).
*   **Top-Right Quadrant (TR):** The input rotated 90° clockwise.
*   **Bottom-Left Quadrant (BL):** The input rotated 180° (or 270° depending on the specific mapping, let's check Example 1).
    *   *Example 1 Check:* Input is $4 \times 4$. Output is $8 \times 8$.
    *   TL: Input.
    *   TR: Input rotated 90° clockwise.
    *   BL: Input rotated 180°.
    *   BR: Input rotated 270° clockwise (or 90° counter-clockwise).

Let's verify with Example 1:
Input:
`Y W B A`
`A Y B A`
`Y A A B`
`B B B A`
TR (90° CW):
`B Y A Y`
`B A A W`
`B B Y B`
`A A B A`
Wait, looking at the output of Example 1:
Row 1: `Y W B A` (TL) + `A A B A` (TR?)
Let's re-examine the rotation direction.
If TR is 90° CW:
Input Col 1 becomes Row 1 (reversed): `B Y A Y`
Input Col 2 becomes Row 2 (reversed): `B A A W`
Input Col 3 becomes Row 3 (reversed): `B B Y B`
Input Col 4 becomes Row 4 (reversed): `A A B A`
Looking at Example 1 Output Row 1: `Y W B A` (TL) then `A A B A`.
`A A B A` is the **last column** of the input.
If the TR quadrant is the input rotated 90° **counter-clockwise**:
Col 1 becomes Row 4: `B Y A Y`
Col 2 becomes Row 3: `B A A W`
Col 3 becomes Row 2: `B B Y B`
Col 4 becomes Row 1: `A A B A`
Yes! If we rotate the input 90° CCW, the first row of the TR quadrant is the last column of the input.

Let's check the quadrants for Example 1 again:
TL: Input
TR: Input rotated 90° CCW
BL: Input rotated 180°
BR: Input rotated 90° CW

Let's test this pattern on Example 1:
Input:
`Y W B A`
`A Y B A`
`Y A A B`
`B B B A`
TL: `Y W B A` / `A Y B A` / `Y A A B` / `B B B A`
TR (90 CCW): `A A B B` / `W Y A B` / `B B A B` / `A A B A` (Wait, let's re-calculate)
Input:
(0,0)Y (0,1)W (0,2)B (0,3)A
(1,0)A (1,1)Y (1,2)B (1,3)A
(2,0)Y (2,1)A (2,2)A (2,3)B
(3,0)B (3,1)B (3,2)B (3,3)A

90 CCW rotation:
New(r,c) = Old(c, N-1-r)
New(0,0) = Old(0,3) = A
New(0,1) = Old(1,3) = A
New(0,2) = Old(2,3) = B
New(0,3) = Old(3,3) = A
Row 1 of TR: `A A B A`. Matches Example 1!

BL (180):
New(r,c) = Old(N-1-r, N-1-c)
New(0,0) = Old(3,3) = A
New(0,1) = Old(3,2) = B
New(0,2) = Old(3,1) = B
New(0,3) = Old(3,0) = B
Row 1 of BL: `A B B B`. Matches Example 1!

BR (90 CW):
New(r,c) = Old(N-1-c, r)
New(0,0) = Old(3,0) = B
New(0,1) = Old(2,0) = Y
New(0,2) = Old(1,0) = A
New(0,3) = Old(0,0) = Y
Row 1 of BR: `B Y A Y`. Matches Example 1!

**Rule Summary:**
- TL: Input
- TR: Input rotated 90° CCW
- BL: Input rotated 180°
- BR: Input rotated 90° CW

### 2. Apply to Test Input

Test Input:
`Y M Y Y`
`Y M Y Y`
`O M O W`
`W Y W O`

**TL (Input):**
`Y M Y Y`
`Y M Y Y`
`O M O W`
`W Y W O`

**TR (90° CCW):**
New(0,0)=Old(0,3)=Y, New(0,1)=Old(1,3)=Y, New(0,2)=Old(2,3)=W, New(0,3)=Old(3,3)=O
New(1,0)=Old(0,2)=Y, New(1,1)=Old(1,2)=Y, New(1,2)=Old(2,2)=O, New(1,3)=Old(3,2)=W
New(2,0)=Old(0,1)=M, New(2,1)=Old(1,1)=M, New(2,2)=Old(2,1)=M, New(2,3)=Old(3,1)=Y
New(3,0)=Old(0,0)=Y, New(3,1)=Old(1,0)=Y, New(3,2)=Old(2,0)=O, New(3,3)=Old(3,0)=W
TR:
`Y Y W O`
`Y Y O W`
`M M M Y`
`Y Y O W`

**BL (180°):**
New(0,0)=Old(3,3)=O, New(0,1)=Old(3,2)=W, New(0,2)=Old(3,1)=Y, New(0,3)=Old(3,0)=W
New(1,0)=Old(2,3)=W, New(1,1)=Old(2,2)=O, New(1,2)=Old(2,1)=M, New(1,3)=Old(2,0)=O
New(2,0)=Old(1,3)=Y, New(2,1)=Old(1,2)=Y, New(2,2)=Old(1,1)=M, New(2,3)=Old(1,0)=Y
New(3,0)=Old(0,3)=Y, New(3,1)=Old(0,2)=Y, New(3,2)=Old(0,1)=M, New(3,3)=Old(0,0)=Y
BL:
`O W Y W`
`W O M O`
`Y Y M Y`
`Y Y M Y`

**BR (90° CW):**
New(0,0)=Old(3,0)=W, New(0,1)=Old(2,0)=O, New(0,2)=Old(1,0)=Y, New(0,3)=Old(0,0)=Y
New(1,0)=Old(3,1)=Y, New(1,1)=Old(2,1)=M, New(1,2)=Old(1,1)=M, New(1,3)=Old(0,1)=M
New(2,0)=Old(3,2)=W, New(2,1)=Old(2,2)=O, New(2,2)=Old(1,2)=Y, New(2,3)=Old(0,2)=Y
New(3,0)=Old(3,3)=O, New(3,1)=Old(2,3)=W, New(3,2)=Old(1,3)=Y, New(3,3)=Old(0,3)=Y
BR:
`W O Y Y`
`Y M M M`
`W O Y Y`
`O W Y Y`

**Combine Quadrants:**
Row 1: `Y M Y Y` (TL) + `Y Y W O` (TR)
Row 2: `Y M Y Y` (TL) + `Y Y O W` (TR)
Row 3: `O M O W` (TL) + `M M M Y` (TR)
Row 4: `W Y W O` (TL) + `Y Y O W` (TR)
Row 5: `O W Y W` (BL) + `W O Y Y` (BR)
Row 6: `W O M O` (BL) + `Y M M M` (BR)
Row 7: `Y Y M Y` (BL) + `W O Y Y` (BR)
Row 8: `Y Y M Y` (BL) + `O W Y Y` (BR)

ANSWER:
Y M Y Y Y Y W O
Y M Y 

... [105 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **The Stage (Input Grid)**: A small $N \times N$ grid representing a single "tableau" or snapshot of the scene.
    *   **The Rotating Stage**: The mechanism that rotates the entire input grid by 90 degrees clockwise for each subsequent "photograph."
    *   **The Panorama (Output Grid)**: A $2N \times 2N$ grid composed of four quadrants.

2.  **Identify the AFFORDANCES**:
    *   The input grid can be rotated 90 degrees clockwise.
    *   The output grid can be divided into four quadrants: Top-Left (TL), Top-Right (TR), Bottom-Left (BL), and Bottom-Right (BR).

3.  **Identify the TRANSFORMATION RULE**:
    *   The output is a $2N \times 2N$ grid.
    *   The four quadrants are filled with rotations of the input grid:
        *   **Top-Left (TL)**: The original input grid (0° rotation).
        *   **Top-Right (TR)**: The input grid rotated 90° clockwise.
        *   **Bottom-Left (BL)**: The input grid rotated 180° clockwise.
        *   **Bottom-Right (BR)**: The input grid rotated 270° clockwise (or 90° counter-clockwise).
    *   *Correction based on Example 1 analysis*: Let's re-verify the rotation order in Example 1.
        Input:
        `Y W B A`
        `A Y B A`
        `Y A A B`
        `B B B A`
        
        Output TL (0°): `Y W B A / A Y B A / Y A A B / B B B A` (Matches)
        Output TR (90° CW): `B Y A A / B A Y W / B A A B / A A B A` (Wait, let's check the example output TR)
        Example Output TR (Rows 1-4, Cols 5-8):
        `A A B A`
        `B B A B`
        `W Y A B`
        `Y A Y B`
        Looking at the example output, the quadrants are:
        TL: Original
        TR: Original rotated 90° clockwise? No. Let's look at the columns.
        Example Output Row 1: `Y W B A` (TL) + `A A B A` (TR).
        If we rotate the input 90 deg CW:
        `B Y A Y`
        `B A Y W`
        `B A B B`
        `A A B A`
        The example TR is `A A B A / B B A B / W Y A B / Y A Y B`. This is the input rotated 90 degrees **counter-clockwise**.
        Let's check BL (Rows 5-8, Cols 1-4): `A B B B / B A A Y / A B Y A / A B W Y`.
        Let's check BR (Rows 5-8, Cols 5-8): `B Y A Y / A Y W B / A B A A / A B A A`.
        
        Actually, the pattern is:
        TL: Input
        TR: Input rotated 90° CW
        BL: Input rotated 180° CW
        BR: Input rotated 270° CW
        Let's re-verify Example 1 TR with 90 CW:
        Input:
        1: Y W B A
        2: A Y B A
        3: Y A A B
        4: B B B A
        90 CW:
        Col 1 becomes Row 1 (reversed): B A Y Y
        Col 2 becomes Row 2 (reversed): B A Y W
        Col 3 becomes Row 3 (reversed): B B B B
        Col 4 becomes Row 4 (reversed): A A A A
        Wait, the example output is complex. Let's look at the quadrants of Example 1 again.
        TL: `Y W B A / A Y B A / Y A A B / B B B A` (Original)
        TR: `A A B A / B B A B / W Y A B / Y A Y B` (This is the input rotated 90 deg CW, but the columns are reversed? No, it's the input rotated 90 deg CW and then flipped? No.)
        
        Let's re-examine the rotation:
        Input:
        `Y W B A`
        `A Y B A`
        `Y A A B`
        `B B B A`
        
        90 CW:
        `B Y A Y`
        `B A Y W`
        `B A B B`
        `A A B A`
        
        180 CW:
        `A B B B`
        `B A A Y`
        `A Y B A`
        `A B W Y`
        
        270 CW:
        `A A A B`
        `A B A B`
        `B Y A B`
        `A W Y B`
        
        Looking at Example 1 Output:
        TL: `Y W B A / A Y B A / Y A A B / B B B A` (Matches 0°)
        TR: `A A B A / B B A B / W Y A B / Y A Y B` (Matches 270° CW)
        BL: `A B B B / B A A Y / A B Y A / A B W Y` (Matches 180° CW)
        BR: `B Y A Y / A Y W B / A B A A / A B A A` (Matches 90° CW)
        
        Wait, the order is:
        TL: 0°
        TR: 270°
        BL: 180°
        BR: 90°
        
        Let's check the rows of Example 1 again.
        Row 1: `Y W B A` (TL) + `A A B A` (TR)
        Row 2: `A Y B A` (TL) + `B B A B` (TR)
        Row 3: `Y A A B` (TL) + `W Y A B` (TR)
        Row 4: `B B B A` (TL) + `Y A Y B` (TR)
        This confirms TR is 270° CW.
        
        Row 5: `A B B B` (BL) + `B Y A Y` (BR)
        Row 6: `B A A Y` (BL) + `A Y W B` (BR)
        Row 7: `A B Y A` (BL) + `A B A A` (BR)
        Row 8: `A B W Y` (BL) + `A B A A` (BR)
        This confirms BL is 180° CW and BR is 90° CW.

### Applying to Test Input

Input:
`Y M Y Y`
`Y M Y Y`
`O M O W`
`W Y W O`

**1. 0° Rotation (TL):**
`Y M Y Y`
`Y M Y Y`
`O M O W`
`W Y W O`

**2. 90° CW Rotation (BR):**
Col 1 (W, O, Y, Y) -> Row 1: `W O Y Y`
Col 2 (Y, M, M, M) -> Row 2: `Y M M M`
Col 3 (W, O, Y, Y) -> Row 3: `W O Y Y`
Col 4 (O, W, Y, Y) -> Row 4: `O W Y Y`
Wait, let's do it carefully:
Input:
(0,0)Y (0,1)M (0,2)Y (0,3)Y
(1,0)Y (1,1)M (1,2

... [1175 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Input Grid**: A small square "planting plan" (size $N \times N$).
    *   **Output Grid**: A larger square "garden" (size $2N \times 2N$) composed of four $N \times N$ quadrants (beds).
    *   **Colors**: Represent different flower varieties.

2.  **Identify the AFFORDANCES**:
    *   The input grid can be **rotated**.
    *   The output grid is formed by tiling four versions of the input grid.

3.  **Identify the TRANSFORMATION RULE**:
    *   The output is a $2 \times 2$ arrangement of the input grid, where each quadrant is a rotated version of the original.
    *   Let the input be $I$. The output $O$ is composed of four quadrants:
        *   **Top-Left (TL)**: The original input $I$.
        *   **Top-Right (TR)**: The input $I$ rotated 90 degrees clockwise.
        *   **Bottom-Left (BL)**: The input $I$ rotated 180 degrees (or 90 degrees counter-clockwise from TR).
        *   **Bottom-Right (BR)**: The input $I$ rotated 270 degrees clockwise (or 90 degrees clockwise from BL).
    *   *Correction based on Example 1 analysis*: Let's re-examine Example 1.
        Input (4x4):
        Row 0: Y W B A
        Row 1: A Y B A
        Row 2: Y A A B
        Row 3: B B B A
        
        Output (8x8) Top-Left (0-3, 0-3): Matches Input.
        Output (0-3, 4-7):
        A A B A
        B B A B
        W Y A B
        Y A Y B
        Looking at the columns of the input: Col 0: Y A Y B; Col 1: W Y A B; Col 2: B B A B; Col 3: A A B A.
        The Top-Right quadrant is the **Transpose** of the input (rows become columns).
        Let's check Bottom-Left (4-7, 0-3):
        A B B B
        B A A Y
        A B Y A
        A B W Y
        This is the input rotated 90 degrees counter-clockwise.
        Let's check Bottom-Right (4-7, 4-7):
        B Y A Y
        A Y W B
        B A B A
        A B A A
        This is the input rotated 180 degrees.

    *   **Refined Rule**:
        1.  **Quadrant 1 (Top-Left)**: Original Input.
        2.  **Quadrant 2 (Top-Right)**: Transpose of Input (Rows $\to$ Columns).
        3.  **Quadrant 3 (Bottom-Left)**: Input rotated 90° counter-clockwise.
        4.  **Quadrant 4 (Bottom-Right)**: Input rotated 180°.

    *Wait, let's re-verify the rotation logic with the description*: "Each bed follows the same planting plan but rotated so the rows that once ran east-west now run north-south". This implies a 90-degree rotation step for each successive bed.
    Sequence:
    Bed 1: Original.
    Bed 2: 90° rotation.
    Bed 3: 180° rotation.
    Bed 4: 270° rotation.

    Let's re-check Example 1 with this:
    Input:
    Y W B A
    A Y B A
    Y A A B
    B B B A
    
    90° CW:
    B Y A Y
    B A Y W
    B A B B
    A A B A
    
    180°:
    A B B B
    B A A Y
    A B Y A
    A B W Y
    
    270° CW:
    A A B A
    B B A B
    A Y A B
    Y A Y B
    
    Looking at Example 1 Output again:
    TL: Y W B A / A Y B A / Y A A B / B B B A (Original)
    TR: A A B A / B B A B / W Y A B / Y A Y B (This is 270° CW / 90° CCW)
    BL: A B B B / B A A Y / A B Y A / A B W Y (This is 180°)
    BR: B Y A Y / A Y W B / B A B B / A A B A (This is 90° CW)

    **Correct Rule**:
    The quadrants are arranged as:
    [Original] [90° CCW]
    [180°] [90° CW]

### Applying to Test Input

Input:
Y M Y Y
Y M Y Y
O M O W
W Y W O

1. **Quadrant 1 (Original)**:
Y M Y Y
Y M Y Y
O M O W
W Y W O

2. **Quadrant 2 (90° CCW)**:
(Row 0 becomes Col 0, but reversed? No. Let's rotate manually)
Original:
(0,0)Y (0,1)M (0,2)Y (0,3)Y
(1,0)Y (1,1)M (1,2)Y (1,3)Y
(2,0)O (2,1)M (2,2)O (2,3)W
(3,0)W (3,1)Y (3,2)W (3,3)O

90° CCW:
(0,3)Y (1,3)Y (2,3)W (3,3)O
(0,2)Y (1,2)Y (2,2)O (3,2)W
(0,1)M (1,1)M (2,1)M (3,1)Y
(0,0)Y (1,0)Y (2,0)O (3,0)W

3. **Quadrant 3 (180°)**:
(3,3)O (3,2)W (3,1)Y (3,0)W
(2,3)W (2,2)O (2,1)M (2,0)O
(1,3)Y (1,2)Y (1,1)M (1,0)Y
(0,3)Y (0,2)Y (0,1)M (0,0)Y

4. **Quadrant 4 (90° CW)**:
(3,0)W (2,0)O (1,0)Y (0,0)Y
(3,1)Y (2,1)M (1,1)M (0,1)M
(3,2)W (2,2)O (1,2)Y (0,2)Y
(3,3)O (2,3)W (1,3)Y (0,3)Y

### Constructing the Output Grid

Row 0: [Q1 R0] [Q2 R0] -> Y M Y Y Y Y W O
Row 1: [Q1 R1] [Q2 R1] -> Y M Y Y Y Y O W
Row 2: [Q1 R2] [Q2 R2] -> O M O W M M M Y
Row 3: [Q1 R3] [Q2 R3] -> W Y W O Y Y O W
Row 4: [Q3 R0] [Q4 R0] -> O W Y W W O Y Y
Row 5: [Q3 R1] [Q4 R1] -> W O M O Y M M M
Row 6: [Q3 R2] [Q4 R2] -> Y Y M Y W O Y Y
Row 7: [Q3 R3] [Q4 R3] -> Y Y M Y O W Y Y

Wait, let me re-verify the 90° CCW rotation for Q2.
Input:
1 2 3 4
5 6 7 8
9 0 a b
c d e f

90 CCW:
4 8 b f
3 7 a e
2 6 0 d
1 5 9 c

Applying to Test Input:
Y M Y Y
Y M Y Y
O M O W
W Y W O

Q2 (90 CCW):
Y Y W O
Y Y O W
M M M Y
Y Y O W

Let's re-check Q3 (180):
O W Y W
W O M O
Y Y M Y
Y Y M Y

Let's re-check Q4 (90 CW):
W O Y Y
Y M M M
W O Y Y
O W Y Y

Re-assembling:
Row 0: Y M Y Y Y Y W O
Row 1: Y M Y Y Y Y O W
Row 2: O M O W M M M Y
Row 3: W Y W O Y Y O W
R

... [3195 chars truncated]

### Conceptual Digestion

**1. Identify the CAST:**
*   **The Azure Cross (A):** A 9x9 framework (the "leading") that divides the grid into four quadrants. It acts as a separator and is removed in the final output.
*   **The Green Tracery (G):** A pattern of green pixels located within one of the four quadrants. This is the "design" to be colored.
*   **The Prism (2x2 block of colors):** A small 2x2 cluster of colored pixels located in one of the corners of the 9x9 grid (outside the central cross). These colors represent the "light" to be cast.
*   **The Darkness (.):** Black pixels that fill the rest of the grid.

**2. Identify the AFFORDANCES:**
*   **The Green Tracery** can be transformed (recolored) by the colors in the prism.
*   **The Prism** can project its colors onto the quadrants. Specifically, the color at the prism's relative position (top-left, top-right, bottom-left, bottom-right) is applied to the corresponding quadrant of the 9x9 grid.
*   **The Azure Cross** can be stripped away, leaving only the 6x6 area (the four quadrants combined).

**3. Identify the TRANSFORMATION RULE:**
1.  Locate the 2x2 prism in one of the corners.
2.  Identify the four colors in the prism:
    *   Top-left color $\rightarrow$ applied to the Top-Left quadrant.
    *   Top-right color $\rightarrow$ applied to the Top-Right quadrant.
    *   Bottom-left color $\rightarrow$ applied to the Bottom-Left quadrant.
    *   Bottom-right color $\rightarrow$ applied to the Bottom-Right quadrant.
3.  Locate the green (G) pattern.
4.  For every green pixel, change its color to the color assigned to its quadrant.
5.  Remove the Azure (A) cross and the original prism/darkness, resulting in a 6x6 grid composed of the four 3x3 quadrants.

### Applying the Rule to Test Input

**Step 1: Locate the Prism**
The prism is in the bottom-right corner:
`R B`
`Y O`
*   Top-left: **R** (Red)
*   Top-right: **B** (Blue)
*   Bottom-left: **Y** (Yellow)
*   Bottom-right: **O** (Orange)

**Step 2: Identify Quadrants and Green Pattern**
The Azure cross is at row 7 and column 7 (index 6).
The quadrants are:
*   **Top-Left (TL):** Rows 1-3, Cols 1-3 (relative to the 6x6 output)
*   **Top-Right (TR):** Rows 1-3, Cols 4-6
*   **Bottom-Left (BL):** Rows 4-6, Cols 1-3
*   **Bottom-Right (BR):** Rows 4-6, Cols 4-6

**Green pattern locations (relative to 6x6):**
*   (1,1): G $\rightarrow$ TL $\rightarrow$ **R**
*   (1,2): G $\rightarrow$ TL $\rightarrow$ **R**
*   (1,4): G $\rightarrow$ TR $\rightarrow$ **B**
*   (1,6): G $\rightarrow$ TR $\rightarrow$ **B**
*   (2,1): G $\rightarrow$ TL $\rightarrow$ **R**
*   (2,3): G $\rightarrow$ TL $\rightarrow$ **R**
*   (2,5): G $\rightarrow$ TR $\rightarrow$ **B**
*   (3,1): G $\rightarrow$ TL $\rightarrow$ **R**
*   (3,2): G $\rightarrow$ TL $\rightarrow$ **R**
*   (3,3): G $\rightarrow$ TL $\rightarrow$ **R**
*   (4,1): G $\rightarrow$ BL $\rightarrow$ **Y**
*   (4,3): G $\rightarrow$ BL $\rightarrow$ **Y**
*   (5,2): G $\rightarrow$ BL $\rightarrow$ **Y**
*   (5,6): G $\rightarrow$ BR $\rightarrow$ **O**
*   (6,3): G $\rightarrow$ BL $\rightarrow$ **Y**

**Step 3: Construct the 6x6 Grid**
Row 1: (1,1)R, (1,2)R, (1,3). | (1,4)B, (1,5)., (1,6)B $\rightarrow$ `R R . B . B`
Row 2: (2,1)R, (2,2)., (2,3)R | (2,4)., (2,5)B, (2,6). $\rightarrow$ `R . R . B .`
Row 3: (3,1)R, (3,2)R, (3,3)R | (3,4)., (3,5)., (3,6). $\rightarrow$ `R R R . . .`
Row 4: (4,1)Y, (4,2)., (4,3)Y | (4,4)., (4,5)., (4,6). $\rightarrow$ `Y . Y . . .`
Row 5: (5,1)., (5,2)Y, (5,3). | (5,4)., (5,5)., (5,6)O $\rightarrow$ `. Y . . . O`
Row 6: (6,1)., (6,2)., (6,3)Y | (6,4)., (6,5)., (6,6). $\rightarrow$ `. . Y . . .`

Wait, let's re-verify the green pattern coordinates from the input:
Input (9x9):
Row 1: `. . . G . . A . .` -> (1,4) is G. In 6x6, this is (1,4) in TR.
Row 2: `G G . G . G A . .` -> (2,1)G, (2,2)G, (2,4)G, (2,6)G. In 6x6: (2,1)TL, (2,2)TL, (2,4)TR, (2,6)TR.
Row 3: `. G . G . G A . .` -> (3,2)G, (3,4)G, (3,6)G. In 6x6: (3,2)TL, (3,4)TR, (3,6)TR.
Row 4: `. G G G . . A . .` -> (4,2)G, (4,3)G, (4,4)G. In 6x6: (4,2)BL, (4,3)BL, (4,4)BR.
Row 5: `. G . . . G A . .` -> (5,2)G, (5,6)G. In 6x6: (5,2)BL, (5,6)BR.
Row 6: `. . G . . . A . .` -> (6,3)G. In 6x6: (6,3)BL.

Let's re-map carefully:
TL (R): (2,1), (2,2), (3,2)
TR (B): (1,4), (2,4), (2,6), (3,4), (3,6)
BL (Y): (4,2), (4,3), (5,2), (6,3)
BR (O): (4,4), (5,6)

Re-constructing 6x6:
R1: `. . . B . .`
R2: `R R . B . B`
R3: `. R . B . B`
R4: `. Y Y O . .`
R5: `. Y . . . O`
R6: `. . Y . . .`

Wait, let's check Row 1 again. Input Row 1 is `. . . G . . A . .`. The G is at index 4. In a 9x9 grid with A at index 7, the quadrants are 1-3, 4-6, 7-9.
Index 4 is the first column of the TR quadrant.
Row 2: `G G . G . G A . .`. Gs at 1, 2, 4, 6.
1,2 are TL. 4,6 are TR.
Row 3: `. G . G . G A . .`. Gs at 2, 4, 6.
2 is TL. 4,6 are TR.
Row 4: `. G G G . . A . .`. Gs at 2, 3, 4.
2,3 are BL. 4 is BR.
Row 5: `. G . . . G A . .`. Gs at 2, 6.
2 is BL. 6 is BR.
Row 6: `. . G . . . A . .`. G at 3.
3 

... [211 chars truncated]

### Conceptual Digestion

**1. Identify the CAST:**
*   **Azure (A):** The "cellular membrane." It acts as a divider/boundary. In the output, it disappears.
*   **Green (G):** The "skeleton of green tissue." This is the core structure that will be transformed.
*   **Pigment Cells (B, R, Y, O, etc.):** A small cluster of colored cells (usually 4) located in a "pocket" (a corner or area separated by the membrane).
*   **Black (.):** Empty space/background.

**2. Identify the AFFORDANCES:**
*   **Green (G):** Can be recolored based on its proximity to the pigment cells.
*   **Azure (A):** Defines the boundaries of the "chambers" and is removed in the final step.
*   **Pigment Cells:** Determine the color of the green tissue in the quadrant they "claim."

**3. Identify the TRANSFORMATION RULE:**
1.  **Identify the Skeleton:** Locate all Green (G) cells.
2.  **Identify the Pigments:** Locate the cluster of non-black, non-azure, non-green cells.
3.  **Divide into Quadrants:** The Azure (A) membrane divides the grid. The pigment cells are located in one area. The rule states each pigment cell claims the "quarter of the body nearest to its origin." Looking at the training example, the grid is effectively divided into four quadrants relative to the center/membrane.
4.  **Recolor:** Each Green (G) cell is replaced by the color of the pigment cell closest to it.
5.  **Cleanup:** Remove all Azure (A) cells and all original Green (G) cells (they are replaced). The output grid contains only the transformed colors and the background (.).

**Applying to Test Input:**
*   **Green (G) locations:**
    Row 0: Col 3
    Row 1: Col 0, 1, 3, 5
    Row 2: Col 1, 3, 5
    Row 3: Col 1, 2, 3
    Row 4: Col 1, 5
    Row 5: Col 2
*   **Pigment Cells (in the bottom right corner):**
    R (Red), B (Blue), Y (Yellow), O (Orange).
*   **Mapping Pigments to Green cells:**
    The pigments are at: (7,7)=R, (7,8)=B, (8,7)=Y, (8,8)=O.
    The Azure membrane (A) is at Row 6 and Col 6.
    The Green cells are all in the top-left area relative to the membrane.
    Wait, let's look at the training example more closely.
    Example 1: Pigments are B, R, Y. Green cells are transformed.
    In the test input, the Green cells are in the top-left. The pigments are in the bottom-right.
    The "quarters" are determined by the position of the pigments relative to the skeleton.
    Looking at the test input:
    The Green cells are in the top-left quadrant.
    The pigments are: R (7,7), B (7,8), Y (8,7), O (8,8).
    The Green cells are:
    (0,3), (1,0), (1,1), (1,3), (1,5), (2,1), (2,3), (2,5), (3,1), (3,2), (3,3), (4,1), (4,5), (5,2).
    
    Let's re-examine the "nearest" rule. The pigments are in the bottom right.
    The Green cells are in the top left.
    The pigments are:
    R: (7,7)
    B: (7,8)
    Y: (8,7)
    O: (8,8)
    
    Wait, the training example shows the output grid is smaller. The output grid is the bounding box of the *transformed* skeleton.
    In Example 1: Input is 9x9. Output is 6x6.
    In Test Input: The Green cells occupy rows 0-5 and columns 0-5.
    The pigments are in the bottom right.
    Let's find the closest pigment for each G:
    (0,3): Closest to R(7,7)? No, let's look at the grid structure.
    The Azure (A) forms a cross/boundary.
    The Green cells are in the top-left.
    The pigments are in the bottom-right.
    The rule "each claiming the quarter of the body nearest to its origin" implies we divide the *skeleton* into four parts based on the pigment positions.
    
    Looking at the pigments in the test input:
    R is top-left of the pigment cluster.
    B is top-right.
    Y is bottom-left.
    O is bottom-right.
    
    However, the Green cells are all in the top-left of the *entire grid*.
    Let's look at the proximity of G cells to the pigments:
    G(0,3) is closest to R(7,7).
    G(1,0) is closest to R(7,7).
    G(1,5) is closest to R(7,7).
    G(5,2) is closest to R(7,7).
    Actually, all G cells are closer to R than to B, Y, or O.
    
    Let's re-read: "Each quarter of the skeleton now displays the color of the pigment cell that claimed it."
    This means we divide the *skeleton* into 4 quadrants.
    The pigments are R, B, Y, O.
    In Example 1: Pigments were B, R, Y.
    The output grid in Example 1 is the bounding box of the G cells.
    In Test Input, the G cells are in rows 0-5, cols 0-5.
    The pigments are R, B, Y, O.
    If we divide the G-area (rows 0-5, cols 0-5) into 4 quadrants:
    Top-Left (TL): (0,0) to (2,2)
    Top-Right (TR): (0,3) to (2,5)
    Bottom-Left (BL): (3,0) to (5,2)
    Bottom-Right (BR): (3,3) to (5,5)
    
    Assigning pigments to quadrants based on their relative position to the skeleton:
    The pigments are in the bottom-right of the grid.
    Relative to the skeleton, the pigments are:
    R (top-left of cluster) -> TL
    B (top-right of cluster) -> TR
    Y (bottom-left of cluster) -> BL
    O (bottom-right of cluster) -> BR
    
    Le

... [5632 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Azure (A) Ridges:** These act as dividers/boundaries. They separate the "weather systems" from the "vegetated valley."
    *   **Green (G) Vegetation:** These are the target pixels that will be recolored.
    *   **Weather Systems (B, R, Y, M, X, O, etc.):** These are the source colors located in a "pocket" (a cluster of non-black, non-azure, non-green pixels).
    *   **Black (.) Ground:** Represents empty space or barren ground.

2.  **Identify the AFFORDANCES:**
    *   **Green (G)** can be recolored based on proximity to weather systems.
    *   **Weather Systems** can "drift" (influence) the green pixels.
    *   **Azure (A)** can be removed (eroded).
    *   **Black (.)** remains unchanged.

3.  **Identify the TRANSFORMATION RULE:**
    *   **Step 1: Identify the Valley and the Weather Systems.** The "valley" is the area containing the Green (G) pixels. The "weather systems" are the colored pixels (not A or G) located in a separate area.
    *   **Step 2: Divide the Valley.** The Azure (A) ridges divide the grid into quadrants/regions.
    *   **Step 3: Map Weather to Green.** Each weather system color is assigned to the nearest quadrant of the green vegetation. Looking at the examples, the "weather systems" are the colored pixels in the corner/pocket. The green pixels are recolored to the color of the weather system that is "closest" to that specific region of green.
    *   **Step 4: Clean up.** Remove all Azure (A) pixels. The output grid size is determined by the bounding box of the green pixels (the valley).

**Refining the Rule via Examples:**
*   **Example 1:** Weather systems are B, R, Y. Green is in a large area. The output shows the green pixels recolored to B, R, or Y based on their position relative to the weather cluster.
*   **Example 2:** Weather systems are R, Y, B, M. Green is divided into quadrants. The output shows the green pixels recolored to match the colors in the corner.
*   **Example 3:** Weather systems are R, Y, M, X. Green is divided by the A-line.

**Applying to Test Input:**
*   **Weather Systems:** R (Red), B (Blue), Y (Yellow), O (Orange).
*   **Green Area (The Valley):**
    Row 1: `. . . G . .`
    Row 2: `G G . G . G`
    Row 3: `. G . G . G`
    Row 4: `. G G G . .`
    Row 5: `. G . . . G`
    Row 6: `. . G . . .`
*   **The Azure Divider:** A horizontal line at Row 7.
*   **The Weather Pocket:** Below the line, we have:
    Row 8: `R B`
    Row 9: `Y O`
*   **Mapping:** The green area is above the line. The weather systems are below the line.
    The weather systems are arranged in a 2x2 block:
    R B
    Y O
    The green area is above them. We divide the green area into four quadrants (top-left, top-right, bottom-left, bottom-right) and assign the colors:
    Top-Left: R
    Top-Right: B
    Bottom-Left: Y
    Bottom-Right: O

**Wait, let's look at the spatial relationship more closely.**
In Example 2: Weather is top-left. Green is bottom-right. The green is split into quadrants.
In Test Input: Weather is bottom-right. Green is top-left.
The green area is:
(1,4)
(2,1)(2,2)(2,4)(2,6)
(3,2)(3,4)(3,6)
(4,2)(4,3)(4,4)
(5,2)(5,6)
(6,3)

Let's map the quadrants of the green area to the weather colors:
The weather colors are:
R (top-left of pocket)
B (top-right of pocket)
Y (bottom-left of pocket)
O (bottom-right of pocket)

The green area is above the pocket. The quadrants of the green area (relative to its own center) will be colored:
Top-Left: R
Top-Right: B
Bottom-Left: Y
Bottom-Right: O

**Let's trace the green pixels in the test input:**
Row 1: Col 4 (G) -> This is in the top-right quadrant of the green area. Color: B.
Row 2: Col 1 (G), Col 2 (G) -> Top-left. Color: R.
Row 2: Col 4 (G) -> Top-right. Color: B.
Row 2: Col 6 (G) -> Top-right. Color: B.
Row 3: Col 2 (G) -> Top-left. Color: R.
Row 3: Col 4 (G) -> Top-right. Color: B.
Row 3: Col 6 (G) -> Top-right. Color: B.
Row 4: Col 2 (G), Col 3 (G), Col 4 (G) -> Bottom-left/center.
Row 5: Col 2 (G) -> Bottom-left. Color: Y.
Row 5: Col 6 (G) -> Bottom-right. Color: O.
Row 6: Col 3 (G) -> Bottom-left. Color: Y.

Let's refine the quadrant division for the green area:
Green bounding box: Rows 1-6, Cols 1-6.
Center of bounding box: Row 3.5, Col 3.5.
Quadrant 1 (Top-Left: R): Rows 1-3, Cols 1-3.
Quadrant 2 (Top-Right: B): Rows 1-3, Cols 4-6.
Quadrant 3 (Bottom-Left: Y): Rows 4-6, Cols 1-3.
Quadrant 4 (Bottom-Right: O): Rows 4-6, Cols 4-6.

**Applying to Green Pixels:**
(1,4): R2, C4 -> Top-Right -> B
(2,1): R2, C1 -> Top-Left -> R
(2,2): R2, C2 -> Top-Left -> R
(2,4): R2, C4 -> Top-Right -> B
(2,6): R2, C6 -> Top-Right -> B
(3,2): R3, C2 -> Top-Left -> R
(3,4): R3, C4 -> Top-Right -> B
(3,6): R3, C6 -> Top-Right -> B
(4,2): R4, C2 -> Bottom-Left -> Y
(4,3): R4, C3 -> Bottom-Left -> Y
(4,4): R4, C4 -> Bottom-Right -> O
(5,2): R5, C2 -> Bottom-Left -> Y
(5,6

... [1290 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Lead (X):** The rigid skeleton/frame. It is static and does not change.
    *   **Jewels (R - Ruby, B - Sapphire):** The color sources. They are embedded in the lead and dictate the color of adjacent empty spaces.
    *   **Empty Panes (. - Black):** The target elements. These are the spaces within the lattice that will be transformed.

2.  **Identify the AFFORDANCES:**
    *   **X and R/B** are "fixed" elements; they provide the structure and the color information.
    *   **.** is a "malleable" element; it "looks" at its immediate neighbors (up, down, left, right) that are part of the frame (X, R, or B) to determine its new color.

3.  **Identify the TRANSFORMATION RULE:**
    *   For every empty cell (`.`), count the number of adjacent (up, down, left, right) colored jewels (`R` or `B`).
    *   If the count of `R` is greater than the count of `B`, the cell becomes `R`.
    *   If the count of `B` is greater than the count of `R`, the cell becomes `B`.
    *   If the counts are equal (including zero), the cell remains `.` (or follows the logic of the majority; in the training examples, if no jewels are adjacent or they are tied, the cell remains black/empty).
    *   *Refinement based on Example 1:* The rule applies to the "panes" (the connected components of `.`). A cell's color is determined by the jewels bordering its specific pane.

### Applying the Rule to Test Input

**Input Grid Analysis:**
Row 0: X B X X X X X X B X
Row 1: . . X . . X . . . .
Row 2: . . X . . B . . . .
Row 3: X X X X B X X R B X
Row 4: . . R . . R . . . .
Row 5: . . X . . X . . . .
Row 6: R X X X R X X B R X
Row 7: . . B . . R . . . .
Row 8: . . X . . X . . . .
Row 9: X R X X X X B X X B

**Pane 1 (Top Left, Rows 1-2, Cols 0-1):**
Neighbors: (0,1)=B, (3,0)=X, (3,1)=X, (0,0)=X, (1,2)=X, (2,2)=X.
Jewels bordering this pane: (0,1) is B.
Result: B B / B B (Wait, let's look at the neighbors of the specific cells).
(1,0) neighbors: (0,0)X, (2,0)., (1,1). -> No jewels.
(1,1) neighbors: (0,1)B, (2,1)., (1,0)., (1,2)X -> One B.
(2,0) neighbors: (1,0)., (3,0)X, (2,1). -> No jewels.
(2,1) neighbors: (1,1)., (3,1)X, (2,0)., (2,2)X -> No jewels.
*Correction:* Looking at Example 1, the entire "pane" (connected component of dots) takes the color of the majority jewel bordering that *entire* pane.
Pane 1 (Top Left): Bordered by (0,1)B. Result: B B / B B.

**Pane 2 (Top Middle, Rows 1-2, Cols 3-4):**
Bordered by: (0,3)X, (0,4)X, (3,4)B, (3,3)X, (1,5)X, (2,5)B, (1,2)X, (2,2)X.
Jewels: (3,4) is B, (2,5) is B.
Result: B B / B B.

**Pane 3 (Top Right, Rows 1-2, Cols 6-9):**
Bordered by: (0,8)B, (3,7)R, (3,8)B, (3,9)X, (0,6)X, (0,7)X, (0,9)X, (1,5)X, (2,5)B.
Jewels: (0,8)B, (3,7)R, (3,8)B, (2,5)B.
Counts: B=3, R=1.
Result: B B / B B / B B.

**Pane 4 (Middle Left, Rows 4-5, Cols 0-1):**
Bordered by: (3,0)X, (3,1)X, (6,0)R, (6,1)X, (4,2)R, (5,2)X.
Jewels: (6,0)R, (4,2)R.
Counts: R=2, B=0.
Result: R R / R R.

**Pane 5 (Middle Center, Rows 4-5, Cols 3-4):**
Bordered by: (3,4)B, (6,4)R, (4,2)R, (4,5)R, (5,5)X, (5,2)X.
Jewels: (3,4)B, (6,4)R, (4,2)R, (4,5)R.
Counts: R=3, B=1.
Result: R R / R R.

**Pane 6 (Middle Right, Rows 4-5, Cols 6-9):**
Bordered by: (3,7)R, (3,8)B, (3,9)X, (6,7)B, (6,8)R, (6,9)X, (4,5)R.
Jewels: (3,7)R, (3,8)B, (6,7)B, (6,8)R, (4,5)R.
Counts: R=3, B=2.
Result: R R / R R.

**Pane 7 (Bottom Left, Rows 7-8, Cols 0-1):**
Bordered by: (6,0)R, (6,1)X, (9,0)X, (9,1)R, (7,2)B, (8,2)X.
Jewels: (6,0)R, (9,1)R, (7,2)B.
Counts: R=2, B=1.
Result: R R / R R.

**Pane 8 (Bottom Middle, Rows 7-8, Cols 3-4):**
Bordered by: (6,4)R, (9,4)X, (7,2)B, (7,5)R, (8,5)X, (8,2)X.
Jewels: (6,4)R, (7,2)B, (7,5)R.
Counts: R=2, B=1.
Result: R R / R R.

**Pane 9 (Bottom Right, Rows 7-8, Cols 6-9):**
Bordered by: (6,7)B, (6,8)R, (6,9)X, (9,6)B, (9,7)X, (9,8)X, (9,9)B, (7,5)R.
Jewels: (6,7)B, (6,8)R, (9,6)B, (9,9)B, (7,5)R.
Counts: B=3, R=2.
Result: B B / B B.

**Final Assembly:**
Row 0: X B X X X X X X B X
Row 1: B B X B B X B B B B
Row 2: B B X B B B B B B B
Row 3: X X X X B X X R B X
Row 4: R R R R R R R R R R (Wait, the R in (4,2) and (4,5) are jewels, they don't change. The dots in (4,0-1) and (4,3-4) and (4,6-9) change.)
Row 4: R R R R R R R R R R (Wait, the dots are at (4,0), (4,1), (4,3), (4,4), (4,6), (4,7), (4,8), (4,9))
Row 5: R R X R R X R R R R
Row 6: R X X X R X X B R X
Row 7: R R B R R R B B B B
Row 8: R R X R R X B B B B
Row 9: X R X X X X B X X B

Let's re-verify Row 4:
(4,0) is ., (4,1) is ., (4,2) is R, (4,3) is ., (4,4) is ., (4,5) is R, (4,6) is ., (4,7) is ., (4,8) is ., (4,9) is .
Row 4: R R R R R R R R R R (Wait, the jewels R stay R).
Row 4: R R R R R R R R R R (Correct).

Let's re-verify Row 7:
(7,0) is ., (7,1) is ., (7,2) is B, (7,3) is ., (7,4) is ., (7,5) is R, (7,6) is ., (7,7) is ., (7,8) is ., (7,9) is .
Row 7: R R B R R R B B B B (Correct).

Let's re-verify Row 8:
(8,0) is ., (8,1) is ., (8,2) is X, (8,3) is ., (8,4) is ., (8,5) is X, (8,6) is ., (8,7) is ., (8,8)

... [810 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Grey fibers (X)**: These act as the boundaries or walls of the "cellular chambers." They are fixed and do not change.
    *   **Receptor proteins (R and B)**: These are the markers located on the boundaries. They determine the color of the interior.
    *   **Hollow cells (.)**: These are the empty spaces (the cytoplasm) that need to be filled.
    *   **Empty interiors**: The regions enclosed by the grey fibers.

2.  **Identify the AFFORDANCES**:
    *   **Grey fibers (X)**: Define the shape and limits of the chambers.
    *   **Receptors (R, B)**: Provide a signal (color) to the adjacent empty spaces.
    *   **Empty spaces (.)**: Can be transformed into either Red (R) or Blue (B) based on the surrounding receptors.

3.  **Identify the TRANSFORMATION RULE**:
    *   Identify each enclosed region of `.` (empty space) bounded by `X` (grey fibers).
    *   For each region, count the number of Red (R) receptors and Blue (B) receptors that touch the perimeter of that specific region.
    *   If the number of Blue (B) receptors is greater than the number of Red (R) receptors, fill the entire region with Blue (B).
    *   If the number of Red (R) receptors is greater than the number of Blue (B) receptors, fill the entire region with Red (R).
    *   If there is a tie (though examples suggest one will dominate), or if the rule is applied strictly to the boundary, the color of the majority wins.

### Applying the Rule to the Test Input

**Input Grid Analysis:**
```
Row 0: X B X X X X X X B X
Row 1: . . X . . X . . . .
Row 2: . . X . . B . . . .
Row 3: X X X X B X X R B X
Row 4: . . R . . R . . . .
Row 5: . . X . . X . . . .
Row 6: R X X X R X X B R X
Row 7: . . B . . R . . . .
Row 8: . . X . . X . . . .
Row 9: X R X X X X B X X B
```

**Chamber 1 (Top Left):**
Cells: (1,0), (1,1), (2,0), (2,1)
Boundary receptors: (0,1) is B.
Count: B=1, R=0.
Result: Fill with **B**.

**Chamber 2 (Top Middle):**
Cells: (1,3), (1,4), (2,3), (2,4)
Boundary receptors: (0,3) is X, (0,4) is X, (3,4) is B, (2,5) is B (Wait, (2,5) is B, but is it part of the boundary? Let's look at the X walls).
Walls for Chamber 2: (0,2)-(0,7), (1,2), (1,5), (2,2), (2,5), (3,0)-(3,4), (3,5)-(3,7).
Wait, let's re-examine the walls for the cell at (1,3), (1,4), (2,3), (2,4).
The walls are: (0,3), (0,4), (1,2), (1,5), (2,2), (2,5), (3,3), (3,4).
Receptors on this boundary: (3,4) is B, (2,5) is B.
Count: B=2, R=0.
Result: Fill with **B**.

**Chamber 3 (Top Right):**
Cells: (1,6), (1,7), (1,8), (1,9), (2,6), (2,7), (2,8), (2,9)
Boundary receptors: (0,8) is B, (3,7) is R, (3,8) is B.
Count: B=2, R=1.
Result: Fill with **B**.

**Chamber 4 (Middle Left):**
Cells: (4,0), (4,1), (5,0), (5,1)
Boundary receptors: (3,0) is X, (3,1) is X, (4,2) is R, (6,0) is R, (6,1) is X.
Wait, (4,2) is R, (6,0) is R.
Count: B=0, R=2.
Result: Fill with **R**.

**Chamber 5 (Middle Middle):**
Cells: (4,3), (4,4), (5,3), (5,4)
Boundary receptors: (3,4) is B, (4,2) is R, (4,5) is R, (6,4) is R.
Count: B=1, R=3.
Result: Fill with **R**.

**Chamber 6 (Middle Right):**
Cells: (4,6), (4,7), (4,8), (4,9), (5,6), (5,7), (5,8), (5,9)
Boundary receptors: (3,7) is R, (3,8) is B, (6,7) is B, (6,8) is R.
Count: B=2, R=2. (Wait, let's re-check).
(3,7) is R, (3,8) is B, (6,7) is B, (6,8) is R.
Looking at Example 1/2, if there's a tie, we look at the surrounding. Let's re-verify the boundary for (4,6)-(5,9).
Walls: (3,5)-(3,7), (3,8), (3,9), (6,5)-(6,9), (5,5), (4,5).
Receptors: (3,7)=R, (3,8)=B, (6,7)=B, (6,8)=R.
In Example 1, the tie-breaker seems to favor the color that fills the space. Let's look at the input again.
Actually, (4,6) is adjacent to (3,7)R and (3,8)B and (6,7)B and (6,8)R.
Wait, let's look at the input grid again.
(3,7) is R. (3,8) is B. (6,7) is B. (6,8) is R.
Let's check the count for Chamber 6 again.
Boundary: (3,5)X, (3,6)X, (3,7)R, (3,8)B, (3,9)X, (4,5)R, (4,9)X, (5,5)X, (5,9)X, (6,5)R, (6,6)X, (6,7)B, (6,8)R, (6,9)X.
Receptors: R: (3,7), (4,5), (6,5), (6,8). B: (3,8), (6,7).
Count: R=4, B=2.
Result: Fill with **R**.

**Chamber 7 (Bottom Left):**
Cells: (7,0), (7,1), (8,0), (8,1)
Boundary receptors: (6,0) is R, (7,2) is B, (9,1) is R.
Count: B=1, R=2.
Result: Fill with **R**.

**Chamber 8 (Bottom Middle):**
Cells: (7,3), (7,4), (8,3), (8,4)
Boundary receptors: (6,4) is R, (7,2) is B, (7,5) is R, (9,3) is X, (9,4) is X.
Wait, (7,2) is B, (6,4) is R, (7,5) is R.
Count: B=1, R=2.
Result: Fill with **R**.

**Chamber 9 (Bottom Right):**
Cells: (7,6), (7,7), (7,8), (7,9), (8,6), (8,7), (8,8), (8,9)
Boundary receptors: (6,7) is B, (6,8) is R, (9,6) is B, (9,9) is B.
Count: B=3, R=1.
Result: Fill with **B**.

**Final Grid Construction:**
Row 0: X B X X X X X X B X
Row 1: B B X B B X B B B B
Row 2: B B X B B B B B B B
Row 3: X X X X B X X R B X
Row 4: R R R R R R R R R R (Wait, (4,2) is R, (4,5) is R. Let's re-check row 4)
Row 4: (4,0)R, (4,1)R, (4,2)R, (4,3)R, (4,4)R, (4,5)R, (4,6)R, (4,7)R, (4,8)R, (4,9)R.
Wait, (4,2) is R

... [2703 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Filaments (X - Grey):** These act as the boundaries or walls that define the shapes of the "voids."
    *   **Stars (R - Red, B - Blue, etc.):** These are the sources of color. They are located on the filaments.
    *   **Voids (. - Black):** These are the empty spaces enclosed by the filaments.
    *   **Output Voids:** The black spaces that will be filled with a color based on the surrounding stars.

2.  **Identify the AFFORDANCES:**
    *   **Filaments (X)** remain static.
    *   **Stars (R, B, etc.)** remain static.
    *   **Voids (.)** can be "illuminated" (filled) with a color.
    *   The color of a void is determined by the majority color of the stars that touch its boundary (the filaments enclosing it).

3.  **Identify the TRANSFORMATION RULE:**
    *   Identify all connected components of black cells (`.`). These are the "voids."
    *   For each void, identify all the star cells (`R`, `B`, etc.) that are adjacent (horizontally or vertically) to the cells within that void.
    *   Count the occurrences of each star color adjacent to the void.
    *   Fill the entire void with the color that appears most frequently. If there is a tie or no stars, the void remains black (though the metaphor implies stars will be present).

---

### Applying the Rule to the Test Input

**Input Grid Analysis:**
```
Row 0: X B X X X X X X B X
Row 1: . . X . . X . . . .
Row 2: . . X . . B . . . .
Row 3: X X X X B X X R B X
Row 4: . . R . . R . . . .
Row 5: . . X . . X . . . .
Row 6: R X X X R X X B R X
Row 7: . . B . . R . . . .
Row 8: . . X . . X . . . .
Row 9: X R X X X X B X X B
```

**Step 1: Identify Voids and adjacent stars.**

*   **Void 1 (Top Left):** Cells (1,0), (1,1), (2,0), (2,1).
    *   Adjacent stars: (0,1) is B.
    *   Count: B=1.
    *   Fill with: **B**.

*   **Void 2 (Top Middle):** Cells (1,3), (1,4), (2,3), (2,4).
    *   Adjacent stars: (3,4) is B, (2,5) is B (Wait, (2,5) is B, but is it adjacent to the void? Let's check connectivity).
    *   Let's re-examine the grid carefully.
    *   Void 2 cells: (1,3), (1,4), (2,3), (2,4).
    *   Neighbors of Void 2: (0,3) is X, (0,4) is X, (1,2) is X, (1,5) is X, (2,2) is X, (2,5) is B, (3,3) is X, (3,4) is B.
    *   Stars touching Void 2: (2,5) is B, (3,4) is B.
    *   Count: B=2.
    *   Fill with: **B**.

*   **Void 3 (Top Right):** Cells (1,6), (1,7), (1,8), (1,9), (2,6), (2,7), (2,8), (2,9).
    *   Neighbors: (0,6)X, (0,7)X, (0,8)B, (0,9)X, (1,5)X, (1,10)X, (2,5)B, (3,6)X, (3,7)R, (3,8)B, (3,9)X.
    *   Stars touching Void 3: (0,8) is B, (2,5) is B, (3,7) is R, (3,8) is B.
    *   Count: B=3, R=1.
    *   Fill with: **B**.

*   **Void 4 (Middle Left):** Cells (4,0), (4,1), (5,0), (5,1).
    *   Neighbors: (3,0)X, (3,1)X, (4,2)R, (5,2)X, (6,0)R, (6,1)X.
    *   Stars touching Void 4: (4,2) is R, (6,0) is R.
    *   Count: R=2.
    *   Fill with: **R**.

*   **Void 5 (Middle Center):** Cells (4,3), (4,4), (5,3), (5,4).
    *   Neighbors: (3,3)X, (3,4)B, (4,2)R, (4,5)R, (5,2)X, (5,5)X, (6,3)X, (6,4)R.
    *   Stars touching Void 5: (3,4) is B, (4,2) is R, (4,5) is R, (6,4) is R.
    *   Count: R=3, B=1.
    *   Fill with: **R**.

*   **Void 6 (Middle Right):** Cells (4,6), (4,7), (4,8), (4,9), (5,6), (5,7), (5,8), (5,9).
    *   Neighbors: (3,6)X, (3,7)R, (3,8)B, (3,9)X, (4,5)R, (5,5)X, (6,6)X, (6,7)B, (6,8)R, (6,9)X.
    *   Stars touching Void 6: (3,7) is R, (3,8) is B, (4,5) is R, (6,7) is B, (6,8) is R.
    *   Count: R=3, B=2.
    *   Fill with: **R**.

*   **Void 7 (Bottom Left):** Cells (7,0), (7,1), (8,0), (8,1).
    *   Neighbors: (6,0)R, (6,1)X, (7,2)B, (8,2)X, (9,0)X, (9,1)R.
    *   Stars touching Void 7: (6,0) is R, (7,2) is B, (9,1) is R.
    *   Count: R=2, B=1.
    *   Fill with: **R**.

*   **Void 8 (Bottom Center):** Cells (7,3), (7,4), (8,3), (8,4).
    *   Neighbors: (6,3)X, (6,4)R, (7,2)B, (7,5)R, (8,2)X, (8,5)X, (9,3)X, (9,4)X.
    *   Stars touching Void 8: (6,4) is R, (7,2) is B, (7,5) is R.
    *   Count: R=2, B=1.
    *   Fill with: **R**.

*   **Void 9 (Bottom Right):** Cells (7,6), (7,7), (7,8), (7,9), (8,6), (8,7), (8,8), (8,9).
    *   Neighbors: (6,6)X, (6,7)B, (6,8)R, (6,9)X, (7,5)R, (8,5)X, (9,6)B, (9,7)X, (9,8)X, (9,9)B.
    *   Stars touching Void 9: (6,7) is B, (6,8) is R, (7,5) is R, (9,6) is B, (9,9) is B.
    *   Count: B=3, R=2.
    *   Fill with: **B**.

**Step 2: Construct the output grid.**

Row 0: X B X X X X X X B X
Row 1: B B X B B X B B B B
Row 2: B B X B B B B B B B
Row 3: X X X X B X X R B X
Row 4: R R R R R R R R R R (Wait, let's re-check row 4)
Row 4: (4,0)R, (4,1)R, (4,2)R, (4,3)R, (4,4)R, (4,5)R, (4,6)R, (4,7)R, (4,8)R, (4,9)R.
Wait, (4,2) is R, (4,5) is R. These are stars, not voids.
Let's re-map carefully.

Row 0: X B X X X X X X B X
Row 1: B B X B B X B B B B
Row 2: B B X B B B B B B B
Row 3: X X X X B X X R B X
Row 4: R R R R R R R R R R (Wait, (4,2) and (4,5) are R. They stay R.)
Row 5: R R X R R X R R R R
Row 6: R X X X R X X B

... [2888 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Reef Walls (X/Grey)**: These act as boundaries or partitions that define the shapes of the "lagoons." They are static and do not change.
    *   **Tide Markers (R/Red, B/Blue, etc.)**: These are specific colored cells located on the reef walls or at specific points. They serve as the "source" of color for the lagoons.
    *   **Lagoons (the '.'/Black areas)**: These are the empty spaces enclosed by the reef walls. They are the areas that will be filled.

2.  **Identify the AFFORDANCES**:
    *   **Reef Walls (X)**: Enclose spaces and provide the structure for the markers.
    *   **Tide Markers (R, B, etc.)**: Provide a color value to be counted.
    *   **Lagoons (.)**: Can be "flooded" (filled) with a single color based on the majority of markers surrounding them.

3.  **Identify the TRANSFORMATION RULE**:
    *   Identify all enclosed regions (lagoons) of `.` cells.
    *   For each enclosed region, look at the colored markers (R, B, etc.) that form the boundary of that specific region.
    *   Count the occurrences of each color on the boundary of that region.
    *   Fill the entire enclosed region with the color that appears most frequently on its boundary.
    *   If a region is bounded by only one color, fill it with that color.
    *   The reef walls (X) and the markers themselves remain unchanged.

### Applying the Rule to Test Input

**Input Analysis:**
The grid is 10x10. Let's identify the enclosed regions (lagoons) of `.` cells:

*   **Region 1 (Top Left):**
    Cells: (1,0), (1,1), (2,0), (2,1)
    Boundary markers: (0,1) is B, (3,0) is X (wall), (3,1) is X (wall), (0,0) is X (wall).
    Wait, let's look closer at the structure:
    Row 0: X B X X X X X X B X
    Row 1: . . X . . X . . . .
    Row 2: . . X . . B . . . .
    Row 3: X X X X B X X R B X
    Row 4: . . R . . R . . . .
    Row 5: . . X . . X . . . .
    Row 6: R X X X R X X B R X
    Row 7: . . B . . R . . . .
    Row 8: . . X . . X . . . .
    Row 9: X R X X X X B X X B

**Identifying Lagoons and Boundaries:**

1.  **Top-Left Lagoon (Rows 1-2, Cols 0-1):**
    Boundary: (0,1)=B, (3,0)=X, (3,1)=X, (0,0)=X, (1,2)=X, (2,2)=X.
    Markers on boundary: B (at 0,1).
    Color: **B**.
    Fill (1,0), (1,1), (2,0), (2,1) with B.

2.  **Top-Middle Lagoon (Rows 1-2, Cols 3-4):**
    Boundary: (0,3)=X, (0,4)=X, (3,4)=B, (3,3)=X, (1,2)=X, (2,2)=X, (1,5)=X, (2,5)=B.
    Markers on boundary: B (at 3,4), B (at 2,5).
    Color: **B**.
    Fill (1,3), (1,4), (2,3), (2,4) with B.

3.  **Top-Right Lagoon (Rows 1-2, Cols 5-9):**
    Boundary: (0,8)=B, (3,8)=B, (3,7)=R, (3,5)=X, (1,5)=X, (2,5)=B, (0,5)=X, (0,9)=X.
    Wait, let's trace the boundary for the area (1,6) to (2,9):
    Boundary: (0,8)=B, (3,8)=B, (3,7)=R, (3,5)=X, (1,5)=X, (0,5)=X, (0,9)=X.
    Markers: B (0,8), B (3,8), R (3,7), B (2,5).
    Count: B=3, R=1.
    Color: **B**.
    Fill (1,6), (1,7), (1,8), (1,9), (2,6), (2,7), (2,8), (2,9) with B.

4.  **Middle-Left Lagoon (Rows 4-5, Cols 0-1):**
    Boundary: (3,0)=X, (3,1)=X, (3,2)=X, (3,3)=X, (6,0)=R, (6,1)=X, (4,2)=R, (5,2)=X.
    Markers: R (6,0), R (4,2).
    Color: **R**.
    Fill (4,0), (4,1), (5,0), (5,1) with R.

5.  **Middle-Center Lagoon (Rows 4-5, Cols 3-4):**
    Boundary: (3,4)=B, (4,5)=R, (5,5)=X, (6,4)=R, (6,3)=X, (3,3)=X, (4,2)=R.
    Markers: B (3,4), R (4,5), R (6,4), R (4,2).
    Count: R=3, B=1.
    Color: **R**.
    Fill (4,3), (4,4), (5,3), (5,4) with R.

6.  **Middle-Right Lagoon (Rows 4-5, Cols 6-9):**
    Boundary: (3,7)=R, (3,8)=B, (3,9)=X, (6,9)=X, (6,7)=B, (6,8)=R, (5,5)=X, (4,5)=R.
    Wait, let's re-examine the boundary for (4,6-9) and (5,6-9):
    Boundary: (3,7)=R, (3,8)=B, (3,9)=X, (6,9)=X, (6,8)=R, (6,7)=B, (5,5)=X, (4,5)=R.
    Markers: R (3,7), B (3,8), R (6,8), B (6,7), R (4,5).
    Count: R=3, B=2.
    Color: **R**.
    Fill (4,6), (4,7), (4,8), (4,9), (5,6), (5,7), (5,8), (5,9) with R.

7.  **Bottom-Left Lagoon (Rows 7-8, Cols 0-1):**
    Boundary: (6,0)=R, (6,1)=X, (9,0)=X, (9,1)=R, (7,2)=B, (8,2)=X.
    Markers: R (6,0), R (9,1), B (7,2).
    Count: R=2, B=1.
    Color: **R**.
    Fill (7,0), (7,1), (8,0), (8,1) with R.

8.  **Bottom-Center Lagoon (Rows 7-8, Cols 3-4):**
    Boundary: (6,4)=R, (7,5)=R, (8,5)=X, (9,4)=X, (9,3)=X, (7,2)=B, (6,3)=X.
    Markers: R (6,4), R (7,5), B (7,2).
    Count: R=2, B=1.
    Color: **R**.
    Fill (7,3), (7,4), (8,3), (8,4) with R.

9.  **Bottom-Right Lagoon (Rows 7-8, Cols 6-9):**
    Boundary: (6,7)=B, (6,8)=R, (6,9)=X, (9,9)=B, (9,8)=X, (9,7)=X, (9,6)=B, (7,5)=R.
    Wait, let's re-check the boundary for (7,6-9) and (8,6-9):
    Boundary: (6,7)=B, (6,8)=R, (6,9)=X, (9,9)=B, (9,8)=X, (9,7)=X, (9,6)=B, (7,5)=R.
    Markers: B (6,7), R (6,8), B (9,9), B (9,6), R (7,5).
    Count: B=3, R=2.
    Color: **B**.
    Fill (7,6), (7,7), (7,8), (7,9), (8,6), (8,7), (8,8), (8,9) with B.

**Final Grid Construction:**
Row 0: X B X X X X X X B X
Row 1: B B X B B X B B B B
Row 2: B B X B B B B B B B
Row 3: X X X X

... [3030 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Iron Framework (X)**: The grey grid lines that act as boundaries. They remain unchanged.
    *   **Seasoning Crystals (R, B)**: Red and Blue cells located on the framework or at specific intersections. They act as the "source" of color.
    *   **Dough Pockets (.)**: The empty (black) cells within the framework. These are the "targets" that will be colored.

2.  **Identify the AFFORDANCES**:
    *   **Framework (X)**: Constrains the movement of color; defines the boundaries of the pockets.
    *   **Crystals (R, B)**: Provide color to adjacent or enclosed pockets.
    *   **Pockets (.)**: Absorb the color of the majority of the crystals that line their immediate boundaries.

3.  **Identify the TRANSFORMATION RULE**:
    *   Identify each contiguous "pocket" of black cells (`.`).
    *   For each pocket, look at the colors of the non-black cells (`R` or `B`) that form its immediate boundary (the cells directly adjacent horizontally or vertically).
    *   Count the occurrences of `R` and `B` around the perimeter of that specific pocket.
    *   The pocket cells are filled with the color that is more frequent among its boundary cells.
    *   If there is a tie or no color is present, the pocket remains black (though the examples suggest the pockets are always surrounded by crystals).

### Applying the Rule to Test Input

**Grid Analysis (Test Input):**
Row 0: `X B X X X X X X B X`
Row 1: `. . X . . X . . . .`
Row 2: `. . X . . B . . . .`
Row 3: `X X X X B X X R B X`
Row 4: `. . R . . R . . . .`
Row 5: `. . X . . X . . . .`
Row 6: `R X X X R X X B R X`
Row 7: `. . B . . R . . . .`
Row 8: `. . X . . X . . . .`
Row 9: `X R X X X X B X X B`

**Pocket Identification and Coloring:**

*   **Pocket 1 (Rows 1-2, Cols 0-1):**
    Boundary cells: (0,1)=B, (1,2)=X, (2,2)=X, (3,0)=X, (3,1)=X, (0,0)=X.
    Wait, let's look at the neighbors of the `.` cells:
    (1,0) neighbors: (0,0)=X, (1,1)=., (2,0)=.
    (1,1) neighbors: (0,1)=B, (1,0)=., (1,2)=X, (2,1)=.
    (2,0) neighbors: (1,0)=., (2,1)=., (3,0)=X
    (2,1) neighbors: (1,1)=., (2,0)=., (2,2)=X, (3,1)=X
    Boundary colors for this pocket: (0,1) is **B**.
    Result: Pocket 1 becomes **B**.

*   **Pocket 2 (Rows 1-2, Cols 3-4):**
    Boundary cells: (0,3)=X, (0,4)=X, (1,2)=X, (1,5)=X, (2,2)=X, (2,5)=B, (3,3)=X, (3,4)=B.
    Boundary colors: (2,5) is **B**, (3,4) is **B**.
    Result: Pocket 2 becomes **B**.

*   **Pocket 3 (Rows 1-2, Cols 6-9):**
    Boundary cells: (0,6)=X, (0,7)=X, (0,8)=B, (0,9)=X, (1,5)=X, (1,9)=., (2,5)=B, (2,9)=., (3,6)=X, (3,7)=R, (3,8)=B, (3,9)=X.
    Boundary colors: (0,8)=B, (2,5)=B, (3,7)=R, (3,8)=B.
    Counts: B=3, R=1.
    Result: Pocket 3 becomes **B**.

*   **Pocket 4 (Rows 4, Cols 0-1):**
    Boundary cells: (3,0)=X, (3,1)=X, (4,2)=R, (5,1)=., (5,2)=X, (5,0)=., (4,0)=.
    Wait, let's re-examine (4,0), (4,1), (5,0), (5,1).
    (4,0) neighbors: (3,0)=X, (4,1)=., (5,0)=.
    (4,1) neighbors: (3,1)=X, (4,0)=., (4,2)=R, (5,1)=.
    (5,0) neighbors: (4,0)=., (5,1)=., (6,0)=R
    (5,1) neighbors: (4,1)=., (5,0)=., (5,2)=X, (6,1)=X
    Boundary colors: (4,2)=R, (6,0)=R.
    Result: Pocket 4 becomes **R**.

*   **Pocket 5 (Rows 4, Cols 3-4):**
    Boundary cells: (3,3)=X, (3,4)=B, (4,2)=R, (4,5)=R, (5,3)=., (5,4)=., (5,5)=X, (6,4)=R.
    Boundary colors: (3,4)=B, (4,2)=R, (4,5)=R, (6,4)=R.
    Counts: R=3, B=1.
    Result: Pocket 5 becomes **R**.

*   **Pocket 6 (Rows 4-5, Cols 6-9):**
    Boundary cells: (3,6)=X, (3,7)=R, (3,8)=B, (3,9)=X, (4,5)=R, (4,9)=., (5,5)=X, (5,9)=., (6,6)=X, (6,7)=B, (6,8)=R, (6,9)=X.
    Boundary colors: (3,7)=R, (3,8)=B, (4,5)=R, (6,7)=B, (6,8)=R.
    Counts: R=3, B=2.
    Result: Pocket 6 becomes **R**.

*   **Pocket 7 (Rows 7-8, Cols 0-1):**
    Boundary cells: (6,0)=R, (6,1)=X, (7,2)=B, (8,2)=X, (9,1)=R, (9,0)=X.
    Boundary colors: (6,0)=R, (7,2)=B, (9,1)=R.
    Counts: R=2, B=1.
    Result: Pocket 7 becomes **R**.

*   **Pocket 8 (Rows 7-8, Cols 3-4):**
    Boundary cells: (6,4)=R, (7,2)=B, (7,5)=R, (8,2)=X, (8,5)=X, (9,3)=X, (9,4)=X.
    Boundary colors: (6,4)=R, (7,2)=B, (7,5)=R.
    Counts: R=2, B=1.
    Result: Pocket 8 becomes **R**.

*   **Pocket 9 (Rows 7-8, Cols 6-9):**
    Boundary cells: (6,7)=B, (6,8)=R, (6,9)=X, (7,5)=R, (7,9)=., (8,5)=X, (8,9)=., (9,6)=B, (9,7)=X, (9,8)=X, (9,9)=B.
    Boundary colors: (6,7)=B, (6,8)=R, (7,5)=R, (9,6)=B, (9,9)=B.
    Counts: B=3, R=2.
    Result: Pocket 9 becomes **B**.

**Final Grid Construction:**
Row 0: X B X X X X X X B X
Row 1: B B X B B X B B B B
Row 2: B B X B B B B B B B
Row 3: X X X X B X X R B X
Row 4: R R R R R R R R R R (Wait, let's re-check Row 4)
Row 4: (4,0)=R, (4,1)=R, (4,2)=R, (4,3)=R, (4,4)=R, (4,5)=R, (4,6)=R, (4,7)=R, (4,8)=R, (4,9)=R.
Wait, (4,2) and (4,5) are R in the input. They are not part of the pockets.
Let's re-verify Row 4:
(4,0) is ., (4,1) is ., (4,2) is R, (4,3) is ., (4,4) is ., (4,5) is R, (4,6) is ., (4,7) is ., (4,8) is ., (4,9) is .
Row 4: R R R R R R R R R R (No

... [3013 chars truncated]

### Conceptual Digestion

**1. Identify the CAST:**
*   **Fences (X - Grey):** These form the boundaries of the "unclaimed zones." They are static and do not change.
*   **Flags (R - Red, B - Blue):** These are the markers placed on the fences that determine the color of the zones.
*   **Unclaimed Zones (. - Black):** These are the empty areas enclosed by the grey fences.
*   **Captured Zones (R or B):** The empty areas once they have been filled with the color of the majority team.

**2. Identify the AFFORDANCES:**
*   **Fences (X)** define the topology of the grid, creating rectangular or irregular cells.
*   **Flags (R, B)** act as weights on the edges of these cells.
*   **Zones (.)** can be "flooded" (filled) with a color based on the sum of the flags on their surrounding boundaries.

**3. Identify the TRANSFORMATION RULE:**
*   The grid is divided into cells by the grey (X) lines.
*   For each cell (a contiguous block of '.' characters), count the number of Red (R) flags and Blue (B) flags that touch its perimeter.
*   If `count(R) > count(B)`, the entire cell becomes Red (R).
*   If `count(B) > count(R)`, the entire cell becomes Blue (B).
*   If `count(R) == count(B)`, the cell remains Black (.).
*   The grey fences (X) and the original flags (R, B) remain in their original positions.

### Applying the Rule to Test Input

**Grid Analysis (Identifying Cells):**
The grey lines (X) create several distinct zones:
1.  **Top-Left Zone:** (1,0) to (2,1) - 2 cells.
    *   Boundaries: (0,1) is B, (1,2) is X, (2,2) is X, (3,0) is X, (3,1) is X.
    *   Flags touching: (0,1) is B.
    *   Count: B=1, R=0. Result: **B**.
2.  **Top-Middle Zone:** (1,3) to (2,4) - 2 cells.
    *   Boundaries: (0,3) is X, (0,4) is X, (1,2) is X, (1,5) is X, (2,2) is X, (2,5) is B, (3,4) is B, (3,5) is X.
    *   Flags touching: (2,5) is B, (3,4) is B.
    *   Count: B=2, R=0. Result: **B**.
3.  **Top-Right Zone:** (1,6) to (2,9) - 4 cells.
    *   Boundaries: (0,8) is B, (0,9) is X, (1,5) is X, (2,5) is B, (3,5) is X, (3,8) is B, (3,9) is X.
    *   Flags touching: (0,8) is B, (2,5) is B, (3,8) is B.
    *   Count: B=3, R=0. Result: **B**.
4.  **Middle-Left Zone:** (4,0) to (5,1) - 2 cells.
    *   Boundaries: (3,0) is X, (3,1) is X, (4,2) is R, (5,2) is X, (6,1) is X, (6,0) is R.
    *   Flags touching: (4,2) is R, (6,0) is R.
    *   Count: R=2, B=0. Result: **R**.
5.  **Middle-Center Zone:** (4,3) to (5,4) - 2 cells.
    *   Boundaries: (3,4) is B, (3,5) is X, (4,5) is R, (5,5) is X, (6,5) is X, (6,4) is R, (6,3) is X, (4,2) is R.
    *   Flags touching: (3,4) is B, (4,5) is R, (6,4) is R, (4,2) is R.
    *   Count: R=3, B=1. Result: **R**.
6.  **Middle-Right Zone:** (4,6) to (5,9) - 4 cells.
    *   Boundaries: (3,7) is R, (3,8) is B, (3,9) is X, (4,5) is R, (5,5) is X, (6,5) is X, (6,7) is B, (6,8) is R, (6,9) is X.
    *   Flags touching: (3,7) is R, (3,8) is B, (4,5) is R, (6,7) is B, (6,8) is R.
    *   Count: R=3, B=2. Result: **R**.
7.  **Bottom-Left Zone:** (7,0) to (8,1) - 2 cells.
    *   Boundaries: (6,0) is R, (6,1) is X, (7,2) is B, (8,2) is X, (9,2) is X, (9,1) is R, (9,0) is X.
    *   Flags touching: (6,0) is R, (7,2) is B, (9,1) is R.
    *   Count: R=2, B=1. Result: **R**.
8.  **Bottom-Middle Zone:** (7,3) to (8,4) - 2 cells.
    *   Boundaries: (6,3) is X, (6,4) is R, (6,5) is X, (7,5) is R, (8,5) is X, (9,5) is X, (9,4) is X, (9,3) is X, (7,2) is B.
    *   Flags touching: (6,4) is R, (7,5) is R, (7,2) is B.
    *   Count: R=2, B=1. Result: **R**.
9.  **Bottom-Right Zone:** (7,6) to (8,9) - 4 cells.
    *   Boundaries: (6,7) is B, (6,8) is R, (6,9) is X, (7,5) is R, (8,5) is X, (9,5) is X, (9,6) is B, (9,7) is X, (9,8) is X, (9,9) is B.
    *   Flags touching: (6,7) is B, (6,8) is R, (7,5) is R, (9,6) is B, (9,9) is B.
    *   Count: B=3, R=2. Result: **B**.

**Constructing the Output Grid:**
Row 0: X B X X X X X X B X (Original)
Row 1: B B X B B X B B B B (Zones 1, 2, 3)
Row 2: B B X B B B B B B B (Zones 1, 2, 3)
Row 3: X X X X B X X R B X (Original)
Row 4: R R R R R R R R R R (Zones 4, 5, 6)
Row 5: R R X R R X R R R R (Zones 4, 5, 6)
Row 6: R X X X R X X B R X (Original)
Row 7: R R B R R R B B B B (Zones 7, 8, 9)
Row 8: R R X R R X B B B B (Zones 7, 8, 9)
Row 9: X R X X X X B X X B (Original)

*Self-Correction on Row 4/5:*
Row 4: (4,0-1) is R, (4,3-4) is R, (4,6-9) is R.
Row 5: (5,0-1) is R, (5,3-4) is R, (5,6-9) is R.
Row 7: (7,0-1) is R, (7,3-4) is R, (7,6-9) is B.
Row 8: (8,0-1) is R, (8,3-4) is R, (8,6-9) is B.

Let's re-verify Row 1/2:
Row 1: (1,0-1) is B, (1,3-4) is B, (1,6-9) is B.
Row 2: (2,0-1) is B, (2,3-4) is B, (2,6-9) is B.

Final check of the grid structure:
Row 0: X B X X X X X X B X
Row 1: B B X B B X B B B B
Row 2: B B X B B B B B B B
Row 3: X X X X B X X R B X
Row 4: R R R R R R R R R R
Row 5: R R X R R X R R R R
Row 6: R X X X R X X B R X
Row 7: R R B R R R B B B B
Row 8: R R X R R X B B B B
Row 9: X R X X X X B X X B

Wait, looking at Row 1/2 again:
Row 1: (1,0) is ., (1,1) is . -

... [2882 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Grey (X):** The "Mountain Ridges" or lattice. These act as boundaries/walls that define enclosed areas.
    *   **Red (R) and Blue (B):** The "Weather Beacons." These are markers placed on the ridges.
    *   **Black (.):** The "Valleys." These are the empty spaces enclosed by the grey ridges.
    *   **Other colors:** Not present in the core logic, but the output fills the black spaces.

2.  **Identify the AFFORDANCES:**
    *   **Grey (X)** elements are static; they define the structure and do not change.
    *   **Red (R) and Blue (B)** elements on the ridges act as "votes" for the color of the adjacent empty space.
    *   **Black (.)** spaces are "fillable." They observe the count of R and B beacons on their immediate boundary (the surrounding X lines) and transform into the color with the majority.

3.  **Identify the TRANSFORMATION RULE:**
    *   Identify all contiguous regions of black (.) cells.
    *   For each region, count the number of Red (R) and Blue (B) beacons that touch the boundary of that specific region.
    *   If `count(R) > count(B)`, fill the entire region with Red (R).
    *   If `count(B) > count(R)`, fill the entire region with Blue (B).
    *   If `count(R) == count(B)`, the region remains black (or follows a default, but usually, the prompt implies a clear majority).
    *   The grey (X) ridges and the beacons themselves remain unchanged.

### Applying the Rule to Test Input

**Input Grid Analysis:**
```
Row 0: X B X X X X X X B X
Row 1: . . X . . X . . . .
Row 2: . . X . . B . . . .
Row 3: X X X X B X X R B X
Row 4: . . R . . R . . . .
Row 5: . . X . . X . . . .
Row 6: R X X X R X X B R X
Row 7: . . B . . R . . . .
Row 8: . . X . . X . . . .
Row 9: X R X X X X B X X B
```

**Identify Regions and Count Beacons:**

*   **Region 1 (Top Left):** Cells (1,0), (1,1), (2,0), (2,1).
    *   Boundary: (0,1) is B, (3,0) is X, (3,1) is X, (0,0) is X, (1,2) is X, (2,2) is X.
    *   Beacons touching: (0,1) is **B**.
    *   Count: R=0, B=1.
    *   Result: Fill with **B**.

*   **Region 2 (Top Middle):** Cells (1,3), (1,4), (2,3), (2,4).
    *   Boundary: (0,3) is X, (0,4) is X, (3,3) is X, (3,4) is B, (1,2) is X, (1,5) is X, (2,2) is X, (2,5) is B.
    *   Beacons touching: (3,4) is **B**, (2,5) is **B**.
    *   Count: R=0, B=2.
    *   Result: Fill with **B**.

*   **Region 3 (Top Right):** Cells (1,5) is X, (1,6), (1,7), (1,8), (1,9), (2,6), (2,7), (2,8), (2,9).
    *   Boundary: (0,6) is X, (0,7) is X, (0,8) is B, (0,9) is X, (3,6) is X, (3,7) is R, (3,8) is B, (3,9) is X, (1,5) is X, (2,5) is B.
    *   Beacons touching: (0,8) is **B**, (3,7) is **R**, (3,8) is **B**, (2,5) is **B**.
    *   Count: R=1, B=3.
    *   Result: Fill with **B**.

*   **Region 4 (Middle Left):** Cells (4,0), (4,1), (5,0), (5,1).
    *   Boundary: (3,0) is X, (3,1) is X, (6,0) is R, (6,1) is X, (4,2) is R, (5,2) is X.
    *   Beacons touching: (6,0) is **R**, (4,2) is **R**.
    *   Count: R=2, B=0.
    *   Result: Fill with **R**.

*   **Region 5 (Middle Center):** Cells (4,3), (4,4), (5,3), (5,4).
    *   Boundary: (3,3) is X, (3,4) is B, (3,5) is X, (4,2) is R, (4,5) is R, (5,2) is X, (5,5) is X, (6,3) is X, (6,4) is R, (6,5) is X.
    *   Beacons touching: (3,4) is **B**, (4,2) is **R**, (4,5) is **R**, (6,4) is **R**.
    *   Count: R=3, B=1.
    *   Result: Fill with **R**.

*   **Region 6 (Middle Right):** Cells (4,6), (4,7), (4,8), (4,9), (5,6), (5,7), (5,8), (5,9).
    *   Boundary: (3,6) is X, (3,7) is R, (3,8) is B, (3,9) is X, (6,6) is X, (6,7) is B, (6,8) is R, (6,9) is X, (4,5) is R, (5,5) is X.
    *   Beacons touching: (3,7) is **R**, (3,8) is **B**, (6,7) is **B**, (6,8) is **R**, (4,5) is **R**.
    *   Count: R=3, B=2.
    *   Result: Fill with **R**.

*   **Region 7 (Bottom Left):** Cells (7,0), (7,1), (8,0), (8,1).
    *   Boundary: (6,0) is R, (6,1) is X, (9,0) is X, (9,1) is R, (7,2) is B, (8,2) is X.
    *   Beacons touching: (6,0) is **R**, (9,1) is **R**, (7,2) is **B**.
    *   Count: R=2, B=1.
    *   Result: Fill with **R**.

*   **Region 8 (Bottom Center):** Cells (7,3), (7,4), (8,3), (8,4).
    *   Boundary: (6,3) is X, (6,4) is R, (6,5) is X, (7,2) is B, (7,5) is R, (8,2) is X, (8,5) is X, (9,3) is X, (9,4) is X.
    *   Beacons touching: (6,4) is **R**, (7,2) is **B**, (7,5) is **R**.
    *   Count: R=2, B=1.
    *   Result: Fill with **R**.

*   **Region 9 (Bottom Right):** Cells (7,6), (7,7), (7,8), (7,9), (8,6), (8,7), (8,8), (8,9).
    *   Boundary: (6,6) is X, (6,7) is B, (6,8) is R, (6,9) is X, (9,6) is B, (9,7) is X, (9,8) is X, (9,9) is B, (7,5) is R, (8,5) is X.
    *   Beacons touching: (6,7) is **B**, (6,8) is **R**, (9,6) is **B**, (9,9) is **B**, (7,5) is **R**.
    *   Count: R=2, B=3.
    *   Result: Fill with **B**.

**Final Grid Construction:**
Row 0: X B X X X X X X B X
Row 1: B B X B B X B B B B
Row 2: B B X B B B B B B B
Row 3: X X X X B X X R B X
Row 4: R R R R R R R R R R
Row 5: R R 

... [2942 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Walls (X/Grey):** These form the boundaries of the "rooms." They are structural and do not change.
    *   **Keystones (R/Red and B/Blue):** These are specific colored markers placed on or near the walls.
    *   **Room Interiors (Dots/Black):** These are the empty spaces enclosed by the walls.
    *   **Output (Colored cells):** The empty spaces (dots) are filled with a color based on the surrounding walls.

2.  **Identify the AFFORDANCES:**
    *   **Walls (X)** define the topology and partition the grid into distinct enclosed regions (rooms).
    *   **Keystones (R, B)** act as "votes" for the color of the room they are associated with.
    *   **Room Interiors (.)** can be transformed (painted) into a solid color.

3.  **Identify the TRANSFORMATION RULE:**
    *   Identify all enclosed regions (rooms) formed by the grey (X) walls.
    *   For each room, look at the colored keystones (R or B) that lie on its boundary walls.
    *   Count the number of Red (R) keystones and Blue (B) keystones associated with that specific room.
    *   Fill the entire interior of that room with the color that has the majority count.
    *   If there is a tie or no keystones, the behavior isn't explicitly defined, but based on examples, the room is filled by the dominant color found on its perimeter.

### Applying the Rule to Test Input

**Grid Analysis (Test Input):**
Walls (X) create several enclosed areas. Let's identify the rooms:

*   **Room 1 (Top Left):**
    Boundaries: (0,0)X, (0,1)B, (0,2)X, (1,2)X, (2,2)X, (3,2)X, (3,1)X, (3,0)X.
    Keystones on boundary: B at (0,1).
    Majority: **B**.
    Cells: (1,0), (1,1), (2,0), (2,1).

*   **Room 2 (Top Middle/Right):**
    Boundaries: (0,2)X, (0,3)X, (0,4)X, (0,5)X, (0,6)X, (0,7)X, (0,8)B, (0,9)X, (1,5)X, (2,5)B, (3,5)X, (3,4)B, (3,3)X, (3,2)X, (2,2)X, (1,2)X.
    Wait, let's look at the structure more carefully.
    The walls are:
    Row 0: X B X X X X X X B X
    Row 1: . . X . . X . . . .
    Row 2: . . X . . B . . . .
    Row 3: X X X X B X X R B X
    Row 4: . . R . . R . . . .
    Row 5: . . X . . X . . . .
    Row 6: R X X X R X X B R X
    Row 7: . . B . . R . . . .
    Row 8: . . X . . X . . . .
    Row 9: X R X X X X B X X B

**Room Identification:**
1.  **Top-Left Room:** (1,0), (1,1), (2,0), (2,1). Boundary: (0,0)X, (0,1)B, (0,2)X, (1,2)X, (2,2)X, (3,2)X, (3,1)X, (3,0)X. Keystones: B. Result: **B**.
2.  **Top-Middle Room:** (1,3), (1,4), (2,3), (2,4). Boundary: (0,2)X, (0,3)X, (0,4)X, (0,5)X, (1,5)X, (2,5)B, (3,5)X, (3,4)B, (3,3)X, (3,2)X, (2,2)X, (1,2)X. Keystones: B (at 2,5) and B (at 3,4). Result: **B**.
3.  **Top-Right Room:** (1,6), (1,7), (1,8), (1,9), (2,6), (2,7), (2,8), (2,9). Boundary: (0,5)X, (0,6)X, (0,7)X, (0,8)B, (0,9)X, (1,9)X, (2,9)X, (3,9)X, (3,8)B, (3,7)R, (3,6)X, (3,5)X, (2,5)B, (1,5)X. Keystones: B (0,8), B (3,8), R (3,7), B (2,5). Count: B=3, R=1. Result: **B**.
4.  **Middle-Left Room:** (4,0), (4,1), (5,0), (5,1). Boundary: (3,0)X, (3,1)X, (3,2)X, (4,2)X (wait, 4,2 is R), (5,2)X, (6,2)X, (6,1)X, (6,0)R. Keystones: R (4,2), R (6,0). Result: **R**.
5.  **Middle-Middle Room:** (4,3), (4,4), (5,3), (5,4). Boundary: (3,3)X, (3,4)B, (3,5)X, (4,5)R, (5,5)X, (6,5)X, (6,4)R, (6,3)X, (6,2)X, (5,2)X, (4,2)R. Keystones: B (3,4), R (4,5), R (6,4), R (4,2). Result: **R**.
6.  **Middle-Right Room:** (4,6), (4,7), (4,8), (4,9), (5,6), (5,7), (5,8), (5,9). Boundary: (3,5)X, (3,6)X, (3,7)R, (3,8)B, (3,9)X, (4,9)X, (5,9)X, (6,9)X, (6,8)R, (6,7)B, (6,6)X, (6,5)X, (5,5)X, (4,5)R. Keystones: R (3,7), B (3,8), R (6,8), B (6,7), R (4,5). Count: R=3, B=2. Result: **R**.
7.  **Bottom-Left Room:** (7,0), (7,1), (8,0), (8,1). Boundary: (6,0)R, (6,1)X, (6,2)X, (7,2)X (wait, 7,2 is B), (8,2)X, (9,2)X, (9,1)R, (9,0)X, (8,0)X, (7,0)X. Keystones: R (6,0), B (7,2), R (9,1). Result: **R**.
8.  **Bottom-Middle Room:** (7,3), (7,4), (8,3), (8,4). Boundary: (6,3)X, (6,4)R, (6,5)X, (7,5)R, (8,5)X, (9,5)X, (9,4)X, (9,3)X, (9,2)X, (8,2)X, (7,2)B, (6,2)X. Keystones: R (6,4), R (7,5), B (7,2). Result: **R**.
9.  **Bottom-Right Room:** (7,6), (7,7), (7,8), (7,9), (8,6), (8,7), (8,8), (8,9). Boundary: (6,5)X, (6,6)X, (6,7)B, (6,8)R, (6,9)X, (7,9)X, (8,9)X, (9,9)B, (9,8)X, (9,7)X, (9,6)B, (8,6)X, (7,6)X. Keystones: B (6,7), R (6,8), B (9,9), B (9,6). Result: **B**.

**Final Grid Construction:**
Row 0: X B X X X X X X B X (Unchanged)
Row 1: B B X B B X B B B B (Room 1: B, Room 2: B, Room 3: B)
Row 2: B B X B B B B B B B (Room 1: B, Room 2: B, Room 3: B)
Row 3: X X X X B X X R B X (Unchanged)
Row 4: R R R R R R R R R R (Room 4: R, Room 5: R, Room 6: R)
Row 5: R R X R R X R R R R (Room 4: R, Room 5: R, Room 6: R)
Row 6: R X X X R X X B R X (Unchanged)
Row 7: R R B R R R B B B B (Room 7: R, Room 8: R, Room 9: B)
Row 8: R R X R R X B B B B (Room 7: R, Room 8: R, Room 9: B)
Row 9: X R X X X X B X X B (Unchanged)

*Self-Correction on Row 4/5/7/8:*
Room 4 (4,0-1, 5,0-1) -> R
Room 5 (4,3-4, 5,3-4) -> R
Room 6 (4,6-9, 5,6

... [2686 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Grey (X):** The "fortification walls" or boundaries. These are fixed and do not change.
    *   **Red (R) and Blue (B):** The "sentries." These are fixed markers placed on the walls.
    *   **Black (.):** The "empty territories." These are the cells that will be colored based on the surrounding sentries.

2.  **Identify the AFFORDANCES:**
    *   **Grey (X)** cells act as boundaries that define the edges of territories.
    *   **Red (R) and Blue (B)** cells act as counters. They are located on the boundaries.
    *   **Black (.)** cells are the targets for transformation. They "surrender" (change color) to either Red or Blue.

3.  **Identify the TRANSFORMATION RULE:**
    *   The grid is divided into enclosed regions (territories) by the grey (X) lines.
    *   For each empty cell (.), identify the "walls" (X cells) that form its immediate boundary.
    *   Count the number of Red (R) sentries and Blue (B) sentries located on the grey walls that enclose that specific territory.
    *   If Red sentries > Blue sentries, the empty cell becomes Red (R).
    *   If Blue sentries > Red sentries, the empty cell becomes Blue (B).
    *   If they are equal, the cell remains black (.).
    *   *Refinement based on Example 1:* The "territory" is the entire connected component of empty cells. All cells in a single connected component of '.' cells are colored based on the total count of R and B sentries on the X-walls surrounding that specific component.

### Applying the Rule to Test Input

**Step 1: Identify connected components of '.' cells.**
*   **Component 1:**
    (1,0), (1,1), (2,0), (2,1)
    (1,3), (1,4), (2,3), (2,4)
    (1,6), (1,7), (1,8), (1,9), (2,6), (2,7), (2,8), (2,9)
    (4,0), (4,1), (5,0), (5,1)
    (4,3), (4,4), (5,3), (5,4)
    (4,6), (4,7), (4,8), (4,9), (5,6), (5,7), (5,8), (5,9)
    (7,0), (7,1), (8,0), (8,1)
    (7,3), (7,4), (8,3), (8,4)
    (7,6), (7,7), (7,8), (7,9), (8,6), (8,7), (8,8), (8,9)

**Step 2: Count sentries for each component.**

*   **Component 1 (Top Left):** (1,0), (1,1), (2,0), (2,1)
    Surrounding walls: (0,0)X, (0,1)B, (0,2)X, (1,2)X, (2,2)X, (3,2)X, (3,1)X, (3,0)X.
    Sentries: B=1, R=0.
    Result: **Blue (B)**.

*   **Component 2 (Top Middle):** (1,3), (1,4), (2,3), (2,4)
    Surrounding walls: (0,3)X, (0,4)X, (1,2)X, (1,5)X, (2,2)X, (2,5)B, (3,5)X, (3,4)B, (3,3)X.
    Sentries: B=2, R=0.
    Result: **Blue (B)**.

*   **Component 3 (Top Right):** (1,6), (1,7), (1,8), (1,9), (2,6), (2,7), (2,8), (2,9)
    Surrounding walls: (0,6)X, (0,7)X, (0,8)X, (0,9)X, (1,5)X, (2,5)B, (3,5)X, (3,6)X, (3,7)R, (3,8)B, (3,9)X.
    Sentries: B=2, R=1.
    Result: **Blue (B)**.

*   **Component 4 (Mid Left):** (4,0), (4,1), (5,0), (5,1)
    Surrounding walls: (3,0)X, (3,1)X, (3,2)X, (4,2)R, (5,2)X, (6,2)X, (6,1)X, (6,0)R.
    Sentries: B=0, R=2.
    Result: **Red (R)**.

*   **Component 5 (Mid Middle):** (4,3), (4,4), (5,3), (5,4)
    Surrounding walls: (3,3)X, (3,4)B, (3,5)X, (4,5)R, (5,5)X, (6,5)X, (6,4)R, (6,3)X.
    Sentries: B=1, R=2.
    Result: **Red (R)**.

*   **Component 6 (Mid Right):** (4,6), (4,7), (4,8), (4,9), (5,6), (5,7), (5,8), (5,9)
    Surrounding walls: (3,5)X, (3,6)X, (3,7)R, (3,8)B, (3,9)X, (4,5)R, (5,5)X, (6,5)X, (6,6)X, (6,7)B, (6,8)R, (6,9)X.
    Sentries: B=2, R=3.
    Result: **Red (R)**.

*   **Component 7 (Bottom Left):** (7,0), (7,1), (8,0), (8,1)
    Surrounding walls: (6,0)R, (6,1)X, (6,2)X, (7,2)X, (8,2)X, (9,2)X, (9,1)R, (9,0)X.
    Sentries: B=0, R=2.
    Result: **Red (R)**.

*   **Component 8 (Bottom Middle):** (7,3), (7,4), (8,3), (8,4)
    Surrounding walls: (6,3)X, (6,4)R, (6,5)X, (7,5)R, (8,5)X, (9,5)X, (9,4)X, (9,3)X.
    Sentries: B=0, R=2.
    Result: **Red (R)**.

*   **Component 9 (Bottom Right):** (7,6), (7,7), (7,8), (7,9), (8,6), (8,7), (8,8), (8,9)
    Surrounding walls: (6,5)X, (6,6)X, (6,7)B, (6,8)R, (6,9)X, (7,5)R, (8,5)X, (9,5)X, (9,6)B, (9,7)X, (9,8)X, (9,9)B.
    Sentries: B=3, R=2.
    Result: **Blue (B)**.

**Step 3: Construct the output grid.**

Row 0: X B X X X X X X B X
Row 1: B B X B B X B B B B
Row 2: B B X B B B B B B B
Row 3: X X X X B X X R B X
Row 4: R R R R R R R R R R (Wait, let's re-check component 4, 5, 6)
*Correction on Row 4/5:*
Component 4 (4,0-1, 5,0-1) -> R
Component 5 (4,3-4, 5,3-4) -> R
Component 6 (4,6-9, 5,6-9) -> R
Row 4: R R R R R R R R R R (Wait, the X's are fixed)
Row 4: R R R R R R R R R R is wrong. The X's must stay.
Row 4: R R R R R R R R R R -> R R R R R R R R R R (No, X is at 4,2 and 4,5)
Row 4: R R X R R R R R R R (Wait, let's look at the input row 4: . . R . . R . . . .)
Input Row 4: . . R . . R . . . . (Indices 0,1,2,3,4,5,6,7,8,9)
Index 2 is R, Index 5 is R.
So Row 4 is: R R R R R R R R R R? No, the R's are sentries on the walls.
The input row 4 is: . . R . . R . . . .
Wait, the input says:
Row 4: . . R . . R . . . .
This means (4,2) is R and (4,5) is R. These are sentries.
The empty cells are (4,0), (4,1), (4,3)

... [3008 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Catwalks (X - Grey):** The structural framework that divides the grid.
    *   **Spotlight Fixtures (R - Red, B - Blue):** The colored markers placed on the catwalks.
    *   **Dark Sections (. - Black):** The empty areas enclosed or bordered by the catwalks.
    *   **Output (Colored cells):** The illuminated dark sections.

2.  **Identify the AFFORDANCES:**
    *   **Catwalks (X):** Act as boundaries. They do not change.
    *   **Fixtures (R, B):** Act as sources of light. They are counted by the adjacent dark sections.
    *   **Dark Sections (.):** Can be "illuminated" (changed from `.` to `R` or `B`) based on the majority color of the fixtures bordering them.

3.  **Identify the TRANSFORMATION RULE:**
    *   Identify each contiguous "dark section" (a connected component of `.` cells).
    *   For each dark section, look at all the colored fixtures (`R` or `B`) that are immediately adjacent (up, down, left, or right) to any cell in that dark section.
    *   Count the number of Red (`R`) fixtures and Blue (`B`) fixtures bordering that specific dark section.
    *   If `count(R) > count(B)`, change all `.` in that dark section to `R`.
    *   If `count(B) > count(R)`, change all `.` in that dark section to `B`.
    *   If `count(R) == count(B)`, the section remains `.`.

### Applying the Rule to Test Input

**Input Grid Analysis:**
Row 0: X B X X X X X X B X
Row 1: . . X . . X . . . .
Row 2: . . X . . B . . . .
Row 3: X X X X B X X R B X
Row 4: . . R . . R . . . .
Row 5: . . X . . X . . . .
Row 6: R X X X R X X B R X
Row 7: . . B . . R . . . .
Row 8: . . X . . X . . . .
Row 9: X R X X X X B X X B

**Identify Dark Sections and their Neighbors:**

*   **Section 1 (Top Left):** Cells (1,0), (1,1), (2,0), (2,1).
    *   Neighbors: (0,1) is B. (3,0) is X, (3,1) is X. (1,2) is X, (2,2) is X.
    *   Fixtures: B (at 0,1).
    *   Counts: B=1, R=0. Result: **B**.

*   **Section 2 (Top Middle):** Cells (1,3), (1,4), (2,3), (2,4).
    *   Neighbors: (0,3) is X, (0,4) is X. (3,3) is X, (3,4) is B. (1,2) is X, (1,5) is X. (2,2) is X, (2,5) is B.
    *   Fixtures: B (at 3,4), B (at 2,5).
    *   Counts: B=2, R=0. Result: **B**.

*   **Section 3 (Top Right):** Cells (1,6), (1,7), (1,8), (1,9), (2,6), (2,7), (2,8), (2,9).
    *   Neighbors: (0,6) is X, (0,7) is X, (0,8) is B, (0,9) is X. (3,6) is X, (3,7) is R, (3,8) is B, (3,9) is X. (1,5) is X, (2,5) is B.
    *   Fixtures: B (at 0,8), R (at 3,7), B (at 3,8), B (at 2,5).
    *   Counts: B=3, R=1. Result: **B**.

*   **Section 4 (Middle Left):** Cells (4,0), (4,1), (5,0), (5,1).
    *   Neighbors: (3,0) is X, (3,1) is X. (6,0) is R, (6,1) is X. (4,2) is R, (5,2) is X.
    *   Fixtures: R (at 6,0), R (at 4,2).
    *   Counts: R=2, B=0. Result: **R**.

*   **Section 5 (Middle Middle):** Cells (4,3), (4,4), (5,3), (5,4).
    *   Neighbors: (3,3) is X, (3,4) is B. (6,3) is X, (6,4) is R. (4,2) is R, (4,5) is R. (5,2) is X, (5,5) is X.
    *   Fixtures: B (at 3,4), R (at 6,4), R (at 4,2), R (at 4,5).
    *   Counts: R=3, B=1. Result: **R**.

*   **Section 6 (Middle Right):** Cells (4,6), (4,7), (4,8), (4,9), (5,6), (5,7), (5,8), (5,9).
    *   Neighbors: (3,6) is X, (3,7) is R, (3,8) is B, (3,9) is X. (6,6) is X, (6,7) is B, (6,8) is R, (6,9) is X. (4,5) is R, (5,5) is X.
    *   Fixtures: R (at 3,7), B (at 3,8), B (at 6,7), R (at 6,8), R (at 4,5).
    *   Counts: R=3, B=2. Result: **R**.

*   **Section 7 (Bottom Left):** Cells (7,0), (7,1), (8,0), (8,1).
    *   Neighbors: (6,0) is R, (6,1) is X. (9,0) is X, (9,1) is R. (7,2) is B, (8,2) is X.
    *   Fixtures: R (at 6,0), R (at 9,1), B (at 7,2).
    *   Counts: R=2, B=1. Result: **R**.

*   **Section 8 (Bottom Middle):** Cells (7,3), (7,4), (8,3), (8,4).
    *   Neighbors: (6,3) is X, (6,4) is R. (9,3) is X, (9,4) is X. (7,2) is B, (7,5) is R. (8,2) is X, (8,5) is X.
    *   Fixtures: R (at 6,4), B (at 7,2), R (at 7,5).
    *   Counts: R=2, B=1. Result: **R**.

*   **Section 9 (Bottom Right):** Cells (7,6), (7,7), (7,8), (7,9), (8,6), (8,7), (8,8), (8,9).
    *   Neighbors: (6,6) is X, (6,7) is B, (6,8) is R, (6,9) is X. (9,6) is B, (9,7) is X, (9,8) is X, (9,9) is B. (7,5) is R, (8,5) is X.
    *   Fixtures: B (at 6,7), R (at 6,8), B (at 9,6), B (at 9,9), R (at 7,5).
    *   Counts: B=3, R=2. Result: **B**.

**Final Grid Construction:**
Row 0: X B X X X X X X B X
Row 1: B B X B B X B B B B
Row 2: B B X B B B B B B B
Row 3: X X X X B X X R B X
Row 4: R R R R R R R R R R
Row 5: R R X R R X R R R R
Row 6: R X X X R X X B R X
Row 7: R R B R R R B B B B
Row 8: R R X R R X B B B B
Row 9: X R X X X X B X X B

Wait, let me re-verify Section 4 and 5.
Section 4 (4,0-5,1): Neighbors are (3,0)X, (3,1)X, (6,0)R, (6,1)X, (4,2)R, (5,2)X. Fixtures: R, R. Correct.
Section 5 (4,3-5,4): Neighbors are (3,3)X, (3,4)B, (6,3)X, (6,4)R, (4,2)R, (4,5)R, (5,2)X, (5,5)X. Fixtures: B, R, R, R. Correct.
Section 6 (4,6-5,9): Neighbors are (3,6)X, (3,7)R, (3,8)B, (3,9)X, (6,6)X, (

... [2783 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Buildings (Green 'G' and Blue 'B'):** Permanent, immovable structures. They are fireproof and remain unchanged.
    *   **Open Ground (Black '.'):** The spaces between buildings. These are the areas subject to change.
    *   **The Edge (Grid Perimeter):** The boundary where the "wildfire" enters the city.
    *   **Fire (Red 'R'):** Represents open ground connected to the perimeter via orthogonal paths (up, down, left, right).
    *   **Smoke (Grey 'X'):** Represents open ground (black '.') that is completely enclosed by buildings and has no orthogonal path to the perimeter.

2.  **Identify the AFFORDANCES:**
    *   **Buildings (G, B):** Act as barriers. They block the spread of fire and smoke.
    *   **Open Ground (.):** Can be transformed into Red (if connected to the edge) or Grey (if enclosed).
    *   **Connectivity:** Fire spreads orthogonally (not diagonally).

3.  **Identify the TRANSFORMATION RULE:**
    *   Identify all black '.' cells.
    *   Perform a flood-fill (orthogonal) starting from all '.' cells located on the grid's perimeter.
    *   All '.' cells reached by this flood-fill become 'R' (Red).
    *   All remaining '.' cells that were not reached by the flood-fill (meaning they are enclosed) become 'X' (Grey).
    *   All 'G' and 'B' cells remain unchanged.

### Applying the Rule to Test Input

**Input Grid Analysis:**
Row 0: B . . B . B B B B G
Row 1: . . . G . G . B . .
Row 2: . B . G G . B G G G
Row 3: G B G B B . G G . B
Row 4: B B G . B G . B B .
Row 5: . G . B G . B B . G
Row 6: B B G . . G . G G G
Row 7: G B B B B G B . G B
Row 8: G . . . G G B . B B
Row 9: B . G B B . . . B .

**Step 1: Identify Perimeter '.' cells (potential fire entry):**
(0,1), (0,2), (0,4), (1,0), (1,1), (1,2), (1,4), (1,6), (1,8), (1,9), (2,0), (2,2), (2,5), (3,5), (3,8), (4,3), (4,6), (4,9), (5,0), (5,2), (5,5), (5,8), (6,3), (6,4), (6,6), (7,7), (8,1), (8,2), (8,3), (8,7), (9,1), (9,5), (9,6), (9,7), (9,9).

**Step 2: Flood Fill (Orthogonal) from Perimeter:**
- (0,1)-(0,2)-(1,2)-(1,1)-(1,0)-(2,0) are connected.
- (0,4)-(1,4) are connected.
- (1,6) is connected.
- (1,8)-(1,9) are connected.
- (2,2) is connected to (1,2).
- (2,5)-(3,5)-(4,6)-(5,5)-(6,6)-(7,7)-(8,7)-(9,7)-(9,6)-(9,5)-(8,5 is G, no)-(9,5)-(9,6)-(9,7)
- (3,8)-(4,9)-(5,8) are connected.
- (4,3)-(5,2)-(6,3)-(6,4)-(7,4 is B, no)-(6,4)-(6,3)-(5,2)-(4,3)
- (8,1)-(8,2)-(8,3)-(9,1) are connected.
- (9,5)-(9,6)-(9,7) are connected.

**Let's trace carefully for enclosed cells (X):**
- (4,3) is connected to (5,2) and (6,3), (6,4). (6,3) is connected to (6,4). (5,2) is connected to (4,3). (4,3) is connected to (5,2).
- Let's check (5,5): (5,5) is connected to (6,5 is G, no). Wait, (5,5) is connected to (4,5 is G, no). (5,5) is connected to (5,4 is G, no). (5,5) is connected to (5,6 is B, no).
- Let's re-evaluate (5,5): Neighbors are (4,5)=G, (6,5)=G, (5,4)=G, (5,6)=B. (5,5) is a '.' cell. It is surrounded by G, G, G, B. It is **enclosed**. -> **X**
- Let's check (4,6): Neighbors are (3,6)=G, (5,6)=B, (4,5)=G, (4,7)=B. (4,6) is a '.' cell. It is surrounded by G, B, G, B. It is **enclosed**. -> **X**
- Let's check (3,5): Neighbors are (2,5), (4,5)=G, (3,4)=B, (3,6)=G. (2,5) is a '.' cell. (2,5) neighbors: (1,5)=G, (3,5), (2,4)=G, (2,6)=B. So (2,5) and (3,5) are connected. Are they connected to the edge? (2,5) neighbors: (1,5)=G, (3,5), (2,4)=G, (2,6)=B. (3,5) neighbors: (2,5), (4,5)=G, (3,4)=B, (3,6)=G. They are enclosed. -> **X, X**
- Let's check (7,7): Neighbors are (6,7)=G, (8,7), (7,6)=B, (7,8)=G. (8,7) is a '.' cell. (8,7) neighbors: (7,7), (9,7), (8,6)=B, (8,8)=B. (9,7) is on the edge. So (7,7), (8,7), (9,7) are connected to the edge. -> **R**
- Let's check (5,8): Neighbors are (4,8)=B, (6,8)=G, (5,7)=B, (5,9)=G. (5,8) is a '.' cell. It is surrounded by B, G, B, G. It is **enclosed**. -> **X**
- Let's check (3,8): Neighbors are (2,8)=G, (4,8)=B, (3,7)=G, (3,9)=B. (3,8) is a '.' cell. It is surrounded by G, B, G, B. It is **enclosed**. -> **X**
- Let's check (4,3): Neighbors are (3,3)=B, (5,3)=B, (4,2)=G, (4,4)=B. (4,3) is a '.' cell. It is surrounded by B, B, G, B. It is **enclosed**. -> **X**
- Let's check (6,3), (6,4): (6,3) neighbors: (5,3)=B, (7,3)=B, (6,2)=G, (6,4). (6,4) neighbors: (5,4)=G, (7,4)=B, (6,3), (6,5)=G. (6,3) and (6,4) are enclosed. -> **X, X**
- Let's check (5,2): Neighbors are (4,2)=G, (6,2)=G, (5,1)=G, (5,3)=B. (5,2) is a '.' cell. It is surrounded by G, G, G, B. It is **enclosed**. -> **X**
- Let's check (8,1), (8,2), (8,3), (9,1): (8,1) is connected to (9,1) (edge). (8,1) neighbors: (7,1)=B, (9,1), (8,0)=G, (8,2). (8,2) neighbors: (7,2)=B, (9,2)=G, (8,1), (8,3). (8,3) neighbors: (7,3)=B, (9,3)=B, (8,2), (8,4)=G. All these are connected to the edge. -> **R**
- Let's check (9,5), (9,6), (9,7): (9,5), (9,6), (9,7) are on the edge. -> **R**

**Final check of all '.' cells:**
(0,1) R, (0,2) R, (0,4) R
(1,0) R, (1,1) R, (1,2) R, (1,4) R,

... [3297 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Fixed Protein Structures (Green 'G' and Blue 'B')**: These act as barriers or walls. They do not change color.
    *   **Interstitial Fluid (Black '.')**: These are the cells that undergo transformation.
    *   **The Grid Edge (Outer boundary)**: This is the trigger for the "inflammatory response."
    *   **Fluid Pools**: Connected components of black cells.

2.  **Identify the AFFORDANCES**:
    *   **Black cells** can be part of a "pool" (a connected component via 4-connectivity or 8-connectivity; looking at the examples, it is 4-connectivity/orthogonal).
    *   **Pools** can either touch the grid boundary (becoming Red 'R') or be completely enclosed by G/B cells (becoming Grey 'X').

3.  **Identify the TRANSFORMATION RULE**:
    *   Identify all connected components (pools) of black cells ('.').
    *   For each pool:
        *   If any cell in the pool is on the edge of the 10x10 grid, change all cells in that pool to Red ('R').
        *   If no cell in the pool is on the edge (the pool is entirely surrounded by G or B), change all cells in that pool to Grey ('X').
    *   Green ('G') and Blue ('B') cells remain unchanged.

### Step-by-Step Execution (Test Input)

**Input Grid Analysis (10x10):**
Row 0: B . . B . B B B B G
Row 1: . . . G . G . B . .
Row 2: . B . G G . B G G G
Row 3: G B G B B . G G . B
Row 4: B B G . B G . B B .
Row 5: . G . B G . B B . G
Row 6: B B G . . G . G G G
Row 7: G B B B B G B . G B
Row 8: G . . . G G B . B B
Row 9: B . G B B . . . B .

**Identify Black Cell Pools:**
*   **Pool 1 (Edge):** (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,2). This pool touches the edge (Row 0 and Col 0). -> **Red (R)**
*   **Pool 2 (Edge):** (0,4). Touches edge. -> **Red (R)**
*   **Pool 3 (Edge):** (1,4). Not touching edge? Wait, let's re-check connectivity. (0,4) is connected to (1,4)? No, (0,4) is B, (1,4) is '.'. (0,4) is surrounded by B, B, G, G. It is a single cell pool. It touches the edge. -> **Red (R)**
*   **Pool 4 (Edge):** (1,6). Surrounded by G, B, G, G. Not touching edge. -> **Grey (X)**
*   **Pool 5 (Edge):** (1,8), (1,9), (2,5), (3,5), (4,3), (4,6), (4,9), (5,0), (5,2), (5,5), (5,8), (6,3), (6,4), (6,6), (7,7), (8,1), (8,2), (8,3), (8,7), (9,1), (9,5), (9,6), (9,7).
    *   Let's trace carefully:
    *   (1,8)-(1,9) is an edge pool. -> **Red (R)**
    *   (2,5)-(3,5) is a pool. (3,5) is connected to (4,3)? No. (3,5) is surrounded by B, G, G, B. It is enclosed. -> **Grey (X)**
    *   (4,3) is surrounded by G, B, B, G. Enclosed. -> **Grey (X)**
    *   (4,6) is surrounded by G, B, B, G. Enclosed. -> **Grey (X)**
    *   (4,9) is an edge cell. -> **Red (R)**
    *   (5,0) is an edge cell. -> **Red (R)**
    *   (5,2) is surrounded by G, B, B, G. Enclosed. -> **Grey (X)**
    *   (5,5) is surrounded by G, G, B, G. Enclosed. -> **Grey (X)**
    *   (5,8) is surrounded by B, G, G, B. Enclosed. -> **Grey (X)**
    *   (6,3)-(6,4) is a pool. (6,3) is surrounded by B, G, G, B. Enclosed. -> **Grey (X)**
    *   (6,6) is surrounded by G, B, G, G. Enclosed. -> **Grey (X)**
    *   (7,7) is surrounded by B, G, B, B. Enclosed. -> **Grey (X)**
    *   (8,1)-(8,2)-(8,3)-(9,1) is a pool. (9,1) is an edge cell. -> **Red (R)**
    *   (8,7) is surrounded by B, B, B, B. Enclosed. -> **Grey (X)**
    *   (9,5)-(9,6)-(9,7) is an edge pool. -> **Red (R)**

**Wait, let's re-verify connectivity for (1,4) and (1,6):**
(0,4) is '.', (1,4) is '.', (2,4) is 'G'. (0,4) is connected to (1,4). (0,4) is on the edge. So (0,4) and (1,4) are **Red (R)**.
(1,6) is '.', (0,6) is 'B', (1,5) is 'G', (1,7) is 'B', (2,6) is 'B'. (1,6) is enclosed. -> **Grey (X)**

**Final Pool Mapping:**
- (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,2) -> R
- (0,4), (1,4) -> R
- (1,6) -> X
- (1,8), (1,9) -> R
- (2,5), (3,5) -> X
- (4,3) -> X
- (4,6) -> X
- (4,9) -> R
- (5,0) -> R
- (5,2) -> X
- (5,5) -> X
- (5,8) -> X
- (6,3), (6,4) -> X
- (6,6) -> X
- (7,7) -> X
- (8,1), (8,2), (8,3), (9,1) -> R
- (8,7) -> X
- (9,5), (9,6), (9,7) -> R

**Re-checking (2,5) and (3,5):**
(2,5) is '.', (3,5) is '.'. (2,5) neighbors: (1,5)G, (2,4)G, (2,6)B, (3,5)'.'. (3,5) neighbors: (2,5)'.', (3,4)B, (3,6)G, (4,5)G.
This pool {(2,5), (3,5)} is enclosed. -> **Grey (X)**

**Re-checking (8,1), (8,2), (8,3), (9,1):**
(8,1) neighbors: (7,1)B, (8,0)G, (8,2)'.', (9,1)'.'.
(8,2) neighbors: (7,2)B, (8,1)'.', (8,3)'.', (9,2)G.
(8,3) neighbors: (7,3)B, (8,2)'.', (8,4)G, (9,3)B.
(9,1) neighbors: (8,1)'.', (9,0)B, (9,2)G.
This pool {(8,1), (8,2), (8,3), (9,1)} touches the edge at (9,1). -> **Red (R)**

**Re-checking (9,5), (9,6), (9,7):**
(9,5) neighbors: (8,5)G, (9,4)B, (9,6)'.'.
(9,6) neighbors: (8,6)B, (9,5)'.', (9,7)'.'.
(9,7) neighbors: (8,7)'.', (9,6)'.', (9,8)B.
(8,7) is also part of this pool? Let's check (8,7) neighbors: (7,7)'.', (8,6)B, (8,8)B, (9,7)'.'.
Yes, (8,7) is connected

... [3454 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Islands (Green 'G' and Blue 'B'):** Static elements that act as boundaries. They do not change color.
    *   **Open Water (Black '.'):** The medium that undergoes transformation.
    *   **Tidal Water (Red 'R'):** Water that is connected to the edges of the grid.
    *   **Landlocked Water (Grey 'X'):** Water that is completely enclosed by islands and cannot reach the grid edges.

2.  **Identify the AFFORDANCES:**
    *   **Islands (G, B):** Block the flow of water.
    *   **Water (.):** Can be "reached" by the tide if there is a path of '.' cells to the edge, or "trapped" if surrounded by G/B.

3.  **Identify the TRANSFORMATION RULE:**
    *   Identify all '.' cells.
    *   Perform a flood-fill starting from all '.' cells located on the perimeter (edges) of the grid.
    *   All '.' cells reachable from the edge via other '.' cells become 'R' (Red).
    *   All '.' cells that are NOT reachable from the edge (trapped by G and B) become 'X' (Grey).
    *   All 'G' and 'B' cells remain unchanged.

### Applying the Rule to Test Input

**Input Grid Analysis:**
Row 0: B . . B . B B B B G
Row 1: . . . G . G . B . .
Row 2: . B . G G . B G G G
Row 3: G B G B B . G G . B
Row 4: B B G . B G . B B .
Row 5: . G . B G . B B . G
Row 6: B B G . . G . G G G
Row 7: G B B B B G B . G B
Row 8: G . . . G G B . B B
Row 9: B . G B B . . . B .

**Step 1: Identify Edge Water (Potential Red):**
- (0,1), (0,2), (0,4)
- (1,0), (1,1), (1,2), (1,4), (1,6), (1,8), (1,9)
- (2,0), (2,2), (2,5)
- (3,5), (3,8)
- (4,3), (4,6), (4,9)
- (5,0), (5,2), (5,5), (5,8)
- (6,3), (6,4), (6,6)
- (7,7)
- (8,1), (8,2), (8,3), (8,7)
- (9,1), (9,5), (9,6), (9,7)

**Step 2: Flood Fill (Red):**
- Start at (0,1) -> (0,2) -> (1,2) -> (1,1) -> (1,0) -> (2,0)
- Start at (0,4) -> (1,4) -> (2,5) -> (3,5) -> (4,6) -> (5,5) -> (6,6) -> (7,7) -> (8,7) -> (9,7) -> (9,6) -> (9,5)
- Start at (1,6) -> (0,6) is B, so no. (1,6) is isolated? Let's check neighbors: (0,6) is B, (2,6) is B, (1,5) is G, (1,7) is B. (1,6) is actually a '.' cell. Is it connected to edge? (1,6) neighbors: (0,6)B, (2,6)B, (1,5)G, (1,7)B. It is trapped!
- Let's re-evaluate (1,6): Neighbors are (0,6)B, (2,6)B, (1,5)G, (1,7)B. It is surrounded by islands. It becomes X.
- Let's re-evaluate (3,8): Neighbors (2,8)G, (4,8)B, (3,7)G, (3,9)B. It is surrounded by islands. It becomes X.
- Let's re-evaluate (4,3): Neighbors (3,3)B, (5,3)B, (4,2)G, (4,4)B. It is surrounded by islands. It becomes X.
- Let's re-evaluate (4,9): Neighbors (3,9)B, (5,9)G, (4,8)B. It is on the edge. It is R.
- Let's re-evaluate (5,0): Edge. R.
- Let's re-evaluate (5,2): Neighbors (4,2)G, (6,2)G, (5,1)G, (5,3)B. It is surrounded by islands. It becomes X.
- Let's re-evaluate (5,8): Neighbors (4,8)B, (6,8)G, (5,7)B, (5,9)G. It is surrounded by islands. It becomes X.
- Let's re-evaluate (6,3), (6,4): Neighbors (5,3)B, (7,3)B, (6,2)G, (6,5)G. (6,3) and (6,4) are connected to each other. Are they connected to edge? (6,3) neighbors: (5,3)B, (7,3)B, (6,2)G, (6,4). (6,4) neighbors: (5,4)G, (7,4)B, (6,3), (6,5)G. They are trapped! They become X.
- Let's re-evaluate (8,1), (8,2), (8,3): (8,1) is connected to (9,1) (edge). (8,2) connected to (8,1). (8,3) connected to (8,2). All R.
- Let's re-evaluate (9,5), (9,6), (9,7): Edge. All R.

**Final Mapping:**
Row 0: B R R B R B B B B G
Row 1: R R R G R G X B R R
Row 2: R B R G G R B G G G
Row 3: G B G B B R G G X B
Row 4: B B G X B G R B B R
Row 5: R G X B G R B B X G
Row 6: B B G X X G R G G G
Row 7: G B B B B G B R G B
Row 8: G R R R G G B R B B
Row 9: B R G B B R R R B R

Wait, let's double check (1,6) and (1,8/9).
(1,8) and (1,9) are edge. (1,8) neighbors: (0,8)B, (2,8)G, (1,7)B, (1,9). (1,9) neighbors: (0,9)G, (2,9)G, (1,8). So (1,8) and (1,9) are R.
(1,6) is surrounded by (0,6)B, (2,6)B, (1,5)G, (1,7)B. It is X.
(3,8) is surrounded by (2,8)G, (4,8)B, (3,7)G, (3,9)B. It is X.
(4,3) is surrounded by (3,3)B, (5,3)B, (4,2)G, (4,4)B. It is X.
(5,2) is surrounded by (4,2)G, (6,2)G, (5,1)G, (5,3)B. It is X.
(5,8) is surrounded by (4,8)B, (6,8)G, (5,7)B, (5,9)G. It is X.
(6,3) and (6,4) are surrounded by (5,3)B, (7,3)B, (6,2)G, (6,5)G. They are X.

Let's re-verify (4,6) and (5,5) and (6,6) and (7,7) and (8,7) and (9,7) and (9,6) and (9,5).
(4,6) neighbors: (3,6)G, (5,6)B, (4,5)G, (4,7)B. Wait, (4,6) is surrounded by G, B, G, B. It is X.
If (4,6) is X, then (5,5) is not connected to (4,6).
(5,5) neighbors: (4,5)G, (6,5)G, (5,4)G, (5,6)B. (5,5) is surrounded by G, G, G, B. It is X.
(6,6) neighbors: (5,6)B, (7,6)B, (6,5)G, (6,7)G. (6,6) is surrounded by B, B, G, G. It is X.
(7,7) neighbors: (6,7)G, (8,7)R, (7,6)B, (7,8)G. (7,7) is connected to (8,7).
(8,7) neighbors: (7,7), (9,7), (8,6)B, (8,8)B. (8,7) is connected to (9,7) (edge).
(9,7) is edge. (9,6) is edge. (9,5) is edge.
So (7,7), (8,7), (9,7), (9,6), (9,5) are R.

Let's re-verify (1,4) and (2,5) and (3,5).
(1,4) neighbors: (0,4)R, (2,4

... [2755 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Stones (G, B)**: Fixed elements that act as barriers.
    *   **Earth (.)**: The element being transformed.
    *   **Boundary**: The edges of the 10x10 grid.
    *   **Curse (R)**: Red color applied to earth connected to the boundary.
    *   **Protection (X)**: Grey color applied to earth completely enclosed by stones/other earth, such that there is no path of earth cells to the boundary.

2.  **Identify the AFFORDANCES**:
    *   **Stones (G, B)**: Block the "spread" of the curse. They are impassable for the connection between earth cells.
    *   **Earth (.)**: Can be "cursed" (R) if it can reach the edge via a path of other earth cells, or "protected" (X) if it is trapped in a pocket of earth surrounded by stones.

3.  **Identify the TRANSFORMATION RULE**:
    *   Identify all cells containing `.` (black).
    *   Perform a flood-fill or connectivity analysis starting from all `.` cells located on the grid boundary.
    *   Any `.` cell reachable from the boundary through a path of other `.` cells becomes `R` (red).
    *   Any `.` cell that is *not* reachable from the boundary (i.e., it is enclosed by G and B cells) becomes `X` (grey).
    *   All other cells (G, B) remain unchanged.

### Applying the Rule to Test Input

**Input Grid Analysis (10x10):**
Row 0: B . . B . B B B B G
Row 1: . . . G . G . B . .
Row 2: . B . G G . B G G G
Row 3: G B G B B . G G . B
Row 4: B B G . B G . B B .
Row 5: . G . B G . B B . G
Row 6: B B G . . G . G G G
Row 7: G B B B B G B . G B
Row 8: G . . . G G B . B B
Row 9: B . G B B . . . B .

**Step 1: Identify Boundary Earth cells:**
(0,1), (0,2), (0,4), (1,0), (1,1), (1,2), (1,4), (1,6), (1,8), (1,9), (2,0), (2,2), (2,5), (3,5), (3,8), (4,3), (4,6), (4,9), (5,0), (5,2), (5,5), (5,8), (6,3), (6,4), (6,6), (7,7), (8,1), (8,2), (8,3), (8,7), (9,1), (9,5), (9,6), (9,7), (9,9).

**Step 2: Trace connectivity (Flood Fill from Boundary):**
- (0,1)-(0,2)-(1,2)-(1,1)-(1,0)-(2,0) are connected to boundary.
- (0,4)-(1,4) are connected to boundary.
- (1,6) is boundary.
- (1,8)-(1,9) are connected to boundary.
- (2,2) is connected to (1,2).
- (2,5)-(3,5)-(4,3)-(5,2)-(6,3)-(6,4)-(5,5)-(4,6)-(3,5)... wait, let's trace carefully.

**Let's map the connectivity of '.' cells:**
- Group A (Boundary connected): (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,2)
- Group B (Boundary connected): (0,4), (1,4)
- Group C (Boundary connected): (1,6)
- Group D (Boundary connected): (1,8), (1,9)
- Group E (Boundary connected): (2,5), (3,5), (4,3), (5,2), (6,3), (6,4), (5,5), (4,6), (3,8), (4,9), (5,8), (6,6), (7,7), (8,7), (9,5), (9,6), (9,7), (8,1), (8,2), (8,3), (9,1)
- Group F (Boundary connected): (5,0)
- Group G (Boundary connected): (9,9)

**Wait, let's check for isolated pockets (X):**
- Look at (4,3): Is it connected to boundary? (4,3) -> (5,2) is not possible (diagonal). (4,3) is surrounded by G, B, B, G. Let's re-examine (4,3).
- (4,3) neighbors: (3,3) is B, (4,2) is G, (4,4) is B, (5,3) is B. **(4,3) is isolated!** It becomes **X**.
- (5,5) neighbors: (4,5) is G, (5,4) is G, (5,6) is B, (6,5) is G. **(5,5) is isolated!** It becomes **X**.
- (6,3) neighbors: (5,3) is B, (6,2) is G, (6,4) is '.', (7,3) is B.
- (6,4) neighbors: (5,4) is G, (6,3) is '.', (6,5) is G, (7,4) is B.
- (6,3)-(6,4) are connected to each other. Are they connected to boundary?
  - (6,3) neighbors: (6,4), (5,3)B, (6,2)G, (7,3)B.
  - (6,4) neighbors: (6,3), (5,4)G, (6,5)G, (7,4)B.
  - (6,3)-(6,4) are trapped! They become **X**.
- (3,5) neighbors: (2,5) is '.', (3,4) is B, (3,6) is G, (4,5) is G.
- (2,5) neighbors: (1,5) is G, (2,4) is G, (2,6) is B, (3,5) is '.'.
- (2,5)-(3,5) are trapped! They become **X**.
- (5,2) neighbors: (4,2) is G, (5,1) is G, (5,3) is B, (6,2) is G. **(5,2) is isolated!** It becomes **X**.
- (8,1)-(8,2)-(8,3)-(9,1) are connected to boundary (9,1 is boundary).
- (7,7) is connected to (8,7) which is connected to (9,7) which is boundary.
- (9,5)-(9,6)-(9,7) are boundary.

**Re-evaluating (4,3), (5,5), (6,3), (6,4), (2,5), (3,5), (5,2):**
- (4,3): Neighbors (3,3)B, (4,2)G, (4,4)B, (5,3)B. **X**
- (5,5): Neighbors (4,5)G, (5,4)G, (5,6)B, (6,5)G. **X**
- (6,3)-(6,4): Neighbors (5,3)B, (5,4)G, (6,2)G, (6,5)G, (7,3)B, (7,4)B. **X**
- (2,5)-(3,5): Neighbors (1,5)G, (2,4)G, (2,6)B, (3,4)B, (3,6)G, (4,5)G. **X**
- (5,2): Neighbors (4,2)G, (5,1)G, (5,3)B, (6,2)G. **X**

**Final check of all '.' cells:**
(0,1) R, (0,2) R, (0,4) R, (1,0) R, (1,1) R, (1,2) R, (1,4) R, (1,6) R, (1,8) R, (1,9) R, (2,0) R, (2,2) R, (2,5) X, (3,5) X, (3,8) R, (4,3) X, (4,6) R, (4,9) R, (5,0) R, (5,2) X, (5,5) X, (5,8) R, (6,3) X, (6,4) X, (6,6) R, (7,7) R, (8,1) R, (8,2) R, (8,3) R, (8,7) R, (9,1) R, (9,5) R, (9,6) R, (9,7) R, (9,9) R.

Wait, (4,6) neighbors: (3,6)G, (4,5)G, (4,7)B, (5,6)B. (4,6) is isolated! It becomes **X**.
(6,6) neighbors: (5,6)B, (6,5)G, (6,7)G, (7,6)B. (6,6) is isolated! It becomes **X**.
(3,8) neighbors: (2,8)G, (3,7)G, (3,9)B, (4,8)B. (3,8) is isol

... [2519 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Dancers (Green 'G' and Blue 'B'):** These are static obstacles/walls that define the boundaries of the "ballroom" and create enclosures.
    *   **Empty Floor (Black '.'):** These are the cells that will be transformed.
    *   **Spotlights (Red 'R'):** These represent empty floor cells that have a direct "sightline" to the grid edges (the walls).
    *   **Shadows (Grey 'X'):** These represent empty floor cells that are completely enclosed by dancers and cannot "see" any edge of the grid.

2.  **Identify the AFFORDANCES:**
    *   **Dancers (G, B):** Block the path of the "spotlight." They act as barriers.
    *   **Empty Floor (.):** Can either be lit up (R) if connected to the edge, or remain in shadow (X) if trapped.
    *   **Grid Edges:** The source of the light.

3.  **Identify the TRANSFORMATION RULE:**
    *   The transformation is a connectivity/flood-fill problem.
    *   Identify all empty cells (`.`) that are part of a connected component (using 4-connectivity or 8-connectivity, though standard grid puzzles usually imply 4-connectivity) that touches the boundary of the 10x10 grid.
    *   All such empty cells connected to the edge become Red (`R`).
    *   All empty cells (`.`) that are *not* connected to the edge (i.e., they are "trapped" inside a perimeter of G and B) become Grey (`X`).
    *   All existing G and B cells remain unchanged.

### Applying the Rule to Test Input

**Input Grid Analysis:**
Row 0: B . . B . B B B B G
Row 1: . . . G . G . B . .
Row 2: . B . G G . B G G G
Row 3: G B G B B . G G . B
Row 4: B B G . B G . B B .
Row 5: . G . B G . B B . G
Row 6: B B G . . G . G G G
Row 7: G B B B B G B . G B
Row 8: G . . . G G B . B B
Row 9: B . G B B . . . B .

**Step 1: Identify all '.' cells.**
**Step 2: Determine connectivity to the edge.**
*   (0,1), (0,2), (0,4), (1,0), (1,1), (1,2), (1,4), (1,6), (1,8), (1,9), (2,0), (2,2), (2,5), (3,5), (3,8), (4,3), (4,6), (4,9), (5,0), (5,2), (5,5), (5,8), (6,3), (6,4), (6,6), (7,7), (8,1), (8,2), (8,3), (8,7), (9,1), (9,5), (9,6), (9,7).

**Step 3: Trace connectivity (Flood Fill from edges):**
*   Edge cells: (0,1), (0,2), (0,4), (1,0), (1,8), (1,9), (2,0), (4,9), (5,0), (8,1), (9,1), (9,5), (9,6), (9,7).
*   From (0,1) -> (0,2), (1,1), (1,2), (1,0), (2,0), (2,2).
*   From (0,4) -> (1,4).
*   From (1,4) -> (2,5) is blocked by G? No, (1,4) is '.', (2,5) is '.', (3,5) is '.'. Let's check (2,5) neighbors: (1,5) is G, (2,4) is G, (2,6) is B, (3,5) is '.'. So (2,5) connects to (3,5).
*   From (3,5) -> (4,5) is G, (3,4) is B, (3,6) is G, (2,5) is '.'. (3,5) is a dead end or connects to (2,5).
*   From (1,6) -> (0,6) is B, (1,5) is G, (1,7) is B, (2,6) is B. (1,6) is isolated? Let's re-check (1,6). (1,6) is '.', neighbors are (0,6)B, (1,5)G, (1,7)B, (2,6)B. (1,6) is trapped! -> **X**
*   From (1,8) -> (1,9), (0,8) is B, (2,8) is G.
*   From (3,8) -> (4,8) is B, (3,7) is G, (3,9) is B, (2,8) is G. (3,8) is trapped! -> **X**
*   From (4,3) -> (4,2) is G, (4,4) is B, (3,3) is B, (5,3) is B. (4,3) is trapped! -> **X**
*   From (4,6) -> (3,6) is G, (5,6) is B, (4,5) is G, (4,7) is B. (4,6) is trapped! -> **X**
*   From (4,9) -> (3,9) is B, (5,9) is G, (4,8) is B. (4,9) is an edge cell? Yes, (4,9) is '.', it is on the edge.
*   From (5,2) -> (5,1) is G, (5,3) is B, (4,2) is G, (6,2) is G. (5,2) is trapped! -> **X**
*   From (5,5) -> (4,5) is G, (6,5) is G, (5,4) is G, (5,6) is B. (5,5) is trapped! -> **X**
*   From (5,8) -> (4,8) is B, (6,8) is G, (5,7) is B, (5,9) is G. (5,8) is trapped! -> **X**
*   From (6,3) -> (6,4) -> (7,3) is B, (6,2) is G, (6,5) is G, (5,3) is B. (6,3) and (6,4) are trapped! -> **X**
*   From (6,6) -> (5,6) is B, (7,6) is B, (6,5) is G, (6,7) is G. (6,6) is trapped! -> **X**
*   From (7,7) -> (8,7) -> (9,7). (7,7) is connected to edge.
*   From (8,1) -> (8,2) -> (8,3) -> (9,3) is B, (7,3) is B, (8,4) is G. (8,1), (8,2), (8,3) are connected to (9,1) which is an edge.
*   From (9,5) -> (9,6) -> (9,7).

**Summary of '.' transformations:**
*   (0,1) R, (0,2) R, (0,4) R, (1,0) R, (1,1) R, (1,2) R, (1,4) R, (1,6) X, (1,8) R, (1,9) R, (2,0) R, (2,2) R, (2,5) R, (3,5) R, (3,8) X, (4,3) X, (4,6) X, (4,9) R, (5,0) R, (5,2) X, (5,5) X, (5,8) X, (6,3) X, (6,4) X, (6,6) X, (7,7) R, (8,1) R, (8,2) R, (8,3) R, (8,7) R, (9,1) R, (9,5) R, (9,6) R, (9,7) R.

**Wait, let's re-verify (2,5) and (3,5):**
(2,5) is '.', neighbors: (1,5)G, (3,5)'.', (2,4)G, (2,6)B.
(3,5) is '.', neighbors: (2,5)'.', (4,5)G, (3,4)B, (3,6)G.
(2,5) and (3,5) are a small pocket. Are they connected to the edge?
(1,5) is G, (2,4) is G, (2,6) is B, (3,4) is B, (3,6) is G, (4,5) is G.
Yes, (2,5) and (3,5) are trapped. They become **X**.

**Re-verifying (1,4):**
(1,4) is '.', neighbors: (0,4)R, (2,4)G, (1,3)G, (1,5)G.
(0,4) is on the edge. So (1,4) is **R**.

**Re-verifying (2,2):**
(2,2) is '.', neighbors: (1,2)R, (3,2)G, (2,

... [1473 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Toppings (Green 'G', Blue 'B')**: Fixed obstacles that block the flow of the glaze.
    *   **Bare Dough (Black '.')**: The substrate that can be transformed.
    *   **Red Glaze (Red 'R')**: The substance that flows from the "pan's rim" (the edges of the grid) into any reachable dough.
    *   **Pale Grey (Grey 'X')**: The state of dough that is completely enclosed/sealed off by toppings and cannot be reached by the glaze from the edges.

2.  **Identify the AFFORDANCES**:
    *   **Toppings (G, B)**: Act as barriers to the flow of 'R'.
    *   **Dough (.)**: Can be reached by 'R' (becoming 'R') or trapped (becoming 'X').
    *   **The Rim (Edges)**: The source of the 'R' glaze. Any '.' connected to the edge via a path of '.' cells will become 'R'.

3.  **Identify the TRANSFORMATION RULE**:
    *   Perform a flood-fill starting from all '.' cells located on the perimeter of the grid.
    *   Any '.' cell reachable from the edge through a path of other '.' cells becomes 'R'.
    *   Any '.' cell that is *not* reachable from the edge (i.e., it is surrounded by 'G' and 'B' cells) becomes 'X'.
    *   All 'G' and 'B' cells remain unchanged.

### Applying the Rule to Test Input

**Input Grid Analysis (10x10):**
Row 0: B . . B . B B B B G
Row 1: . . . G . G . B . .
Row 2: . B . G G . B G G G
Row 3: G B G B B . G G . B
Row 4: B B G . B G . B B .
Row 5: . G . B G . B B . G
Row 6: B B G . . G . G G G
Row 7: G B B B B G B . G B
Row 8: G . . . G G B . B B
Row 9: B . G B B . . . B .

**Step 1: Identify all '.' cells.**
**Step 2: Flood-fill from the edges.**
- (0,1), (0,2), (0,4), (1,0), (1,1), (1,2), (1,4), (1,6), (1,8), (1,9), (2,0), (2,2), (2,5), (3,5), (3,8), (4,3), (4,6), (4,9), (5,0), (5,2), (5,5), (5,8), (6,3), (6,4), (6,6), (7,7), (8,1), (8,2), (8,3), (8,7), (9,1), (9,5), (9,6), (9,7), (9,9).

**Tracing connectivity to edges:**
- (0,1), (0,2) are on edge -> R.
- (0,4) is on edge -> R.
- (1,0), (1,1), (1,2) connect to (0,1)/(0,2) -> R.
- (1,4) connects to (0,4) -> R.
- (1,6) connects to (0,6)? No, (0,6) is B. (1,6) connects to (2,6)? No, (2,6) is B. Let's check (1,6) neighbors: (0,6)B, (1,5)G, (1,7)B, (2,6)B. (1,6) is isolated. Wait, (1,6) is not on the edge. Is it connected to an edge? No. (1,6) is surrounded by G, B, B, B. It is **X**.
- (1,8), (1,9) are on edge -> R.
- (2,0) is on edge -> R.
- (2,2) connects to (1,2) -> R.
- (2,5) connects to (3,5)? (3,5) connects to (4,5)? No, (4,5) is G. (3,5) neighbors: (2,5), (3,4)B, (3,6)G, (4,5)G. (2,5) neighbors: (1,5)G, (2,4)G, (2,6)B, (3,5). (2,5) and (3,5) are a small pocket. Are they connected to edge? (2,5) neighbors are G, G, B, (3,5). (3,5) neighbors are (2,5), B, G, G. This pocket {(2,5), (3,5)} is surrounded by G and B. It is **X**.
- (3,8) connects to (4,8)? No, (4,8) is B. (3,8) neighbors: (2,8)G, (3,7)G, (3,9)B, (4,8)B. (3,8) is surrounded. It is **X**.
- (4,3) connects to (5,3)? No, (5,3) is B. (4,3) neighbors: (3,3)B, (4,2)G, (4,4)B, (5,3)B. (4,3) is surrounded. It is **X**.
- (4,6) connects to (5,6)? No, (5,6) is B. (4,6) neighbors: (3,6)G, (4,5)G, (4,7)B, (5,6)B. (4,6) is surrounded. It is **X**.
- (4,9) is on edge -> R.
- (5,0) is on edge -> R.
- (5,2) connects to (6,2)? No, (6,2) is G. (5,2) neighbors: (4,2)G, (5,1)G, (5,3)B, (6,2)G. (5,2) is surrounded. It is **X**.
- (5,5) connects to (6,5)? No, (6,5) is G. (5,5) neighbors: (4,5)G, (5,4)G, (5,6)B, (6,5)G. (5,5) is surrounded. It is **X**.
- (5,8) connects to (4,8)? No, (4,8) is B. (5,8) neighbors: (4,8)B, (5,7)B, (5,9)G, (6,8)G. (5,8) is surrounded. It is **X**.
- (6,3), (6,4) connect to (7,3)? No, (7,3) is B. (6,3) neighbors: (5,3)B, (6,2)G, (6,4), (7,3)B. (6,4) neighbors: (5,4)G, (6,3), (6,5)G, (7,4)B. (6,3) and (6,4) are a pocket. It is **X**.
- (6,6) connects to (7,6)? No, (7,6) is B. (6,6) neighbors: (5,6)B, (6,5)G, (6,7)G, (7,6)B. (6,6) is surrounded. It is **X**.
- (7,7) connects to (8,7)? (8,7) connects to (9,7)? (9,7) is on edge. So (7,7), (8,7), (9,7) are R.
- (8,1), (8,2), (8,3) connect to (9,1)? (9,1) is on edge. So (8,1), (8,2), (8,3), (9,1) are R.
- (8,7) connects to (9,7) -> R.
- (9,5), (9,6), (9,7) are on edge -> R.
- (9,9) is on edge -> R.

**Final check of '.' cells:**
(0,1) R, (0,2) R, (0,4) R
(1,0) R, (1,1) R, (1,2) R, (1,4) R, (1,6) X, (1,8) R, (1,9) R
(2,0) R, (2,2) R, (2,5) X
(3,5) X, (3,8) X
(4,3) X, (4,6) X, (4,9) R
(5,0) R, (5,2) X, (5,5) X, (5,8) X
(6,3) X, (6,4) X, (6,6) X
(7,7) R
(8,1) R, (8,2) R, (8,3) R, (8,7) R
(9,1) R, (9,5) R, (9,6) R, (9,7) R, (9,9) R

**Constructing the grid:**
Row 0: B R R B R B B B B G
Row 1: R R R G R G X B R R
Row 2: R B R G G X B G G G
Row 3: G B G B B X G G X B
Row 4: B B G X B G X B B R
Row 5: R G X B G X B B X G
Row 6: B B G X X G X G G G
Row 7: G B B B B G B R G B
Row 8: G R R R G G B R B B
Row 9: B R G B B R R R B R

ANSWER:
B R R B R B B B B G
R R R G R G X B R R
R B R G G X B G G G
G B G B B X G G X B
B B G X B G X B B R
R G X B G X B B X 

... [81 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Instruments (Green 'G' and Blue 'B'):** These are static elements that act as barriers or walls. They do not change.
    *   **Rests (Black '.'):** These are the elements being transformed.
    *   **Entrances (Perimeter):** Any black cell ('.') located on the edge of the 10x10 grid is an "entrance" where noise enters.
    *   **Noise (Red 'R'):** Represents black cells that are connected to the perimeter via a path of other black cells.
    *   **Soundproofed (Grey 'X'):** Represents black cells that are completely enclosed by instruments (G and B) and have no path of black cells leading to the perimeter.

2.  **Identify the AFFORDANCES:**
    *   **Instruments (G, B):** Block the flow of noise. They define the boundaries of "rooms."
    *   **Rests (.):** Can either "resonate red" (if connected to the edge) or "remain grey" (if isolated from the edge).

3.  **Identify the TRANSFORMATION RULE:**
    *   Identify all black cells ('.') on the perimeter.
    *   Perform a flood-fill (or BFS/DFS) starting from all perimeter black cells, moving only through other black cells. All cells reached in this flood-fill become Red ('R').
    *   All remaining black cells ('.') that were not reached by the flood-fill are enclosed by instruments and must become Grey ('X').
    *   All other cells (G, B) remain unchanged.

### Applying the Rule to Test Input

**Input Grid Analysis:**
Row 0: B . . B . B B B B G (Perimeter: (0,1), (0,2), (0,4) are '.')
Row 1: . . . G . G . B . . (Perimeter: (1,0), (1,9) are '.')
Row 2: . B . G G . B G G G (Perimeter: (2,0) is '.')
Row 3: G B G B B . G G . B (Perimeter: (3,9) is 'B', no '.')
Row 4: B B G . B G . B B . (Perimeter: (4,9) is '.')
Row 5: . G . B G . B B . G (Perimeter: (5,0) is '.')
Row 6: B B G . . G . G G G (Perimeter: (6,9) is 'G', no '.')
Row 7: G B B B B G B . G B (Perimeter: (7,9) is 'B', no '.')
Row 8: G . . . G G B . B B (Perimeter: (8,0) is 'G', no '.')
Row 9: B . G B B . . . B . (Perimeter: (9,1), (9,5), (9,6), (9,7), (9,9) are '.')

**Step 1: Identify Perimeter Rests (Entrances):**
(0,1), (0,2), (0,4), (1,0), (1,1), (1,2), (1,4), (1,6), (1,8), (1,9), (2,0), (2,2), (2,5), (3,5), (3,8), (4,3), (4,6), (4,9), (5,0), (5,2), (5,5), (5,8), (6,3), (6,4), (6,6), (7,7), (8,1), (8,2), (8,3), (8,7), (9,1), (9,5), (9,6), (9,7), (9,9).

**Step 2: Flood Fill from Perimeter Rests:**
- Path 1: (0,1)-(0,2)-(1,2)-(1,1)-(1,0)-(2,0) ... and (0,1)-(1,1)-(1,2)...
- Path 2: (0,4)-(1,4)-(2,5)-(3,5)-(4,6)-(5,5)-(6,6)-(7,7)-(8,7)-(9,7)-(9,6)-(9,5)-(8,5 is G, stop)
- Path 3: (1,6)-(1,5 is G, stop)
- Path 4: (1,8)-(1,9)-(2,9 is G, stop)
- Path 5: (4,9)-(5,9 is G, stop)
- Path 6: (5,0)-(6,0 is B, stop)
- Path 7: (5,8)-(4,8 is B, stop)
- Path 8: (6,3)-(6,4)-(5,4 is G, stop)
- Path 9: (8,1)-(8,2)-(8,3)-(9,3 is B, stop)
- Path 10: (9,1)-(8,1)...
- Path 11: (9,9)-(8,9 is B, stop)

**Let's trace carefully:**
- (0,1), (0,2), (1,2), (1,1), (1,0), (2,0) are R.
- (0,4), (1,4) are R.
- (1,6) is R.
- (1,8), (1,9) are R.
- (2,2) is R? Let's check: (2,2) is connected to (1,2). Yes.
- (2,5) is R? (2,5) is connected to (1,5 is G, no), (3,5) is connected to (2,5). (3,5) is connected to (4,5 is G, no). Wait, (2,5) is connected to (3,5). Is (3,5) connected to perimeter? (3,5) is not on perimeter. (2,5) is not on perimeter. (1,5) is G. (3,5) is connected to (4,5 is G). (3,5) is connected to (3,4 is B). (3,5) is connected to (3,6 is G). (3,5) is connected to (2,5). (2,5) is connected to (1,5 is G). (2,5) is connected to (2,4 is G). (2,5) is connected to (2,6 is B). So (2,5) and (3,5) are isolated from perimeter? Let's re-check (2,5). (2,5) is surrounded by (1,5)=G, (2,4)=G, (2,6)=B, (3,5)=. So (2,5) and (3,5) are a cluster. Are they connected to perimeter? (1,5) is G, (2,4) is G, (2,6) is B, (3,4) is B, (3,6) is G, (4,5) is G. Yes, (2,5) and (3,5) are enclosed. They become X.
- (3,8) is connected to (4,8 is B), (3,7 is G), (3,9 is B), (2,8 is G). (3,8) is isolated. It becomes X.
- (4,3) is connected to (5,3 is B), (4,2 is G), (4,4 is B), (3,3 is B). (4,3) is isolated. It becomes X.
- (4,6) is connected to (5,6 is B), (4,5 is G), (4,7 is B), (3,6 is G). (4,6) is isolated. It becomes X.
- (4,9) is on perimeter. It becomes R.
- (5,2) is connected to (6,2 is G), (5,1 is G), (5,3 is B), (4,2 is G). (5,2) is isolated. It becomes X.
- (5,5) is connected to (6,5 is G), (5,4 is G), (5,6 is B), (4,5 is G). (5,5) is isolated. It becomes X.
- (5,8) is connected to (6,8 is G), (5,7 is B), (5,9 is G), (4,8 is B). (5,8) is isolated. It becomes X.
- (6,3), (6,4) are connected to (7,3 is B), (6,2 is G), (6,5 is G), (5,3 is B), (5,4 is G), (7,4 is B). (6,3) and (6,4) are isolated. They become X.
- (6,6) is connected to (7,6 is B), (6,5 is G), (6,7 is G), (5,6 is B). (6,6) is isolated. It becomes X.
- (7,7) is connected to (8,7), (9,7), (9,6), (9,5). (8,7) is connected to (8,6 is B), (8,8 is B), (7,7), (9,7). (9,7) is on perimeter. So (7,7), (8,7), (9,7), (9,6), (9,5) are R

... [3423 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Terrain (Fixed):** Green (G) and Blue (B) cells. These act as barriers that block the "wind."
    *   **Clearings (Transformable):** Black (.) cells. These are the only cells that change.
    *   **Storm (Wind):** A force that travels from the edges of the grid into the clearings.
    *   **Red (R):** Clearings that are connected to the edge of the grid through a path of other clearings (exposed to the storm).
    *   **Grey (X):** Clearings that are completely enclosed by terrain (G or B) and have no path to the edge.

2.  **Identify the AFFORDANCES:**
    *   **Terrain (G, B):** Blocks the movement of the storm.
    *   **Clearings (.):** Can be transformed into Red (R) if they have a path to the boundary, or Grey (X) if they are trapped.

3.  **Identify the TRANSFORMATION RULE:**
    *   The transformation is a connectivity problem on a graph where nodes are black cells (`.`).
    *   A black cell becomes **Red (R)** if there exists a path of black cells from that cell to any cell on the grid's boundary (the first/last row or first/last column).
    *   A black cell becomes **Grey (X)** if it is not connected to the boundary via other black cells.
    *   All other cells (G, B) remain unchanged.

### Applying the Rule to Test Input

**Input Grid:**
```
0: B . . B . B B B B G
1: . . . G . G . B . .
2: . B . G G . B G G G
3: G B G B B . G G . B
4: B B G . B G . B B .
5: . G . B G . B B . G
6: B B G . . G . G G G
7: G B B B B G B . G B
8: G . . . G G B . B B
9: B . G B B . . . B .
```

**Step 1: Identify all black cells (.) and their connectivity to the boundary.**

*   **Boundary-connected clearings (Red - R):**
    *   Row 0: (0,1), (0,2), (0,4)
    *   Row 1: (1,0), (1,1), (1,2), (1,4), (1,6), (1,8), (1,9)
    *   Row 2: (2,0), (2,2), (2,5)
    *   Row 3: (3,5), (3,8)
    *   Row 4: (4,3), (4,6), (4,9)
    *   Row 5: (5,0), (5,2), (5,5), (5,8)
    *   Row 6: (6,3), (6,4), (6,6)
    *   Row 7: (7,7)
    *   Row 8: (8,1), (8,2), (8,3), (8,7)
    *   Row 9: (9,1), (9,5), (9,6), (9,7)

*   **Tracing paths to boundary:**
    *   (0,1)-(0,2)-(1,2)-(1,1)-(1,0)-(2,0) are all connected to boundary.
    *   (0,4)-(1,4)-(2,5)-(3,5)-(4,6)-(5,5)-(6,6)-(7,7)-(8,7)-(9,7)-(9,6)-(9,5) are all connected to boundary.
    *   (1,6) is connected to (0,4) via (1,4)? No, (1,5) is G. Let's re-check (1,6). (1,6) is adjacent to (0,6) which is B. (1,6) is adjacent to (1,5) which is G. (1,6) is adjacent to (1,7) which is B. (1,6) is adjacent to (2,6) which is B. (1,6) is isolated? Wait, (1,6) is adjacent to (0,6) [B], (1,5) [G], (1,7) [B], (2,6) [B]. (1,6) is a black cell. Is it connected to the edge? It is at (1,6). It is not on the edge. Its neighbors are (0,6) [B], (2,6) [B], (1,5) [G], (1,7) [B]. It is trapped! So (1,6) is **Grey (X)**.
    *   Let's re-evaluate (1,8)-(1,9)-(2,9) is G. (1,8) is adjacent to (0,8) [B], (2,8) [G], (1,7) [B], (1,9) [.] -> (1,9) is on edge. So (1,8) and (1,9) are **Red (R)**.
    *   (2,2) is adjacent to (1,2) [.] -> (2,2) is **Red (R)**.
    *   (3,8) is adjacent to (4,8) [B], (2,8) [G], (3,7) [G], (3,9) [B]. (3,8) is isolated? Neighbors: (2,8) G, (4,8) B, (3,7) G, (3,9) B. (3,8) is **Grey (X)**.
    *   (4,3) is adjacent to (3,3) [B], (5,3) [B], (4,2) [G], (4,4) [B]. (4,3) is **Grey (X)**.
    *   (4,9) is on edge. (4,9) is **Red (R)**.
    *   (5,0) is on edge. (5,0) is **Red (R)**.
    *   (5,2) is adjacent to (4,2) [G], (6,2) [G], (5,1) [G], (5,3) [B]. (5,2) is **Grey (X)**.
    *   (5,8) is adjacent to (4,8) [B], (6,8) [G], (5,7) [B], (5,9) [G]. (5,8) is **Grey (X)**.
    *   (6,3)-(6,4) are adjacent to (5,3) [B], (7,3) [B], (6,2) [G], (6,5) [G]. (6,3) and (6,4) are **Grey (X)**.
    *   (8,1)-(8,2)-(8,3)-(9,1) are connected to (9,1) which is on edge. So (8,1), (8,2), (8,3), (9,1) are **Red (R)**.
    *   (9,5)-(9,6)-(9,7)-(8,7)-(7,7) are connected to (9,5) which is on edge. So (9,5), (9,6), (9,7), (8,7), (7,7) are **Red (R)**.
    *   (8,7) is connected to (7,7) and (9,7).

**Refined Connectivity Check:**
1.  (0,1), (0,2), (0,4), (1,0), (1,1), (1,2), (1,4), (1,8), (1,9), (2,0), (2,2), (2,5), (3,5), (4,6), (4,9), (5,0), (5,5), (7,7), (8,1), (8,2), (8,3), (8,7), (9,1), (9,5), (9,6), (9,7) are **Red (R)**.
2.  (1,6), (3,8), (4,3), (5,2), (5,8), (6,3), (6,4), (6,6) are **Grey (X)**.
    *   Wait, (6,6) neighbors: (5,6) [B], (7,6) [B], (6,5) [G], (6,7) [G]. (6,6) is **Grey (X)**.
    *   Wait, (5,5) neighbors: (4,5) [G], (6,5) [G], (5,4) [G], (5,6) [B]. (5,5) is **Grey (X)**.
    *   Let's re-trace (5,5): (5,5) is adjacent to (4,5) [G], (6,5) [G], (5,4) [G], (5,6) [B]. It is trapped. So (5,5) is **Grey (X)**.
    *   Let's re-trace (4,6): (4,6) is adjacent to (3,6) [G], (5,6) [B], (4,5) [G], (4,7) [B]. (4,6) is trapped. So (4,6) is **Grey (X)**.
    *   Let's re-trace (3,5): (3,5) is adjacent to (2,5) [.] and (4,5) [G]. (2,5) is adjacent to (1,5) [G], (3,5) [.], (2,4) [G], (2,6) [B]. (2,5) is trapped. So (2,5) and (3,5) are **Gre

... [2804 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Fortifications (Green 'G' and Blue 'B')**: These act as barriers or walls. They are static and define the boundaries of different regions.
    *   **No-man's-land (Black '.')**: These are the areas subject to change.
    *   **Frontier (Edges of the grid)**: The source of "fire." Any black cell connected to the edge of the grid (directly or through a path of other black cells) is exposed.
    *   **Red ('R')**: Represents no-man's-land that is exposed to the frontier.
    *   **Grey ('X')**: Represents no-man's-land that is completely enclosed by fortifications and cannot reach the edge.

2.  **Identify the AFFORDANCES**:
    *   **'G' and 'B'** afford containment; they block the "fire" from spreading.
    *   **'.'** affords connectivity; it can either connect to the edge (becoming 'R') or be trapped (becoming 'X').

3.  **Identify the TRANSFORMATION RULE**:
    *   Identify all black cells ('.').
    *   Perform a flood-fill or connectivity analysis starting from all black cells located on the grid edges.
    *   Any black cell reachable from the edge via other black cells becomes Red ('R').
    *   Any black cell that is *not* reachable from the edge (i.e., it is surrounded by 'G' or 'B') becomes Grey ('X').
    *   All other cells ('G', 'B') remain unchanged.

### Test Input Analysis

Input Grid:
Row 0: B . . B . B B B B G
Row 1: . . . G . G . B . .
Row 2: . B . G G . B G G G
Row 3: G B G B B . G G . B
Row 4: B B G . B G . B B .
Row 5: . G . B G . B B . G
Row 6: B B G . . G . G G G
Row 7: G B B B B G B . G B
Row 8: G . . . G G B . B B
Row 9: B . G B B . . . B .

**Step 1: Identify Black Cells on Edges (Potential Red)**
(0,1), (0,2), (0,4), (1,0), (1,1), (1,2), (1,4), (1,6), (1,8), (1,9), (2,0), (2,2), (2,5), (3,5), (3,8), (4,3), (4,6), (4,9), (5,0), (5,2), (5,5), (5,8), (6,3), (6,4), (6,6), (7,7), (8,1), (8,2), (8,3), (8,7), (9,1), (9,5), (9,6), (9,7), (9,9).

**Step 2: Trace Connectivity (Flood Fill from Edges)**
- (0,1)-(0,2)-(1,2)-(1,1)-(1,0)-(2,0) are connected to edge.
- (0,4)-(1,4) are connected to edge.
- (1,6) is connected to edge? No, (1,6) is surrounded by G, G, B, G. Wait, let's re-check (1,6). Row 1: `. . . G . G . B . .`. (1,6) is a dot. Neighbors: (0,6) is B, (2,6) is B, (1,5) is G, (1,7) is B. (1,6) is trapped! It becomes **X**.
- (1,8)-(1,9)-(2,9) is not possible, (2,9) is G. (1,8)-(1,9) are connected to edge.
- (2,2) is connected to (1,2) which is connected to edge.
- (2,5) is connected to (3,5) which is connected to (4,5)? No, (4,5) is G. (3,5) neighbors: (2,5), (4,5), (3,4) is B, (3,6) is G. (2,5) is connected to (1,5)? No, (1,5) is G. Let's re-examine (2,5). Neighbors: (1,5) is G, (3,5) is ., (2,4) is G, (2,6) is B. (3,5) neighbors: (2,5), (4,5) is G, (3,4) is B, (3,6) is G. So (2,5) and (3,5) are a small pocket. Are they connected to edge? (2,5) -> (3,5). Neither connects to edge. They become **X**.
- (3,8) neighbors: (2,8) is G, (4,8) is B, (3,7) is G, (3,9) is B. (3,8) is trapped! It becomes **X**.
- (4,3) neighbors: (3,3) is B, (5,3) is B, (4,2) is G, (4,4) is B. (4,3) is trapped! It becomes **X**.
- (4,6) neighbors: (3,6) is G, (5,6) is B, (4,5) is G, (4,7) is B. (4,6) is trapped! It becomes **X**.
- (4,9) is edge. (4,9) is R.
- (5,0) is edge. (5,0) is R.
- (5,2) neighbors: (4,2) is G, (6,2) is G, (5,1) is G, (5,3) is B. (5,2) is trapped! It becomes **X**.
- (5,5) neighbors: (4,5) is G, (6,5) is G, (5,4) is G, (5,6) is B. (5,5) is trapped! It becomes **X**.
- (5,8) neighbors: (4,8) is B, (6,8) is G, (5,7) is B, (5,9) is G. (5,8) is trapped! It becomes **X**.
- (6,3)-(6,4) neighbors: (5,3) is B, (5,4) is G, (7,3) is B, (7,4) is B, (6,2) is G, (6,5) is G. (6,3)-(6,4) are trapped! They become **X**.
- (6,6) neighbors: (5,6) is B, (7,6) is B, (6,5) is G, (6,7) is G. (6,6) is trapped! It becomes **X**.
- (7,7) neighbors: (6,7) is G, (8,7) is ., (7,6) is B, (7,8) is G. (7,7) connects to (8,7).
- (8,7) neighbors: (7,7), (9,7), (8,6) is B, (8,8) is B. (8,7) connects to (9,7).
- (9,7) is edge. So (7,7), (8,7), (9,7) are **R**.
- (9,5)-(9,6)-(9,7) are edge. (9,5), (9,6), (9,7) are **R**.
- (8,1)-(8,2)-(8,3) neighbors: (7,1) is B, (7,2) is B, (7,3) is B, (9,1) is ., (9,2) is G, (9,3) is B. (8,1)-(8,2)-(8,3) connects to (9,1). (9,1) is edge. So (8,1), (8,2), (8,3), (9,1) are **R**.
- (9,9) is edge. (9,9) is R.

**Final Check of Black Cells:**
(0,1) R, (0,2) R, (0,4) R
(1,0) R, (1,1) R, (1,2) R, (1,4) R, (1,6) X, (1,8) R, (1,9) R
(2,0) R, (2,2) R, (2,5) X
(3,5) X, (3,8) X
(4,3) X, (4,6) X, (4,9) R
(5,0) R, (5,2) X, (5,5) X, (5,8) X
(6,3) X, (6,4) X, (6,6) X
(7,7) R
(8,1) R, (8,2) R, (8,3) R, (8,7) R
(9,1) R, (9,5) R, (9,6) R, (9,7) R, (9,9) R

**Constructing the Grid:**
Row 0: B R R B R B B B B G
Row 1: R R R G R G X B R R
Row 2: R B R G G X B G G G
Row 3: G B G B B X G G X B
Row 4: B B G X B G X B B R
Row 5: R G X B G X B B X G
Row 6: B B G X X G X G G G
Row 7: G B B B B G B R G B
Row 8: G R R R G G B R B B
Row 9: B R G B B R

... [3027 chars truncated]