MARC2 Variant Comparison March 26, 2026

### Conceptual Digestion

1.  **CAST**:
    *   **Mineral Seeds**: Single colored pixels in the input grid.
    *   **Strata**: Horizontal sections of the grid assigned to a specific seed.
    *   **Solid Veins**: Full horizontal lines of color.
    *   **Hollow Corridors**: Rows with color only on the left and right edges.

2.  **AFFORDANCES**:
    *   The **y-coordinate (row)** of a seed determines which color controls which section of the grid.
    *   The **x-coordinate (column)** of a seed is ignored.
    *   The **top row (0)** and **bottom row (14)** are always solid veins of the color controlling that section.

3.  **TRANSFORMATION RULE**:
    *   Identify the row indices of all seeds in order: $S_1, S_2, \dots, S_n$.
    *   The grid is divided into $n$ regions.
    *   The boundary between region $i$ and region $i+1$ is the midpoint between seed $S_i$ and seed $S_{i+1}$: $\text{Boundary}_i = \lfloor (S_i + S_{i+1}) / 2 \rfloor$.
    *   **Region 1** (Seed $S_1$): Rows $0$ to $\text{Boundary}_1$.
    *   **Region $i$** (Seed $S_i$): Rows $\text{Boundary}_{i-1} + 1$ to $\text{Boundary}_i$.
    *   **Region $n$** (Seed $S_n$): Rows $\text{Boundary}_{n-1} + 1$ to $14$.
    *   **Within each region**:
        *   The row containing the seed $S_i$ is a **solid line** of color $C_i$.
        *   The very first row (0) and very last row (14) of the entire grid are **solid lines** of the color of the region they belong to.
        *   All other rows in the region are **hollow** (color $C_i$ at index 0 and 14, black `.` in between).

### Applying the Rule to Test Input

**Seeds**:
1.  Azure (A) at row 2
2.  Red (R) at row 4
3.  Blue (B) at row 9
4.  Green (G) at row 12

**Boundaries**:
*   B1 (A to R): $\lfloor (2+4)/2 \rfloor = 3$
*   B2 (R to B): $\lfloor (4+9)/2 \rfloor = 6$
*   B3 (B to G): $\lfloor (9+12)/2 \rfloor = 10$

**Regions**:
*   **Region A (Rows 0-3)**: Seed at row 2. Solid: 0, 2. Hollow: 1, 3.
*   **Region R (Rows 4-6)**: Seed at row 4. Solid: 4. Hollow: 5, 6.
*   **Region B (Rows 7-10)**: Seed at row 9. Solid: 9. Hollow: 7, 8, 10.
*   **Region G (Rows 11-14)**: Seed at row 12. Solid: 12, 14. Hollow: 11, 13.

**Row-by-Row Construction**:
0: A A A A A A A A A A A A A A A (Top edge)
1: A . . . . . . . . . . . . . A
2: A A A A A A A A A A A A A A A (Seed A)
3: A . . . . . . . . . . . . . A
4: R R R R R R R R R R R R R R R (Seed R)
5: R . . . . . . . . . . . . . R
6: R . . . . . . . . . . . . . R
7: B . . . . . . . . . . . . . B
8: B . . . . . . . . . . . . . B
9: B B B B B B B B B B B B B B B (Seed B)
10: B . . . . . . . . . . . . . B
11: G . . . . . . . . . . . . . G
12: G G G G G G G G G G G G G G G (Seed G)
13: G . . . . . . . . . . . . . G
14: G G G G G G G G G G G G G G G (Bottom edge)

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

### Conceptual Digestion

1.  **CAST**:
    *   **Colored Cells**: Act as "seeds" or "anchors" for horizontal zones.
    *   **Black Cells (.)**: The background tissue.
    *   **Grid Boundaries**: The "outer skin" of the organism.

2.  **AFFORDANCES**:
    *   A colored cell at position `(row, col)` defines a zone of influence.
    *   The color of the cell determines the color of the frame/membrane created.
    *   The row index of the cell determines the position of a horizontal "membrane wall."

3.  **TRANSFORMATION RULE**:
    *   The grid is partitioned into horizontal bands based on the row indices of the colored cells.
    *   Let the sorted row indices of the colored cells be $r_1, r_2, \dots, r_n$.
    *   **Membrane Walls**: For every colored cell at row $r_i$, the entire row $r_i$ in the output becomes that cell's color.
    *   **Outer Skin**: For the rows *between* the membrane walls (and the rows before the first and after the last), the leftmost column (0) and rightmost column (14) take the color of the "controlling" cell for that zone.
    *   **Zone Control**:
        *   Rows $0$ to $r_1$ are controlled by the color of the cell at $r_1$.
        *   Rows $r_1+1$ to $r_2$ are controlled by the color of the cell at $r_2$.
        *   ...
        *   Rows $r_{n-1}+1$ to $r_n$ are controlled by the color of the cell at $r_n$.
        *   Rows $r_n+1$ to $14$ are controlled by the color of the cell at $r_n$.
    *   **Interior**: All other cells remain black (.).

### Applying the Rule to Test Input

**Input Cells:**
1. Azure (A) at row 2
2. Red (R) at row 4
3. Blue (B) at row 9
4. Green (G) at row 12

**Zone Mapping:**
*   **Rows 0-2**: Controlled by Azure (A). Row 2 is a full wall of A. Rows 0-1 have A at col 0 and 14.
*   **Rows 3-4**: Controlled by Red (R). Row 4 is a full wall of R. Row 3 has R at col 0 and 14.
*   **Rows 5-9**: Controlled by Blue (B). Row 9 is a full wall of B. Rows 5-8 have B at col 0 and 14.
*   **Rows 10-12**: Controlled by Green (G). Row 12 is a full wall of G. Rows 10-11 have G at col 0 and 14.
*   **Rows 13-14**: Controlled by Green (G). Rows 13-14 have G at col 0 and 14. (Note: The last cell's influence extends to the bottom).

**Wait, checking Example 1 again:**
Example 1 cells: R(3), O(7), A(12).
Output:
Rows 0-3: R (Wall at 3, edges at 0-2)
Rows 4-7: O (Wall at 7, edges at 4-6)
Rows 8-12: A (Wall at 12, edges at 8-11)
Rows 13-14: A (Edges at 13-14, Wall at 14? No, Example 1 has a wall at 14 too).
Looking closer at Example 1 output:
Row 0: RRR... (Wall)
Row 1: R...R
Row 2: R...R
Row 3: RRR... (Wall)
Row 4: R...R
Row 5: R...R
Row 6: O...O
Row 7: OOO... (Wall)
Row 8: O...O
Row 9: O...O
Row 10: A...A
Row 11: A...A
Row 12: AAA... (Wall)
Row 13: A...A
Row 14: AAA... (Wall)

Correction: The first row (0) and last row (14) are always walls of the color controlling that zone.

**Revised Mapping for Test Input:**
*   Row 0: A A A A A A A A A A A A A A A (Wall - Start)
*   Row 1: A . . . . . . . . . . . . . A
*   Row 2: A A A A A A A A A A A A A A A (Wall - Cell A)
*   Row 3: R . . . . . . . . . . . . . R
*   Row 4: R R R R R R R R R R R R R R R (Wall - Cell R)
*   Row 5: R . . . . . . . . . . . . . R
*   Row 6: R . . . . . . . . . . . . . R
*   Row 7: B . . . . . . . . . . . . . B
*   Row 8: B . . . . . . . . . . . . . B
*   Row 9: B B B B B B B B B B B B B B B (Wall - Cell B)
*   Row 10: B . . . . . . . . . . . . . B
*   Row 11: G . . . . . . . . . . . . . G
*   Row 12: G G G G G G G G G G G G G G G (Wall - Cell G)
*   Row 13: G . . . . . . . . . . . . . G
*   Row 14: G G G G G G G G G G G G G G G (Wall - End)

Wait, looking at Example 1 again:
Row 0: R (Wall)
Row 1: R...R
Row 2: R...R
Row 3: R (Wall)
Row 4: R...R
Row 5: R...R
Row 6: O...O
Row 7: O (Wall)
Row 8: O...O
Row 9: O...O
Row 10: A...A
Row 11: A...A
Row 12: A (Wall)
Row 13: A...A
Row 14: A (Wall)

The transition from R to O happens at row 6. The cell O is at row 7. So the zone for O starts at row 6.
The transition from O to A happens at row 10. The cell A is at row 12. So the zone for A starts at row 10.
Rule: The boundary between zone $i$ and $i+1$ is the midpoint between $r_i$ and $r_{i+1}$.
Example 1: $r_1=3, r_2=7, r_3=12$.
Midpoint(3, 7) = 5. Zone R ends at 5, Zone O starts at 6.
Midpoint(7, 12) = 9.5. Zone O ends at 9, Zone A starts at 10.

Test Input: $r_1=2, r_2=4, r_3=9, r_4=12$.
Midpoint(2, 4) = 3. Zone A ends at 3, Zone R starts at 4? No, that doesn't fit.
Let's re-examine:
Example 1:
R: 0, 1, 2, 3, 4, 5 (6 rows)
O: 6, 7, 8, 9 (4 rows)
A: 10, 11, 12, 13, 14 (5 rows)
The cell R is at 3. The cell O is at 7. The cell A is at 12.
The distance from 3 to 7 is 4. Half is 2. $3+2=5$ (end of R), $7-2=5$ (start of O is 6).
The distance from 7 to 12 is 5. Half is 2.5. $7+2.5=9.5$ (end of O is 9), $12-2.5=9.5$ (start of A is 10).

Test Input:
$r_1=2, r_2=4, r_3=9, r_4=12$.
Mid(2, 4) = 3. Zone A: 0-3, Zone R: 4-?
Mid(4, 9) = 6.5. Zone R: 4-6, Zone B: 7-?
Mid(9, 12) = 10.5. Zone B: 7-10, Zone G: 11-14.

Let's chec

... [557 chars truncated]

### Conceptual Digestion

1.  **CAST**:
    *   **Stars**: Isolated colored pixels in the input grid.
    *   **Emission Lines**: Full horizontal rows of the star's color.
    *   **Frame Edges**: The leftmost (column 0) and rightmost (column 14) pixels of every row.
    *   **Void**: The interior pixels (columns 1-13) that are not part of an emission line.

2.  **AFFORDANCES**:
    *   Each star defines a "zone of influence" based on its vertical position (row index).
    *   The star's color is projected horizontally across its own row.
    *   The star's color is projected to the left and right edges of the grid for all rows between it and the previous/next star.

3.  **TRANSFORMATION RULE**:
    *   **Step 1: Identify Stars**. Find all colored pixels $(r, c)$ and their colors $C$.
    *   **Step 2: Horizontal Lines**. For every row $r$ containing a star of color $C$, the entire row $r$ becomes color $C$.
    *   **Step 3: Frame Borders**. For every row $i$ from $0$ to $14$:
        *   Find the nearest star above or below row $i$.
        *   If multiple stars are equidistant, the one above usually takes precedence or the logic follows a "nearest neighbor" approach. Looking at the examples:
            *   Rows above the first star take the first star's color.
            *   Rows between star $S_1$ (row $r_1$) and star $S_2$ (row $r_2$) take the color of the star that is closer to that row.
            *   Rows below the last star take the last star's color.
        *   Apply this color to the first and last columns (index 0 and 14).
    *   **Step 4: Special Case (Edges)**: The very top row (0) and very bottom row (14) are always fully filled with the color of the governing star (the first and last stars respectively).

### Applying Rule to Test Input
*   Star 1: Row 2, Color A (Azure)
*   Star 2: Row 4, Color R (Red)
*   Star 3: Row 9, Color B (Blue)
*   Star 4: Row 12, Color G (Green)

**Row-by-row analysis:**
- Row 0: Top edge. Governed by Star 1 (A). $\rightarrow$ **Full A**
- Row 1: Nearest is Star 1 (dist 1). $\rightarrow$ **A . . . . . . . . . . . . . A**
- Row 2: Star 1 is here. $\rightarrow$ **Full A**
- Row 3: Nearest is Star 1 (dist 1) or Star 2 (dist 1). Example 1 shows that if equidistant, the upper star's color is used for the border? No, let's check Example 1: Star R (row 3), Star O (row 7). Row 5 is dist 2 from R, dist 2 from O. Output row 5 is R. So upper star wins. $\rightarrow$ **A . . . . . . . . . . . . . A**
- Row 4: Star 2 is here. $\rightarrow$ **Full R**
- Row 5: Nearest is Star 2 (dist 1). $\rightarrow$ **R . . . . . . . . . . . . . R**
- Row 6: Nearest is Star 2 (dist 2) or Star 3 (dist 3). $\rightarrow$ **R . . . . . . . . . . . . . R**
- Row 7: Nearest is Star 2 (dist 3) or Star 3 (dist 2). $\rightarrow$ **B . . . . . . . . . . . . . B**
- Row 8: Nearest is Star 3 (dist 1). $\rightarrow$ **B . . . . . . . . . . . . . B**
- Row 9: Star 3 is here. $\rightarrow$ **Full B**
- Row 10: Nearest is Star 3 (dist 1). $\rightarrow$ **B . . . . . . . . . . . . . B**
- Row 11: Nearest is Star 3 (dist 2) or Star 4 (dist 1). $\rightarrow$ **G . . . . . . . . . . . . . G**
- Row 12: Star 4 is here. $\rightarrow$ **Full G**
- Row 13: Nearest is Star 4 (dist 1). $\rightarrow$ **G . . . . . . . . . . . . . G**
- Row 14: Bottom edge. Governed by Star 4 (G). $\rightarrow$ **Full G**

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

### Conceptual Digestion

1.  **CAST**:
    *   **Colored Pixels (Buoys)**: Act as anchors for horizontal lines (nets) and define the color of the vertical boundaries (harbor walls) in their vicinity.
    *   **Black Pixels (.)**: The "water" that is either replaced by walls or remains empty.
    *   **Grid Boundaries**: The top and bottom edges act as potential boundaries for the first and last buoy zones.

2.  **AFFORDANCES**:
    *   A buoy at `(row, col)` creates a full horizontal line of its color across the entire row.
    *   The space between two buoys (at `row1` and `row2`) is filled on the left and right edges (column 0 and column 14) with the color of the *nearest* buoy.
    *   The space above the first buoy is filled on the edges and the top row with the first buoy's color.
    *   The space below the last buoy is filled on the edges and the bottom row with the last buoy's color.

3.  **TRANSFORMATION RULE**:
    *   **Horizontal Nets**: For every row $i$ containing a colored pixel of color $C$, the entire row $i$ becomes color $C$.
    *   **Harbor Walls**: For every row $j$ that does *not* contain a colored pixel:
        *   Find the nearest row $i$ (above or below) that contains a colored pixel.
        *   Set the pixels at `(j, 0)` and `(j, 14)` to the color of the buoy at row $i$.
    *   **Surface/Seabed**:
        *   Row 0 becomes the color of the first buoy.
        *   Row 14 becomes the color of the last buoy.

### Applying the Rule to Test Input

**Buoys identified:**
- Row 2: Azure (A)
- Row 4: Red (R)
- Row 9: Blue (B)
- Row 12: Green (G)

**Step-by-step Row Construction:**
- Row 0: First buoy is A $\rightarrow$ `A A A A A A A A A A A A A A A`
- Row 1: Nearest buoy is A (Row 2) $\rightarrow$ `A . . . . . . . . . . . . . A`
- Row 2: Buoy A $\rightarrow$ `A A A A A A A A A A A A A A A`
- Row 3: Nearest buoy is A (Row 2) or R (Row 4). Distance is equal. Looking at Example 1, the color changes at the midpoint or favors the closer one. In Example 1, between R (row 3) and O (row 7), the transition happens at row 6 (O). Let's check: Row 4, 5 are R; Row 6 is O. Midpoint is 5. Row 6 is closer to O. So, Row 3 is closer to A (dist 1) and R (dist 1). Usually, the "nearest" rule in these puzzles splits the difference or favors the one below. Let's re-examine Example 1: R(3), O(7). Rows 4, 5 are R. Row 6 is O. (3 to 4=1, 3 to 5=2, 3 to 6=3; 7 to 6=1, 7 to 5=2, 7 to 4=3). So Row 6 is O because it's closer to O. Row 5 is R because it's closer to R. Row 4 is R because it's closer to R.
- Row 3: Dist to A(2) is 1, Dist to R(4) is 1. If we split, Row 3 is A.
- Row 4: Buoy R $\rightarrow$ `R R R R R R R R R R R R R R R`
- Row 5: Dist to R(4) is 1, Dist to B(9) is 4 $\rightarrow$ `R . . . . . . . . . . . . . R`
- Row 6: Dist to R(4) is 2, Dist to B(9) is 3 $\rightarrow$ `R . . . . . . . . . . . . . R`
- Row 7: Dist to R(4) is 3, Dist to B(9) is 2 $\rightarrow$ `B . . . . . . . . . . . . . B`
- Row 8: Dist to R(4) is 4, Dist to B(9) is 1 $\rightarrow$ `B . . . . . . . . . . . . . B`
- Row 9: Buoy B $\rightarrow$ `B B B B B B B B B B B B B B B`
- Row 10: Dist to B(9) is 1, Dist to G(12) is 2 $\rightarrow$ `B . . . . . . . . . . . . . B`
- Row 11: Dist to B(9) is 2, Dist to G(12) is 1 $\rightarrow$ `G . . . . . . . . . . . . . G`
- Row 12: Buoy G $\rightarrow$ `G G G G G G G G G G G G G G G`
- Row 13: Dist to G(12) is 1 $\rightarrow$ `G . . . . . . . . . . . . . G`
- Row 14: Last buoy is G $\rightarrow$ `G G G G G G G G G G G G G G G`

*Self-Correction on Row 3*: In Example 1, R is at 3, O is at 7. Rows 4, 5 are R. Row 6 is O.
Row 4: dist R=1, O=3 $\rightarrow$ R
Row 5: dist R=2, O=2 $\rightarrow$ R (favors above)
Row 6: dist R=3, O=1 $\rightarrow$ O
Applying "favors above" to Row 3: Dist A=1, R=1 $\rightarrow$ A.

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

### Conceptual Digestion

1.  **CAST**:
    *   **Charges**: Isolated colored pixels in the input grid.
    *   **Traces**: Full horizontal lines of the same color as the charge.
    *   **Edge Pins**: The leftmost and rightmost columns of the grid, which act as vertical connectors.
    *   **Package Boundaries**: The top and bottom edges of the grid.

2.  **AFFORDANCES**:
    *   A charge at `(row, col)` dictates that `row` becomes a full horizontal trace of that color.
    *   The space between two traces (or between a trace and a boundary) is filled at the edges (column 0 and column 14) by the color of the *nearest* charge.
    *   The top boundary (row 0) and bottom boundary (row 14) are always filled as full horizontal traces of the color of the nearest charge (the first and last charges, respectively).

3.  **TRANSFORMATION RULE**:
    *   **Step 1**: Identify all charges and their positions $(r, c)$ and colors $C$.
    *   **Step 2**: For every row $r$ that contains a charge, fill the entire row with color $C$.
    *   **Step 3**: For every row $r$ that does *not* contain a charge:
        *   Find the nearest row $r_{near}$ that contains a charge.
        *   Set the cells at `(r, 0)` and `(r, 14)` to the color of the charge at $r_{near}$.
        *   Keep the interior cells `(r, 1)` to `(r, 13)` as black (`.`).
    *   **Step 4**: Special Boundary Rule: The very first row (row 0) and the very last row (row 14) are always full horizontal traces of the color of the nearest charge (the first charge for row 0, the last charge for row 14).

### Applying the Rule to Test Input

**Charges identified:**
1.  Row 2: Azure (A)
2.  Row 4: Red (R)
3.  Row 9: Blue (B)
4.  Row 12: Green (G)

**Row-by-row construction:**
*   Row 0: Nearest charge is A (Row 2). Full trace: **A A A A A A A A A A A A A A A**
*   Row 1: Nearest charge is A (Row 2). Edges: **A . . . . . . . . . . . . . A**
*   Row 2: Charge A. Full trace: **A A A A A A A A A A A A A A A**
*   Row 3: Nearest charge is A (Row 2) or R (Row 4). Both distance 1. Usually, the "governing" charge is the one it's moving toward or the one above. Looking at Example 1 (Row 4 A, Row 7 O), Row 5 and 6 are O. Looking at Example 2 (Row 1 A, Row 3 B), Row 2 is A. It seems the rule is: if equidistant, use the one above? No, let's re-examine Example 1: Row 3 is R, Row 7 is O. Row 4, 5, 6 are between. Row 4, 5 are R, Row 6 is O. Distance to R is 1, 2, 3. Distance to O is 3, 2, 1. So it's strictly nearest.
    *   Row 3: Distance to A (Row 2) is 1. Distance to R (Row 4) is 1. In Example 2, Row 2 is between Row 1 (A) and Row 3 (B). Row 2 is A. So for Row 3, we use A. **A . . . . . . . . . . . . . A**
*   Row 4: Charge R. Full trace: **R R R R R R R R R R R R R R R**
*   Row 5: Nearest charge R (Row 4). Edges: **R . . . . . . . . . . . . . R**
*   Row 6: Nearest charge R (Row 4). Edges: **R . . . . . . . . . . . . . R**
*   Row 7: Nearest charge R (Row 4) dist 3, B (Row 9) dist 2. Nearest is B. Edges: **B . . . . . . . . . . . . . B**
*   Row 8: Nearest charge B (Row 9). Edges: **B . . . . . . . . . . . . . B**
*   Row 9: Charge B. Full trace: **B B B B B B B B B B B B B B B**
*   Row 10: Nearest charge B (Row 9). Edges: **B . . . . . . . . . . . . . B**
*   Row 11: Nearest charge B (Row 9) dist 2, G (Row 12) dist 1. Nearest is G. Edges: **G . . . . . . . . . . . . . G**
*   Row 12: Charge G. Full trace: **G G G G G G G G G G G G G G G**
*   Row 13: Nearest charge G (Row 12). Edges: **G . . . . . . . . . . . . . G**
*   Row 14: Nearest charge G (Row 12). Full trace: **G G G G G G G G G G G G G G G**

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

### Conceptual Digestion

1.  **The CAST**:
    *   **Runes**: Isolated colored pixels in the input grid.
    *   **Bands**: Full horizontal lines of a specific color.
    *   **Edge-Pillars**: The leftmost and rightmost columns of the grid.
    *   **Monolith**: The 15x15 grid area.

2.  **The AFFORDANCES**:
    *   A rune at row $y$ forces the entire row $y$ to become that color.
    *   The edge-pillars (column 0 and column 14) are filled based on the proximity to the nearest rune's row.
    *   The top edge (row 0) and bottom edge (row 14) are filled based on the nearest rune above/below them.

3.  **The TRANSFORMATION RULE**:
    *   **Horizontal Bands**: For every row $y$ containing a rune of color $C$, set all pixels in row $y$ to color $C$.
    *   **Edge-Pillars**: For every cell $(y, 0)$ and $(y, 14)$, assign the color of the rune whose row is closest to $y$. If two runes are equidistant, the one above takes precedence (based on Example 1's behavior where the gap between R and O is split, and the gap between O and A is split).
    *   **Refining the Proximity Rule**:
        *   Let the sorted rune rows be $r_1, r_2, \dots, r_n$ with colors $c_1, c_2, \dots, c_n$.
        *   For $y < r_1$: Color is $c_1$.
        *   For $y > r_n$: Color is $c_n$.
        *   For $r_i < y < r_{i+1}$:
            *   If $y - r_i < r_{i+1} - y$, color is $c_i$.
            *   If $y - r_i > r_{i+1} - y$, color is $c_{i+1}$.
            *   If $y - r_i = r_{i+1} - y$, the color is $c_i$ (based on Example 1: R at 3, O at 7. Midpoint is 5. Row 5 is R, Row 6 is O. Wait, let's re-examine Example 1).

    *   **Re-examining Example 1**:
        *   Rune R at row 3, O at row 7, A at row 12.
        *   Rows 0-3: R (Row 3 is band)
        *   Rows 4-6: Row 4 is R, Row 5 is R, Row 6 is O. (Distance to R: 1, 2, 3. Distance to O: 3, 2, 1).
        *   Rows 7-11: Row 7 is band O. Row 8 is O, Row 9 is O, Row 10 is A, Row 11 is A. (Distance to O: 1, 2, 3, 4. Distance to A: 5, 4, 3, 2).
        *   Rows 12-14: Row 12 is band A. Row 13 is A, Row 14 is A.
        *   *Correction*: The edge pillars follow the nearest rune. If equidistant, the one below takes precedence? No, let's check row 5: $5-3=2$ and $7-5=2$. Row 5 is R. So if equidistant, the one *above* takes precedence.

### Applying to Test Input
Runes:
- Row 2: Azure (A)
- Row 4: Red (R)
- Row 9: Blue (B)
- Row 12: Green (G)

**Row-by-row analysis for edges (Col 0 and 14):**
- Row 0: Nearest is A (row 2). Color: A
- Row 1: Nearest is A (row 2). Color: A
- Row 2: Band A. Color: A
- Row 3: Nearest A (dist 1) or R (dist 1). Equidistant $\rightarrow$ Above (A). Color: A
- Row 4: Band R. Color: R
- Row 5: Nearest R (dist 1). Color: R
- Row 6: Nearest R (dist 2) or B (dist 3). Color: R
- Row 7: Nearest R (dist 3) or B (dist 2). Color: B
- Row 8: Nearest B (dist 1). Color: B
- Row 9: Band B. Color: B
- Row 10: Nearest B (dist 1). Color: B
- Row 11: Nearest B (dist 2) or G (dist 1). Color: G
- Row 12: Band G. Color: G
- Row 13: Nearest G (dist 1). Color: G
- Row 14: Nearest G (dist 2). Color: G

**Final Grid Construction:**
- Row 0: A . . . . . . . . . . . . . A
- Row 1: A . . . . . . . . . . . . . A
- Row 2: A A A A A A A A A A A A A A A
- Row 3: A . . . . . . . . . . . . . A
- Row 4: R R R R R R R R R R R R R R R
- Row 5: R . . . . . . . . . . . . . R
- Row 6: R . . . . . . . . . . . . . R
- Row 7: B . . . . . . . . . . . . . B
- Row 8: B . . . . . . . . . . . . . B
- Row 9: B B B B B B B B B B B B B B B
- Row 10: B . . . . . . . . . . . . . B
- Row 11: G . . . . . . . . . . . . . G
- Row 12: G G G G G G G G G G G G G G G
- Row 13: G . . . . . . . . . . . . . G
- Row 14: G G G G G G G G G G G G G G G (Wait, Example 1 row 14 is a band? No, Example 1 row 14 is A A A... because the last rune was A at row 12. Let's check: Row 12 band A, Row 13 A, Row 14 A. No, Example 1 row 14 is a band. Let me re-read: "The crown and base solidify under their governing rune." This means Row 0 and Row 14 are always full bands of the first and last rune's colors.)

**Revised Row 0 and 14:**
- Row 0: A A A A A A A A A A A A A A A
- Row 14: G G G G G G G G G G G G G G G

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

### Conceptual Digestion

1.  **CAST**:
    *   **Dancers**: The isolated colored dots in the input grid.
    *   **Stage**: The 15x15 grid.
    *   **Horizontal Lines**: Full rows of a dancer's color.
    *   **Wing-poses**: The leftmost and rightmost columns of the grid.

2.  **AFFORDANCES**:
    *   A dancer at row $y$ forces row $y$ to be entirely filled with that dancer's color.
    *   The space between two dancers (rows $y_1$ and $y_2$) is filled at the edges (column 0 and column 14) by the color of the nearest dancer.
    *   The space above the first dancer is filled at the edges (and potentially fully) by the first dancer's color.
    *   The space below the last dancer is filled at the edges (and potentially fully) by the last dancer's color.

3.  **TRANSFORMATION RULE**:
    *   Identify all rows containing a colored dot.
    *   For every row $y$ that contains a dot of color $C$, set all cells in that row to $C$.
    *   For every row $y$ that does *not* contain a dot:
        *   Find the nearest row $y_{near}$ that contains a dot.
        *   Set the cells at $(y, 0)$ and $(y, 14)$ to the color of the dot at $y_{near}$.
        *   Keep the interior cells $(y, 1 \dots 13)$ as black (.).
    *   **Boundary Condition**: The very first row and very last row of the grid are always filled completely if they are governed by the first/last dancer (as seen in Example 1 and 2). Specifically, if the first dancer is at row $y_1$, rows $0 \dots y_1$ are filled at the edges, but row 0 is filled completely. Similarly, if the last dancer is at row $y_n$, rows $y_n \dots 14$ are filled at the edges, but row 14 is filled completely.

### Applying the Rule to Test Input

*   **Dancers**:
    *   Row 2: Azure (A)
    *   Row 4: Red (R)
    *   Row 9: Blue (B)
    *   Row 12: Green (G)

*   **Row-by-row construction**:
    *   Row 0: First dancer is A. Row 0 is boundary $\rightarrow$ **A A A A A A A A A A A A A A A**
    *   Row 1: Nearest is A $\rightarrow$ **A . . . . . . . . . . . . . A**
    *   Row 2: Dancer A $\rightarrow$ **A A A A A A A A A A A A A A A**
    *   Row 3: Nearest is A or R (equidistant, but usually follows the "governing" logic of the segment). In Example 1, row 4 is R, row 7 is O. Row 5, 6 are R. Here, Row 3 is between A (row 2) and R (row 4). It is equidistant. Looking at Example 2, row 2 is A, row 3 is B. Row 2 is A. Row 3 is B. Let's check Example 1 again: Row 3 is R, Row 7 is O. Rows 4, 5, 6 are R. Row 4 is closer to R. Row 6 is closer to O. Wait, Row 6 in Example 1 is O. Row 3 is R, Row 7 is O. Distance from 6 to 3 is 3, distance from 6 to 7 is 1. So Row 6 is O.
    *   Row 3: Nearest is A (dist 1) and R (dist 1). In Example 2, Row 2 is A, Row 3 is B. Row 2 is A. Row 3 is B. Let's re-evaluate: Row 3 is between A (2) and R (4). It's equidistant. Usually, the lower dancer takes over or the upper. Let's look at Example 1: Row 3 (R), Row 7 (O). Row 4, 5 are R. Row 6 is O. Row 4 is dist 1 from R, dist 3 from O. Row 5 is dist 2 from R, dist 2 from O. Row 5 is R. Row 6 is dist 3 from R, dist 1 from O. So it's simple proximity. For Row 3: dist to A is 1, dist to R is 1. In ties, the upper dancer (A) typically governs.
    *   Row 4: Dancer R $\rightarrow$ **R R R R R R R R R R R R R R R**
    *   Row 5: Nearest R $\rightarrow$ **R . . . . . . . . . . . . . R**
    *   Row 6: Nearest R $\rightarrow$ **R . . . . . . . . . . . . . R**
    *   Row 7: Nearest R (dist 3) or B (dist 2) $\rightarrow$ **B . . . . . . . . . . . . . B**
    *   Row 8: Nearest B $\rightarrow$ **B . . . . . . . . . . . . . B**
    *   Row 9: Dancer B $\rightarrow$ **B B B B B B B B B B B B B B B**
    *   Row 10: Nearest B $\rightarrow$ **B . . . . . . . . . . . . . B**
    *   Row 11: Nearest B (dist 2) or G (dist 1) $\rightarrow$ **G . . . . . . . . . . . . . G**
    *   Row 12: Dancer G $\rightarrow$ **G G G G G G G G G G G G G G G**
    *   Row 13: Nearest G $\rightarrow$ **G . . . . . . . . . . . . . G**
    *   Row 14: Boundary $\rightarrow$ **G G G G G G G G G G G G G G G**

*   **Refining Row 3**: Between A (2) and R (4). Distances are 1 and 1. In Example 2, Row 1 is A, Row 3 is B. Row 2 is A. Row 2 is dist 1 from A and dist 1 from B. Row 2 is A. So ties go to the upper dancer. Row 3 will be A.

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

### Conceptual Digestion

1.  **CAST**:
    *   **Mineral Points**: Isolated colored cells in the input grid.
    *   **Seams**: Full horizontal lines of the same color as the mineral point.
    *   **Cliff Edges**: The leftmost (column 0) and rightmost (column 14) cells of the grid.
    *   **Dark Cliff**: The black cells (`.`) filling the interior.

2.  **AFFORDANCES**:
    *   A mineral point at row $y$ transforms the entire row $y$ into a solid line of that color.
    *   The cliff edges (columns 0 and 14) are "stained" by the nearest mineral point. This means for any row $i$, the cells at $(i, 0)$ and $(i, 14)$ take the color of the mineral point whose row $y$ is closest to $i$.
    *   If a row is equidistant between two mineral points, the one below (higher index) typically takes precedence or the boundary is split. Looking at Example 1: Row 3 is R, Row 7 is O. Row 5 is equidistant (dist 2 from R, dist 2 from O). In Example 1, Row 5 is R, Row 6 is O. This suggests the boundary is the midpoint; if the distance is equal, the lower color starts. Wait, let's re-examine Example 1:
        *   R at row 3, O at row 7. Midpoint is row 5. Row 5 is R, Row 6 is O.
        *   O at row 7, A at row 12. Midpoint is row 9.5. Row 9 is O, Row 10 is A.
        *   Top edge (row 0) is R because R is the first mineral.
        *   Bottom edge (row 14) is A because A is the last mineral.

3.  **TRANSFORMATION RULE**:
    *   **Step 1**: Identify all mineral points $(y_n, x_n)$ and their colors $C_n$.
    *   **Step 2**: For every row $i$ from 0 to 14:
        *   If row $i$ contains a mineral point $C_n$, the entire row $i$ becomes $C_n$.
        *   Otherwise, the cells at $(i, 0)$ and $(i, 14)$ are colored $C_n$ where $y_n$ is the row of the mineral point closest to $i$.
        *   If $i$ is exactly halfway between $y_n$ and $y_{n+1}$, the color $C_n$ is used for the row above the midpoint and $C_{n+1}$ for the row below.
        *   All other cells $(i, 1 \dots 13)$ remain black (`.`).
    *   **Step 3**: Special case for boundaries: The very top row (0) and very bottom row (14) become full seams if they are the closest to a mineral point (which they always are). *Correction based on Example 1*: Row 0 is a full seam of R, and Row 14 is a full seam of A.

### Applying to Test Input
Mineral points:
1. Row 2: A (Azure)
2. Row 4: R (Red)
3. Row 9: B (Blue)
4. Row 12: G (Green)

**Row-by-row analysis**:
- Row 0: Closest to Row 2 (A). Full seam A.
- Row 1: Closest to Row 2 (A). Edges A.
- Row 2: Mineral A. Full seam A.
- Row 3: Midpoint between Row 2 (A) and Row 4 (R). Row 3 is equidistant. Based on Example 1 (Row 5 was R, Row 6 was O for points at 3 and 7), the split is at the center. For 2 and 4, Row 3 is the center. Let's check Example 2: Row 1 (A) and Row 3 (B). Row 2 is equidistant. Output Row 2 is A. So for 2 and 4, Row 3 is A. Edges A.
- Row 4: Mineral R. Full seam R.
- Row 5: Closest to Row 4 (R). Edges R.
- Row 6: Midpoint between Row 4 (R) and Row 9 (B). Distances: $|6-4|=2, |6-9|=3$. Closest is R. Edges R.
- Row 7: Midpoint between Row 4 (R) and Row 9 (B). Distances: $|7-4|=3, |7-9|=2$. Closest is B. Edges B.
- Row 8: Closest to Row 9 (B). Edges B.
- Row 9: Mineral B. Full seam B.
- Row 10: Closest to Row 9 (B). Edges B.
- Row 11: Midpoint between Row 9 (B) and Row 12 (G). Distances: $|11-9|=2, |11-12|=1$. Closest is G. Edges G.
- Row 12: Mineral G. Full seam G.
- Row 13: Closest to Row 12 (G). Edges G.
- Row 14: Closest to Row 12 (G). Full seam G.

**Refining Row 3**: In Example 2, points are at 1 and 3. Row 2 is equidistant. Output Row 2 is A. So the "top" color wins the tie.
Row 3: Dist to A(2) is 1, Dist to R(4) is 1. Tie goes to A.

**Final Grid Construction**:
0: A A A A A A A A A A A A A A A
1: A . . . . . . . . . . . . . A
2: A A A A A A A A A A A A A A A
3: A . . . . . . . . . . . . . A
4: R R R R R R R R R R R R R R R
5: R . . . . . . . . . . . . . R
6: R . . . . . . . . . . . . . R
7: B . . . . . . . . . . . . . B
8: B . . . . . . . . . . . . . B
9: B B B B B B B B B B B B B B B
10: B . . . . . . . . . . . . . B
11: G . . . . . . . . . . . . . G
12: G G G G G G G G G G G G G G G
13: G . . . . . . . . . . . . . G
14: G G G G G G G G G G G G G G G

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

### Conceptual Digestion

1.  **CAST**:
    *   **Spice Drops**: Single colored pixels in the input grid.
    *   **Horizontal Bands**: Full rows of the same color as the spice drop.
    *   **Glass Walls**: The leftmost and rightmost columns of the grid.
    *   **Rims**: The top (row 0) and bottom (row 14) rows of the grid.

2.  **AFFORDANCES**:
    *   A spice drop at row $y$ transforms that entire row $y$ into a solid band of its color.
    *   The glass walls (column 0 and column 14) are colored based on the nearest spice drop (nearest row $y$).
    *   The top and bottom rims are colored based on the spice drop governing that region.

3.  **TRANSFORMATION RULE**:
    *   **Step 1 (Bands)**: For every colored pixel at $(y, x)$, fill the entire row $y$ with that color.
    *   **Step 2 (Walls/Rims)**: For every row $i$ from 0 to 14, determine which spice drop row $y$ is closest to $i$.
        *   If there is a tie (equidistant), the one above (smaller $y$) usually takes precedence or the logic follows a Voronoi-like partition on the Y-axis.
        *   Looking at Example 1:
            *   Drops at $y=3 (R), 7 (O), 12 (A)$.
            *   Rows 0-5 are closest to $y=3 \rightarrow$ Walls are $R$.
            *   Rows 6-9 are closest to $y=7 \rightarrow$ Walls are $O$.
            *   Rows 10-14 are closest to $y=12 \rightarrow$ Walls are $A$.
        *   Looking at Example 2:
            *   Drops at $y=1 (A), 3 (B), 7 (R), 9 (G)$.
            *   Rows 0-2: closest to $y=1 \rightarrow A$.
            *   Rows 3-5: closest to $y=3 \rightarrow B$.
            *   Rows 6-8: closest to $y=7 \rightarrow R$.
            *   Rows 9-14: closest to $y=9 \rightarrow G$.
    *   **Step 3 (Final Assembly)**:
        *   If row $i$ is a "band" row, it is solid color.
        *   If row $i$ is not a band row, only column 0 and column 14 are colored with the color of the nearest band.

### Applying to Test Input
*   **Spice Drops**:
    *   $y=2$: Azure (A)
    *   $y=4$: Red (R)
    *   $y=9$: Blue (B)
    *   $y=12$: Green (G)

*   **Partitioning the Y-axis**:
    *   $y=0, 1, 2, 3$: Closest to $y=2 \rightarrow$ Color A.
    *   $y=4, 5, 6$: Closest to $y=4 \rightarrow$ Color R.
    *   $y=7, 8, 9, 10$: Closest to $y=9 \rightarrow$ Color B. (Note: $y=7$ is 2 away from 9 and 3 away from 4).
    *   $y=11, 12, 13, 14$: Closest to $y=12 \rightarrow$ Color G.

*   **Constructing the Grid**:
    *   Row 0: Wall A, Rim A $\rightarrow$ A A A A A A A A A A A A A A A
    *   Row 1: Wall A $\rightarrow$ A . . . . . . . . . . . . . A
    *   Row 2: Band A $\rightarrow$ A A A A A A A A A A A A A A A
    *   Row 3: Wall A $\rightarrow$ A . . . . . . . . . . . . . A
    *   Row 4: Band R $\rightarrow$ R R R R R R R R R R R R R R R
    *   Row 5: Wall R $\rightarrow$ R . . . . . . . . . . . . . R
    *   Row 6: Wall R $\rightarrow$ R . . . . . . . . . . . . . R
    *   Row 7: Wall B $\rightarrow$ B . . . . . . . . . . . . . B
    *   Row 8: Wall B $\rightarrow$ B . . . . . . . . . . . . . B
    *   Row 9: Band B $\rightarrow$ B B B B B B B B B B B B B B B
    *   Row 10: Wall B $\rightarrow$ B . . . . . . . . . . . . . B
    *   Row 11: Wall G $\rightarrow$ G . . . . . . . . . . . . . G
    *   Row 12: Band G $\rightarrow$ G G G G G G G G G G G G G G G
    *   Row 13: Wall G $\rightarrow$ G . . . . . . . . . . . . . G
    *   Row 14: Wall G, Rim G $\rightarrow$ G G G G G G G G G G G G G G G

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

### Conceptual Digestion

1.  **CAST**:
    *   **Stones**: Single colored pixels in the input grid.
    *   **Beams**: Full horizontal rows of a specific color.
    *   **Walls**: The leftmost (column 0) and rightmost (column 14) columns of the grid.
    *   **Void**: The black space (`.`) inside the tower.

2.  **AFFORDANCES**:
    *   A stone at `(row, col)` dictates that the entire `row` becomes a beam of that stone's color.
    *   The walls (column 0 and 14) take the color of the nearest beam above or below them.
    *   The top row (row 0) and bottom row (row 14) act as boundaries; if they fall within the influence of a color zone, they become solid beams.

3.  **TRANSFORMATION RULE**:
    *   **Step 1 (Beams)**: For every row $r$ containing a colored pixel $C$, set all cells in row $r$ to color $C$.
    *   **Step 2 (Walls)**: For every cell in column 0 and column 14, assign it the color of the nearest beam (the beam in the row $r'$ that minimizes $|r - r'|$). If there is a tie (equidistant), the rule in Example 1 suggests the color of the beam *below* takes precedence or the zone is split. Looking at Example 1:
        *   Beam R at row 3. Beam O at row 7.
        *   Rows 0, 1, 2 are closer to R (row 3).
        *   Rows 4, 5, 6 are closer to R (row 3) or O (row 7). Row 4 is 1 away from R, 3 from O. Row 5 is 2 away from R, 2 from O. Row 6 is 3 away from R, 1 from O.
        *   Wait, looking at Example 1 output:
            *   Row 0: R (Beam)
            *   Row 1: R . . . R
            *   Row 2: R . . . R
            *   Row 3: R R R R R (Beam)
            *   Row 4: R . . . R
            *   Row 5: R . . . R
            *   Row 6: O . . . O
            *   Row 7: O O O O O (Beam)
            *   Row 8: O . . . O
            *   Row 9: O . . . O
            *   Row 10: A . . . A
            *   Row 11: A . . . A
            *   Row 12: A A A A A (Beam)
            *   Row 13: A . . . A
            *   Row 14: A A A A A (Beam)
    *   **Refined Rule**:
        1.  Identify rows with colors: $R_1, R_2, \dots, R_n$.
        2.  Fill those rows completely.
        3.  Fill the top row (0) and bottom row (14) completely with the color of the nearest beam.
        4.  For all other rows $r$, set cells $(r, 0)$ and $(r, 14)$ to the color of the nearest beam. If equidistant, the lower beam wins (as seen in row 6 of Example 1: distance to R(3) is 3, distance to O(7) is 1; actually, row 5 is distance 2 from R and 2 from O, and it stayed R. Let's re-check: Row 5 is index 5. $5-3=2$, $7-5=2$. It stayed R. So the *upper* beam wins ties).
        5.  Wait, let's re-calculate Example 1:
            *   R at 3, O at 7, A at 12.
            *   Row 0: Nearest is R(3). Result: Beam R.
            *   Row 1: Nearest is R(3). Result: R . . R.
            *   Row 2: Nearest is R(3). Result: R . . R.
            *   Row 3: Beam R.
            *   Row 4: Nearest is R(3). Result: R . . R.
            *   Row 5: Nearest is R(3) [dist 2] or O(7) [dist 2]. Result: R . . R. (Upper wins).
            *   Row 6: Nearest is O(7). Result: O . . O.
            *   Row 7: Beam O.
            *   Row 8: Nearest is O(7). Result: O . . O.
            *   Row 9: Nearest is O(7) [dist 2] or A(12) [dist 3]. Result: O . . O.
            *   Row 10: Nearest is A(12) [dist 2] or O(7) [dist 3]. Result: A . . A.
            *   Row 11: Nearest is A(12). Result: A . . A.
            *   Row 12: Beam A.
            *   Row 13: Nearest is A(12). Result: A . . A.
            *   Row 14: Nearest is A(12). Result: Beam A.

### Applying to Test Input
*   Stones: A at row 2, R at row 4, B at row 9, G at row 12.
*   Beams: Row 2 (A), Row 4 (R), Row 9 (B), Row 12 (G).
*   Top/Bottom: Row 0 (nearest A at 2) -> Beam A. Row 14 (nearest G at 12) -> Beam G.

**Wall Colors (Col 0 and 14):**
*   Row 0: A (Beam)
*   Row 1: Nearest A(2) [dist 1] or R(4) [dist 3] -> A
*   Row 2: A (Beam)
*   Row 3: Nearest A(2) [dist 1] or R(4) [dist 1] -> A (Upper wins)
*   Row 4: R (Beam)
*   Row 5: Nearest R(4) [dist 1] or B(9) [dist 4] -> R
*   Row 6: Nearest R(4) [dist 2] or B(9) [dist 3] -> R
*   Row 7: Nearest R(4) [dist 3] or B(9) [dist 2] -> B
*   Row 8: Nearest B(9) [dist 1] or R(4) [dist 4] -> B
*   Row 9: B (Beam)
*   Row 10: Nearest B(9) [dist 1] or G(12) [dist 2] -> B
*   Row 11: Nearest B(9) [dist 2] or G(12) [dist 1] -> G
*   Row 12: G (Beam)
*   Row 13: Nearest G(12) [dist 1] or B(9) [dist 4] -> G
*   Row 14: G (Beam)

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

### Conceptual Digestion

1.  **The CAST**:
    *   **Flares**: The colored pixels in the input grid (A, R, B, G).
    *   **Defensive Lines**: Horizontal lines across the full width of the grid.
    *   **Flanks**: The leftmost and rightmost columns of the grid.
    *   **Field**: The 15x15 grid.

2.  **The AFFORDANCES**:
    *   A flare's **row index** determines where a solid horizontal line of its color is drawn.
    *   A flare's **color** determines the color of its corresponding line and the surrounding flank segments.

3.  **The TRANSFORMATION RULE**:
    *   **Horizontal Lines**: For every colored pixel (flare) at row $y$, the entire row $y$ in the output grid becomes that color.
    *   **Flanks (Columns 0 and 14)**: The perimeter edges (left and right columns) are filled based on the nearest flare.
        *   Rows above the first flare take the color of the first flare.
        *   Rows between flare $i$ and flare $i+1$ take the color of the nearest flare (or the one that "governs" that zone). Looking at Example 1:
            *   Flare 1 (R) at row 3: Rows 0-3 are R.
            *   Flare 2 (O) at row 7: Rows 4-7 are O? No, let's re-examine Example 1.
            *   Example 1 Analysis:
                *   R at row 3 $\rightarrow$ Row 3 is all R. Rows 0, 1, 2, 3, 4, 5 are R on the flanks.
                *   O at row 7 $\rightarrow$ Row 7 is all O. Rows 6, 7, 8, 9, 10 are O on the flanks.
                *   A at row 12 $\rightarrow$ Row 12 is all A. Rows 11, 12, 13, 14 are A on the flanks.
                *   Wait, the boundary is the midpoint between flares.
                *   R(3) and O(7): Midpoint is $(3+7)/2 = 5$. Row 5 is R, Row 6 is O.
                *   O(7) and A(12): Midpoint is $(7+12)/2 = 9.5$. Row 9 is O, Row 10 is A.
                *   Wait, Example 1 output:
                    *   Rows 0-5: Flanks are R.
                    *   Rows 6-10: Flanks are O.
                    *   Rows 11-14: Flanks are A.
                    *   Row 0 is also all R (top perimeter).
                    *   Row 14 is also all A (bottom perimeter).
    *   **Refined Rule**:
        1.  Identify flares: $(r_1, c_1, color_1), (r_2, c_2, color_2), \dots$
        2.  Fill rows $r_i$ with $color_i$.
        3.  Fill top row (0) and bottom row (14) with the color of the first and last flare respectively.
        4.  Fill columns 0 and 14:
            *   From row 0 to $\lfloor(r_1+r_2)/2\rfloor$, use $color_1$.
            *   From $\lfloor(r_1+r_2)/2\rfloor + 1$ to $\lfloor(r_2+r_3)/2\rfloor$, use $color_2$.
            *   And so on.
            *   Example 1 check: $r_1=3, r_2=7, r_3=12$.
                *   Mid 1: $(3+7)/2 = 5$. Rows 0-5: R.
                *   Mid 2: $(7+12)/2 = 9.5 \rightarrow 9$. Rows 6-9: O.
                *   Rows 10-14: A.
                *   Wait, Example 1 output row 10 flank is A. So Mid 2 is 9. Rows 6-9 are O, 10-14 are A.
                *   Let's re-verify: Row 6 flank is O, Row 7 is all O, Row 8 flank is O, Row 9 flank is O, Row 10 flank is A. Correct.

### Applying to Test Input
Flares:
1. A at row 2
2. R at row 4
3. B at row 9
4. G at row 12

**Horizontal Lines**:
Row 2: All A
Row 4: All R
Row 9: All B
Row 12: All G
Row 0: All A (Top perimeter)
Row 14: All G (Bottom perimeter)

**Flanks (Col 0 and 14)**:
1. Mid(2, 4) = 3. Rows 0-3: A.
2. Mid(4, 9) = 6.5 $\rightarrow$ 6. Rows 4-6: R.
3. Mid(9, 12) = 10.5 $\rightarrow$ 10. Rows 7-10: B.
4. Rows 11-14: G.

**Final Grid Construction**:
Row 0: A A A A A A A A A A A A A A A
Row 1: A . . . . . . . . . . . . . A
Row 2: A A A A A A A A A A A A A A A
Row 3: A . . . . . . . . . . . . . A
Row 4: R R R R R R R R R R R R R R R
Row 5: R . . . . . . . . . . . . . R
Row 6: R . . . . . . . . . . . . . R
Row 7: B . . . . . . . . . . . . . B
Row 8: B . . . . . . . . . . . . . B
Row 9: B B B B B B B B B B B B B B B
Row 10: B . . . . . . . . . . . . . B
Row 11: G . . . . . . . . . . . . . G
Row 12: G G G G G G G G G G G G G G G
Row 13: G . . . . . . . . . . . . . G
Row 14: G G G G G G G G G G G G G G G

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

### Conceptual Digestion

1.  **CAST**:
    *   **Colored Dots (Bulbs)**: These are the seeds of the transformation. Each dot defines a "zone" of influence.
    *   **Black Space (.)**: The soil.
    *   **Grid Borders**: The top row, bottom row, left column, and right column.

2.  **AFFORDANCES**:
    *   A bulb at row `r` (0-indexed) creates a horizontal line (mat) across the entire row `r`.
    *   The space between two bulbs (or between a bulb and the grid boundary) is partitioned.
    *   The "border" (left and right columns) is colored based on the nearest bulb's row.

3.  **TRANSFORMATION RULE**:
    *   **Horizontal Mats**: For every input cell `(r, c)` that contains a color `C`, the entire output row `r` becomes color `C`.
    *   **Vertical Borders**: For every cell `(r, 0)` and `(r, 14)`, the color is determined by the bulb closest to row `r` in terms of vertical distance.
    *   **Top and Bottom Edges**: Row 0 and Row 14 are always filled with the color of the bulb closest to them.
    *   **Interior**: Cells that are not part of a horizontal mat or a vertical border remain black (`.`).

**Applying to Test Input**:
*   Bulbs: `A` at (2, 3), `R` at (4, 6), `B` at (9, 11), `G` at (12, 9).
*   **Horizontal Mats**: Rows 2, 4, 9, and 12 are fully colored `A`, `R`, `B`, and `G` respectively.
*   **Vertical Borders (Col 0 and 14)**:
    *   Rows 0, 1, 2: Closest bulb is `A` (row 2). Color: `A`.
    *   Rows 3, 4: Closest bulb is `R` (row 4). Color: `R`.
    *   Rows 5, 6, 7: Closest bulb is `R` (row 4). Color: `R`. (Distance to R is 1, 2, 3; distance to B is 4, 3, 2. Wait, let's re-calculate).
        *   Row 5: dist(R)=1, dist(B)=4 -> `R`
        *   Row 6: dist(R)=2, dist(B)=3 -> `R`
        *   Row 7: dist(R)=3, dist(B)=2 -> `B`
        *   Row 8: dist(R)=4, dist(B)=1 -> `B`
        *   Row 9: `B`
        *   Row 10: dist(B)=1, dist(G)=2 -> `B`
        *   Row 11: dist(B)=2, dist(G)=1 -> `G`
        *   Row 12: `G`
        *   Row 13: dist(G)=1 -> `G`
        *   Row 14: dist(G)=2 -> `G`
*   **Top/Bottom Edges**:
    *   Row 0: Closest is `A` (row 2). Color: `A`.
    *   Row 14: Closest is `G` (row 12). Color: `G`.

**Refining Vertical Borders**:
*   R0: A (dist 2 to A)
*   R1: A (dist 1 to A)
*   R2: A (mat)
*   R3: A or R? dist(A)=1, dist(R)=1. Looking at Example 1: Row 3 is R (mat), Row 2 is R (border). In Example 1, bulb R is at row 3. Row 2 is R. This suggests if distance is equal, it might favor the bulb below or the one that creates the mat. Let's check Example 2: A at 1, B at 3. Row 2 is A. dist(A)=1, dist(B)=1. It chose A. So it favors the bulb *above*.
*   R3: dist(A)=1, dist(R)=1 -> `A`
*   R4: R (mat)
*   R5: R (dist 1 to R)
*   R6: R (dist 2 to R, 3 to B)
*   R7: B (dist 3 to R, 2 to B)
*   R8: B (dist 4 to R, 1 to B)
*   R9: B (mat)
*   R10: B (dist 1 to B)
*   R11: G (dist 2 to B, 1 to G)
*   R12: G (mat)
*   R13: G (dist 1 to G)
*   R14: G (mat/border)

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

**Conceptual Digestion**

1.  **CAST**:
    *   **Night Sky**: The black background (`.`).
    *   **Scattered Loners (Diaspora)**: Colors that appear in multiple isolated locations across the grid.
    *   **Loyal Flock (Constellation)**: A single color that appears only in one contiguous cluster (connected orthogonally or diagonally).
    *   **Frame**: The smallest rectangular bounding box that contains the entire "Loyal Flock."

2.  **AFFORDANCES**:
    *   The grid can be scanned to count the number of separate clusters for each color.
    *   The "Loyal Flock" is the unique color that forms exactly one connected component.
    *   The output is a crop of the original grid, specifically the bounding box of that single cluster.

3.  **TRANSFORMATION RULE**:
    *   Identify all colors present in the grid.
    *   For each color, determine if all pixels of that color form a single connected component.
    *   The color that forms exactly one connected component is the target.
    *   Find the minimum and maximum row and column indices that contain this color.
    *   Extract the subgrid defined by these boundaries.

**Applying the Rule to Test Input**

*   **Blue (B)**: Scattered throughout (e.g., (0,2), (1,27), (2,0), (3,12), etc.). Multiple clusters.
*   **Green (G)**: Scattered throughout (e.g., (0,4), (0,19), (1,3), (3,19), etc.). Multiple clusters.
*   **Azure (A)**: Scattered throughout (e.g., (0,6), (0,20), (1,17), (1,19), etc.). Multiple clusters.
*   **Red (R)**: Located at:
    (12, 20), (12, 21), (12, 22)
    (13, 19), (13, 20), (13, 22)
    (14, 19), (14, 20), (14, 22)
    (15, 20), (15, 21), (15, 22)
    Checking connectivity: All Red pixels are connected to each other (forming a 4x3 block with some holes). There are no other Red pixels elsewhere in the grid.
*   **Red is the Loyal Flock.**

**Bounding Box for Red (R):**
*   Rows: Min = 12, Max = 15
*   Cols: Min = 19, Max = 22

**Extracting the subgrid:**
Row 12: (12,19)=`.`, (12,20)=`R`, (12,21)=`R`, (12,22)=`R`
Row 13: (13,19)=`R`, (13,20)=`R`, (13,21)=`.`, (13,22)=`R`
Row 14: (14,19)=`R`, (14,20)=`R`, (14,21)=`.`, (14,22)=`R`
Row 15: (15,19)=`.`, (15,20)=`R`, (15,21)=`R`, (15,22)=`R`

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

**Conceptual Digestion**

1.  **Identify the CAST:**
    *   **Background:** The black dots (`.`) represent the empty sky.
    *   **Stars:** Colored dots (B, R, G, Y, X, M, O, A, W) represent stars.
    *   **Asterism:** A small, dense cluster of stars of a single color.
    *   **Stray Stars:** Stars of various colors scattered sparsely across the grid.

2.  **Identify the AFFORDANCES:**
    *   The grid can be scanned to find the density of specific colors.
    *   A "compact asterism" is defined as a small bounding box containing a high concentration of a single color, whereas other colors are spread out (large distances between dots of the same color).

3.  **Identify the TRANSFORMATION RULE:**
    *   Scan the input grid for the color that forms the tightest, most compact cluster (the asterism).
    *   Determine the minimum bounding box (the smallest rectangle) that contains all the stars of that specific color in that cluster.
    *   Crop the grid to this bounding box.
    *   The output is the content of this bounding box.

**Applying the rule to the Test Input:**

*   **Blue (B):** Scattered throughout (e.g., (0,2), (1,28), (2,0), (3,12), etc.).
*   **Green (G):** Scattered throughout (e.g., (0,4), (0,19), (3,19), (4,16), etc.).
*   **Azure (A):** Scattered throughout (e.g., (0,6), (0,21), (1,17), (2,19), etc.).
*   **Red (R):** There is a very tight cluster located around rows 12-15 and columns 20-22.
    *   (12, 20): R
    *   (12, 21): R
    *   (12, 22): R
    *   (13, 20): R
    *   (13, 21): .
    *   (13, 22): R
    *   (14, 20): R
    *   (14, 21): R
    *   (14, 22): .
    *   (14, 23): R (Wait, let's re-examine the coordinates)

Let's re-map the Red (R) cluster carefully:
Row 12: `. . . . . . . . . . . G . G . . . . . . R R R . . . . . . .` -> Col 20, 21, 22 are R.
Row 13: `. . B . . . . . . . A . . . . . . . . R R . R . . . B . . .` -> Col 19, 20 are R, Col 22 is R.
Row 14: `. . . . . . . . B . . . . . . . . . . R R . R . . . . . . .` -> Col 19, 20 are R, Col 22 is R.
Row 15: `. . . G . . . . . . . . B . . . . G . . R R R . A G . . . .` -> Col 20, 21, 22 are R.

Wait, let's look at the columns again for the Red cluster:
Row 12: Col 20=R, 21=R, 22=R
Row 13: Col 19=R, 20=R, 22=R
Row 14: Col 19=R, 20=R, 22=R
Row 15: Col 20=R, 21=R, 22=R

Bounding box for R:
Rows: 12 to 15
Cols: 19 to 22

Extracting the grid:
Row 12: (Col 19=., 20=R, 21=R, 22=R) -> `. R R R`
Row 13: (Col 19=R, 20=R, 21=., 22=R) -> `R R . R`
Row 14: (Col 19=R, 20=R, 21=., 22=R) -> `R R . R`
Row 15: (Col 19=., 20=R, 21=R, 22=R) -> `. R R R`

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

To solve this puzzle, we need to follow the metaphorical description and the patterns observed in the training examples.

### Conceptual Digestion

1.  **CAST**:
    *   **Background**: The black cells (`.`) represent the terrain.
    *   **Scattered Minerals**: Various colored cells (B, G, A, etc.) that appear sporadically across the grid.
    *   **Coherent Outcrop**: A specific color that forms a dense, localized cluster (a contiguous or nearly contiguous block) while other colors of the same type are either absent or widely dispersed.

2.  **AFFORDANCES**:
    *   The grid allows us to identify clusters of colors.
    *   The "outcrop" is defined as the only color that forms a significant, localized rectangle/block rather than being scattered.

3.  **TRANSFORMATION RULE**:
    *   Scan the input grid for all colors present.
    *   Identify which color forms a dense, localized "outcrop" (a small, tight group of cells).
    *   Determine the bounding box (the smallest rectangle that contains all cells of that specific color cluster).
    *   Extract and output only the contents of that bounding box.

### Applying the Rule to Test Input

1.  **Analyze Colors**:
    *   **Blue (B)**: Scattered throughout the grid.
    *   **Green (G)**: Scattered throughout the grid.
    *   **Azure (A)**: Scattered throughout the grid.
    *   **Red (R)**: Located in a tight cluster around rows 12-15 and columns 19-22.

2.  **Locate the Red (R) Outcrop**:
    *   Row 12: `. . . . . . . . . . . G . G . . . . . . . R R R . . . . . . .` (Cols 19, 20, 21)
    *   Row 13: `. . B . . . . . . . A . . . . . . . . R R . R . . . B . . .` (Cols 18, 19, 21)
    *   Row 14: `. . . . . . . . B . . . . . . . . . . R R . R . . . . . . .` (Cols 18, 19, 21)
    *   Row 15: `. . . G . . . . . . . . B . . . . G . . R R R . A G . . . .` (Cols 19, 20, 21)

3.  **Determine Bounding Box for Red (R)**:
    *   Min Row: 12, Max Row: 15
    *   Min Col: 18, Max Col: 21
    *   Let's re-examine the coordinates carefully:
        *   Row 12: Col 19=R, 20=R, 21=R
        *   Row 13: Col 18=R, 19=R, 21=R
        *   Row 14: Col 18=R, 19=R, 21=R
        *   Row 15: Col 19=R, 20=R, 21=R

4.  **Extract the Grid**:
    *   Row 12: `R R R` (Cols 18 is `.`, 19 is `R`, 20 is `R`, 21 is `R`) -> `. R R R`
    *   Row 13: `R R . R`
    *   Row 14: `R R . R`
    *   Row 15: `. R R R`

Wait, let's re-count the columns for Row 12:
`. . . B . . B . . . G . G . . . . . . . R R R . . . . . . .`
Indices: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
Row 12: Col 19=R, 20=R, 21=R. Col 18 is `.`.
Row 13: Col 18=R, 19=R, 20=`.`, 21=R.
Row 14: Col 18=R, 19=R, 20=`.`, 21=R.
Row 15: Col 18=`.`, 19=R, 20=R, 21=R.

The bounding box is from Row 12 to 15 and Col 18 to 21.

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

**Conceptual Digestion**

1.  **CAST**:
    *   **Isolated Dots**: Single colored pixels scattered across the grid. These represent the "dissonant notes" or "soloists."
    *   **Ensemble**: A contiguous cluster of pixels of the same color. This represents the "harmonious ensemble."
    *   **Background**: The black pixels (`.`).

2.  **AFFORDANCES**:
    *   The grid can be scanned to find connected components of the same color.
    *   The "ensemble" is defined as the only group of pixels of the same color that are adjacent (orthogonally or diagonally) to each other, forming a distinct shape, while other colors are either singletons or very small, disconnected fragments.

3.  **TRANSFORMATION RULE**:
    *   Scan the input grid for the largest contiguous cluster of a single color.
    *   In Example 1, there are many red (`R`) dots, but only one cluster of yellow (`Y`) pixels.
    *   The output is a bounding box (crop) containing only that specific cluster.
    *   The bounding box is the smallest rectangle that encompasses all pixels of that specific ensemble color.

**Applying the rule to the Test Input**:
1.  Scan for clusters:
    *   Blue (`B`), Green (`G`), and Azure (`A`) are scattered throughout the grid as singletons or pairs.
    *   Red (`R`) appears in a concentrated area:
        *   Row 12: Col 20, 21, 22 (`R R R`)
        *   Row 13: Col 19, 20, 22 (`R R . R`)
        *   Row 14: Col 19, 20, 22 (`R R . R`)
        *   Row 15: Col 19, 20, 21 (`R R R`)
2.  This forms a 4x4 block (with some holes) of Red pixels.
3.  Identify the bounding box for the Red ensemble:
    *   Top row: 12
    *   Bottom row: 15
    *   Left column: 19
    *   Right column: 22
4.  Extract the content of this bounding box:
    *   Row 12: `R R R` (Cols 20-22) -> Wait, let's re-count.
    *   Looking at the test input:
        Row 12: `. . . B . . B . . . G . G . . . . . . . R R R . . . . . . .` (R is at index 19, 20, 21 - 0 indexed)
        Row 13: `. . B . . . . . . . A . . . . . . . . R R . R . . . B . . .` (R is at index 18, 19, 21)
        Row 14: `. . . . . . . . B . . . . . . . . . . R R . R . . . . . . .` (R is at index 18, 19, 21)
        Row 15: `. . . G . . . . . . . . B . . . . G . . R R R . A G . . . .` (R is at index 19, 20, 21)

    Let's re-examine the indices carefully:
    Row 12: `... . . . . R R R . . .` (Indices 19, 20, 21)
    Row 13: `... . . R R . R . . .` (Indices 18, 19, 21)
    Row 14: `... . . R R . R . . .` (Indices 18, 19, 21)
    Row 15: `... . R R R . . . .` (Indices 19, 20, 21)

    Wait, let's look at the raw text again:
    Row 12: `. . . B . . B . . . G . G . . . . . . . R R R . . . . . . .`
    Count: 0(.), 1(.), 2(.), 3(B), 4(.), 5(.), 6(B), 7(.), 8(.), 9(.), 10(G), 11(.), 12(G), 13(.), 14(.), 15(.), 16(.), 17(.), 18(.), 19(R), 20(R), 21(R)
    Row 13: `. . B . . . . . . . A . . . . . . . . R R . R . . . B . . .`
    Count: 0(.), 1(.), 2(B), ... 18(R), 19(R), 20(.), 21(R)
    Row 14: `. . . . . . . . B . . . . . . . . . . R R . R . . . . . . .`
    Count: 0(.), ... 18(R), 19(R), 20(.), 21(R)
    Row 15: `. . . G . . . . . . . . B . . . . G . . R R R . A G . . . .`
    Count: 0(.), ... 18(.), 19(R), 20(R), 21(R)

    Bounding box:
    Rows: 12 to 15
    Cols: 18 to 21

    Grid:
    Row 12: `. R R R`
    Row 13: `R R . R`
    Row 14: `R R . R`
    Row 15: `. R R R`

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

### Conceptual Digestion

1.  **CAST**:
    *   **Rooftop/Lightning Rods**: Three adjacent colored cells at the top of the grid (row 0).
    *   **Storm Clouds**: Single colored cells scattered in the grid.
    *   **Lightning Bolts**: Paths of the same color connecting a rod to its matching cloud.
    *   **Ground**: The bottom edge of the grid.

2.  **AFFORDANCES**:
    *   Each rod (color $C$) searches for a cloud of the same color $C$.
    *   If a cloud of color $C$ exists at $(row, col)$, the rod at $(0, rod\_col)$ creates a path:
        *   Vertical segment from $(0, rod\_col)$ down to $(row, rod\_col)$.
        *   Horizontal segment from $(row, rod\_col)$ to $(row, col)$.
    *   If no cloud of color $C$ exists, the rod creates a vertical pillar from $(0, rod\_col)$ all the way to the bottom of the grid.

3.  **TRANSFORMATION RULE**:
    *   Identify the three rods at the top (adjacent colors).
    *   For each rod color $C$ at column $j$:
        *   Find the cell $(i, k)$ where the color is $C$ (excluding the rod itself).
        *   If found: Fill all cells $(r, j)$ for $0 \le r \le i$ with color $C$, and all cells $(i, c)$ between $j$ and $k$ (inclusive) with color $C$.
        *   If not found: Fill all cells $(r, j)$ for $0 \le r < \text{height}$ with color $C$.

### Applying the Rule to Test Input

**Input Analysis**:
*   Grid size: 11 rows x 11 columns.
*   Rods at row 0: Column 3 (Y), Column 4 (A), Column 5 (X).
*   Clouds:
    *   Yellow (Y) at (7, 0).
    *   Azure (A) - None found.
    *   Grey (X) at (9, 10).

**Path Construction**:
1.  **Yellow (Y)**: Rod at (0, 3), Cloud at (7, 0).
    *   Vertical: (0, 3) to (7, 3).
    *   Horizontal: (7, 3) to (7, 0).
2.  **Azure (A)**: Rod at (0, 4), No cloud.
    *   Vertical: (0, 4) to (10, 4).
3.  **Grey (X)**: Rod at (0, 5), Cloud at (9, 10).
    *   Vertical: (0, 5) to (9, 5).
    *   Horizontal: (9, 5) to (9, 10).

**Final Grid Construction**:
*   Row 0: `. . . Y A X . . . . .`
*   Row 1: `. . . Y A X . . . . .`
*   Row 2: `. . . Y A X . . . . .`
*   Row 3: `. . . Y A X . . . . .`
*   Row 4: `. . . Y A X . . . . .`
*   Row 5: `. . . Y A X . . . . .`
*   Row 6: `. . . Y A X . . . . .`
*   Row 7: `Y Y Y Y A X . . . . .`
*   Row 8: `. . . . A X . . . . .`
*   Row 9: `. . . . A X X X X X X`
*   Row 10: `. . . . A . . . . . .`

ANSWER:
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
Y Y Y Y A X . . . . .
. . . . A X . . . . .
. . . . A X X X X X X
. . . . A . . . . . .

### Conceptual Digestion

1.  **CAST**:
    *   **Nerve Cells**: The three adjacent colored cells at the top of the grid (Row 0). In the test input, these are `Y` (Yellow) at (0, 3), `A` (Azure) at (0, 4), and `X` (Grey) at (0, 5).
    *   **Organ Cells**: Isolated cells of the same colors scattered in the grid. In the test input, these are `Y` (Yellow) at (7, 0) and `X` (Grey) at (9, 10).
    *   **Dendrites**: The paths created to connect Nerve Cells to Organ Cells.
    *   **Tissue Matrix**: The black cells (`.`) that serve as the background.

2.  **AFFORDANCES**:
    *   Nerve cells act as starting points for vertical growth.
    *   Organ cells act as targets for the dendrites.
    *   The grid allows vertical movement (downward) and horizontal movement (sideways) to bridge the gap between a nerve and its organ.

3.  **TRANSFORMATION RULE**:
    *   For each nerve cell at `(0, col)`:
        1.  Identify the organ cell of the same color at `(row_target, col_target)`.
        2.  If an organ exists:
            *   Draw a vertical line of that color from `(0, col)` down to `(row_target, col)`.
            *   Draw a horizontal line of that color from `(row_target, col)` to `(row_target, col_target)`.
        3.  If no organ exists:
            *   Draw a vertical line of that color from `(0, col)` down to the bottom edge of the grid.

### Applying the Rule to Test Input

*   **Nerve 1: Yellow (Y) at (0, 3)**
    *   Organ Y is at (7, 0).
    *   Vertical path: (0, 3) $\rightarrow$ (7, 3).
    *   Horizontal path: (7, 3) $\rightarrow$ (7, 0).
*   **Nerve 2: Azure (A) at (0, 4)**
    *   No Organ A exists in the grid.
    *   Vertical path: (0, 4) $\rightarrow$ (10, 4).
*   **Nerve 3: Grey (X) at (0, 5)**
    *   Organ X is at (9, 10).
    *   Vertical path: (0, 5) $\rightarrow$ (9, 5).
    *   Horizontal path: (9, 5) $\rightarrow$ (9, 10).

### Final Grid Construction

Row 0: `. . . Y A X . . . . .`
Row 1: `. . . Y A X . . . . .`
Row 2: `. . . Y A X . . . . .`
Row 3: `. . . Y A X . . . . .`
Row 4: `. . . Y A X . . . . .`
Row 5: `. . . Y A X . . . . .`
Row 6: `. . . Y A X . . . . .`
Row 7: `Y Y Y Y A X . . . . .` (Y path from col 3 to 0)
Row 8: `. . . . A X . . . . .`
Row 9: `. . . . A X X X X X X` (X path from col 5 to 10)
Row 10: `. . . . A . . . . . .` (A path ends at bottom)

ANSWER:
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
Y Y Y Y A X . . . . .
. . . . A X . . . . .
. . . . A X X X X X X
. . . . A . . . . . .

### Conceptual Digestion

1.  **CAST**:
    *   **Lighthouses**: The three colored cells at the top of the grid (in the test input: `Y`, `A`, `X` at columns 3, 4, and 5).
    *   **Shipwrecks**: Single colored cells of the same colors as the lighthouses, scattered elsewhere in the grid (in the test input: `Y` at (7, 0) and `X` at (9, 10)).
    *   **Light Beams**: The paths created by the transformation.

2.  **AFFORDANCES**:
    *   Lighthouses act as sources.
    *   Wrecks act as targets.
    *   The "beam" travels vertically down from the lighthouse and then horizontally to the wreck.

3.  **TRANSFORMATION RULE**:
    *   For each lighthouse at `(row 0, col C)` with color `K`:
        1.  Search for a wreck of the same color `K` at `(row R, col W)`.
        2.  If a wreck exists:
            *   Fill the vertical path from `(0, C)` down to `(R, C)` with color `K`.
            *   Fill the horizontal path from `(R, C)` to `(R, W)` with color `K`.
        3.  If no wreck exists:
            *   Fill the vertical path from `(0, C)` down to the bottom of the grid with color `K`.

### Applying the Rule to Test Input

*   **Lighthouse 1 (Yellow 'Y')**: Position (0, 3).
    *   Wreck 'Y' is at (7, 0).
    *   Vertical path: (0, 3) to (7, 3) filled with `Y`.
    *   Horizontal path: (7, 3) to (7, 0) filled with `Y`.
*   **Lighthouse 2 (Azure 'A')**: Position (0, 4).
    *   No wreck 'A' exists in the grid.
    *   Vertical path: (0, 4) to (11, 4) filled with `A`. (Wait, the grid is 11 rows high, indices 0-10). So (0, 4) to (10, 4).
*   **Lighthouse 3 (Grey 'X')**: Position (0, 5).
    *   Wreck 'X' is at (9, 10).
    *   Vertical path: (0, 5) to (9, 5) filled with `X`.
    *   Horizontal path: (9, 5) to (9, 10) filled with `X`.

### Final Grid Construction

Row 0: `. . . Y A X . . . . .`
Row 1: `. . . Y A X . . . . .`
Row 2: `. . . Y A X . . . . .`
Row 3: `. . . Y A X . . . . .`
Row 4: `. . . Y A X . . . . .`
Row 5: `. . . Y A X . . . . .`
Row 6: `. . . Y A X . . . . .`
Row 7: `Y Y Y Y A . . . . . .`
Row 8: `. . . . A X . . . . .` (Wait, X vertical path goes to row 9) -> `. . . . A X . . . . .`
Row 9: `. . . . A X X X X X X`
Row 10: `. . . . A . . . . . .`

Wait, let's re-verify Row 8.
Lighthouse X is at col 5. Wreck X is at row 9. So vertical path for X is (0,5) to (9,5).
Lighthouse A is at col 4. No wreck. Vertical path is (0,4) to (10,4).
Lighthouse Y is at col 3. Wreck Y is at row 7. Vertical path is (0,3) to (7,3). Horizontal path is (7,3) to (7,0).

Row 0: . . . Y A X . . . . .
Row 1: . . . Y A X . . . . .
Row 2: . . . Y A X . . . . .
Row 3: . . . Y A X . . . . .
Row 4: . . . Y A X . . . . .
Row 5: . . . Y A X . . . . .
Row 6: . . . Y A X . . . . .
Row 7: Y Y Y Y A X . . . . . (X is still going down)
Row 8: . . . . A X . . . . .
Row 9: . . . . A X X X X X X
Row 10: . . . . A . . . . . .

ANSWER:
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
Y Y Y Y A X . . . . .
. . . . A X . . . . .
. . . . A X X X X X X
. . . . A . . . . . .

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Deities (Thrones):** The colored cells located at the top of the grid (Row 0). In the test input, these are `Y` (Yellow) at (0, 3), `A` (Azure) at (0, 4), and `X` (Grey) at (0, 5).
    *   **Champions:** Colored cells of the same color as the deities, located elsewhere in the grid. In the test input, these are `Y` (Yellow) at (7, 0) and `X` (Grey) at (9, 10).
    *   **Divine Chains:** The paths created to connect a deity to their champion.

2.  **Identify the AFFORDANCES:**
    *   **Vertical Descent:** A deity's chain always drops straight down from its column.
    *   **Horizontal Binding:** Once the chain reaches the row of its corresponding champion, it extends horizontally to reach that champion.
    *   **Underworld Fall:** If a deity has no champion, the chain simply drops straight down to the bottom of the grid.

3.  **Identify the TRANSFORMATION RULE:**
    *   For each deity at `(0, col_d)` with color `C`:
        1.  Find the champion of color `C` at `(row_c, col_c)`.
        2.  If a champion exists:
            *   Fill all cells from `(0, col_d)` down to `(row_c, col_d)` with color `C`.
            *   Fill all cells from `(row_c, col_d)` horizontally to `(row_c, col_c)` with color `C`.
        3.  If no champion exists:
            *   Fill all cells from `(0, col_d)` down to the bottom of the grid with color `C`.

### Applying the Rule to Test Input

*   **Deity 1: Yellow (Y) at (0, 3)**
    *   Champion: Yellow (Y) at (7, 0).
    *   Vertical chain: (0, 3) $\rightarrow$ (7, 3).
    *   Horizontal chain: (7, 3) $\rightarrow$ (7, 0).
*   **Deity 2: Azure (A) at (0, 4)**
    *   Champion: None.
    *   Vertical chain: (0, 4) $\rightarrow$ (10, 4).
*   **Deity 3: Grey (X) at (0, 5)**
    *   Champion: Grey (X) at (9, 10).
    *   Vertical chain: (0, 5) $\rightarrow$ (9, 5).
    *   Horizontal chain: (9, 5) $\rightarrow$ (9, 10).

### Final Grid Construction

Row 0: `. . . Y A X . . . . .`
Row 1: `. . . Y A X . . . . .`
Row 2: `. . . Y A X . . . . .`
Row 3: `. . . Y A X . . . . .`
Row 4: `. . . Y A X . . . . .`
Row 5: `. . . Y A X . . . . .`
Row 6: `. . . Y A X . . . . .`
Row 7: `Y Y Y Y A X . . . . .`
Row 8: `. . . . A X . . . . .`
Row 9: `. . . . A X X X X X X`
Row 10: `. . . . A . . . . . .`

ANSWER:
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
Y Y Y Y A X . . . . .
. . . . A X . . . . .
. . . . A X X X X X X
. . . . A . . . . . .

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Lead Dancers:** The three colored cells located at the top of the grid (row 0). In the test input, these are `Y` (yellow) at (0, 3), `A` (azure) at (0, 4), and `X` (grey) at (0, 5).
    *   **Solo Partners:** Cells of the same color located elsewhere in the grid. In the test input, these are `Y` (yellow) at (7, 0) and `X` (grey) at (9, 10).
    *   **Ballroom Floor:** The black cells (`.`).

2.  **Identify the AFFORDANCES:**
    *   Lead dancers can move vertically (down) and horizontally (left/right).
    *   The path they take is filled with their respective color.
    *   The movement is triggered by the row position of their partner.

3.  **Identify the TRANSFORMATION RULE:**
    *   For each lead dancer at `(0, col)`:
        1.  **Vertical Descent:** Draw a line of the dancer's color from `(0, col)` down to the row of their partner `(partner_row, partner_col)`.
        2.  **Horizontal Chassé:** At `partner_row`, draw a line of the dancer's color from `(partner_row, col)` to `(partner_row, partner_col)`.
        3.  **No Partner Case:** If a lead dancer has no partner in the grid, they glide straight down to the bottom wall (last row).

### Applying the Rule to Test Input

*   **Lead 1 (Yellow - Y):**
    *   Start: (0, 3). Partner: (7, 0).
    *   Vertical: (0, 3) down to (7, 3).
    *   Horizontal: (7, 3) left to (7, 0).
*   **Lead 2 (Azure - A):**
    *   Start: (0, 4). Partner: None.
    *   Vertical: (0, 4) down to (10, 4).
*   **Lead 3 (Grey - X):**
    *   Start: (0, 5). Partner: (9, 10).
    *   Vertical: (0, 5) down to (9, 5).
    *   Horizontal: (9, 5) right to (9, 10).

### Final Grid Construction

Row 0: `. . . Y A X . . . . .`
Row 1: `. . . Y A X . . . . .`
Row 2: `. . . Y A X . . . . .`
Row 3: `. . . Y A X . . . . .`
Row 4: `. . . Y A X . . . . .`
Row 5: `. . . Y A X . . . . .`
Row 6: `. . . Y A X . . . . .`
Row 7: `Y Y Y Y A X . . . . .`
Row 8: `. . . . A X . . . . .`
Row 9: `. . . . A X X X X X X`
Row 10: `. . . . A . . . . . .`

ANSWER:
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
Y Y Y Y A X . . . . .
. . . . A X . . . . .
. . . . A X X X X X X
. . . . A . . . . . .

### Conceptual Digestion

1.  **CAST**:
    *   **Mineral Veins**: The colored cells located at the top of the grid (Row 0). In the test input, these are `Y` (Yellow) at (0, 3), `A` (Azure) at (0, 4), and `X` (Grey) at (0, 5).
    *   **Magma Pockets**: Single colored cells of the same color as a vein, located elsewhere in the grid. In the test input, these are `Y` at (7, 0) and `X` at (9, 10).
    *   **Solid Rock**: Black cells (`.`).

2.  **AFFORDANCES**:
    *   Veins can extend vertically downwards from their starting position.
    *   Veins can extend horizontally to connect to their matching magma pocket.
    *   The path is a "Manhattan" connection: Down from the top, then across to the pocket.

3.  **TRANSFORMATION RULE**:
    *   For each vein color at the top (Row 0, Col $C$):
        1.  Identify the matching magma pocket at (Row $R$, Col $P$).
        2.  Draw a vertical line of that color from $(0, C)$ down to $(R, C)$.
        3.  Draw a horizontal line of that color from $(R, C)$ to $(R, P)$.
        4.  If no matching pocket exists for a vein color, the vein simply extends vertically from $(0, C)$ to the bottom of the grid.
    *   In the test input:
        *   **Yellow (Y)**: Top at (0, 3). Pocket at (7, 0). Path: (0, 3) $\rightarrow$ (7, 3) $\rightarrow$ (7, 0).
        *   **Azure (A)**: Top at (0, 4). No pocket. Path: (0, 4) $\rightarrow$ (10, 4).
        *   **Grey (X)**: Top at (0, 5). Pocket at (9, 10). Path: (0, 5) $\rightarrow$ (9, 5) $\rightarrow$ (9, 10).

### Step-by-Step Application

*   **Yellow (Y)**:
    *   Vertical: (0,3), (1,3), (2,3), (3,3), (4,3), (5,3), (6,3), (7,3)
    *   Horizontal: (7,3), (7,2), (7,1), (7,0)
*   **Azure (A)**:
    *   Vertical: (0,4), (1,4), (2,4), (3,4), (4,4), (5,4), (6,4), (7,4), (8,4), (9,4), (10,4)
*   **Grey (X)**:
    *   Vertical: (0,5), (1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (7,5), (8,5), (9,5)
    *   Horizontal: (9,5), (9,6), (9,7), (9,8), (9,9), (9,10)

ANSWER:
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
Y Y Y Y A X . . . . .
. . . . A X . . . . .
. . . . A X X X X X X
. . . . A . . . . . .

### Conceptual Digestion

1.  **CAST**:
    *   **Shelf Bottles**: The three adjacent colored cells at the top of the grid (Row 0). In the test input, these are `Y` (Yellow), `A` (Azure), and `X` (Grey) at columns 3, 4, and 5.
    *   **Plates**: Single colored cells scattered in the lower part of the grid. In the test input, these are `Y` (Yellow) at (7, 0) and `X` (Grey) at (9, 10).
    *   **Sauce Streams**: Vertical lines of the same color extending downwards from the bottles.
    *   **Sauce Pools**: Horizontal lines of the same color extending from the stream to the plate.

2.  **AFFORDANCES**:
    *   Bottles act as sources of color.
    *   Plates act as targets/anchors for the sauce.
    *   The space between them is filled based on the "pour and pool" logic.

3.  **TRANSFORMATION RULE**:
    *   For each bottle at `(0, col)` with color `C`:
        1.  Check if there is a plate of color `C` anywhere in the grid at `(row_p, col_p)`.
        2.  If a plate exists:
            *   Draw a vertical line of color `C` from `(0, col)` down to `(row_p, col)`.
            *   Draw a horizontal line of color `C` from `(row_p, col)` to `(row_p, col_p)`.
        3.  If no plate exists:
            *   Draw a vertical line of color `C` from `(0, col)` all the way to the bottom of the grid.

### Applying the Rule to Test Input

*   **Bottle 1: Yellow (Y) at (0, 3)**
    *   Plate found: Yellow (Y) at (7, 0).
    *   Vertical stream: (0, 3) to (7, 3).
    *   Horizontal pool: (7, 3) to (7, 0).
*   **Bottle 2: Azure (A) at (0, 4)**
    *   Plate found: None.
    *   Vertical stream: (0, 4) to (10, 4).
*   **Bottle 3: Grey (X) at (0, 5)**
    *   Plate found: Grey (X) at (9, 10).
    *   Vertical stream: (0, 5) to (9, 5).
    *   Horizontal pool: (9, 5) to (9, 10).

### Final Grid Construction

Row 0: `. . . Y A X . . . . .`
Row 1: `. . . Y A X . . . . .`
Row 2: `. . . Y A X . . . . .`
Row 3: `. . . Y A X . . . . .`
Row 4: `. . . Y A X . . . . .`
Row 5: `. . . Y A X . . . . .`
Row 6: `. . . Y A X . . . . .`
Row 7: `Y Y Y Y A X . . . . .`
Row 8: `. . . . A X . . . . .`
Row 9: `. . . . A X X X X X X`
Row 10: `. . . . A . . . . . .`

ANSWER:
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
Y Y Y Y A X . . . . .
. . . . A X . . . . .
. . . . A X X X X X X
. . . . A . . . . . .

### Conceptual Digestion

1.  **CAST**:
    *   **Tuning Forks**: The three adjacent colored cells at the top of the grid (in the test input: `Y` at (0,3), `A` at (0,4), and `X` at (0,5)).
    *   **Sympathetic Strings**: Lone cells of the same color located elsewhere in the grid (in the test input: `Y` at (7,0) and `X` at (9,10)).
    *   **Resonance Paths**: The filled-in paths of the same color connecting the forks to the strings.
    *   **The Hall**: The black background (`.`).

2.  **AFFORDANCES**:
    *   Forks act as sources of vertical resonance.
    *   Strings act as targets that "pull" the resonance horizontally.
    *   If no string exists for a fork, the resonance simply travels vertically to the bottom of the grid.

3.  **TRANSFORMATION RULE**:
    *   For each fork at `(0, col)` with color `C`:
        1.  Draw a vertical line of color `C` from `(0, col)` downwards.
        2.  If there is a string of color `C` at `(row, target_col)`:
            *   The vertical line extends from `row 0` to `row`.
            *   At `row`, a horizontal line of color `C` is drawn from `col` to `target_col`.
            *   The vertical line stops at `row` (unless the string is below the fork, but the rule says "at the depth where its sympathetic string waits, the vibration travels horizontally"). *Correction based on Example 1*: The vertical line continues down to the bottom of the grid if it's the leftmost fork or if it's the "main" path, but looking closely at Example 1:
                *   Fork X (0,4) -> String X (4,0): Vertical from (0,4) to (4,4), then horizontal from (4,4) to (4,0).
                *   Fork G (0,5) -> String G (8,0): Vertical from (0,5) to (8,5), then horizontal from (8,5) to (8,0).
                *   Fork A (0,6) -> String A (5,11): Vertical from (0,6) to (5,6), then horizontal from (5,6) to (5,11).
                *   Crucially, the vertical lines for X and G continue down to the row of the lowest string (row 8), and the vertical line for A stops at its string's row (row 5). Wait, looking again: Fork G's line goes to row 8. Fork X's line goes to row 4, but the output shows it stops at row 4? No, Example 1 output: X is at (0,4) to (4,4), then (4,4) to (4,0). G is at (0,5) to (8,5), then (8,5) to (8,0). A is at (0,6) to (5,6), then (5,6) to (5,11).
    *   **Refined Rule**: For each fork at `(0, col)`:
        1.  Find the string of the same color at `(row, target_col)`.
        2.  Fill cells from `(0, col)` to `(row, col)` (vertical).
        3.  Fill cells from `(row, col)` to `(row, target_col)` (horizontal).
        4.  If no string exists, fill from `(0, col)` to the bottom of the grid.

### Applying to Test Input
*   **Fork Y** at (0,3). String Y at (7,0).
    *   Vertical: (0,3) to (7,3).
    *   Horizontal: (7,3) to (7,0).
*   **Fork A** at (0,4). No string A.
    *   Vertical: (0,4) to (10,4).
*   **Fork X** at (0,5). String X at (9,10).
    *   Vertical: (0,5) to (9,5).
    *   Horizontal: (9,5) to (9,10).

### Final Grid Construction
Row 0: . . . Y A X . . . . .
Row 1: . . . Y A X . . . . .
Row 2: . . . Y A X . . . . .
Row 3: . . . Y A X . . . . .
Row 4: . . . Y A X . . . . .
Row 5: . . . Y A X . . . . .
Row 6: . . . Y A X . . . . .
Row 7: Y Y Y Y A X . . . . .
Row 8: . . . . A X . . . . .
Row 9: . . . . A X X X X X X
Row 10: . . . . A . . . . . .

ANSWER:
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
Y Y Y Y A X . . . . .
. . . . A X . . . . .
. . . . A X X X X X X
. . . . A . . . . . .

### Conceptual Digestion

1.  **CAST**:
    *   **Runners**: The three colored cells located adjacently in the top row (Row 0). In the test input, these are at (0, 3)=Y, (0, 4)=A, and (0, 5)=X.
    *   **Teammates**: Single cells of the same color scattered elsewhere in the grid. In the test input, these are (7, 0)=Y and (9, 10)=X. (Note: Azure 'A' has no teammate).
    *   **Track**: The black cells ('.') that will be filled by the runners' paths.

2.  **AFFORDANCES**:
    *   Runners move vertically down their column.
    *   Runners move horizontally to reach their teammate.
    *   If no teammate exists, the runner continues to the bottom of the grid (or the far end of the track).

3.  **TRANSFORMATION RULE**:
    *   For each runner at `(0, col)` with color `C`:
        1.  Identify the teammate of color `C` at `(row_t, col_t)`.
        2.  If a teammate exists:
            *   Fill the column from `(0, col)` down to `(row_t, col)`.
            *   Fill the row from `(row_t, col)` to `(row_t, col_t)`.
        3.  If no teammate exists:
            *   Fill the column from `(0, col)` down to the bottom of the grid (Row 10).

### Applying the Rule to Test Input

*   **Runner 1 (Yellow 'Y')**:
    *   Start: (0, 3). Teammate: (7, 0).
    *   Vertical path: (0, 3) $\rightarrow$ (7, 3).
    *   Horizontal path: (7, 3) $\rightarrow$ (7, 0).
*   **Runner 2 (Azure 'A')**:
    *   Start: (0, 4). Teammate: None.
    *   Vertical path: (0, 4) $\rightarrow$ (10, 4).
*   **Runner 3 (Grey 'X')**:
    *   Start: (0, 5). Teammate: (9, 10).
    *   Vertical path: (0, 5) $\rightarrow$ (9, 5).
    *   Horizontal path: (9, 5) $\rightarrow$ (9, 10).

### Final Grid Construction

Row 0: . . . Y A X . . . . .
Row 1: . . . Y A X . . . . .
Row 2: . . . Y A X . . . . .
Row 3: . . . Y A X . . . . .
Row 4: . . . Y A X . . . . .
Row 5: . . . Y A X . . . . .
Row 6: . . . Y A X . . . . .
Row 7: Y Y Y Y A X . . . . .
Row 8: . . . . A X . . . . .
Row 9: . . . . A X X X X X X
Row 10: . . . . A . . . . . .

ANSWER:
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
Y Y Y Y A X . . . . .
. . . . A X . . . . .
. . . . A X X X X X X
. . . . A . . . . . .

### Conceptual Digestion

1.  **The CAST**:
    *   **Spotlights**: The three colored cells located at the top of the grid (Row 0). In the test input, these are `Y` (Yellow) at (0, 3), `A` (Azure) at (0, 4), and `X` (Grey) at (0, 5).
    *   **Actors**: Cells of the same color as the spotlights located elsewhere in the grid. In the test input, these are `Y` (Yellow) at (7, 0) and `X` (Grey) at (9, 10).
    *   **Beams**: The paths filled with the spotlight's color.

2.  **The AFFORDANCES**:
    *   **Vertical Beam**: A spotlight at `(0, col)` casts a beam straight down the column `col`.
    *   **Horizontal Sweep**: When the vertical beam reaches the row `row` where an actor of the same color is located, it fills the row from the column `col` to the actor's column `actor_col`.
    *   **Orchestra Pit**: If no actor of that color exists, the beam simply continues vertically to the bottom of the grid.

3.  **The TRANSFORMATION RULE**:
    *   For each spotlight at `(0, col)` with color `C`:
        1.  Find the actor of color `C` at `(row, actor_col)`.
        2.  Fill all cells from `(0, col)` down to `(row, col)` with color `C`.
        3.  Fill all cells from `(row, col)` to `(row, actor_col)` with color `C`.
        4.  If no actor of color `C` exists, fill all cells from `(0, col)` to the bottom of the grid with color `C`.

### Applying the Rule to Test Input

*   **Spotlight 1: Yellow (Y) at (0, 3)**
    *   Actor Y is at (7, 0).
    *   Vertical beam: (0, 3) to (7, 3) becomes `Y`.
    *   Horizontal sweep: (7, 3) to (7, 0) becomes `Y`.
*   **Spotlight 2: Azure (A) at (0, 4)**
    *   No actor A is present in the grid.
    *   Vertical beam: (0, 4) to (10, 4) becomes `A`.
*   **Spotlight 3: Grey (X) at (0, 5)**
    *   Actor X is at (9, 10).
    *   Vertical beam: (0, 5) to (9, 5) becomes `X`.
    *   Horizontal sweep: (9, 5) to (9, 10) becomes `X`.

### Final Grid Construction

Row 0: `. . . Y A X . . . . .`
Row 1: `. . . Y A X . . . . .`
Row 2: `. . . Y A X . . . . .`
Row 3: `. . . Y A X . . . . .`
Row 4: `. . . Y A X . . . . .`
Row 5: `. . . Y A X . . . . .`
Row 6: `. . . Y A X . . . . .`
Row 7: `Y Y Y Y A X . . . . .`
Row 8: `. . . . A X . . . . .`
Row 9: `. . . . A X X X X X X`
Row 10: `. . . . A . . . . . .`

ANSWER:
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
. . . Y A X . . . . .
Y Y Y Y A X . . . . .
. . . . A X . . . . .
. . . . A X X X X X X
. . . . A . . . . . .

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Sticks:** Contiguous lines of a single color (vertical or horizontal) on a black background.
    *   **Length:** The number of cells occupied by a stick.
    *   **Shells:** Concentric square rings of color in the output grid.

2.  **Identify the AFFORDANCES:**
    *   Sticks can be measured for length.
    *   Sticks can be sorted by length (descending).
    *   The sorted colors determine the order of the nested shells from the outermost layer to the innermost core.

3.  **Identify the TRANSFORMATION RULE:**
    *   **Step 1: Find all sticks.** Identify every contiguous line of color and record its color and length.
    *   **Step 2: Sort sticks.** Order the colors from the longest stick to the shortest stick.
    *   **Step 3: Determine output size.** The output is a square grid. The size is determined by the number of unique colors (shells). If there are $N$ colors, the grid size is $(2N - 1) \times (2N - 1)$ if $N$ is the number of layers to reach a center point, or more simply, the smallest square that can accommodate the nesting. Looking at Example 1: 3 colors (B=6, G=5, M=2) $\rightarrow$ $6 \times 6$ output. Wait, let's re-examine Example 1.
    *   **Example 1 Analysis:**
        *   B: length 6
        *   G: length 5
        *   M: length 2
        *   Output: $6 \times 6$. Outer ring B, next ring G, center $2 \times 2$ M.
    *   **Test Input Analysis:**
        *   A (Azure): length 7 (rows 12-18, col 10)
        *   M (Magenta): length 5 (row 11, cols 2-6)
        *   O (Orange): length 3 (rows 6-8, col 10)
        *   G (Green): length 1 (row 4, col 6)
    *   **Sorting by length:** A (7), M (5), O (3), G (1).
    *   **Output Construction:**
        *   The longest stick (A) has length 7. The output grid will be $7 \times 7$.
        *   Layer 1 (Outer): A
        *   Layer 2: M
        *   Layer 3: O
        *   Layer 4 (Center): G

### Applying the Rule to Test Input

1.  **Colors sorted by length:** A (7), M (5), O (3), G (1).
2.  **Grid Size:** $7 \times 7$.
3.  **Layering:**
    *   Outer ring (Row 0, Row 6, Col 0, Col 6): **A**
    *   Second ring (Row 1, Row 5, Col 1, Col 5): **M**
    *   Third ring (Row 2, Row 4, Col 2, Col 4): **O**
    *   Center (Row 3, Col 3): **G**

**Grid Layout:**
Row 0: A A A A A A A
Row 1: A M M M M M A
Row 2: A M O O O M A
Row 3: A M O G O M A
Row 4: A M O O O M A
Row 5: A M M M M M A
Row 6: A A A A A A A

ANSWER:
A A A A A A A
A M M M M M A
A M O O O M A
A M O G O M A
A M O O O M A
A M M M M M A
A A A A A A A

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Colored Pixels:** These represent "tones." Each distinct color is a unique tone.
    *   **Pixel Count (Duration):** The number of pixels of a specific color represents the "duration" or "length" of that tone.
    *   **Black Pixels (.):** Background/silence.

2.  **Identify the AFFORDANCES:**
    *   The colors can be counted to determine their relative lengths.
    *   The colors can be arranged into concentric squares (rings) based on their length.

3.  **Identify the TRANSFORMATION RULE:**
    *   **Step 1: Count the pixels of each color.**
        *   Azure (A): 7 pixels
        *   Magenta (M): 5 pixels
        *   Orange (O): 3 pixels
        *   Green (G): 1 pixel
    *   **Step 2: Order the colors by length (descending).**
        *   Longest: Azure (A)
        *   Second: Magenta (M)
        *   Third: Orange (O)
        *   Shortest: Green (G)
    *   **Step 3: Create concentric rings.**
        *   The longest color (A) forms the outermost frame.
        *   The next longest (M) forms the frame inside that.
        *   The next (O) forms the frame inside that.
        *   The shortest (G) sits at the center.
    *   **Step 4: Determine the grid size.**
        *   With 4 colors, the smallest possible square to accommodate 4 concentric layers (including a center) is a $7 \times 7$ grid (since $2n-1$ for $n$ layers).
        *   Layer 1 (A): $7 \times 7$ perimeter.
        *   Layer 2 (M): $5 \times 5$ perimeter.
        *   Layer 3 (O): $3 \times 3$ perimeter.
        *   Layer 4 (G): $1 \times 1$ center.

### Applying the Rule to Test Input

*   **A (Azure):** 7 pixels $\rightarrow$ Outermost ring ($7 \times 7$)
*   **M (Magenta):** 5 pixels $\rightarrow$ Second ring ($5 \times 5$)
*   **O (Orange):** 3 pixels $\rightarrow$ Third ring ($3 \times 3$)
*   **G (Green):** 1 pixel $\rightarrow$ Center ($1 \times 1$)

**Grid Construction:**
Row 1: A A A A A A A
Row 2: A M M M M M A
Row 3: A M O O O M A
Row 4: A M O G O M A
Row 5: A M O O O M A
Row 6: A M M M M M A
Row 7: A A A A A A A

ANSWER:
A A A A A A A
A M M M M M A
A M O O O M A
A M O G O M A
A M O O O M A
A M M M M M A
A A A A A A A

### Conceptual Digestion

1.  **CAST**:
    *   **Baton**: A contiguous line of colored cells in the input grid.
    *   **Length**: The number of cells making up a baton.
    *   **Trophy/Monument**: The final output grid, which is a square composed of concentric colored rings.

2.  **AFFORDANCES**:
    *   Each baton's color is preserved.
    *   The length of the baton determines its position in the nesting order (longest = outermost, shortest = innermost).

3.  **TRANSFORMATION RULE**:
    *   Identify all colored segments (batons) in the input grid and measure their lengths.
    *   Sort the batons by length in descending order.
    *   The longest baton defines the size of the square output grid. If the longest baton has length $L$, the output grid is an $L \times L$ square.
    *   The longest baton forms the outermost perimeter of the square.
    *   The second longest baton forms the perimeter of the square immediately inside the first.
    *   This process continues, nesting each subsequent baton one layer deeper, until the shortest baton fills the remaining center.

### Applying the Rule to Test Input

1.  **Identify Batons and Lengths**:
    *   Green (G): 1 cell (Length = 1)
    *   Orange (O): 3 cells (Length = 3)
    *   Magenta (M): 7 cells (Length = 7)
    *   Azure (A): 7 cells (Length = 7) - *Wait, let's re-count.*
        *   M: `M M M M M` (Wait, looking closer at the test input: `. . M M M M M . .`) -> Length = 5.
        *   A: `A` (rows 13-19) -> Length = 7.
        *   O: `O` (rows 7-9) -> Length = 3.
        *   G: `G` (row 5) -> Length = 1.

    *   **Recounting carefully**:
        *   M: `M M M M M` (5 cells)
        *   A: `A` (7 cells)
        *   O: `O` (3 cells)
        *   G: `G` (1 cell)

2.  **Sort by Length (Descending)**:
    *   Azure (A): 7
    *   Magenta (M): 5
    *   Orange (O): 3
    *   Green (G): 1

3.  **Construct the Square**:
    *   The longest is A (7), so the grid is $7 \times 7$.
    *   Layer 1 (Outer): Azure (A)
    *   Layer 2: Magenta (M)
    *   Layer 3: Orange (O)
    *   Layer 4 (Center): Green (G)

    *   **Layer 1 (7x7 perimeter)**:
        A A A A A A A
        A . . . . . A
        A . . . . . A
        A . . . . . A
        A . . . . . A
        A . . . . . A
        A A A A A A A

    *   **Layer 2 (5x5 perimeter)**:
        A A A A A A A
        A M M M M M A
        A M . . . M A
        A M . . . M A
        A M . . . M A
        A M M M M M A
        A A A A A A A

    *   **Layer 3 (3x3 perimeter)**:
        A A A A A A A
        A M M M M M A
        A M O O O M A
        A M O . O M A
        A M O O O M A
        A M M M M M A
        A A A A A A A

    *   **Layer 4 (1x1 center)**:
        A A A A A A A
        A M M M M M A
        A M O O O M A
        A M O G O M A
        A M O O O M A
        A M M M M M A
        A A A A A A A

ANSWER:
A A A A A A A
A M M M M M A
A M O O O M A
A M O G O M A
A M O O O M A
A M M M M M A
A A A A A A A

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Beams:** Contiguous lines of a single color in the input grid.
    *   **Length:** The number of cells occupied by a beam.
    *   **Nested Rooms:** Concentric square borders in the output grid.

2.  **Identify the AFFORDANCES:**
    *   The length of each beam determines the size of the square it forms.
    *   The relative length determines the order of nesting (longest = outermost, shortest = innermost).

3.  **Identify the TRANSFORMATION RULE:**
    *   Find all colored beams in the input and record their lengths and colors.
    *   Sort the beams by length in descending order.
    *   The longest beam (length $L_1$) creates a square perimeter of size $L_1 \times L_1$.
    *   The second longest beam (length $L_2$) creates a square perimeter of size $L_2 \times L_2$ centered inside the first.
    *   Continue this process for all beams.
    *   The output grid is a square of size $L_{max} \times L_{max}$.

### Applying the Rule to Test Input

1.  **Identify Beams and Lengths:**
    *   Green (G): 1 cell (length 1)
    *   Orange (O): 3 cells (length 3)
    *   Magenta (M): 7 cells (length 7)
    *   Azure (A): 8 cells (length 8) - *Wait, let's re-count A: rows 12 to 18 is 7 cells? No, let's check indices: row 12, 13, 14, 15, 16, 17, 18. That is 7 cells. Let's re-count M: row 11, cols 2, 3, 4, 5, 6, 7, 8. That is 7 cells. Let's re-count O: row 6, 7, 8. That is 3 cells. Let's re-count G: row 4. That is 1 cell.*
    *   **Correction on counts:**
        *   A: (12,10) to (18,10) $\rightarrow$ 7 cells.
        *   M: (11,2) to (11,8) $\rightarrow$ 7 cells.
        *   O: (6,10) to (8,10) $\rightarrow$ 3 cells.
        *   G: (4,6) $\rightarrow$ 1 cell.
    *   *Wait, if A and M are both 7, the prompt says "the longest beam defines the outer wall". If they are equal, they likely share the same layer or the input implies a specific order. However, looking at the training example: B=6, G=5, M=2. Output is 6x6. B is outer, G is inner, M is center. The lengths are 6, 5, 2. The output is a 6x6 square. The G square is 4x4 (since it's inside B), and M is 2x2. This means the length of the beam corresponds to the side length of the square it forms.*

2.  **Determine Square Sizes:**
    *   Longest: A (7) or M (7). Let's assume the outer wall is 7x7.
    *   Next: A/M (7). If both are 7, they might be the same wall or one is slightly different. Let's re-count A again: row 12, 13, 14, 15, 16, 17, 18. That is 7. M: col 2, 3, 4, 5, 6, 7, 8. That is 7.
    *   Next: O (3).
    *   Shortest: G (1).

3.  **Construct the Grid (7x7):**
    *   Layer 1 (Outer, 7x7): Color A or M. Let's use A (usually the one appearing later or based on a specific priority, but let's check the training example again. B was 6, output was 6x6. G was 5, output had a 4x4 G square. M was 2, output had a 2x2 M square).
    *   Wait, in Example 1: B=6, G=5, M=2. Output: B(6x6), G(4x4), M(2x2).
    *   The rule is: Beam length $L$ $\rightarrow$ Square of side $L$.
    *   Our lengths: 7, 7, 3, 1.
    *   If we have two 7s, the outer is 7x7. The next is also 7x7? That's impossible for nesting. Let's re-count A one more time.
    *   A: row 12, 13, 14, 15, 16, 17, 18. (18-12)+1 = 7.
    *   M: col 2, 3, 4, 5, 6, 7, 8. (8-2)+1 = 7.
    *   Perhaps one is 8? Let's look at the dots.
    *   A: row 12, 13, 14, 15, 16, 17, 18. (7 cells).
    *   M: col 2, 3, 4, 5, 6, 7, 8. (7 cells).
    *   If two are 7, the outer is 7x7. The next longest is 7... this suggests the beams are A=8 or M=8. Let's look closer.
    *   M: . . M M M M M . . (col 2, 3, 4, 5, 6, 7, 8). That is 7.
    *   A: row 12, 13, 14, 15, 16, 17, 18. That is 7.
    *   Wait, let's re-read the training example. B=6, G=5, M=2. Output: B is 6x6, G is 4x4, M is 2x2.
    *   The side length of the square is $L$ if $L$ is the beam length? No.
    *   B=6 $\rightarrow$ 6x6.
    *   G=5 $\rightarrow$ 4x4.
    *   M=2 $\rightarrow$ 2x2.
    *   It seems the side length is the largest even number $\le L$? Or $L$ if $L$ is even, $L-1$ if $L$ is odd?
    *   Let's check: 6 $\rightarrow$ 6, 5 $\rightarrow$ 4, 2 $\rightarrow$ 2. Yes.
    *   Our lengths: 7, 7, 3, 1.
    *   Side lengths: 6, 6, 2, 0? No.
    *   Let's try: $L$ is the side length. 7, 7, 3, 1.
    *   Outer: 7x7 (A). Next: 7x7 (M) - impossible.
    *   Let's re-count A again. Row 12, 13, 14, 15, 16, 17, 18. That is 7.
    *   Let's re-count M again. Col 2, 3, 4, 5, 6, 7, 8. That is 7.
    *   Is it possible A is 8? Row 12, 13, 14, 15, 16, 17, 18... no, that's 7.
    *   Is it possible M is 8? Col 2, 3, 4, 5, 6, 7, 8... no, that's 7.
    *   Wait! Look at the test input again.
    *   A: row 12, 13, 14, 15, 16, 17, 18. (7 cells).
    *   M: col 2, 3, 4, 5, 6, 7, 8. (7 cells).
    *   O: row 6, 7, 8. (3 cells).
    *   G: row 4. (1 cell).
    *   If the lengths are 7, 7, 3, 1, and the output must be a square, and the training example used $L$ for the outer wall...
    *   Maybe the two 

... [1023 chars truncated]

### Conceptual Digestion

1.  **CAST**:
    *   **Sky (Background)**: The dominant color filling the grid (Azure 'A' in the test input).
    *   **Books**: Clusters of non-background colors. Each book consists of two halves: a **Top Cover** (upper block of color) and a **Bottom Cover** (lower block of color).
    *   **Spine**: The horizontal boundary where the Top Cover and Bottom Cover meet.

2.  **AFFORDANCES**:
    *   The **Top Cover** can shift one unit to the left.
    *   The **Bottom Cover** can shift one unit to the right.
    *   The **Sky** fills the void left by the shifting covers.

3.  **TRANSFORMATION RULE**:
    *   Identify each "book" (a contiguous vertical stack of two different colors).
    *   For each book:
        *   Identify the rows belonging to the Top Cover and the rows belonging to the Bottom Cover.
        *   Shift all cells of the Top Cover one column to the left.
        *   Shift all cells of the Bottom Cover one column to the right.
        *   Fill the vacated cells with the background color ('A').

### Applying the Rule to Test Input

**Background**: A (Azure)

**Book 1**:
*   Top Cover (Blue 'B'): Rows 1-3.
*   Bottom Cover (Red 'R'): Row 4.
*   Transformation:
    *   Rows 1-3: Shift 'B's left by 1.
    *   Row 4: Shift 'R's right by 1.

**Book 2**:
*   Top Cover (Red 'R'): Rows 9-10.
*   Bottom Cover (Yellow 'Y'): Rows 11-12.
*   Transformation:
    *   Rows 9-10: Shift 'R's left by 1.
    *   Rows 11-12: Shift 'Y's right by 1.

**Step-by-step Grid Construction**:
*   Row 0: `A A A A A A A A A A A A A` (No change)
*   Row 1: `A B A A B A A A A A A A A` $\rightarrow$ `A B A A B A A A A A A A A` (Wait, shift left: `A B A A B A...` becomes `A B A A B A...` No, the 'B' at col 2 moves to col 1, 'B' at col 5 moves to col 4). $\rightarrow$ `A B A A B A A A A A A A A` (Wait, let's re-examine: Input Row 1 is `A A B A A B A A A A A A A`. Shift left: `A B A A B A A A A A A A A`)
*   Row 2: `A A B B B B A A A A A A A` $\rightarrow$ `A B B B B A A A A A A A A`
*   Row 3: `A A B B B B A A A A A A A` $\rightarrow$ `A B B B B A A A A A A A A`
*   Row 4: `A A R R R R A A A A A A A` $\rightarrow$ `A A A R R R R A A A A A A`
*   Rows 5-8: No change.
*   Row 9: `A A A A A A R R R A A A A` $\rightarrow$ `A A A A A R R R A A A A A`
*   Row 10: `A A A A A A R R R A A A A` $\rightarrow$ `A A A A A R R R A A A A A`
*   Row 11: `A A A A A A Y Y Y A A A A` $\rightarrow$ `A A A A A A A Y Y Y A A A`
*   Row 12: `A A A A A A Y A Y A A A A` $\rightarrow$ `A A A A A A A Y A Y A A A`
*   Row 13: No change.

ANSWER:
A A A A A A A A A A A A A
A B A A B A A A A A A A A
A B B B B A A A A A A A A
A B B B B A A A A A A A A
A A A R R R R A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A R R R A A A A A
A A A A A R R R A A A A A
A A A A A A A Y Y Y A A A
A A A A A A A Y A Y A A A
A A A A A A A A A A A A A

### Conceptual Digestion

1.  **CAST**:
    *   **Cytoplasm (Background)**: The dominant color filling the grid (e.g., `B` in Example 1, `A` in Example 2 and Test Input).
    *   **Composite Organisms**: Clusters of non-background colors. Each organism consists of two distinct colored tissue layers (upper and lower) fused horizontally.
    *   **Tissue Layers**: The upper layer is the top set of colored pixels; the lower layer is the bottom set of colored pixels.

2.  **AFFORDANCES**:
    *   The upper tissue layer can shift one unit to the left.
    *   The lower tissue layer can shift one unit to the right.
    *   Vacated cells (where the tissue used to be) are filled by the background color.
    *   Cells that the tissue moves into are overwritten by the tissue color.

3.  **TRANSFORMATION RULE**:
    *   Identify each distinct "organism" (a contiguous block of non-background colors).
    *   Split the organism into two parts: the upper part (all rows of the organism above the lowest row of the top color) and the lower part (all rows of the organism starting from the first row of the bottom color).
    *   **Shift Left**: Move every pixel of the upper tissue layer one column to the left.
    *   **Shift Right**: Move every pixel of the lower tissue layer one column to the right.
    *   **Fill**: Any cell that becomes empty or is vacated reverts to the background color.

### Applying the Rule to Test Input

**Background**: `A` (Azure)

**Organism 1**:
*   Upper Layer (Blue `B`): Rows 1-3, Cols 2-5 (approx).
    *   Row 1: `A A B A A B A...` -> `A B A A B A A...` (Shifted left)
    *   Row 2: `A A B B B B A...` -> `A B B B B A A...` (Shifted left)
    *   Row 3: `A A B B B B A...` -> `A B B B B A A...` (Shifted left)
*   Lower Layer (Red `R`): Row 4, Cols 2-5.
    *   Row 4: `A A R R R R A...` -> `A A A R R R R A...` (Shifted right)

**Organism 2**:
*   Upper Layer (Red `R`): Rows 9-10, Cols 6-8.
    *   Row 9: `A A A A A A R R R A...` -> `A A A A A R R R A A...` (Shifted left)
    *   Row 10: `A A A A A A R R R A...` -> `A A A A A R R R A A...` (Shifted left)
*   Lower Layer (Yellow `Y`): Rows 11-12, Cols 6-8.
    *   Row 11: `A A A A A A Y Y Y A...` -> `A A A A A A A Y Y Y A...` (Shifted right)
    *   Row 12: `A A A A A A Y A Y A...` -> `A A A A A A A Y A Y A...` (Shifted right)

### Final Grid Construction

Row 0: `A A A A A A A A A A A A A`
Row 1: `A B A A B A A A A A A A A`
Row 2: `A B B B B A A A A A A A A`
Row 3: `A B B B B A A A A A A A A`
Row 4: `A A A R R R R A A A A A A`
Row 5: `A A A A A A A A A A A A A`
Row 6: `A A A A A A A A A A A A A`
Row 7: `A A A A A A A A A A A A A`
Row 8: `A A A A A A A A A A A A A`
Row 9: `A A A A A R R R A A A A A`
Row 10: `A A A A A R R R A A A A A`
Row 11: `A A A A A A A Y Y Y A A A`
Row 12: `A A A A A A A Y A Y A A A`
Row 13: `A A A A A A A A A A A A A`

ANSWER:
A A A A A A A A A A A A A
A B A A B A A A A A A A A
A B B B B A A A A A A A A
A B B B B A A A A A A A A
A A A R R R R A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A R R R A A A A A
A A A A A R R R A A A A A
A A A A A A A Y Y Y A A A
A A A A A A A Y A Y A A A
A A A A A A A A A A A A A

### Conceptual Digestion

1.  **Identify the CAST**:
    *   **Void (Background)**: The dominant color filling the grid (in Example 1 it was Blue `B`, in the Test Input it is Azure `A`).
    *   **Stellar Bodies (Objects)**: Contiguous blocks of a different color.
    *   **Binary Systems**: Pairs of objects where one is positioned above the other.

2.  **Identify the AFFORDANCES**:
    *   The objects can be shifted horizontally.
    *   The background color fills any space vacated by the shift.

3.  **Identify the TRANSFORMATION RULE**:
    *   Identify pairs of colored objects (one above, one below).
    *   The **upper body** (the object higher up in the grid) moves **one unit to the left**.
    *   The **lower body** (the object lower down in the grid) moves **one unit to the right**.
    *   The background color (`A` in the test case) fills the gaps.

### Applying the Rule to Test Input

**Input Analysis**:
*   Background: `A`
*   System 1:
    *   Upper body: Blue (`B`) block at rows 1-3, cols 2-5.
    *   Lower body: Red (`R`) block at row 4, cols 2-5.
*   System 2:
    *   Upper body: Red (`R`) block at rows 9-10, cols 6-8.
    *   Lower body: Yellow (`Y`) block at rows 11-12, cols 6-8.

**Transformation**:
*   **System 1 Upper (Blue)**: Shift left by 1.
    *   Row 1: `A A B A A B` $\rightarrow$ `A B A A B A` (Wait, looking at the block: Col 2 is B, Col 5 is B. Shift left: Col 1 becomes B, Col 4 becomes B).
    *   Rows 2-3: `A A B B B B` $\rightarrow$ `A B B B B A`.
*   **System 1 Lower (Red)**: Shift right by 1.
    *   Row 4: `A A R R R R` $\rightarrow$ `A A A R R R R`.
*   **System 2 Upper (Red)**: Shift left by 1.
    *   Rows 9-10: `A A A A A A R R R` $\rightarrow$ `A A A A A R R R A`.
*   **System 2 Lower (Yellow)**: Shift right by 1.
    *   Row 11: `A A A A A A Y Y Y` $\rightarrow$ `A A A A A A A Y Y Y`.
    *   Row 12: `A A A A A A Y A Y` $\rightarrow$ `A A A A A A A Y A Y`.

**Refining the coordinates**:
Input System 1:
R1: C2=B, C5=B
R2: C2=B, C3=B, C4=B, C5=B
R3: C2=B, C3=B, C4=B, C5=B
R4: C2=R, C3=R, C4=R, C5=R
Output System 1:
R1: C1=B, C4=B
R2: C1=B, C2=B, C3=B, C4=B
R3: C1=B, C2=B, C3=B, C4=B
R4: C3=R, C4=R, C5=R, C6=R

Input System 2:
R9: C6=R, C7=R, C8=R
R10: C6=R, C7=R, C8=R
R11: C6=Y, C7=Y, C8=Y
R12: C6=Y, C8=Y
Output System 2:
R9: C5=R, C6=R, C7=R
R10: C5=R, C6=R, C7=R
R11: C7=Y, C8=Y, C9=Y
R12: C7=Y, C9=Y

ANSWER:
A A A A A A A A A A A A A
A B A A B A A A A A A A A
A B B B B A A A A A A A A
A B B B B A A A A A A A A
A A A R R R R A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A R R R A A A A A
A A A A A R R R A A A A A
A A A A A A A Y Y Y A A A
A A A A A A A Y A Y A A A
A A A A A A A A A A A A A

### Conceptual Digestion

1.  **CAST**:
    *   **Board Background**: The dominant color of the grid (Azure 'A' in the test input).
    *   **Relay Assembly**: A contiguous block of two different colors (Upper Plate and Lower Plate) that share a boundary line.
    *   **Upper Plate**: The top color of a relay assembly.
    *   **Lower Plate**: The bottom color of a relay assembly.

2.  **AFFORDANCES**:
    *   The **Upper Plate** can shift one position to the left.
    *   The **Lower Plate** can shift one position to the right.
    *   The **Board Background** fills any gaps created by the shift.

3.  **TRANSFORMATION RULE**:
    *   Identify distinct relay assemblies (two colors touching horizontally).
    *   For every row containing the **Upper Plate** color of an assembly, shift all pixels of that color in that row one cell to the left.
    *   For every row containing the **Lower Plate** color of an assembly, shift all pixels of that color in that row one cell to the right.
    *   If a pixel moves out of bounds, it is removed.
    *   The background color ('A') fills the vacated spots.

### Applying the Rule to Test Input

**Relay 1:**
*   Upper Plate: Blue (B) - Rows 1, 2, 3
*   Lower Plate: Red (R) - Row 4
*   Shift B left, Shift R right.
    *   Row 1: `A A B A A B A A A A A A A` $\rightarrow$ `A B A A B A A A A A A A A`
    *   Row 2: `A A B B B B A A A A A A A` $\rightarrow$ `A B B B B A A A A A A A A`
    *   Row 3: `A A B B B B A A A A A A A` $\rightarrow$ `A B B B B A A A A A A A A`
    *   Row 4: `A A R R R R A A A A A A A` $\rightarrow$ `A A A R R R R A A A A A A`

**Relay 2:**
*   Upper Plate: Red (R) - Rows 9, 10
*   Lower Plate: Yellow (Y) - Rows 11, 12
*   Shift R left, Shift Y right.
    *   Row 9: `A A A A A A R R R A A A A` $\rightarrow$ `A A A A A R R R A A A A A`
    *   Row 10: `A A A A A A R R R A A A A` $\rightarrow$ `A A A A A R R R A A A A A`
    *   Row 11: `A A A A A A Y Y Y A A A A` $\rightarrow$ `A A A A A A A Y Y Y A A A`
    *   Row 12: `A A A A A A Y A Y A A A A` $\rightarrow$ `A A A A A A A Y A Y A A A`

**Other Rows:**
*   Rows 0, 5, 6, 7, 8, 13 remain all 'A'.

### Final Grid Construction

Row 0: A A A A A A A A A A A A A
Row 1: A B A A B A A A A A A A A
Row 2: A B B B B A A A A A A A A
Row 3: A B B B B A A A A A A A A
Row 4: A A A R R R R A A A A A A
Row 5: A A A A A A A A A A A A A
Row 6: A A A A A A A A A A A A A
Row 7: A A A A A A A A A A A A A
Row 8: A A A A A A A A A A A A A
Row 9: A A A A A R R R A A A A A
Row 10: A A A A A R R R A A A A A
Row 11: A A A A A A A Y Y Y A A A
Row 12: A A A A A A A Y A Y A A A
Row 13: A A A A A A A A A A A A A

ANSWER:
A A A A A A A A A A A A A
A B A A B A A A A A A A A
A B B B B A A A A A A A A
A B B B B A A A A A A A A
A A A R R R R A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A R R R A A A A A
A A A A A R R R A A A A A
A A A A A A A Y Y Y A A A
A A A A A A A Y A Y A A A
A A A A A A A A A A A A A

### Conceptual Digestion

1.  **CAST**:
    *   **Staff (Background)**: The uniform color filling the grid (Azure 'A' in the test input).
    *   **Composite Tone Clusters**: Groups of colored blocks that are vertically adjacent.
    *   **Upper Register Block**: The top part of a cluster (e.g., Blue 'B' in the first cluster, Red 'R' in the second).
    *   **Lower Register Block**: The bottom part of a cluster (e.g., Red 'R' in the first cluster, Yellow 'Y' in the second).
    *   **Harmonic Boundary**: The horizontal line where the upper and lower colors meet.

2.  **AFFORDANCES**:
    *   The upper block can shift horizontally.
    *   The lower block can shift horizontally.
    *   Vacated spaces are filled with the background color (Staff).

3.  **TRANSFORMATION RULE**:
    *   Identify distinct clusters (contiguous blocks of non-background colors).
    *   Split each cluster into its "Upper Voice" (the color(s) appearing in the top rows of the cluster) and "Lower Voice" (the color(s) appearing in the bottom rows of the cluster).
    *   **Upper Voice $\rightarrow$ Shift Left by 1**: Every cell of the upper color moves one column to the left.
    *   **Lower Voice $\rightarrow$ Shift Right by 1**: Every cell of the lower color moves one column to the right.
    *   If a shift moves a block outside the grid boundaries, it is clipped.
    *   The background color 'A' fills all cells not occupied by the shifted blocks.

### Applying the Rule to Test Input

**Cluster 1:**
*   Upper Voice (Blue 'B'): Rows 1-3, Cols 2-5 (approx).
    *   Row 1: Col 2, Col 5 $\rightarrow$ Shift Left $\rightarrow$ Col 1, Col 4.
    *   Row 2: Col 2, 3, 4, 5 $\rightarrow$ Shift Left $\rightarrow$ Col 1, 2, 3, 4.
    *   Row 3: Col 2, 3, 4, 5 $\rightarrow$ Shift Left $\rightarrow$ Col 1, 2, 3, 4.
*   Lower Voice (Red 'R'): Row 4, Cols 2-5.
    *   Row 4: Col 2, 3, 4, 5 $\rightarrow$ Shift Right $\rightarrow$ Col 3, 4, 5, 6.

**Cluster 2:**
*   Upper Voice (Red 'R'): Rows 9-10, Cols 6-8.
    *   Row 9: Col 6, 7, 8 $\rightarrow$ Shift Left $\rightarrow$ Col 5, 6, 7.
    *   Row 10: Col 6, 7, 8 $\rightarrow$ Shift Left $\rightarrow$ Col 5, 6, 7.
*   Lower Voice (Yellow 'Y'): Rows 11-12, Cols 6-8.
    *   Row 11: Col 6, 7, 8 $\rightarrow$ Shift Right $\rightarrow$ Col 7, 8, 9.
    *   Row 12: Col 6, 8 $\rightarrow$ Shift Right $\rightarrow$ Col 7, 9.

### Final Grid Construction

Row 0: A A A A A A A A A A A A A
Row 1: A B A A B A A A A A A A A
Row 2: A B B B B A A A A A A A A
Row 3: A B B B B A A A A A A A A
Row 4: A A A R R R R A A A A A A
Row 5: A A A A A A A A A A A A A
Row 6: A A A A A A A A A A A A A
Row 7: A A A A A A A A A A A A A
Row 8: A A A A A A A A A A A A A
Row 9: A A A A A R R R A A A A A
Row 10: A A A A A R R R A A A A A
Row 11: A A A A A A A Y Y Y A A A
Row 12: A A A A A A A Y A Y A A A
Row 13: A A A A A A A A A A A A A

ANSWER:
A A A A A A A A A A A A A
A B A A B A A A A A A A A
A B B B B A A A A A A A A
A B B B B A A A A A A A A
A A A R R R R A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A R R R A A A A A
A A A A A R R R A A A A A
A A A A A A A Y Y Y A A A
A A A A A A A Y A Y A A A
A A A A A A A A A A A A A

### Conceptual Digestion

1.  **CAST**:
    *   **Sky (Background)**: The dominant color of the grid (Azure 'A' in the test input).
    *   **Storm Formations**: Clusters of non-background colors.
    *   **Upper Cloud Mass**: The top part of a formation (the color appearing in the upper rows of a cluster).
    *   **Lower Precipitation Mass**: The bottom part of a formation (the color appearing in the lower rows of a cluster).
    *   **Boundary**: The horizontal line where the color changes within a formation.

2.  **AFFORDANCES**:
    *   The **Upper Cloud Mass** can slide one cell to the left.
    *   The **Lower Precipitation Mass** can slide one cell to the right.
    *   **Empty spaces** created by the movement are filled with the background color ('A').
    *   **Overlapping** cells are overwritten by the moving mass.

3.  **TRANSFORMATION RULE**:
    *   Identify distinct "storm" objects (contiguous blocks of non-background colors).
    *   For each object, find the row index where the color changes from the "upper" color to the "lower" color.
    *   Shift all cells of the upper color (and any cells above the boundary within that object) one cell to the left.
    *   Shift all cells of the lower color (and any cells below the boundary within that object) one cell to the right.
    *   Fill the resulting gaps with the background color 'A'.

### Applying the Rule to Test Input

**Formation 1 (Top Left):**
*   Upper Mass (Blue 'B'): Rows 1-3.
*   Lower Mass (Red 'R'): Row 4.
*   Shift 'B's left:
    *   Row 1: `A A B A A B` $\rightarrow$ `A B A A B A`
    *   Row 2: `A A B B B B` $\rightarrow$ `A B B B B A`
    *   Row 3: `A A B B B B` $\rightarrow$ `A B B B B A`
*   Shift 'R's right:
    *   Row 4: `A A R R R R` $\rightarrow$ `A A A R R R R` (Wait, the input is 13 wide. `A A R R R R A...` becomes `A A A R R R R A...`)

**Formation 2 (Bottom Right):**
*   Upper Mass (Red 'R'): Rows 9-10.
*   Lower Mass (Yellow 'Y'): Rows 11-12.
*   Shift 'R's left:
    *   Row 9: `A A A A A A R R R A A A A` $\rightarrow$ `A A A A A R R R A A A A A`
    *   Row 10: `A A A A A A R R R A A A A` $\rightarrow$ `A A A A A R R R A A A A A`
*   Shift 'Y's right:
    *   Row 11: `A A A A A A Y Y Y A A A A` $\rightarrow$ `A A A A A A A Y Y Y A A A A`
    *   Row 12: `A A A A A A Y A Y A A A A` $\rightarrow$ `A A A A A A A Y A Y A A A A` (Wait, shift right: `A A A A A A A Y A Y A A A A` is wrong. Original: `A A A A A A Y A Y A A A A`. Shift right: `A A A A A A A Y A Y A A A`)

**Refining the shifts:**
*   Row 1: `A A B A A B A A A A A A A` $\rightarrow$ `A B A A B A A A A A A A A`
*   Row 2: `A A B B B B A A A A A A A` $\rightarrow$ `A B B B B A A A A A A A A`
*   Row 3: `A A B B B B A A A A A A A` $\rightarrow$ `A B B B B A A A A A A A A`
*   Row 4: `A A R R R R A A A A A A A` $\rightarrow$ `A A A R R R R A A A A A A`
*   Row 9: `A A A A A A R R R A A A A` $\rightarrow$ `A A A A A R R R A A A A A`
*   Row 10: `A A A A A A R R R A A A A` $\rightarrow$ `A A A A A R R R A A A A A`
*   Row 11: `A A A A A A Y Y Y A A A A` $\rightarrow$ `A A A A A A A Y Y Y A A A`
*   Row 12: `A A A A A A Y A Y A A A A` $\rightarrow$ `A A A A A A A Y A Y A A A`

ANSWER:
A A A A A A A A A A A A A
A B A A B A A A A A A A A
A B B B B A A A A A A A A
A B B B B A A A A A A A A
A A A R R R R A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A R R R A A A A A
A A A A A R R R A A A A A
A A A A A A A Y Y Y A A A
A A A A A A A Y A Y A A A
A A A A A A A A A A A A A

### Conceptual Digestion

1.  **CAST**:
    *   **Background Terrain**: The dominant color filling the grid (e.g., Azure 'A' in the test input).
    *   **Formations**: Groups of colored pixels that differ from the background.
    *   **Front Rank**: The upper part of a formation (the color appearing higher in the grid).
    *   **Rear Rank**: The lower part of a formation (the color appearing lower in the grid).
    *   **Battle Line**: The horizontal boundary where the two colors of a formation meet.

2.  **AFFORDANCES**:
    *   The **Front Rank** (upper color) can shift one position to the left.
    *   The **Rear Rank** (lower color) can shift one position to the right.
    *   The **Background Terrain** fills the gaps created by these shifts.

3.  **TRANSFORMATION RULE**:
    *   Identify distinct multi-colored formations.
    *   For each formation:
        *   Identify the top color (Front Rank) and the bottom color (Rear Rank).
        *   Shift all pixels of the top color one cell to the left.
        *   Shift all pixels of the bottom color one cell to the right.
    *   If a formation consists of only one color, it is treated as a single rank (though the examples show formations always come in pairs of colors).
    *   The background color remains constant and fills any vacated cells.

### Applying the Rule to Test Input

**Formation 1 (Top):**
*   Front Rank (Blue 'B'): Rows 1-3.
*   Rear Rank (Red 'R'): Row 4.
*   Shift 'B' left: (1,2)->(1,1), (1,5)->(1,4), (2,2)->(2,1), (2,2-5)->(2,1-4), (3,2)->(3,1), (3,2-5)->(3,1-4).
*   Shift 'R' right: (4,2-5)->(4,3-6).

**Formation 2 (Bottom):**
*   Front Rank (Red 'R'): Rows 9-10.
*   Rear Rank (Yellow 'Y'): Rows 11-12.
*   Shift 'R' left: (9,6-8)->(9,5-7), (10,6-8)->(10,5-7).
*   Shift 'Y' right: (11,6-8)->(11,7-9), (12,6)->(12,7), (12,8)->(12,9).

**Step-by-step Grid Construction:**
*   Row 0: A A A A A A A A A A A A A (No change)
*   Row 1: A B A A B A A A A A A A A (B at 2->1, 5->4)
*   Row 2: A B B B B A A A A A A A A (B at 2-5 -> 1-4)
*   Row 3: A B B B B A A A A A A A A (B at 2-5 -> 1-4)
*   Row 4: A A R R R R A A A A A A A (R at 2-5 -> 3-6)
*   Row 5-8: All A
*   Row 9: A A A A A R R R A A A A (R at 6-8 -> 5-7)
*   Row 10: A A A A A R R R A A A A (R at 6-8 -> 5-7)
*   Row 11: A A A A A A A Y Y Y A A A (Y at 6-8 -> 7-9)
*   Row 12: A A A A A A A Y A Y A A A (Y at 6->7, 8->9)
*   Row 13: All A

ANSWER:
A A A A A A A A A A A A A
A B A A B A A A A A A A A
A B B B B A A A A A A A A
A B B B B A A A A A A A A
A A A R R R R A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A R R R A A A A A
A A A A A R R R A A A A A
A A A A A A A Y Y Y A A A
A A A A A A A Y A Y A A A
A A A A A A A A A A A A A

### Conceptual Digestion

1.  **CAST**:
    *   **Bare Stage**: The background color of the grid (the most frequent color, in this case, Azure 'A').
    *   **Scenic Pieces (Flats)**: Contiguous blocks of non-background colors.
    *   **Upper Panel**: The top half of a scenic piece.
    *   **Lower Panel**: The bottom half of a scenic piece.
    *   **Horizontal Seam**: The dividing line between the upper and lower panels of a flat.

2.  **AFFORDANCES**:
    *   The upper panel can shift one position to the left (stage-left).
    *   The lower panel can shift one position to the right (stage-right).
    *   The background color fills the gaps created by the movement.

3.  **TRANSFORMATION RULE**:
    *   Identify distinct "flats" (connected components of non-background colors).
    *   For each flat, find the horizontal midline (the seam).
    *   Shift all cells above the midline one column to the left.
    *   Shift all cells below the midline one column to the right.
    *   Cells on the midline itself are treated as part of the upper or lower panel based on the split. Looking at Example 1:
        *   Input Y-block (rows 2-5) and R-block (rows 6-7).
        *   Y-block (rows 2, 3, 4, 5): Rows 2-3 shift left, rows 4-5 shift left? No, wait.
        *   Let's re-examine Example 1:
            *   Y-block: Rows 2-5. Output: Rows 2-5 shift left.
            *   R-block: Rows 6-7. Output: Rows 6-7 shift right.
        *   Correction: The "flats" are the distinct colored objects. The "upper panel" is the top color block, and the "lower panel" is the bottom color block.
        *   In Example 1: The Yellow block is the upper panel (shifts left), and the Red block is the lower panel (shifts right).
        *   In the Test Input:
            *   Flat 1: Blue (rows 1-3) and Red (row 4). Blue is upper (left), Red is lower (right).
            *   Flat 2: Red (rows 9-10) and Yellow (rows 11-12). Red is upper (left), Yellow is lower (right).

### Applying the Rule to Test Input

**Flat 1 (Top Left):**
*   Upper (Blue): Rows 1, 2, 3.
    *   Row 1: `A A B A A B A...` $\rightarrow$ `A B A A B A A...`
    *   Row 2: `A A B B B B A...` $\rightarrow$ `A B B B B A A...`
    *   Row 3: `A A B B B B A...` $\rightarrow$ `A B B B B A A...`
*   Lower (Red): Row 4.
    *   Row 4: `A A R R R R A...` $\rightarrow$ `A A A R R R R A...` (Wait, shift right: `A A R R R R A` $\rightarrow$ `A A A R R R R`)

**Flat 2 (Bottom Right):**
*   Upper (Red): Rows 9, 10.
    *   Row 9: `A A A A A A R R R A A A A` $\rightarrow$ `A A A A A R R R A A A A A`
    *   Row 10: `A A A A A A R R R A A A A` $\rightarrow$ `A A A A A R R R A A A A A`
*   Lower (Yellow): Rows 11, 12.
    *   Row 11: `A A A A A A Y Y Y A A A A` $\rightarrow$ `A A A A A A A Y Y Y A A A`
    *   Row 12: `A A A A A A Y A Y A A A A` $\rightarrow$ `A A A A A A A Y A Y A A A`

**Final Grid Construction:**
Row 0: `A A A A A A A A A A A A A`
Row 1: `A B A A B A A A A A A A A`
Row 2: `A B B B B A A A A A A A A`
Row 3: `A B B B B A A A A A A A A`
Row 4: `A A A R R R R A A A A A A`
Row 5: `A A A A A A A A A A A A A`
Row 6: `A A A A A A A A A A A A A`
Row 7: `A A A A A A A A A A A A A`
Row 8: `A A A A A A A A A A A A A`
Row 9: `A A A A A R R R A A A A A`
Row 10: `A A A A A R R R A A A A A`
Row 11: `A A A A A A A Y Y Y A A A`
Row 12: `A A A A A A A Y A Y A A A`
Row 13: `A A A A A A A A A A A A A`

ANSWER:
A A A A A A A A A A A A A
A B A A B A A A A A A A A
A B B B B A A A A A A A A
A B B B B A A A A A A A A
A A A R R R R A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A R R R A A A A A
A A A A A R R R A A A A A
A A A A A A A Y Y Y A A A
A A A A A A A Y A Y A A A
A A A A A A A A A A A A A

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Trellis (Background):** The dominant color filling the grid (e.g., `B` in Example 1, `A` in Example 2 and Test Input).
    *   **Plants:** Contiguous blocks of colors different from the background.
    *   **Foliage (Upper part):** The top section of a plant (the color block appearing higher in the grid).
    *   **Roots (Lower part):** The bottom section of a plant (the color block appearing lower in the grid).
    *   **Graft Line:** The boundary row where the foliage color meets the root color.

2.  **Identify the AFFORDANCES:**
    *   **Foliage:** Can shift one cell to the left.
    *   **Roots:** Can shift one cell to the right.
    *   **Trellis:** Fills in any gaps created by the shifts and remains the background.

3.  **Identify the TRANSFORMATION RULE:**
    *   Locate distinct "plants" (clusters of non-background colors).
    *   A plant consists of an upper color (foliage) and a lower color (roots).
    *   **Rule:**
        *   Every cell of the **upper color** shifts left by 1 column.
        *   Every cell of the **lower color** shifts right by 1 column.
        *   If a shift moves a cell outside the grid boundaries, it is clipped.
        *   The original positions are filled with the background color (Trellis).

### Applying the Rule to Test Input

**Background Color:** `A` (Azure)

**Plant 1:**
*   Foliage (Blue `B`): Rows 1-3.
    *   Row 1: `(1, 2), (1, 5)` $\rightarrow$ shift left $\rightarrow$ `(1, 1), (1, 4)`
    *   Row 2: `(2, 2), (2, 3), (2, 4), (2, 5)` $\rightarrow$ shift left $\rightarrow$ `(2, 1), (2, 2), (2, 3), (2, 4)`
    *   Row 3: `(3, 2), (3, 3), (3, 4), (3, 5)` $\rightarrow$ shift left $\rightarrow$ `(3, 1), (3, 2), (3, 3), (3, 4)`
*   Roots (Red `R`): Row 4.
    *   Row 4: `(4, 2), (4, 3), (4, 4), (4, 5)` $\rightarrow$ shift right $\rightarrow$ `(4, 3), (4, 4), (4, 5), (4, 6)`

**Plant 2:**
*   Foliage (Red `R`): Rows 9-10.
    *   Row 9: `(9, 6), (9, 7), (9, 8)` $\rightarrow$ shift left $\rightarrow$ `(9, 5), (9, 6), (9, 7)`
    *   Row 10: `(10, 6), (10, 7), (10, 8)` $\rightarrow$ shift left $\rightarrow$ `(10, 5), (10, 6), (10, 7)`
*   Roots (Yellow `Y`): Rows 11-12.
    *   Row 11: `(11, 6), (11, 7), (11, 8)` $\rightarrow$ shift right $\rightarrow$ `(11, 7), (11, 8), (11, 9)`
    *   Row 12: `(12, 6), (12, 8)` $\rightarrow$ shift right $\rightarrow$ `(12, 7), (12, 9)`

**Final Grid Construction:**
*   Rows 0, 5, 6, 7, 8, 13: All `A`
*   Row 1: `A B A A B A A A A A A A A`
*   Row 2: `A B B B B A A A A A A A A`
*   Row 3: `A B B B B A A A A A A A A`
*   Row 4: `A A A R R R R A A A A A A`
*   Row 9: `A A A A A R R R A A A A A`
*   Row 10: `A A A A A R R R A A A A A`
*   Row 11: `A A A A A A A Y Y Y A A A`
*   Row 12: `A A A A A A A Y A Y A A A`

ANSWER:
A A A A A A A A A A A A A
A B A A B A A A A A A A A
A B B B B A A A A A A A A
A B B B B A A A A A A A A
A A A R R R R A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A A A A A A A A A
A A A A A R R R A A A A A
A A A A A R R R A A A A A
A A A A A A A Y Y Y A A A
A A A A A A A Y A Y A A A
A A A A A A A A A A A A A

### Conceptual Digestion

1.  **CAST**:
    *   **Black (.)**: The empty space (void).
    *   **Red (R)**: The "cliff" (a reflective wall).
    *   **Azure (A)**: The "sparks" representing the ball's initial trajectory.
    *   **Green (G)**: The "phosphorescent trail" representing the ball's future path.

2.  **AFFORDANCES**:
    *   The **Azure sparks** define a vector (direction and speed).
    *   The **Red cliff** acts as a mirror; when the ball hits it, the perpendicular component of its velocity is reversed.
    *   The **Green trail** marks every cell the ball enters after the last Azure spark.

3.  **TRANSFORMATION RULE**:
    *   **Step 1: Determine Velocity**. Look at the sequence of Azure (A) cells. In the test input, A is at (0, 3) and (1, 4). The movement is $\Delta r = (+1, +1)$ per step.
    *   **Step 2: Project Forward**. Continue the path from the last A (1, 4) using the vector $(+1, +1)$.
    *   **Step 3: Handle Collision**. The path is: (2, 5), (3, 6), (4, 7), (5, 8).
    *   **Step 4: Reflection**. At (5, 8), the next step would be (6, 9). However, the Red cliff (R) starts at column 8. The ball hits the wall at the boundary between column 7 and 8.
    *   Wait, let's re-examine the grid:
        Columns 0-7 are black (.), Columns 8-11 are red (R).
        Last A is at (1, 4).
        Path:
        (2, 5) -> G
        (3, 6) -> G
        (4, 7) -> G
        At (5, 8), it hits the Red wall.
        The reflection rule: Angle of incidence = Angle of reflection.
        Incoming vector: $(+1, +1)$.
        Wall is vertical (x-axis constant).
        Reflected vector: $(+1, -1)$.
    *   **Step 5: Continue Path**. From the point of impact (approx 5, 7.5), the ball moves in direction $(+1, -1)$.
        (5, 7) is where it bounces? No, let's look at the training examples.
        Example 2: A at (10, 3), (11, 2). Vector $(-1, -1)$. Hits wall at col 10. Bounces to $(+1, -1)$? No, it bounces back to the left.
        Let's re-evaluate Example 2:
        A: (10, 3), (11, 2). Vector is $\Delta r = (-1, -1)$ per step (moving up and left).
        Wait, the A's are at (10, 3) and (11, 2). The ball is moving from (11, 2) to (10, 3).
        Vector: $\Delta r = (-1, +1)$.
        Path: (9, 4), (8, 5), (7, 6), (6, 7), (5, 8), (4, 9).
        At (4, 9), it hits the wall at col 10.
        Reflected vector: $(-1, -1)$.
        Path: (5, 8), (6, 7), (7, 6), (8, 5), (9, 4).
        Wait, the output for Example 2 shows G at (4, 9), then (5, 8), (6, 7)...
        Let's trace Example 2 again:
        A: (11, 2), (10, 3). Vector: $(-1, 1)$.
        G: (9, 4), (8, 5), (7, 6), (6, 7), (5, 8), (4, 9) [Hits wall], (3, 8), (2, 7), (1, 6), (0, 5).
        This matches the output!

    *   **Applying to Test Input**:
        A: (0, 3), (1, 4). Vector: $(+1, +1)$.
        Path:
        (2, 5) -> G
        (3, 6) -> G
        (4, 7) -> G
        (5, 8) -> Hits wall (R is at col 8).
        Wait, the wall is at col 8. The ball is at (4, 7). The next step is (5, 8).
        Since (5, 8) is Red, it bounces.
        The reflection of $(+1, +1)$ off a vertical wall is $(+1, -1)$.
        The bounce occurs at the boundary. The cell (5, 8) is not entered; it reflects back into the black area.
        The path becomes:
        (2, 5) G
        (3, 6) G
        (4, 7) G
        (5, 6) G
        (6, 5) G
        (7, 4) G
        (8, 3) G
        (9, 2) G
        (10, 1) G
        (11, 0) G

### Final Grid Calculation
Input:
(0, 3)=A, (1, 4)=A
Wall: Col 8, 9, 10, 11.
Path:
(2, 5) G
(3, 6) G
(4, 7) G
(5, 6) G
(6, 5) G
(7, 4) G
(8, 3) G
(9, 2) G
(10, 1) G
(11, 0) G

ANSWER:
. . . A . . . . R R R R
. . . . A . . . R R R R
. . . . . G . . R R R R
. . . . . . G . R R R R
. . . . . . . G R R R R
. . . . . . G . R R R R
. . . . . G . . R R R R
. . . . G . . . R R R R
. . . G . . . . R R R R
. . G . . . . . R R R R
. G . . . . . . R R R R
G . . . . . . . R R R R

### Conceptual Digestion

1.  **CAST**:
    *   **Azure (A)**: The initial wavefront. It defines the starting position and the direction of travel.
    *   **Red (R)**: The magma body (a solid wall). It acts as a reflective boundary.
    *   **Black (.)**: Stable rock (empty space) through which the wave travels.
    *   **Green (G)**: The seismic trace. It marks the path the wavefront takes as it moves and reflects.

2.  **AFFORDANCES**:
    *   The wavefront (A) moves diagonally. In Example 1, the wavefront is at (2,0), (3,1), (4,2). The leading edge is (4,2), moving in the direction (+1, +1).
    *   The magma body (R) reflects the wave. If the wave hits a vertical wall, the horizontal component of its velocity reverses. If it hits a horizontal wall, the vertical component reverses.

3.  **TRANSFORMATION RULE**:
    *   **Step 1: Propagation**. Extend the wavefront from its leading edge in its current direction.
    *   **Step 2: Reflection**. When the wavefront hits the red block, it reflects.
    *   **Step 3: Marking**. Every cell the wavefront passes through (excluding the original A cells and the R cells) is colored Green (G).
    *   **Test Input Analysis**:
        *   Wavefront (A) is at (0,3) and (1,4).
        *   Leading edge is (1,4). Direction is (+1, +1).
        *   Path: (2,5) $\rightarrow$ (3,6) $\rightarrow$ (4,7) $\rightarrow$ (5,8).
        *   At (5,8), the wave hits the red wall (which starts at column 8).
        *   The wall is vertical. The horizontal direction (+1) reverses to (-1). The vertical direction (+1) remains.
        *   Reflected Path: (6,7) $\rightarrow$ (7,6) $\rightarrow$ (8,5) $\rightarrow$ (9,4) $\rightarrow$ (10,3) $\rightarrow$ (11,2).

### Step-by-Step Application

1.  **Initial A positions**: (0,3), (1,4).
2.  **Forward Path (G)**:
    *   (2,5)
    *   (3,6)
    *   (4,7)
    *   (5,8) - *Wait, the red wall is at column 8. The wave hits the wall at (5,8). Since the wall is at index 8, the wave reflects at the boundary between 7 and 8.*
    *   Let's re-evaluate: The wave is at (1,4). It moves to (2,5), (3,6), (4,7).
    *   At (5,8), it hits the red block. The reflection happens.
    *   The path is: (2,5), (3,6), (4,7).
    *   Reflection at (5,8) is not possible because (5,8) is Red. The reflection occurs at the boundary.
    *   The wave reaches (4,7). The next step would be (5,8), but (5,8) is Red.
    *   The reflection reverses the x-direction: (+1, +1) becomes (-1, +1).
    *   From (4,7), the next reflected point is (5,6), then (6,5), (7,4), (8,3), (9,2), (10,1), (11,0).

Let's double check Example 1:
A at (2,0), (3,1), (4,2). Leading edge (4,2). Direction (+1, +1).
Hits R at row 9.
Path: (5,3), (6,4), (7,5), (8,6).
At row 9, it reflects. Vertical direction (+1) becomes (-1).
Reflected path: (8,7), (7,8), (6,9), (5,10), (4,11).
Looking at Example 1 output:
(5,3)G, (6,4)G, (7,5)G, (8,6)G.
(8,7)G, (7,8)G, (6,9)G, (5,10)G, (4,11)G.
This matches perfectly.

Applying to Test Input:
A at (0,3), (1,4). Leading edge (1,4). Direction (+1, +1).
Red wall at column 8.
Forward path: (2,5), (3,6), (4,7).
At (5,8), it hits the wall. Horizontal direction (+1) becomes (-1).
Reflected path: (5,6), (6,5), (7,4), (8,3), (9,2), (10,1), (11,0).

ANSWER:
. . . A . . . . R R R R
. . . . A . . . R R R R
. . . . . G . . R R R R
. . . . . . G . R R R R
. . . . . . . G R R R R
. . . . . G . . R R R R
. . . . G . . . R R R R
. . . G . . . . R R R R
. . G . . . . . R R R R
. G . . . . . . R R R R
G . . . . . . . R R R R
. . . . . . . . R R R R
Wait, I miscounted the reflection start.
Forward: (2,5), (3,6), (4,7).
Reflection: (5,6), (6,5), (7,4), (8,3), (9,2), (10,1), (11,0).
Let's re-verify the coordinates.
(0,3) A
(1,4) A
(2,5) G
(3,6) G
(4,7) G
(5,6) G
(6,5) G
(7,4) G
(8,3) G
(9,2) G
(10,1) G
(11,0) G

ANSWER:
. . . A . . . . R R R R
. . . . A . . . R R R R
. . . . . G . . R R R R
. . . . . . G . R R R R
. . . . . . . G R R R R
. . . . . G . . R R R R
. . . . G . . . R R R R
. . . G . . . . R R R R
. . G . . . . . R R R R
. G . . . . . . R R R R
G . . . . . . . R R R R
. . . . . . . . R R R R
Wait, (11,0) is row 11, col 0.
Row 0: . . . A . . . . R R R R
Row 1: . . . . A . . . R R R R
Row 2: . . . . . G . . R R R R
Row 3: . . . . . . G . R R R R
Row 4: . . . . . . . G R R R R
Row 5: . . . . . G . . R R R R
Row 6: . . . . G . . . R R R R
Row 7: . . . G . . . . R R R R
Row 8: . . G . . . . . R R R R
Row 9: . G . . . . . . R R R R
Row 10: G . . . . . . . R R R R
Row 11: . . . . . . . . R R R R
Wait, (11,0) is row 11, col 0.
Row 11: G . . . . . . . R R R R
Let's re-count.
(5,6) is Row 5, Col 6.
(6,5) is Row 6, Col 5.
(7,4) is Row 7, Col 4.
(8,3) is Row 8, Col 3.
(9,2) is Row 9, Col 2.
(10,1) is Row 10, Col 1.
(11,0) is Row 11, Col 0.

Corrected Grid:
Row 0: . . . A . . . . R R R R
Row 1: . . . . A . . . R R R R
Row 2: . . . . . G . . R R R R
Row 3: . . . . . . G . R R R R
Row 4: . . . . . . . G R R R R
Row 5: . . . . . G . . R R R R
Row 6: . . . . G . . . R R R R
Row 7: . . . G . . 

... [2350 chars truncated]

### Conceptual Digestion

1.  **CAST**:
    *   **Azure (A)**: The "glaze" or the starting point/direction of the drizzle.
    *   **Red (R)**: The "rim" or the reflective boundary.
    *   **Black (.)**: The "pan surface" or empty space.
    *   **Green (G)**: The "caramelized trail" created by the movement of the glaze.

2.  **AFFORDANCES**:
    *   The Azure pixels define a vector (direction and slope).
    *   The Red pixels act as a mirror/wall that reflects the path of the vector.
    *   The Green pixels fill the path from the end of the Azure line, to the Red wall, and then reflect away.

3.  **TRANSFORMATION RULE**:
    *   **Step 1: Determine Direction**: Find the slope of the Azure (A) line. In the test input, A is at (0, 3) and (1, 4). The slope is $\Delta y=1, \Delta x=1$ (down-right).
    *   **Step 2: Project Forward**: Continue this path (G) from the last A (1, 4) until it hits the Red (R) boundary.
        *   (2, 5), (3, 6), (4, 7), (5, 8) $\rightarrow$ hits R at (5, 8) because R starts at column 8.
    *   **Step 3: Reflect**: Upon hitting the vertical Red wall at column 8, the x-direction reverses ($\Delta x = -1$), while the y-direction remains the same ($\Delta y = 1$).
        *   The path reflects from (5, 8) to (6, 7), (7, 6), (8, 5), (9, 4), (10, 3), (11, 2).
    *   **Step 4: Finalize**: Keep original A and R pixels. Fill the calculated path with G.

### Applying the Rule to Test Input

*   **Azure points**: (0, 3), (1, 4)
*   **Red boundary**: Columns 8, 9, 10, 11 (all rows)
*   **Forward Path (G)**:
    *   (2, 5)
    *   (3, 6)
    *   (4, 7)
    *   (5, 8) $\rightarrow$ This is the collision point. Since the rim is a block, the "hit" happens at the boundary. Let's check Example 1: A is at (2,0), (3,1), (4,2). Path goes to (5,3), (6,4), (7,5), (8,6) then hits R at row 9. It reflects to (8,7), (7,8), (6,9), (5,10), (4,11).
    *   In our test case, the wall is vertical.
    *   Path: (2, 5), (3, 6), (4, 7).
    *   Collision at (5, 8) - but (5, 8) is Red. The trail stops *before* the red or *at* the red? In Example 1, the G trail does not replace R.
    *   Reflection: From (4, 7), the next step would be (5, 8) [Red]. The reflection angle is $\Delta y=1, \Delta x=-1$.
    *   Reflected Path: (5, 7), (6, 6), (7, 5), (8, 4), (9, 3), (10, 2), (11, 1).

Wait, let's re-verify the reflection point.
Example 1: A(2,0), A(3,1), A(4,2). Path: (5,3), (6,4), (7,5), (8,6). Then it hits R at (9,6). The reflection starts from (8,6) and goes to (8,7), (7,8), (6,9), (5,10), (4,11).
So the point of contact is the "pivot".
Test Input: A(0,3), A(1,4).
Path: (2,5), (3,6), (4,7).
Pivot: (5,8) is Red.
Reflected path starts from (4,7) and moves $\Delta y=1, \Delta x=-1$:
(5,6), (6,5), (7,4), (8,3), (9,2), (10,1), (11,0).

Let's double check the coordinates:
Row 0: . . . A . . . . R R R R
Row 1: . . . . A . . . R R R R
Row 2: . . . . . G . . R R R R
Row 3: . . . . . . G . R R R R
Row 4: . . . . . . . G R R R R
Row 5: . . . . . . G . R R R R
Row 6: . . . . . G . . R R R R
Row 7: . . . . G . . . R R R R
Row 8: . . . G . . . . R R R R
Row 9: . . G . . . . . R R R R
Row 10: . G . . . . . . R R R R
Row 11: G . . . . . . . R R R R

ANSWER:
. . . A . . . . R R R R
. . . . A . . . R R R R
. . . . . G . . R R R R
. . . . . . G . R R R R
. . . . . . . G R R R R
. . . . . . G . R R R R
. . . . . G . . R R R R
. . . . G . . . R R R R
. . . G . . . . R R R R
. . G . . . . . R R R R
. G . . . . . . R R R R
G . . . . . . . R R R R

### Conceptual Digestion

1.  **CAST**:
    *   **Input Grid**: A "workshop table" (background color `O` = orange) containing scattered "doll pieces" (blocks of other colors).
    *   **Doll Pieces**: Contiguous blocks of the same color.
    *   **Weight/Size**: The "heaviness" of a doll is determined by the total number of pixels (count) of that color in the input grid.
    *   **Output Grid**: A square grid representing concentric shells of colors.

2.  **AFFORDANCES**:
    *   The colors can be counted to determine their relative size.
    *   The colors can be ordered from largest (heaviest) to smallest (lightest).
    *   The output size is determined by the number of unique colors present (excluding the background `O`). If there are $N$ unique colors, the output is a square of size $(2N-1) \times (2N-1)$.

3.  **TRANSFORMATION RULE**:
    *   **Step 1**: Identify all unique colors in the input grid, excluding the background color `O`.
    *   **Step 2**: Count the number of pixels for each of these colors.
    *   **Step 3**: Sort the colors by count in descending order (Heaviest $\rightarrow$ Lightest).
    *   **Step 4**: Construct a square grid where the heaviest color forms the outermost perimeter, the second heaviest forms the next inner perimeter, and so on, until the lightest color occupies the center pixel.

---

### Applying the Rule to Test Input

**1. Identify and Count Colors (excluding `O`):**
*   `W` (maroon): 2 pixels
*   `R` (red): 4 pixels
*   `X` (grey): 4 pixels
*   `A` (azure): 1 pixel
*   `.` (black): 4 pixels
*   `B` (blue): 8 pixels

**2. Sort by Count (Descending):**
1.  `B` (blue): 8
2.  `R` (red): 4 (Tie - order usually follows appearance or specific logic, but let's check others)
3.  `X` (grey): 4
4.  `.` (black): 4
5.  `W` (maroon): 2
6.  `A` (azure): 1

*Wait, let's re-count carefully:*
*   `W`: (0,2), (1,2) $\rightarrow$ 2
*   `R`: (0,5), (0,6), (0,7), (1,5), (1,7) $\rightarrow$ 5
*   `X`: (2,0), (3,0), (3,1), (4,0) $\rightarrow$ 4
*   `A`: (3,3) $\rightarrow$ 1
*   `.`: (4,5), (5,5), (5,6), (6,6), (7,6), (7,7) $\rightarrow$ 6 (Wait, let me re-read the grid)

*Re-counting Test Input Grid:*
Row 0: O O W O O R R R (W:1, R:3)
Row 1: O O W O O R O R (W:1, R:2)
Row 2: X O O O O O O O (X:1)
Row 3: X X O A O O O O (X:2, A:1)
Row 4: X O O O O . O O (X:1, .:1)
Row 5: O O B O O . . O (B:1, .:2)
Row 6: O B B B O O . . (B:3, .:2)
Row 7: B B B B B O O . (B:5, .:1)

**Totals:**
*   `B` (blue): 1 + 3 + 5 = 9
*   `.` (black): 1 + 2 + 2 + 1 = 6
*   `R` (red): 3 + 2 = 5
*   `X` (grey): 1 + 2 + 1 = 4
*   `W` (maroon): 1 + 1 = 2
*   `A` (azure): 1 = 1

**3. Sorted Order (Heaviest to Lightest):**
1. `B` (9)
2. `.` (6)
3. `R` (5)
4. `X` (4)
5. `W` (2)
6. `A` (1)

**4. Construct Output Grid:**
Number of colors = 6. Output size = $(2 \times 6) - 1 = 11 \times 11$.
*   Layer 1 (Outer): `B`
*   Layer 2: `.`
*   Layer 3: `R`
*   Layer 4: `X`
*   Layer 5: `W`
*   Layer 6 (Center): `A`

ANSWER:
B B B B B B B B B B B
B . . . . . . . . . B
B . R R R R R R R . B
B . R X X X X X R . B
B . R X W W W X R . B
B . R X W A W X R . B
B . R X W W W X R . B
B . R X X X X X R . B
B . R R R R R R R . B
B . . . . . . . . . B
B B B B B B B B B B B

### Conceptual Digestion

**1. Identify the CAST:**
*   **Species (Colors):** Each unique color (excluding the background color `O` = orange) represents a different species of colony.
*   **Abundance (Count):** The number of cells occupied by each color determines its rank.
*   **Organism (Output):** A concentric, symmetric square/rectangular structure where layers are determined by abundance.

**2. Identify the AFFORDANCES:**
*   The background color `O` is discarded.
*   The counts of the remaining colors determine the order of the layers.
*   The most abundant color forms the outermost boundary (cell wall).
*   The next most abundant forms the next inner layer, and so on.
*   The least abundant color forms the center (nucleus).

**3. Identify the TRANSFORMATION RULE:**
1.  **Count** the occurrences of each non-`O` color in the input grid.
2.  **Rank** the colors from most frequent to least frequent.
3.  **Determine the size** of the output grid. The total number of non-`O` cells is not the driver; rather, the output is a square/rectangle that can fit the layers. Looking at Example 2, the output size is determined by the smallest square/rectangle that can encapsulate the layers based on the number of distinct species. Specifically, if there are $N$ species, the output is a $(2N-1) \times (2N-1)$ square.
    *   Example 1: 2 species (Y, M) $\rightarrow$ $3 \times 3$ grid.
    *   Example 2: 3 species (B, A, X) $\rightarrow$ $5 \times 5$ grid.
    *   Test Input: 5 species (B, R, X, W, A, .) $\rightarrow$ Wait, let's count the colors in the test input.

**Test Input Analysis:**
*   `B` (Blue): (5,2), (6,1), (6,2), (6,3), (7,0), (7,1), (7,2), (7,3), (7,4) $\rightarrow$ Count: 9
*   `R` (Red): (0,5), (0,6), (0,7), (1,5), (1,7) $\rightarrow$ Count: 5
*   `X` (Grey): (2,0), (3,0), (3,1), (4,0) $\rightarrow$ Count: 4
*   `.` (Black): (4,5), (5,5), (5,6), (6,6), (7,6), (7,7) $\rightarrow$ Count: 6
*   `W` (Maroon): (0,2), (1,2) $\rightarrow$ Count: 2
*   `A` (Azure): (3,3) $\rightarrow$ Count: 1

**Ranking (Most to Least):**
1.  `B` (9) - Outermost
2.  `.` (6) - Layer 2
3.  `R` (5) - Layer 3
4.  `X` (4) - Layer 4
5.  `W` (2) - Layer 5
6.  `A` (1) - Nucleus

**Output Grid Size:**
With 6 species, the output grid will be $(2 \times 6 - 1) \times (2 \times 6 - 1) = 11 \times 11$.

**Layer Construction:**
*   Layer 1 (Edge): `B`
*   Layer 2: `.`
*   Layer 3: `R`
*   Layer 4: `X`
*   Layer 5: `W`
*   Layer 6 (Center): `A`

ANSWER:
B B B B B B B B B B B
B . . . . . . . . . B
B . R R R R R R R . B
B . R X X X X X R . B
B . R X W W W X R . B
B . R X W A W X R . B
B . R X W W W X R . B
B . R X X X X X R . B
B . R R R R R R R . B
B . . . . . . . . . B
B B B B B B B B B B B

### Conceptual Digestion

1.  **CAST**:
    *   **Debris**: Colored cells (non-orange `O` cells).
    *   **Ocean**: Orange cells (`O`), which act as the background and are discarded in the final output.
    *   **Reef**: The final output grid, consisting of concentric square rings of colors.

2.  **AFFORDANCES**:
    *   Each color's frequency (count of cells) determines its position in the concentric structure.
    *   The most frequent color forms the outermost ring.
    *   The least frequent color forms the center.

3.  **TRANSFORMATION RULE**:
    *   **Step 1: Count frequencies.** Count every cell that is not orange (`O`).
    *   **Step 2: Rank colors.** Order colors from most frequent to least frequent.
    *   **Step 3: Determine grid size.** The total number of non-orange cells determines the size of the square reef. However, looking at the examples, the output is always a square with a side length $N$ such that $N^2$ is the total number of non-orange cells (or the smallest square that can contain them in concentric rings).
    *   **Step 4: Construct rings.**
        *   The most frequent color fills the perimeter of the $N \times N$ square.
        *   The next most frequent color fills the perimeter of the $(N-2) \times (N-2)$ square inside it.
        *   This continues until the rarest color is placed in the center.

### Applying the Rule to Test Input

**1. Count Frequencies (excluding 'O'):**
*   `B` (Blue): (5,2), (6,1), (6,2), (6,3), (7,0), (7,1), (7,2), (7,3), (7,4) $\rightarrow$ **9 cells**
*   `X` (Grey): (2,0), (3,0), (3,1), (4,0) $\rightarrow$ **4 cells**
*   `R` (Red): (0,5), (0,6), (0,7), (1,5), (1,7) $\rightarrow$ **5 cells**
*   `W` (Maroon): (0,2), (1,2) $\rightarrow$ **2 cells**
*   `A` (Azure): (3,3) $\rightarrow$ **1 cell**
*   `.` (Black): (4,5), (5,5), (5,6), (6,6), (7,6), (7,7) $\rightarrow$ **6 cells**

**Total count:** $9 + 4 + 5 + 2 + 1 + 6 = 27$.
Wait, let me re-count carefully.
Input:
Row 0: O O W O O R R R (W:1, R:3)
Row 1: O O W O O R O R (W:1, R:2)
Row 2: X O O O O O O O (X:1)
Row 3: X X O A O O O O (X:2, A:1)
Row 4: X O O O O . O O (X:1, .:1)
Row 5: O O B O O . . O (B:1, .:2)
Row 6: O B B B O O . . (B:3, .:2)
Row 7: B B B B B O O . (B:5, .:1)

Recount:
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 = 1
Total: $9 + 6 + 5 + 4 + 2 + 1 = 27$.

Wait, the output must be a square. $27$ is not a perfect square. Let's re-examine the examples.
Example 1: Input colors: Y(4), M(3). Total = 7. Output: $3 \times 3$ (9 cells).
Example 2: Input colors: B(8), A(5), X(4). Total = 17. Output: $5 \times 5$ (25 cells).
Example 3: Input colors: A(8), G(8), W(1), .(4). Total = 21. Output: $7 \times 7$ (49 cells).

Observation: The output size is the smallest odd square $N \times N$ that can accommodate the colors in concentric rings. The number of rings is $\lceil N/2 \rceil$.
The number of distinct colors in the test input is 6 (B, ., R, X, W, A).
To have 6 concentric rings, the outermost ring is ring 1, then 2, 3, 4, 5, and the 6th is the center.
A square with 6 rings (where the 6th is a $1 \times 1$ center) must have a side length of $2(6-1) + 1 = 11$.

**Ranking Colors (Most to Least):**
1. B (9)
2. . (6)
3. R (5)
4. X (4)
5. W (2)
6. A (1)

**Constructing the $11 \times 11$ Grid:**
Ring 1 (Outermost, $11 \times 11$): B
Ring 2 ($9 \times 9$): .
Ring 3 ($7 \times 7$): R
Ring 4 ($5 \times 5$): X
Ring 5 ($3 \times 3$): W
Ring 6 (Center, $1 \times 1$): A

ANSWER:
B B B B B B B B B B B
B . . . . . . . . . B
B . R R R R R R R . B
B . R X X X X X R . B
B . R X W W W X R . B
B . R X W A W X R . B
B . R X W W W X R . B
B . R X X X X X R . B
B . R R R R R R R . B
B . . . . . . . . . B
B B B B B B B B B B B

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Dancers/Companies:** Groups of connected or identical colored pixels.
    *   **Company Size:** The total count of pixels of a specific color.
    *   **The Stage:** The final output grid, which is a square formation.

2.  **Identify the AFFORDANCES:**
    *   Each color group can be counted to determine its rank (largest to smallest).
    *   The colors are mapped to concentric rings of a square.

3.  **Identify the TRANSFORMATION RULE:**
    *   **Step 1: Count the occurrences of each color** (excluding the background/empty space if applicable, though here we count all distinct colors present).
    *   **Step 2: Rank the colors** by frequency (count) from largest to smallest.
    *   **Step 3: Determine the output grid size.** The total number of pixels in the output is the sum of the counts of the colors used. Since it's a square formation, the size is $\sqrt{\text{Total Pixels}} \times \sqrt{\text{Total Pixels}}$.
    *   **Step 4: Fill the square.**
        *   The most frequent color forms the outermost ring.
        *   The second most frequent color forms the next ring inside.
        *   This continues until the least frequent color (a single pixel) occupies the center.

### Applying the Rule to Test Input

**1. Count the colors:**
*   `O` (Orange): 31
*   `B` (Blue): 8
*   `X` (Grey): 4
*   `R` (Red): 4
*   `W` (Maroon): 2
*   `A` (Azure): 1
*   `.` (Black): 4

**Wait, let's re-count carefully:**
Input Grid (8x8):
Row 0: O O W O O R R R (O:4, W:1, R:3)
Row 1: O O W O O R O R (O:5, W:1, R:2)
Row 2: X O O O O O O O (X:1, O:7)
Row 3: X X O A O O O O (X:2, O:5, A:1)
Row 4: X O O O O . O O (X:1, O:6, .:1)
Row 5: O O B O O . . O (O:5, B:1, .:2)
Row 6: O B B B O O . . (O:3, B:3, .:2)
Row 7: B B B B B O O . (B:5, O:2, .:1)

**Totals:**
*   `O`: 4 + 5 + 7 + 5 + 6 + 5 + 3 + 2 = 37
*   `B`: 0 + 0 + 0 + 0 + 0 + 1 + 3 + 5 = 9
*   `X`: 0 + 0 + 1 + 2 + 1 + 0 + 0 + 0 = 4
*   `R`: 3 + 2 + 0 + 0 + 0 + 0 + 0 + 0 = 5
*   `W`: 1 + 1 + 0 + 0 + 0 + 0 + 0 + 0 = 2
*   `.`: 0 + 0 + 0 + 0 + 1 + 2 + 2 + 1 = 6
*   `A`: 0 + 0 + 0 + 1 + 0 + 0 + 0 + 0 = 1

**Total pixels = 37 + 9 + 4 + 5 + 2 + 6 + 1 = 64.**
The output grid is $8 \times 8$.

**2. Rank the colors (Largest to Smallest):**
1.  `O` (37) - Wait, the rule says "the largest company lines the outer edge".
    Let's check the math for a $8 \times 8$ square:
    *   Outer ring (8x8): $8+8+6+6 = 28$ pixels.
    *   Second ring (6x6): $6+6+4+4 = 20$ pixels.
    *   Third ring (4x4): $4+4+2+2 = 12$ pixels.
    *   Center (2x2): $2+2+0+0 = 4$ pixels.
    Total = $28+20+12+4 = 64$.

**Wait, the counts don't match the ring sizes perfectly.** Let's re-read: "the largest company lines the outer edge, the next largest the ring inside... until the smallest company — a single dancer — takes the center position."
This implies the output size is determined by the number of colors/companies, and the rings are filled by the colors in order of size.

Let's re-examine Example 1:
Input: O(18), Y(4), M(3). Total = 25.
Output: 3x3.
Outer ring (3x3): 8 pixels.
Center: 1 pixel.
Wait, Example 1 output is:
Y Y Y
Y M Y
Y Y Y
Here, Y is the outer ring (8 pixels) and M is the center (1 pixel).
Input counts for Example 1: O=18, Y=4, M=3.
The output uses Y and M. O is ignored?
Looking at Example 1 again:
Input:
O O O O O O
O O O Y Y O
O O O Y Y O
M O O O O O
M M O O O O
O O O O O O
Counts: O=25, Y=4, M=3.
Output:
Y Y Y
Y M Y
Y Y Y
The output is 3x3. The colors used are Y (8 pixels) and M (1 pixel).
Wait, the output uses Y for the outer ring and M for the center.
Y was the 2nd largest (4), M was the 3rd largest (3).
The largest (O=25) was the background? No, O is a color.
Maybe the background is the most frequent color?
In Example 1: O is the most frequent. It is not in the output.
The remaining colors are Y(4) and M(3).
The output is a 3x3 square.
Outer ring (8 pixels) = Y.
Center (1 pixel) = M.
Wait, Y only had 4 pixels, but it fills 8. M only had 3, but it fills 1.
The rule is: **The colors are used in order of size to fill the rings, regardless of their original count.**

**Applying this to Test Input:**
Colors: O(37), B(9), .(6), R(5), X(4), W(2), A(1).
Most frequent (O) is background.
Remaining colors in order: B, ., R, X, W, A.
Number of colors remaining = 6.
A square with 6 concentric rings (including center) would be $11 \times 11$.
But the total number of pixels in the input is 64.
If the output is a square, and we use the remaining colors:
B (Outer), . (Ring 2), R (Ring 3), X (Ring 4), W (Ring 5), A (Center).
A square with 6 layers (including center) is $11 \times 11$.
Wait, if the center is a single pixel (1x1), the rings are:
1x1 (Center)
3x3 (Ring 2)
5x5 (Ring 3)
7x7 (Ring 4)
9x9 (Ring 5)
11x11 (Ring 6)
This matches the 6 colors!

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

**Wait, let's check the output size again.**
Example 1: Input 

... [861 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Input Grid:** A field of colors representing "mineral deposits."
    *   **Colors:** Each unique color (excluding the background/filler, though the metaphor suggests all colors present are minerals) is a mineral.
    *   **Abundance:** The number of times a color appears in the input grid.

2.  **Identify the AFFORDANCES:**
    *   The colors can be counted to determine a hierarchy of abundance.
    *   The output is a square "geode" (concentric squares).
    *   The size of the output square is determined by the number of unique colors present (since each color forms a band or the center point).

3.  **Identify the TRANSFORMATION RULE:**
    *   **Step 1: Count the frequency** of every unique color in the input grid.
    *   **Step 2: Rank the colors** from most abundant to least abundant.
    *   **Step 3: Determine output size.** If there are $N$ unique colors, the output is an $N \times N$ square (if $N$ is even, the center is a $2 \times 2$ block; if $N$ is odd, the center is a $1 \times 1$ block). Wait, looking at Example 1: 3 colors (O, Y, M). Output is $3 \times 3$. Example 2: 4 colors (O, A, X, B). Output is $5 \times 5$.
    *   **Correction on Size:** The output size is $2k-1$ where $k$ is the number of unique colors. This allows for a single center point.
    *   **Step 4: Map colors to bands.**
        *   Most abundant $\rightarrow$ Outermost shell.
        *   2nd most abundant $\rightarrow$ Next inner band.
        *   ...
        *   Least abundant $\rightarrow$ Center point.
    *   **Wait, let's re-verify Example 1:**
        Input: O (23), Y (4), M (3).
        Colors: O, Y, M.
        Output: $3 \times 3$.
        Shell: Y, Center: M.
        Wait, the output is Y Y Y / Y M Y / Y Y Y.
        The most abundant (O) is ignored? Let's check Example 2.
    *   **Re-verify Example 2:**
        Input: O (45), B (8), A (4), X (4). (Wait, A and X are tied).
        Output: $5 \times 5$.
        Shell: B, Next: A, Center: X.
        Most abundant (O) is ignored.
    *   **Revised Rule:**
        1. Count all colors.
        2. Ignore the most abundant color (the "sediment/background").
        3. Rank the remaining colors by abundance (descending).
        4. The number of remaining colors $k$ determines the output size: $(2k-1) \times (2k-1)$.
        5. Most abundant of the remainder $\rightarrow$ Outer shell, and so on, until the rarest $\rightarrow$ center.

### Applying to Test Input

**1. Count Frequencies:**
*   O (orange): 36
*   B (blue): 9
*   R (red): 4
*   X (grey): 4
*   . (black): 4
*   W (maroon): 2
*   A (azure): 1

**2. Ignore most abundant (O):**
Remaining: B (9), R (4), X (4), . (4), W (2), A (1).

**3. Rank remaining colors:**
1. B (9)
2. R (4) / X (4) / . (4) - (Tied)
3. W (2)
4. A (1)

*Handling Ties:* In Example 2, A and X were tied (4 each). The output had A as the band and X as the center. This suggests that if there is a tie, the order they appear in the input or a specific priority is used. However, looking at Example 2 again: B(8), A(4), X(4). Output: B shell, A band, X center. A appears before X in the input grid (reading top-to-bottom, left-to-right).

*Test Input Ties:* R, X, . all have 4.
Order of appearance:
R (row 0, col 5)
X (row 2, col 0)
. (row 4, col 5)
So the order is R, then X, then .

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

**5. Determine Output Size:**
$k = 6$ colors.
Size = $(2 \times 6) - 1 = 11 \times 11$.

**6. Construct Geode:**
*   Shell 1 (Outer): B
*   Shell 2: R
*   Shell 3: X
*   Shell 4: .
*   Shell 5: W
*   Shell 6 (Center): A

**Wait, let me re-count the colors in the test input carefully.**
O: 36
B: 9 (row 5 col 2; row 6 col 1,2,3; row 7 col 0,1,2,3,4) -> 1+3+5 = 9. Correct.
R: 4 (row 0 col 5,6,7; row 1 col 5,7) -> 3+2 = 5. (Wait, R is 5).
X: 4 (row 2 col 0; row 3 col 0,1; row 4 col 0) -> 1+2+1 = 4. Correct.
.: 6 (row 4 col 5; row 5 col 5,6; row 6 col 5,6; row 7 col 6,7) -> 1+2+2+2 = 7. (Wait, . is 7).
W: 2 (row 0 col 2; row 1 col 2) -> 2. Correct.
A: 1 (row 3 col 3) -> 1. Correct.

**Recounting again:**
O: 36
B: 9
.: 7
R: 5
X: 4
W: 2
A: 1

**Ranking:**
1. B (9)
2. . (7)
3. R (5)
4. X (4)
5. W (2)
6. A (1)

**Output Size:** $11 \times 11$.
Shell 1: B
Shell 2: .
Shell 3: R
Shell 4: X
Shell 5: W
Shell 6: A

ANSWER:
B B B B B B B B B B B
B . . . . . . . . . B
B . R R R R R R R . B
B . R X X X X X R . B
B . R X . . . X R . B
B . R X . W . X R . B
B . R X . . . X R . B
B . R X X X X X R . B
B . R R R R R R R . B
B . . . . . . . . . B
B B B B B B B B B B B

Wait, the center is A.
B . R X . W . X R . B -> The center point is the 6th element.
Row 6: B . R X . A . X R . B

Let's refine the center:
Shell 1: B (index 0)
Shell 2: . (index 1)
Shell 3: R (index 2)
Shell 4: X (index 3)
Shell 5: W (index 4)
Shell 6: A (index 5)

Row 6 (center row):
Col 0: B
Col 1: .
Col 2: R
Col 3: X
Col 4: W
Col 5: A
Col 6: W
Col 7: X
Col 8: R
Col 9: .


... [2255 chars truncated]

### Conceptual Digestion

1.  **CAST**:
    *   **Musicians (Colored Pixels)**: The non-black/non-background pixels. Each color represents a "section" of musicians.
    *   **Practice Rooms (Input Grid)**: The initial layout where musicians are scattered.
    *   **Stage (Output Grid)**: A square grid where musicians are arranged in concentric rings.

2.  **AFFORDANCES**:
    *   **Counting**: We can count the number of pixels of each color to determine the size of each section.
    *   **Ordering**: We can rank the colors from most frequent to least frequent.
    *   **Placement**: We can place colors into a square grid starting from the outermost perimeter and moving inward.

3.  **TRANSFORMATION RULE**:
    *   **Step 1: Count frequencies.** Count the occurrences of every unique color in the input grid (excluding the background color, which is usually the most frequent or explicitly defined as empty space; however, in Example 1, 'O' was the background/empty space).
    *   **Step 2: Rank sections.** Order the colors by frequency (descending).
    *   **Step 3: Determine output size.** The output is a square grid. The size is determined by the total number of unique colors (sections). If there are $N$ unique colors, the output is an $N \times N$ grid (or the smallest square that can accommodate the concentric rings). Looking at Example 1: Colors were Y (4), M (3), O (background). Output was $3 \times 3$. Wait, let's re-evaluate Example 1.
    *   **Example 1 Re-evaluation**:
        Input: O (22), Y (4), M (3).
        Output: $3 \times 3$ grid.
        Outer ring: Y (8 pixels). Inner: M (1 pixel).
        Wait, the rule says "the largest section forms the outermost ring".
        Y count = 4, M count = 3.
        Output:
        Y Y Y
        Y M Y
        Y Y Y
        Here, Y is the outer ring (8 spots) and M is the center (1 spot). The counts in the input are used to *rank* the colors, but the output fills the rings completely.
    *   **Step 4: Fill the rings.**
        - Largest section $\rightarrow$ Outermost ring.
        - Second largest $\rightarrow$ Next ring inside.
        - ...
        - Smallest section $\rightarrow$ Center pixel.

### Applying the Rule to Test Input

**1. Count Frequencies:**
*   O (Orange): 33
*   B (Blue): 8
*   X (Grey): 4
*   R (Red): 4
*   W (Maroon): 2
*   A (Azure): 1
*   . (Black): 4

*Wait, is 'O' the background?* In Example 1, 'O' was the most frequent and acted as the background. In the test input, 'O' is by far the most frequent. Let's treat 'O' as the background.

**2. Rank Sections (excluding 'O'):**
1.  B (Blue): 8
2.  X (Grey): 4
3.  R (Red): 4
4.  W (Maroon): 2
5.  . (Black): 4 (Wait, . is black, usually background, but here it's a color in the key. Let's count it).
6.  A (Azure): 1

Let's re-count carefully:
B: (5,2), (6,1), (6,2), (6,3), (7,0), (7,1), (7,2), (7,3), (7,4) $\rightarrow$ actually:
(5,2) B
(6,1) B, (6,2) B, (6,3) B
(7,0) B, (7,1) B, (7,2) B, (7,3) B, (7,4) B
Total B = 9.

X: (2,0), (3,0), (3,1), (4,0) $\rightarrow$ Total X = 4.
R: (0,5), (0,6), (0,7), (1,5), (1,7) $\rightarrow$ Total R = 5.
W: (0,2), (1,2) $\rightarrow$ Total W = 2.
A: (3,3) $\rightarrow$ Total A = 1.
.: (4,5), (5,5), (5,6), (6,6), (6,7), (7,6), (7,7) $\rightarrow$ Total . = 7.

**Ranked List:**
1. B (9)
2. . (7)
3. R (5)
4. X (4)
5. W (2)
6. A (1)

**3. Determine Output Size:**
There are 6 unique colors. A $5 \times 5$ grid has rings of size 16, 8, 1. A $7 \times 7$ grid has rings of 24, 16, 8, 1.
Wait, the number of colors is 6. For a square grid to have concentric rings ending in a single center point, the dimension must be odd ($1, 3, 5, 7...$).
If $N=5$, we have 3 rings (Outer: 16, Mid: 8, Center: 1).
If $N=7$, we have 4 rings (Outer: 24, Mid1: 16, Mid2: 8, Center: 1).
Wait, the number of colors is 6. Let's check the counts again.
B(9), .(7), R(5), X(4), W(2), A(1).
If the output is $5 \times 5$:
Ring 1 (Outer): B
Ring 2 (Mid): .
Ring 3 (Center): R? No, that doesn't use all colors.

Let's re-read: "the largest section forms the outermost ring, the next largest forms the ring inside it, continuing until the smallest section — the soloist — occupies the single center spot."
This implies the number of rings equals the number of unique colors.
If there are 6 colors, the grid must be $11 \times 11$ (Rings: 40, 32, 24, 16, 8, 1). That seems too large.

Let's re-examine the colors. Is 'O' a musician?
If O is a musician:
O: 33
B: 9
.: 7
R: 5
X: 4
W: 2
A: 1
Total colors = 7.
For 7 colors, the grid is $13 \times 13$. Still too large.

Let's look at Example 1 again.
Input: O(22), Y(4), M(3).
Output: $3 \times 3$.
Colors used in output: Y, M.
'O' was ignored. 'O' is the background.
So we ignore 'O'.
Remaining colors: B(9), .(7), R(5), X(4), W(2), A(1).
Total = 6 colors.
Wait, if the output is $5 \times 5$, we have 3 rings. If $7 \times 7$, 4 rings.
Is it possible some colors are grouped? Or is the output size based on the number of colors?
If there are 6 colors, and we 

... [3059 chars truncated]

### Conceptual Digestion

1.  **CAST**:
    *   **Players**: Non-orange (`O`) cells. Each unique color represents a "team".
    *   **Field**: Orange (`O`) cells, which act as background/empty space.
    *   **Teams**: Groups of cells of the same color.

2.  **AFFORDANCES**:
    *   The number of cells of each color determines the "team size".
    *   The total number of players (excluding background `O`) determines the size of the final square formation.
    *   The final output is a square grid where the side length $S$ is the smallest odd integer such that $S^2 \ge \text{Total Players}$. (Wait, looking at Example 1: 4 Y, 3 M, 18 O. Total players = 7. Output is 3x3. $3^2=9$. Example 2: 7 B, 4 A, 4 X. Total = 15. Output is 5x5. $5^2=25$. Example 3: 8 A, 3 G, 1 W, 4 black. Total = 16. Output is 7x7. $7^2=49$. Actually, the output size is determined by the team with the most players needing to form the outermost ring. A ring of size $S$ requires $4(S-1)$ cells. If the largest team has $N$ players, $4(S-1) \le N$ is not the rule; rather, the largest team *fills* the outermost ring, the next fills the next, etc.)

3.  **TRANSFORMATION RULE**:
    *   **Step 1: Count the players** for each color (excluding `O`).
    *   **Step 2: Rank the teams** by population (descending).
    *   **Step 3: Determine the output grid size**. The output is a square of size $S \times S$. The outermost ring has $4(S-1)$ cells, the next $4(S-3)$, and so on, until the center cell (1 cell).
    *   **Step 4: Fill the rings**. Start from the outermost ring and fill it with the most populous team. If the team has more players than the ring size, they still only occupy that ring (or the ring is filled and the remainder is discarded/ignored). If they have fewer, the ring is filled with that color (based on the examples, the rings are filled completely by the team assigned to them).
    *   **Step 5: Repeat** for the next largest team in the next inner ring.

### Applying the Rule to Test Input

**1. Count the players:**
*   `B` (blue): 8 cells
*   `X` (grey): 4 cells
*   `R` (red): 4 cells
*   `W` (maroon): 2 cells
*   `A` (azure): 1 cell
*   `.` (black): 4 cells
*   `O` (orange): Background

**2. Rank the teams:**
1.  `B` (8)
2.  `X` (4) / `R` (4) / `.` (4) - (Tie)
3.  `W` (2)
4.  `A` (1)

**3. Determine Grid Size:**
The largest team is `B` with 8 players. A ring of size $S=3$ has $4(3-1) = 8$ cells.
If $S=3$:
*   Outer ring: 8 cells (Team `B`)
*   Inner ring (center): 1 cell (Team `X`, `R`, or `.`)
This doesn't account for all teams. Let's look at the training examples again.
Example 1: Y(4), M(3). Output 3x3. Outer ring (8) is Y, center (1) is M.
Example 2: B(7), A(4), X(4). Output 5x5. Outer ring (16) is B, next ring (8) is A, center (1) is X.
Example 3: A(8), G(3), W(1), .(4). Output 7x7. Outer ring (24) is A, next ring (16) is ., next ring (8) is G, center (1) is W.

Wait, the rule is: The team with the most players takes the outermost ring, the next largest the next ring, and so on. The size of the grid is determined by the number of teams.
Number of teams in Test Input: `B`, `X`, `R`, `.`, `W`, `A` (6 teams).
If there are $N$ teams, the grid size is $S = 2N - 1$.
For 6 teams, $S = 2(6) - 1 = 11$.
But wait, let's re-examine the counts.
Example 1: 2 teams (Y, M) -> 3x3.
Example 2: 3 teams (B, A, X) -> 5x5.
Example 3: 4 teams (A, G, W, .) -> 7x7.
Test Input: 6 teams (B, X, R, ., W, A) -> 11x11.

Wait, let's re-count the test input colors:
B: (5,2), (6,1), (6,2), (6,3), (7,0), (7,1), (7,2), (7,3), (7,4) -> 9 cells.
X: (2,0), (3,0), (3,1), (4,0) -> 4 cells.
R: (0,5), (0,6), (0,7), (1,5), (1,7) -> 5 cells.
W: (0,2), (1,2) -> 2 cells.
A: (3,3) -> 1 cell.
.: (4,5), (5,5), (5,6), (6,6), (6,7), (7,7) -> 6 cells.

Recounting:
B: 9
.: 6
R: 5
X: 4
W: 2
A: 1
Total teams = 6. Grid size = 11x11.

Wait, let's check the counts again.
B: (5,2), (6,1), (6,2), (6,3), (7,0), (7,1), (7,2), (7,3), (7,4) = 9
.: (4,5), (5,5), (5,6), (6,6), (6,7), (7,7) = 6
R: (0,5), (0,6), (0,7), (1,5), (1,7) = 5
X: (2,0), (3,0), (3,1), (4,0) = 4
W: (0,2), (1,2) = 2
A: (3,3) = 1

Ranking: B(9), .(6), R(5), X(4), W(2), A(1).
Grid size: 11x11.
Ring 1 (Outer): B
Ring 2: .
Ring 3: R
Ring 4: X
Ring 5: W
Ring 6 (Center): A

ANSWER:
B B B B B B B B B B B
B . . . . . . . . . B
B . R R R R R R R . B
B . R X X X X X R . B
B . R X W W W X R . B
B . R X W A W X R . B
B . R X W W W X R . B
B . R X X X X X R . B
B . R R R R R R R . B
B . . . . . . . . . B
B B B B B B B B B B B

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Regiments:** Contiguous blocks of the same color (excluding the background color 'O' - orange).
    *   **Size:** The number of cells occupied by a specific color block.
    *   **Fortress Plan:** The output grid, which is a square concentric structure.

2.  **Identify the AFFORDANCES:**
    *   Each unique color (excluding 'O') represents a "regiment".
    *   The size of each regiment determines its placement in the concentric rings of the output grid.
    *   The output grid size is determined by the number of unique colors found in the input. If there are $N$ unique colors, the output is an $N \times N$ grid (if $N$ is odd) or $(N+1) \times (N+1)$ (if $N$ is even) to allow for a center point. However, looking at Example 1: 2 colors (Y, M) $\rightarrow$ $3 \times 3$ output. This suggests the output size is $(2 \times \text{number of colors} - 1)$ if we consider the center as the smallest. Wait, Example 1 has Y (4 cells) and M (3 cells). The output is $3 \times 3$. The largest (Y) forms the outer ring, the smallest (M) forms the center.

3.  **Identify the TRANSFORMATION RULE:**
    *   **Step 1:** Identify all unique colors in the input grid (excluding the background color 'O').
    *   **Step 2:** Count the number of cells for each color to determine the size of each "regiment".
    *   **Step 3:** Rank the colors by size (largest to smallest).
    *   **Step 4:** Create a square grid. The size of the grid is $(2k - 1)$ where $k$ is the number of unique colors.
    *   **Step 5:** Place the largest regiment on the outermost perimeter.
    *   **Step 6:** Place the next largest on the next inner perimeter.
    *   **Step 7:** Place the smallest regiment in the center cell.

### Applying the Rule to Test Input

1.  **Identify Regiments and Sizes:**
    *   **B (Blue):** (5,2), (6,1), (6,2), (6,3), (7,0), (7,1), (7,2), (7,3), (7,4) $\rightarrow$ 9 cells.
    *   **X (Grey):** (2,0), (3,0), (3,1), (4,0) $\rightarrow$ 4 cells.
    *   **R (Red):** (0,5), (0,6), (0,7), (1,5), (1,7) $\rightarrow$ 5 cells.
    *   **W (Maroon):** (0,2), (1,2) $\rightarrow$ 2 cells.
    *   **A (Azure):** (3,3) $\rightarrow$ 1 cell.
    *   **. (Black):** (4,5), (5,5), (5,6), (6,6), (6,7), (7,6), (7,7) $\rightarrow$ 7 cells.

2.  **Rank by Size:**
    1.  B (9) - Largest
    2.  . (7)
    3.  R (5)
    4.  X (4)
    5.  W (2)
    6.  A (1) - Smallest

3.  **Determine Output Grid Size:**
    There are 6 unique colors. To form concentric squares with a single center point, we need an odd-dimensioned grid. For 6 colors, the outermost is ring 1, then 2, 3, and the 6th is the center. This requires a $11 \times 11$ grid (Ring 1: $11 \times 11$, Ring 2: $9 \times 9$, Ring 3: $7 \times 7$, Ring 4: $5 \times 5$, Ring 5: $3 \times 3$, Ring 6: $1 \times 1$).

    *Wait, let's re-verify Example 1:*
    Input: Y (4), M (3). 2 colors. Output: $3 \times 3$.
    Ring 1 (Outer): Y. Center: M.
    This follows the formula: Output size = $(2 \times \text{number of colors} - 1)$.
    For 6 colors: $(2 \times 6) - 1 = 11$.

4.  **Construct the Grid:**
    *   Ring 1 (Outer $11 \times 11$): B
    *   Ring 2 ($9 \times 9$): .
    *   Ring 3 ($7 \times 7$): R
    *   Ring 4 ($5 \times 5$): X
    *   Ring 5 ($3 \times 3$): W
    *   Ring 6 (Center $1 \times 1$): A

ANSWER:
B B B B B B B B B B B
B . . . . . . . . . B
B . R R R R R R R . B
B . R X X X X X R . B
B . R X W W W X R . B
B . R X W A W X R . B
B . R X W W W X R . B
B . R X X X X X R . B
B . R R R R R R R . B
B . . . . . . . . . B
B B B B B B B B B B B

### Conceptual Digestion

**1. Identify the CAST:**
*   **Troupes (Clusters):** Groups of connected cells of the same color (excluding the background color 'O').
*   **Background:** The color 'O' (orange) acts as the empty stage and is discarded in the final output.
*   **Performers:** The individual cells within each troupe.

**2. Identify the AFFORDANCES:**
*   **Size:** Each troupe has a specific count of cells.
*   **Hierarchy:** Troupes are ranked by size (largest to smallest).
*   **Placement:** The rank determines the "layer" of the concentric square output.

**3. Identify the TRANSFORMATION RULE:**
*   **Step 1: Identify and Count.** Find all contiguous clusters of non-'O' colors.
*   **Step 2: Rank.** Sort the clusters by size (number of cells) in descending order.
*   **Step 3: Determine Output Size.** The total number of cells in the final output is a square. The size of the square is determined by the number of distinct troupes. If there are $N$ troupes, the output is a $(2N-1) \times (2N-1)$ square.
*   **Step 4: Assemble Concentric Borders.**
    *   The largest troupe forms the outermost border.
    *   The second largest forms the next inner border.
    *   This continues until the smallest troupe (which must be a single cell) occupies the center spot.

---

### Applying the Rule to Test Input

**Step 1: Identify and Count Troupes**
*   **Blue (B):** (5,2), (6,1), (6,2), (6,3), (7,0), (7,1), (7,2), (7,3), (7,4) $\rightarrow$ **9 cells**
*   **Grey (X):** (2,0), (3,0), (3,1), (4,0) $\rightarrow$ **4 cells**
*   **Red (R):** (0,5), (0,6), (0,7), (1,5), (1,7) $\rightarrow$ **5 cells**
*   **Black (.):** (4,5), (5,5), (5,6), (6,6), (6,7), (7,7) $\rightarrow$ **6 cells**
*   **Maroon (W):** (0,2), (1,2) $\rightarrow$ **2 cells**
*   **Azure (A):** (3,3) $\rightarrow$ **1 cell**

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

**Step 3: Determine Output Size**
There are 6 troupes. The output grid will be $(2 \times 6 - 1) = 11 \times 11$.
Wait, let's re-verify the training examples.
Example 1: Troupes are Y(4), M(3). Output is $3 \times 3$. (2 troupes $\rightarrow 3 \times 3$).
Example 2: Troupes are B(8), A(4), X(4). Wait, X and A are both 4. Output is $5 \times 5$. (3 troupes $\rightarrow 5 \times 5$).
The rule is: Output size is $(2 \times \text{number of troupes} - 1)$.
For the test input, there are 6 troupes $\rightarrow 11 \times 11$.

**Step 4: Assemble the Grid**
*   Layer 1 (Outer): Blue (B)
*   Layer 2: Black (.)
*   Layer 3: Red (R)
*   Layer 4: Grey (X)
*   Layer 5: Maroon (W)
*   Layer 6 (Center): Azure (A)

**Construction:**
Row 0: B B B B B B B B B B B
Row 1: B . . . . . . . . . B
Row 2: B . R R R R R R R . B
Row 3: B . R X X X X X R . B
Row 4: B . R X W W W X R . B
Row 5: B . R X W A W X R . B
Row 6: B . R X W W W X R . B
Row 7: B . R X X X X X R . B
Row 8: B . R R R R R R R . B
Row 9: B . . . . . . . . . B
Row 10: B B B B B B B B B B B

ANSWER:
B B B B B B B B B B B
B . . . . . . . . . B
B . R R R R R R R . B
B . R X X X X X R . B
B . R X W W W X R . B
B . R X W A W X R . B
B . R X W W W X R . B
B . R X X X X X R . B
B . R R R R R R R . B
B . . . . . . . . . B
B B B B B B B B B B B

### Conceptual Digestion

**1. Identify the CAST:**
*   **Seeds (Input Colors):** The distinct colors present in the input grid (excluding the background color `O` = orange).
*   **Counts:** The number of times each color appears in the input grid.
*   **Garden Bed (Output Grid):** A square grid formed by concentric rings of colors.

**2. Identify the AFFORDANCES:**
*   The colors can be counted to determine their frequency.
*   The frequencies determine the order of the concentric rings (most frequent = outermost, least frequent = center).
*   The total number of unique colors (excluding `O`) determines the size of the output square. If there are $N$ unique colors, the output grid is a square of size $(2N-1) \times (2N-1)$.

**3. Identify the TRANSFORMATION RULE:**
1.  Count the occurrences of every color in the input grid, ignoring the background color `O`.
2.  Sort the colors by frequency in descending order.
3.  Create a square grid of size $(2N-1) \times (2N-1)$, where $N$ is the number of unique colors found.
4.  Fill the outermost ring with the most frequent color.
5.  Fill the next ring inward with the second most frequent color, and so on.
6.  The least frequent color becomes the single center pixel.

---

### Applying the Rule to Test Input

**Step 1: Count the colors (excluding O)**
*   `W` (maroon): 2
*   `R` (red): 4
*   `X` (grey): 3
*   `A` (azure): 1
*   `.` (black): 4
*   `B` (blue): 7

**Step 2: Sort by frequency (Descending)**
1.  `B` (blue): 7
2.  `R` (red): 4
3.  `.` (black): 4 (Tie-break: usually follows input order or alphabetical, but let's check the counts again. R=4, .=4. In Example 2, B=8, A=4, X=2. Here we have a tie. Let's re-verify counts: R: (0,5), (0,6), (0,7), (1,5), (1,7) -> Wait, let me recount carefully.)

**Recounting Test Input:**
*   `W`: (0,2), (1,2) $\rightarrow$ **2**
*   `R`: (0,5), (0,6), (0,7), (1,5), (1,7) $\rightarrow$ **5**
*   `X`: (2,0), (3,0), (3,1), (4,0) $\rightarrow$ **4**
*   `A`: (3,3) $\rightarrow$ **1**
*   `.`: (4,5), (5,5), (5,6), (6,6), (7,6), (7,7) $\rightarrow$ **6**
*   `B`: (5,2), (6,1), (6,2), (6,3), (7,0), (7,1), (7,2), (7,3), (7,4) $\rightarrow$ **9**

**Sorted Frequencies:**
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: Determine Grid Size**
There are $N=6$ unique colors.
Grid size = $(2 \times 6) - 1 = 11 \times 11$.

**Step 4: Construct Concentric Rings**
*   Ring 1 (Outer): `B`
*   Ring 2: `.`
*   Ring 3: `R`
*   Ring 4: `X`
*   Ring 5: `W`
*   Ring 6 (Center): `A`

**Construction:**
Row 0: B B B B B B B B B B B
Row 1: B . . . . . . . . . B
Row 2: B . R R R R R R R . B
Row 3: B . R X X X X X R . B
Row 4: B . R X W W W X R . B
Row 5: B . R X W A W X R . B
Row 6: B . R X W W W X R . B
Row 7: B . R X X X X X R . B
Row 8: B . R R R R R R R . B
Row 9: B . . . . . . . . . B
Row 10: B B B B B B B B B B B

ANSWER:
B B B B B B B B B B B
B . . . . . . . . . B
B . R R R R R R R . B
B . R X X X X X R . B
B . R X W W W X R . B
B . R X W A W X R . B
B . R X W W W X R . B
B . R X X X X X R . B
B . R R R R R R R . B
B . . . . . . . . . B
B B B B B B B B B B B

### Conceptual Digestion

1.  **CAST**:
    *   **Dark Landscape (`.`)**: The background/empty space.
    *   **Grey Animals (`X`)**: The objects to be transformed. They form "herds" (connected components of `X`s).
    *   **Species Tags (Colors)**: The new colors assigned to the `X`s based on the size of their connected component.

2.  **AFFORDANCES**:
    *   The `X` cells can be grouped into connected components (orthogonal adjacency).
    *   The size of each component determines its new color.
    *   The background `.` remains unchanged.

3.  **TRANSFORMATION RULE**:
    *   Identify all connected components of `X`.
    *   Count the number of cells in each component.
    *   Assign colors based on the size (referencing Example 1):
        *   Size 1: Not explicitly defined in Example 1, but looking at the logic:
            *   Example 1: Size 2 $\rightarrow$ Blue (`B`), Size 3 $\rightarrow$ Red (`R`), Size 4 $\rightarrow$ Blue (`B`)? Wait, let's re-examine Example 1.
            *   Example 1 Analysis:
                *   (2,2)-(2,4) is size 3 $\rightarrow$ Red (`R`)
                *   (2,7)-(3,7) is size 2 $\rightarrow$ Green (`G`)? No, (2,7) and (3,7) are `G`.
                *   (5,3)-(5,4) and (6,3) is size 3 $\rightarrow$ Red (`R`)
                *   (7,7) is size 1 $\rightarrow$ Blue (`B`)
                *   (8,1)-(9,2) is size 4 $\rightarrow$ Blue (`B`)
                *   (8,6)-(8,8) is size 3 $\rightarrow$ Blue (`B`)? No, let's look closer.
                *   Example 1 Output:
                    *   (2,2-4): R R R (Size 3)
                    *   (2,7-3,7): G G (Size 2)
                    *   (5,3-6,3): R R R (Size 3)
                    *   (7,7): B (Size 1)
                    *   (8,1-9,2): B B B B (Size 4)
                    *   (8,6-8,8): B B B (Size 3) - *Wait, this contradicts the first size 3.*

    *   **Re-evaluating Example 1**:
        *   Size 1: (7,7) $\rightarrow$ B
        *   Size 2: (2,7-3,7) $\rightarrow$ G
        *   Size 3: (2,2-4) $\rightarrow$ R, (5,3-6,3) $\rightarrow$ R, (8,6-8,8) $\rightarrow$ B? No, (8,6-8,8) is size 3 and is B.
        *   Size 4: (8,1-9,2) $\rightarrow$ B.
        *   Let's look at the colors again:
            *   Size 1: B
            *   Size 2: G
            *   Size 3: R (mostly)
            *   Size 4: B
            *   Wait, (8,6-8,8) is size 3 and is B. (8,1-9,2) is size 4 and is B. (7,7) is size 1 and is B.
            *   Maybe the rule is: Size 1, 4, and some others are B?
            *   Let's look at the "Metaphorical Description": "smallest pairs share one color, the middle-sized trios another, and the largest bands of four a third."
            *   Pairs (2) $\rightarrow$ Color A
            *   Trios (3) $\rightarrow$ Color B
            *   Bands of four (4) $\rightarrow$ Color C
            *   Looking at Example 1 again:
                *   Size 2: (2,7-3,7) $\rightarrow$ G. (Pairs = Green)
                *   Size 3: (2,2-4) $\rightarrow$ R, (5,3-6,3) $\rightarrow$ R. (Trios = Red)
                *   Size 4: (8,1-9,2) $\rightarrow$ B. (Fours = Blue)
                *   What about (7,7) and (8,6-8,8)? (7,7) is size 1. (8,6-8,8) is size 3.
                *   Wait, (8,6-8,8) is `B B B`. That's size 3. But (2,2-4) is `R R R`.
                *   Let me re-read the grid carefully.
                *   Example 1 Input: (8,6) is X, (8,7) is X, (8,8) is X. Output: (8,6) is B, (8,7) is B, (8,8) is B.
                *   Example 1 Input: (2,2) is X, (2,3) is X, (2,4) is X. Output: (2,2) is R, (2,3) is R, (2,4) is R.
                *   This is contradictory unless the color depends on something else.
                *   Let's re-read: "smallest pairs share one color, the middle-sized trios another, and the largest bands of four a third."
                *   Pairs (2) $\rightarrow$ G
                *   Trios (3) $\rightarrow$ R
                *   Fours (4) $\rightarrow$ B
                *   Wait, (7,7) is size 1. (8,6-8,8) is size 3. (8,1-9,2) is size 4.
                *   If (8,6-8,8) is B, and (8,1-9,2) is B, and (7,7) is B... maybe B is for everything else?
                *   Let's check the test input.

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

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

ANSWER:
. . . . . . . . . B
. . . . . .

... [168 chars truncated]

### Conceptual Digestion

1.  **CAST**:
    *   `.` (Black): The background (the "dark slide").
    *   `X` (Grey): The "pale, unstained cell colonies". These are connected components of `X` cells (orthogonal connectivity).

2.  **AFFORDANCES**:
    *   The `X` cells can be grouped into distinct colonies based on adjacency.
    *   Each colony can be recolored based on its total cell count (size).

3.  **TRANSFORMATION RULE**:
    *   Identify all connected components of `X`.
    *   Count the number of cells in each component.
    *   Assign a color based on the size:
        *   Looking at Example 1:
            *   Colony size 1: `G` (Green) - *Wait, let's re-evaluate.*
            *   Colony size 2: `R` (Red) - *Wait, let's check again.*
            *   Example 1 sizes:
                *   (2,2)-(2,4) is size 3 -> `R`
                *   (2,7)-(3,7) is size 2 -> `G`
                *   (5,3)-(5,4) and (6,3) is size 3 -> `R`
                *   (7,7) is size 1 -> `B`
                *   (8,1)-(9,2) is size 4 -> `B`
                *   (8,6)-(8,8) is size 3 -> `B`
                *   *Correction*: Let's look at the sizes again.
                *   Example 1:
                    *   Size 1: (7,7) -> `B`
                    *   Size 2: (2,7)-(3,7) -> `G`
                    *   Size 3: (2,2)-(2,4) -> `R`, (5,3)-(6,3) -> `R`, (8,6)-(8,8) -> `B` (Wait, (8,6)-(8,8) is size 3 and is `B`).
                    *   Size 4: (8,1)-(9,2) -> `B`
                *   Let's re-examine Example 2:
                    *   Size 2: (1,8)-(1,9), (2,1)-(2,2), (3,2)-(3,3), (6,5)-(6,6), (8,4)-(9,4) -> `B`, `B`, `B`, `G`, `R`
                    *   Size 4: (1,8)-(2,9) is size 4 -> `B`
                    *   Size 2: (6,5)-(6,6) is size 2 -> `G`
                    *   Size 3: (5,9)-(7,9) is size 3 -> `R`
                    *   Size 2: (7,0)-(8,0) is size 2 -> `G`
                    *   Size 2: (8,4)-(9,4) is size 2 -> `R`
                *   *Wait, the rule is simpler: The colors are assigned based on the size of the colony relative to other colonies in the same grid.*
                *   Example 1:
                    *   Sizes: 1, 2, 3, 3, 3, 4.
                    *   Smallest (1) -> `B`
                    *   Medium (2) -> `G`
                    *   Largest (3, 4) -> `R`? No.
                *   Let's try again.
                *   Example 1:
                    *   Size 1: `B`
                    *   Size 2: `G`
                    *   Size 3: `R`
                    *   Size 4: `B` (Wait, this is confusing).

    *   **Let's look at the colors again.**
        Example 1:
        - (2,2-4): size 3 -> `R`
        - (2,7-3,7): size 2 -> `G`
        - (5,3-6,3): size 3 -> `R`
        - (7,7): size 1 -> `B`
        - (8,1-9,2): size 4 -> `B`
        - (8,6-8): size 3 -> `B`
        *Wait, (8,6-8) is size 3 and is `B`. (2,2-4) is size 3 and is `R`.*
        *Is it based on position? No, "based on the number of cells it contains".*

    *   **Let's re-count Example 1 carefully.**
        - Colony 1: (2,2), (2,3), (2,4) -> Size 3 -> `R`
        - Colony 2: (2,7), (3,7) -> Size 2 -> `G`
        - Colony 3: (5,3), (5,4), (6,3) -> Size 3 -> `R`
        - Colony 4: (7,7) -> Size 1 -> `B`
        - Colony 5: (8,1), (8,2), (9,1), (9,2) -> Size 4 -> `B`
        - Colony 6: (8,6), (8,7), (8,8) -> Size 3 -> `B`
        *Wait, (8,6-8) is size 3 and is `B`. (2,2-4) is size 3 and is `R`. This contradicts "based on the number of cells".*

    *   **Let's re-read the metaphor: "Smaller fusions glow one hue, mid-sized another, and the largest yet another."**
        This implies:
        Smallest size(s) -> Color 1
        Medium size(s) -> Color 2
        Largest size(s) -> Color 3

    *   **Re-evaluating Example 1 with this logic:**
        Sizes: 1, 2, 3, 3, 3, 4.
        Smallest: 1 -> `B`
        Medium: 2 -> `G`
        Largest: 3, 4 -> `R`?
        Wait, in Example 1:
        Size 1: `B`
        Size 2: `G`
        Size 3: `R` (at 2,2 and 5,3)
        Size 3: `B` (at 8,6)
        Size 4: `B` (at 8,1)
        *This is still not fitting.*

    *   **Let's look at Example 2:**
        - (1,8-2,9): size 4 -> `B`
        - (2,1-3,3): size 4 -> `B`
        - (6,5-6,6): size 2 -> `G`
        - (5,9-7,9): size 3 -> `R`
        - (7,0-8,0): size 2 -> `G`
        - (8,4-9,4): size 2 -> `R` (Wait, (8,4-9,4) is size 2 and is `R`).

    *   **Wait! Let's look at the colors again. Maybe it's not size, but the *number* of colonies of that size?**
        No, the metaphor says "based on the number of cells it contains".

    *   **Let's try a different approach. Maybe the colors are assigned to the sizes present in the grid?**
        Example 1 sizes: 1, 2, 3, 4.
        Example 2 sizes: 2, 3, 4.
        In Example 2:
        Size 4 -> `B`
        Size 3 -> `R`
        Size 2 -> `G` (mostly)

... [2676 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   `.` (Black): The background/empty space.
    *   `X` (Grey): The "stars" forming clusters.
    *   Output Colors (`B`, `R`, `G`, etc.): The "false-color palette" based on the size of the cluster.

2.  **Identify the AFFORDANCES:**
    *   Clusters are groups of adjacent `X` cells (connected horizontally or vertically).
    *   The size of the cluster (number of `X` cells) determines the color of all cells within that cluster in the output.

3.  **Identify the TRANSFORMATION RULE:**
    *   Analyze Example 1 to determine the color mapping:
        *   Cluster 1 (top left): 3 cells $\rightarrow$ Red (`R`)
        *   Cluster 2 (top right): 2 cells $\rightarrow$ Green (`G`)
        *   Cluster 3 (middle left): 3 cells $\rightarrow$ Red (`R`)
        *   Cluster 4 (bottom right): 1 cell $\rightarrow$ Blue (`B`)
        *   Cluster 5 (bottom left): 4 cells $\rightarrow$ Blue (`B`)
        *   Cluster 6 (bottom rightmost): 3 cells $\rightarrow$ Blue (`B`)
    *   *Correction on Example 1 analysis:*
        *   Cluster (2,2)-(2,4): size 3 $\rightarrow$ Red
        *   Cluster (2,7)-(3,7): size 2 $\rightarrow$ Green
        *   Cluster (5,3)-(6,3) and (5,4): size 3 $\rightarrow$ Red
        *   Cluster (7,7): size 1 $\rightarrow$ Blue
        *   Cluster (8,1)-(9,2): size 4 $\rightarrow$ Blue
        *   Cluster (8,6)-(8,8): size 3 $\rightarrow$ Blue
    *   *Wait, let's re-evaluate the mapping:*
        *   Size 1: Blue (`B`)
        *   Size 2: Green (`G`)
        *   Size 3: Red (`R`)
        *   Size 4+: Blue (`B`)? No, that's inconsistent.
    *   *Re-evaluating Example 1 again:*
        *   Cluster 1: (2,2), (2,3), (2,4) $\rightarrow$ size 3 $\rightarrow$ Red
        *   Cluster 2: (2,7), (3,7) $\rightarrow$ size 2 $\rightarrow$ Green
        *   Cluster 3: (5,3), (5,4), (6,3) $\rightarrow$ size 3 $\rightarrow$ Red
        *   Cluster 4: (7,7) $\rightarrow$ size 1 $\rightarrow$ Blue
        *   Cluster 5: (8,1), (8,2), (9,1), (9,2) $\rightarrow$ size 4 $\rightarrow$ Blue
        *   Cluster 6: (8,6), (8,7), (8,8) $\rightarrow$ size 3 $\rightarrow$ Blue? No, looking at the output: (8,6), (8,7), (8,8) is Blue.
    *   *Let's look at the colors again:*
        *   Size 1: Blue
        *   Size 2: Green
        *   Size 3: Red (sometimes Blue?)
        *   Wait, look at the output grid for Example 1 again.
        *   (2,2-4) is R R R (size 3)
        *   (2,7-3,7) is G G (size 2)
        *   (5,3-6,4) is R R R (size 3)
        *   (7,7) is B (size 1)
        *   (8,1-9,2) is B B B B (size 4)
        *   (8,6-8,8) is B B B (size 3)
    *   *Hypothesis:* The colors are based on size, but maybe the mapping is: 1=B, 2=G, 3=R, 4=B, 5=B... or perhaps it's a cycle? Or maybe the mapping is: 1=B, 2=G, 3=R, 4=B, 5=G, 6=R?
    *   Let's check the clusters in the Test Input:
        *   Cluster 1 (top right): (0,9), (1,9), (2,9), (3,9) $\rightarrow$ size 4
        *   Cluster 2 (top left): (2,2), (2,3), (3,1), (3,2) $\rightarrow$ size 4
        *   Cluster 3 (top mid): (2,5), (2,6), (3,6) $\rightarrow$ size 3
        *   Cluster 4 (bottom left): (6,0), (7,0), (8,0) $\rightarrow$ size 3
        *   Cluster 5 (bottom mid): (7,3), (7,4) $\rightarrow$ size 2
        *   Cluster 6 (bottom right): (7,7), (8,7) $\rightarrow$ size 2

    *   *Refining the mapping:*
        *   Size 1: Blue (B)
        *   Size 2: Green (G)
        *   Size 3: Red (R)
        *   Size 4: Blue (B)
        *   (This follows a pattern: 1=B, 2=G, 3=R, 4=B, 5=G, 6=R...)

### Applying to Test Input

1.  Cluster 1: (0,9), (1,9), (2,9), (3,9) $\rightarrow$ size 4 $\rightarrow$ Blue (B)
2.  Cluster 2: (2,2), (2,3), (3,1), (3,2) $\rightarrow$ size 4 $\rightarrow$ Blue (B)
3.  Cluster 3: (2,5), (2,6), (3,6) $\rightarrow$ size 3 $\rightarrow$ Red (R)
4.  Cluster 4: (6,0), (7,0), (8,0) $\rightarrow$ size 3 $\rightarrow$ Red (R)
5.  Cluster 5: (7,3), (7,4) $\rightarrow$ size 2 $\rightarrow$ Green (G)
6.  Cluster 6: (7,7), (8,7) $\rightarrow$ size 2 $\rightarrow$ Green (G)

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

### Conceptual Digestion

1.  **CAST**:
    *   `.` (Black): Open water (background).
    *   `X` (Grey): Boats.
    *   Clusters of `X` (connected orthogonally): Rafts.
    *   Colors (R, G, B, etc.): Harbor signal colors assigned based on the size of the raft.

2.  **AFFORDANCES**:
    *   The grey cells (`X`) can be grouped into connected components (rafts).
    *   The size of each component (number of cells) determines the new color of all cells in that component.

3.  **TRANSFORMATION RULE**:
    *   Identify all connected components of `X`.
    *   Count the number of cells in each component.
    *   Map the size to a color. Looking at Example 1:
        *   Size 1: `G` (Green) - *Wait, let's re-examine Example 1.*
        *   Example 1 Analysis:
            *   Raft 1: (2,1), (2,2), (2,3) -> Size 3. Color: `R` (Red).
            *   Raft 2: (2,7), (3,7) -> Size 2. Color: `G` (Green). *Wait, (2,7) and (3,7) is size 2.*
            *   Raft 3: (5,3), (5,4), (6,3) -> Size 3. Color: `R` (Red).
            *   Raft 4: (7,7) -> Size 1. Color: `B` (Blue).
            *   Raft 5: (8,1), (8,2), (9,1), (9,2) -> Size 4. Color: `B` (Blue).
            *   Raft 6: (8,6), (8,7), (8,8) -> Size 3. Color: `B` (Blue).
        *   *Correction on Example 1 mapping*:
            *   Size 1: `B` (Blue)
            *   Size 2: `G` (Green)
            *   Size 3: `R` (Red)
            *   Size 4: `B` (Blue)
            *   Wait, the mapping is not strictly linear. Let's look at the colors again.
            *   Rafts of size 3: (2,1-3) is Red, (5,3-6,3) is Red, (8,6-8) is Blue.
            *   Let's re-count Example 1 carefully.
            *   Raft A: (2,1), (2,2), (2,3) -> Size 3. Output: `R R R`
            *   Raft B: (2,7), (3,7) -> Size 2. Output: `G G`
            *   Raft C: (5,3), (5,4), (6,3) -> Size 3. Output: `R R R`
            *   Raft D: (7,7) -> Size 1. Output: `B`
            *   Raft E: (8,1), (8,2), (9,1), (9,2) -> Size 4. Output: `B B B B`
            *   Raft F: (8,6), (8,7), (8,8) -> Size 3. Output: `B B B`
            *   Wait, Raft F is size 3 and is Blue. Raft A and C are size 3 and are Red.
            *   Let's check connectivity again. Raft F (8,6-8) is size 3. Raft E (8,1-2, 9,1-2) is size 4.
            *   Is it possible the color depends on the size *and* something else? Or maybe the mapping is: 1=B, 2=G, 3=R, 4=B, 5=G, 6=R? (Modulo 3).
            *   Let's test: Size 1 mod 3 = 1 (B), Size 2 mod 3 = 2 (G), Size 3 mod 3 = 0 (R), Size 4 mod 3 = 1 (B).
            *   This fits! 1->B, 2->G, 3->R, 4->B, 5->G, 6->R...

4.  **Applying to Test Input**:
    *   Raft 1: (0,9), (1,9), (2,9), (3,9) -> Size 4. Color: 4 mod 3 = 1 -> `B`.
    *   Raft 2: (2,2), (2,3), (3,1), (3,2) -> Size 4. Color: 4 mod 3 = 1 -> `B`.
    *   Raft 3: (2,5), (2,6), (3,6) -> Size 3. Color: 3 mod 3 = 0 -> `R`.
    *   Raft 4: (6,0), (7,0), (8,0) -> Size 3. Color: 3 mod 3 = 0 -> `R`.
    *   Raft 5: (7,3), (7,4) -> Size 2. Color: 2 mod 3 = 2 -> `G`.
    *   Raft 6: (7,7), (8,7) -> Size 2. Color: 2 mod 3 = 2 -> `G`.

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

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

### Conceptual Digestion

1.  **CAST**:
    *   `.` (Black): The substrate/background. It remains unchanged.
    *   `X` (Grey): Solder blobs. These are connected components of `X` cells.
    *   Output Colors: Diagnostic colors based on the size (number of cells) of each connected component of `X`.

2.  **AFFORDANCES**:
    *   Connected components of `X` (adjacent horizontally or vertically) can be identified.
    *   The size (area) of each component determines its replacement color.

3.  **TRANSFORMATION RULE**:
    *   Identify all connected components of `X`.
    *   Count the number of cells in each component.
    *   Map the count to a color based on the training examples:
        *   Example 1 analysis:
            *   Component 1: (2,1), (2,2), (2,3) -> Size 3 -> Color `R` (Red)
            *   Component 2: (2,7), (3,7) -> Size 2 -> Color `G` (Green) - *Wait, let's re-examine.*
            *   Example 1 components:
                *   (2,1)-(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` (Blue)
                *   (8,1)-(8,2), (9,1)-(9,2): size 4 -> `B` (Blue)
                *   (8,6)-(8,8): size 3 -> `B` (Blue) - *Wait, the mapping is inconsistent.*

    *   **Re-evaluating Example 1 Mapping**:
        *   Component 1: (2,1), (2,2), (2,3) [Size 3] -> `R`
        *   Component 2: (2,7), (3,7) [Size 2] -> `G`
        *   Component 3: (5,3), (5,4), (6,3) [Size 3] -> `R`
        *   Component 4: (7,7) [Size 1] -> `B`
        *   Component 5: (8,1), (8,2), (9,1), (9,2) [Size 4] -> `B`
        *   Component 6: (8,6), (8,7), (8,8) [Size 3] -> `B` - *Wait, (8,6-8) is size 3, but it's Blue. (2,1-3) is size 3 and it's Red.*

    *   **Correcting the Rule**: Let's look at the components again.
        *   (2,1-3): 3 cells -> `R`
        *   (2,7-3,7): 2 cells -> `G`
        *   (5,3-6,3): 3 cells -> `R`
        *   (7,7): 1 cell -> `B`
        *   (8,1-9,2): 4 cells -> `B`
        *   (8,6-8): 3 cells -> `B`
        *   Wait, look at the output again. (8,6-8) is `B B B`. (8,1-9,2) is `B B / B B`. (7,7) is `B`.
        *   It seems:
            *   Size 1: `B`
            *   Size 2: `G`
            *   Size 3: `R`
            *   Size 4: `B`
            *   Wait, if Size 1 and Size 4 are both `B`, maybe the rule is based on something else?
            *   Let's check the "bridge count" metaphor. "Count the pads each solder bridge connects".
            *   In a grid, "pads" are usually the edges or corners. But here, the `X`s *are* the bridge.
            *   Maybe it's the number of *distinct* rows or columns the component spans?
            *   (2,1-3): 1 row, 3 cols. Total 3. -> `R`
            *   (2,7-3,7): 2 rows, 1 col. Total 2. -> `G`
            *   (5,3-6,3): 2 rows, 2 cols. Total 3. -> `R`
            *   (7,7): 1 row, 1 col. Total 1. -> `B`
            *   (8,1-9,2): 2 rows, 2 cols. Total 4. -> `B`
            *   (8,6-8): 1 row, 3 cols. Total 3. -> `B`
            *   This is still confusing. Let's look at the colors again.
            *   `B` = Blue, `R` = Red, `G` = Green.
            *   Example 1 Output:
                (2,1-3) = R R R (Size 3)
                (2,7-3,7) = G G (Size 2)
                (5,3-6,3) = R R R (Size 3)
                (7,7) = B (Size 1)
                (8,1-9,2) = B B B B (Size 4)
                (8,6-8) = B B B (Size 3)
            *   Wait! Look at the coordinates of the `B` components: (7,7), (8,1-9,2), (8,6-8). They are all in the bottom section of the grid. The `R` and `G` are in the top/middle.
            *   No, that's unlikely. Let's look at the shapes.
            *   Size 1: B
            *   Size 2: G
            *   Size 3: R (if it's a line or L-shape)
            *   Size 3: B (if it's a line) - No.
            *   Let's try: Size 1=B, 2=G, 3=R, 4=B, 5=G, 6=R... (Modulo 3).
            *   Size 1: 1 mod 3 = 1 -> B
            *   Size 2: 2 mod 3 = 2 -> G
            *   Size 3: 3 mod 3 = 0 -> R
            *   Size 4: 4 mod 3 = 1 -> B
            *   Size 5: 5 mod 3 = 2 -> G
            *   Size 6: 6 mod 3 = 0 -> R
            *   Let's test this on Example 1:
                (2,1-3): Size 3 -> 0 mod 3 -> R (Correct)
                (2,7-3,7): Size 2 -> 2 mod 3 -> G (Correct)
                (5,3-6,3): Size 3 -> 0 mod 3 -> R (Correct)
                (7,7): Size 1 -> 1 mod 3 -> B (Correct)
                (8,1-9,2): Size 4 -> 1 mod 3 -> B (Correct)
                (8,6-8): Size 3 -> 0 mod 3 -> R... wait, the output says (8,6-8) is `B B B`.
                Let me re-read the output for Example 1 carefully.
                Output (8,6-8) is `B B B`. Size is 3. 3 mod 3 is 0. Should be R.
                Wait, (8,6-8) is `B B B`. (8,1-9,2) is `B B / B B`. (7,7) is `B`.
                All these are `B`. Si

... [4573 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   `.` (Black): The "dark divination cloth" (background).
    *   `X` (Grey): The "pale bone fragments" (foreground objects).
    *   Clusters: Groups of `X` cells connected orthogonally (up, down, left, right).

2.  **Identify the AFFORDANCES:**
    *   The `X` cells can be grouped into clusters.
    *   The size of each cluster (number of `X` cells) determines a specific color.
    *   The background `.` remains unchanged.

3.  **Identify the TRANSFORMATION RULE:**
    *   Analyze Example 1 to find the mapping between cluster size and color:
        *   Cluster 1 (top left): 3 cells $\rightarrow$ Red (R)
        *   Cluster 2 (top right): 2 cells $\rightarrow$ Green (G)
        *   Cluster 3 (middle left): 3 cells $\rightarrow$ Red (R)
        *   Cluster 4 (bottom right): 1 cell $\rightarrow$ Blue (B)
        *   Cluster 5 (bottom left): 4 cells $\rightarrow$ Blue (B)
        *   Cluster 6 (bottom right-most): 3 cells $\rightarrow$ Blue (B)
    *   *Wait, let's re-evaluate Example 1 carefully:*
        *   Cluster (2,2)-(2,4): size 3 $\rightarrow$ R
        *   Cluster (2,7)-(3,7): size 2 $\rightarrow$ G
        *   Cluster (5,3)-(6,3) and (5,4): size 3 $\rightarrow$ R
        *   Cluster (7,7): size 1 $\rightarrow$ B
        *   Cluster (8,1)-(9,2): size 4 $\rightarrow$ B
        *   Cluster (8,6)-(8,8): size 3 $\rightarrow$ B
    *   *Correction on Example 1 mapping:*
        *   Size 1: Blue (B)
        *   Size 2: Green (G)
        *   Size 3: Red (R) or Blue (B)? Let's look again.
        *   (2,2)-(2,4) is size 3 $\rightarrow$ R. (5,3)-(6,3) and (5,4) is size 3 $\rightarrow$ R. (8,6)-(8,8) is size 3 $\rightarrow$ B.
        *   Wait, (8,6)-(8,8) is size 3, but (8,1)-(9,2) is size 4. Both are Blue.
        *   Let's re-count:
            *   Cluster A: (2,2), (2,3), (2,4) $\rightarrow$ size 3 $\rightarrow$ R
            *   Cluster B: (2,7), (3,7) $\rightarrow$ size 2 $\rightarrow$ G
            *   Cluster C: (5,3), (5,4), (6,3) $\rightarrow$ size 3 $\rightarrow$ R
            *   Cluster D: (7,7) $\rightarrow$ size 1 $\rightarrow$ B
            *   Cluster E: (8,1), (8,2), (9,1), (9,2) $\rightarrow$ size 4 $\rightarrow$ B
            *   Cluster F: (8,6), (8,7), (8,8) $\rightarrow$ size 3 $\rightarrow$ B
    *   *Re-evaluating the pattern:*
        *   Size 1: B
        *   Size 2: G
        *   Size 3: R (usually), but Cluster F is size 3 and is B.
        *   Let's look at the positions. Maybe it's based on size modulo something? Or a specific scale?
        *   Let's check the sizes again: 3(R), 2(G), 3(R), 1(B), 4(B), 3(B).
        *   Actually, looking at the output of Example 1 again:
            (8,6)-(8,8) is `B B B`. (8,1)-(9,2) is `B B / B B`. (7,7) is `B`.
            All clusters with size $\ge 4$ or size 1 are Blue? No.
            Let's try: Size 1=B, 2=G, 3=R, 4=B, 5=B...
            Wait, look at the output again: (8,6)-(8,8) is size 3. It is Blue. (2,2)-(2,4) is size 3. It is Red.
            Is it based on the *shape* or *location*?
            Let's re-read: "Count the touching shards... determine which deity claims the omen".
            Maybe the rule is: Size 1=B, 2=G, 3=R, 4=B, 5=Y, 6=M... (a cycle of colors).
            Wait, if Size 3 can be both R and B, the rule must be different.
            Let's re-examine Example 1 output:
            (2,2-4) R R R (size 3)
            (2,7-3,7) G G (size 2)
            (5,3-6,3, 5,4) R R R (size 3)
            (7,7) B (size 1)
            (8,1-9,2) B B B B (size 4)
            (8,6-8,8) B B B (size 3)
            Wait, (8,6-8,8) is size 3. (2,2-4) is size 3. One is R, one is B.
            Is it the number of clusters of that size?
            Size 3: 3 clusters. Size 2: 1 cluster. Size 1: 1 cluster. Size 4: 1 cluster.
            This is confusing. Let's look at the test input.

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

    If we use the mapping: 1=B, 2=G, 3=R, 4=B...
    Cluster 1 (4) $\rightarrow$ B
    Cluster 2 (4) $\rightarrow$ B
    Cluster 3 (3) $\rightarrow$ R
    Cluster 4 (3) $\rightarrow$ R
    Cluster 5 (2) $\rightarrow$ G
    Cluster 6 (2) $\rightarrow$ G

    Let's check if this fits Example 1:
    Size 1: B
    Size 2: G
    Size 3: R
    Size 4: B
    Size 5: ?
    Example 1:
    Size 3 $\rightarrow$ R (Correct)
    Size 2 $\rightarrow$ G (Correct)
    Size 3 $\rightarrow$ R (Correct)
    Size 1 $\rightarrow$ B (Correct)
    Size 4 $\rightarrow$ B (Correct)
    Size 3 $\rightarrow$ B (Wait, (8,6-8,8) is size 3 and is B).
    Let me re-count (8,6-8,8) in Example 1.
    Input: `. X X X .` at (8,6), (8,7), (8,8). Tha

... [1905 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   `.` (Black): The "dark stage floor" (background).
    *   `X` (Grey): The "dancers" forming clusters.
    *   Colors (`R`, `G`, `B`, etc.): The "costume colors" assigned based on the size of the cluster.

2.  **Identify the AFFORDANCES:**
    *   The grey cells (`X`) form connected components (clusters).
    *   The size of each cluster (number of `X` cells) determines the color it will be changed to.

3.  **Identify the TRANSFORMATION RULE:**
    *   Analyze Example 1 to find the mapping:
        *   Cluster 1: (2,2), (2,3), (2,4) -> 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).
        *   Cluster 5: (8,1), (8,2), (9,1), (9,2) -> Size 4. Color: `B` (Blue).
        *   Cluster 6: (8,6), (8,7), (8,8) -> Size 3. Color: `B` (Blue).
    *   *Wait, let's re-evaluate the mapping from Example 1:*
        *   Size 1: (7,7) -> `B`
        *   Size 2: (2,7), (3,7) -> `G`
        *   Size 3: (2,2-4) -> `R`; (5,3-4, 6,3) -> `R`; (8,6-8) -> `B` (Wait, (8,6-8) is size 3 and is `B`).
        *   Size 4: (8,1-2, 9,1-2) -> `B`.
    *   *Correction:* Let's look closer at the clusters in Example 1.
        *   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) [Size 1] -> `B`
        *   Cluster E: (8,1), (8,2), (9,1), (9,2) [Size 4] -> `B`
        *   Cluster F: (8,6), (8,7), (8,8) [Size 3] -> `B`
    *   *Re-evaluating the rule:* It seems the colors are assigned by size, but there might be a pattern.
        *   Size 1: `B`
        *   Size 2: `G`
        *   Size 3: `R` (mostly)
        *   Size 4: `B`
        *   Wait, Cluster F (Size 3) is `B`. Let's check the coordinates again.
        *   Example 1 Output:
            (2,2-4) = R R R (Size 3)
            (2,7), (3,7) = G G (Size 2)
            (5,3-4), (6,3) = R R R (Size 3)
            (7,7) = B (Size 1)
            (8,1-2), (9,1-2) = B B B B (Size 4)
            (8,6-8) = B B B (Size 3)
    *   *Hypothesis:* The colors are assigned based on size: 1=B, 2=G, 3=R, 4=B... No, that's not consistent.
    *   *Alternative Hypothesis:* Look at the colors again.
        Size 1: B
        Size 2: G
        Size 3: R (Top/Middle), B (Bottom)
        Size 4: B
    *   *Let's look at the 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

    *   *Re-examining Example 1 mapping:*
        Size 1 -> B
        Size 2 -> G
        Size 3 -> R
        Size 4 -> B
        Wait, Cluster F (8,6-8) is size 3 and is B. Let me re-read the output grid.
        Example 1 Output row 8: `. B B . . . B B B .`
        Indices: 0 1 2 3 4 5 6 7 8 9
        (8,1), (8,2) are B. (8,6), (8,7), (8,8) are B.
        (9,1), (9,2) are B.
        So Cluster E (8,1-2, 9,1-2) is size 4 -> B.
        Cluster F (8,6-8) is size 3 -> B.
        Cluster A (2,2-4) is size 3 -> R.
        Cluster C (5,3-4, 6,3) is size 3 -> R.
        This means size 3 can be R or B. This is confusing.

    *   *Let's look at the positions:*
        Cluster A (Size 3) is at row 2.
        Cluster C (Size 3) is at row 5.
        Cluster F (Size 3) is at row 8.
        Maybe the color depends on the row?
        Row 2: R
        Row 5: R
        Row 8: B
        Or maybe the color depends on the size AND the row?
        Size 1: B
        Size 2: G
        Size 3: R (if row < 7), B (if row >= 7)
        Size 4: B

    *   *Let's test this on the Test Input:*
        1. Size 4 (0,9...): B
        2. Size 4 (2,2...): B
        3. Size 3 (2,5...): R (row 2 < 7)
        4. Size 3 (6,0...): R (row 6 < 7)
        5. Size 2 (7,3...): G
        6. Size 2 (7,7...): G

### Applying the Rule to Test Input

*   Cluster 1: (0,9), (1,9), (2,9), (3,9) [Size 4] -> B
*   Cluster 2: (2,2), (2,3), (3,1), (3,2) [Size 4] -> B
*   Cluster 3: (2,5), (2,6), (3,6) [Size 3] -> R
*   Cluster 4: (6,0), (7,0), (8,0) [Size 3] -> R
*   Cluster 5: (7,3), (7,4) [Size 2] -> G
*   Cluster 6: (7,7), (8,7) [Size 2] -> G

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

### Conceptual Digestion

1.  **CAST**:
    *   `.` (Black): Bare rock (background).
    *   `X` (Grey): Mineral veins (connected components of `X`).
    *   `B, R, G, Y, M, O, A, W`: Survey colors based on the size of the connected vein.

2.  **AFFORDANCES**:
    *   Connected components of `X` (adjacent horizontally or vertically) can be identified and their size (number of cells) counted.
    *   Each connected component is then recolored based on its total size.

3.  **TRANSFORMATION RULE**:
    *   Identify all connected components of `X`.
    *   Count the number of cells in each component.
    *   Apply the color mapping based on the training examples:
        *   Size 1: `B` (Blue) - *Wait, let's re-examine Example 1.*
            *   Ex 1: Vein at (2,7)-(3,7) size 2 -> `G`. Vein at (7,7) size 1 -> `B`. Vein at (8,6)-(8,8) size 3 -> `B`. Vein at (8,1)-(9,2) size 4 -> `B`. Vein at (2,1)-(2,3) size 3 -> `R`. Vein at (5,3)-(6,3) size 3 -> `R`.
            *   *Correction*: Let's look closer at Example 1 and 2.
            *   Ex 1:
                *   (2,1)-(2,3): size 3 -> `R`
                *   (5,3)-(6,3): size 3 -> `R`
                *   (2,7)-(3,7): size 2 -> `G`
                *   (7,7): size 1 -> `B`
                *   (8,6)-(8,8): size 3 -> `B` (Wait, (8,6)-(8,8) is size 3, but (8,1)-(9,2) is size 4. Both are `B`?)
                *   (8,1)-(9,2): size 4 -> `B`
            *   Ex 2:
                *   (1,8)-(2,9): size 4 -> `B`
                *   (2,1)-(3,3): size 4 -> `B`
                *   (5,9)-(7,9): size 3 -> `R`
                *   (6,5)-(6,6): size 2 -> `G`
                *   (7,1)-(8,1): size 2 -> `G`
                *   (8,4)-(9,5): size 3 -> `R`

    *   **Revised Color Mapping**:
        *   Size 1: `B` (Blue)
        *   Size 2: `G` (Green)
        *   Size 3: `R` (Red)
        *   Size 4: `B` (Blue)
        *   Size 5+: (Not seen, but likely follows a pattern or continues).
        *   *Wait*, let's re-count Example 1 (8,6)-(8,8) is size 3. (8,1)-(9,2) is size 4. Both are `B`.
        *   Let's re-count Example 1 (2,1)-(2,3) is size 3. (5,3)-(6,3) is size 3. Both are `R`.
        *   Wait, (8,6)-(8,8) is size 3, but it's `B`. (2,1)-(2,3) is size 3, but it's `R`.
        *   Let's look at the shapes.
        *   Ex 1: (2,1)-(2,3) is a line of 3. (5,3)-(6,3) is a line of 2? No, (5,3)-(5,4) and (6,3). That's 3.
        *   Ex 1: (8,6)-(8,8) is a line of 3. (8,1)-(9,2) is a 2x2 block.
        *   Let's re-evaluate:
            *   Size 1: `B`
            *   Size 2: `G`
            *   Size 3: `R`
            *   Size 4: `B`
            *   Wait, in Ex 1, (8,6)-(8,8) is size 3 and is `B`. (2,1)-(2,3) is size 3 and is `R`.
            *   Let's look at the coordinates again.
            *   Ex 1: (2,1),(2,2),(2,3) -> `R`. (5,3),(5,4),(6,3) -> `R`. (2,7),(3,7) -> `G`. (7,7) -> `B`. (8,6),(8,7),(8,8) -> `B`. (8,1),(8,2),(9,1),(9,2) -> `B`.
            *   Ex 2: (1,8),(1,9),(2,8),(2,9) -> `B`. (2,1),(2,2),(3,2),(3,3) -> `B`. (5,9),(6,9),(7,9) -> `R`. (6,5),(6,6) -> `G`. (7,1),(8,1) -> `G`. (8,4),(9,4),(9,5) -> `R`.

    *   **Corrected Mapping**:
        *   Size 1: `B`
        *   Size 2: `G`
        *   Size 3: `R`
        *   Size 4: `B`
        *   Wait, Ex 1 (8,6)-(8,8) is size 3 and is `B`. Let me re-read the grid.
        *   Ex 1 Output: `(8,6) B, (8,7) B, (8,8) B`. That is size 3.
        *   Ex 1 Output: `(2,1) R, (2,2) R, (2,3) R`. That is size 3.
        *   Is it based on position? No. Is it based on shape?
        *   (2,1)-(2,3) is horizontal. (8,6)-(8,8) is horizontal.
        *   Let's re-count Ex 1 (8,6)-(8,8) again. `X X X` at the end of row 8. That's 3.
        *   Wait! Look at Ex 1 Output row 8: `. . . . . . . B . .` (index 7 is B). Row 9: `. B B . . . B B B .` (indices 1,2 and 6,7,8 are B).
        *   So (7,7) is `B` (size 1). (8,6),(8,7),(8,8) is `B` (size 3). (8,1),(8,2),(9,1),(9,2) is `B` (size 4).
        *   This is confusing. Let's look at the colors again.
        *   Ex 1: Size 1=B, 2=G, 3=R, 4=B. But (8,6-8) is size 3 and is B.
        *   Wait, (8,6-8) is `B B B`. (7,7) is `B`. (8,1-2, 9,1-2) is `B B B B`.
        *   Maybe the rule is: Size 1=B, 2=G, 3=R, 4=B, 5=G, 6=R... (Modulo 3)?
        *   Size 1: 1 mod 3 = 1 (B)
        *   Size 2: 2 mod 3 = 2 (G)
        *   Size 3: 3 mod 3 = 0 (R)
        *   Size 4: 4 mod 3 = 1 (B)
        *   Size 5: 5 mod 3 = 2 (G)
        *   Size 6: 6 mod 3 = 0 (R)
        *   Let's check Ex 1:
            *   (2,1-3): size 3 -> 0 mod 3 -> `R`. Correct.
            *   (5,3-6,3): size 3 -> 0 mod 3 -> `R`. Correct.
            *   (2,7-3,7): size 2 -> 2 mod 3 -> `G`. Correct.
            *   (7,7): size 1 -> 1 mod 3 -> `B`. Correct.
            *   (8,6-8): size 3 -> 0 mod 3 -> `R`. **Wait**, the output says `B B B`.
            *   Let me re-read Ex 1 Output row 9: `. B B . . . B B B .`
            

... [2551 chars truncated]

### Conceptual Digestion

1.  **CAST**:
    *   **Background (`.`)**: The cutting board (remains unchanged).
    *   **Mushrooms (`X`)**: Connected clusters of grey pixels.
    *   **Sauce (Colors)**: The final color assigned to a cluster based on its size (number of connected `X` pixels).

2.  **AFFORDANCES**:
    *   Clusters are defined by orthogonal connectivity (up, down, left, right).
    *   Each cluster's size determines its color transformation.

3.  **TRANSFORMATION RULE**:
    *   Analyze Example 1 to determine the weight-to-color mapping:
        *   Cluster 1: (2,2) to (2,4) $\rightarrow$ size 3 $\rightarrow$ **Red (R)**
        *   Cluster 2: (2,7) to (3,7) $\rightarrow$ size 2 $\rightarrow$ **Green (G)**
        *   Cluster 3: (5,3) to (6,3) and (5,4) $\rightarrow$ size 3 $\rightarrow$ **Red (R)**
        *   Cluster 4: (7,7) $\rightarrow$ size 1 $\rightarrow$ **Blue (B)**
        *   Cluster 5: (8,1) to (9,2) $\rightarrow$ size 4 $\rightarrow$ **Blue (B)**
        *   Cluster 6: (8,6) to (8,8) $\rightarrow$ size 3 $\rightarrow$ **Blue (B)**
    *   *Wait, let's re-evaluate Example 1 carefully:*
        *   Cluster (2,2)-(2,4): size 3 $\rightarrow$ R
        *   Cluster (2,7)-(3,7): size 2 $\rightarrow$ G
        *   Cluster (5,3)-(6,3) and (5,4): size 3 $\rightarrow$ R
        *   Cluster (7,7): size 1 $\rightarrow$ B
        *   Cluster (8,1)-(9,2): size 4 $\rightarrow$ B
        *   Cluster (8,6)-(8,8): size 3 $\rightarrow$ B
    *   *Correction on Example 1 mapping:*
        *   Size 1: Blue (B)
        *   Size 2: Green (G)
        *   Size 3: Red (R) or Blue (B)? Let's look again.
        *   (2,2)-(2,4) is size 3 $\rightarrow$ R.
        *   (5,3)-(6,3) and (5,4) is size 3 $\rightarrow$ R.
        *   (8,6)-(8,8) is size 3 $\rightarrow$ B.
    *   *Re-counting Example 1 clusters:*
        *   (2,2), (2,3), (2,4) = 3 $\rightarrow$ R
        *   (2,7), (3,7) = 2 $\rightarrow$ G
        *   (5,3), (5,4), (6,3) = 3 $\rightarrow$ R
        *   (7,7) = 1 $\rightarrow$ B
        *   (8,1), (8,2), (9,1), (9,2) = 4 $\rightarrow$ B
        *   (8,6), (8,7), (8,8) = 3 $\rightarrow$ B.
    *   *Wait, (8,6)-(8,8) is size 3 and is Blue. (2,2)-(2,4) is size 3 and is Red. This suggests the rule is not just size, or I miscounted.*
    *   *Let's re-examine Example 1 Output:*
        *   (2,2-4) = R R R (3)
        *   (2,7-3,7) = G G (2)
        *   (5,3-6,3) = R R R (3)
        *   (7,7) = B (1)
        *   (8,1-9,2) = B B B B (4)
        *   (8,6-8,8) = B B B (3)
    *   *Looking at the colors again: B is used for sizes 1, 3, and 4. R for 3. G for 2. This is inconsistent. Let's look at the positions.*
    *   *Maybe the rule is: Size 1=B, Size 2=G, Size 3=R, Size 4=B? But (8,6-8,8) is size 3 and is B. Let me re-read the grid.*
    *   *Example 1 Input (8,6-8,8) is X X X. Output is B B B. (2,2-2,4) is X X X. Output is R R R. (5,3-6,3) is X X X. Output is R R R.*
    *   *Wait! (8,6-8,8) is actually (8,6), (8,7), (8,8). That is size 3. (2,2-2,4) is size 3. (5,3-6,3) is size 3. Why is one B and others R?*
    *   *Let's look at the coordinates. (8,6-8,8) is at the bottom right. (2,2-2,4) is top left. (5,3-6,3) is middle. Maybe it's based on the number of clusters of that size? No.*
    *   *Let's try: Size 1=B, Size 2=G, Size 3=R, Size 4=B. If (8,6-8,8) is actually size 3, why is it B? Let me re-count the input again.*
    *   *Example 1 Input: (8,6) is X, (8,7) is X, (8,8) is X. That's 3. (8,1) is X, (8,2) is X, (9,1) is X, (9,2) is X. That's 4. (7,7) is X. That's 1. All these are B. So Size 1, 3, 4 = B? No, that's unlikely.*
    *   *Let's re-examine the output for (8,6-8,8). It is B B B. (8,1-9,2) is B B B B. (7,7) is B. (2,2-2,4) is R R R. (5,3-6,3) is R R R. (2,7-3,7) is G G.*
    *   *Hypothesis: Size 1=B, Size 2=G, Size 3=R, Size 4=B. But (8,6-8,8) is size 3. Is it connected to something else? (7,7) is X. (8,7) is X. YES! (7,7) and (8,6-8,8) are connected!*
    *   *Recounting Example 1 clusters with connectivity:*
        *   Cluster A: (2,2), (2,3), (2,4) $\rightarrow$ Size 3 $\rightarrow$ **Red (R)**
        *   Cluster B: (2,7), (3,7) $\rightarrow$ Size 2 $\rightarrow$ **Green (G)**
        *   Cluster C: (5,3), (5,4), (6,3) $\rightarrow$ Size 3 $\rightarrow$ **Red (R)**
        *   Cluster D: (7,7), (8,6), (8,7), (8,8) $\rightarrow$ Size 4 $\rightarrow$ **Blue (B)**
        *   Cluster E: (8,1), (8,2), (9,1), (9,2) $\rightarrow$ Size 4 $\rightarrow$ **Blue (B)**
    *   *Revised Rule: Size 1= ?, Size 2=G, Size 3=R, Size 4=B.*

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


... [559 chars truncated]

### Conceptual Digestion

1.  **Identify the CAST:**
    *   `.` (Black): The background/silent measures.
    *   `X` (Grey): The "notes" forming "phrases" (connected components of grey cells).
    *   Colors (R, G, B, etc.): The "instrument sections" assigned based on the length of the phrase.

2.  **Identify the AFFORDANCES:**
    *   Grey cells (`X`) can be grouped into connected components (phrases) using 4-connectivity (up, down, left, right).
    *   The size (number of cells) of each connected component determines its new color.

3.  **Identify the TRANSFORMATION RULE:**
    *   Analyze Example 1 to find the mapping of length $\rightarrow$ color:
        *   Phrase 1: (2,2), (2,3), (2,4) $\rightarrow$ Length 3. Output: Red (R).
        *   Phrase 2: (2,7), (3,7) $\rightarrow$ Length 2. Output: Green (G).
        *   Phrase 3: (5,3), (5,4), (6,3) $\rightarrow$ Length 3. Output: Red (R).
        *   Phrase 4: (7,7) $\rightarrow$ Length 1. Output: Blue (B).
        *   Phrase 5: (8,1), (8,2), (9,1), (9,2) $\rightarrow$ Length 4. Output: Blue (B).
        *   Phrase 6: (8,6), (8,7), (8,8) $\rightarrow$ Length 3. Output: Blue (B).
        *   *Wait, let's re-evaluate Example 1 mapping:*
            *   Length 1: (7,7) $\rightarrow$ B
            *   Length 2: (2,7), (3,7) $\rightarrow$ G
            *   Length 3: (2,2-4) $\rightarrow$ R; (5,3-4, 6,3) $\rightarrow$ R; (8,6-8) $\rightarrow$ B.
            *   Length 4: (8,1-2, 9,1-2) $\rightarrow$ B.
        *   *Correction:* Let's look at the lengths again.
            *   Phrase (2,2-4): 3 cells $\rightarrow$ R
            *   Phrase (2,7), (3,7): 2 cells $\rightarrow$ G
            *   Phrase (5,3-4), (6,3): 3 cells $\rightarrow$ R
            *   Phrase (7,7): 1 cell $\rightarrow$ B
            *   Phrase (8,1-2), (9,1-2): 4 cells $\rightarrow$ B
            *   Phrase (8,6-8): 3 cells $\rightarrow$ B
        *   *Wait, the mapping is likely based on the specific lengths present in the test set or a fixed scale.*
        *   Let's re-examine Example 1 output:
            *   (2,2-4) is R (3)
            *   (2,7-3,7) is G (2)
            *   (5,3-4, 6,3) is R (3)
            *   (7,7) is B (1)
            *   (8,1-2, 9,1-2) is B (4)
            *   (8,6-8) is B (3)
        *   Actually, looking closer at Example 1:
            *   Length 1: B
            *   Length 2: G
            *   Length 3: R (mostly), but (8,6-8) is B.
            *   Length 4: B.
        *   Let's re-count Example 1 carefully.
            *   (2,2), (2,3), (2,4) = 3 cells $\rightarrow$ R
            *   (2,7), (3,7) = 2 cells $\rightarrow$ G
            *   (5,3), (5,4), (6,3) = 3 cells $\rightarrow$ R
            *   (7,7) = 1 cell $\rightarrow$ B
            *   (8,1), (8,2), (9,1), (9,2) = 4 cells $\rightarrow$ B
            *   (8,6), (8,7), (8,8) = 3 cells $\rightarrow$ B
        *   This suggests: Length 1=B, 2=G, 3=R (usually), 4=B. But (8,6-8) is length 3 and is B.
        *   Let's look at the *position* or *shape*? No, the metaphor says "phrase length".
        *   Let's re-read: "assign it to the instrument section that owns that phrase length".
        *   Maybe: Length 1: B, Length 2: G, Length 3: R, Length 4: B, Length 5: ?
        *   Wait, (8,6-8) in Example 1 is (8,6), (8,7), (8,8). That is length 3. Why is it B?
        *   Let's re-count (8,1-2, 9,1-2). That is 4 cells. It is B.
        *   Let's re-count (7,7). That is 1 cell. It is B.
        *   So Lengths 1, 3 (sometimes), and 4 are B? That's unlikely.
        *   Let's re-examine Example 1 Output again.
            (8,6), (8,7), (8,8) is B B B.
            (8,1), (8,2), (9,1), (9,2) is B B / B B.
            (7,7) is B.
            All these are in the bottom right/bottom left.
            (2,2-4) is R R R.
            (5,3-4, 6,3) is R R / R.
            (2,7-3,7) is G G.
            Mapping: Length 2 $\rightarrow$ G, Length 3 $\rightarrow$ R, Length 1 $\rightarrow$ B, Length 4 $\rightarrow$ B.
            Wait, (8,6-8) is length 3. If it's B, then Length 3 can be R or B?
            Let's look at the grid again. (8,6), (8,7), (8,8) is length 3. (2,2), (2,3), (2,4) is length 3.
            Maybe the rule is: Length 1=B, 2=G, 3=R, 4=B, 5=...
            Wait! (8,6), (8,7), (8,8) is length 3. Is it possible it's connected to (7,7)?
            (7,7) is row 7, col 7. (8,6) is row 8, col 6. (8,7) is row 8, col 7. (8,8) is row 8, col 8.
            Yes! (7,7) is adjacent to (8,7).
            So the phrase is {(7,7), (8,6), (8,7), (8,8)}. Total length = 4.
            Now the mapping is: Length 2 $\rightarrow$ G, Length 3 $\rightarrow$ R, Length 4 $\rightarrow$ B.
            Let's check the other B: (8,1), (8,2), (9,1), (9,2). Length = 4.
            Mapping: **Length 2 = G, Length 3 = R, Length 4 = B.**

4.  **Apply to Test Input:**
    *   Phrase 1: (0,9), (1,9), (2,9), (3,9) $\rightarrow$ Length 4 $\rightarrow$ **B**
    *   Phrase 2: (2,2), (2,3), (3,1), (3,2) $\rightarrow$ Length 4 $\rightarrow$ **B**
  

... [753 chars truncated]

### Conceptual Digestion

1.  **CAST**:
    *   `.` (Black): Empty field/background.
    *   `X` (Grey): Unassigned players.
    *   `B, R, G, Y, M, O, A, W`: Jersey colors assigned based on group size.

2.  **AFFORDANCES**:
    *   Grey cells (`X`) form connected components (groups) via orthogonal adjacency (up, down, left, right).
    *   Each group is analyzed for its total count of cells.
    *   The color of the group is determined by the size of that group.

3.  **TRANSFORMATION RULE**:
    *   Identify all connected components of `X`.
    *   Count the number of cells in each component.
    *   Assign colors based on the training examples:
        *   Example 1:
            *   Group size 1: `G` (at 2,7) - *Wait, let's re-examine.*
            *   Example 1 analysis:
                *   (2,1)-(2,3): size 3 -> `R`
                *   (2,7)-(3,7): size 2 -> `G`? No, (2,7) is size 1? No, (2,7) and (3,7) are connected. Size 2.
                *   (5,3)-(5,4) and (6,3): size 3 -> `R`
                *   (7,7): size 1 -> `B`
                *   (8,1)-(9,2): size 4 -> `B`
                *   (8,6)-(8,8): size 3 -> `B`?
                *   *Correction*: Let's look closer at Example 1.
                *   (2,1-3): 3 cells -> `R`
                *   (2,7-3,7): 2 cells -> `G`
                *   (5,3-4, 6,3): 3 cells -> `R`
                *   (7,7): 1 cell -> `B`
                *   (8,1-9,2): 4 cells -> `B`
                *   (8,6-8): 3 cells -> `B`? No, (8,6-8) is 3 cells.
                *   Wait, let's re-count Example 1 carefully.
                *   Group 1: (2,1), (2,2), (2,3) -> Size 3 -> `R`
                *   Group 2: (2,7), (3,7) -> Size 2 -> `G`
                *   Group 3: (5,3), (5,4), (6,3) -> Size 3 -> `R`
                *   Group 4: (7,7) -> Size 1 -> `B`
                *   Group 5: (8,1), (8,2), (9,1), (9,2) -> Size 4 -> `B`
                *   Group 6: (8,6), (8,7), (8,8) -> Size 3 -> `B`? No, the output says `B B B`.
                *   Wait, looking at Example 1 output: (8,6-8) is `B B B`. (8,1-9,2) is `B B / B B`. (7,7) is `B`.
                *   It seems size 1, 3, and 4 are all `B`? No.
                *   Let's re-evaluate Example 2.
                *   (1,8-9, 2,8-9): Size 4 -> `B`
                *   (2,1-2, 3,2-3): Size 4 -> `B`
                *   (5,9, 6,9, 7,9): Size 3 -> `R`
                *   (6,5-6, 7,0-1): No. (6,5-6) is size 2 -> `G`. (7,0-8,0) is size 2 -> `G`.
                *   (8,4, 9,3-4): Size 3 -> `R`.
                *   Summary of Example 2: Size 4 = `B`, Size 3 = `R`, Size 2 = `G`.
                *   Re-checking Example 1 with this: Size 3 = `R`, Size 2 = `G`, Size 1 = `B`, Size 4 = `B`.
                *   Wait, in Example 1: (8,6-8) is size 3, but it's `B`. (2,1-3) is size 3, it's `R`.
                *   Let's look at the coordinates again. (8,6), (8,7), (8,8) is size 3. (2,1), (2,2), (2,3) is size 3. Why is one `B` and one `R`?
                *   Maybe it's not size, but shape? Or maybe I miscounted.
                *   Example 1: (8,6-8) is size 3. (2,1-3) is size 3. (5,3-4, 6,3) is size 3.
                *   Wait! (8,6-8) is `B B B`. (2,1-3) is `R R R`. (5,3-4, 6,3) is `R R R`.
                *   Is it possible that size 3 is `R` and size 1 is `B`, and (8,6-8) is actually size 1? No, they are adjacent.
                *   Let's look at Example 1 again. (7,7) is `B`. (8,6-8) is `B B B`. (8,1-9,2) is `B B B B`.
                *   Maybe the rule is: Size 1 = `B`, Size 2 = `G`, Size 3 = `R`, Size 4 = `B`?
                *   Let's check Example 2: Size 4 = `B`, Size 3 = `R`, Size 2 = `G`.
                *   This fits! Size 1=B, 2=G, 3=R, 4=B.

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

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

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

### Conceptual Digestion

1.  **CAST**:
    *   `.` (Black): Background/Clear sky.
    *   `X` (Grey): Storm cells. Connected groups of `X` form a "weather system".
    *   Colors (`B`, `R`, `G`, etc.): Alert levels based on the size of the system.

2.  **AFFORDANCES**:
    *   Connected components of `X` (orthogonal adjacency) can be measured by their total cell count.
    *   Each size category maps to a specific color.

3.  **TRANSFORMATION RULE**:
    *   Analyze Example 1 to determine the size-to-color mapping:
        *   System 1: (2,2), (2,3), (2,4) -> Size 3. Color: `R` (Red).
        *   System 2: (2,7), (3,7) -> Size 2. Color: `G` (Green).
        *   System 3: (5,3), (5,4), (6,3) -> Size 3. Color: `R` (Red).
        *   System 4: (7,7) -> Size 1. Color: `B` (Blue).
        *   System 5: (8,1), (8,2), (9,1), (9,2) -> Size 4. Color: `B` (Blue).
        *   System 6: (8,6), (8,7), (8,8) -> Size 3. Color: `B` (Blue).
    *   *Wait, let's re-evaluate Example 1 carefully:*
        *   (2,2)-(2,4): Size 3 -> Red
        *   (2,7)-(3,7): Size 2 -> Green
        *   (5,3)-(6,3): Size 3 -> Red
        *   (7,7): Size 1 -> Blue
        *   (8,1)-(9,2): Size 4 -> Blue
        *   (8,6)-(8,8): Size 3 -> Blue
    *   *Correction on Example 1 analysis:*
        *   Size 1: Blue
        *   Size 2: Green
        *   Size 3: Red (Wait, (8,6)-(8,8) is size 3 and is Blue).
    *   *Let's re-count Example 1 again:*
        *   System A: (2,2), (2,3), (2,4) -> Size 3. Output: `R R R`
        *   System B: (2,7), (3,7) -> Size 2. Output: `G G`
        *   System C: (5,3), (5,4), (6,3) -> Size 3. Output: `R R R`
        *   System D: (7,7) -> Size 1. Output: `B`
        *   System E: (8,1), (8,2), (9,1), (9,2) -> Size 4. Output: `B B B B`
        *   System F: (8,6), (8,7), (8,8) -> Size 3. Output: `B B B`
    *   *Wait, System F is size 3 and is Blue. System A and C are size 3 and are Red. Let's check connectivity again.*
        *   System A: (2,2), (2,3), (2,4) - Horizontal line.
        *   System C: (5,3), (5,4), (6,3) - L-shape.
        *   System F: (8,6), (8,7), (8,8) - Horizontal line.
    *   *Let's look at the sizes again:*
        *   Size 1: Blue
        *   Size 2: Green
        *   Size 3: Red (A, C)
        *   Size 3: Blue (F) - *Wait, (8,6), (8,7), (8,8) is size 3. Why is it Blue?*
        *   Size 4: Blue (E)
    *   *Let's re-examine the output of Example 1:*
        (8,1)(8,2)(9,1)(9,2) is Blue. (8,6)(8,7)(8,8) is Blue. (7,7) is Blue.
        These are all in the bottom section.
        (2,2)(2,3)(2,4) is Red. (5,3)(5,4)(6,3) is Red.
        (2,7)(3,7) is Green.
    *   *Hypothesis:*
        Size 1: Blue
        Size 2: Green
        Size 3: Red
        Size 4+: Blue
    *   *Check System F again:* (8,6), (8,7), (8,8) is size 3. If the rule is Size 3 = Red, it should be Red. But it is Blue.
    *   *Alternative Hypothesis:*
        Size 1: Blue
        Size 2: Green
        Size 3: Red
        Size 4: Blue
        Wait, maybe the rule is based on the *number* of cells?
        Size 1: Blue
        Size 2: Green
        Size 3: Red
        Size 4: Blue
        Size 5: ?
        Let's look at System F again. (8,6), (8,7), (8,8). That is 3 cells. Why is it Blue?
        Looking closer at Example 1 Output:
        Row 8: `. B B . . . B B B .`
        Row 9: `. B B . . . . . . .`
        Cells at (8,6), (8,7), (8,8) are indeed Blue.
        Cells at (8,1), (8,2), (9,1), (9,2) are Blue.
        Cell at (7,7) is Blue.
        All these are Blue. Sizes are 3, 4, and 1.
        Wait! Maybe the rule is:
        Size 1: Blue
        Size 2: Green
        Size 3: Red
        ...but if it's in the bottom half? No.
        Let's look at the shapes.
        Size 3 (Line): Red (A), Blue (F).
        Size 3 (L-shape): Red (C).
        This is confusing. Let's try:
        Size 1: Blue
        Size 2: Green
        Size 3: Red
        Size 4: Blue
        Wait, if Size 3 is Red, why is System F Blue? Let me re-count System F.
        (8,6), (8,7), (8,8). That is 3 cells.
        Is it possible System F is connected to (7,7)?
        (7,7) is Blue. (8,7) is Blue. Yes! They are connected!
        System (7,7), (8,6), (8,7), (8,8) = Size 4.
        Now the rule is:
        Size 1: ?
        Size 2: Green
        Size 3: Red
        Size 4: Blue
        Let's check Example 1 again.
        System A: Size 3 -> Red
        System B: Size 2 -> Green
        System C: Size 3 -> Red
        System D+F: Size 4 -> Blue
        System E: Size 4 -> Blue
        Rule: Size 2 = Green, Size 3 = Red, Size 4 = Blue.

4.  **Applying to Test Input**:
    *   System 1: (0,9), (1,9), (2,9), (3,9) -> Size 4. Color: Blue (`B`).
    *   System 2: (2,2), (2,3), (3,1), (3,2) -> Size 4. Color: Blue (`B`).
    *   System 3: (2,5), (2,6), (3,6) -> Size 3. Color: Red (`R`).
    *   System 4: (6,0), (7,0), (8,0) -> Size 3. Color: Red (`R`).
    *   System 5: (7,3), (

... [534 chars truncated]

### Conceptual Digestion

1.  **CAST**:
    *   `.` (Black): Unmarked terrain (background).
    *   `X` (Grey): Enemy markers forming units (connected components).
    *   Colored Pins (`B`, `R`, `G`, etc.): Strength-coded replacements for the grey markers.

2.  **AFFORDANCES**:
    *   Grey markers (`X`) can be grouped into "units" based on connectivity (orthogonal/adjacent).
    *   The size (number of `X` cells) of each unit determines the color it will be replaced with.

3.  **TRANSFORMATION RULE**:
    *   Identify all connected components of `X`.
    *   Count the number of cells in each component.
    *   Map the size to a color based on the training example:
        *   Size 1 $\rightarrow$ `B` (Blue) - *Wait, let's re-examine Example 1.*
        *   Example 1 analysis:
            *   Unit 1: (2,1), (2,2), (2,3) $\rightarrow$ Size 3 $\rightarrow$ `R` (Red)
            *   Unit 2: (2,7), (3,7) $\rightarrow$ Size 2 $\rightarrow$ `G` (Green)
            *   Unit 3: (5,3), (5,4), (6,3) $\rightarrow$ Size 3 $\rightarrow$ `R` (Red)
            *   Unit 4: (7,7) $\rightarrow$ Size 1 $\rightarrow$ `B` (Blue)
            *   Unit 5: (8,1), (8,2), (9,1), (9,2) $\rightarrow$ Size 4 $\rightarrow$ `B` (Blue) - *Wait, this is confusing.*
            *   Unit 6: (8,6), (8,7), (8,8) $\rightarrow$ Size 3 $\rightarrow$ `B` (Blue) - *Wait, let me re-count.*

    *   **Re-evaluating Example 1 carefully**:
        *   Unit A: (2,1), (2,2), (2,3) [Size 3] $\rightarrow$ `R`
        *   Unit B: (2,7), (3,7) [Size 2] $\rightarrow$ `G`
        *   Unit C: (5,3), (5,4), (6,3) [Size 3] $\rightarrow$ `R`
        *   Unit D: (7,7) [Size 1] $\rightarrow$ `B`
        *   Unit E: (8,1), (8,2), (9,1), (9,2) [Size 4] $\rightarrow$ `B`
        *   Unit F: (8,6), (8,7), (8,8) [Size 3] $\rightarrow$ `B`
        *   *Correction*: Looking at the output of Example 1 again:
            *   (2,1-3) is `R R R` (Size 3)
            *   (2,7), (3,7) is `G G` (Size 2)
            *   (5,3-4), (6,3) is `R R R` (Size 3)
            *   (7,7) is `B` (Size 1)
            *   (8,1-2), (9,1-2) is `B B B B` (Size 4)
            *   (8,6-8) is `B B B` (Size 3)
        *   *Wait*, the mapping is not consistent by size alone. Let's look at the colors again.
        *   Size 1: `B`
        *   Size 2: `G`
        *   Size 3: `R` (sometimes) or `B` (sometimes).
        *   Size 4: `B`
        *   Let's look at the *position* or *shape*? No, the metaphor says "counting its connected positions".
        *   Let's re-read the output of Example 1 one more time.
            *   (8,6), (8,7), (8,8) is `B B B`. (Size 3)
            *   (2,1), (2,2), (2,3) is `R R R`. (Size 3)
            *   (5,3), (5,4), (6,3) is `R R R`. (Size 3)
        *   Is it possible the color depends on the size, but the training example output provided in the prompt has a typo or a different logic?
        *   Let's check the sizes again:
            *   Size 1: `B`
            *   Size 2: `G`
            *   Size 3: `R`
            *   Size 4: `B`
            *   Wait, (8,6-8) is size 3 and it is `B B B`. (8,1-2, 9,1-2) is size 4 and it is `B B B B`.
            *   Maybe the rule is: Size 1=B, 2=G, 3=R, 4=B, 5=... ? Or maybe Size $\le 2$ is B/G, Size $\ge 3$ is R? No.
            *   Let's look at the colors again: B=blue, R=red, G=green.
            *   Size 1: B
            *   Size 2: G
            *   Size 3: R
            *   Size 4: B (Wait, if it cycles B, G, R, B, G, R... then 1=B, 2=G, 3=R, 4=B).
            *   Let's test this cycle: 1=B, 2=G, 3=R, 4=B, 5=G, 6=R...
            *   Example 1:
                *   Size 3 $\rightarrow$ R (Correct)
                *   Size 2 $\rightarrow$ G (Correct)
                *   Size 3 $\rightarrow$ R (Correct)
                *   Size 1 $\rightarrow$ B (Correct)
                *   Size 4 $\rightarrow$ B (Correct)
                *   Size 3 $\rightarrow$ B (Wait, (8,6-8) is size 3, but output says `B B B`. Let me re-count).
                *   Input: `. X X . . . X X X .` (Row 8). `X` at indices 6, 7, 8. That is size 3.
                *   Output: `. B B . . . B B B .` (Row 8).
                *   Wait, (8,1-2) and (9,1-2) are also `B`. That's a size 4 block.
                *   So Size 3 can be R or B? Size 4 is B?
                *   Let's look at the coordinates again. (8,6-8) is a horizontal line. (2,1-3) is a horizontal line. (5,3-4, 6,3) is an L-shape.
                *   Maybe the rule is: Size 1=B, 2=G, 3=R, 4=B, 5=G, 6=R... and I misread the output?
                *   Example 1 Output Row 8: `. B B . . . B B B .`
                *   Example 1 Output Row 9: `. B B . . . . . . .`
                *   Wait, (8,1-2) and (9,1-2) are `B`. That's size 4. (8,6-8) is size 3. Both are `B`.
                *   This means Size 3 can be `R` or `B`. This is contradictory.
                *   Let's re-examine the input for Example 1 one more time.
                *   Row 2: `. X X X . . . X . .` (X at 1,2,3 and 7)
                *   Row 3: `.

... [4302 chars truncated]

### Conceptual Digestion

1.  **The CAST**:
    *   `.` (Black): The "dark stage" (background).
    *   `X` (Grey): The "performers" (actors).
    *   Groups of `X` (connected orthogonally): The "spotlight pools".

2.  **The AFFORDANCES**:
    *   The grey actors (`X`) are grouped into contiguous clusters.
    *   Each cluster is counted to determine its size.
    *   The color of the entire cluster is changed based on that size.

3.  **The TRANSFORMATION RULE**:
    *   Analyze Example 1 to find the size-to-color mapping:
        *   Cluster 1: (2,2) to (2,4) -> size 3. Color: `R` (Red).
        *   Cluster 2: (2,7) to (3,7) -> size 2. Color: `G` (Green). Wait, let's re-count.
        *   Example 1 re-evaluation:
            *   (2,1), (2,2), (2,3) -> size 3. Output: `R R R`.
            *   (2,7), (3,7) -> size 2. Output: `G G`.
            *   (5,3), (5,4), (6,3) -> size 3. Output: `R R R`.
            *   (7,7) -> size 1. Output: `B`.
            *   (8,1), (8,2), (9,1), (9,2) -> size 4. Output: `B B B B`.
            *   (8,6), (8,7), (8,8) -> size 3. Output: `B B B`.
        *   Wait, the mapping is inconsistent if I just use size. Let's look closer at Example 1.
        *   Cluster sizes: 3, 2, 3, 1, 4, 3.
        *   Colors: R, G, R, B, B, B.
        *   Actually, looking at the output of Example 1 again:
            *   (2,1-3) size 3 -> Red
            *   (2,7-3,7) size 2 -> Green
            *   (5,3-6,3) size 3 -> Red
            *   (7,7) size 1 -> Blue
            *   (8,1-9,2) size 4 -> Blue
            *   (8,6-8) size 3 -> Blue.
        *   Wait, (8,6-8) is size 3 and is Blue, but (2,1-3) is size 3 and is Red. This suggests the mapping might be based on something else or I miscounted.
        *   Let's re-examine Example 1 Output:
            (2,1-3) size 3: R R R
            (2,7-3,7) size 2: G G
            (5,3-6,3) size 3: R R R
            (7,7) size 1: B
            (8,1-9,2) size 4: B B B B
            (8,6-8) size 3: B B B
        *   Wait, (8,6-8) is size 3. (2,1-3) is size 3. (5,3-6,3) is size 3. Why are some Red and some Blue?
        *   Let's look at the positions. Reds are in the top/middle. Blue is at the bottom.
        *   Let's re-read: "switch the gel to the color assigned to that grouping size."
        *   Maybe the mapping is: 1=B, 2=G, 3=R, 4=B? No.
        *   Let's re-count Example 1 carefully.
            Input:
            (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:
            (2,1-3) -> R (3)
            (2,7-3,7) -> G (2)
            (5,3-6,3) -> R (3)
            (7,7) -> B (1)
            (8,1-9,2) -> B (4)
            (8,6-8) -> B (3)
        *   Wait, (8,6-8) is size 3. (2,1-3) is size 3. (5,3-6,3) is size 3.
        *   Is it possible the mapping is: 1=B, 2=G, 3=R, 4=B, 5=... and the last one (8,6-8) is actually size 3 but colored B?
        *   Let's look at the test input.
            Cluster 1: (0,9), (1,9), (2,9), (3,9) -> size 4
            Cluster 2: (2,2), (2,3), (3,1), (3,2) -> size 4
            Cluster 3: (2,5), (2,6), (3,6) -> size 3
            Cluster 4: (6,0), (7,0), (8,0) -> size 3
            Cluster 5: (7,3), (7,4) -> size 2
            Cluster 6: (7,7), (8,7) -> size 2

        *   If 1=B, 2=G, 3=R, 4=B:
            Cluster 1 (4) -> B
            Cluster 2 (4) -> B
            Cluster 3 (3) -> R
            Cluster 4 (3) -> R
            Cluster 5 (2) -> G
            Cluster 6 (2) -> G

### Applying the Rule to Test Input

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

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

### Conceptual Digestion

1.  **CAST**:
    *   `.` (Black): The "dark garden bed" (background).
    *   `X` (Grey): The "seedlings" (objects to be grouped and colored).
    *   Colors (B, R, G, etc.): The "ribbons" (the output colors assigned based on cluster size).

2.  **AFFORDANCES**:
    *   `X` cells can be grouped into clusters based on adjacency (up, down, left, right).
    *   Each cluster has a size (total number of `X` cells).
    *   The size of the cluster determines the replacement color for all `X` cells in that cluster.

3.  **TRANSFORMATION RULE**:
    *   Identify all connected components (clusters) of `X`.
    *   Count the number of cells in each cluster.
    *   Map the count to a specific color.
    *   Looking at the training examples:
        *   Example 1:
            *   Cluster 1 (size 3): `R` (Red)
            *   Cluster 2 (size 2): `R` (Red) - *Wait, let's re-examine.*
            *   Example 1 clusters:
                *   (2,1)-(2,3): size 3 -> `R`
                *   (5,3)-(6,3): size 2 -> `R` (Wait, (5,3),(5,4),(6,3) is size 3) -> `R`
                *   (2,7)-(3,7): size 2 -> `G` (Wait, (2,7),(3,7) is size 2)
                *   (7,7): size 1 -> `B`
                *   (8,1)-(9,2): size 4 -> `B`
                *   (8,6)-(8,8): size 3 -> `B`
        *   *Correction on Color Mapping*: Let's look closer at the counts.
        *   Example 1:
            *   Size 3: `R` (at 2,1-3), `R` (at 5,3-6,3), `B` (at 8,6-8) -> This is inconsistent.
        *   *Re-evaluating 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`
            *   Cluster 5: (8,1), (8,2), (9,1), (9,2) -> Size 4 -> `B`
            *   Cluster 6: (8,6), (8,7), (8,8) -> Size 3 -> `B`
        *   *Wait, let's look at Example 2*:
            *   Cluster 1: (1,8), (1,9), (2,8), (2,9) -> Size 4 -> `B`
            *   Cluster 2: (2,1), (2,2), (3,2), (3,3) -> Size 4 -> `B`
            *   Cluster 3: (5,9), (6,9), (7,9) -> Size 3 -> `R`
            *   Cluster 4: (6,5), (6,6) -> Size 2 -> `G`
            *   Cluster 5: (7,1), (8,1) -> Size 2 -> `G`
            *   Cluster 6: (8,4), (9,3), (9,4) -> Size 3 -> `R`
        *   *Consistent Mapping found in Example 2*:
            *   Size 4 -> `B` (Blue)
            *   Size 3 -> `R` (Red)
            *   Size 2 -> `G` (Green)
            *   Size 1 -> ? (Example 1 had size 1 as `B`, but Example 1 is very messy. Let's re-check Example 1 again).
        *   *Re-checking Example 1 with the mapping (4:B, 3:R, 2:G, 1:B?)*:
            *   Size 3: (2,1-3) -> `R` (Correct)
            *   Size 2: (2,7-3,7) -> `G` (Correct)
            *   Size 3: (5,3-6,3) -> `R` (Correct)
            *   Size 1: (7,7) -> `B` (Wait, 1 is `B`?)
            *   Size 4: (8,1-9,2) -> `B` (Correct)
            *   Size 3: (8,6-8) -> `B` (Wait, this is size 3 but colored `B`?)
        *   *Let's look at the colors again*. Maybe the mapping is based on the *number* of clusters of that size? No.
        *   Let's look at the colors provided in the key: B=blue, R=red, G=green, Y=yellow, M=magenta, O=orange, A=azure, W=maroon.
        *   Example 2 is perfectly consistent: 4=B, 3=R, 2=G.
        *   Example 1: 3=R, 2=G, 1=B, 4=B, 3=B. (Wait, the last one is size 3 but B).
        *   Let's re-count Example 1, Cluster 6: (8,6), (8,7), (8,8). That is size 3. It is colored `B`.
        *   Wait! Look at the coordinates of Example 1, Cluster 6: (8,6), (8,7), (8,8). And Cluster 4: (7,7). They are adjacent diagonally. If diagonal counts, Cluster 4+6 = size 4.
        *   If diagonal counts:
            *   Ex 1: (2,1-3) size 3 `R`; (2,7-3,7) size 2 `G`; (5,3-6,3) size 3 `R`; (7,7 + 8,6-8) size 4 `B`; (8,1-9,2) size 4 `B`.
            *   Ex 2: (1,8-2,9) size 4 `B`; (2,1-3,3) size 4 `B`; (5,9-7,9) size 3 `R`; (6,5-6,6) size 2 `G`; (7,1-8,1) size 2 `G`; (8,4-9,4) size 3 `R`.
        *   *Confirmed Mapping*:
            *   Size 4 -> `B` (Blue)
            *   Size 3 -> `R` (Red)
            *   Size 2 -> `G` (Green)
            *   Size 1 -> ? (Not seen in Ex 2, but if we follow the pattern, maybe `Y` or something else, but let's see the test).

### Applying to Test Input

1.  **Identify Clusters (including diagonals)**:
    *   Cluster A: (0,9), (1,9), (2,9), (3,9) -> Size 4.
    *   Cluster B: (2,2), (2,3), (3,1), (3,2) -> Size 4.
    *   Cluster C: (2,5), (2,6), (3,6) -> Size 3.
    *   Cluster D: (6,0), (7,0), (8,0) -> Size 3.
    *   Cluster E: (7,3), (7,4) -> Size 2.
    *   Cluster F: (7,7), (8,7) -> Size 2.

2.  **Assign Colors**:
    *   Cluster A (Size 4) -> `B`
    *   Cluster B (Size 4) -> `B`
    *   Cluster C (Size 3) -> `R`
    *   Cluster D (Siz

... [526 chars truncated]

### Conceptual Digestion

1.  **CAST**:
    *   **Night Sky**: The black cells (`.`).
    *   **Launchpad**: The row containing the blue runway (`B`).
    *   **Engines**: The colored cells embedded in the launchpad: Azure (`A`) and Red (`R`).
    *   **Plumes**: The vertical columns of blue (`B`) exhaust trailing up from the engines.
    *   **Flares**: The top-most cell of a plume, which matches the engine's color (`A` or `R`).

2.  **AFFORDANCES**:
    *   The launchpad row remains static.
    *   Engines (`A` and `R`) trigger a vertical growth upwards.
    *   The height of the plume depends on the engine color.

3.  **TRANSFORMATION RULE**:
    *   Locate the row containing the blue runway (`B`).
    *   For every `A` (Azure) engine in that row: Create a plume extending upwards. Based on Example 1, an `A` engine at row 3 (0-indexed) creates a plume reaching row 0. The plume consists of the engine itself, blue cells (`B`) above it, and is capped by an `A` at the top. In Example 1, the engine is at row 3, the plume is at row 1 and 2, and the flare is at row 0. Total height from runway = 3 cells.
    *   For every `R` (Red) engine in that row: Create a plume extending upwards. Based on Example 2, an `R` engine at row 5 creates a plume reaching row 1. The plume consists of the engine itself, blue cells (`B`) above it, and is capped by an `R` at the top. In Example 2, the engine is at row 5, the plume is at rows 2, 3, 4, and the flare is at row 1. Total height from runway = 4 cells.
    *   **Height Summary**:
        *   `A` (Azure) $\rightarrow$ Plume height = 3 cells (including engine).
        *   `R` (Red) $\rightarrow$ Plume height = 4 cells (including engine).
    *   The plumes stop if they hit the top edge of the grid.

### Applying the Rule to Test Input

*   **Grid Dimensions**: 9 rows x 11 columns.
*   **Launchpad Row**: Row 6 (0-indexed).
*   **Engines found at Row 6**:
    *   Col 1: `R` (Red) $\rightarrow$ Height 4. Plume cells: Row 5 (`B`), Row 4 (`B`), Row 3 (`B`), Row 2 (`R`).
    *   Col 3: `A` (Azure) $\rightarrow$ Height 3. Plume cells: Row 5 (`B`), Row 4 (`B`), Row 3 (`A`).
    *   Col 7: `A` (Azure) $\rightarrow$ Height 3. Plume cells: Row 5 (`B`), Row 4 (`B`), Row 3 (`A`).
    *   Col 9: `R` (Red) $\rightarrow$ Height 4. Plume cells: Row 5 (`B`), Row 4 (`B`), Row 3 (`B`), Row 2 (`R`).

**Mapping the coordinates (Row, Col):**
*   Col 1 (R): (5,1)=B, (4,1)=B, (3,1)=B, (2,1)=R
*   Col 3 (A): (5,3)=B, (4,3)=B, (3,3)=A
*   Col 7 (A): (5,7)=B, (4,7)=B, (3,7)=A
*   Col 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 . 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.  **The CAST**:
    *   **The Strand**: The horizontal line of colored cells (B, R, A, etc.) in the input grid.
    *   **Enzyme Markers**: The specific colored cells within that strand.
    *   **Protein Chains**: The vertical lines of cells that grow from the enzyme markers.
    *   **Cap Molecule**: The cell at the end of the protein chain, which matches the color of the enzyme that started it.

2.  **The AFFORDANCES**:
    *   The strand acts as the anchor.
    *   The colors of the enzymes determine the length of the chain they synthesize.
    *   The chains grow perpendicular (upwards, based on the examples) from the strand.

3.  **The TRANSFORMATION RULE**:
    *   Identify the horizontal strand.
    *   For each colored cell (enzyme) in the strand:
        *   Determine the chain length based on the color.
        *   Looking at Example 1: `A` (Azure) produced a chain of length 3 (including the cap) extending upwards. The cells are: `A` (cap), `B` (body), `B` (body). Wait, looking closer: The strand is at row 3. The output has cells at row 0, 1, 2. Row 0 is `A`, Row 1 is `B`, Row 2 is `B`. The body of the chain is always the color of the strand's primary material (`B` in both examples).
        *   Looking at Example 2: `R` (Red) produced a chain of length 4 (including the cap) extending upwards. The cells are: `R` (cap), `B` (body), `B` (body), `B` (body).
    *   **Refined Rule**:
        *   The "strand material" is the most common color in the strand (Blue `B`).
        *   The "enzyme" is any color different from the strand material.
        *   The chain grows upwards from the enzyme.
        *   The chain consists of the strand material (`B`) and is capped by the enzyme's color.
        *   **Length Mapping**:
            *   `A` (Azure) $\rightarrow$ Length 3 (Cap `A` + 2 `B`s).
            *   `R` (Red) $\rightarrow$ Length 4 (Cap `R` + 3 `B`s).
    *   **Test Input Application**:
        *   Strand is at row 6.
        *   Enzymes are at:
            *   Col 1: `R` (Red) $\rightarrow$ Length 4. Cells at (2,1)=R, (3,1)=B, (4,1)=B, (5,1)=B.
            *   Col 3: `A` (Azure) $\rightarrow$ Length 3. Cells at (3,3)=A, (4,3)=B, (5,3)=B.
            *   Col 7: `A` (Azure) $\rightarrow$ Length 3. Cells at (3,7)=A, (4,7)=B, (5,7)=B.
            *   Col 9: `R` (Red) $\rightarrow$ Length 4. Cells at (2,9)=R, (3,9)=B, (4,9)=B, (5,9)=B.

### 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 . . . . . . . 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.  **CAST**:
    *   **Galactic Arm**: A horizontal line of colored cells (mostly Blue 'B' in the examples).
    *   **Stellar Anomalies**: Cells within the arm that have a color different from the primary arm color (e.g., Azure 'A', Red 'R').
    *   **Plasma Jets**: Vertical lines extending from the anomalies.
    *   **Void**: The black background ('.').

2.  **AFFORDANCES**:
    *   The arm provides the base coordinates for the anomalies.
    *   The anomalies determine the position and the color of the "tip" of the plasma jet.
    *   The "spectral class" (the specific color of the anomaly) determines the length and direction of the jet.

3.  **TRANSFORMATION RULE**:
    *   Identify the horizontal arm.
    *   Identify anomalies (non-B cells) in that arm.
    *   For each anomaly:
        *   A jet of color 'B' (the arm's primary color) extends perpendicularly (upwards) from the anomaly.
        *   The jet terminates with the anomaly's original color.
        *   **Length Determination**:
            *   In Example 1: Anomaly 'A' at (3, 1) produced a jet of length 3 (cells at (2,1), (1,1), (0,1)). The tip (0,1) is 'A', and the body (1,1), (2,1) is 'B'.
            *   In Example 2: Anomaly 'R' at (5, 3) produced a jet of length 4 (cells at (4,3), (3,3), (2,3), (1,3)). The tip (1,3) is 'R', and the body (2,3), (3,3), (4,3) is 'B'.
        *   **Refined Rule**: The jet extends upwards. The length seems to be based on the color:
            *   'A' (Azure) $\rightarrow$ length 3.
            *   'R' (Red) $\rightarrow$ length 4.
        *   The cell at the very end of the jet takes the anomaly's color; the cells between the arm and the tip take the arm's color ('B').

### Applying Rule to Test Input
Input Arm: `B R B A B B B A B R B` at row 6.
Anomalies:
1.  Col 1: 'R' $\rightarrow$ Length 4. Cells: (5,1)=B, (4,1)=B, (3,1)=B, (2,1)=R.
2.  Col 3: 'A' $\rightarrow$ Length 3. Cells: (5,3)=B, (4,3)=B, (3,3)=A.
3.  Col 7: 'A' $\rightarrow$ Length 3. Cells: (5,7)=B, (4,7)=B, (3,7)=A.
4.  Col 9: 'R' $\rightarrow$ Length 4. Cells: (5,9)=B, (4,9)=B, (3,9)=B, (2,9)=R.

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 . . . . . . . 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.  **CAST**:
    *   **The Chain**: The horizontal line of `B` (blue) cells.
    *   **The Anchors**: Any cell in that horizontal line that is *not* `B` (e.g., `A` for azure, `R` for red).
    *   **The Tethers**: Vertical lines of `B` (blue) cells extending upwards from an anchor.
    *   **The Buoys**: The colored cell at the very top of a tether, matching the color of the anchor.

2.  **AFFORDANCES**:
    *   The horizontal line acts as the base.
    *   Non-blue cells in the base trigger the creation of a vertical column.
    *   The height of the column is determined by the specific color of the anchor.

3.  **TRANSFORMATION RULE**:
    *   Identify the row containing the chain (the row with `B`s and other colors).
    *   For every cell in that row that is not `B` (the anchor):
        *   Determine the height of the tether based on the color:
            *   `A` (Azure) $\rightarrow$ Height = 3 (including the anchor). The column consists of `A` at the top, followed by `B`s down to the anchor.
            *   `R` (Red) $\rightarrow$ Height = 4 (including the anchor). The column consists of `R` at the top, followed by `B`s down to the anchor.
        *   Draw the tether upwards from the anchor's position.
    *   *Verification from examples*:
        *   Ex 1: `A` at index 1. Height 3. Row 3 is base. Row 2=B, Row 1=B, Row 0=A. (Wait, looking closer: Row 3 is base. Row 2=B, Row 1=B, Row 0=A. That's 4 cells. Let's re-count).
        *   Ex 1: Base row 3. `A` is at (3,1). Output has `A` at (0,1), `B` at (1,1), `B` at (2,1). Total height = 4.
        *   Ex 2: Base row 5. `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 = 5.
        *   Ex 3: Base row 7. `A` at (7,1) $\rightarrow$ `A` at (4,1), `B` at (5,1), `B` at (6,1). Height = 4. `A` at (7,5) $\rightarrow$ `A` at (4,5), `B` at (5,5), `B` at (6,5). Height = 4. `R` at (7,7) $\rightarrow$ `R` at (3,7), `B` at (4,7), `B` at (5,7), `B` at (6,7). Height = 5.
    *   **Refined Rule**:
        *   If anchor is `A`: Tether height is 4 (Anchor + 2 `B`s + 1 `A` top).
        *   If anchor is `R`: Tether height is 5 (Anchor + 3 `B`s + 1 `R` top).

### Applying to Test Input
Input Base Row: 6
`B R B A B B B A B R B`
Indices of anchors:
- Index 1: `R` $\rightarrow$ Height 5. Top at row $6-4 = 2$. (Row 2: `R`, Row 3: `B`, Row 4: `B`, Row 5: `B`, Row 6: `R`)
- Index 3: `A` $\rightarrow$ Height 4. Top at row $6-3 = 3$. (Row 3: `A`, Row 4: `B`, Row 5: `B`, Row 6: `A`)
- Index 7: `A` $\rightarrow$ Height 4. Top at row $6-3 = 3$. (Row 3: `A`, Row 4: `B`, Row 5: `B`, Row 6: `A`)
- Index 9: `R` $\rightarrow$ Height 5. Top at row $6-4 = 2$. (Row 2: `R`, Row 3: `B`, Row 4: `B`, Row 5: `B`, Row 6: `R`)

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.  **CAST**:
    *   **Bus Line**: A horizontal row containing colored components (B, R, A, etc.) and a background color (B in these examples, though the "bus" is the line itself).
    *   **Components**: The non-blue (or distinct) colors on the bus line (e.g., A, R).
    *   **Signal Traces**: Vertical lines extending upwards from the components.

2.  **AFFORDANCES**:
    *   The bus line acts as the origin for the signal traces.
    *   The color of the component determines the color of the "cap" (the end of the trace).
    *   The "bus" color (B) fills the trace between the component and the cap.

3.  **TRANSFORMATION RULE**:
    *   Identify the horizontal line (the bus).
    *   For every cell in that line that is **not** the bus color (B), a vertical trace is drawn upwards.
    *   The length of the trace is determined by the component's position or a fixed logic relative to the grid height. Looking at the examples:
        *   Example 1: 'A' at (3,1) -> Trace ends at (0,1). Length = 3 cells above.
        *   Example 2: 'R' at (5,3) -> Trace ends at (1,3). Length = 4 cells above.
        *   Example 3: 'A' at (7,1) -> Trace ends at (4,1). 'A' at (7,5) -> Trace ends at (4,5). 'R' at (7,7) -> Trace ends at (3,7).
    *   **Refined Rule**: The trace extends upwards. The distance from the bus to the cap is determined by the "value" or "index" of the color? No, looking closer:
        *   Example 1: A is at index 1. Trace length is 3.
        *   Example 2: R is at index 3. Trace length is 4.
        *   Example 3: A (index 1) length 3; A (index 5) length 3; R (index 7) length 4.
    *   **Pattern**: The trace length is based on the color.
        *   A (Azure) $\rightarrow$ length 3.
        *   R (Red) $\rightarrow$ length 4.
    *   The trace consists of the bus color (B) for all cells except the topmost cell, which takes the component's color.

### Applying to Test Input
Bus line is at row 6 (0-indexed).
Components:
- Col 1: R (Red) $\rightarrow$ Length 4. Trace: (5,1)=B, (4,1)=B, (3,1)=B, (2,1)=R.
- Col 3: A (Azure) $\rightarrow$ Length 3. Trace: (5,3)=B, (4,3)=B, (3,3)=A.
- Col 7: A (Azure) $\rightarrow$ Length 3. Trace: (5,7)=B, (4,7)=B, (3,7)=A.
- Col 9: R (Red) $\rightarrow$ Length 4. Trace: (5,9)=B, (4,9)=B, (3,9)=B, (2,9)=R.

Grid dimensions: 9 rows x 11 columns.
Bus row: 6.

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

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.  **CAST**:
    *   **The Root**: A horizontal line of colored cells (mostly one color, e.g., Blue 'B').
    *   **The Runes**: Cells on the root that have a different color than the dominant root color (e.g., Azure 'A', Red 'R').
    *   **The Pillars**: Vertical lines of cells extending upwards from the runes.

2.  **AFFORDANCES**:
    *   The root provides the base coordinates for the pillars.
    *   The color of the rune determines the color of the "crown" (the top cell of the pillar).
    *   The dominant root color (Blue 'B') forms the body of the pillar.

3.  **TRANSFORMATION RULE**:
    *   Identify the horizontal root line.
    *   Identify the "runes" (cells with a color different from the root's dominant color).
    *   For each rune, create a vertical pillar extending upwards.
    *   The height of the pillar is determined by the specific rune color (based on training examples):
        *   Example 1: Azure 'A' $\rightarrow$ Pillar height of 3 (including the rune itself). The top is 'A', the middle is 'B', the bottom is 'B' (the rune). Wait, looking closer at Example 1: Root is at row 3. Output has 'A' at row 0, 'B' at row 1, 'B' at row 2. That is a height of 4 cells total from the root? No, row 0, 1, 2 are above row 3. So the pillar is 3 cells *above* the root.
        *   Example 2: Red 'R' $\rightarrow$ Pillar height of 4 cells *above* the root. Row 5 is root. Output has 'R' at row 1, 'B' at row 2, 'B' at row 3, 'B' at row 4.
    *   **Refined Height Rule**:
        *   Azure 'A': 3 cells above the root. Top is 'A', others are 'B'.
        *   Red 'R': 4 cells above the root. Top is 'R', others are 'B'.
    *   The root itself remains unchanged.

### Applying Rule to Test Input

*   **Root**: Row 6 (0-indexed).
*   **Dominant Color**: Blue 'B'.
*   **Runes**:
    *   Col 1: Red 'R' $\rightarrow$ Pillar height 4 above. Cells (2,1)=R, (3,1)=B, (4,1)=B, (5,1)=B.
    *   Col 3: Azure 'A' $\rightarrow$ Pillar height 3 above. Cells (3,3)=A, (4,3)=B, (5,3)=B.
    *   Col 7: Azure 'A' $\rightarrow$ Pillar height 3 above. Cells (3,7)=A, (4,7)=B, (5,7)=B.
    *   Col 9: Red 'R' $\rightarrow$ Pillar height 4 above. Cells (2,9)=R, (3,9)=B, (4,9)=B, (5,9)=B.

**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 . . . . . . . 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.  **The CAST**:
    *   **Chorus Line**: A horizontal row of colored cells (the "dancers").
    *   **Uniform Costume**: The dominant color in the row (Blue 'B').
    *   **Distinct Costumes**: Colors that differ from the dominant color (Red 'R', Azure 'A').
    *   **Motion Trail**: A vertical line of the dominant color ('B') extending upwards from a distinct costume.
    *   **Apex**: The top cell of the motion trail, which takes the color of the distinct costume.

2.  **The AFFORDANCES**:
    *   The distinct costumes trigger a "leap" (a vertical column of cells).
    *   The height of the leap is determined by the specific color of the distinct costume.

3.  **The TRANSFORMATION RULE**:
    *   Identify the row containing the chorus line.
    *   Identify the dominant color (B).
    *   For every cell in that row that is NOT the dominant color:
        *   Determine the leap height based on the color:
            *   **Azure (A)**: Height = 3 cells above the line (including the apex).
            *   **Red (R)**: Height = 4 cells above the line (including the apex).
        *   Draw a vertical line from the cell upwards.
        *   The cells between the line and the apex are filled with the dominant color (B).
        *   The cell at the apex is filled with the distinct costume's color (A or R).

**Applying to Test Input:**
*   Chorus line is at row 6 (0-indexed).
*   Dominant color: B.
*   Distinct costumes at:
    *   Col 1: R (Height 4) $\rightarrow$ Apex at row 2 (R), trail at rows 3, 4, 5 (B).
    *   Col 3: A (Height 3) $\rightarrow$ Apex at row 3 (A), trail at rows 4, 5 (B).
    *   Col 7: A (Height 3) $\rightarrow$ Apex at row 3 (A), trail at rows 4, 5 (B).
    *   Col 9: R (Height 4) $\rightarrow$ Apex at row 2 (R), trail at rows 3, 4, 5 (B).

**Mapping the coordinates (Row, Col):**
*   (6, 1) is R $\rightarrow$ (2, 1)=R, (3, 1)=B, (4, 1)=B, (5, 1)=B
*   (6, 3) is A $\rightarrow$ (3, 3)=A, (4, 3)=B, (5, 3)=B
*   (6, 7) is A $\rightarrow$ (3, 7)=A, (4, 7)=B, (5, 7)=B
*   (6, 9) is R $\rightarrow$ (2, 9)=R, (3, 9)=B, (4, 9)=B, (5, 9)=B

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.  **CAST**:
    *   **Stratum**: The horizontal line of colored cells (the "bedrock").
    *   **Mineral Deposits**: The specific colors within that stratum that differ from the primary stratum color (Blue 'B').
    *   **Intrusions**: Vertical lines extending upwards from the mineral deposits.

2.  **AFFORDANCES**:
    *   The stratum acts as the base.
    *   The mineral deposits (A, R, etc.) trigger the creation of a vertical column.
    *   The length of the column is determined by the specific color of the mineral.

3.  **TRANSFORMATION RULE**:
    *   Identify the horizontal line (stratum).
    *   For every cell in the stratum that is NOT the base color (B), create a vertical intrusion extending upwards.
    *   **Intrusion Lengths (based on training examples)**:
        *   **Azure (A)**: Length = 3 cells total (including the base). The tip is 'A', and the cells between the tip and the base are 'B'. (Example 1: A is at row 3, tip at row 0. Length = 4? No, let's re-examine).
        *   *Correction on Lengths*:
            *   Example 1: Stratum at row 3. 'A' is at (3, 1). Output has 'A' at (0, 1), 'B' at (1, 1), 'B' at (2, 1). Total height from base = 4 cells.
            *   Example 2: Stratum at row 5. '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 from base = 5 cells.
            *   Example 3: Stratum at row 7.
                *   'A' at (7, 1) $\rightarrow$ Tip 'A' at (4, 1), 'B' at (5, 1), 'B' at (6, 1). Height = 4.
                *   'A' at (7, 5) $\rightarrow$ Tip 'A' at (4, 5), 'B' at (5, 5), 'B' at (6, 5). Height = 4.
                *   'R' at (7, 7) $\rightarrow$ Tip 'R' at (3, 7), 'B' at (4, 7), 'B' at (5, 7), 'B' at (6, 7). Height = 5.
    *   **Rule Summary**:
        *   If color is **A (Azure)**: Intrusion height is 4 (Tip 'A' is 3 cells above the stratum).
        *   If color is **R (Red)**: Intrusion height is 5 (Tip 'R' is 4 cells above the stratum).
        *   The cells between the tip and the stratum are filled with the base color **B (Blue)**.

### Applying to Test Input
*   Stratum is at row 6.
*   Mineral deposits:
    *   Col 1: **R** $\rightarrow$ Height 5. Tip 'R' at row (6-4) = 2. Cells (3, 4, 5) = 'B'.
    *   Col 3: **A** $\rightarrow$ Height 4. Tip 'A' at row (6-3) = 3. Cells (4, 5) = 'B'.
    *   Col 7: **A** $\rightarrow$ Height 4. Tip 'A' at row (6-3) = 3. Cells (4, 5) = 'B'.
    *   Col 9: **R** $\rightarrow$ Height 5. Tip 'R' at row (6-4) = 2. Cells (3, 4, 5) = 'B'.

**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 . . . . . . . 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.  **CAST**:
    *   **The Dough**: The black cells (`.`) and the base line of colored cells.
    *   **The Spice Plugs**: The colored cells in the input grid.
    *   **The Infused Columns**: The vertical extensions created above the spice plugs.

2.  **AFFORDANCES**:
    *   The base line of colors acts as the anchor.
    *   Specific colors (spices) trigger vertical growth upwards.
    *   The height of the growth depends on the "strength" of the spice.

3.  **TRANSFORMATION RULE**:
    *   Identify the row containing the colored line (the "roll of dough").
    *   For every colored cell in that row, a column of cells rises vertically upwards.
    *   The column is topped by the color of the spice plug.
    *   The cells *below* the top spice (but above the base line) are filled with the color of the "dough" (which, based on the examples, is the dominant color of the roll, Blue `B`).
    *   **Height Determination**:
        *   In Example 1: `A` (Azure) creates a column of height 3 (including the base). The base is at row 3, the column reaches row 0. (3 cells total: A at row 0, B at row 1, B at row 2, A at row 3). Wait, looking closer: Base is row 3. Output has A at row 0, B at row 1, B at row 2. That is 3 cells *above* the base.
        *   In Example 2: `R` (Red) creates a column of height 4 above the base. Base is row 5. Output has R at row 1, B at row 2, B at row 3, B at row 4. That is 4 cells *above* the base.
    *   **Refined Height Rule**:
        *   `A` (Azure) $\rightarrow$ 3 cells high above the base.
        *   `R` (Red) $\rightarrow$ 4 cells high above the base.
        *   `B` (Blue) $\rightarrow$ 0 cells high above the base (it is the dough itself).

### Applying the Rule to Test Input

*   **Base Line**: Row 6 (0-indexed).
*   **Colors at Row 6**:
    *   Col 0: B (Height 0)
    *   Col 1: R (Height 4) $\rightarrow$ Cells at (5,1), (4,1), (3,1), (2,1). Top (2,1) is R, others are B.
    *   Col 2: B (Height 0)
    *   Col 3: A (Height 3) $\rightarrow$ Cells at (5,3), (4,3), (3,3). Top (3,3) is A, others are B.
    *   Col 4: B (Height 0)
    *   Col 5: B (Height 0)
    *   Col 6: B (Height 0)
    *   Col 7: A (Height 3) $\rightarrow$ Cells at (5,7), (4,7), (3,7). Top (3,7) is A, others are B.
    *   Col 8: B (Height 0)
    *   Col 9: R (Height 4) $\rightarrow$ Cells at (5,9), (4,9), (3,9), (2,9). Top (2,9) is R, others are B.
    *   Col 10: B (Height 0)

**Mapping the columns:**
*   Col 1: Row 2=R, Row 3=B, Row 4=B, Row 5=B
*   Col 3: Row 3=A, Row 4=B, Row 5=B
*   Col 7: Row 3=A, Row 4=B, Row 5=B
*   Col 9: Row 2=R, Row 3=B, Row 4=B, Row 5=B

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.  **CAST**:
    *   **The Drone**: The single horizontal line containing colors.
    *   **Tonal Signatures**: The specific colored cells on that line (A, R, etc.).
    *   **Harmonic Overtones**: The vertical stacks of colors projecting upwards from the tonal signatures.
    *   **The Background**: The black cells (`.`).

2.  **AFFORDANCES**:
    *   The drone line acts as the anchor.
    *   The colors on the drone line determine where stacks are built and how high they go.
    *   The "timbre" (color) determines the height of the stack.

3.  **TRANSFORMATION RULE**:
    *   Identify the horizontal line containing colors.
    *   For every cell on that line that is **not** the drone's base color (Blue `B`), project a vertical stack upwards.
    *   **Stack Composition**:
        *   The top-most cell of the stack is the original color (the "tonal signature").
        *   The cells between the drone line and the top cell are filled with the drone's base color (Blue `B`).
    *   **Stack Height (Timbre)**:
        *   Azure (`A`): Height of 3 cells (including the drone line). This means 2 cells are added above the drone line (1 `B`, 1 `A`).
        *   Red (`R`): Height of 5 cells (including the drone line). This means 4 cells are added above the drone line (3 `B`s, 1 `R`).
    *   *Verification from Examples*:
        *   Ex 1: `A` at index 1 $\rightarrow$ Output has `B` at (2,1) and `A` at (1,1). Total height 3.
        *   Ex 2: `R` at index 3 $\rightarrow$ Output has `B` at (5,3), (4,3), (3,3), (2,3) and `R` at (1,3). Wait, let's re-count. Drone is at row 5. Output has `R` at row 1, `B` at 2, 3, 4. Total height 5.
        *   Ex 3: `A` at index 1 $\rightarrow$ `B` at row 6, `A` at row 4. (Height 4? No, drone is row 7. `B` at 6, `A` at 4. There is a gap at row 5? No, looking closer: `A` at (4,1), `B` at (5,1), `B` at (6,1). Total height 4. Let's re-evaluate).
        *   *Correction on Height*:
            *   `A` (Azure): Height 4 (3 cells above drone: `B`, `B`, `A`).
            *   `R` (Red): Height 5 (4 cells above drone: `B`, `B`, `B`, `R`).
        *   *Re-checking Ex 1*: Drone row 3. `A` at (3,1). Output: `A` at (0,1), `B` at (1,1), `B` at (2,1). Total height 4.
        *   *Re-checking Ex 2*: Drone row 5. `R` at (5,3). Output: `R` at (1,3), `B` at (2,3), (3,3), (4,3). Total height 5.
        *   *Re-checking Ex 3*: Drone row 7. `A` at (7,1) $\rightarrow$ `A` at (4,1), `B` at (5,1), (6,1). Total height 4. `A` at (7,5) $\rightarrow$ `A` at (4,5), `B` at (5,5), (6,5). Total height 4. `R` at (7,7) $\rightarrow$ `R` at (3,7), `B` at (4,7), (5,7), (6,7). Total height 5.

4.  **Applying to Test Input**:
    *   Drone line: Row 6.
    *   Positions of non-B colors:
        *   Index 1: `R` (Height 5 $\rightarrow$ cells at row 2, 3, 4, 5. Row 2=`R`, 3,4,5=`B`)
        *   Index 3: `A` (Height 4 $\rightarrow$ cells at row 3, 4, 5. Row 3=`A`, 4,5=`B`)
        *   Index 7: `A` (Height 4 $\rightarrow$ cells at row 3, 4, 5. Row 3=`A`, 4,5=`B`)
        *   Index 9: `R` (Height 5 $\rightarrow$ cells at row 2, 3, 4, 5. Row 2=`R`, 3,4,5=`B`)

### 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 . . . . . . . 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.  **CAST**:
    *   **Scrimmage Line**: The single row containing colored cells (the "players").
    *   **Uniform Players**: The most frequent color in that line (in the examples, Blue 'B').
    *   **Uniquely Jersied Players**: The colors that differ from the uniform color (e.g., Azure 'A', Red 'R').
    *   **Trails**: The path of uniform players ('B') created when a unique player "sprints."

2.  **AFFORDANCES**:
    *   The unique players act as "heads" of a column.
    *   The uniform players ('B') act as the "fill" for the trail.
    *   The direction of the sprint is perpendicular to the line (upwards, based on Example 1).
    *   The distance of the sprint is determined by the color of the unique jersey.

3.  **TRANSFORMATION RULE**:
    *   Identify the row containing the line of players.
    *   Identify the uniform color (B).
    *   For every cell in that row that is NOT the uniform color (B) or empty (.):
        *   Determine the sprint distance (length of the trail).
        *   In Example 1: 'A' (Azure) at (3, 1) moved up to (0, 1). The distance from row 3 to row 0 is 3 units. The cell at (0, 1) becomes 'A', and the cells between (1, 1) and (2, 1) become 'B'.
        *   Looking at the colors: 'A' (Azure) seems to trigger a sprint of 3 units upward.
        *   In the Test Input, we have 'R' (Red) and 'A' (Azure).
        *   Based on the pattern of "sprinting perpendicular," and the provided example where 'A' moved 3 spaces up, we apply the same logic. If 'A' moves 3 spaces, we must determine 'R'. In many such puzzles, colors correspond to values (e.g., A=1, B=2... or specific distance mappings). However, looking at the grid height, the line is at row 6. A sprint of 3 for 'A' reaches row 3.
        *   Let's re-evaluate Example 1: Line is at row 3. 'A' is at (3, 1). Output 'A' is at (0, 1). Distance = 3.
        *   In Test Input: Line is at row 6. 'A' is at (6, 3) and (6, 7). 'R' is at (6, 1) and (6, 9).
        *   If 'A' distance = 3, 'A' moves to row 3.
        *   If 'R' distance = 2 (following a common logic where colors have associated values, or simply observing the available space), 'R' moves to row 4.
        *   Wait, let's look at the colors again. A=Azure, R=Red. In the alphabet, A=1, R=18? No. In common color-value puzzles, Red is often 2 or 3. Let's check the grid: the line is at row 6. If 'A' moves 3 units (to row 3), and 'R' moves, say, 5 units, it would reach row 1.
        *   Actually, looking at Example 1 again: The line is at index 3. The 'A' moves to index 0. That is a distance of 3.
        *   In the test input, the line is at index 6. If 'A' moves 3 units, it reaches index 3. If 'R' moves a different distance, say 4 units, it reaches index 2.
        *   Let's assume the distance is tied to the color: A=3, R=4. (This is a common pattern in these puzzles).

    *   **Applying to Test Input**:
        *   Line at row 6: `B R B A B B B A B R B`
        *   Col 1: 'R' -> Sprint 4 units up to row 2. Trail of 'B' at rows 3, 4, 5.
        *   Col 3: 'A' -> Sprint 3 units up to row 3. Trail of 'B' at rows 4, 5.
        *   Col 7: 'A' -> Sprint 3 units up to row 3. Trail of 'B' at rows 4, 5.
        *   Col 9: 'R' -> Sprint 4 units up to row 2. Trail of 'B' at rows 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 . . . . . . . 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.  **The CAST**:
    *   **The Front**: A horizontal line of colored cells (the "pressure front").
    *   **Thermal Anomalies**: Cells within that line that differ from the base color of the front (the base color is Blue 'B').
    *   **Updraft Columns**: Vertical lines of cells extending upwards from the anomalies.

2.  **The AFFORDANCES**:
    *   The base color 'B' acts as the background of the front.
    *   Any color other than 'B' (e.g., 'A', 'R') acts as a trigger for a column.
    *   The height of the column is determined by the "thermal signature" (the specific color of the anomaly).

3.  **The TRANSFORMATION RULE**:
    *   Identify the row containing the colored line.
    *   For every cell in that row that is NOT 'B' (the anomaly):
        *   Create a vertical column extending upwards from that cell.
        *   The top cell of the column must be the color of the anomaly.
        *   The cells between the anomaly and the top cell must be the base color 'B'.
        *   **Determining Height**:
            *   Example 1: 'A' (Azure) $\rightarrow$ Height 3 (Anomaly at row 3, top at row 0. Total cells: 3). Wait, let's re-count.
                *   Ex 1: Row 3 is the front. 'A' is at (3,1). Output has 'A' at (0,1), 'B' at (1,1), 'B' at (2,1). Total height = 4 cells including the anomaly.
            *   Example 2: 'R' (Red) $\rightarrow$ Height 5 (Anomaly at row 5, top at row 1. Total cells: 5).
            *   Example 3:
                *   'A' (Azure) $\rightarrow$ Height 4 (Anomaly at row 7, top at row 4. Total cells: 4).
                *   'R' (Red) $\rightarrow$ Height 5 (Anomaly at row 7, top at row 3. Total cells: 5).
    *   **Refined Height Rule**:
        *   Azure ('A') $\rightarrow$ Column height = 4 cells (including the anomaly).
        *   Red ('R') $\rightarrow$ Column height = 5 cells (including the anomaly).

### Applying the Rule to Test Input

*   **Front Row**: Row 6 (0-indexed).
*   **Anomalies**:
    *   Col 1: 'R' (Red) $\rightarrow$ Height 5. Top at $6 - (5-1) = 2$. Cells: (2,1)=R, (3,1)=B, (4,1)=B, (5,1)=B, (6,1)=R.
    *   Col 3: 'A' (Azure) $\rightarrow$ Height 4. Top at $6 - (4-1) = 3$. Cells: (3,3)=A, (4,3)=B, (5,3)=B, (6,3)=A.
    *   Col 7: 'A' (Azure) $\rightarrow$ Height 4. Top at $6 - (4-1) = 3$. Cells: (3,7)=A, (4,7)=B, (5,7)=B, (6,7)=A.
    *   Col 9: 'R' (Red) $\rightarrow$ Height 5. Top at $6 - (5-1) = 2$. Cells: (2,9)=R, (3,9)=B, (4,9)=B, (5,9)=B, (6,9)=R.

**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 . . . . . . . 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.  **CAST**:
    *   **Foundation Wall**: The horizontal line of colored cells (B, R, A, etc.) across the grid.
    *   **Keystones**: The specific cells within that wall that differ from the primary wall color (B). In Example 1, 'A' is the keystone. In Example 2, 'R' is the keystone. In the Test Input, 'R' and 'A' are the keystones.
    *   **Columns**: Vertical lines of the primary wall color (B) rising upwards from the keystones.
    *   **Capstones**: The cell at the very top of the column, matching the color of the keystone.

2.  **AFFORDANCES**:
    *   The position of a keystone determines the x-coordinate of a column.
    *   The color of the keystone determines the height of the column (based on a mapping) and the color of the capstone.

3.  **TRANSFORMATION RULE**:
    *   Identify the "wall" row.
    *   Identify "keystones" (cells in the wall row that are not the primary wall color 'B').
    *   For each keystone:
        *   Determine height $H$. Looking at Example 1: Keystone 'A' (Azure) produced a column of height 3 (including capstone). Looking at Example 2: Keystone 'R' (Red) produced a column of height 5 (including capstone).
        *   Wait, let's re-evaluate height based on the examples:
            *   Example 1: Wall at row 3 (0-indexed). Keystone 'A' at (3, 1). Output has cells at (2, 1), (1, 1), (0, 1). Total height above wall = 3. Capstone 'A' at (0, 1).
            *   Example 2: Wall at row 5. Keystone 'R' at (5, 3). Output has cells at (4, 3), (3, 3), (2, 3), (1, 3). Total height above wall = 4. Capstone 'R' at (1, 3).
        *   Let's check the color mapping for height:
            *   'A' (Azure) $\rightarrow$ Height 3.
            *   'R' (Red) $\rightarrow$ Height 4.
    *   The column consists of the primary wall color ('B') for $H-1$ cells, topped by the keystone color.

### Applying Rule to Test Input

*   **Wall Row**: Row 6.
*   **Primary Wall Color**: B.
*   **Keystones**:
    1.  'R' at (6, 1)
    2.  'A' at (6, 3)
    3.  'A' at (6, 7)
    4.  'R' at (6, 9)

*   **Column 1 (R at 6, 1)**: Height 4.
    *   (5, 1) = B
    *   (4, 1) = B
    *   (3, 1) = B
    *   (2, 1) = R (Capstone)
*   **Column 2 (A at 6, 3)**: Height 3.
    *   (5, 3) = B
    *   (4, 3) = B
    *   (3, 3) = A (Capstone)
*   **Column 3 (A at 6, 7)**: Height 3.
    *   (5, 7) = B
    *   (4, 7) = B
    *   (3, 7) = A (Capstone)
*   **Column 4 (R at 6, 9)**: Height 4.
    *   (5, 9) = B
    *   (4, 9) = B
    *   (3, 9) = B
    *   (2, 9) = R (Capstone)

### 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 . . . . . . . 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.  **CAST**:
    *   **Trench**: The horizontal line of colored cells (the "fortified trench line").
    *   **Flares**: Specific colored cells within that line that differ from the primary trench color (or are designated as signal points). In the examples, the "trench" is primarily Blue (B), and any cell that is *not* Blue (or is a specific signal color) acts as a flare.
    *   **Smoke Trails**: A vertical line of the trench's primary color (Blue) extending from the flare.
    *   **Apex**: The original flare color placed at the top of the smoke trail.

2.  **AFFORDANCES**:
    *   The trench line provides the base coordinates.
    *   The color of the flare determines the "altitude" (height) of the smoke trail.
    *   The primary color of the trench (Blue) fills the gap between the trench and the apex.

3.  **TRANSFORMATION RULE**:
    *   Identify the horizontal line (trench).
    *   Identify the primary color of the trench (B).
    *   For every cell in the trench that is NOT the primary color (the flares):
        *   Determine the height $H$ based on the color.
        *   Looking at Example 1: Azure (A) is at index 1. The output shows a trail of length 3 (including the apex). The apex (A) is at row 0, and B fills rows 1 and 2. Trench is at row 3. Height = 3.
        *   Looking at Example 2: Red (R) is at index 3. The output shows a trail of length 4 (including the apex). The apex (R) is at row 1, and B fills rows 2, 3, 4. Trench is at row 5. Height = 4.
    *   **Color-to-Height Mapping**:
        *   Azure (A) $\rightarrow$ Height 3.
        *   Red (R) $\rightarrow$ Height 4.
    *   The smoke trail extends *upwards* from the trench. The cell at `(trench_row - height, col)` becomes the flare color, and cells from `(trench_row - 1)` up to `(trench_row - height + 1)` become the trench color (B).

### Applying to Test Input

*   **Trench Row**: 6
*   **Trench Color**: B
*   **Flares**:
    *   Col 1: Red (R) $\rightarrow$ Height 4. Apex at row $6-4=2$. Trail (B) at rows 3, 4, 5.
    *   Col 3: Azure (A) $\rightarrow$ Height 3. Apex at row $6-3=3$. Trail (B) at rows 4, 5.
    *   Col 7: Azure (A) $\rightarrow$ Height 3. Apex at row $6-3=3$. Trail (B) at rows 4, 5.
    *   Col 9: Red (R) $\rightarrow$ Height 4. Apex at row $6-4=2$. Trail (B) at rows 3, 4, 5.

**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 . . . . . . . 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.  **CAST**:
    *   **Stage Floor**: The row containing colored markers (B, R, A, etc.) and the surrounding black space (.).
    *   **Trapdoor Markers**: The specific colored cells in the row.
    *   **Columns**: The vertical extensions created from the markers.
    *   **Spotlight**: The marker color appearing at the top of the column.

2.  **AFFORDANCES**:
    *   The colored markers act as anchors.
    *   The space above the markers can be filled to create a vertical column.
    *   The height of the column is determined by the specific color of the marker.

3.  **TRANSFORMATION RULE**:
    *   Identify the row containing the colored markers.
    *   For each colored marker, a column is built upwards from that cell.
    *   The height of the column is determined by the color:
        *   **B (Blue)**: Height 0 (stays as is, no column extends above it).
        *   **A (Azure)**: Height 3 (extends 3 cells up, including the base).
        *   **R (Red)**: Height 5 (extends 5 cells up, including the base).
    *   The column consists of the base color (B) for all cells except the top cell, which takes the color of the marker (the "spotlight").
    *   *Correction based on Example 1*: In Example 1, the marker 'A' at (3, 1) produced a column of height 3. The cells at (2, 1) and (1, 1) were filled. (2, 1) became 'B' and (1, 1) became 'A'.
    *   *Correction based on Example 2*: In Example 2, the marker 'R' at (5, 3) produced a column of height 5. The cells at (4, 3), (3, 3), (2, 3), and (1, 3) were filled. (4, 3), (3, 3), (2, 3) became 'B' and (1, 3) became 'R'.
    *   **Refined Rule**:
        *   If marker is **B**, height = 1 (only the base).
        *   If marker is **A**, height = 3 (base + 2 cells above). Top cell is 'A', middle is 'B'.
        *   If marker is **R**, height = 5 (base + 4 cells above). Top cell is 'R', others are 'B'.

### Applying Rule to Test Input
Input row 6 (index 6): `B R B A B B B A B R B`
Columns:
- Col 0: B -> Height 1 (No change)
- Col 1: R -> Height 5. Cells (5,1), (4,1), (3,1), (2,1) = B; Cell (1,1) = R.
- Col 2: B -> Height 1 (No change)
- Col 3: A -> Height 3. Cell (5,3) = B; Cell (4,3) = A. Wait, let's re-check Example 1.
  Example 1: Marker A at (3,1). Output has A at (0,1) and B at (1,1), (2,1). That is height 4? No, row 0, 1, 2, 3. That's 4 cells.
  Let's re-count Example 1:
  Input row 3: B A B B B B
  Output:
  Row 0: . A . . . .
  Row 1: . B . . . .
  Row 2: . B . . . .
  Row 3: B A B B B B
  Height = 4.
  Example 2:
  Input row 5: B B B R B B B
  Output:
  Row 1: . . . R . . .
  Row 2: . . . B . . .
  Row 3: . . . B . . .
  Row 4: . . . B . . .
  Row 5: B B B R B B B
  Height = 5.

  **Revised Height Rule**:
  - A = 4 cells total.
  - R = 5 cells total.
  - B = 1 cell total.

Test Input Row 6: `B R B A B B B A B R B`
- Col 1 (R): Height 5. Row 6=R, Row 5=B, Row 4=B, Row 3=B, Row 2=R.
- Col 3 (A): Height 4. Row 6=A, Row 5=B, Row 4=B, Row 3=A.
- Col 7 (A): Height 4. Row 6=A, Row 5=B, Row 4=B, Row 3=A.
- Col 9 (R): Height 5. Row 6=R, Row 5=B, Row 4=B, Row 3=B, Row 2=R.

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.  **CAST**:
    *   **The Channel**: The row containing colored cells (seeds).
    *   **The Seeds**: The colored cells within that row.
    *   **The Stems**: Green (G) cells that grow upwards from the seeds.
    *   **The Flowers**: The original color of the seed appearing at the top of the stem.
    *   **The Background**: Black (.) cells.

2.  **AFFORDANCES**:
    *   A colored cell in a row of otherwise similar colors (or a specific row) acts as a trigger for vertical growth.
    *   The height of the growth is determined by the specific color of the seed.

3.  **TRANSFORMATION RULE**:
    *   Identify the row containing the "seeds" (the channel).
    *   For every cell in that row that is **not** the base color of the channel (in the examples, the base color is Blue 'B'), a stem grows upwards.
    *   **Stem Height Logic**:
        *   In Example 1: Seed 'A' (Azure) at index 1. The output shows a stem of height 3 (including the flower). The flower 'A' is at row 0, and 'B' (Blue/Green stem) is at rows 1 and 2. Wait, looking closer at Example 1: The input row is `B A B B B B`. The output has `A` at (0,1), `B` at (1,1), `B` at (2,1). The stem is Blue.
        *   In Example 2: Seed 'R' (Red) at index 3. The output shows a stem of height 4 (including the flower). The flower 'R' is at row 1, and 'B' (Blue/Green stem) is at rows 2, 3, 4.
    *   **Correction on Stem Color**: The metaphorical description says "stems of uniform green rise", but the training examples use 'B' (Blue) for the stems. Looking at the input, the channel base color is 'B'. The stems take the color of the channel base ('B').
    *   **Height Determination**:
        *   Example 1: 'A' is at index 1. Height = 3.
        *   Example 2: 'R' is at index 3. Height = 4.
        *   Wait, let's look at the coordinates.
        *   Ex 1: Channel at row 3. Flower 'A' at row 0. Distance = 3.
        *   Ex 2: Channel at row 5. Flower 'R' at row 1. Distance = 4.
        *   Is the height constant for the color?
        *   'A' (Azure) $\rightarrow$ height 3.
        *   'R' (Red) $\rightarrow$ height 4.

### Applying to Test Input
*   **Channel Row**: Row 6.
*   **Base Color**: B (Blue).
*   **Seeds**:
    *   Index 1: R (Red) $\rightarrow$ Height 4. Flower at row $6-4 = 2$. Stem at rows 3, 4, 5.
    *   Index 3: A (Azure) $\rightarrow$ Height 3. Flower at row $6-3 = 3$. Stem at rows 4, 5.
    *   Index 7: A (Azure) $\rightarrow$ Height 3. Flower at row $6-3 = 3$. Stem at rows 4, 5.
    *   Index 9: R (Red) $\rightarrow$ Height 4. Flower at row $6-4 = 2$. Stem at rows 3, 4, 5.

**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 . . . . . . . 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.  **CAST**:
    *   **Yellow (Y)**: Raindrops.
    *   **Red (R)**: The drumhead (a horizontal line acting as a boundary).
    *   **Black (.)**: The sky/void.

2.  **AFFORDANCES**:
    *   **Raindrops (Y)**: Can move downward by one cell.
    *   **Drumhead (R)**: Stationary. Acts as a trigger for a "splash" if a raindrop lands on it.
    *   **Sky (.)**: Space for movement and splash patterns.

3.  **TRANSFORMATION RULE**:
    *   **Gravity**: Every yellow cell `(r, c)` moves to `(r+1, c)`.
    *   **The Splash**: If a raindrop's destination `(r+1, c)` is occupied by a red cell `R`, it triggers a V-shaped splash.
    *   **Splash Pattern**: The splash consists of yellow cells extending diagonally upwards and outwards from the point of impact `(r+1, c)`.
        *   The impact point itself remains `R`.
        *   The spray goes up-left and up-right: `(r, c-1), (r-1, c-2), (r-2, c-3)...` and `(r, c+1), (r-1, c+2), (r-2, c+3)...` until the edge of the grid is reached.
    *   **Non-Impact**: If the destination is black `(.)`, the drop simply moves there. If the drop is already below the red line, it simply moves down one cell.

### Applying the Rule to Test Input

**Input Grid Analysis:**
*   Grid size: 12 rows x 8 columns.
*   Red line (Drumhead): Row 3 (0-indexed).
*   Yellow drops:
    1.  `(0, 4)`
    2.  `(1, 0)`
    3.  `(6, 1)`

**Step-by-Step Transformation:**
1.  **Drop at (0, 4)**: Moves to `(1, 4)`. Destination is `.`. Result: `(1, 4) = Y`.
2.  **Drop at (1, 0)**: Moves to `(2, 0)`. Destination is `.`. Result: `(2, 0) = Y`.
3.  **Drop at (6, 1)**: Moves to `(7, 1)`. Destination is `.`. Result: `(7, 1) = Y`.

**Wait, let's re-examine the "Splash" trigger in the training examples.**
Example 1: Red line at row 9. Drop at (7, 4) moves to (8, 4). No splash.
Example 2: Red line at row 5. Drop at (2, 1) moves to (3, 1). Drop at (3, 4) moves to (4, 4).
Wait, looking at Example 2 output:
Input: `(2, 1)` and `(3, 4)`.
Output: `(0, 0), (1, 1), (2, 2), (3, 3), (4, 4)` and `(3, 0), (3, 2), (3, 4), (3, 6)`.
This suggests a different trigger. Let's re-read: "Any drop that lands right on the drumhead's edge erupts into a V-shaped splash".
In Example 2, the red line is at row 5. A drop at row 4 would land on row 5.
In Example 2, the drops are at `(2, 1)` and `(3, 4)`. Neither is at row 4.
Let's look at Example 4: Red line at row 6. Drop at `(4, 2)`.
Output: `(3, 0), (3, 4), (4, 1), (4, 3), (5, 2)`.
This is a V-shape centered at `(5, 2)`? No, the drop was at `(4, 2)`, it moves to `(5, 2)`. The splash is `(4, 1), (4, 3), (3, 0), (3, 4)`.
The splash happens when the drop *would* land on the red line.
In Example 4, the drop is at `(4, 2)`. It moves to `(5, 2)`. The red line is at `(6, 2)`.
Wait, the rule says "lands right on the drumhead's edge". If the red line is at row 6, a drop at row 5 lands on it.
In Example 4, the drop is at `(4, 2)`. It moves to `(5, 2)`. It hasn't hit the red line yet.
Let's re-read Example 4 again. Input: `(4, 2)`. Output: `(5, 2)` is Y, and `(4, 1), (4, 3), (3, 0), (3, 4)` are Y.
This means the splash happens when the drop *moves to the cell immediately above* the red line?
No, look at Example 2 again. Red line row 5. Drops at `(2, 1)` and `(3, 4)`.
Output: `(0, 0), (1, 1), (2, 2), (3, 3), (4, 4)` and `(3, 0), (3, 2), (3, 4), (3, 6)`.
This is very confusing. Let's look at the V-shape again.
Example 4: Drop `(4, 2)` -> `(5, 2)`. Splash: `(4, 1), (4, 3), (3, 0), (3, 4)`.
This is a V-shape centered at `(5, 2)` extending upwards.
Wait, the rule says "Any drop that lands right on the drumhead's edge erupts".
If the red line is at row 6, and the drop is at row 5, it lands on row 6.
In Example 4, the drop is at `(4, 2)`. It moves to `(5, 2)`. It is now *on the edge* (the cell above the line).
So, if a drop moves to `(row_of_red - 1, col)`, it splashes.
Let's check Example 2: Red line row 5. Drops at `(2, 1)` and `(3, 4)`.
Drop `(2, 1)` moves to `(3, 1)`. Not on edge.
Drop `(3, 4)` moves to `(4, 4)`. This IS on the edge (row 4).
Splash from `(4, 4)`: `(3, 3), (2, 2), (1, 1), (0, 0)` and `(3, 5), (2, 6)...`
Wait, the output of Example 2 has `(3, 0), (3, 2), (3, 4), (3, 6)`. That's a horizontal line.
Let me re-read: "twin arcs of yellow spray flying up-left and up-right".
Example 4: Drop `(4, 2)` moves to `(5, 2)`. Splash: `(4, 1), (3, 0)` and `(4, 3), (3, 4)`.
This matches!
Example 2: Drop `(3, 4)` moves to `(4, 4)`. Splash: `(3, 3), (2, 2), (1, 1), (0, 0)` and `(3, 5), (2, 6)...`
Wait, the Example 2 output provided in the prompt is:
`Y . . . . .`
`. Y . . . .`
`. . Y . . .`
`. Y . Y . Y`
`. . . . Y .`
`R R R R R R`
The Ys are at: `(0,0), (1,1), (2,2), (3,3), (4,4)` and `(3,0), (3,2), (3,4), (3,6)`.
This is a V-shape and a horizontal line? No, look closer.
`(3,0), (3,2), (3,4), (3,6)` and `(0,0), (1,1), (2,2), (3,3), (4,4)`.
This is very strange. Let's re-evaluate the test input.

Test Input:
Red line: Row 3.
Drops: `(0, 4)`, `(1, 0)`, `(6, 1)`.
1. `(0, 4)` moves to `

... [460 chars truncated]

### Conceptual Digestion

1.  **CAST**:
    *   **Red (R)**: The magma intrusion. It acts as a stationary boundary/trigger.
    *   **Yellow (Y)**: Boulders (initially) and pyroclastic flows (after fracturing).
    *   **Black (.)**: Solid rock (empty space).

2.  **AFFORDANCES**:
    *   **Boulders (Y)**: Can move down one row (one stratum).
    *   **Magma (R)**: If a boulder lands on the row immediately above the magma, it triggers a "fracture."
    *   **Fracture**: Creates two diagonal lines of Yellow (Y) starting from the point of impact and moving upwards and outwards until they hit the top edge of the grid.

3.  **TRANSFORMATION RULE**:
    *   **Step 1 (Gravity)**: Every Yellow (Y) cell moves down exactly one row.
    *   **Step 2 (Collision Check)**: Check if any Yellow cell landed on the row immediately above the Red (R) line.
    *   **Step 3 (Pyroclastic Flow)**: For every Yellow cell that landed on the row above the Red line:
        *   The cell itself remains Yellow.
        *   Generate a diagonal line of Yellow cells moving up-left (row-1, col-1) until the top of the grid is reached.
        *   Generate a diagonal line of Yellow cells moving up-right (row-1, col+1) until the top of the grid is reached.
    *   **Step 4 (Persistence)**: Boulders that did not land above the magma simply stay in their new shifted position.

### Applying the Rule to Test Input

**Input Grid Analysis:**
*   Magma (R) is at Row 3 (0-indexed).
*   Boulders (Y) are at: (0, 4), (1, 0), (6, 1).

**Step 1: Gravity (Move Y down one row)**
*   (0, 4) $\rightarrow$ (1, 4)
*   (1, 0) $\rightarrow$ (2, 0)
*   (6, 1) $\rightarrow$ (7, 1)

**Step 2: Collision Check**
*   Magma is at Row 3.
*   The row immediately above magma is Row 2.
*   Boulder at (2, 0) has landed on the row above the magma. **Trigger Fracture!**

**Step 3: Pyroclastic Flow (from (2, 0))**
*   Impact point: (2, 0)
*   Up-Left diagonal: (1, -1) $\rightarrow$ Out of bounds.
*   Up-Right diagonal: (1, 1), (0, 2).

**Step 4: Final Positions**
*   Boulders shifted: (1, 4), (2, 0), (7, 1).
*   Flows generated: (1, 1), (0, 2).
*   Magma remains: Row 3.

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

### Conceptual Digestion

1.  **CAST**:
    *   **Red String (R)**: A horizontal line acting as a trigger/boundary.
    *   **Yellow Notes (Y)**: Moving particles.
    *   **Black Space (.)**: Empty space/silence.

2.  **AFFORDANCES**:
    *   **Movement**: Every `Y` moves down one row (row index increases by 1).
    *   **Resonance**: If a `Y` moving from row `r` to `r+1` lands on the row immediately above the `R` line (meaning it was at `R_row - 1` and moves to `R_row`), it triggers a resonance. Wait, let's re-read: "A note arriving directly above the string resonates".
    *   **Correction on Resonance**: Looking at Example 1: `R` is at row 9. A `Y` at row 8 moves to row 9? No, the `R` line stays. If a `Y` is at `R_row - 1`, it moves to `R_row`? No, the `R` line is a physical barrier or a trigger.
    *   **Re-evaluating Example 1**:
        Input `Y` positions: (0,1), (1,5), (4,2), (7,4), (10,3).
        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 (7,4) moved to (8,4). The `R` line is at row 9.
        The `Y` at (10,3) moved to (11,3).
        The `Y` at (4,2) moved to (5,2).
        The `Y` at (0,1) moved to (1,1).
        The `Y` at (1,5) moved to (2,5).
        What happened at row 5? There are `Y`s at (5,1), (5,2), (5,7).
        Let's look at Example 2: `R` is at row 5. `Y` at (2,1) moves to (3,1). `Y` at (3,4) moves to (4,4).
        Wait, the "harmonic overtones" are the key. "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 2: `Y` at (3,4) moves to (4,4). (4,4) is directly above `R` (row 5). This triggers harmonics.
        Harmonics from (4,4) ascend diagonally:
        (3,3), (2,2), (1,1), (0,0)
        (3,5), (2,6)... (out of bounds)
        Wait, Example 2 output has `Y` at (0,0), (1,1), (2,2), (3,3), (4,4), (3,5), (2,6) - no, (3,5) is there, (2,6) is out.
        Let's check Example 2 output again:
        Row 0: (0,0)
        Row 1: (1,1)
        Row 2: (2,2)
        Row 3: (3,1), (3,3), (3,5)
        Row 4: (4,4)
        The `Y` at (2,1) moved to (3,1). The `Y` at (3,4) moved to (4,4) and created harmonics at (3,3), (2,2), (1,1), (0,0) and (3,5), (2,6)... wait, the grid is 6 wide. (3,5) is the edge.
        So: If a note moves to `R_row - 1`, it creates a "V" shape (or rather an inverted V) of `Y`s going upwards and outwards.

3.  **TRANSFORMATION RULE**:
    1.  Every `Y` moves down one row: `(r, c) -> (r+1, c)`.
    2.  If the new position `(r+1, c)` is exactly one row above the `R` line (`r+1 == R_row - 1`), it generates harmonics.
    3.  Harmonics are placed at `(r+1 - k, c - k)` and `(r+1 - k, c + k)` for `k = 1, 2, 3...` until they leave the grid boundaries.

### Applying to Test Input

Input:
Row 0: . . . . Y . . . (0,4)
Row 1: Y . . . . . . . (1,0)
Row 2: . . . . . . . .
Row 3: R R R R R R R R (R_row = 3)
Row 4: . . . . . . . .
Row 5: . . . . . . . .
Row 6: . Y . . . . . . (6,1)
Row 7: . . . . . . . .
Row 8: . . . . . . . .
Row 9: . . . . . . . .
Row 10: . . . . . . . .
Row 11: . . . . . . . .

Step 1: Move all `Y` down one.
(0,4) -> (1,4)
(1,0) -> (2,0)
(6,1) -> (7,1)

Step 2: Check for resonance.
`R_row` is 3. Resonance occurs if a note moves to `R_row - 1 = 2`.
Note (1,0) moved to (2,0). This is a resonance!

Step 3: Generate harmonics from (2,0).
`k=1`: (2-1, 0-1) = (1, -1) [Out], (2-1, 0+1) = (1, 1)
`k=2`: (2-2, 0-2) = (0, -2) [Out], (2-2, 0+2) = (0, 2)
`k=3`: (2-3, 0-3) = (-1, -3) [Out], (2-3, 0+3) = (-1, 3) [Out]

Final positions of `Y`:
(1,4)
(2,0)
(1,1)
(0,2)
(7,1)

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: . . . . . . . .

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

### Conceptual Digestion

1.  **CAST**:
    *   **Red (R)**: The "Inversion Layer" (a horizontal boundary).
    *   **Yellow (Y)**: "Hailstones" (moving objects) and "Shards" (result of impact).
    *   **Black (.)**: "Clear Sky" (empty space).

2.  **AFFORDANCES**:
    *   **Hailstones** can move down one row (descend).
    *   **Hailstones** can "shatter" if they move from the row immediately above the Red layer into the Red layer's position (or effectively, if they are at `row = RedRow - 1` and move to `row = RedRow`).
    *   **Shards** are created upon shattering and move diagonally upwards and outwards from the point of impact.

3.  **TRANSFORMATION RULE**:
    *   **Step 1: Descent.** Every Yellow cell `(r, c)` moves to `(r+1, c)`.
    *   **Step 2: Collision Check.** If a Yellow cell was at `(RedRow - 1, c)`, it hits the Red layer.
    *   **Step 3: Shattering.** When a hailstone hits at `(RedRow, c)`, it disappears (or is consumed by the layer), and shards `Y` are generated. The shards spread diagonally upwards from the impact point `(RedRow, c)`.
    *   **Shard Pattern**: From the impact point `(RedRow, c)`, shards appear at `(RedRow-1, c-1)`, `(RedRow-1, c+1)`, `(RedRow-2, c-2)`, `(RedRow-2, c)`, `(RedRow-2, c+2)`... wait, let's re-examine Example 1.
    *   *Example 1 Analysis*: Red is at row 9. Hailstone at (4, 2) moves to (5, 2). Hailstone at (0, 1) moves to (1, 1). Hailstone at (1, 5) moves to (2, 5). Hailstone at (7, 4) is at `RedRow - 2`, it moves to (8, 4).
    *   Wait, look at the shards in Example 1: The hailstone at (7, 4) moves to (8, 4). But the output shows a V-shape spreading *upwards* from (8, 4)? No, the output shows shards at (5, 1), (5, 2), (5, 7), (6, 2), (6, 6), (7, 3), (7, 5), (8, 4).
    *   Let's re-read: "A hailstone arriving just above the inversion layer shatters on contact and launches shards along two diagonal updrafts spreading apart".
    *   In Example 1, the hailstone at (7, 4) is 2 rows above Red (row 9). It moves to (8, 4). Since it is now "just above" the layer, it shatters? No, the rule says "arriving just above... shatters on contact". This implies the movement *into* the layer triggers it.
    *   Let's re-evaluate Example 1: Hailstone at (7, 4) moves to (8, 4). The shards are at (8, 4), (7, 3), (7, 5), (6, 2), (6, 6), (5, 1), (5, 7). This is a perfect V-shape originating from (8, 4) and going up.
    *   Wait, the hailstone at (7, 4) moved to (8, 4), and *then* it shattered? Or did it shatter because it reached the "just above" position? "A hailstone arriving just above the inversion layer shatters on contact".
    *   If Red is at row 9, "just above" is row 8. When the hailstone at (7, 4) moves to (8, 4), it shatters.
    *   The shards spread diagonally up: `(8, 4) -> (7, 3), (7, 5) -> (6, 2), (6, 6) -> (5, 1), (5, 7)`.

### Applying to 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: . . . . . . . .
Row 8: . . . . . . . .
Row 9: . . . . . . . .
Row 10: . . . . . . . .
Row 11: . . . . . . . .

1.  **Identify Red Layer**: Row 3.
2.  **Identify Hailstones**: (0, 4), (1, 0), (6, 1).
3.  **Move Hailstones**:
    *   (0, 4) -> (1, 4)
    *   (1, 0) -> (2, 0)
    *   (6, 1) -> (7, 1)
4.  **Check for Shattering**:
    *   Hailstone at (2, 0) is now "just above" the inversion layer (Row 3).
    *   It shatters at (2, 0).
    *   Shards spread diagonally up from (2, 0):
        *   (2, 0)
        *   (1, -1) [Out], (1, 1) [In]
        *   (0, -2) [Out], (0, 2) [In]
    *   Shards: (2, 0), (1, 1), (0, 2).
5.  **Other Hailstones**:
    *   (1, 4) stays (it's not at Row 2).
    *   (7, 1) stays.

Final 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 . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .

### Conceptual Digestion

1.  **CAST**:
    *   **Red (R)**: The structural I-beam. It acts as a floor/barrier.
    *   **Yellow (Y)**: Bricks. They are the moving objects.
    *   **Black (.)**: Empty space/wall.
    *   **Yellow (Y) in Output**: Can represent either a shifted brick or a "crack line" resulting from a fracture.

2.  **AFFORDANCES**:
    *   **Bricks (Y)**: Can move down by one row.
    *   **Beam (R)**: If a brick lands on the beam (moves into the row immediately above the beam), it triggers a fracture.
    *   **Fracture**: Creates diagonal lines of Yellow (Y) extending upwards and outwards from the point of impact until the top edge of the grid is reached.

3.  **TRANSFORMATION RULE**:
    *   **Step 1 (Drop)**: Every Yellow (Y) cell moves down exactly one row.
    *   **Step 2 (Collision Check)**: If a Yellow cell's new position is the row immediately above the Red (R) beam, it "fractures."
    *   **Step 3 (Fracture Propagation)**: For every brick that landed on the beam (at row $r$, col $c$), create diagonal lines of Yellow (Y) starting from $(r, c)$. The lines move up-left $(r-1, c-1), (r-2, c-2)...$ and up-right $(r-1, c+1), (r-2, c+2)...$ until they hit the boundary of the grid.
    *   **Step 4 (Non-colliding Bricks)**: Bricks that move but do not land on the beam simply stay in their new position.

### Applying the Rule to Test Input

**Input Grid Analysis:**
*   Grid size: 12 rows x 8 columns.
*   Beam (R) is at Row 3 (0-indexed).
*   Bricks (Y) are at:
    *   (0, 4)
    *   (1, 0)
    *   (6, 1)

**Step 1: Drop Bricks**
*   (0, 4) $\rightarrow$ (1, 4)
*   (1, 0) $\rightarrow$ (2, 0)
*   (6, 1) $\rightarrow$ (7, 1)

**Step 2: Collision Check**
*   The beam is at Row 3.
*   Brick at (2, 0) is now directly above the beam. **Collision!**
*   Brick at (1, 4) is not on the beam.
*   Brick at (7, 1) is not on the beam.

**Step 3: Fracture Propagation**
*   Collision point: (2, 0).
*   Up-Left: (1, -1) $\rightarrow$ Out of bounds.
*   Up-Right: (1, 1), (0, 2).
*   Resulting crack cells: (1, 1), (0, 2).

**Step 4: Final Positions**
*   Brick 1: (1, 4)
*   Brick 2: (2, 0)
*   Brick 3: (7, 1)
*   Cracks: (1, 1), (0, 2)
*   Beam: Row 3 (all R)

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

### Conceptual Digestion

1.  **CAST**:
    *   **Red (R)**: The "proscenium arch line." This is a static boundary/trigger line.
    *   **Yellow (Y)**: "Spotlights" (single dots) and "light beams" (the resulting lines).
    *   **Black (.)**: The empty stage.

2.  **AFFORDANCES**:
    *   Yellow dots move downward (downstage) by one row.
    *   If a Yellow dot moves into a row occupied by the Red line, it "refracts."
    *   Refraction creates two diagonal beams (Yellow) moving upward and outward from the point of impact until they hit the grid boundary.

3.  **TRANSFORMATION RULE**:
    *   **Step 1 (Shift)**: Every `Y` in the input moves down one row (`row + 1`).
    *   **Step 2 (Refract)**: If a `Y` was at `row` and the Red line is at `row + 1`, the `Y` at `row + 1` is replaced by the Red line (the line stays Red), but it triggers the creation of two diagonal lines starting from `(row + 1, col)` and moving upwards:
        *   Beam 1: `(row, col-1), (row-1, col-2), (row-2, col-3)...`
        *   Beam 2: `(row, col+1), (row-1, col+2), (row-2, col+3)...`
    *   **Step 3 (Persistence)**: Yellow dots that do not hit the Red line simply shift down one row.

### Applying Rule to Test Input

**Input Analysis**:
*   Grid size: 12 rows x 8 columns.
*   Red line: Row 3 (0-indexed).
*   Yellow dots:
    1.  (0, 4)
    2.  (1, 0)
    3.  (6, 1)

**Processing**:
1.  **Dot (0, 4)**: Moves to (1, 4). No refraction.
2.  **Dot (1, 0)**: Moves to (2, 0). No refraction.
3.  **Dot (6, 1)**: Moves to (7, 1). No refraction.
4.  **Wait, check for refraction**:
    *   Are there any dots at Row 2 (the row immediately above the Red line at Row 3)?
    *   Input has no dots at Row 2.
    *   Wait, let's re-examine the training examples.
    *   Example 1: Red line at Row 9. Input Y at (7, 4) moves to (8, 4). No refraction. Input Y at (4, 2) moves to (5, 2). No refraction.
    *   Wait, look at Example 1 Output: There are beams! Where did they come from?
    *   Example 1 Input Ys: (0, 1), (1, 5), (4, 2), (7, 4), (10, 3).
    *   Example 1 Output Ys: (1, 0), (2, 5), (5, 0), (5, 1), (5, 7), (6, 2), (6, 6), (7, 3), (7, 5), (8, 4), (11, 3).
    *   Let's re-evaluate: The "spotlight" is the Y. It moves down. If it *arrives* at the arch, it refracts.
    *   In Example 1, the Y at (7, 4) moves to (8, 4). It hasn't hit the arch (Row 9) yet.
    *   Wait, look at the output of Example 1 again. The Y at (7, 4) moved to (8, 4). The Y at (10, 3) moved to (11, 3).
    *   The beams in Example 1 are at (5, 0), (5, 1), (5, 7), (6, 2), (6, 6), (7, 3), (7, 5), (8, 4).
    *   These beams originate from (9, 4) - the position of the Y at (8, 4) if it moved one more step.
    *   Wait, the rule says "A spotlight arriving at the arch refracts". In Example 1, the Y at (7, 4) is 2 steps above the arch.
    *   Let's look at Example 2: Red line at Row 5. Ys at (2, 1) and (3, 4).
    *   Y(2, 1) moves to (3, 1). Y(3, 4) moves to (4, 4).
    *   Output Ys: (0, 0), (1, 1), (2, 2), (3, 1), (3, 3), (3, 5), (4, 4).
    *   The beams originate from (4, 4) moving up: (3, 3), (2, 2), (1, 1), (0, 0) AND (3, 5), (2, 6)...
    *   This means if a Y moves to the row *immediately above* the Red line, it triggers the refraction *immediately*? No, "arriving at the arch".
    *   If Y is at `row`, and `row + 1` is the Red line, the Y moves to `row + 1` (the arch), disappears (or is covered by Red), and shoots beams back up.
    *   In Example 2: Y at (3, 4) moves to (4, 4). Row 5 is Red. So (4, 4) is the row *above* the arch.
    *   Wait, if Y is at (3, 4), it moves to (4, 4). If (4, 4) is the row above the arch, it doesn't refract yet.
    *   Let's re-read: "A spotlight arriving at the arch refracts".
    *   Example 2: Y at (3, 4) moves to (4, 4). The arch is at Row 5. The Y at (4, 4) is now "arriving" at the arch?
    *   Let's check Example 1 again. Y at (7, 4) moves to (8, 4). Arch is at Row 9. (8, 4) is the row above the arch.
    *   Beams from (8, 4) go to (7, 3), (7, 5), (6, 2), (6, 6), (5, 1), (5, 7), (4, 0), (4, 8).
    *   This matches the output!
    *   Rule: If a Y moves to `row` and `row + 1` is the Red line, it creates beams starting from `(row, col)` and going diagonally up.

**Applying to Test Input**:
*   Red line: Row 3.
*   Y1: (0, 4) -> moves to (1, 4). Is (1+1) Red? No (Row 2 is black).
*   Y2: (1, 0) -> moves to (2, 0). Is (2+1) Red? Yes (Row 3 is Red).
    *   Refract from (2, 0):
        *   Beam 1: (1, -1) -> Out of bounds.
        *   Beam 2: (1, 1), (0, 2).
*   Y3: (6, 1) -> moves to (7, 1). Is (7+1) Red? No.

**Final Positions**:
*   Y1: (1, 4)
*   Y2: (2, 0)
*   Y3: (7, 1)
*   Beams from Y2: (1, 1), (0, 2)

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

### Conceptual Digestion

1.  **CAST**:
    *   **The Heart (Embryo)**: A 2x2 block of colors located at a specific coordinate in the input grid.
    *   **The Echoes (Shadows)**: Scaled-up versions of the 2x2 heart (4x4, 8x8, 16x16, etc.) that appear to the right of the original.
    *   **The Field**: The background grid (black `.`).

2.  **AFFORDANCES**:
    *   The heart's colors are mapped to quadrants: Top-Left (TL), Top-Right (TR), Bottom-Left (BL), Bottom-Right (BR).
    *   Each echo preserves these quadrants. For a $2n \times 2n$ echo, each quadrant is an $n \times n$ block of the corresponding color.
    *   Echoes are centered vertically relative to the original heart's center.
    *   Echoes are placed sequentially to the right.

3.  **TRANSFORMATION RULE**:
    *   **Identify Heart**: Find the 2x2 block. In the test input, it is at row 7, col 0 (0-indexed).
        *   TL: B, TR: B, BL: A, BR: A. (Wait, looking at the test input: `B B` / `A A`. This means TL=B, TR=B, BL=A, BR=A).
    *   **Center Point**: The center of the 2x2 heart is between row 7/8 and col 0/1.
    *   **Echo 1 (4x4)**:
        *   Size: 4x4.
        *   Vertical range: Centered on the heart. Heart rows are 7, 8. Center is 7.5. A 4x4 block centered at 7.5 spans rows 6, 7, 8, 9.
        *   Horizontal range: Placed to the right. In Example 1, the 2x2 was at (4,0)-(5,1). The 4x4 echo started at col 2. So the 4x4 echo starts immediately after the heart.
        *   Colors: TL=B, TR=B, BL=A, BR=A. (Each quadrant is 2x2).
    *   **Echo 2 (8x8)**:
        *   Size: 8x8.
        *   Vertical range: Centered at 7.5. Spans rows 4, 5, 6, 7, 8, 9, 10, 11.
        *   Horizontal range: Starts after the 4x4 echo (col 6).
        *   Colors: Each quadrant is 4x4.
    *   **Echo 3 (16x16)**:
        *   Size: 16x16.
        *   Vertical range: Centered at 7.5. Spans rows 0 to 15.
        *   Horizontal range: Starts after the 8x8 echo (col 14).
        *   Colors: Each quadrant is 8x8.

### Applying to Test Input
*   **Heart (2x2)**: Rows 7-8, Cols 0-1. Colors: `B B` / `A A`.
*   **Echo 1 (4x4)**: Rows 6-9, Cols 2-5.
    *   Rows 6-7, Cols 2-3: B | Rows 6-7, Cols 4-5: B
    *   Rows 8-9, Cols 2-3: A | Rows 8-9, Cols 4-5: A
*   **Echo 2 (8x8)**: Rows 4-11, Cols 6-13.
    *   Rows 4-7, Cols 6-9: B | Rows 4-7, Cols 10-13: B
    *   Rows 8-11, Cols 6-9: A | Rows 8-11, Cols 10-13: A
*   **Echo 3 (16x16)**: Rows 0-15, Cols 14-29.
    *   Rows 0-7, Cols 14-21: B | Rows 0-7, Cols 22-29: B
    *   Rows 8-15, Cols 14-21: A | Rows 8-15, Cols 22-29: A

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

### Conceptual Digestion

1.  **CAST**:
    *   **Dancers**: The $2 \times 2$ block of colored pixels in the input grid.
    *   **Stage**: The black background (`.`).
    *   **Formation**: The resulting pattern of expanded colored blocks.

2.  **AFFORDANCES**:
    *   The $2 \times 2$ block can be scaled (doubled) in size.
    *   The pattern can be repeated horizontally.
    *   The vertical position is centered around the original $2 \times 2$ block's height.

3.  **TRANSFORMATION RULE**:
    *   The input is a $2 \times 2$ square at coordinates $(r, c)$ to $(r+1, c+1)$.
    *   The transformation creates a sequence of "echoes" moving to the right.
    *   The first echo is the original $2 \times 2$ block.
    *   The second echo is a $4 \times 4$ block (each pixel of the original $2 \times 2$ becomes a $2 \times 2$ block of the same color), placed immediately to the right of the first.
    *   The third echo is an $8 \times 8$ block (each pixel of the original $2 \times 2$ becomes a $4 \times 4$ block of the same color), placed immediately to the right of the second.
    *   The process continues until the grid width is filled.
    *   **Vertical Centering**: The blocks are centered on the original vertical midpoint. For a $2 \times 2$ block at rows $r, r+1$, the center is between $r$ and $r+1$.
        *   $2 \times 2$ block: rows $r, r+1$.
        *   $4 \times 4$ block: rows $r-1, r, r+1, r+2$.
        *   $8 \times 8$ block: rows $r-3, r-2, r-1, r, r+1, r+2, r+3, r+4$.
    *   **Horizontal Placement**:
        *   Block 1 (size 2): cols $c$ to $c+1$.
        *   Block 2 (size 4): cols $c+2$ to $c+5$.
        *   Block 3 (size 8): cols $c+6$ to $c+13$.
        *   Block 4 (size 16): cols $c+14$ to $c+29$.

### Applying the Rule to Test Input
*   **Input**: $2 \times 2$ block at row 7, col 0.
    *   (7,0)=B, (7,1)=B
    *   (8,0)=A, (8,1)=A
*   **Block 1 (2x2)**:
    *   Rows 7-8, Cols 0-1.
    *   (7,0)=B, (7,1)=B, (8,0)=A, (8,1)=A.
*   **Block 2 (4x4)**:
    *   Rows 6-9, Cols 2-5.
    *   Rows 6-7, Cols 2-3: B; Rows 6-7, Cols 4-5: B.
    *   Rows 8-9, Cols 2-3: A; Rows 8-9, Cols 4-5: A.
*   **Block 3 (8x8)**:
    *   Rows 4-11, Cols 6-13.
    *   Rows 4-7, Cols 6-9: B; Rows 4-7, Cols 10-13: B.
    *   Rows 8-11, Cols 6-9: A; Rows 8-11, Cols 10-13: A.
*   **Block 4 (16x16)**:
    *   Rows 0-15, Cols 14-29.
    *   Rows 0-7, Cols 14-21: B; Rows 0-7, Cols 22-29: B.
    *   Rows 8-15, Cols 14-21: A; Rows 8-15, Cols 22-29: A.

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

### Conceptual Digestion

1.  **CAST**:
    *   **The Chord**: A 2x2 block of colored pixels (the "four colored voices").
    *   **The Staff**: The grid.
    *   **Silence**: Black pixels (`.`).

2.  **AFFORDANCES**:
    *   The 2x2 chord can be scaled (duplicated) in both width and height.
    *   The scaled versions are appended to the right.
    *   The growth is centered vertically on the original chord's position.

3.  **TRANSFORMATION RULE**:
    *   **Initial State**: A 2x2 block at position $(r, c)$ to $(r+1, c+1)$.
    *   **Sequence of Movements**:
        *   Movement 1: The original 2x2 block.
        *   Movement 2: A 2x2 scaled version (each pixel becomes 2x2), resulting in a 4x4 block.
        *   Movement 3: A 4x4 scaled version (each pixel becomes 4x4), resulting in an 8x8 block.
    *   **Placement**:
        *   Movement 1: Starts at $(r, c)$.
        *   Movement 2: Starts at $(r-1, c+2)$. (Centered vertically: $r$ becomes $r-1$ to $r+2$).
        *   Movement 3: Starts at $(r-3, c+6)$. (Centered vertically: $r$ becomes $r-3$ to $r+4$).
    *   **Scaling Logic**:
        *   The 2x2 block consists of colors:
            Top-Left: $C_{00}$, Top-Right: $C_{01}$
            Bottom-Left: $C_{10}$, Bottom-Right: $C_{11}$
        *   In the $2^n \times 2^n$ block, each original pixel $C_{ij}$ becomes a block of size $2^{n-1} \times 2^{n-1}$.

### Applying to Test Input
*   **Input Chord**:
    (7, 0): B, (7, 1): B
    (8, 0): A, (8, 1): A
*   **Movement 1 (2x2)**:
    (7, 0): B, (7, 1): B
    (8, 0): A, (8, 1): A
*   **Movement 2 (4x4)**:
    Rows 6-7, Cols 2-3: B | Rows 6-7, Cols 4-5: B
    Rows 8-9, Cols 2-3: A | Rows 8-9, Cols 4-5: A
*   **Movement 3 (8x8)**:
    Rows 4-7, Cols 6-9: B | Rows 4-7, Cols 10-13: B
    Rows 8-11, Cols 6-9: A | Rows 8-11, Cols 10-13: A
    *Wait, let's check the width of the grid.* The grid is 30 columns wide.
    Movement 1: width 2 (0-1)
    Movement 2: width 4 (2-5)
    Movement 3: width 8 (6-13)
    Movement 4: width 16 (14-29)
    Rows for Movement 4: $r-7$ to $r+6$. $7-7=0$ to $7+6=13$.
    Rows 0-7, Cols 14-21: B | Rows 0-7, Cols 22-29: B
    Rows 8-15, Cols 14-21: A | Rows 8-15, Cols 22-29: A

### Final Grid Construction
Rows 0-3: Cols 14-29 are B
Rows 4-7: Cols 6-29 are B
Row 8: Cols 0-1 are A, Cols 2-5 are A, Cols 6-29 are A (All A)
Row 9: Cols 2-29 are A
Rows 10-11: Cols 6-29 are A
Rows 12-15: Cols 14-29 are A

Wait, let's re-verify the vertical centering.
Original: rows 7, 8.
M1 (2x2): rows 7, 8.
M2 (4x4): rows 6, 7, 8, 9.
M3 (8x8): rows 4, 5, 6, 7, 8, 9, 10, 11.
M4 (16x16): rows 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15.

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

### Conceptual Digestion

1.  **CAST**:
    *   **Athletes**: The 2x2 block of colored pixels in the input grid.
    *   **Field**: The black background (`.`).
    *   **Legs of the Relay**: Successive iterations of the pattern expanding to the right.

2.  **AFFORDANCES**:
    *   The 2x2 block can be scaled (doubled in size).
    *   The pattern can be repeated and shifted rightward.
    *   The vertical position of the scaled blocks expands symmetrically around the original midline.

3.  **TRANSFORMATION RULE**:
    *   The input is a 2x2 block at coordinates $(r, c), (r, c+1), (r+1, c), (r+1, c+1)$.
    *   The transformation creates a sequence of "legs" moving right.
    *   **Leg 1 (Original)**: Size 2x2. Position: $(r, c)$ to $(r+1, c+1)$.
    *   **Leg 2**: Size 4x4. Position: $(r-1, c+2)$ to $(r+2, c+5)$.
    *   **Leg 3**: Size 8x8. Position: $(r-3, c+6)$ to $(r+4, c+13)$.
    *   **Leg 4**: Size 16x16. Position: $(r-7, c+14)$ to $(r+8, c+29)$.
    *   Within each leg, the 2x2 color distribution of the original block is preserved as quadrants. For example, if the original block is $\begin{pmatrix} C1 & C2 \\ C3 & C4 \end{pmatrix}$, a leg of size $2N \times 2N$ will have four $N \times N$ blocks of colors $C1, C2, C3, C4$ respectively.

### Applying to Test Input
*   **Input Block**: 2x2 at row 7, col 0.
    *   (7,0)=B, (7,1)=B
    *   (8,0)=A, (8,1)=A
*   **Leg 1 (2x2)**:
    *   (7,0)=B, (7,1)=B
    *   (8,0)=A, (8,1)=A
*   **Leg 2 (4x4)**:
    *   Rows 6-7, Cols 2-3: B; Rows 6-7, Cols 4-5: B
    *   Rows 8-9, Cols 2-3: A; Rows 8-9, Cols 4-5: A
*   **Leg 3 (8x8)**:
    *   Rows 4-7, Cols 6-9: B; Rows 4-7, Cols 10-13: B
    *   Rows 8-11, Cols 6-9: A; Rows 8-11, Cols 10-13: A
*   **Leg 4 (16x16)**:
    *   Rows 0-7, Cols 14-21: B; Rows 0-7, Cols 22-29: B
    *   Rows 8-15, Cols 14-21: A; Rows 8-15, Cols 22-29: A

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

### Conceptual Digestion

1.  **CAST**:
    *   **The Dam**: Red cells (`R`). These act as the boundary/divider.
    *   **Sediment**: Any non-red, non-black cells (e.g., `B` in Example 1, `X` in Example 2, `W` in Test Input).
    *   **Empty Air/Sky**: Black cells (`.`).

2.  **AFFORDANCES**:
    *   The Red cells are immovable; they stay in their exact positions.
    *   The area *above* the dam is cleared (becomes black `.`).
    *   The area *below* the dam is filled with the sediment color found in the input.

3.  **TRANSFORMATION RULE**:
    *   For each column:
    *   Find the **highest** red cell (`R`) in that column.
    *   Every cell **above** that first `R` becomes black (`.`).
    *   The red cell itself stays `R`.
    *   Every cell **below** that first `R` (including other red cells) is filled. However, looking at Example 2, the rule is more nuanced: the dam is the *entire set* of red cells.
    *   **Refined Rule**:
        1.  Identify all `R` cells. They remain `R`.
        2.  For each column, find the **topmost** `R`.
        3.  All cells in that column **above** the topmost `R` become `.`.
        4.  All cells in that column **below** the topmost `R` that are not `R` become the "sediment" color.
        5.  Wait, looking at Example 2 again: The red cells form a "staircase". The area *below* the red boundary is filled with the sediment color (`X`). The red cells themselves are preserved.
        6.  **Final Rule**: For each column, find the row index of the first `R`. All cells above this index become `.`. All cells at or below this index become either `R` (if they were `R` in the input) or the sediment color (if they were anything else).

### Applying to Test Input

**Sediment Color**: Maroon (`W`)

**Input Grid Analysis (Column by Column):**
Col 0: R at row 5. (Rows 0-4 become `.`, Row 5 is `R`, Rows 6-9 become `W`)
Col 1: R at row 4. (Rows 0-3 become `.`, Row 4 is `R`, Row 5 is `R`, Rows 6-9 become `W`)
Col 2: R at row 4. (Rows 0-3 become `.`, Row 4 is `R`, Row 5 is `.`, Rows 6-9 become `W`) -> Wait, Row 5 is `.` in input, but it's below the first `R`, so it becomes `W`.
Col 3: R at row 4. (Rows 0-3 become `.`, Row 4 is `R`, Row 5 is `R`, Rows 6-9 become `W`)
Col 4: R at row 4. (Rows 0-3 become `.`, Row 4 is `.`, Row 5 is `R`, Rows 6-9 become `W`) -> Wait, Col 4 first R is at row 5. (Rows 0-4 become `.`, Row 5 is `R`, Rows 6-9 become `W`)
Col 5: R at row 4. (Rows 0-3 become `.`, Row 4 is `R`, Row 5 is `R`, Rows 6-9 become `W`)
Col 6: R at row 4. (Rows 0-3 become `.`, Row 4 is `R`, Row 5 is `.`, Rows 6-9 become `W`)
Col 7: R at row 4. (Rows 0-3 become `.`, Row 4 is `R`, Row 5 is `R`, Rows 6-9 become `W`)
Col 8: R at row 4. (Rows 0-3 become `.`, Row 4 is `W`, Row 5 is `R`, Rows 6-9 become `W`) -> First R is at row 5. (Rows 0-4 become `.`, Row 5 is `R`, Rows 6-9 become `W`)
Col 9: R at row 4. (Rows 0-3 become `.`, Row 4 is `R`, Row 5 is `R`, Rows 6-9 become `W`)

**Let's re-verify the "Dam" positions (R):**
Row 4: . R R R . R R R W R (Cols 1, 2, 3, 5, 6, 7, 9)
Row 5: R R . R R R W R R R (Cols 0, 1, 3, 4, 5, 7, 8, 9)

**Topmost R per column:**
Col 0: Row 5
Col 1: Row 4
Col 2: Row 4
Col 3: Row 4
Col 4: Row 5
Col 5: Row 4
Col 6: Row 4
Col 7: Row 4
Col 8: Row 5
Col 9: Row 4

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

Wait, looking at Example 2 again:
Input Row 3: R . X X R . X . R X
Input Row 4: X R . R . R . R . R
Output Row 3: R . . . R . . . R .
Output Row 4: X R . R X R . R X R
In Example 2, the cells *above* the dam are cleared, but the cells *at the same level* as the dam that are not red are also cleared?
Let's re-examine Example 2:
Col 0: First R at Row 3. Output Col 0: Row 0-2 `.`, Row 3 `R`, Row 4-9 `X`.
Col 1: First R at Row 4. Output Col 1: Row 0-3 `.`, Row 4 `R`, Row 5-9 `X`.
Col 2: First R at Row 4. Output Col 2: Row 0-3 `.`, Row 4 `.`, Row 5-9 `X`. (Wait, Input Row 4 Col 2 is `.`, Output is `.`).
Actually, the rule is:
1. For each column, find the first `R`.
2. Everything above that `R` is `.`.
3. The `R` itself stays `R`.
4. Everything below that `R` is filled with sediment (`X` or `W`), UNLESS it was an `R` in the original grid, in which case it stays `R`.
5. What about the cells on the same row as the first `R` but in other columns?
In Example 2, Col 1 first R is at Row 4. Row 3 Col 1 is `.`. It stays `.`.
In Example 2, Col 0 first R is at Row 3. Row 3 Col 1 is `.`. It stays `.`.
This means the "clear sky" is everything above the first `R` *of that specific column*.

Let's re-apply to Test Input:
Col 0: First R at Row 5. Rows 0-4: `.`, Row 5: `R`, Rows 6-9: `W`.
Col 1: First R at Row 4. Rows 0-3: `.`, Row 4: `R`, Rows 5-9: `R` if input was `R` else `W`.
Col 2: First R at Row 4. Rows 0-3: `.`, Row 4: `R`, Rows 5-9:

... [978 chars truncated]

### Conceptual Digestion

1.  **CAST**:
    *   **The Membrane**: The red (R) cells. They act as a boundary/divider.
    *   **The Cells**: The non-red, non-black cells (in the test case, maroon 'W').
    *   **The Void**: Black cells (.).

2.  **AFFORDANCES**:
    *   The red membrane is static; it does not change position or color.
    *   The area above the membrane is subject to "death" (becomes black).
    *   The area below the membrane is subject to "regeneration" (fills with the dominant cell color).

3.  **TRANSFORMATION RULE**:
    *   **Step 1: Identify the Membrane**. Find the row(s) containing red (R). In the training examples, the membrane is a horizontal band. In Example 1, it's a solid line. In Example 2, it's a jagged band.
    *   **Step 2: Clear Above**. Every cell located in a row *above* the highest red cell (or specifically, any cell that is not part of the red membrane and is above the "lowest" point of the membrane's top edge) becomes black (.).
    *   **Step 3: Fill Below**. Every cell located in a row *below* the lowest red cell (or specifically, any cell that is not part of the red membrane and is below the "highest" point of the membrane's bottom edge) becomes the dominant cell color (W in the test case).
    *   **Refining the Rule based on Example 2**: In Example 2, the red cells are scattered across rows 3, 4, and 5.
        *   Rows 0-2 (above the first R) become black.
        *   Rows 3-5 (the membrane rows) keep their R cells; other cells in those rows that were not R are cleared if they are "above" the membrane's logic, but the example shows the R cells stay exactly where they are.
        *   Rows 6-9 (below the last R) become solid X (the dominant color).
        *   Wait, looking closer at Example 2:
            Input Row 3: `R . X X R . X . R X` -> Output Row 3: `R . . . R . . . R .`
            Input Row 4: `X R . R . R . R . R` -> Output Row 4: `X R . R X R . R X R` (Wait, this is different).
            Let's re-examine Example 2 carefully.
            Input Row 3: `R . X X R . X . R X` -> Output Row 3: `R . . . R . . . R .`
            Input Row 4: `X R . R . R . R . R` -> Output Row 4: `X R . R X R . R X R` (No, the output is `X R . R X R . R X R`? No, looking at the provided text: `R . . . R . . . R .` then `X R . R X R . R X R` then `X X R X X X R X X X`).
            Actually, the rule is simpler:
            1. Rows above the first row containing R become black.
            2. Rows below the last row containing R become solid dominant color.
            3. For rows containing R: The R cells stay. Any cell *above* the R cells in that specific column becomes black. Any cell *below* the R cells in that specific column becomes the dominant color.

    *   **Applying to Test Input**:
        *   Dominant color: W (maroon).
        *   Red cells (R) are in rows 4 and 5.
        *   Rows 0, 1, 2, 3: Become black (.).
        *   Rows 4 and 5:
            *   Col 0: R at row 5. Row 4 is above it -> black.
            *   Col 1: R at row 4, 5. Both stay R.
            *   Col 2: R at row 4. Row 5 is below it -> W.
            *   Col 3: R at row 4, 5. Both stay R.
            *   Col 4: R at row 5. Row 4 is above it -> black.
            *   Col 5: R at row 4, 5. Both stay R.
            *   Col 6: R at row 4. Row 5 is below it -> W.
            *   Col 7: R at row 4, 5. Both stay R.
            *   Col 8: R at row 4, 5. Both stay R.
            *   Col 9: R at row 4. Row 5 is below it -> W.
        *   Rows 6, 7, 8, 9: Become solid W.

### Final Grid Calculation
Row 0: . . . . . . . . . .
Row 1: . . . . . . . . . .
Row 2: . . . . . . . . . .
Row 3: . . . . . . . . . .
Row 4: . R R R . R R R R R (Wait, let's re-check Col 9: Input R4C9 is R, R5C9 is R. So R4C9=R, R5C9=R)
Let's re-evaluate the R positions in Test Input:
Row 4: `. R R R . R R R W R` (R at indices 1, 2, 3, 5, 6, 7, 9)
Row 5: `R R . R R R W R R R` (R at indices 0, 1, 3, 4, 5, 7, 8, 9)

Column by Column for Rows 4 & 5:
Col 0: R4=., R5=R -> R4=., R5=R
Col 1: R4=R, R5=R -> R4=R, R5=R
Col 2: R4=R, R5=. -> R4=R, R5=W
Col 3: R4=R, R5=R -> R4=R, R5=R
Col 4: R4=., R5=R -> R4=., R5=R
Col 5: R4=R, R5=R -> R4=R, R5=R
Col 6: R4=R, R5=W -> R4=R, R5=W (Wait, R5C6 is W, not R. So R4C6=R, R5C6=W)
Col 7: R4=R, R5=R -> R4=R, R5=R
Col 8: R4=W, R5=R -> R4=., R5=R
Col 9: R4=R, R5=R -> R4=R, R5=R

Row 4: . R R R . R R R . R
Row 5: R R W R R R W R R R
Rows 6-9: W W W W W W W W W W

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

**Conceptual Digestion**

1.  **Identify the CAST:**
    *   **Red (R):** The "Accretion Ring." It acts as a boundary and is immovable.
    *   **Matter (W, B, X, etc.):** Any color that is not Red or Black. In the test input, the matter is Maroon (W).
    *   **Void (.):** Black space.

2.  **Identify the AFFORDANCES:**
    *   The area **above** the red ring is subject to "vanishing" (becoming void).
    *   The area **below** the red ring is subject to "accretion" (filling up with the matter color).
    *   The red ring itself stays exactly where it is.

3.  **Identify the TRANSFORMATION RULE:**
    *   **Step 1:** Locate all Red (R) cells. These cells remain unchanged in the output.
    *   **Step 2:** For every cell $(r, c)$ where $r$ is above the *highest* row containing a Red cell, it becomes Black (.). *Correction based on Example 2:* The rule is more specific. The "ring" is the set of all Red cells. Everything *above* the topmost Red cell becomes void. Everything *below* the bottommost Red cell becomes the matter color.
    *   **Refining the Rule based on Example 2:** In Example 2, the Red cells are scattered across rows 3, 4, and 5.
        *   Rows 0, 1, 2 (above the first R) become `.`.
        *   Rows 3, 4, 5 (containing R) keep their R positions. The non-R cells in these rows are filled with the matter color (X) *only if* they are below the "ring" logic? No, looking closer:
        *   Example 1: R is a solid line at row 3. Rows 0-2 become `.`. Rows 4-9 become `B`.
        *   Example 2: R cells are at (3,0), (3,4), (3,8), (4,1), (4,3), (4,5), (4,7), (4,9), (5,2), (5,6).
            *   Rows 0-2 become `.`.
            *   Row 3: R stays, others become `.`.
            *   Row 4: R stays, others become `X` (matter).
            *   Row 5: R stays, others become `X` (matter).
            *   Rows 6-9: All become `X`.
    *   **Precise Rule:**
        1.  Identify the row index of the first (topmost) Red cell ($R_{min}$).
        2.  Identify the row index of the last (bottommost) Red cell ($R_{max}$).
        3.  For rows $r < R_{min}$: All cells become `.`.
        4.  For rows $r > R_{max}$: All cells become the matter color.
        5.  For rows $R_{min} \le r \le R_{max}$:
            *   If cell $(r, c)$ is `R`, it stays `R`.
            *   If cell $(r, c)$ is not `R`, it becomes the matter color *if* it is below the "top" of the ring structure? Let's re-examine Example 2 Row 3. Row 3 is $R_{min}$. In Row 3, only the R's remained, others became `.`. In Row 4 and 5, R's remained and others became `X`.
            *   **Revised Rule for $R_{min} \le r \le R_{max}$:** If $r = R_{min}$, non-R cells become `.`. If $r > R_{min}$, non-R cells become the matter color.

**Applying to Test Input:**
*   Matter color: Maroon (W).
*   Red cells (R) are at:
    *   Row 4: (4,1), (4,2), (4,3), (4,5), (4,6), (4,7), (4,9)
    *   Row 5: (5,0), (5,1), (5,3), (5,4), (5,5), (5,7), (5,8), (5,9)
*   $R_{min} = 4$, $R_{max} = 5$.
*   Rows 0, 1, 2, 3: Become `.`.
*   Row 4 ($R_{min}$): R's stay, others become `.`.
*   Row 5 ($R_{max}$): R's stay, others become `W`.
*   Rows 6, 7, 8, 9: All become `W`.

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

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

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Thermocline:** The Red (R) cells. These act as the boundary/divider.
    *   **Brine:** The non-red, non-black color present in the grid (in the test case, Maroon 'W').
    *   **Clear Water:** The Black (.) cells.

2.  **Identify the AFFORDANCES:**
    *   The Red cells are fixed; they do not change position or color.
    *   The area *above* the thermocline (all cells in rows above the first occurrence of R, and cells in the same row as R that are above the "line") is cleared to black.
    *   The area *below* the thermocline is filled entirely with the brine color.

3.  **Identify the TRANSFORMATION RULE:**
    *   **Step 1: Locate the Thermocline.** The thermocline is the set of all Red (R) cells.
    *   **Step 2: Define "Above" and "Below".** Looking at the training examples, the "thermocline layer" is the set of rows containing Red. However, Example 2 shows Red cells scattered. Let's re-examine:
        *   Example 1: Red is a solid line at row 4. Rows 0-3 become black. Rows 4-9 (below the line) become Blue (the brine).
        *   Example 2: Red cells are scattered. The "thermocline" seems to be the *highest* row containing a Red cell. Wait, looking closer at Example 2: The Red cells themselves stay. Everything *above* the highest Red cell becomes black. Everything *below* the lowest Red cell becomes the brine color (Grey 'X'). What happens to the rows containing Red? The cells in those rows that are not Red are treated based on their position relative to the "layer".
        *   Actually, looking at Example 2 again:
            Input Row 3: `R . X X R . X . R X` -> Output Row 3: `R . . . R . . . R .` (Non-reds become black)
            Input Row 4: `X R . R . R . R . R` -> Output Row 4: `X R . R X R . R X R` (Wait, this is complex).
            Let's re-read: "The thermocline remains fixed. Water above it clears to transparency. Water below it saturates fully with the colored brine."
            In Example 2, the Red cells are in rows 3, 4, and 5.
            Rows 0, 1, 2 become all `.`.
            Row 3 (contains R): Non-R cells become `.`.
            Row 4 (contains R): Non-R cells... wait, the output for Row 4 is `X R . R X R . R X R`. This looks like the original row but with some changes.
            Let's look at Row 5 (contains R): `X X R X X X R X X X`.
            Rows 6, 7, 8, 9 become all `X`.
            
            Correction on Example 2:
            Row 3: `R . X X R . X . R X` -> `R . . . R . . . R .` (Non-R become black)
            Row 4: `X R . R . R . R . R` -> `X R . R X R . R X R` (Wait, the X's are preserved? No, the X's are the brine).
            Actually, the rule is simpler:
            1. Any row *entirely above* the first row containing R becomes all `.`.
            2. Any row *entirely below* the last row containing R becomes all Brine.
            3. For rows *containing* R:
               - If it's the *first* row containing R, non-R cells become `.`.
               - If it's the *last* row containing R, non-R cells become Brine.
               - If it's a row *between* the first and last R-row, non-R cells... stay as they were? No, in Example 2 Row 4, the `.` became `X` and `X` stayed `X`. So non-R cells become Brine.
            
            Let's re-verify Example 2 Row 4:
            Input: `X R . R . R . R . R`
            Output: `X R . R X R . R X R`
            Wait, the `.` stayed `.` and `X` stayed `X`? No, the `.` at index 2 stayed `.`, but the `.` at index 4 became `X`. This is confusing.
            
            Let's try a simpler interpretation:
            The "Thermocline" is the region from the first row containing R to the last row containing R.
            - Rows above the first R-row $\rightarrow$ all `.`
            - Rows below the last R-row $\rightarrow$ all Brine.
            - Within the R-rows:
                - First R-row: Non-R $\rightarrow$ `.`
                - Last R-row: Non-R $\rightarrow$ Brine.
                - Middle R-rows: Non-R $\rightarrow$ Brine.
            
            Let's check Example 2 again with this:
            Row 3 (First R): `R . X X R . X . R X` $\rightarrow$ `R . . . R . . . R .` (Correct)
            Row 4 (Middle R): `X R . R . R . R . R` $\rightarrow$ `X R X R X R X R X R` (Wait, the output provided was `X R . R X R . R X R`. Let me re-count the dots).
            Example 2 Output Row 4: `X R . R X R . R X R`. There are dots at index 2 and 6.
            Input Row 4: `X R . R . R . R . R`. Dots are at index 2, 4, 6, 8.
            Output Row 4: `X R . R X R . R X R`. Dots at 2 and 6. X's at 0, 4, 8.
            This means the brine (X) filled the dots at 4 and 8, but not 2 and 6? This is inconsistent.
            
            Let's look at Example 2 Row 5 (Last R):
            Input: `. . R X X X R . X .`
            Output: `X X R X X X R X X X` (Correct, all non-R become X).
            
            Wait! Look at Example 2 Ro

... [4005 chars truncated]

**Conceptual Digestion**

1.  **CAST:**
    *   **Junction:** Red cells (`R`). These act as the boundary/divider.
    *   **Charge:** Any non-red, non-black color (in the test case, Maroon `W`).
    *   **Zero State:** Black cells (`.`).

2.  **AFFORDANCES:**
    *   The junction (`R`) is fixed and does not change position or color.
    *   Cells above the junction (relative to the row index) are cleared to the zero state (`.`).
    *   Cells below the junction are filled with the "charge" color.

3.  **TRANSFORMATION RULE:**
    *   Identify the "charge" color (the non-red, non-black color present in the grid). In the test input, this is Maroon (`W`).
    *   Identify the junction (`R`).
    *   For every cell `(row, col)`:
        *   If `grid[row][col] == 'R'`, it remains `'R'`.
        *   If the cell is "above" the junction: This is interpreted as any cell in a row that is entirely above the *lowest* red cell, or more accurately based on the examples, any cell in a row that does not contain the "lowest" part of the junction.
        *   Looking closer at Example 3: The junction is a diagonal of `R`s. Cells to the left/below the `R`s are filled, cells to the right/above are cleared.
        *   Re-evaluating the rule: The junction `R` acts as a mask. For any cell `(r, c)`, if there is an `R` at `(r, c)`, it stays `R`. If the cell is "above" the junction (meaning for that column `c`, the row `r` is less than the row of the `R` in that column), it becomes `.`. If the cell is "below" the junction (meaning for that column `c`, the row `r` is greater than the row of the `R` in that column), it becomes the charge color.
        *   Wait, Example 1: Junction is a straight line at row 3. Rows 0-2 become `.`, row 3 stays `R`, rows 4-9 become `B`.
        *   Example 2: Junction is a complex shape. The output shows that for each column, the cells above the *last* `R` in that column are cleared (unless they are `R`), and cells below the *last* `R` in that column are filled with the charge color `X`.
        *   Example 3: For each column, cells above the `R` are cleared, and cells below the `R` are filled with `A`.

**Applying to Test Input:**
Charge color: `W` (Maroon).
Junction `R` positions:
Col 0: Row 5
Col 1: Row 4, 5
Col 2: Row 4, 5
Col 3: Row 4, 5
Col 4: Row 5
Col 5: Row 4, 5
Col 6: Row 4
Col 7: Row 4, 5
Col 8: Row 4, 5
Col 9: Row 4, 5

Last `R` per column:
Col 0: 5
Col 1: 5
Col 2: 5
Col 3: 5
Col 4: 5
Col 5: 5
Col 6: 4
Col 7: 5
Col 8: 5
Col 9: 5

Transformation:
- Rows 0-3: All become `.`
- Row 4: Col 0: `.`, Col 1: `R`, Col 2: `R`, Col 3: `R`, Col 4: `.`, Col 5: `R`, Col 6: `R`, Col 7: `R`, Col 8: `W` (Wait, Col 8 last R is 5, so Row 4 is above it -> `.`), Col 9: `R` (Wait, Col 9 last R is 5, so Row 4 is above it -> `.`).
Let's re-check Row 4: `. R R R . R R R . .` (Wait, Col 6 last R is 4, so Row 4 Col 6 is `R`).
- Row 5: Col 0: `R`, Col 1: `R`, Col 2: `.`, Col 3: `R`, Col 4: `R`, Col 5: `R`, Col 6: `W` (Below last R at 4), Col 7: `R`, Col 8: `R`, Col 9: `R`.
- Rows 6-9: All become `W`.

Wait, let's re-examine Example 2 carefully.
Input Row 3: `R . X X R . X . R X`
Input Row 4: `X R . R . R . R . R`
Input Row 5: `. . R X X X R . X .`
Output Row 3: `R . . . R . . . R .`
Output Row 4: `X R . R X R . R X R`
Output Row 5: `X X R X X X R X X X`
In Example 2, the `R`s stay exactly where they were. Any cell `(r, c)` becomes `X` if there is an `R` at some `(r_j, c)` where `r_j < r`. Any cell `(r, c)` becomes `.` if there is no `R` at or above it.

Applying this logic to Test Input:
Charge: `W`
R positions: (4,1), (4,2), (4,3), (4,5), (4,6), (4,7), (4,9), (5,0), (5,1), (5,3), (5,4), (5,5), (5,7), (5,8), (5,9)

Row 0-3: No `R` above or at these rows for any column except where `R` might be. No `R`s here. All `.`.
Row 4:
Col 0: No `R` at or above. `.`
Col 1: `R` at (4,1). `R`
Col 2: `R` at (4,2). `R`
Col 3: `R` at (4,3). `R`
Col 4: No `R` at or above. `.`
Col 5: `R` at (4,5). `R`
Col 6: `R` at (4,6). `R`
Col 7: `R` at (4,7). `R`
Col 8: No `R` at or above. `.`
Col 9: `R` at (4,9). `R`
Row 5:
Col 0: `R` at (5,0). `R`
Col 1: `R` at (4,1). `R`
Col 2: No `R` at (5,2), but `R` at (4,2). `W`
Col 3: `R` at (4,3). `R`
Col 4: `R` at (5,4). `R`
Col 5: `R` at (4,5). `R`
Col 6: No `R` at (5,6), but `R` at (4,6). `W`
Col 7: `R` at (4,7). `R`
Col 8: `R` at (5,8). `R`
Col 9: `R` at (4,9). `R`
Rows 6-9:
All columns have an `R` above them. All `W`.

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

### Conceptual Digestion

1.  **CAST**:
    *   **The River of Fire**: Red cells (`R`). These are static and do not move or change.
    *   **The Spirits**: Cells of a single color other than Red or Black (in Example 1: Blue `B`, Example 2: Grey `X`, Test Input: Maroon `W`).
    *   **The Void**: Black cells (`.`).

2.  **AFFORDANCES**:
    *   The grid is divided into two regions by the "river" (the Red cells).
    *   Cells above the river (rows index < river index) can be cleared.
    *   Cells below the river (rows index > river index) can be filled.

3.  **TRANSFORMATION RULE**:
    *   **Identify the River**: Locate all Red (`R`) cells. These remain exactly where they are.
    *   **Annihilation (Above)**: Any cell located in a row that is *above* the highest Red cell, or any non-red cell in a row that contains Red cells but is "above" the river's flow (though the examples suggest a simpler row-based split), is turned to Black (`.`). Specifically, looking at Example 1, the river is a solid line at row 3. Everything above row 3 becomes `.`.
    *   **Packing (Below)**: Count all "spirit" cells (the non-red, non-black color) present in the *entire* input grid. These spirits "sink" to the bottom of the grid, filling the available non-red spaces from the bottom up, row by row, left to right.
    *   **Refining the Rule based on Example 2**: In Example 2, the river is not a straight line. The Red cells stay. All non-red cells *above* the lowest Red cell are cleared? No, that's not it. Let's re-examine:
        *   Example 1: River at row 3. Rows 0-2 become `.`. Rows 4-9 fill with `B` until all `B`s from the input are used.
        *   Example 2: Red cells are scattered. Red cells stay. All non-red cells *above* the lowest Red cell are cleared? No.
        *   Wait, look at Example 2 again: The Red cells are at (3,0), (3,4), (3,8), (4,1), (4,3), (4,5), (4,7), (4,9), (5,2), (5,6). The lowest Red cell is at row 5.
        *   Output Example 2: Rows 0-2 are `.`. Row 3 and 4 keep their `R`s, others become `.`. Row 5 keeps `R`s, others become `X`. Rows 6-9 are all `X`.
        *   **Correct Rule**: 
            1. Keep all `R` cells.
            2. Count total number of spirit cells (`S`) in the input.
            3. Clear all non-`R` cells.
            4. Fill the remaining non-`R` cells with `S` starting from the bottom-right (or bottom-left) moving upwards.
            5. Let's check Example 2 count: Input has 43 `X`s. Output has `X`s at: Row 4 (3 cells), Row 5 (4 cells), Row 6-9 (40 cells). Total = 47. Wait, the count is different.
            6. Let's re-read: "Below it, the dead pack together until the underworld is full."
            7. In Example 1: River is row 3. Rows 4-9 are all `B`. (60 cells). Input had 35 `B`s.
            8. In Example 2: Lowest Red cell is row 5. Rows 6-9 are all `X` (40 cells). Row 5 has 4 `X`s. Row 4 has 3 `X`s. Total = 47.
            9. **Revised Rule**: Find the lowest row containing a Red cell. Every non-Red cell *below* that row becomes the spirit color. Every non-Red cell *above* that row becomes Black. For the row containing the lowest Red cell, fill non-Red cells with the spirit color? No, Example 2: Row 5 is the lowest Red row. Row 5 has `X`s. Row 4 also has `X`s.
            10. Let's look at the "River" as a boundary. In Example 1, the river is row 3. In Example 2, the river is the collection of Red cells. The "Underworld" is everything below the *highest* Red cell? No.
            11. Let's try: Everything *above* the first row containing a Red cell becomes `.`. Everything *below* the last row containing a Red cell becomes the spirit color. For rows *containing* Red cells, only the Red cells remain? No, Example 2 row 4 and 5 have `X`s.
            12. Let's look at Example 2 again. Red cells are in rows 3, 4, 5.
                Row 3: `R . . . R . . . R .` (Only R's kept)
                Row 4: `X R . R X R . R X R` (R's kept, some X's)
                Row 5: `X X R X X X R X X X` (R's kept, some X's)
                Rows 6-9: All `X`.
                This looks like the spirit color fills from the bottom up, skipping `R` cells.
                Total `X` cells in Example 2 output: Row 4 (3), Row 5 (6), Row 6-9 (40). Total = 49.
                Total `X` cells in Example 2 input: 43.
                Wait, the rule is simpler: **Everything above the first Red row is void. Everything below the first Red row is filled with the spirit color, except where Red cells are.**
                Example 1: First Red row is 3. Rows 0-2 = `.`. Rows 4-9 = `B`. Row 3 = `R`.
                Example 2: First Red row is 3. Rows 0-2 = `.`. Row 3 = `R . . . R . . . R .`. Rows 4-9 = `X` (except `R`s).
                Let's check Example 2 Row 3: `R . . . R . . . R .`. This matches the input Red positions!
                So:
                1. Find the index of the first row containing at least one `R`.
                2. All rows before this index become all `.`.
                3. 

... [1307 chars truncated]

**Conceptual Digestion**

1.  **Identify the CAST:**
    *   **Ribbon Dancers (R):** The red cells. They act as the "divider" or "freeze line."
    *   **Chorus Dancers (C):** The other non-black color present in the grid (Example 1: Blue B, Example 2: Grey X, Test Input: Maroon W).
    *   **Empty Marks (.):** Black cells.

2.  **Identify the AFFORDANCES:**
    *   The Red cells (R) are stationary.
    *   The space "ahead" (above) the Red cells is cleared (becomes black).
    *   The space "behind" (below) the Red cells is filled with the Chorus color (C) wherever there was a mark (either a Chorus dancer or an empty mark).

3.  **Identify the TRANSFORMATION RULE:**
    *   **Step 1: Locate the Red (R) cells.** These cells remain exactly where they are in the output.
    *   **Step 2: Clear the "Ahead" space.** Any cell located in a row *above* the highest row containing a Red cell is cleared to black (.). However, looking closer at Example 2, the "ahead" space is relative to the *specific* Red cells. Wait, let's re-examine Example 2.
    *   **Re-evaluating Example 2:**
        Input R positions: (3,0), (3,4), (3,8), (4,1), (4,3), (4,5), (4,7), (4,9), (5,2), (5,6).
        Output R positions: Same.
        Output above the *lowest* R: The rows above the lowest R are cleared *except* for the R cells themselves.
        Output below the *lowest* R: All cells are filled with the chorus color (X).
        Wait, looking at Example 2 again:
        Row 3: `R . . . R . . . R .` (Only R's kept)
        Row 4: `X R . R X R . R X R` (R's kept, others filled with X)
        Row 5: `X X R X X X R X X X` (R's kept, others filled with X)
        Row 6-9: All `X`.
        
        **Correct Rule:**
        1. Identify the Chorus color (C).
        2. Identify all Red (R) cells.
        3. For every cell $(r, c)$ in the grid:
           - If grid$(r, c) == R$, it stays $R$.
           - If grid$(r, c)$ is in a row *above* the first row that contains any $R$, it becomes black (.).
           - If grid$(r, c)$ is in a row *below* the first row that contains any $R$:
             - If it is in the same row as the *first* $R$ but not an $R$ itself, it becomes black (.).
             - If it is in any row *below* the first row containing an $R$, it becomes the Chorus color (C), unless it is an $R$.

    *   **Refining Rule based on Example 1:**
        First R row: Row 3.
        Rows 0-2: All black.
        Row 3: All R.
        Rows 4-9: All B.
        This matches.

    *   **Refining Rule based on Example 2:**
        First R row: Row 3.
        Rows 0-2: All black.
        Row 3: Only R's kept, others black.
        Rows 4-9: All X (except where R's are).
        This matches perfectly.

**Applying to Test Input:**
Chorus color: Maroon (W).
First row containing R: Row 4.
- Rows 0, 1, 2, 3: All become black (.).
- Row 4: Keep R's, others become black (.).
- Rows 5, 6, 7, 8, 9: All become W, except where R's are.

Test Input Row 4: `. R R R . R R R W R` -> Output: `. R R R . R R R . R`
Test Input Row 5: `R R . R R R W R R R` -> Output: `W W W W W W W W W W` (Wait, R's stay!) -> `R R W R R R W R R R` (Wait, the R's are at (5,0), (5,1), (5,3), (5,4), (5,5), (5,7), (5,8), (5,9). The cell (5,2) and (5,6) become W).
Correct Row 5: `R R W R R R W R R R` (Wait, (5,6) was W, now it's W. (5,2) was ., now it's W).
Actually, the rule is: if it's R, it stays R. If it's not R and it's below the first R-row, it becomes W.

Row 4: `. R R R . R R R . R`
Row 5: `R R W R R R W R R R`
Row 6: `W W W W W W W W W W`
Row 7: `W W W W W W W W W W`
Row 8: `W W W W W W W W W W`
Row 9: `W W W W W W W W W W`

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

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Magma Intrusion (R):** The fixed boundary/divider.
    *   **Sediment (W, B, X, etc.):** The material that can either be eroded or lithified. In the test input, the sediment color is Maroon (W).
    *   **Voids (.):** Empty spaces.

2.  **Identify the AFFORDANCES:**
    *   **Above the Magma:** Everything (sediment and voids) is subject to erosion.
    *   **The Magma itself:** Remains exactly where it is.
    *   **Below the Magma:** Voids are filled by the sediment color (lithification).

3.  **Identify the TRANSFORMATION RULE:**
    *   **Step 1 (Above):** For every row index $i$ that is strictly above the *first* occurrence of a Red (R) cell in any column, or more simply, any cell $(i, j)$ where there is no Red cell at or above it in that column? No, looking at Example 1: The Red line is at row 3 (0-indexed). Rows 0, 1, and 2 become all black (.).
    *   **Step 2 (The Magma):** Any cell that is Red (R) in the input remains Red (R) in the output.
    *   **Step 3 (Below):** For any row $i$ that is below the *highest* Red cell? No, let's look at Example 2. The Red cells are scattered across rows 3, 4, and 5.
        *   Rows 0, 1, 2 (above the first R) $\rightarrow$ all black (.).
        *   Rows 3, 4, 5 (containing R) $\rightarrow$ R stays R, other cells stay as they were? No, in Example 2, row 3: `R . X X R . X . R X` stays `R . . . R . . . R .`. Wait, looking closer at Example 2:
            *   Input Row 3: `R . X X R . X . R X` $\rightarrow$ Output Row 3: `R . . . R . . . R .` (Only R remains).
            *   Input Row 4: `X R . R . R . R . R` $\rightarrow$ Output Row 4: `X R . R X R . R X R` (Wait, this is confusing).
            *   Let's re-examine Example 1: Red is a solid line at row 3. Rows 0-2 $\rightarrow$ `.`. Row 3 $\rightarrow$ `R`. Rows 4-9 $\rightarrow$ `B`.
            *   Let's re-examine Example 2:
                *   Row 0-2: All `.`
                *   Row 3: `R . . . R . . . R .` (Only R's kept)
                *   Row 4: `X R . R X R . R X R` (R's kept, X's kept, .'s stay .) - No, that's not it.
                *   Let's look at the "Below" rule: "Below it, sediment lithifies and fills every pore until the rock is solid."
                *   In Example 2, the "magma intrusion" is the set of all Red cells.
                *   The "Below" region starts from the *lowest* row containing a Red cell.
                *   Lowest Red cell in Example 2 is in Row 5.
                *   Rows 6, 7, 8, 9 in Example 2 become all `X` (the sediment color).
                *   What happens to the rows *containing* Red?
                *   Example 2, Row 3: `R . X X R . X . R X` $\rightarrow$ `R . . . R . . . R .` (X's removed)
                *   Example 2, Row 4: `X R . R . R . R . R` $\rightarrow$ `X R . R X R . R X R` (Wait, the X's are appearing where they weren't? No, look at the input: `X R . R . R . R . R`. Output: `X R . R X R . R X R`. The X at index 4 was a `.` in input).
                *   Actually, look at Example 2 again. The Red cells are at: (3,0), (3,4), (3,8), (4,1), (4,3), (4,5), (4,7), (4,9), (5,2), (5,6).
                *   The output for rows 3, 4, 5 is simply the input but with the sediment color (X) filling in? No.
                *   Let's re-read: "The magma intrusion is fixed." This means R stays R.
                *   "Above it, all material erodes away." $\rightarrow$ Rows above the *first* R become `.`.
                *   "Below it, sediment lithifies and fills every pore." $\rightarrow$ Rows below the *last* R become the sediment color.
                *   What about the rows *containing* R? In Example 1, the R's are all in one row. In Example 2, they span rows 3, 4, 5.
                *   Example 2 Output Rows 3-5:
                    Row 3: `R . . . R . . . R .`
                    Row 4: `X R . R X R . R X R`
                    Row 5: `X X R X X X R X X X`
                *   Observation: In the rows containing R, the sediment (X) only stays/appears if it is *below* a Red cell in that column.
                *   Wait, look at Example 2 Row 5: `X X R X X X R X X X`. Input was `. . R X X X R . X .`. The `.` at (5,0), (5,1), (5,7), (5,9) became `X`.
                *   Rule for rows containing R: If a cell is R, it stays R. If it is not R, it becomes the sediment color IF there is a Red cell *above* it in the same column (including the current row? No).
                *   Let's try this: For any cell $(i, j)$:
                    1. If $i <$ index of first row containing any R $\rightarrow$ `.`
                    2. If $i >$ index of last row containing any R $\rightarrow$ sediment color.
                    3. If $i$ is a row containing R:
                       - If cell $(i, j)$ is R $\rightarrow$ R.
                       - If cell $(i, j)$ is not R:
                         - If there is an R at some $(k, j)$ where $k \le i$ $\rightarrow$ sediment color.
                         - Else $

... [3697 chars truncated]

**Conceptual Digestion:**

1.  **CAST:**
    *   **Red (R):** The "sustained chord." It acts as the dividing line/anchor of the transformation.
    *   **Other Color (e.g., B, X, W):** The "orchestral color." This is the fill color used for the bottom section.
    *   **Black (.):** Silence/Empty space.

2.  **AFFORDANCES:**
    *   The Red cells are immutable (they stay exactly where they are).
    *   Cells *above* the highest Red cell (or above the Red line in a specific column) are cleared to Black.
    *   Cells *below* the Red cells are filled with the orchestral color.

3.  **TRANSFORMATION RULE:**
    *   Identify the "Red chord" (the row(s) containing R).
    *   Any cell $(r, c)$ where $r$ is above the first occurrence of $R$ in column $c$ becomes Black (.).
    *   Any cell $(r, c)$ where $r$ is below the last occurrence of $R$ in column $c$ becomes the orchestral color (in the test case, Maroon 'W').
    *   Wait, looking closer at Example 1 and 2:
        *   Example 1: Red is a solid line at row 3. Everything above row 3 becomes `.`. Everything below row 3 becomes `B`.
        *   Example 2: Red is scattered across rows 3, 4, 5. Everything above the *first* Red in a column is cleared? No, looking at Example 2 output: Row 0, 1, 2 are all `.`. Row 3 (the first row containing R) keeps its R's and clears others. Row 4 keeps its R's and fills others with X? No, Row 4 in output is `X R . R X R . R X R`. This means the R's are preserved, and the non-R's are filled with X *if they are below the first R of that column*.
        *   Let's re-evaluate Example 2:
            Input Row 3: `R . X X R . X . R X` -> Output Row 3: `R . . . R . . . R .` (Only R's kept)
            Input Row 4: `X R . R . R . R . R` -> Output Row 4: `X R . R X R . R X R` (R's kept, others filled with X)
            Input Row 5: `. . R X X X R . X .` -> Output Row 5: `X X R X X X R X X X` (R's kept, others filled with X)
            Input Row 6-9: All filled with X.
        *   **Refined Rule:**
            1. Find the row index of the *first* Red cell in the entire grid (let's call this `min_r`).
            2. For all rows $r < min\_r$: Set all cells to `.`.
            3. For the row $r = min\_r$: Keep `R`, set all others to `.`.
            4. For all rows $r > min\_r$: Keep `R`, set all others to the orchestral color (W).

**Applying to Test Input:**
*   Orchestral color: W (Maroon).
*   First row containing R: Row 4 (`. R R R . R R R W R`).
*   Rows 0, 1, 2, 3: Become all `.`.
*   Row 4: Keep R, others become `.` -> `. R R R . R R R . R`
*   Rows 5, 6, 7, 8, 9: Keep R, others become `W`.

Row 5: `R R . R R R W R R R` -> `R R W R R R W R R R`
Row 6: `W . . W W W . W W .` -> `W W W W W W W W W W`
Row 7: `. . . . W . W . . W` -> `W W W W W W W W W W`
Row 8: `. W W . . . . W W .` -> `W W W W W W W W W W`
Row 9: `W . W . . W . W . .` -> `W W W W W W W W W W`

Wait, let's check Example 2 Row 4 again.
Input Row 4: `X R . R . R . R . R`
Output Row 4: `X R . R X R . R X R`
Wait, the `.` in the input remained `.` in the output.
So the rule is: If it's `R`, keep it. If it's the orchestral color, keep it. If it's `.`, it stays `.`? No, that doesn't fit Example 1.
Let's look at Example 2 Row 4 again: `X R . R . R . R . R` -> `X R . R X R . R X R`.
The `X`s were kept, the `.`s stayed `.`.
Example 2 Row 5: `. . R X X X R . X .` -> `X X R X X X R X X X`.
The `.`s became `X`.
This means: Below the first row of R, any cell that is NOT R becomes the orchestral color, UNLESS it was a `.` and the rule is different?
No, in Example 2 Row 4, the `.` at index 2 stayed `.`. In Row 5, the `.` at index 0 became `X`.
Looking closer at Example 2 Row 4: `X R . R . R . R . R` -> `X R . R X R . R X R`.
The `.` at index 2 was a `.` in the input. The `.` at index 4 became `X`.
Wait, the input Row 4 index 4 was `.`. The output Row 4 index 4 is `X`.
The input Row 4 index 2 was `.`. The output Row 4 index 2 is `.`.
Is it possible the `.` in the output of Example 2 Row 4 index 2 is actually a typo in my reading and it should be `X`?
Let's re-read: `X R . R X R . R X R`.
Input Row 4: `X R . R . R . R . R`.
Indices of `.`: 2, 4, 6, 8.
Output Row 4: `X R . R X R . R X R`.
Indices of `.`: 2, 6.
This is confusing. Let's look at the "Metaphorical Description": "Every beat below the chord is filled with the orchestra's color."
This implies all non-R cells below the chord become the orchestral color.

Let's re-verify Example 2 Row 4: `X R . R X R . R X R`.
If the rule is "Fill all non-R with X", it should be `X R X R X R X R X R`.
Why are there `.` at index 2 and 6?
Input Row 4: `X R . R . R . R . R`
Input Row 3: `R . X X R . X . R X`
Wait! The `.` in the output of Row 4 are at the same positions as the `.` in Row 3!
Row 3: `R . X X R . X . R X`
Row 4: `X R . R X R . R X R`
Indices of `.` in Row 3: 1, 5, 7.
Indices of `.` in Row 4: 2, 6.
No, that's not it.

Let's try the simplest interpretation:
1. Rows above the first R-row: Al

... [311 chars truncated]

### Conceptual Digestion

1.  **CAST**:
    *   **Defensive Line**: Red (R) cells. These act as the boundary/pivot for the transformation.
    *   **Players**: Any color other than Red (R) or Black (.). In the test input, this is Maroon (W).
    *   **Open Field**: Black (.) cells.

2.  **AFFORDANCES**:
    *   The Red (R) cells are static; they do not move or change.
    *   The area "in front" of the defensive line (above the highest R or in the gaps before the line) is cleared to black (.).
    *   The area "behind" the defensive line (below the R cells) is filled with the player color (W).

3.  **TRANSFORMATION RULE**:
    *   The "defensive line" is defined by the presence of Red (R) cells.
    *   Looking at the training examples, the transformation is based on the vertical position of the Red cells.
    *   In Example 1, the Red line is a solid horizontal line at row 3 (0-indexed). Everything above row 3 becomes `.`, and everything below row 3 becomes `B`.
    *   In Example 2, the Red cells are scattered. However, the rule is: for every column, find the *first* (topmost) Red cell. Everything above that first Red cell in that column becomes `.`. Everything below that first Red cell (including the Red cell itself) is treated as the "defensive zone". Specifically, the Red cells stay Red, and every cell *below* the topmost Red cell in that column becomes the player color (X).
    *   Wait, let's re-examine Example 2 more closely:
        Input Row 3: `R . X X R . X . R X`
        Input Row 4: `X R . R . R . R . R`
        Input Row 5: `. . R X X X R . X .`
        Output Row 3: `R . . . R . . . R .`
        Output Row 4: `X R . R X R . R X R`
        Output Row 5: `X X R X X X R X X X`
        Output Row 6-9: All `X`.
        Observation: In column 0, the first R is at row 3. Row 0-2 become `.`, row 3 stays `R`, rows 4-9 become `X`. In column 1, the first R is at row 4. Row 0-3 become `.`, row 4 stays `R`, rows 5-9 become `X`.
    *   **Refined Rule**: For each column $j$:
        1. Find the index $i$ of the first occurrence of 'R'.
        2. For all rows $k < i$, set cell $(k, j) = '.'$.
        3. For row $k = i$, set cell $(k, j) = 'R'$.
        4. For all rows $k > i$, set cell $(k, j) = \text{Player Color}$ (unless cell $(k, j)$ was already 'R', in which case it stays 'R').

### Applying Rule to Test Input
Player Color: Maroon (W)

Columns:
0: First R at row 5. Rows 0-4 $\to$ `.`, Row 5 $\to$ `R`, Rows 6-9 $\to$ `W`.
1: First R at row 4. Rows 0-3 $\to$ `.`, Row 4 $\to$ `R`, Rows 5-9 $\to$ `W` (Row 5 is R, so stays R).
2: First R at row 4. Rows 0-3 $\to$ `.`, Row 4 $\to$ `R`, Rows 5-9 $\to$ `W` (Row 5 is `.`, becomes W).
3: First R at row 4. Rows 0-3 $\to$ `.`, Row 4 $\to$ `R`, Rows 5-9 $\to$ `W` (Row 5 is R, so stays R).
4: First R at row 5. Rows 0-4 $\to$ `.`, Row 5 $\to$ `R`, Rows 6-9 $\to$ `W`.
5: First R at row 4. Rows 0-3 $\to$ `.`, Row 4 $\to$ `R`, Rows 5-9 $\to$ `W` (Row 5 is R, so stays R).
6: First R at row 4. Rows 0-3 $\to$ `.`, Row 4 $\to$ `R`, Rows 5-9 $\to$ `W` (Row 5 is `.`, becomes W).
7: First R at row 4. Rows 0-3 $\to$ `.`, Row 4 $\to$ `R`, Rows 5-9 $\to$ `W` (Row 5 is R, so stays R).
8: First R at row 4. Rows 0-3 $\to$ `.`, Row 4 $\to$ `R`, Rows 5-9 $\to$ `W` (Row 5 is R, so stays R).
9: First R at row 4. Rows 0-3 $\to$ `.`, Row 4 $\to$ `R`, Rows 5-9 $\to$ `W` (Row 5 is R, so stays R).

Wait, let's re-check the test input grid for R positions:
Row 4: `. R R R . R R R W R` (Cols 1, 2, 3, 5, 6, 7, 9 are R)
Row 5: `R R . R R R W R R R` (Cols 0, 1, 3, 4, 5, 7, 8, 9 are R)

Correcting Column Analysis:
Col 0: First R at row 5.
Col 1: First R at row 4.
Col 2: First R at row 4.
Col 3: First R at row 4.
Col 4: First R at row 5.
Col 5: First R at row 4.
Col 6: First R at row 4.
Col 7: First R at row 4.
Col 8: First R at row 5.
Col 9: First R at row 4.

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

Wait, looking at Row 5 again: `R R . R R R W R R R`.
Col 2 is `.` at Row 5. Col 6 is `W` at Row 5. Col 8 is `R` at Row 5.
Let's re-verify Col 8: Row 4 is `W`, Row 5 is `R`. First R is at row 5.
So Row 4 Col 8 is `.`. Row 5 Col 8 is `R`. Row 6-9 Col 8 is `W`.

Let's re-verify Row 5:
Col 0: R (First R)
Col 1: R (First R was row 4) $\to$ R
Col 2: . (First R was row 4) $\to$ W
Col 3: R (First R was row 4) $\to$ R
Col 4: R (First R)
Col 5: R (First R was row 4) $\to$ R
Col 6: W (First R was row 4) $\to$ W
Col 7: R (First R was row 4) $\to$ R
Col 8: R (First R)
Col 9: R (First R was row 4) $\to$ R

Row 5 becomes: `R R W R R R W R R R`

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

**Conceptual Digestion**

1.  **CAST**:
    *   **Fortifications**: Red (R) cells. They act as a boundary/divider.
    *   **Troops**: A single non-red, non-black color (in the test case, Maroon 'W').
    *   **Empty Ground**: Black (.) cells.

2.  **AFFORDANCES**:
    *   The Red (R) cells are static; they do not move or change.
    *   The "forward side" (the side of the red line that is "exposed" or "outside" the shelter) is cleared.
    *   The "sheltered side" (the side of the red line that is "protected") is filled completely with the troop color.

3.  **TRANSFORMATION RULE**:
    *   Identify the boundary formed by the Red (R) cells.
    *   The Red cells divide the grid into two regions. In the training examples, the "forward side" is the area "above/left" of the red line's progression, and the "sheltered side" is "below/right".
    *   Specifically:
        *   Any cell $(r, c)$ that is "above" or "to the left" of the red boundary (where the red boundary is the frontier) is set to black (.).
        *   Any cell $(r, c)$ that is "below" or "to the right" of the red boundary is filled with the troop color (W), *unless* that cell is already occupied by a Red (R) cell.
    *   Looking at Example 3: The red line moves diagonally. Everything above/left of the red line is cleared; everything below/right is filled with Azure (A).
    *   In the Test Input: The red line is roughly in the middle (rows 4 and 5).
        *   Row 4: `. R R R . R R R W R`
        *   Row 5: `R R . R R R W R R R`
        *   The "forward side" is the top part of the grid.
        *   The "sheltered side" is the bottom part of the grid.
        *   Everything above the red line is cleared.
        *   Everything below the red line is filled with Maroon (W), preserving the Red (R) cells.

**Applying to Test Input**:
*   Rows 0-3: All become `.`
*   Row 4: `. R R R . R R R W R` -> The `W` is on the "forward" side of the red line's boundary at that specific column? No, the red line is the boundary. Let's look at the red cells: (4,1), (4,2), (4,3), (4,5), (4,6), (4,7), (4,9) and (5,0), (5,1), (5,3), (5,4), (5,5), (5,7), (5,8), (5,9).
*   The red line effectively separates the grid. Everything "behind" the red line (rows 5-9 and the gaps in row 4) becomes W.
*   Wait, looking at Example 2: The red cells are scattered. The output shows that any cell $(r, c)$ is filled if there is a red cell at $(r', c')$ where $r' < r$ and $c' < c$? No.
*   Let's re-examine Example 2: Red cells are at (3,0), (3,4), (3,8), (4,1), (4,3), (4,5), (4,7), (4,9), (5,2), (5,6).
    The output fills everything below and to the right of these red cells.
    Specifically, if a cell $(r, c)$ is "sheltered" by the red line, it becomes the troop color.
    In Example 2, the red line is a jagged diagonal. Everything below the "lowest" red cell in a column or to the right of the "rightmost" red cell in a row is filled.
    Actually, the simplest rule: If a cell $(r, c)$ is "below" the red line (meaning for that column $c$, there is a red cell at some $r' \le r$), it is filled.
    Let's check Example 3: Red cells at (1,0), (2,0), (2,1), (3,1), (3,2), (4,2), (4,3), (5,3), (5,4), (6,4), (6,5), (7,5), (7,6), (8,6), (8,7), (9,7), (9,8).
    Output: Cell $(r, c)$ is A if $r \ge$ the row of the red cell in that column.
    Example 1: Red line at row 3. Rows 0-2 become `.`, row 3 stays `R`, rows 4-9 become `B`.

**Applying to Test Input**:
Red cells are at:
Col 0: Row 5
Col 1: Row 4, 5
Col 2: Row 4
Col 3: Row 4, 5
Col 4: Row 5
Col 5: Row 4, 5
Col 6: Row 4
Col 7: Row 4, 5
Col 8: Row 5
Col 9: Row 4, 5

The "lowest" (minimum row) red cell for each column:
Col 0: 5
Col 1: 4
Col 2: 4
Col 3: 4
Col 4: 5
Col 5: 4
Col 6: 4
Col 7: 4
Col 8: 5
Col 9: 4

Everything above these rows is `.`. Everything at or below these rows is `W` (unless it's `R`).
Row 0: . . . . . . . . . .
Row 1: . . . . . . . . . .
Row 2: . . . . . . . . . .
Row 3: . . . . . . . . . .
Row 4: . R R R . R R R . R (Col 0 is above row 5, Col 4 is above row 5, Col 8 is above row 5)
Row 5: R R W R R R W R R R (Col 2 is row 4, so row 5 is W. Col 6 is row 4, so row 5 is W)
Row 6: W W W W W W W W W W
Row 7: W W W W W W W W W W
Row 8: W W W W W W W W W W
Row 9: W W W W W W W W W W

Wait, in Example 2, the red cells themselves are preserved.
Row 4: `. R R R . R R R . R` (The W at (4,8) becomes . because it's above the red cell at (5,8))
Row 5: `R R W R R R W R R R` (The . at (5,2) becomes W because it's below the red cell at (4,2))

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

### Conceptual Digestion

**1. Identify the CAST:**
*   **Azure Cross (A):** The "leading" or scaffolding. It divides the 9x9 grid into four 4x4 quadrants (with the center row/column being the dividers).
*   **Green Glass (G):** The pattern to be colored. It exists within one of the 4x4 quadrants.
*   **The Prism (2x2 block of colors):** A small square located in one of the dark quadrants. It contains four colors (one for each quadrant: Top-Left, Top-Right, Bottom-Left, Bottom-Right).
*   **Darkness (.):** Empty space.

**2. Identify the AFFORDANCES:**
*   The **Azure Cross** defines the boundaries.
*   The **Prism** provides the color mapping. The color at position (0,0) of the prism corresponds to the Top-Left quadrant, (0,1) to Top-Right, (1,0) to Bottom-Left, and (1,1) to Bottom-Right.
*   The **Green Glass** acts as a mask; it takes the color assigned to its quadrant.

**3. Identify the TRANSFORMATION RULE:**
1.  Locate the 2x2 prism. In the test input, it is at the bottom right:
    *   Top-Left of prism: `R` (Red)
    *   Top-Right of prism: `B` (Blue)
    *   Bottom-Left of prism: `Y` (Yellow)
    *   Bottom-Right of prism: `O` (Orange)
2.  Identify the quadrant containing the green pattern. In the test input, the green pattern is in the Top-Left quadrant (rows 0-5, cols 0-5, excluding the azure lines).
3.  The Top-Left quadrant of the pattern is mapped to the Top-Left color of the prism.
    *   Top-Left Prism Color = `R` (Red).
4.  Replace all `G` in the Top-Left quadrant with `R`.
5.  Remove the Azure scaffolding (the 7th row and 7th column in the 9x9 grid, though the prompt says the result is 6x6, implying we remove the dividers and the empty padding).
6.  Wait, let's re-examine the grid dimensions. The input is 9x9. The output is 6x6.
    *   Input: 9x9. Azure lines are at index 6 (column) and index 6 (row).
    *   This divides the grid into a 6x6 area and some margins.
    *   Looking at Example 1: Azure is at col 6 and row 2. The output is 6x6.
    *   In Example 1: Prism is at (0,7) to (1,8). Colors: (0,7)=B, (0,8)=R, (1,7)=Y, (1,8)=B.
    *   Pattern is in the bottom-left quadrant (rows 3-8, cols 0-5).
    *   Bottom-Left color of prism is `Y`.
    *   Wait, the example output shows the pattern colored with `B` and `R` and `Y`.
    *   Let's re-read: "Each of the prism's four facets shines toward the quadrant of the pattern that mirrors its own position".
    *   The 6x6 pattern itself is divided into four 3x3 quadrants.
    *   Prism (2x2):
        Top-Left color $\rightarrow$ Top-Left 3x3 of the 6x6 pattern.
        Top-Right color $\rightarrow$ Top-Right 3x3 of the 6x6 pattern.
        Bottom-Left color $\rightarrow$ Bottom-Left 3x3 of the 6x6 pattern.
        Bottom-Right color $\rightarrow$ Bottom-Right 3x3 of the 6x6 pattern.

**Applying to Test Input:**
1.  **Prism (bottom right):**
    (0,0) of prism: `R`
    (0,1) of prism: `B`
    (1,0) of prism: `Y`
    (1,1) of prism: `O`
2.  **Pattern (top left 6x6):**
    Row 0: `. . . G . .`
    Row 1: `G G . G . G`
    Row 2: `. G . G . G`
    Row 3: `. G G G . .`
    Row 4: `. G . . . G`
    Row 5: `. . G . . .`
3.  **Apply colors to 3x3 quadrants of the 6x6:**
    *   Top-Left (0-2, 0-2) $\rightarrow$ `R`
    *   Top-Right (0-2, 3-5) $\rightarrow$ `B`
    *   Bottom-Left (3-5, 0-2) $\rightarrow$ `Y`
    *   Bottom-Right (3-5, 3-5) $\rightarrow$ `O`

**Mapping:**
Row 0: `.` `.` `.` | `B` `.` `.`
Row 1: `R` `R` `.` | `B` `.` `B`
Row 2: `.` `R` `.` | `B` `.` `B`
Row 3: `.` `Y` `Y` | `O` `.` `.`
Row 4: `.` `Y` `.` | `.` `.` `O`
Row 5: `.` `.` `Y` | `.` `.` `.`

ANSWER:
. . . B . .
R R . B . B
. R . B . B
. Y Y O . .
. Y . . . O
. . Y . . .

### Conceptual Digestion

**1. Identify the CAST:**
*   **Azure (A) bus traces:** These form a cross (horizontal and vertical lines) that divide the grid into four quadrants. They are markers and are removed in the output.
*   **Green (G) active nodes:** These are the "pads" on the chip. They are the only elements that persist in the output, but their color changes.
*   **Voltage Sources (B, R, Y, O, etc.):** Four colored pixels located in a small square (usually at the end of the bus traces). Each color corresponds to one of the four quadrants.
*   **Black (.) pads:** Inactive areas that remain black.

**2. Identify the AFFORDANCES:**
*   The **Azure cross** defines the boundaries of the four quadrants.
*   The **Voltage Sources** provide the color for the green nodes within their respective quadrants.
*   The **Green nodes** act as targets that "light up" based on their location relative to the azure cross.

**3. Identify the TRANSFORMATION RULE:**
1.  Locate the Azure (A) cross. This divides the "chip area" (the area containing green nodes) into four equal quadrants: Top-Left, Top-Right, Bottom-Left, and Bottom-Right.
2.  Locate the 2x2 square of voltage sources.
    *   The source at the top-left of the 2x2 square powers the Top-Left quadrant.
    *   The source at the top-right of the 2x2 square powers the Top-Right quadrant.
    *   The source at the bottom-left of the 2x2 square powers the Bottom-Left quadrant.
    *   The source at the bottom-right of the 2x2 square powers the Bottom-Right quadrant.
3.  For every Green (G) node in the input:
    *   If it is in the Top-Left quadrant, change its color to the Top-Left voltage source color.
    *   If it is in the Top-Right quadrant, change its color to the Top-Right voltage source color.
    *   If it is in the Bottom-Left quadrant, change its color to the Bottom-Left voltage source color.
    *   If it is in the Bottom-Right quadrant, change its color to the Bottom-Right voltage source color.
4.  Remove all Azure (A) traces and the voltage source square. The output grid dimensions are trimmed to only include the "chip area" (the area bounded by the cross).

### Applying the Rule to Test Input

**Input Analysis:**
*   **Azure Cross:** Vertical line at column 6 (index 6), Horizontal line at row 6 (index 6).
*   **Chip Area:** Rows 0-5 and Columns 0-5.
*   **Voltage Sources (at bottom right):**
    *   (7, 7): R (Top-Left of the 2x2 square) -> Powers Top-Left quadrant.
    *   (7, 8): B (Top-Right of the 2x2 square) -> Powers Top-Right quadrant.
    *   (8, 7): Y (Bottom-Left of the 2x2 square) -> Powers Bottom-Left quadrant.
    *   (8, 8): O (Bottom-Right of the 2x2 square) -> Powers Bottom-Right quadrant.

**Wait, let's re-examine the quadrant mapping based on the 2x2 source position relative to the cross:**
In Example 1:
Cross: Col 6, Row 2.
Sources: (0, 7)=B, (0, 8)=R, (1, 7)=Y, (1, 8)=B.
Output:
Top-Left (Rows 3-8, Cols 0-5) is split.
Actually, looking at Example 1 again:
Input: Azure at Col 6, Row 2.
Sources: (0, 7)=B, (0, 8)=R, (1, 7)=Y, (1, 8)=B.
Output:
Top-Left quadrant (Rows 3-5, Cols 0-2) -> B
Top-Right quadrant (Rows 3-5, Cols 3-5) -> R
Bottom-Left quadrant (Rows 6-8, Cols 0-2) -> Y
Bottom-Right quadrant (Rows 6-8, Cols 3-5) -> B
This confirms: Top-Left source -> Top-Left quadrant, etc.

**Applying to Test Input:**
*   **Cross:** Row 6, Col 6.
*   **Chip Area:** Rows 0-5, Cols 0-5.
*   **Voltage Sources:** (7, 7)=R, (7, 8)=B, (8, 7)=Y, (8, 8)=O.
*   **Quadrants (within the 6x6 chip area):**
    *   Top-Left: Rows 0-2, Cols 0-2. Color: R
    *   Top-Right: Rows 0-2, Cols 3-5. Color: B
    *   Bottom-Left: Rows 3-5, Cols 0-2. Color: Y
    *   Bottom-Right: Rows 3-5, Cols 3-5. Color: O

**Mapping Green Nodes:**
Row 0: `. . . G . .` -> Col 3 is Top-Right -> `B`
Row 1: `G G . G . G` -> Col 0,1 (TL) -> `R R`, Col 3 (TR) -> `B`, Col 5 (TR) -> `B`
Row 2: `. G . G . G` -> Col 1 (TL) -> `R`, Col 3 (TR) -> `B`, Col 5 (TR) -> `B`
Row 3: `. G G G . .` -> Col 1,2 (BL) -> `Y Y`, Col 3 (BR) -> `O`
Row 4: `. G . . . G` -> Col 1 (BL) -> `Y`, Col 5 (BR) -> `O`
Row 5: `. . G . . .` -> Col 2 (BL) -> `Y`

**Final Grid Construction (6x6):**
Row 0: `. . . B . .`
Row 1: `R R . B . B`
Row 2: `. R . B . B`
Row 3: `. Y Y O . .`
Row 4: `. Y . . . O`
Row 5: `. . Y . . .`

ANSWER:
. . . B . .
R R . B . B
. R . B . B
. Y Y O . .
. Y . . . O
. . Y . . .

### Conceptual Digestion

**1. Identify the CAST:**
*   **Azure (A):** These act as dividers/grid lines, splitting the input into quadrants.
*   **Green (G):** These represent the "rhythm" or the pattern to be transformed.
*   **Colored Notes (B, R, Y, O, etc.):** These are the "instrument assignments." They are located in a small cluster (usually a 2x2 area) and assign a specific color to a specific quadrant of the green pattern.
*   **Black (.):** These are "rests" or empty space.

**2. Identify the AFFORDANCES:**
*   The Azure lines define the boundaries of the quadrants.
*   The Green notes are the targets for recoloring.
*   The Colored Notes provide the mapping: each color in the cluster corresponds to the quadrant it is closest to (or logically mapped to).

**3. Identify the TRANSFORMATION RULE:**
*   **Step 1: Isolate the Pattern.** Identify the area containing the Green (G) notes. In the test input, the pattern is in the top-left section (rows 0-5, cols 0-5).
*   **Step 2: Identify the Instrument Map.** Look at the colored cluster. In the test input, the cluster is at the bottom right:
    *   (7, 7): R (Red)
    *   (7, 8): B (Blue)
    *   (8, 7): Y (Yellow)
    *   (8, 8): O (Orange)
*   **Step 3: Map Colors to Quadrants.** The pattern is divided into four quadrants (top-left, top-right, bottom-left, bottom-right).
    *   Top-Left quadrant $\rightarrow$ Red (R)
    *   Top-Right quadrant $\rightarrow$ Blue (B)
    *   Bottom-Left quadrant $\rightarrow$ Yellow (Y)
    *   Bottom-Right quadrant $\rightarrow$ Orange (O)
*   **Step 4: Apply Transformation.** Replace every 'G' in the pattern with the color assigned to its quadrant.
    *   Pattern size: 6 rows x 6 columns.
    *   Quadrant size: 3 rows x 3 columns.
    *   Rows 0-2, Cols 0-2: G $\rightarrow$ R
    *   Rows 0-2, Cols 3-5: G $\rightarrow$ B
    *   Rows 3-5, Cols 0-2: G $\rightarrow$ Y
    *   Rows 3-5, Cols 3-5: G $\rightarrow$ O

**Applying to Test Input:**
Pattern (Rows 0-5, Cols 0-5):
Row 0: `. . . G . .`
Row 1: `G G . G . G`
Row 2: `. G . G . G`
Row 3: `. G G G . .`
Row 4: `. G . . . G`
Row 5: `. . G . . .`

Transformation:
Row 0: `. . .` (R) | `B . .` (B) $\rightarrow$ `. . . B . .`
Row 1: `R R .` (R) | `B . B` (B) $\rightarrow$ `R R . B . B`
Row 2: `. R .` (R) | `B . B` (B) $\rightarrow$ `. R . B . B`
Row 3: `. Y Y` (Y) | `O . .` (O) $\rightarrow$ `. Y Y O . .`
Row 4: `. Y .` (Y) | `. . O` (O) $\rightarrow$ `. Y . . . O`
Row 5: `. . Y` (Y) | `. . .` (O) $\rightarrow$ `. . Y . . .`

ANSWER:
. . . B . .
R R . B . B
. R . B . B
. Y Y O . .
. Y . . . O
. . Y . . .

### Conceptual Digestion

1.  **The CAST**:
    *   **Azure (A)**: The "chalk lines." These act as dividers that split the grid into four quadrants.
    *   **Green (G)**: The "player markers." These are the target cells that will change color.
    *   **Other Colors (R, B, Y, O, M, X)**: The "captains." These are located in a small 2x2 huddle in one of the corners created by the azure lines.
    *   **Black (.)**: Empty space.

2.  **The AFFORDANCES**:
    *   The Azure lines define the boundaries.
    *   The Captains' colors are mapped to the four quadrants of the formation area.
    *   Green markers are replaced by the color of the captain assigned to their quadrant.
    *   Azure lines and empty spaces (black) are removed/remain black.

3.  **The TRANSFORMATION RULE**:
    *   **Step 1: Identify the Dividers.** Find the row and column of Azure (A) cells. This splits the grid into four quadrants.
    *   **Step 2: Identify the Captains.** Find the 2x2 block of non-azure, non-green colors.
    *   **Step 3: Map Captains to Quadrants.** The 2x2 huddle of captains corresponds spatially to the four quadrants of the formation.
        *   Top-Left Captain $\rightarrow$ Top-Left Quadrant
        *   Top-Right Captain $\rightarrow$ Top-Right Quadrant
        *   Bottom-Left Captain $\rightarrow$ Bottom-Left Quadrant
        *   Bottom-Right Captain $\rightarrow$ Bottom-Right Quadrant
    *   **Step 4: Recolor.** For every Green (G) cell in the formation area, change its color to the corresponding captain's color.
    *   **Step 5: Clean up.** Remove the Azure lines and the Captains' huddle. Only the recolored formation remains.

### Applying the Rule to Test Input

**Input Analysis:**
*   Azure lines: Column 6 (index 6) and Row 6 (index 6).
*   Formation Area: Rows 0-5, Columns 0-5.
*   Captains' Huddle: Rows 7-8, Columns 7-8.
    *   (7, 7): R (Red) $\rightarrow$ Top-Left Quadrant
    *   (7, 8): B (Blue) $\rightarrow$ Top-Right Quadrant
    *   (8, 7): Y (Yellow) $\rightarrow$ Bottom-Left Quadrant
    *   (8, 8): O (Orange) $\rightarrow$ Bottom-Right Quadrant

**Mapping the Formation (Rows 0-5, Cols 0-5):**
*   **Top-Left Quadrant (Rows 0-2, Cols 0-2):**
    (0, 3) is outside. (1, 0)=G, (1, 1)=G, (2, 1)=G.
    *Wait, let's re-evaluate the quadrants based on the Azure lines.*
    Azure line is at Row 6 and Col 6.
    Formation is Rows 0-5 and Cols 0-5.
    The "quarters" of the formation are split by the *center* of the formation area.
    Formation area: 6x6. Center is between row 2/3 and col 2/3.
    *   Top-Left: Rows 0-2, Cols 0-2
    *   Top-Right: Rows 0-2, Cols 3-5
    *   Bottom-Left: Rows 3-5, Cols 0-2
    *   Bottom-Right: Rows 3-5, Cols 3-5

**Recoloring:**
*   **Top-Left (R):**
    (1, 0) G $\rightarrow$ R
    (1, 1) G $\rightarrow$ R
    (2, 1) G $\rightarrow$ R
*   **Top-Right (B):**
    (0, 3) G $\rightarrow$ B
    (1, 3) G $\rightarrow$ B
    (1, 5) G $\rightarrow$ B
    (2, 3) G $\rightarrow$ B
    (2, 5) G $\rightarrow$ B
*   **Bottom-Left (Y):**
    (3, 1) G $\rightarrow$ Y
    (3, 2) G $\rightarrow$ Y
    (4, 1) G $\rightarrow$ Y
    (5, 2) G $\rightarrow$ Y
*   **Bottom-Right (O):**
    (3, 3) G $\rightarrow$ O
    (4, 5) G $\rightarrow$ O

**Final Grid Construction (6x6):**
Row 0: . . . B . .
Row 1: R R . B . B
Row 2: . R . B . B
Row 3: . Y Y O . .
Row 4: . Y . . . O
Row 5: . . Y . . .

ANSWER:
. . . B . .
R R . B . B
. R . B . B
. Y Y O . .
. Y . . . O
. . Y . . .

### Conceptual Digestion

**1. Identify the CAST:**
*   **Azure Ridges (A):** These act as dividers (a cross shape) that split the grid into four quadrants. They are removed in the final output.
*   **Green Vegetation (G):** The target pattern. This is the "land" that will be recolored.
*   **Weather Systems (B, R, Y, O, M, X, etc.):** Four distinct colored pixels located in one of the quadrants. These provide the new colors for the vegetation.
*   **Barren Ground (.):** Black space that remains unchanged.

**2. Identify the AFFORDANCES:**
*   The Azure ridges define the boundaries of four quadrants.
*   The Green pixels are "recolorable" based on which quadrant they reside in.
*   The four colored pixels (weather systems) are mapped to the four quadrants.

**3. Identify the TRANSFORMATION RULE:**
*   **Step 1: Locate the Ridges.** Find the horizontal and vertical lines of 'A'.
*   **Step 2: Define Quadrants.** The ridges divide the grid into four areas: Top-Left, Top-Right, Bottom-Left, and Bottom-Right.
*   **Step 3: Map Weather to Quadrants.** Find the four colored pixels. Each one is located in one of the four quadrants.
    *   In the test input: The colors are R, B, Y, O. They are all located in the Bottom-Right quadrant (relative to the ridges). Wait, looking closer at the test input:
        *   Ridges: Row 6 (index 6) and Column 6 (index 6).
        *   Weather pixels: (7, 7)=R, (7, 8)=B, (8, 7)=Y, (8, 8)=O.
        *   Wait, the metaphor says "Each weather system drifts into the nearest quarter". In Example 1, the weather is Top-Right, and it fills the Top-Left, Top-Right, Bottom-Left, and Bottom-Right of the *vegetated area*.
        *   Let's re-examine the mapping:
            *   Example 1: Weather (0,7)=B, (0,8)=R, (1,7)=Y, (1,8)=B. Vegetation is in Top-Left.
            *   Actually, the weather pixels are always in a 2x2 block.
            *   Top-Left pixel of 2x2 $\rightarrow$ Top-Left quadrant of vegetation.
            *   Top-Right pixel of 2x2 $\rightarrow$ Top-Right quadrant of vegetation.
            *   Bottom-Left pixel of 2x2 $\rightarrow$ Bottom-Left quadrant of vegetation.
            *   Bottom-Right pixel of 2x2 $\rightarrow$ Bottom-Right quadrant of vegetation.
*   **Step 4: Recolor.**
    *   Vegetation in Top-Left quadrant $\rightarrow$ Color of Top-Left weather pixel.
    *   Vegetation in Top-Right quadrant $\rightarrow$ Color of Top-Right weather pixel.
    *   Vegetation in Bottom-Left quadrant $\rightarrow$ Color of Bottom-Left weather pixel.
    *   Vegetation in Bottom-Right quadrant $\rightarrow$ Color of Bottom-Right weather pixel.
*   **Step 5: Crop.** The output grid size is the size of the vegetated area (the area bounded by the ridges).

### Applying to Test Input:
1.  **Ridges:** Row 6, Column 6.
2.  **Vegetation Area:** Rows 0-5, Columns 0-5.
3.  **Weather System (2x2 block):**
    *   (7, 7) = R (Top-Left of block)
    *   (7, 8) = B (Top-Right of block)
    *   (8, 7) = Y (Bottom-Left of block)
    *   (8, 8) = O (Bottom-Right of block)
4.  **Mapping to Vegetation (6x6 area):**
    *   Top-Left (Rows 0-2, Cols 0-2) $\rightarrow$ R
    *   Top-Right (Rows 0-2, Cols 3-5) $\rightarrow$ B
    *   Bottom-Left (Rows 3-5, Cols 0-2) $\rightarrow$ Y
    *   Bottom-Right (Rows 3-5, Cols 3-5) $\rightarrow$ O
5.  **Processing the Grid (0-5, 0-5):**
    *   (0, 3): G $\rightarrow$ B
    *   (1, 0): G $\rightarrow$ R
    *   (1, 1): G $\rightarrow$ R
    *   (1, 3): G $\rightarrow$ B
    *   (1, 5): G $\rightarrow$ B
    *   (2, 1): G $\rightarrow$ R
    *   (2, 3): G $\rightarrow$ B
    *   (2, 5): G $\rightarrow$ B
    *   (3, 1): G $\rightarrow$ Y
    *   (3, 2): G $\rightarrow$ Y
    *   (3, 3): G $\rightarrow$ O
    *   (4, 1): G $\rightarrow$ Y
    *   (4, 5): G $\rightarrow$ O
    *   (5, 2): G $\rightarrow$ Y

**Final Grid Construction (6x6):**
Row 0: . . . B . .
Row 1: R R . B . B
Row 2: . R . B . B
Row 3: . Y Y O . .
Row 4: . Y . . . O
Row 5: . . Y . . .

ANSWER:
. . . B . .
R R . B . B
. R . B . B
. Y Y O . .
. Y . . . O
. . Y . . .

### Conceptual Digestion

**1. Identify the CAST:**
*   **Azure (A):** The "inner walls" (dividing lines). They form a cross (horizontal and vertical lines) that splits the grid into four quadrants.
*   **Green (G):** The "fortifications" (the target area). These are the pixels that will be recolored.
*   **Colored Pixels (R, B, Y, O, M, X, etc.):** The "battalion banners." These are located in a small 2x2 cluster (the "gatehouse") and determine the replacement colors for the quadrants.
*   **Black (.):** "Open ground." These remain black.

**2. Identify the AFFORDANCES:**
*   The Azure lines act as boundaries to define four quadrants.
*   The 2x2 banner cluster maps its colors to the four quadrants based on relative position (Top-Left, Top-Right, Bottom-Left, Bottom-Right).
*   Green pixels are "claimed" (changed to the banner color) if they fall within a specific quadrant.
*   Azure lines are removed ("crumble") in the final output.

**3. Identify the TRANSFORMATION RULE:**
1.  Locate the Azure (A) cross. It divides the grid into four quadrants.
2.  Locate the 2x2 banner cluster.
3.  Map the banner colors to the quadrants:
    *   Top-Left Banner $\rightarrow$ Top-Left Quadrant
    *   Top-Right Banner $\rightarrow$ Top-Right Quadrant
    *   Bottom-Left Banner $\rightarrow$ Bottom-Left Quadrant
    *   Bottom-Right Banner $\rightarrow$ Bottom-Right Quadrant
4.  For every Green (G) pixel in the input:
    *   If it is in the Top-Left quadrant, change it to the Top-Left banner color.
    *   If it is in the Top-Right quadrant, change it to the Top-Right banner color.
    *   If it is in the Bottom-Left quadrant, change it to the Bottom-Left banner color.
    *   If it is in the Bottom-Right quadrant, change it to the Bottom-Right banner color.
5.  Remove all Azure (A) lines and keep Black (.) as is.
6.  The output grid dimensions are the dimensions of the "courtyard" (the area excluding the Azure cross).

### Applying the Rule to Test Input

**Input Analysis:**
*   Azure Cross: Column 6 (index 6) and Row 6 (index 6).
*   Courtyard size: 6x6 (Rows 0-5, Cols 0-5).
*   Banners (at Row 7-8, Col 7-8):
    *   Top-Left (7,7): R (Red)
    *   Top-Right (7,8): B (Blue)
    *   Bottom-Left (8,7): Y (Yellow)
    *   Bottom-Right (8,8): O (Orange)

**Mapping Banners to Quadrants:**
*   Top-Left Quadrant (Rows 0-5, Cols 0-5) $\rightarrow$ Wait, the banners are in a 2x2 block. The rule is: the 2x2 block's internal layout maps to the 4 quadrants of the courtyard.
    *   Banner (0,0) [R] $\rightarrow$ Top-Left Quadrant (Rows 0-2, Cols 0-2)
    *   Banner (0,1) [B] $\rightarrow$ Top-Right Quadrant (Rows 0-2, Cols 3-5)
    *   Banner (1,0) [Y] $\rightarrow$ Bottom-Left Quadrant (Rows 3-5, Cols 0-2)
    *   Banner (1,1) [O] $\rightarrow$ Bottom-Right Quadrant (Rows 3-5, Cols 3-5)

**Processing Green Pixels (G):**
*   (0,3): Top-Right $\rightarrow$ B
*   (1,0): Top-Left $\rightarrow$ R
*   (1,1): Top-Left $\rightarrow$ R
*   (1,3): Top-Right $\rightarrow$ B
*   (1,5): Top-Right $\rightarrow$ B
*   (2,1): Top-Left $\rightarrow$ R
*   (2,3): Top-Right $\rightarrow$ B
*   (2,5): Top-Right $\rightarrow$ B
*   (3,1): Bottom-Left $\rightarrow$ Y
*   (3,2): Bottom-Left $\rightarrow$ Y
*   (3,3): Bottom-Right $\rightarrow$ O
*   (4,1): Bottom-Left $\rightarrow$ Y
*   (4,5): Bottom-Right $\rightarrow$ O
*   (5,2): Bottom-Left $\rightarrow$ Y

**Constructing the 6x6 Output:**
Row 0: . . . B . .
Row 1: R R . B . B
Row 2: . R . B . B
Row 3: . Y Y O . .
Row 4: . Y . . . O
Row 5: . . Y . . .

ANSWER:
. . . B . .
R R . B . B
. R . B . B
. Y Y O . .
. Y . . . O
. . Y . . .

### Conceptual Digestion

**1. Identify the CAST:**
*   **Azure (A):** The "figures" or "souls" on the stage.
*   **Black (.):** The "darkness" or "empty ground."
*   **Red (R):** The result of two azure figures mirroring each other (recognition).
*   **Grey (X):** The result of an azure figure mirroring an empty space (loneliness).

**2. Identify the AFFORDANCES:**
*   The grid is $6 \times 6$.
*   The "lake" is a horizontal mirror line running through the middle.
*   A cell at $(row, col)$ mirrors the cell at $(5-row, col)$.
*   Only Azure (A) cells are transformed. Black (.) cells remain Black.

**3. Identify the TRANSFORMATION RULE:**
For every cell $(r, c)$ in the grid:
*   If the cell is Black (.), it remains Black (.).
*   If the cell is Azure (A):
    *   Check the mirrored cell $(5-r, c)$.
    *   If the mirrored cell is also Azure (A), the current cell becomes Red (R).
    *   If the mirrored cell is Black (.), the current cell becomes Grey (X).

---

### Applying the Rule to Test Input

**Test Input:**
Row 0: `. . . A . A`
Row 1: `A A A . A A`
Row 2: `A A A A . A`
Row 3: `A . . . A A`
Row 4: `. A . . . A`
Row 5: `A A A . A A`

**Step-by-step Mirroring:**
*   **Col 0:**
    *   (0,0) is `.` $\rightarrow$ `.`
    *   (1,0) is `A`, (4,0) is `.` $\rightarrow$ `X`
    *   (2,0) is `A`, (3,0) is `A` $\rightarrow$ `R`
    *   (3,0) is `A`, (2,0) is `A` $\rightarrow$ `R`
    *   (4,0) is `.` $\rightarrow$ `.`
    *   (5,0) is `A`, (0,0) is `.` $\rightarrow$ `X`

*   **Col 1:**
    *   (0,1) is `.` $\rightarrow$ `.`
    *   (1,1) is `A`, (4,1) is `A` $\rightarrow$ `R`
    *   (2,1) is `A`, (3,1) is `.` $\rightarrow$ `X`
    *   (3,1) is `.` $\rightarrow$ `.`
    *   (4,1) is `A`, (1,1) is `A` $\rightarrow$ `R`
    *   (5,1) is `A`, (0,1) is `.` $\rightarrow$ `X`

*   **Col 2:**
    *   (0,2) is `.` $\rightarrow$ `.`
    *   (1,2) is `A`, (4,2) is `.` $\rightarrow$ `X`
    *   (2,2) is `A`, (3,2) is `.` $\rightarrow$ `X`
    *   (3,2) is `.` $\rightarrow$ `.`
    *   (4,2) is `.` $\rightarrow$ `.`
    *   (5,2) is `A`, (0,2) is `.` $\rightarrow$ `X`

*   **Col 3:**
    *   (0,3) is `A`, (5,3) is `.` $\rightarrow$ `X`
    *   (1,3) is `.` $\rightarrow$ `.`
    *   (2,3) is `A`, (3,3) is `.` $\rightarrow$ `X`
    *   (3,3) is `.` $\rightarrow$ `.`
    *   (4,3) is `.` $\rightarrow$ `.`
    *   (5,3) is `.` $\rightarrow$ `.`

*   **Col 4:**
    *   (0,4) is `.` $\rightarrow$ `.`
    *   (1,4) is `A`, (4,4) is `.` $\rightarrow$ `X`
    *   (2,4) is `.` $\rightarrow$ `.`
    *   (3,4) is `A`, (2,4) is `.` $\rightarrow$ `X`
    *   (4,4) is `.` $\rightarrow$ `.`
    *   (5,4) is `A`, (0,4) is `.` $\rightarrow$ `X`

*   **Col 5:**
    *   (0,5) is `A`, (5,5) is `A` $\rightarrow$ `R`
    *   (1,5) is `A`, (4,5) is `A` $\rightarrow$ `R`
    *   (2,5) is `A`, (3,5) is `A` $\rightarrow$ `R`
    *   (3,5) is `A`, (2,5) is `A` $\rightarrow$ `R`
    *   (4,5) is `A`, (1,5) is `A` $\rightarrow$ `R`
    *   (5,5) is `A`, (0,5) is `A` $\rightarrow$ `R`

**Final Grid Construction:**
Row 0: `. . . X . R`
Row 1: `X R X . X R`
Row 2: `R X X X . R`
Row 3: `R . . . X R`
Row 4: `. R . . . R`
Row 5: `X X X . X R`

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

### Conceptual Digestion

**1. Identify the CAST:**
*   **Stars (A - Azure):** The active elements being evaluated.
*   **Void (. - Black):** The background; remains unchanged.
*   **Red (R):** The result of a "paired" star.
*   **Grey (X):** The result of a "lone" star.

**2. Identify the AFFORDANCES:**
*   The grid is 6x6.
*   The "galactic midplane" is the horizontal center. This means we compare cells in the top half (rows 0, 1, 2) with their mirrored counterparts in the bottom half (rows 5, 4, 3).
*   Specifically, cell $(r, c)$ is paired with cell $(5-r, c)$.

**3. Identify the TRANSFORMATION RULE:**
*   If a cell $(r, c)$ is Void (`.`), it remains Void (`.`).
*   If a cell $(r, c)$ is a Star (`A`):
    *   Check the mirrored cell $(5-r, c)$.
    *   If the mirrored cell is also a Star (`A`), both cells become Red (`R`).
    *   If the mirrored cell is Void (`.`), the cell becomes Grey (`X`).

---

### Applying the Rule to Test Input

**Input Grid:**
Row 0: `. . . A . A`
Row 1: `A A A . A A`
Row 2: `A A A A . A`
Row 3: `A . . . A A`
Row 4: `. A . . . A`
Row 5: `A A A . A A`

**Step-by-step Mirror Comparison:**

*   **Col 0:**
    *   (0,0) `.` vs (5,0) `A` $\rightarrow$ (0,0) stays `.`, (5,0) becomes `X`
    *   (1,0) `A` vs (4,0) `.` $\rightarrow$ (1,0) becomes `X`, (4,0) stays `.`
    *   (2,0) `A` vs (3,0) `A` $\rightarrow$ (2,0) becomes `R`, (3,0) becomes `R`

*   **Col 1:**
    *   (0,1) `.` vs (5,1) `A` $\rightarrow$ (0,1) stays `.`, (5,1) becomes `X`
    *   (1,1) `A` vs (4,1) `A` $\rightarrow$ (1,1) becomes `R`, (4,1) becomes `R`
    *   (2,1) `A` vs (3,1) `.` $\rightarrow$ (2,1) becomes `X`, (3,1) stays `.`

*   **Col 2:**
    *   (0,2) `.` vs (5,2) `A` $\rightarrow$ (0,2) stays `.`, (5,2) becomes `X`
    *   (1,2) `A` vs (4,2) `.` $\rightarrow$ (1,2) becomes `X`, (4,2) stays `.`
    *   (2,2) `A` vs (3,2) `.` $\rightarrow$ (2,2) becomes `X`, (3,2) stays `.`

*   **Col 3:**
    *   (0,3) `A` vs (5,3) `.` $\rightarrow$ (0,3) becomes `X`, (5,3) stays `.`
    *   (1,3) `.` vs (4,3) `.` $\rightarrow$ (1,3) stays `.`, (4,3) stays `.`
    *   (2,3) `A` vs (3,3) `.` $\rightarrow$ (2,3) becomes `X`, (3,3) stays `.`

*   **Col 4:**
    *   (0,4) `.` vs (5,4) `A` $\rightarrow$ (0,4) stays `.`, (5,4) becomes `X`
    *   (1,4) `A` vs (4,4) `.` $\rightarrow$ (1,4) becomes `X`, (4,4) stays `.`
    *   (2,4) `.` vs (3,4) `A` $\rightarrow$ (2,4) stays `.`, (3,4) becomes `X`

*   **Col 5:**
    *   (0,5) `A` vs (5,5) `A` $\rightarrow$ (0,5) becomes `R`, (5,5) becomes `R`
    *   (1,5) `A` vs (4,5) `A` $\rightarrow$ (1,5) becomes `R`, (4,5) becomes `R`
    *   (2,5) `A` vs (3,5) `A` $\rightarrow$ (2,5) becomes `R`, (3,5) becomes `R`

**Constructing the Output Grid:**
Row 0: `. . . X . R`
Row 1: `X R X . X R`
Row 2: `R X X X . R`
Row 3: `R . . . X R`
Row 4: `. R . . . R`
Row 5: `X X X . X R`

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

### Conceptual Digestion

**1. Identify the CAST:**
*   **Azure (A):** Solder pads. These are the active elements that interact during the fold.
*   **Black (.):** Bare substrate. This is the background/empty space.
*   **Red (R):** The result of two Azure pads aligning perfectly after the fold.
*   **Grey (X):** The result of an Azure pad landing on a Black space (or vice versa) during the fold.

**2. Identify the AFFORDANCES:**
*   The grid is 6x6. The "fold axis" is the horizontal midline (between row 3 and row 4).
*   Row 1 maps to Row 6.
*   Row 2 maps to Row 5.
*   Row 3 maps to Row 4.
*   Columns remain the same.

**3. Identify the TRANSFORMATION RULE:**
For each cell $(r, c)$ in the grid:
*   Let $Cell_{top}$ be the cell at $(r, c)$ and $Cell_{bottom}$ be the cell at $(6-r+1, c)$.
*   If both $Cell_{top}$ and $Cell_{bottom}$ are Azure (A), both become Red (R).
*   If one is Azure (A) and the other is Black (.), both become Grey (X).
*   If both are Black (.), both remain Black (.).

---

### Applying the Rule to Test Input

**Input Grid:**
Row 1: `. . . A . A`
Row 2: `A A A . A A`
Row 3: `A A A A . A`
Row 4: `A . . . A A`
Row 5: `. A . . . A`
Row 6: `A A A . A A`

**Step-by-step Mapping:**

*   **Row 1 vs Row 6:**
    *   Col 1: `.` vs `A` $\rightarrow$ **X**
    *   Col 2: `.` vs `A` $\rightarrow$ **X**
    *   Col 3: `.` vs `A` $\rightarrow$ **X**
    *   Col 4: `A` vs `.` $\rightarrow$ **X**
    *   Col 5: `.` vs `A` $\rightarrow$ **X**
    *   Col 6: `A` vs `A` $\rightarrow$ **R**
    *   *Result Row 1: X X X X X R*
    *   *Result Row 6: X X X X X R* (Wait, let's re-verify the logic from Example 1).

*Re-evaluating Example 1 logic:*
Input R1: `. A . A A A` | R6: `A A . . . A`
Output R1: `. R . X X R` | R6: `X R . . . R`
Wait, the example output is different. Let's look closer.
Example 1 R1C1: `.` and R6C1: `A` $\rightarrow$ Output R1C1: `.`, R6C1: `X`.
Example 1 R1C2: `A` and R6C2: `A` $\rightarrow$ Output R1C2: `R`, R6C2: `R`.
Example 1 R1C4: `A` and R6C4: `.` $\rightarrow$ Output R1C4: `X`, R6C4: `.`.

**Corrected Rule:**
*   If both are `A`: Both become `R`.
*   If top is `A` and bottom is `.`: Top becomes `X`, bottom stays `.`.
*   If top is `.` and bottom is `A`: Top stays `.`, bottom becomes `X`.
*   If both are `.`: Both stay `.`.

**Applying Corrected Rule to Test Input:**

*   **Row 1 vs Row 6:**
    *   C1: `.` vs `A` $\rightarrow$ R1: `.`, R6: `X`
    *   C2: `.` vs `A` $\rightarrow$ R1: `.`, R6: `X`
    *   C3: `.` vs `A` $\rightarrow$ R1: `.`, R6: `X`
    *   C4: `A` vs `.` $\rightarrow$ R1: `X`, R6: `.`
    *   C5: `.` vs `A` $\rightarrow$ R1: `.`, R6: `X`
    *   C6: `A` vs `A` $\rightarrow$ R1: `R`, R6: `R`

*   **Row 2 vs Row 5:**
    *   C1: `A` vs `.` $\rightarrow$ R2: `X`, R5: `.`
    *   C2: `A` vs `A` $\rightarrow$ R2: `R`, R5: `R`
    *   C3: `A` vs `.` $\rightarrow$ R2: `X`, R5: `.`
    *   C4: `.` vs `.` $\rightarrow$ R2: `.`, R5: `.`
    *   C5: `A` vs `.` $\rightarrow$ R2: `X`, R5: `.`
    *   C6: `A` vs `A` $\rightarrow$ R2: `R`, R5: `R`

*   **Row 3 vs Row 4:**
    *   C1: `A` vs `A` $\rightarrow$ R3: `R`, R4: `R`
    *   C2: `A` vs `.` $\rightarrow$ R3: `X`, R4: `.`
    *   C3: `A` vs `.` $\rightarrow$ R3: `X`, R4: `.`
    *   C4: `A` vs `.` $\rightarrow$ R3: `X`, R4: `.`
    *   C5: `.` vs `A` $\rightarrow$ R3: `.`, R4: `X`
    *   C6: `A` vs `A` $\rightarrow$ R3: `R`, R4: `R`

**Final Grid Assembly:**
R1: `. . . X . R`
R2: `X R X . X R`
R3: `R X X X . R`
R4: `R . . . X R`
R5: `. R . . . R`
R6: `X X X . X R`

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

### Conceptual Digestion

**1. Identify the CAST:**
*   **Azure (A):** Mineral deposits.
*   **Black (.):** Hollow caverns.
*   **Red (R):** Crystallized deposits (result of A meeting A).
*   **Grey (X):** Altered deposits (result of A meeting .).

**2. Identify the AFFORDANCES:**
*   The grid is a 6x6 cross-section.
*   The "fault line" is the horizontal midline (between row 3 and row 4).
*   The transformation involves "folding" the strata. This means the top half (rows 1-3) is reflected/folded onto the bottom half (rows 4-6), or vice versa. Looking at the training example, the output is a 6x6 grid, implying the fold creates a superposition of the top and bottom halves.
*   Specifically, row 1 interacts with row 6, row 2 with row 5, and row 3 with row 4.

**3. Identify the TRANSFORMATION RULE:**
For each cell $(r, c)$ in the top half (rows 1-3) and its corresponding mirrored cell $(r', c)$ in the bottom half (where $r' = 7-r$):
*   If both cells are Azure (A + A) $\rightarrow$ Result is Red (R).
*   If one is Azure (A) and one is Black (.) $\rightarrow$ Result is Grey (X).
*   If both are Black (. + .) $\rightarrow$ Result is Black (.).
*   The output grid maintains the 6x6 shape. The top half and bottom half of the output grid are identical because they represent the result of the "press" (the superposition).

**Wait, let's re-verify with Example 1:**
Input Row 1: `. A . A A A` | Row 6: `A A . . . A`
Result Row 1: `. R . X X R` (Wait, let's check: .+A=X, A+A=R, .+ .=., A+.=X, A+.=X, A+A=R. Result: `X R . X X R`. The example says `. R . X X R`. Let me re-examine).

Example 1 again:
R1: `. A . A A A`
R6: `A A . . . A`
Out R1: `. R . X X R`
Out R6: `X R . . . R`
Wait, the output is not identical top and bottom. Let's look closer.
Out R1: `. R . X X R`
Out R6: `X R . . . R`
Looking at R1 and R6:
Col 1: . and A -> Out R1: ., Out R6: X.
Col 2: A and A -> Out R1: R, Out R6: R.
Col 3: . and . -> Out R1: ., Out R6: ..
Col 4: A and . -> Out R1: X, Out R6: ..
Col 5: A and . -> Out R1: X, Out R6: ..
Col 6: A and A -> Out R1: R, Out R6: R.

Correct Rule:
If both are A $\rightarrow$ both become R.
If one is A and one is . $\rightarrow$ the A becomes X, the . remains ..
If both are . $\rightarrow$ both remain ..

Let's test this on Example 1, Row 2 and 5:
R2: `A A A A A .`
R5: `A A . A A .`
Col 1: A, A -> R, R
Col 2: A, A -> R, R
Col 3: A, . -> X, .
Col 4: A, A -> R, R
Col 5: A, A -> R, R
Col 6: ., . -> ., .
Out R2: `R R X R R .`
Out R5: `R R . R R .`
Matches Example 1!

Let's test Row 3 and 4:
R3: `A . A . A .`
R4: `. A A A . A`
Col 1: A, . -> X, .
Col 2: ., A -> ., X
Col 3: A, A -> R, R
Col 4: ., A -> ., X
Col 5: A, . -> X, .
Col 6: ., A -> ., X
Out R3: `X . R . X .`
Out R4: `. X R X . X`
Matches Example 1!

### Applying to Test Input:
Input:
R1: `. . . A . A`
R2: `A A A . A A`
R3: `A A A A . A`
R4: `A . . . A A`
R5: `. A . . . A`
R6: `A A A . A A`

**Pair R1 & R6:**
C1: . & A $\rightarrow$ . & X
C2: . & A $\rightarrow$ . & X
C3: . & A $\rightarrow$ . & X
C4: A & . $\rightarrow$ X & .
C5: . & A $\rightarrow$ . & X
C6: A & A $\rightarrow$ R & R
R1: `. . . X . R`
R6: `X X X . X R`

**Pair R2 & R5:**
C1: A & . $\rightarrow$ X & .
C2: A & A $\rightarrow$ R & R
C3: A & . $\rightarrow$ X & .
C4: . & . $\rightarrow$ . & .
C5: A & . $\rightarrow$ X & .
C6: A & A $\rightarrow$ R & R
R2: `X R X . X R`
R5: `. R . . . R`

**Pair R3 & R4:**
C1: A & A $\rightarrow$ R & R
C2: A & . $\rightarrow$ X & .
C3: A & . $\rightarrow$ X & .
C4: A & . $\rightarrow$ X & .
C5: . & A $\rightarrow$ . & X
C6: A & A $\rightarrow$ R & R
R3: `R X X X . R`
R4: `R . . . X R`

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

### Conceptual Digestion

**1. Identify the CAST:**
*   **Azure (A):** "Batter" - The active element that interacts during the fold.
*   **Black (.):** "Bare spots" - The passive background.
*   **Red (R):** "Caramelized batter" - The result of Azure overlapping Azure.
*   **Grey (X):** "Cooled batter" - The result of Azure overlapping Black.

**2. Identify the AFFORDANCES:**
*   The grid is 6x6.
*   The transformation is a **horizontal fold** along the midline (Row 0 maps to Row 5, Row 1 to Row 4, Row 2 to Row 3).
*   The interaction happens between the cell at `(row, col)` and its mirrored cell at `(5-row, col)`.

**3. Identify the TRANSFORMATION RULE:**
For each cell `(r, c)` in the grid:
*   If `Input[r, c] == '.'` AND `Input[5-r, c] == '.'`, the result is `.` (Bare spot stays bare).
*   If `Input[r, c] == 'A'` AND `Input[5-r, c] == 'A'`, the result is `R` (Batter overlaps batter).
*   If `Input[r, c] == 'A'` AND `Input[5-r, c] == '.'`, the result is `X` (Batter folds onto bare spot).
*   If `Input[r, c] == '.'` AND `Input[5-r, c] == 'A'`, the result is `X` (Bare spot receives batter).

*Wait, let's double-check the logic against Example 1:*
Input (0,0)=`.`, (5,0)=`A` -> Output (0,0)=`.`? No, Example 1 Output (0,0) is `.`.
Let's re-read: "Where batter overlaps batter, it caramelizes red. Where batter folds onto a bare spot, it cools grey. Bare spots stay bare."
This implies the "top" half is folded *onto* the "bottom" half, or vice versa, but the result is applied to both? No, looking at Example 1:
Input (0,0)=`.`, (5,0)=`A`. Output (0,0)=`.`, (5,0)=`X`.
Input (0,1)=`A`, (5,1)=`A`. Output (0,1)=`R`, (5,1)=`R`.
Input (0,2)=`.`, (5,2)=`.`. Output (0,2)=`.`, (5,2)=`.`.
Input (0,3)=`A`, (5,3)=`.`. Output (0,3)=`X`, (5,3)=`.`.
Wait, let's look closer at Example 1 again.
Input:
Row 0: . A . A A A
Row 5: A A . . . A
Output:
Row 0: . R . X X R
Row 5: X R . . . R

Analysis of Example 1 Col 0: In(0,0)=`.`, In(5,0)=`A` -> Out(0,0)=`.`, Out(5,0)=`X`.
Analysis of Example 1 Col 1: In(0,1)=`A`, In(5,1)=`A` -> Out(0,1)=`R`, Out(5,1)=`R`.
Analysis of Example 1 Col 3: In(0,3)=`A`, In(5,3)=`.` -> Out(0,3)=`X`, Out(5,3)=`.`.
Analysis of Example 1 Col 5: In(0,5)=`A`, In(5,5)=`A` -> Out(0,5)=`R`, Out(5,5)=`R`.

**Revised Rule:**
The "fold" happens from top to bottom.
1. If `Input[r, c] == 'A'` and `Input[5-r, c] == 'A'`, both become `R`.
2. If `Input[r, c] == 'A'` and `Input[5-r, c] == '.'`, `Input[r, c]` becomes `X` and `Input[5-r, c]` stays `.`.
3. If `Input[r, c] == '.'` and `Input[5-r, c] == 'A'`, `Input[r, c]` stays `.` and `Input[5-r, c]` becomes `X`.
4. If both are `.`, both stay `.`.

Essentially:
- `A` + `A` = `R`
- `A` (top) + `.` (bottom) = `X` (top) + `.` (bottom)
- `.` (top) + `A` (bottom) = `.` (top) + `X` (bottom)
- `.` + `.` = `.`

### Applying Rule to Test Input

Input:
Row 0: . . . A . A
Row 1: A A A . A A
Row 2: A A A A . A
Row 3: A . . . A A
Row 4: . A . . . A
Row 5: A A A . A A

**Column 0:**
(0,0)=`.`, (5,0)=`A` -> Out(0,0)=`.`, Out(5,0)=`X`
(1,0)=`A`, (4,0)=`.` -> Out(1,0)=`X`, Out(4,0)=`.`
(2,0)=`A`, (3,0)=`A` -> Out(2,0)=`R`, Out(3,0)=`R`

**Column 1:**
(0,1)=`.`, (5,1)=`A` -> Out(0,1)=`.`, Out(5,1)=`X`
(1,1)=`A`, (4,1)=`A` -> Out(1,1)=`R`, Out(4,1)=`R`
(2,1)=`A`, (3,1)=`.` -> Out(2,1)=`X`, Out(3,1)=`.`

**Column 2:**
(0,2)=`.`, (5,2)=`A` -> Out(0,2)=`.`, Out(5,2)=`X`
(1,2)=`A`, (4,2)=`.` -> Out(1,2)=`X`, Out(4,2)=`.`
(2,2)=`A`, (3,2)=`.` -> Out(2,2)=`X`, Out(3,2)=`.`

**Column 3:**
(0,3)=`A`, (5,3)=`.` -> Out(0,3)=`X`, Out(5,3)=`.`
(1,3)=`.`, (4,3)=`.` -> Out(1,3)=`.`, Out(4,3)=`.`
(2,3)=`A`, (3,3)=`.` -> Out(2,3)=`X`, Out(3,3)=`.`

**Column 4:**
(0,4)=`.`, (5,4)=`A` -> Out(0,4)=`.`, Out(5,4)=`X`
(1,4)=`A`, (4,4)=`.` -> Out(1,4)=`X`, Out(4,4)=`.`
(2,4)=`.`, (3,4)=`A` -> Out(2,4)=`.`, Out(3,4)=`X`

**Column 5:**
(0,5)=`A`, (5,5)=`A` -> Out(0,5)=`R`, Out(5,5)=`R`
(1,5)=`A`, (4,5)=`A` -> Out(1,5)=`R`, Out(4,5)=`R`
(2,5)=`A`, (3,5)=`A` -> Out(2,5)=`R`, Out(3,5)=`R`

**Constructing the Grid:**
Row 0: . . . X . R
Row 1: X R X . X R
Row 2: R X X X . R
Row 3: R . . . X R
Row 4: . R . . . R
Row 5: X X X . X R

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

### Conceptual Digestion

**1. Identify the CAST:**
*   **Azure (A):** Represents "Walls" in the architectural plan.
*   **Black (.):** Represents "Open Space" (rooms/corridors).
*   **Red (R):** Represents "Aligned Walls" (where walls overlap after folding).
*   **Grey (X):** Represents "Unmatched Walls" (where a wall exists on one side of the fold but not the other).

**2. Identify the AFFORDANCES:**
*   The grid is 6x6.
*   The "fold" is the horizontal midline. This means Row 0 maps to Row 5, Row 1 to Row 4, and Row 2 to Row 3.
*   The transformation is a comparison between the top half and the bottom half of the grid.

**3. Identify the TRANSFORMATION RULE:**
For each cell $(r, c)$ in the grid:
*   Let $r'$ be the mirrored row index: $r' = 5 - r$.
*   If the cell is open space (`.`), it remains open space (`.`).
*   If the cell is a wall (`A`):
    *   Check the mirrored cell $(r', c)$.
    *   If the mirrored cell is also a wall (`A`), the cell becomes Red (`R`).
    *   If the mirrored cell is open space (`.`), the cell becomes Grey (`X`).

---

### Applying the Rule to Test Input

**Test Input:**
Row 0: `. . . A . A`
Row 1: `A A A . A A`
Row 2: `A A A A . A`
Row 3: `A . . . A A`
Row 4: `. A . . . A`
Row 5: `A A A . A A`

**Step-by-step Mapping:**

*   **Row 0 vs Row 5:**
    *   (0,0) `.` vs (5,0) `A` $\rightarrow$ `.`
    *   (0,1) `.` vs (5,1) `A` $\rightarrow$ `.`
    *   (0,2) `.` vs (5,2) `A` $\rightarrow$ `.`
    *   (0,3) `A` vs (5,3) `.` $\rightarrow$ `X`
    *   (0,4) `.` vs (5,4) `A` $\rightarrow$ `.`
    *   (0,5) `A` vs (5,5) `A` $\rightarrow$ `R`
    *   *Row 0 Result: `. . . X . R`*

*   **Row 1 vs Row 4:**
    *   (1,0) `A` vs (4,0) `.` $\rightarrow$ `X`
    *   (1,1) `A` vs (4,1) `A` $\rightarrow$ `R`
    *   (1,2) `A` vs (4,2) `.` $\rightarrow$ `X`
    *   (1,3) `.` vs (4,3) `.` $\rightarrow$ `.`
    *   (1,4) `A` vs (4,4) `.` $\rightarrow$ `X`
    *   (1,5) `A` vs (4,5) `A` $\rightarrow$ `R`
    *   *Row 1 Result: `X R X . X R`*

*   **Row 2 vs Row 3:**
    *   (2,0) `A` vs (3,0) `A` $\rightarrow$ `R`
    *   (2,1) `A` vs (3,1) `.` $\rightarrow$ `X`
    *   (2,2) `A` vs (3,2) `.` $\rightarrow$ `X`
    *   (2,3) `A` vs (3,3) `.` $\rightarrow$ `X`
    *   (2,4) `.` vs (3,4) `A` $\rightarrow$ `.`
    *   (2,5) `A` vs (3,5) `A` $\rightarrow$ `R`
    *   *Row 2 Result: `R X X X . R`*

*   **Row 3 vs Row 2:**
    *   (3,0) `A` vs (2,0) `A` $\rightarrow$ `R`
    *   (3,1) `.` vs (2,1) `A` $\rightarrow$ `.`
    *   (3,2) `.` vs (2,2) `A` $\rightarrow$ `.`
    *   (3,3) `.` vs (2,3) `A` $\rightarrow$ `.`
    *   (3,4) `A` vs (2,4) `.` $\rightarrow$ `X`
    *   (3,5) `A` vs (2,5) `A` $\rightarrow$ `R`
    *   *Row 3 Result: `R . . . X R`*

*   **Row 4 vs Row 1:**
    *   (4,0) `.` vs (1,0) `A` $\rightarrow$ `.`
    *   (4,1) `A` vs (1,1) `A` $\rightarrow$ `R`
    *   (4,2) `.` vs (1,2) `A` $\rightarrow$ `.`
    *   (4,3) `.` vs (1,3) `.` $\rightarrow$ `.`
    *   (4,4) `.` vs (1,4) `A` $\rightarrow$ `.`
    *   (4,5) `A` vs (1,5) `A` $\rightarrow$ `R`
    *   *Row 4 Result: `. R . . . R`*

*   **Row 5 vs Row 0:**
    *   (5,0) `A` vs (0,0) `.` $\rightarrow$ `X`
    *   (5,1) `A` vs (0,1) `.` $\rightarrow$ `X`
    *   (5,2) `A` vs (0,2) `.` $\rightarrow$ `X`
    *   (5,3) `.` vs (0,3) `A` $\rightarrow$ `.`
    *   (5,4) `A` vs (0,4) `.` $\rightarrow$ `X`
    *   (5,5) `A` vs (0,5) `A` $\rightarrow$ `R`
    *   *Row 5 Result: `X X X . X R`*

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

### Conceptual Digestion

1.  **CAST**:
    *   **Azure (A)**: "Seeds". These are the active elements that change based on the fold.
    *   **Black (.)**: "Bare soil". This is the background/passive element.
    *   **Red (R)**: "Bloomed seeds". Result of Azure meeting Azure.
    *   **Grey (X)**: "Withered seeds". Result of Azure meeting Black.

2.  **AFFORDANCES**:
    *   The grid is 6x6.
    *   The "fold" is along the horizontal center. This means row 0 maps to row 5, row 1 to row 4, and row 2 to row 3.
    *   The transformation is symmetric: the result at `(row, col)` depends on the interaction between `(row, col)` and `(5-row, col)`.

3.  **TRANSFORMATION RULE**:
    For each cell `(r, c)` in the 6x6 grid:
    *   If `Input[r, c] == A` AND `Input[5-r, c] == A`, then `Output[r, c] = R` (Red).
    *   If `Input[r, c] == A` AND `Input[5-r, c] == .`, then `Output[r, c] = X` (Grey).
    *   If `Input[r, c] == .` AND `Input[5-r, c] == A`, then `Output[r, c] = .` (Black) - *Wait, let's re-verify with Example 1.*

    *Re-verifying Example 1:*
    Input Row 0: `. A . A A A`
    Input Row 5: `A A . . . A`
    Output Row 0: `. R . X X R`
    - (0,0): . and A -> . (Correct)
    - (0,1): A and A -> R (Correct)
    - (0,2): . and . -> . (Correct)
    - (0,3): A and . -> X (Correct)
    - (0,4): A and . -> X (Correct)
    - (0,5): A and A -> R (Correct)

    *Rule Refined:*
    - If `Input[r, c] == A` and `Input[5-r, c] == A` $\rightarrow$ `R`
    - If `Input[r, c] == A` and `Input[5-r, c] == .` $\rightarrow$ `X`
    - If `Input[r, c] == .` $\rightarrow$ `.` (Regardless of the other side)

### Applying Rule to Test Input

Input:
Row 0: `. . . A . A`
Row 1: `A A A . A A`
Row 2: `A A A A . A`
Row 3: `A . . . A A`
Row 4: `. A . . . A`
Row 5: `A A A . A A`

**Row 0 (vs Row 5):**
- (0,0): . vs A $\rightarrow$ .
- (0,1): . vs A $\rightarrow$ .
- (0,2): . vs A $\rightarrow$ .
- (0,3): A vs . $\rightarrow$ X
- (0,4): . vs A $\rightarrow$ .
- (0,5): A vs A $\rightarrow$ R
Result: `. . . X . R`

**Row 1 (vs Row 4):**
- (1,0): A vs . $\rightarrow$ X
- (1,1): A vs A $\rightarrow$ R
- (1,2): A vs . $\rightarrow$ X
- (1,3): . vs . $\rightarrow$ .
- (1,4): A vs . $\rightarrow$ X
- (1,5): A vs A $\rightarrow$ R
Result: `X R X . X R`

**Row 2 (vs Row 3):**
- (2,0): A vs A $\rightarrow$ R
- (2,1): A vs . $\rightarrow$ X
- (2,2): A vs . $\rightarrow$ X
- (2,3): A vs . $\rightarrow$ X
- (2,4): . vs A $\rightarrow$ .
- (2,5): A vs A $\rightarrow$ R
Result: `R X X X . R`

**Row 3 (vs Row 2):**
- (3,0): A vs A $\rightarrow$ R
- (3,1): . vs A $\rightarrow$ .
- (3,2): . vs A $\rightarrow$ .
- (3,3): . vs A $\rightarrow$ .
- (3,4): A vs . $\rightarrow$ X
- (3,5): A vs A $\rightarrow$ R
Result: `R . . . X R`

**Row 4 (vs Row 1):**
- (4,0): . vs A $\rightarrow$ .
- (4,1): A vs A $\rightarrow$ R
- (4,2): . vs A $\rightarrow$ .
- (4,3): . vs . $\rightarrow$ .
- (4,4): . vs A $\rightarrow$ .
- (4,5): A vs A $\rightarrow$ R
Result: `. R . . . R`

**Row 5 (vs Row 0):**
- (5,0): A vs . $\rightarrow$ X
- (5,1): A vs . $\rightarrow$ X
- (5,2): A vs . $\rightarrow$ X
- (5,3): . vs A $\rightarrow$ .
- (5,4): A vs . $\rightarrow$ X
- (5,5): A vs A $\rightarrow$ R
Result: `X X X . X R`

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

### Conceptual Digestion

1.  **CAST**:
    *   **The Lake**: The background grid of orange (`O`) cells.
    *   **The Pebble**: A single cell of a different color (the "seed") located within the orange lake.
    *   **The Waves**: Concentric diamond-shaped rings radiating from the pebble.

2.  **AFFORDANCES**:
    *   The position of the pebble $(r, c)$ serves as the center of the transformation.
    *   The color of the pebble determines the starting color of the sequence.
    *   The distance from the center determines which color in the sequence is applied.

3.  **TRANSFORMATION RULE**:
    *   **Distance Metric**: The "diamond-shaped waves" indicate the use of **Manhattan Distance**: $dist = |r - r_{pebble}| + |c - c_{pebble}|$.
    *   **Color Sequence**: The "prayer wheel of nine hues" refers to a specific cyclic order of the colors provided in the key. Looking at Example 1:
        *   Pebble is Green (`G`) at (0, 6).
        *   At $dist=0$, color is `G`.
        *   At $dist=1$, color is `Y`.
        *   At $dist=2$, color is `.`.
        *   At $dist=3$, color is `X`.
        *   At $dist=4$, color is `R`.
        *   At $dist=5$, color is `A`.
        *   At $dist=6$, color is `W`.
        *   At $dist=7$, color is `M`.
        *   At $dist=8$, color is `B`.
        *   Sequence: `G` $\rightarrow$ `Y` $\rightarrow$ `.` $\rightarrow$ `X` $\rightarrow$ `R` $\rightarrow$ `A` $\rightarrow$ `W` $\rightarrow$ `M` $\rightarrow$ `B` $\rightarrow$ (back to `G`).
    *   **The Master Sequence**: Based on the order provided in the color key (`. B R G Y X M O A W`), but adjusted to match the example's cycle:
        The sequence in Example 1 is: `G` (Green) $\rightarrow$ `Y` (Yellow) $\rightarrow$ `.` (Black) $\rightarrow$ `X` (Grey) $\rightarrow$ `R` (Red) $\rightarrow$ `A` (Azure) $\rightarrow$ `W` (Maroon) $\rightarrow$ `M` (Magenta) $\rightarrow$ `B` (Blue).
        Let's verify the sequence: `G, Y, ., X, R, A, W, M, B`.
        This is a cycle of 9 colors.

### Applying to Test Input

*   **Pebble Position**: (2, 1) - Row 2, Col 1 (0-indexed).
*   **Pebble Color**: `.` (Black).
*   **Sequence starting from `.`**: `.` $\rightarrow$ `X` $\rightarrow$ `R` $\rightarrow$ `A` $\rightarrow$ `W` $\rightarrow$ `M` $\rightarrow$ `B` $\rightarrow$ `G` $\rightarrow$ `Y` $\rightarrow$ (back to `.`).

**Distance Map and Color Assignment**:
*   $dist=0$: `.`
*   $dist=1$: `X`
*   $dist=2$: `R`
*   $dist=3$: `A`
*   $dist=4$: `W`
*   $dist=5$: `M`
*   $dist=6$: `B`
*   $dist=7$: `G`
*   $dist=8$: `Y`
*   $dist=9$: `.`
*   $dist=10$: `X`
*   $dist=11$: `R`
*   $dist=12$: `A`
*   $dist=13$: `W`

**Calculating for each cell (r, c)**:
Row 0:
(0,0): dist=3 $\rightarrow$ A | (0,1): dist=2 $\rightarrow$ R | (0,2): dist=3 $\rightarrow$ A | (0,3): dist=4 $\rightarrow$ W | (0,4): dist=5 $\rightarrow$ M | (0,5): dist=6 $\rightarrow$ B | (0,6): dist=7 $\rightarrow$ G | (0,7): dist=8 $\rightarrow$ Y
Row 1:
(1,0): dist=2 $\rightarrow$ R | (1,1): dist=1 $\rightarrow$ X | (1,2): dist=2 $\rightarrow$ R | (1,3): dist=3 $\rightarrow$ A | (1,4): dist=4 $\rightarrow$ W | (1,5): dist=5 $\rightarrow$ M | (1,6): dist=6 $\rightarrow$ B | (1,7): dist=7 $\rightarrow$ G
Row 2:
(2,0): dist=1 $\rightarrow$ X | (2,1): dist=0 $\rightarrow$ . | (2,2): dist=1 $\rightarrow$ X | (2,3): dist=2 $\rightarrow$ R | (2,4): dist=3 $\rightarrow$ A | (2,5): dist=4 $\rightarrow$ W | (2,6): dist=5 $\rightarrow$ M | (2,7): dist=6 $\rightarrow$ B
Row 3:
(3,0): dist=2 $\rightarrow$ R | (3,1): dist=1 $\rightarrow$ X | (3,2): dist=2 $\rightarrow$ R | (3,3): dist=3 $\rightarrow$ A | (3,4): dist=4 $\rightarrow$ W | (3,5): dist=5 $\rightarrow$ M | (3,6): dist=6 $\rightarrow$ B | (3,7): dist=7 $\rightarrow$ G
Row 4:
(4,0): dist=3 $\rightarrow$ A | (4,1): dist=2 $\rightarrow$ R | (4,2): dist=3 $\rightarrow$ A | (4,3): dist=4 $\rightarrow$ W | (4,4): dist=5 $\rightarrow$ M | (4,5): dist=6 $\rightarrow$ B | (4,6): dist=7 $\rightarrow$ G | (4,7): dist=8 $\rightarrow$ Y
Row 5:
(5,0): dist=4 $\rightarrow$ W | (5,1): dist=3 $\rightarrow$ A | (5,2): dist=4 $\rightarrow$ W | (5,3): dist=5 $\rightarrow$ M | (5,4): dist=6 $\rightarrow$ B | (5,5): dist=7 $\rightarrow$ G | (5,6): dist=8 $\rightarrow$ Y | (5,7): dist=9 $\rightarrow$ .
Row 6:
(6,0): dist=5 $\rightarrow$ M | (6,1): dist=4 $\rightarrow$ W | (6,2): dist=5 $\rightarrow$ M | (6,3): dist=6 $\rightarrow$ B | (6,4): dist=7 $\rightarrow$ G | (6,5): dist=8 $\rightarrow$ Y | (6,6): dist=9 $\rightarrow$ . | (6,7): dist=10 $\rightarrow$ X
Row 7:
(7,0): dist=6 $\rightarrow$ B | (7,1): dist=5 $\rightarrow$ M | (7,2): dist=6 $\rightarrow$ B | (7,3): dist=7 $\rightarrow$ G | (7,4): dist=8 $\rightarrow$ Y | (7,5): dist=9 $\rightarrow$ . | (7,6): dist=10 $\rightarrow$ X | (7,7): dist=11 $\rightarrow$ R

ANSWER:
A R A W M B G Y
R X R A W M B G
X . X R A W M B
R X R A W M B G
A R A W M B G Y
W A W M B G Y .
M W M B G Y . X
B M B G Y . X R

### Conceptual Digestion

1.  **The CAST**:
    *   **Quiescent Tissue**: The background cells (represented by `O` or `.` in the input).
    *   **Aberrant Cell**: The single cell with a distinct color (the "mutant cell").
    *   **Differentiation Spectrum**: A looping sequence of colors that radiate from the mutant cell.

2.  **The AFFORDANCES**:
    *   The mutant cell acts as the origin $(r_0, c_0)$.
    *   The distance from the origin is measured by the **Manhattan distance**: $d = |r - r_0| + |c - c_0|$.
    *   Each distance $d$ corresponds to a specific color in a repeating sequence.

3.  **The TRANSFORMATION RULE**:
    *   **Identify the Origin**: Find the coordinates $(r_0, c_0)$ of the cell that differs from the background.
    *   **Identify the Sequence**: Looking at Example 1, the origin is at $(0, 6)$ with color `G`.
        *   $d=0$: `G`
        *   $d=1$: `Y`
        *   $d=2$: `.` (black)
        *   $d=3$: `X` (grey)
        *   $d=4$: `R` (red)
        *   $d=5$: `A` (azure)
        *   $d=6$: `W` (maroon)
        *   $d=7$: `M` (magenta)
        *   $d=8$: `B` (blue)
        *   $d=9$: `G` (loop back to start)
        *   Sequence: `G` $\rightarrow$ `Y` $\rightarrow$ `.` $\rightarrow$ `X` $\rightarrow$ `R` $\rightarrow$ `A` $\rightarrow$ `W` $\rightarrow$ `M` $\rightarrow$ `B` $\rightarrow$ (repeat)
    *   **Apply to Grid**: For every cell $(r, c)$, calculate $d = |r - r_0| + |c - c_0|$. The color is the $d$-th element of the sequence starting from the mutant cell's color.

### Applying to Test Input

*   **Origin**: The mutant cell is at $(2, 1)$ (row 2, col 1, 0-indexed). Its color is `.` (black).
*   **Sequence**: Since the origin is `.`, the sequence shifts. Based on the established order (`G Y . X R A W M B`), the sequence starting from `.` is:
    $d=0: .$
    $d=1: X$
    $d=2: R$
    $d=3: A$
    $d=4: W$
    $d=5: M$
    $d=6: B$
    $d=7: G$
    $d=8: Y$
    $d=9: .$ (repeat)

*   **Calculating Distances and Colors**:
    Origin $(2, 1)$
    Row 0: (2,1) $\rightarrow$ d=3(A), (2,0) $\rightarrow$ d=3(A) - wait, (0,0) is $|0-2|+|0-1|=3(A)$, (0,1) is $|0-2|+|1-1|=2(R)$, (0,2) is $|0-2|+|2-1|=3(A)$, (0,3) is $|0-2|+|3-1|=4(W)$, (0,4) is $|0-2|+|4-1|=5(M)$, (0,5) is $|0-2|+|5-1|=6(B)$, (0,6) is $|0-2|+|6-1|=7(G)$, (0,7) is $|0-2|+|7-1|=8(Y)$.
    Row 1: (1,0) is $|1-2|+|0-1|=2(R)$, (1,1) is $|1-2|+|1-1|=1(X)$, (1,2) is $|1-2|+|2-1|=2(R)$, (1,3) is $|1-2|+|3-1|=3(A)$, (1,4) is $|1-2|+|4-1|=4(W)$, (1,5) is $|1-2|+|5-1|=5(M)$, (1,6) is $|1-2|+|6-1|=6(B)$, (1,7) is $|1-2|+|7-1|=7(G)$.
    Row 2: (2,0) is $|2-2|+|0-1|=1(X)$, (2,1) is $|2-2|+|1-1|=0(.)$, (2,2) is $|2-2|+|2-1|=1(X)$, (2,3) is $|2-2|+|3-1|=2(R)$, (2,4) is $|2-2|+|4-1|=3(A)$, (2,5) is $|2-2|+|5-1|=4(W)$, (2,6) is $|2-2|+|6-1|=5(M)$, (2,7) is $|2-2|+|7-1|=6(B)$.
    Row 3: (3,0) is $|3-2|+|0-1|=2(R)$, (3,1) is $|3-2|+|1-1|=1(X)$, (3,2) is $|3-2|+|2-1|=2(R)$, (3,3) is $|3-2|+|3-1|=3(A)$, (3,4) is $|3-2|+|4-1|=4(W)$, (3,5) is $|3-2|+|5-1|=5(M)$, (3,6) is $|3-2|+|6-1|=6(B)$, (3,7) is $|3-2|+|7-1|=7(G)$.
    Row 4: (4,0) is $|4-2|+|0-1|=3(A)$, (4,1) is $|4-2|+|1-1|=2(R)$, (4,2) is $|4-2|+|2-1|=3(A)$, (4,3) is $|4-2|+|3-1|=4(W)$, (4,4) is $|4-2|+|4-1|=5(M)$, (4,5) is $|4-2|+|5-1|=6(B)$, (4,6) is $|4-2|+|6-1|=7(G)$, (4,7) is $|4-2|+|7-1|=8(Y)$.
    Row 5: (5,0) is $|5-2|+|0-1|=4(W)$, (5,1) is $|5-2|+|1-1|=3(A)$, (5,2) is $|5-2|+|2-1|=4(W)$, (5,3) is $|5-2|+|3-1|=5(M)$, (5,4) is $|5-2|+|4-1|=6(B)$, (5,5) is $|5-2|+|5-1|=7(G)$, (5,6) is $|5-2|+|6-1|=8(Y)$, (5,7) is $|5-2|+|7-1|=9(.)$.
    Row 6: (6,0) is $|6-2|+|0-1|=5(M)$, (6,1) is $|6-2|+|1-1|=4(W)$, (6,2) is $|6-2|+|2-1|=5(M)$, (6,3) is $|6-2|+|3-1|=6(B)$, (6,4) is $|6-2|+|4-1|=7(G)$, (6,5) is $|6-2|+|5-1|=8(Y)$, (6,6) is $|6-2|+|6-1|=9(.)$, (6,7) is $|6-2|+|7-1|=10(X)$.
    Row 7: (7,0) is $|7-2|+|0-1|=6(B)$, (7,1) is $|7-2|+|1-1|=5(M)$, (7,2) is $|7-2|+|2-1|=6(B)$, (7,3) is $|7-2|+|3-1|=7(G)$, (7,4) is $|7-2|+|4-1|=8(Y)$, (7,5) is $|7-2|+|5-1|=9(.)$, (7,6) is $|7-2|+|6-1|=10(X)$, (7,7) is $|7-2|+|7-1|=11(R)$.

ANSWER:
A R A W M B G Y
R X R A W M B G
X . X R A W M B
R X R A W M B G
A R A W M B G Y
W A W M B G Y .
M W M B G Y . X
B M B G Y . X R

### Conceptual Digestion

1.  **CAST**:
    *   **The Sea**: The 8x8 grid.
    *   **The Buoy**: The single colored cell in the input grid (the only cell that is not 'O').
    *   **The Swells**: Diamond-shaped rings of color radiating from the buoy.

2.  **AFFORDANCES**:
    *   The buoy's position $(r, c)$ serves as the origin $(0, 0)$ for calculating Manhattan distance: $d = |r - r_0| + |c - c_0|$.
    *   The buoy's color determines the starting point in the maritime color cycle.

3.  **TRANSFORMATION RULE**:
    *   **The Color Cycle**: By analyzing the training examples, we can deduce the sequence.
        Example 1: Buoy is Green (G) at (0, 6).
        Distance 0: G
        Distance 1: Y
        Distance 2: . (Black)
        Distance 3: X (Grey)
        Distance 4: R (Red)
        Distance 5: A (Azure)
        Distance 6: W (Maroon)
        Distance 7: M (Magenta)
        Distance 8: B (Blue)
        Cycle: **G $\rightarrow$ Y $\rightarrow$ . $\rightarrow$ X $\rightarrow$ R $\rightarrow$ A $\rightarrow$ W $\rightarrow$ M $\rightarrow$ B $\rightarrow$ (back to G)**.
    *   **The Mapping**: For every cell $(r, c)$ in the 8x8 grid, calculate the Manhattan distance $d$ from the buoy $(r_0, c_0)$. The color of the cell is the color at index $(d \pmod 9)$ in the cycle starting from the buoy's color.

### Applying the Rule to Test Input

*   **Buoy Position**: (2, 1) - Row 2, Col 1 (0-indexed).
*   **Buoy Color**: . (Black).
*   **Cycle starting from .**: . $\rightarrow$ X $\rightarrow$ R $\rightarrow$ A $\rightarrow$ W $\rightarrow$ M $\rightarrow$ B $\rightarrow$ G $\rightarrow$ Y $\rightarrow$ (back to .).

**Distance Map from (2, 1):**
Row 0: |0-2|+|0-1|=3 |0-2|+|1-1|=2 |0-2|+|2-1|=3 |0-2|+|3-1|=4 |0-2|+|4-1|=5 |0-2|+|5-1|=6 |0-2|+|6-1|=7 |0-2|+|7-1|=8
Row 1: |1-2|+|0-1|=2 |1-2|+|1-1|=1 |1-2|+|2-1|=2 |1-2|+|3-1|=3 |1-2|+|4-1|=4 |1-2|+|5-1|=5 |1-2|+|6-1|=6 |1-2|+|7-1|=7
Row 2: |2-2|+|0-1|=1 |2-2|+|1-1|=0 |2-2|+|2-1|=1 |2-2|+|3-1|=2 |2-2|+|4-1|=3 |2-2|+|5-1|=4 |2-2|+|6-1|=5 |2-2|+|7-1|=6
Row 3: |3-2|+|0-1|=2 |3-2|+|1-1|=1 |3-2|+|2-1|=2 |3-2|+|3-1|=3 |3-2|+|4-1|=4 |3-2|+|5-1|=5 |3-2|+|6-1|=6 |3-2|+|7-1|=7
Row 4: |4-2|+|0-1|=3 |4-2|+|1-1|=2 |4-2|+|2-1|=3 |4-2|+|3-1|=4 |4-2|+|4-1|=5 |4-2|+|5-1|=6 |4-2|+|6-1|=7 |4-2|+|7-1|=8
Row 5: |5-2|+|0-1|=4 |5-2|+|1-1|=3 |5-2|+|2-1|=4 |5-2|+|3-1|=5 |5-2|+|4-1|=6 |5-2|+|5-1|=7 |5-2|+|6-1|=8 |5-2|+|7-1|=9(0)
Row 6: |6-2|+|0-1|=5 |6-2|+|1-1|=4 |6-2|+|2-1|=5 |6-2|+|3-1|=6 |6-2|+|4-1|=7 |6-2|+|5-1|=8 |6-2|+|6-1|=9(0) |6-2|+|7-1|=10(1)
Row 7: |7-2|+|0-1|=6 |7-2|+|1-1|=5 |7-2|+|2-1|=6 |7-2|+|3-1|=7 |7-2|+|4-1|=8 |7-2|+|5-1|=9(0) |7-2|+|6-1|=10(1) |7-2|+|7-1|=11(2)

**Color Mapping (Cycle: 0:., 1:X, 2:R, 3:A, 4:W, 5:M, 6:B, 7:G, 8:Y):**
Row 0: A R A W M B G Y
Row 1: R X R A W M B G
Row 2: X . X R A W M B
Row 3: R X R A W M B G
Row 4: A R A W M B G Y
Row 5: W A W M B G Y .
Row 6: M W M B G Y . X
Row 7: B M B G Y . X R

ANSWER:
A R A W M B G Y
R X R A W M B G
X . X R A W M B
R X R A W M B G
A R A W M B G Y
W A W M B G Y .
M W M B G Y . X
B M B G Y . X R

### Conceptual Digestion

1.  **CAST**:
    *   **Dormant Surface**: The background color (Orange 'O' or Black '.').
    *   **Energized Node**: The single cell that differs from the background in the input grid.
    *   **Voltage Ripple**: The resulting pattern of colors radiating from the node.

2.  **AFFORDANCES**:
    *   The energized node acts as the center $(r_0, c_0)$ of a Manhattan distance calculation.
    *   The distance $d = |r - r_0| + |c - c_0|$ determines the color of the cell $(r, c)$.

3.  **TRANSFORMATION RULE**:
    *   **Distance Calculation**: For every cell $(r, c)$ in the 8x8 grid, calculate the Manhattan distance $d$ from the energized node.
    *   **Color Cycle**: The colors follow a specific 9-LED cycle. Looking at Example 1:
        *   Node is at (0, 6), color G.
        *   $d=0$: G (Green)
        *   $d=1$: Y (Yellow)
        *   $d=2$: . (Black)
        *   $d=3$: X (Grey)
        *   $d=4$: R (Red)
        *   $d=5$: A (Azure)
        *   $d=6$: W (Maroon)
        *   $d=7$: M (Magenta)
        *   $d=8$: B (Blue)
        *   Cycle: G $\rightarrow$ Y $\rightarrow$ . $\rightarrow$ X $\rightarrow$ R $\rightarrow$ A $\rightarrow$ W $\rightarrow$ M $\rightarrow$ B $\rightarrow$ (repeat G)
    *   **Mapping**: The color of a cell is the color at index $(d \pmod 9)$ in the sequence starting with the energized node's color.

### Applying to Test Input

*   **Energized Node**: Position (2, 1), Color '.' (Black).
*   **Color Cycle starting from '.'**:
    0: . (Black)
    1: X (Grey)
    2: R (Red)
    3: A (Azure)
    4: W (Maroon)
    5: M (Magenta)
    6: B (Blue)
    7: G (Green)
    8: Y (Yellow)

*   **Calculating Distances from (2, 1)**:
    Row 0: |0-2|+|0-1|=3(A), |0-2|+|1-1|=2(R), |0-2|+|2-1|=3(A), |0-2|+|3-1|=4(W), |0-2|+|4-1|=5(M), |0-2|+|5-1|=6(B), |0-2|+|6-1|=7(G), |0-2|+|7-1|=8(Y)
    Row 1: |1-2|+|0-1|=2(R), |1-2|+|1-1|=1(X), |1-2|+|2-1|=2(R), |1-2|+|3-1|=3(A), |1-2|+|4-1|=4(W), |1-2|+|5-1|=5(M), |1-2|+|6-1|=6(B), |1-2|+|7-1|=7(G)
    Row 2: |2-2|+|0-1|=1(X), |2-2|+|1-1|=0(.), |2-2|+|2-1|=1(X), |2-2|+|3-1|=2(R), |2-2|+|4-1|=3(A), |2-2|+|5-1|=4(W), |2-2|+|6-1|=5(M), |2-2|+|7-1|=6(B)
    Row 3: |3-2|+|0-1|=2(R), |3-2|+|1-1|=1(X), |3-2|+|2-1|=2(R), |3-2|+|3-1|=3(A), |3-2|+|4-1|=4(W), |3-2|+|5-1|=5(M), |3-2|+|6-1|=6(B), |3-2|+|7-1|=7(G)
    Row 4: |4-2|+|0-1|=3(A), |4-2|+|1-1|=2(R), |4-2|+|2-1|=3(A), |4-2|+|3-1|=4(W), |4-2|+|4-1|=5(M), |4-2|+|5-1|=6(B), |4-2|+|6-1|=7(G), |4-2|+|7-1|=8(Y)
    Row 5: |5-2|+|0-1|=4(W), |5-2|+|1-1|=3(A), |5-2|+|2-1|=4(W), |5-2|+|3-1|=5(M), |5-2|+|4-1|=6(B), |5-2|+|5-1|=7(G), |5-2|+|6-1|=8(Y), |5-2|+|7-1|=9$\rightarrow$0(.)
    Row 6: |6-2|+|0-1|=5(M), |6-2|+|1-1|=4(W), |6-2|+|2-1|=5(M), |6-2|+|3-1|=6(B), |6-2|+|4-1|=7(G), |6-2|+|5-1|=8(Y), |6-2|+|6-1|=9$\rightarrow$0(.), |6-2|+|7-1|=10$\rightarrow$1(X)
    Row 7: |7-2|+|0-1|=6(B), |7-2|+|1-1|=5(M), |7-2|+|2-1|=6(B), |7-2|+|3-1|=7(G), |7-2|+|4-1|=8(Y), |7-2|+|5-1|=9$\rightarrow$0(.), |7-2|+|6-1|=10$\rightarrow$1(X), |7-2|+|7-1|=11$\rightarrow$2(R)

ANSWER:
A R A W M B G Y
R X R A W M B G
X . X R A W M B
R X R A W M B G
A R A W M B G Y
W A W M B G Y .
M W M B G Y . X
B M B G Y . X R

### Conceptual Digestion

1.  **The CAST**:
    *   **The Void**: The background grid (represented by 'O' in the input).
    *   **The Rune-stone**: The single cell that differs from the background (the "divine color"). In Example 1, it is 'G' at (0, 6). In the Test Input, it is '.' at (2, 1).
    *   **The Divine Cycle**: A sequence of 9 colors. By analyzing Example 1, we can deduce the sequence.

2.  **The AFFORDANCES**:
    *   The rune-stone acts as the origin (center) of a Manhattan distance expansion.
    *   The distance $d = |r_1 - r_2| + |c_1 - c_2|$ determines which color from the cycle is placed in a cell.

3.  **The TRANSFORMATION RULE**:
    *   **Identify the Cycle**: In Example 1, the center is 'G' at (0, 6).
        *   $d=0$: G
        *   $d=1$: Y (at 0, 5; 0, 7; 1, 6)
        *   $d=2$: . (at 0, 4; 1, 5; 1, 7; 2, 6)
        *   $d=3$: X (at 0, 3; 1, 4; 2, 5; 3, 6)
        *   $d=4$: R (at 0, 2; 1, 3; 2, 4; 3, 5; 4, 6)
        *   $d=5$: A (at 0, 1; 1, 2; 2, 3; 3, 4; 4, 5; 5, 6)
        *   $d=6$: W (at 0, 0; 1, 1; 2, 2; 3, 3; 4, 4; 5, 5; 6, 6)
        *   $d=7$: M (at 1, 0; 2, 1; 3, 2; 4, 3; 5, 4; 6, 5; 7, 6)
        *   $d=8$: B (at 2, 0; 3, 1; 4, 2; 5, 3; 6, 4; 7, 5)
        *   $d=9$: G (at 3, 0; 4, 1; 5, 2; 6, 3; 7, 4)
    *   **The Cycle Sequence**: G $\rightarrow$ Y $\rightarrow$ . $\rightarrow$ X $\rightarrow$ R $\rightarrow$ A $\rightarrow$ W $\rightarrow$ M $\rightarrow$ B $\rightarrow$ (repeat G).
    *   **Application**: For every cell $(r, c)$, calculate $d = |r - r_{stone}| + |c - c_{stone}|$. The color is the $d$-th element of the cycle starting from the stone's color.

### Applying to Test Input
*   **Rune-stone**: '.' at (2, 1).
*   **Cycle starting from '.'**: . $\rightarrow$ X $\rightarrow$ R $\rightarrow$ A $\rightarrow$ W $\rightarrow$ M $\rightarrow$ B $\rightarrow$ G $\rightarrow$ Y $\rightarrow$ (repeat .).

**Distance Map from (2, 1):**
Row 0: |0-2|+|0-1|=3 |0-2|+|1-1|=2 |0-2|+|2-1|=3 |0-2|+|3-1|=4 |0-2|+|4-1|=5 |0-2|+|5-1|=6 |0-2|+|6-1|=7 |0-2|+|7-1|=8
Row 1: |1-2|+|0-1|=2 |1-2|+|1-1|=1 |1-2|+|2-1|=2 |1-2|+|3-1|=3 |1-2|+|4-1|=4 |1-2|+|5-1|=5 |1-2|+|6-1|=6 |1-2|+|7-1|=7
Row 2: |2-2|+|0-1|=1 |2-2|+|1-1|=0 |2-2|+|2-1|=1 |2-2|+|3-1|=2 |2-2|+|4-1|=3 |2-2|+|5-1|=4 |2-2|+|6-1|=5 |2-2|+|7-1|=6
Row 3: |3-2|+|0-1|=2 |3-2|+|1-1|=1 |3-2|+|2-1|=2 |3-2|+|3-1|=3 |3-2|+|4-1|=4 |3-2|+|5-1|=5 |3-2|+|6-1|=6 |3-2|+|7-1|=7
Row 4: |4-2|+|0-1|=3 |4-2|+|1-1|=2 |4-2|+|2-1|=3 |4-2|+|3-1|=4 |4-2|+|4-1|=5 |4-2|+|5-1|=6 |4-2|+|6-1|=7 |4-2|+|7-1|=8
Row 5: |5-2|+|0-1|=4 |5-2|+|1-1|=3 |5-2|+|2-1|=4 |5-2|+|3-1|=5 |5-2|+|4-1|=6 |5-2|+|5-1|=7 |5-2|+|6-1|=8 |5-2|+|7-1|=9
Row 6: |6-2|+|0-1|=5 |6-2|+|1-1|=4 |6-2|+|2-1|=5 |6-2|+|3-1|=6 |6-2|+|4-1|=7 |6-2|+|5-1|=8 |6-2|+|6-1|=9 |6-2|+|7-1|=10
Row 7: |7-2|+|0-1|=6 |7-2|+|1-1|=5 |7-2|+|2-1|=6 |7-2|+|3-1|=7 |7-2|+|4-1|=8 |7-2|+|5-1|=9 |7-2|+|6-1|=10 |7-2|+|7-1|=11

**Mapping Distances to Colors (. X R A W M B G Y):**
0: . | 1: X | 2: R | 3: A | 4: W | 5: M | 6: B | 7: G | 8: Y | 9: . | 10: X | 11: R

Row 0: A R A W M B G Y
Row 1: R X R A W M B G
Row 2: X . X R A W M B
Row 3: R X R A W M B G
Row 4: A R A W M B G Y
Row 5: W A W M B G Y .
Row 6: M W M B G Y . X
Row 7: B M B G Y . X R

ANSWER:
A R A W M B G Y
R X R A W M B G
X . X R A W M B
R X R A W M B G
A R A W M B G Y
W A W M B G Y .
M W M B G Y . X
B M B G Y . X R

### Conceptual Digestion

1.  **Identify the CAST:**
    *   **Magma Field:** The background color (in the examples, `O` for orange).
    *   **Crystal Seed:** A single cell with a different color (e.g., `G` in Example 1, `.` in Test Input).
    *   **Crystallization Bands:** Diamond-shaped rings expanding from the seed.

2.  **Identify the AFFORDANCES:**
    *   The seed's position $(r, c)$ acts as the origin $(0,0)$ for calculating Manhattan distance: $d = |r - r_i| + |c - c_i|$.
    *   The seed's color determines the starting point in a specific 9-color cycle.

3.  **Identify the TRANSFORMATION RULE:**
    *   **The Cycle:** By analyzing Example 1 (Seed `G` at (0, 6)), the output colors follow a sequence based on distance $d$.
        *   $d=0$: `G` (Green)
        *   $d=1$: `Y` (Yellow)
        *   $d=2$: `.` (Black)
        *   $d=3$: `X` (Grey)
        *   $d=4$: `R` (Red)
        *   $d=5$: `A` (Azure)
        *   $d=6$: `W` (Maroon)
        *   $d=7$: `M` (Magenta)
        *   $d=8$: `B` (Blue)
        *   (Cycle repeats: G $\rightarrow$ Y $\rightarrow$ . $\rightarrow$ X $\rightarrow$ R $\rightarrow$ A $\rightarrow$ W $\rightarrow$ M $\rightarrow$ B $\rightarrow$ G...)
    *   **The Mapping:** For every cell $(i, j)$, calculate the Manhattan distance $d$ to the seed. The color of the cell is the color at index $(start\_index + d) \pmod 9$ in the cycle.
    *   **The Cycle Order:** G (0), Y (1), . (2), X (3), R (4), A (5), W (6), M (7), B (8).

### Applying the Rule to Test Input

*   **Seed Position:** (2, 1) - Row 2, Col 1 (0-indexed).
*   **Seed Color:** `.` (Black).
*   **Seed Index in Cycle:** `.` is index 2.
*   **Color Sequence starting from `.`:**
    *   $d=0$: `.` (Black)
    *   $d=1$: `X` (Grey)
    *   $d=2$: `R` (Red)
    *   $d=3$: `A` (Azure)
    *   $d=4$: `W` (Maroon)
    *   $d=5$: `M` (Magenta)
    *   $d=6$: `B` (Blue)
    *   $d=7$: `G` (Green)
    *   $d=8$: `Y` (Yellow)
    *   $d=9$: `.` (Black) ... and so on.

**Distance Map and Color Assignment for Seed at (2, 1):**
Row 0: (2,1) $\rightarrow$ d=2+0=2(R), 2+1=3(A), 2+2=4(W), 2+3=5(M), 2+4=6(B), 2+5=7(G), 2+6=8(Y), 2+7=9(.)
Row 1: (1,1) $\rightarrow$ d=1+0=1(X), 1+1=2(R), 1+2=3(A), 1+3=4(W), 1+4=5(M), 1+5=6(B), 1+6=7(G), 1+7=8(Y)
Row 2: (0,1) $\rightarrow$ d=2+0=2(R), 0+0=0(.), 0+1=1(X), 0+2=2(R), 0+3=3(A), 0+4=4(W), 0+5=5(M), 0+6=6(B)
Row 3: (1,1) $\rightarrow$ d=1+0=1(X), 1+1=2(R), 1+2=3(A), 1+3=4(W), 1+4=5(M), 1+5=6(B), 1+6=7(G), 1+7=8(Y)
Row 4: (2,1) $\rightarrow$ d=2+0=2(R), 2+1=3(A), 2+2=4(W), 2+3=5(M), 2+4=6(B), 2+5=7(G), 2+6=8(Y), 2+7=9(.)
Row 5: (3,1) $\rightarrow$ d=3+0=3(A), 3+1=4(W), 3+2=5(M), 3+3=6(B), 3+4=7(G), 3+5=8(Y), 3+6=9(.), 3+7=10(X)
Row 6: (4,1) $\rightarrow$ d=4+0=4(W), 4+1=5(M), 4+2=6(B), 4+3=7(G), 4+4=8(Y), 4+5=9(.), 4+6=10(X), 4+7=11(R)
Row 7: (5,1) $\rightarrow$ d=5+0=5(M), 5+1=6(B), 5+2=7(G), 5+3=8(Y), 5+4=9(.), 5+5=10(X), 5+6=11(R), 5+7=12(A)

Wait, let's re-calculate Row 2 carefully:
(2,0): $|2-2| + |0-1| = 1 \rightarrow$ X
(2,1): $|2-2| + |1-1| = 0 \rightarrow$ .
(2,2): $|2-2| + |2-1| = 1 \rightarrow$ X
(2,3): $|2-2| + |3-1| = 2 \rightarrow$ R
(2,4): $|2-2| + |4-1| = 3 \rightarrow$ A
(2,5): $|2-2| + |5-1| = 4 \rightarrow$ W
(2,6): $|2-2| + |6-1| = 5 \rightarrow$ M
(2,7): $|2-2| + |7-1| = 6 \rightarrow$ B

Recalculating all rows:
R0: (2+1)=3(A), (2+0)=2(R), (2+1)=3(A), (2+2)=4(W), (2+3)=5(M), (2+4)=6(B), (2+5)=7(G), (2+6)=8(Y)
Wait, Col 0 is $|0-2| + |0-1| = 3$.
R0: (0,0): 2+1=3(A), (0,1): 2+0=2(R), (0,2): 2+1=3(A), (0,3): 2+2=4(W), (0,4): 2+3=5(M), (0,5): 2+4=6(B), (0,6): 2+5=7(G), (0,7): 2+6=8(Y)
R1: (1,0): 1+1=2(R), (1,1): 1+0=1(X), (1,2): 1+1=2(R), (1,3): 1+2=3(A), (1,4): 1+3=4(W), (1,5): 1+4=5(M), (1,6): 1+5=6(B), (1,7): 1+6=7(G)
R2: (2,0): 0+1=1(X), (2,1): 0+0=0(.), (2,2): 0+1=1(X), (2,3): 0+2=2(R), (2,4): 0+3=3(A), (2,5): 0+4=4(W), (2,6): 0+5=5(M), (2,7): 0+6=6(B)
R3: (3,0): 1+1=2(R), (3,1): 1+0=1(X), (3,2): 1+1=2(R), (3,3): 1+2=3(A), (3,4): 1+3=4(W), (3,5): 1+4=5(M), (3,6): 1+5=6(B), (3,7): 1+6=7(G)
R4: (4,0): 2+1=3(A), (4,1): 2+0=2(R), (4,2): 2+1=3(A), (4,3): 2+2=4(W), (4,4): 2+3=5(M), (4,5): 2+4=6(B), (4,6): 2+5=7(G), (4,7): 2+6=8(Y)
R5: (5,0): 3+1=4(W), (5,1): 3+0=3(A), (5,2): 3+1=4(W), (5,3): 3+2=5(M), (5,4): 3+3=6(B), (5,5): 3+4=7(G), (5,6): 3+5=8(Y), (5,7): 3+6=9(.)
R6: (6,0): 4+1=5(M), (6,1): 4+0=4(W), (6,2): 4+1=5(M), (6,3): 4+2=6(B), (6,4): 4+3=7(G), (6,5): 4+4=8(Y), (6,6): 4+5=9(.), (6,7): 4+6=10(X)
R7: (7,0): 5+1=6(B), (7,1): 5+0=5(M), (7,2): 5+1=6(B), (7,3): 5+2=7(G), (7,4): 5+3=8(Y), (7,5): 5+4=9(.), (7,6): 5+5=10(X), (7,7): 5+6=11(R)

ANSWER:
A R A W M B G Y
R X R A W M B G
X . X R A W M B
R X R A W M B G
A R A W M B G Y
W A W M B G Y .
M W M B G Y . X
B M B G Y . X R

### Conceptual Digestion

1.  **CAST**:
    *   **Plain dough (O)**: The background grid.
    *   **Spice dot (G, X, .)**: The source point (seed) of the transformation.
    *   **Seasonings (W, A, R, X, ., Y, G, B, M)**: The colors that form the radiating diamond rings.

2.  **AFFORDANCES**:
    *   The source point determines the center of the "diamond waves."
    *   The distance from the source point is measured by the **Manhattan distance**: $d = |x_1 - x_2| + |y_1 - y_2|$.
    *   The color of a cell depends solely on its Manhattan distance from the source point.

3.  **TRANSFORMATION RULE**:
    *   Identify the coordinates $(r, c)$ of the non-O cell in the input.
    *   The sequence of colors is a repeating cycle. Let's analyze the distance $d$ from the source:
        *   $d=0$: The color of the source dot itself.
        *   $d=1, 2, 3, \dots$: The colors cycle through a specific sequence.
    *   Looking at Example 1: Source is at (0, 6) and is 'G'.
        *   $d=0$: G
        *   $d=1$: Y
        *   $d=2$: .
        *   $d=3$: X
        *   $d=4$: R
        *   $d=5$: A
        *   $d=6$: W
        *   $d=7$: M
        *   $d=8$: B
        *   $d=9$: G (cycle repeats)
    *   Looking at Example 2: Source is at (5, 2) and is 'X'.
        *   $d=0$: X
        *   $d=1$: R
        *   $d=2$: .
        *   $d=3$: Y
        *   $d=4$: G
        *   $d=5$: B
        *   $d=6$: M
        *   $d=7$: W
        *   $d=8$: A
        *   $d=9$: R (cycle repeats)
    *   **The Sequence**: The sequence is a fixed loop: **G $\to$ Y $\to$ . $\to$ X $\to$ R $\to$ A $\to$ W $\to$ M $\to$ B $\to$ (back to G)**.
    *   The color at distance $d$ is the color $d$ steps forward in this sequence from the source color.

### Applying to Test Input
*   **Source**: (2, 1) is '.' (black).
*   **Sequence**: G $\to$ Y $\to$ . $\to$ X $\to$ R $\to$ A $\to$ W $\to$ M $\to$ B $\to$ G
*   **Source Color**: '.' (Index 2 in the sequence)
*   **Mapping**:
    *   $d=0$: .
    *   $d=1$: X
    *   $d=2$: R
    *   $d=3$: A
    *   $d=4$: W
    *   $d=5$: M
    *   $d=6$: B
    *   $d=7$: G
    *   $d=8$: Y
    *   $d=9$: .
    *   $d=10$: X
    *   $d=11$: R
    *   $d=12$: A
    *   $d=13$: W

**Calculating distances from (2, 1):**
Row 0: (0,0)d=3:A, (0,1)d=2:R, (0,2)d=3:A, (0,3)d=4:W, (0,4)d=5:M, (0,5)d=6:B, (0,6)d=7:G, (0,7)d=8:Y
Row 1: (1,0)d=2:R, (1,1)d=1:X, (1,2)d=2:R, (1,3)d=3:A, (1,4)d=4:W, (1,5)d=5:M, (1,6)d=6:B, (1,7)d=7:G
Row 2: (2,0)d=1:X, (2,1)d=0:., (2,2)d=1:X, (2,3)d=2:R, (2,4)d=3:A, (2,5)d=4:W, (2,6)d=5:M, (2,7)d=6:B
Row 3: (3,0)d=2:R, (3,1)d=1:X, (3,2)d=2:R, (3,3)d=3:A, (3,4)d=4:W, (3,5)d=5:M, (3,6)d=6:B, (3,7)d=7:G
Row 4: (4,0)d=3:A, (4,1)d=2:R, (4,2)d=3:A, (4,3)d=4:W, (4,4)d=5:M, (4,5)d=6:B, (4,6)d=7:G, (4,7)d=8:Y
Row 5: (5,0)d=4:W, (5,1)d=3:A, (5,2)d=4:W, (5,3)d=5:M, (5,4)d=6:B, (5,5)d=7:G, (5,6)d=8:Y, (5,7)d=9:.
Row 6: (6,0)d=5:M, (6,1)d=4:W, (6,2)d=5:M, (6,3)d=6:B, (6,4)d=7:G, (6,5)d=8:Y, (6,6)d=9:., (6,7)d=10:X
Row 7: (7,0)d=6:B, (7,1)d=5:M, (7,2)d=6:B, (7,3)d=7:G, (7,4)d=8:Y, (7,5)d=9:., (7,6)d=10:X, (7,7)d=11:R

ANSWER:
A R A W M B G Y
R X R A W M B G
X . X R A W M B
R X R A W M B G
A R A W M B G Y
W A W M B G Y .
M W M B G Y . X
B M B G Y . X R

### Conceptual Digestion

1.  **CAST**:
    *   **The Plucked String**: The single non-orange (or non-background) cell in the input grid. In Example 1, it is the Green (G) cell at (0, 6). In the Test Input, it is the Black (.) cell at (2, 1).
    *   **The Soundboard**: The 8x8 grid.
    *   **The Scale (Pitches)**: A cycling sequence of colors. Looking at Example 1, the center (plucked point) is G. The cells at distance 1 are Y, and distance 2 are B, etc. Wait, let's look closer at the sequence.
    *   **The Wavefronts**: Manhattan distance from the plucked point.

2.  **AFFORDANCES**:
    *   The distance $d = |r_1 - r_2| + |c_1 - c_2|$ determines the color of the cell.
    *   The sequence of colors cycles. Let's determine the sequence from Example 1:
        Plucked point (0, 6) is G.
        Distance 1: (0, 5) is Y, (0, 7) is Y, (1, 6) is . (Wait, let's re-examine).
        Example 1 Output:
        (0, 6) = G (dist 0)
        (0, 5) = Y, (0, 7) = Y, (1, 6) = . (dist 1) -> Wait, (1, 6) is '.', (0, 5) is Y, (0, 7) is Y.
        Let's check the sequence along a line:
        (0, 6) G $\rightarrow$ (0, 5) Y $\rightarrow$ (0, 4) X $\rightarrow$ (0, 3) R $\rightarrow$ (0, 2) A $\rightarrow$ (0, 1) W $\rightarrow$ (0, 0) M (Wait, (0,0) is W).
        Let's re-read the output of Example 1 carefully:
        Row 0: W A R X . Y G Y
        Col 6 is G.
        Dist 0: (0, 6) = G
        Dist 1: (0, 5)=Y, (0, 7)=Y, (1, 6)=.
        Dist 2: (0, 4)=X, (1, 5)=X, (1, 7)=X, (2, 6)=X, (0, 8)N/A
        Dist 3: (0, 3)=R, (1, 4)=R, (2, 5)=R, (3, 6)=R...
        Dist 4: (0, 2)=A, (1, 3)=A, (2, 4)=A, (3, 5)=A, (4, 6)=A...
        Dist 5: (0, 1)=W, (1, 2)=W, (2, 3)=W, (3, 4)=W, (4, 5)=W, (5, 6)=W...
        Dist 6: (0, 0)=M (Wait, (0,0) is W), (1, 1)=M, (2, 2)=M, (3, 3)=M, (4, 4)=M, (5, 5)=M, (6, 6)=M...
        Let's re-evaluate the sequence from the center G:
        Dist 0: G
        Dist 1: Y
        Dist 2: .
        Dist 3: X
        Dist 4: R
        Dist 5: A
        Dist 6: W
        Dist 7: M
        Dist 8: B
        Dist 9: G (cycle)
        Wait, let's check the sequence again: G (0), Y (1), . (2), X (3), R (4), A (5), W (6), M (7), B (8).
        Let's check (7, 7) in Example 1: Dist from (0, 6) is $|7-0| + |7-6| = 7+1 = 8$. Color is B. Correct.
        Let's check (7, 0) in Example 1: Dist from (0, 6) is $|7-0| + |0-6| = 7+6 = 13$.
        $13 \pmod 9 = 4$. Color at index 4 is R. Correct.
        Let's check (0, 0) in Example 1: Dist from (0, 6) is $|0-0| + |0-6| = 6$. Color at index 6 is W. Correct.

3.  **TRANSFORMATION RULE**:
    1.  Find the coordinates $(r_0, c_0)$ of the unique non-orange cell.
    2.  Define the color sequence: `S = [G, Y, ., X, R, A, W, M, B]` (where G is the color of the plucked string).
    3.  Wait, the sequence starts with the color of the plucked string. If the plucked string is '.', the sequence starts with '.'.
    4.  The sequence of colors is: `[PluckedColor, Y, ., X, R, A, W, M, B]`? No, the sequence is a fixed cycle: `G -> Y -> . -> X -> R -> A -> W -> M -> B -> G`.
    5.  For every cell $(r, c)$, calculate $d = |r - r_0| + |c - c_0|$.
    6.  The color is $S[(index\_of\_plucked\_color + d) \pmod 9]$.

    *   Test Input Plucked Point: (2, 1) is color '.'.
    *   Sequence: `G, Y, ., X, R, A, W, M, B`
    *   Index of '.' is 2.
    *   Color at distance $d$ is $S[(2 + d) \pmod 9]$.

    *   $d=0: S[2] = .$
    *   $d=1: S[3] = X$
    *   $d=2: S[4] = R$
    *   $d=3: S[5] = A$
    *   $d=4: S[6] = W$
    *   $d=5: S[7] = M$
    *   $d=6: S[8] = B$
    *   $d=7: S[0] = G$
    *   $d=8: S[1] = Y$
    *   $d=9: S[2] = .$
    *   $d=10: S[3] = X$
    *   $d=11: S[4] = R$
    *   $d=12: S[5] = A$
    *   $d=13: S[6] = W$
    *   $d=14: S[7] = M$

    *   Row 0:
        (0,0): d=3 -> A | (0,1): d=2 -> R | (0,2): d=3 -> A | (0,3): d=4 -> W | (0,4): d=5 -> M | (0,5): d=6 -> B | (0,6): d=7 -> G | (0,7): d=8 -> Y
    *   Row 1:
        (1,0): d=2 -> R | (1,1): d=1 -> X | (1,2): d=2 -> R | (1,3): d=3 -> A | (1,4): d=4 -> W | (1,5): d=5 -> M | (1,6): d=6 -> B | (1,7): d=7 -> G
    *   Row 2:
        (2,0): d=1 -> X | (2,1): d=0 -> . | (2,2): d=1 -> X | (2,3): d=2 -> R | (2,4): d=3 -> A | (2,5): d=4 -> W | (2,6): d=5 -> M | (2,7): d=6 -> B
    *   Row 3:
        (3,0): d=2 -> R | (3,1): d=1 -> X | (3,2): d=2 -> R | (3,3): d=3 -> A | (3,4): d=4 -> W | (3,5): d=5 -> M | (3,6): d=6 -> B | (3,7): d=7 -> G
    *   Row 4:
        (4,0): d=3 -> A | (4,1): d=2 -> R | (4,2): d=3 -> A | (4,3): d=4 -> W | (4,4): d=5 -> M | (4,5): d=6 -> B | (4,6): d=7 -> G | (4,7): d=8 -> Y
    *   Row 5:
        (5,0): d=4 -> W | (5,1): d=3 -> A | (5,2): d=4 -> W | (5,3): d=5 -> M | (5,4): d=6 -> B | (5,5): d=7 -> G | (5,6): d=8 -> Y | (5,7): d=9 -> .
    *   Row 6:
        (6,0): d=5 -> M | (6,1): d=4 -> W | (6,2): d=5 -> M | (6,

... [2945 chars truncated]

### Conceptual Digestion

1.  **CAST**:
    *   **The Pitch**: The 8x8 grid.
    *   **The Athlete**: The single cell that differs from the background color (the "seed").
    *   **The Background**: The filler color (Orange 'O' in the examples).
    *   **The Palette**: A cycling sequence of colors.

2.  **AFFORDANCES**:
    *   The athlete's position $(r, c)$ acts as the center of a Manhattan distance calculation.
    *   The distance $d = |r - r_0| + |c - c_0|$ determines which color from the palette is assigned to a cell.

3.  **TRANSFORMATION RULE**:
    *   **Identify Seed**: Find the coordinates $(r_0, c_0)$ of the non-orange cell.
    *   **Determine Palette**: Looking at Example 1:
        Seed is at (0, 6) and is Green (G).
        Distance 0: G
        Distance 1: Y (at 0,5; 0,7; 1,6)
        Distance 2: . (at 0,4; 1,5; 1,7; 2,6)
        Distance 3: X (at 0,3; 1,4; 2,5; 2,7; 3,6)
        Distance 4: R (at 0,2; 1,3; 2,4; 3,5; 3,7; 4,6)
        Distance 5: A (at 0,1; 1,2; 2,3; 3,4; 4,5; 4,7; 5,6)
        Distance 6: W (at 0,0; 1,1; 2,2; 3,3; 4,4; 5,5; 5,7; 6,6)
        Distance 7: M (at 1,0; 2,1; 3,2; 4,3; 5,4; 6,5; 6,7; 7,6)
        Distance 8: B (at 2,0; 3,1; 4,2; 5,3; 6,4; 7,5; 7,7)
        Distance 9: G (at 3,0; 4,1; 5,2; 6,3; 7,4)
        Distance 10: Y (at 4,0; 5,1; 6,2; 7,3)
        Distance 11: . (at 5,0; 6,1; 7,2)
        Distance 12: X (at 6,0; 7,1)
        Distance 13: R (at 7,0)
    *   **Palette Sequence**: G $\rightarrow$ Y $\rightarrow$ . $\rightarrow$ X $\rightarrow$ R $\rightarrow$ A $\rightarrow$ W $\rightarrow$ M $\rightarrow$ B $\rightarrow$ (repeat G).
    *   **Application**: For every cell $(r, c)$, calculate $d = |r - r_0| + |c - c_0|$. The color is the $d$-th element in the sequence starting with the seed color.

### Applying to Test Input
*   **Seed**: The non-orange cell is at (2, 1) and is Black (.).
*   **Palette starting from '.'**: . $\rightarrow$ X $\rightarrow$ R $\rightarrow$ A $\rightarrow$ W $\rightarrow$ M $\rightarrow$ B $\rightarrow$ G $\rightarrow$ Y $\rightarrow$ (repeat .).
*   **Calculations**:
    (0,0): d=3 $\rightarrow$ A | (0,1): d=2 $\rightarrow$ R | (0,2): d=3 $\rightarrow$ A | (0,3): d=4 $\rightarrow$ W | (0,4): d=5 $\rightarrow$ M | (0,5): d=6 $\rightarrow$ B | (0,6): d=7 $\rightarrow$ G | (0,7): d=8 $\rightarrow$ Y
    (1,0): d=2 $\rightarrow$ R | (1,1): d=1 $\rightarrow$ X | (1,2): d=2 $\rightarrow$ R | (1,3): d=3 $\rightarrow$ A | (1,4): d=4 $\rightarrow$ W | (1,5): d=5 $\rightarrow$ M | (1,6): d=6 $\rightarrow$ B | (1,7): d=7 $\rightarrow$ G
    (2,0): d=1 $\rightarrow$ X | (2,1): d=0 $\rightarrow$ . | (2,2): d=1 $\rightarrow$ X | (2,3): d=2 $\rightarrow$ R | (2,4): d=3 $\rightarrow$ A | (2,5): d=4 $\rightarrow$ W | (2,6): d=5 $\rightarrow$ M | (2,7): d=6 $\rightarrow$ B
    (3,0): d=2 $\rightarrow$ R | (3,1): d=1 $\rightarrow$ X | (3,2): d=2 $\rightarrow$ R | (3,3): d=3 $\rightarrow$ A | (3,4): d=4 $\rightarrow$ W | (3,5): d=5 $\rightarrow$ M | (3,6): d=6 $\rightarrow$ B | (3,7): d=7 $\rightarrow$ G
    (4,0): d=3 $\rightarrow$ A | (4,1): d=2 $\rightarrow$ R | (4,2): d=3 $\rightarrow$ A | (4,3): d=4 $\rightarrow$ W | (4,4): d=5 $\rightarrow$ M | (4,5): d=6 $\rightarrow$ B | (4,6): d=7 $\rightarrow$ G | (4,7): d=8 $\rightarrow$ Y
    (5,0): d=4 $\rightarrow$ W | (5,1): d=3 $\rightarrow$ A | (5,2): d=4 $\rightarrow$ W | (5,3): d=5 $\rightarrow$ M | (5,4): d=6 $\rightarrow$ B | (5,5): d=7 $\rightarrow$ G | (5,6): d=8 $\rightarrow$ Y | (5,7): d=9 $\rightarrow$ .
    (6,0): d=5 $\rightarrow$ M | (6,1): d=4 $\rightarrow$ W | (6,2): d=5 $\rightarrow$ M | (6,3): d=6 $\rightarrow$ B | (6,4): d=7 $\rightarrow$ G | (6,5): d=8 $\rightarrow$ Y | (6,6): d=9 $\rightarrow$ . | (6,7): d=10 $\rightarrow$ X
    (7,0): d=6 $\rightarrow$ B | (7,1): d=5 $\rightarrow$ M | (7,2): d=6 $\rightarrow$ B | (7,3): d=7 $\rightarrow$ G | (7,4): d=8 $\rightarrow$ Y | (7,5): d=9 $\rightarrow$ . | (7,6): d=10 $\rightarrow$ X | (7,7): d=11 $\rightarrow$ R

ANSWER:
A R A W M B G Y
R X R A W M B G
X . X R A W M B
R X R A W M B G
A R A W M B G Y
W A W M B G Y .
M W M B G Y . X
B M B G Y . X R

### Conceptual Digestion

1.  **CAST**:
    *   **Atmosphere**: The uniform background color (in Example 1, it's Orange 'O'; in the Test Input, it's Orange 'O').
    *   **Strike Point**: The single cell that differs from the background (in Example 1, it's Green 'G' at (0, 6); in the Test Input, it's Black '.' at (2, 1)).
    *   **Shockwave**: The resulting pattern of colors radiating from the strike point.

2.  **AFFORDANCES**:
    *   The strike point acts as the origin $(r_0, c_0)$ for a Manhattan distance calculation.
    *   The distance $d = |r - r_0| + |c - c_0|$ determines the color of the cell $(r, c)$.

3.  **TRANSFORMATION RULE**:
    *   The colors follow a specific nine-hue loop. By analyzing Example 1:
        Strike point (0, 6) is Green (G).
        Distance 0: G
        Distance 1: Y (at 0, 5; 0, 7; 1, 6)
        Distance 2: . (at 0, 4; 1, 5; 1, 7; 2, 6)
        Distance 3: X (at 0, 3; 1, 4; 2, 5; 3, 6)
        Distance 4: R (at 0, 2; 1, 3; 2, 4; 3, 5; 4, 6)
        Distance 5: A (at 0, 1; 1, 2; 2, 3; 3, 4; 4, 5; 5, 6)
        Distance 6: W (at 0, 0; 1, 1; 2, 2; 3, 3; 4, 4; 5, 5; 6, 6)
        Distance 7: M (at 1, 0; 2, 1; 3, 2; 4, 3; 5, 4; 6, 5; 7, 6)
        Distance 8: B (at 2, 0; 3, 1; 4, 2; 5, 3; 6, 4; 7, 5)
        Distance 9: G (at 3, 0; 4, 1; 5, 2; 6, 3; 7, 4)
        ... and so on.
    *   **The Color Loop**: G $\rightarrow$ Y $\rightarrow$ . $\rightarrow$ X $\rightarrow$ R $\rightarrow$ A $\rightarrow$ W $\rightarrow$ M $\rightarrow$ B $\rightarrow$ (back to G).
    *   The color of a cell is determined by: `Loop[(Distance + LoopIndex(StrikeColor)) % 9]`.
    *   Wait, looking at Example 1 again: Strike point (0, 6) is G. Distance 0 is G.
        Distance 0: G
        Distance 1: Y
        Distance 2: .
        Distance 3: X
        Distance 4: R
        Distance 5: A
        Distance 6: W
        Distance 7: M
        Distance 8: B
        Distance 9: G
    *   The loop is: **G, Y, ., X, R, A, W, M, B**.

### Applying to Test Input
*   **Strike Point**: (2, 1) is Black ('.').
*   **Strike Color**: '.' is the 3rd element in the loop (index 2).
*   **Loop**: G(0), Y(1), .(2), X(3), R(4), A(5), W(6), M(7), B(8).
*   **Formula**: Color at $(r, c) = \text{Loop}[(\text{dist} + 2) \pmod 9]$.

**Distance Map from (2, 1):**
Row 0: |0-2|+|0-1|=3, |0-2|+|1-1|=2, |0-2|+|2-1|=3, |0-2|+|3-1|=4, |0-2|+|4-1|=5, |0-2|+|5-1|=6, |0-2|+|6-1|=7, |0-2|+|7-1|=8
Row 1: |1-2|+|0-1|=2, |1-2|+|1-1|=1, |1-2|+|2-1|=2, |1-2|+|3-1|=3, |1-2|+|4-1|=4, |1-2|+|5-1|=5, |1-2|+|6-1|=6, |1-2|+|7-1|=7
Row 2: |2-2|+|0-1|=1, |2-2|+|1-1|=0, |2-2|+|2-1|=1, |2-2|+|3-1|=2, |2-2|+|4-1|=3, |2-2|+|5-1|=4, |2-2|+|6-1|=5, |2-2|+|7-1|=6
Row 3: |3-2|+|0-1|=2, |3-2|+|1-1|=1, |3-2|+|2-1|=2, |3-2|+|3-1|=3, |3-2|+|4-1|=4, |3-2|+|5-1|=5, |3-2|+|6-1|=6, |3-2|+|7-1|=7
Row 4: |4-2|+|0-1|=3, |4-2|+|1-1|=2, |4-2|+|2-1|=3, |4-2|+|3-1|=4, |4-2|+|4-1|=5, |4-2|+|5-1|=6, |4-2|+|6-1|=7, |4-2|+|7-1|=8
Row 5: |5-2|+|0-1|=4, |5-2|+|1-1|=3, |5-2|+|2-1|=4, |5-2|+|3-1|=5, |5-2|+|4-1|=6, |5-2|+|5-1|=7, |5-2|+|6-1|=8, |5-2|+|7-1|=9
Row 6: |6-2|+|0-1|=5, |6-2|+|1-1|=4, |6-2|+|2-1|=5, |6-2|+|3-1|=6, |6-2|+|4-1|=7, |6-2|+|5-1|=8, |6-2|+|6-1|=9, |6-2|+|7-1|=10
Row 7: |7-2|+|0-1|=6, |7-2|+|1-1|=5, |7-2|+|2-1|=6, |7-2|+|3-1|=7, |7-2|+|4-1|=8, |7-2|+|5-1|=9, |7-2|+|6-1|=10, |7-2|+|7-1|=11

**Mapping Distances to Colors (Loop index = (dist + 2) % 9):**
0: (0+2)%9 = 2 (.)
1: (1+2)%9 = 3 (X)
2: (2+2)%9 = 4 (R)
3: (3+2)%9 = 5 (A)
4: (4+2)%9 = 6 (W)
5: (5+2)%9 = 7 (M)
6: (6+2)%9 = 8 (B)
7: (7+2)%9 = 0 (G)
8: (8+2)%9 = 1 (Y)
9: (9+2)%9 = 2 (.)
10: (10+2)%9 = 3 (X)
11: (11+2)%9 = 4 (R)

**Final Grid Construction:**
Row 0: A R A W M B G Y
Row 1: R X R A W M B G
Row 2: X . X R A W M B
Row 3: R X R A W M B G
Row 4: A R A W M B G Y
Row 5: W A W M B G Y .
Row 6: M W M B G Y . X
Row 7: B M B G Y . X R

ANSWER:
A R A W M B G Y
R X R A W M B G
X . X R A W M B
R X R A W M B G
A R A W M B G Y
W A W M B G Y .
M W M B G Y . X
B M B G Y . X R

### Conceptual Digestion

1.  **CAST**:
    *   **Cornerstone**: The single colored tile in the input grid (the one that is not 'O').
    *   **Plaza**: The 8x8 grid to be filled.
    *   **Palette**: A fixed sequence of nine colors.

2.  **AFFORDANCES**:
    *   The cornerstone's position $(r, c)$ acts as the origin $(0,0)$ for calculating Manhattan distance.
    *   The distance $d = |r - r_0| + |c - c_0|$ determines which color from the palette is assigned to a cell.

3.  **TRANSFORMATION RULE**:
    *   **Identify the Palette**: Looking at Example 1 and 2, the colors cycle. Let's trace the distance from the cornerstone.
    *   Example 1: Cornerstone at (0, 6) is 'G'.
        *   $d=0$: G
        *   $d=1$: Y (at 0,5; 0,7; 1,6)
        *   $d=2$: . (black) (at 0,4; 1,5; 1,7; 2,6)
        *   $d=3$: X (grey)
        *   $d=4$: R (red)
        *   $d=5$: A (azure)
        *   $d=6$: W (maroon)
        *   $d=7$: M (magenta)
        *   $d=8$: B (blue)
        *   $d=9$: G (green) ... and so on.
    *   **Palette Sequence**: G $\rightarrow$ Y $\rightarrow$ . $\rightarrow$ X $\rightarrow$ R $\rightarrow$ A $\rightarrow$ W $\rightarrow$ M $\rightarrow$ B $\rightarrow$ (repeat G).
    *   **Test Input**: Cornerstone is at (2, 1) and the color is '.' (black).
    *   **Mapping for Test Input**:
        *   $d=0$: .
        *   $d=1$: X
        *   $d=2$: R
        *   $d=3$: A
        *   $d=4$: W
        *   $d=5$: M
        *   $d=6$: B
        *   $d=7$: G
        *   $d=8$: Y
        *   $d=9$: . (repeat)

### Applying the Rule to Test Input
Cornerstone at (2, 1), Color = '.'
Palette: . (0), X (1), R (2), A (3), W (4), M (5), B (6), G (7), Y (8)

*   (0,0): $d=|0-2|+|0-1|=3 \rightarrow$ A
*   (0,1): $d=|0-2|+|1-1|=2 \rightarrow$ R
*   (0,2): $d=|0-2|+|2-1|=3 \rightarrow$ A
*   (0,3): $d=|0-2|+|3-1|=4 \rightarrow$ W
*   (0,4): $d=|0-2|+|4-1|=5 \rightarrow$ M
*   (0,5): $d=|0-2|+|5-1|=6 \rightarrow$ B
*   (0,6): $d=|0-2|+|6-1|=7 \rightarrow$ G
*   (0,7): $d=|0-2|+|7-1|=8 \rightarrow$ Y

*   (1,0): $d=|1-2|+|0-1|=2 \rightarrow$ R
*   (1,1): $d=|1-2|+|1-1|=1 \rightarrow$ X
*   (1,2): $d=|1-2|+|2-1|=2 \rightarrow$ R
*   (1,3): $d=|1-2|+|3-1|=3 \rightarrow$ A
*   (1,4): $d=|1-2|+|4-1|=4 \rightarrow$ W
*   (1,5): $d=|1-2|+|5-1|=5 \rightarrow$ M
*   (1,6): $d=|1-2|+|6-1|=6 \rightarrow$ B
*   (1,7): $d=|1-2|+|7-1|=7 \rightarrow$ G

*   (2,0): $d=|2-2|+|0-1|=1 \rightarrow$ X
*   (2,1): $d=|2-2|+|1-1|=0 \rightarrow$ .
*   (2,2): $d=|2-2|+|2-1|=1 \rightarrow$ X
*   (2,3): $d=|2-2|+|3-1|=2 \rightarrow$ R
*   (2,4): $d=|2-2|+|4-1|=3 \rightarrow$ A
*   (2,5): $d=|2-2|+|5-1|=4 \rightarrow$ W
*   (2,6): $d=|2-2|+|6-1|=5 \rightarrow$ M
*   (2,7): $d=|2-2|+|7-1|=6 \rightarrow$ B

*   (3,0): $d=|3-2|+|0-1|=2 \rightarrow$ R
*   (3,1): $d=|3-2|+|1-1|=1 \rightarrow$ X
*   (3,2): $d=|3-2|+|2-1|=2 \rightarrow$ R
*   (3,3): $d=|3-2|+|3-1|=3 \rightarrow$ A
*   (3,4): $d=|3-2|+|4-1|=4 \rightarrow$ W
*   (3,5): $d=|3-2|+|5-1|=5 \rightarrow$ M
*   (3,6): $d=|3-2|+|6-1|=6 \rightarrow$ B
*   (3,7): $d=|3-2|+|7-1|=7 \rightarrow$ G

*   (4,0): $d=|4-2|+|0-1|=3 \rightarrow$ A
*   (4,1): $d=|4-2|+|1-1|=2 \rightarrow$ R
*   (4,2): $d=|4-2|+|2-1|=3 \rightarrow$ A
*   (4,3): $d=|4-2|+|3-1|=4 \rightarrow$ W
*   (4,4): $d=|4-2|+|4-1|=5 \rightarrow$ M
*   (4,5): $d=|4-2|+|5-1|=6 \rightarrow$ B
*   (4,6): $d=|4-2|+|6-1|=7 \rightarrow$ G
*   (4,7): $d=|4-2|+|7-1|=8 \rightarrow$ Y

*   (5,0): $d=|5-2|+|0-1|=4 \rightarrow$ W
*   (5,1): $d=|5-2|+|1-1|=3 \rightarrow$ A
*   (5,2): $d=|5-2|+|2-1|=4 \rightarrow$ W
*   (5,3): $d=|5-2|+|3-1|=5 \rightarrow$ M
*   (5,4): $d=|5-2|+|4-1|=6 \rightarrow$ B
*   (5,5): $d=|5-2|+|5-1|=7 \rightarrow$ G
*   (5,6): $d=|5-2|+|6-1|=8 \rightarrow$ Y
*   (5,7): $d=|5-2|+|7-1|=9 \rightarrow$ .

*   (6,0): $d=|6-2|+|0-1|=5 \rightarrow$ M
*   (6,1): $d=|6-2|+|1-1|=4 \rightarrow$ W
*   (6,2): $d=|6-2|+|2-1|=5 \rightarrow$ M
*   (6,3): $d=|6-2|+|3-1|=6 \rightarrow$ B
*   (6,4): $d=|6-2|+|4-1|=7 \rightarrow$ G
*   (6,5): $d=|6-2|+|5-1|=8 \rightarrow$ Y
*   (6,6): $d=|6-2|+|6-1|=9 \rightarrow$ .
*   (6,7): $d=|6-2|+|7-1|=10 \rightarrow$ X

*   (7,0): $d=|7-2|+|0-1|=6 \rightarrow$ B
*   (7,1): $d=|7-2|+|1-1|=5 \rightarrow$ M
*   (7,2): $d=|7-2|+|2-1|=6 \rightarrow$ B
*   (7,3): $d=|7-2|+|3-1|=7 \rightarrow$ G
*   (7,4): $d=|7-2|+|4-1|=8 \rightarrow$ Y
*   (7,5): $d=|7-2|+|5-1|=9 \rightarrow$ .
*   (7,6): $d=|7-2|+|6-1|=10 \rightarrow$ X
*   (7,7): $d=|7-2|+|7-1|=11 \rightarrow$ R

ANSWER:
A R A W M B G Y
R X R A W M B G
X . X R A W M B
R X R A W M B G
A R A W M B G Y
W A W M B G Y .
M W M B G Y . X
B M B G Y . X R

### Conceptual Digestion

1.  **CAST**:
    *   **Barren Ground**: The background color (Orange 'O' in the examples).
    *   **Flare**: The single cell with a different color (Green 'G' in Example 1, Black '.' in the Test Input). This is the origin point $(r, c)$.
    *   **Blast/Wavefronts**: The resulting pattern of colors radiating from the flare.

2.  **AFFORDANCES**:
    *   The flare's position determines the center of the expansion.
    *   The distance from the flare is measured by the **Manhattan Distance**: $d = |r_1 - r_2| + |c_1 - c_2|$.
    *   The color of each cell in the output depends on its distance $d$ from the flare.

3.  **TRANSFORMATION RULE**:
    *   The output is a sequence of colors cycling based on the distance $d$.
    *   Looking at Example 1: Flare is at $(0, 6)$.
        *   $d=0$: Color is 'G' (the flare color).
        *   $d=1$: Color is 'Y'.
        *   $d=2$: Color is '.'.
        *   $d=3$: Color is 'X'.
        *   $d=4$: Color is 'R'.
        *   $d=5$: Color is 'A'.
        *   $d=6$: Color is 'W'.
        *   $d=7$: Color is 'M'.
        *   $d=8$: Color is 'B'.
        *   $d=9$: Color is 'G'.
        *   $d=10$: Color is 'Y'.
        *   $d=11$: Color is '.'.
        *   $d=12$: Color is 'X'.
        *   $d=13$: Color is 'R'.
        *   $d=14$: Color is 'A'.
    *   The sequence is: **G $\rightarrow$ Y $\rightarrow$ . $\rightarrow$ X $\rightarrow$ R $\rightarrow$ A $\rightarrow$ W $\rightarrow$ M $\rightarrow$ B $\rightarrow$ (repeat G)**.
    *   The flare color in the input is the starting point ($d=0$) of this sequence.

### Applying to Test Input

*   **Flare Position**: $(2, 1)$ (row 2, col 1, 0-indexed).
*   **Flare Color**: '.' (Black).
*   **Sequence starting from '.'**: . $\rightarrow$ X $\rightarrow$ R $\rightarrow$ A $\rightarrow$ W $\rightarrow$ M $\rightarrow$ B $\rightarrow$ G $\rightarrow$ Y $\rightarrow$ (repeat .).

**Distance Map for (2, 1):**
Row 0: |0-2|+|0-1|=3 |0-2|+|1-1|=2 |0-2|+|2-1|=3 |0-2|+|3-1|=4 |0-2|+|4-1|=5 |0-2|+|5-1|=6 |0-2|+|6-1|=7 |0-2|+|7-1|=8
Row 1: |1-2|+|0-1|=2 |1-2|+|1-1|=1 |1-2|+|2-1|=2 |1-2|+|3-1|=3 |1-2|+|4-1|=4 |1-2|+|5-1|=5 |1-2|+|6-1|=6 |1-2|+|7-1|=7
Row 2: |2-2|+|0-1|=1 |2-2|+|1-1|=0 |2-2|+|2-1|=1 |2-2|+|3-1|=2 |2-2|+|4-1|=3 |2-2|+|5-1|=4 |2-2|+|6-1|=5 |2-2|+|7-1|=6
Row 3: |3-2|+|0-1|=2 |3-2|+|1-1|=1 |3-2|+|2-1|=2 |3-2|+|3-1|=3 |3-2|+|4-1|=4 |3-2|+|5-1|=5 |3-2|+|6-1|=6 |3-2|+|7-1|=7
Row 4: |4-2|+|0-1|=3 |4-2|+|1-1|=2 |4-2|+|2-1|=3 |4-2|+|3-1|=4 |4-2|+|4-1|=5 |4-2|+|5-1|=6 |4-2|+|6-1|=7 |4-2|+|7-1|=8
Row 5: |5-2|+|0-1|=4 |5-2|+|1-1|=3 |5-2|+|2-1|=4 |5-2|+|3-1|=5 |5-2|+|4-1|=6 |5-2|+|5-1|=7 |5-2|+|6-1|=8 |5-2|+|7-1|=9
Row 6: |6-2|+|0-1|=5 |6-2|+|1-1|=4 |6-2|+|2-1|=5 |6-2|+|3-1|=6 |6-2|+|4-1|=7 |6-2|+|5-1|=8 |6-2|+|6-1|=9 |6-2|+|7-1|=10
Row 7: |7-2|+|0-1|=6 |7-2|+|1-1|=5 |7-2|+|2-1|=6 |7-2|+|3-1|=7 |7-2|+|4-1|=8 |7-2|+|5-1|=9 |7-2|+|6-1|=10 |7-2|+|7-1|=11

**Mapping Distances to Colors:**
0: . | 1: X | 2: R | 3: A | 4: W | 5: M | 6: B | 7: G | 8: Y | 9: . | 10: X | 11: R

Row 0: A R A W M B G Y
Row 1: R X R A W M B G
Row 2: X . X R A W M B
Row 3: R X R A W M B G
Row 4: A R A W M B G Y
Row 5: W A W M B G Y .
Row 6: M W M B G Y . X
Row 7: B M B G Y . X R

ANSWER:
A R A W M B G Y
R X R A W M B G
X . X R A W M B
R X R A W M B G
A R A W M B G Y
W A W M B G Y .
M W M B G Y . X
B M B G Y . X R

### Conceptual Digestion

1.  **CAST**:
    *   **The Stage**: An 8x8 grid.
    *   **The Spotlight**: A single cell in the input grid that differs from the background (the "source"). In Example 1, the source is at (0, 6) with color 'G'. In the Test Input, the source is at (2, 1) with color '.'.
    *   **The Gel Wheel**: A sequence of 9 colors. By analyzing Example 1, we can deduce the sequence. The source is 'G'. The cells at distance 1 (Manhattan distance) are 'Y', distance 2 are 'X', distance 3 are '.', distance 4 are 'W', distance 5 are 'A', distance 6 are 'R', distance 7 are 'X' (wait, let's re-examine).

2.  **AFFORDANCES**:
    *   The source cell acts as the origin (distance 0).
    *   The distance is measured using the **Manhattan Distance**: $d = |r_1 - r_2| + |c_1 - c_2|$.
    *   The color of a cell depends on its distance $d$ from the source.

3.  **TRANSFORMATION RULE**:
    *   Identify the source cell $(r_0, c_0)$ and its color $C_0$.
    *   The color sequence (the "gel wheel") is a cycle of 9 colors.
    *   Looking at Example 1: Source (0, 6) is 'G'.
        *   $d=0$: G
        *   $d=1$: Y
        *   $d=2$: X
        *   $d=3$: .
        *   $d=4$: W
        *   $d=5$: A
        *   $d=6$: R
        *   $d=7$: X (Wait, let's check the output again).
        *   Let's re-map Example 1 Output:
            (0,6) is G.
            (0,7) is Y ($d=1$), (0,5) is Y ($d=1$), (1,6) is Y ($d=1$).
            (0,4) is X ($d=2$), (1,5) is X ($d=2$), (1,7) is . ($d=2$ - No, (1,7) is .).
            Let's re-evaluate the sequence from the output:
            (0,6): G (d=0)
            (0,7): Y (d=1)
            (0,5): Y (d=1)
            (1,6): Y (d=1)
            (0,4): X (d=2)
            (1,5): X (d=2)
            (1,7): . (d=2) -> Wait, (1,7) is '.', (0,7) is 'Y'. Distance from (0,6) to (1,7) is $|1-0| + |7-6| = 2$.
            So $d=2$ is '.'.
            Let's check $d=3$: (0,3) is R, (1,4) is R, (2,5) is R, (2,7) is X.
            Wait, let's look at the diagonal:
            (0,6) G
            (1,5) X (d=2)
            (2,4) A (d=4)
            (3,3) M (d=6)
            (4,2) B (d=8)
            (5,1) Y (d=10)
            (6,0) R (d=12)
            
            Let's look at the sequence again:
            $d=0: G$
            $d=1: Y$
            $d=2: .$
            $d=3: X$
            $d=4: R$
            $d=5: A$
            $d=6: W$
            $d=7: M$
            $d=8: B$
            $d=9: G$ (cycle repeats)
            
            Let's test this sequence on Example 1:
            Source (0,6) = G.
            (0,7): $d=1 \rightarrow Y$. Correct.
            (0,5): $d=1 \rightarrow Y$. Correct.
            (1,6): $d=1 \rightarrow Y$. Correct.
            (0,4): $d=2 \rightarrow .$. Correct.
            (1,5): $d=2 \rightarrow .$. Correct.
            (1,7): $d=2 \rightarrow .$. Correct.
            (0,3): $d=3 \rightarrow X$. Correct.
            (1,4): $d=3 \rightarrow X$. Correct.
            (2,5): $d=3 \rightarrow X$. Correct.
            (2,6): $d=2 \rightarrow .$. Correct.
            (2,7): $d=3 \rightarrow X$. Correct.
            (0,2): $d=4 \rightarrow R$. Correct.
            (1,3): $d=4 \rightarrow R$. Correct.
            (2,4): $d=4 \rightarrow R$. Correct.
            (3,5): $d=4 \rightarrow R$. Correct.
            (3,6): $d=3 \rightarrow X$. Correct.
            (3,7): $d=4 \rightarrow R$. Correct.
            
            The sequence is: **G $\rightarrow$ Y $\rightarrow$ . $\rightarrow$ X $\rightarrow$ R $\rightarrow$ A $\rightarrow$ W $\rightarrow$ M $\rightarrow$ B $\rightarrow$ (repeat G)**.
            
            Wait, the source in Example 1 is G. The sequence starts at G.
            In the Test Input, the source is at (2, 1) and the color is '.'.
            The sequence must start from the source color.
            The wheel is: **G $\rightarrow$ Y $\rightarrow$ . $\rightarrow$ X $\rightarrow$ R $\rightarrow$ A $\rightarrow$ W $\rightarrow$ M $\rightarrow$ B $\rightarrow$ G**
            If the source is '.', the sequence is:
            $d=0: .$
            $d=1: X$
            $d=2: R$
            $d=3: A$
            $d=4: W$
            $d=5: M$
            $d=6: B$
            $d=7: G$
            $d=8: Y$
            $d=9: .$

### Applying to Test Input
Source: (2, 1), Color: '.'
Sequence: . (0), X (1), R (2), A (3), W (4), M (5), B (6), G (7), Y (8), . (9)

Row 0:
(0,0): d=3 -> A
(0,1): d=2 -> R
(0,2): d=3 -> A
(0,3): d=4 -> W
(0,4): d=5 -> M
(0,5): d=6 -> B
(0,6): d=7 -> G
(0,7): d=8 -> Y

Row 1:
(1,0): d=2 -> R
(1,1): d=1 -> X
(1,2): d=2 -> R
(1,3): d=3 -> A
(1,4): d=4 -> W
(1,5): d=5 -> M
(1,6): d=6 -> B
(1,7): d=7 -> G

Row 2:
(2,0): d=1 -> X
(2,1): d=0 -> .
(2,2): d=1 -> X
(2,3): d=2 -> R
(2,4): d=3 -> A
(2,5): d=4 -> W
(2,6): d=5 -> M
(2,7): d=6 -> B

Row 3:
(3,0): d=2 -> R
(3,1): d=1 -> X
(3,2): d=2 -> R
(3,3): d=3 -> A
(3,4): d=4 -> W
(3,5): d=5 -&

... [823 chars truncated]

### Conceptual Digestion

1.  **CAST**:
    *   **Bare Soil**: The background color (represented by 'O' in the input, but the description says it "vanishes").
    *   **Seed**: The single colored pixel that differs from the background (in Example 1, it's 'G' at (0, 6); in the Test Input, it's '.' at (2, 1)).
    *   **Flowers**: The resulting colors filling the grid.

2.  **AFFORDANCES**:
    *   The seed acts as the origin $(r_0, c_0)$ for a distance-based coloring algorithm.
    *   The distance metric is the **Manhattan Distance**: $d = |r - r_0| + |c - c_0|$.

3.  **TRANSFORMATION RULE**:
    *   **The Cycle**: A nine-color sequence. Looking at Example 1:
        Seed is 'G' at (0, 6).
        Distance 0: G
        Distance 1: Y (at 0, 5; 0, 7; 1, 6)
        Distance 2: . (at 0, 4; 0, 8(out); 1, 5; 1, 7; 2, 6)
        Distance 3: X (at 0, 3; 1, 4; 2, 5; 3, 6...)
        Distance 4: R
        Distance 5: A
        Distance 6: W
        Distance 7: M
        Distance 8: B
        Distance 9: G (cycle repeats)
        Sequence: **G $\rightarrow$ Y $\rightarrow$ . $\rightarrow$ X $\rightarrow$ R $\rightarrow$ A $\rightarrow$ W $\rightarrow$ M $\rightarrow$ B $\rightarrow$ (G)**.
    *   **Application**: For every cell $(r, c)$ in the 8x8 grid, calculate $d = |r - r_0| + |c - c_0|$. The color is the $d$-th element in the sequence starting from the seed color.

### Applying to Test Input

*   **Seed**: '.' (black) at $(r_0, c_0) = (2, 1)$.
*   **Sequence starting from '.'**:
    0: .
    1: X
    2: R
    3: A
    4: W
    5: M
    6: B
    7: G
    8: Y
    9: . (cycle)

*   **Calculations**:
    Row 0:
    (0,0): |0-2|+|0-1|=3 $\rightarrow$ A
    (0,1): |0-2|+|1-1|=2 $\rightarrow$ R
    (0,2): |0-2|+|2-1|=3 $\rightarrow$ A
    (0,3): |0-2|+|3-1|=4 $\rightarrow$ W
    (0,4): |0-2|+|4-1|=5 $\rightarrow$ M
    (0,5): |0-2|+|5-1|=6 $\rightarrow$ B
    (0,6): |0-2|+|6-1|=7 $\rightarrow$ G
    (0,7): |0-2|+|7-1|=8 $\rightarrow$ Y

    Row 1:
    (1,0): |1-2|+|0-1|=2 $\rightarrow$ R
    (1,1): |1-2|+|1-1|=1 $\rightarrow$ X
    (1,2): |1-2|+|2-1|=2 $\rightarrow$ R
    (1,3): |1-2|+|3-1|=3 $\rightarrow$ A
    (1,4): |1-2|+|4-1|=4 $\rightarrow$ W
    (1,5): |1-2|+|5-1|=5 $\rightarrow$ M
    (1,6): |1-2|+|6-1|=6 $\rightarrow$ B
    (1,7): |1-2|+|7-1|=7 $\rightarrow$ G

    Row 2:
    (2,0): |2-2|+|0-1|=1 $\rightarrow$ X
    (2,1): |2-2|+|1-1|=0 $\rightarrow$ .
    (2,2): |2-2|+|2-1|=1 $\rightarrow$ X
    (2,3): |2-2|+|3-1|=2 $\rightarrow$ R
    (2,4): |2-2|+|4-1|=3 $\rightarrow$ A
    (2,5): |2-2|+|5-1|=4 $\rightarrow$ W
    (2,6): |2-2|+|6-1|=5 $\rightarrow$ M
    (2,7): |2-2|+|7-1|=6 $\rightarrow$ B

    Row 3:
    (3,0): |3-2|+|0-1|=2 $\rightarrow$ R
    (3,1): |3-2|+|1-1|=1 $\rightarrow$ X
    (3,2): |3-2|+|2-1|=2 $\rightarrow$ R
    (3,3): |3-2|+|3-1|=3 $\rightarrow$ A
    (3,4): |3-2|+|4-1|=4 $\rightarrow$ W
    (3,5): |3-2|+|5-1|=5 $\rightarrow$ M
    (3,6): |3-2|+|6-1|=6 $\rightarrow$ B
    (3,7): |3-2|+|7-1|=7 $\rightarrow$ G

    Row 4:
    (4,0): |4-2|+|0-1|=3 $\rightarrow$ A
    (4,1): |4-2|+|1-1|=2 $\rightarrow$ R
    (4,2): |4-2|+|2-1|=3 $\rightarrow$ A
    (4,3): |4-2|+|3-1|=4 $\rightarrow$ W
    (4,4): |4-2|+|4-1|=5 $\rightarrow$ M
    (4,5): |4-2|+|5-1|=6 $\rightarrow$ B
    (4,6): |4-2|+|6-1|=7 $\rightarrow$ G
    (4,7): |4-2|+|7-1|=8 $\rightarrow$ Y

    Row 5:
    (5,0): |5-2|+|0-1|=4 $\rightarrow$ W
    (5,1): |5-2|+|1-1|=3 $\rightarrow$ A
    (5,2): |5-2|+|2-1|=4 $\rightarrow$ W
    (5,3): |5-2|+|3-1|=5 $\rightarrow$ M
    (5,4): |5-2|+|4-1|=6 $\rightarrow$ B
    (5,5): |5-2|+|5-1|=7 $\rightarrow$ G
    (5,6): |5-2|+|6-1|=8 $\rightarrow$ Y
    (5,7): |5-2|+|7-1|=9 $\rightarrow$ .

    Row 6:
    (6,0): |6-2|+|0-1|=5 $\rightarrow$ M
    (6,1): |6-2|+|1-1|=4 $\rightarrow$ W
    (6,2): |6-2|+|2-1|=5 $\rightarrow$ M
    (6,3): |6-2|+|3-1|=6 $\rightarrow$ B
    (6,4): |6-2|+|4-1|=7 $\rightarrow$ G
    (6,5): |6-2|+|5-1|=8 $\rightarrow$ Y
    (6,6): |6-2|+|6-1|=9 $\rightarrow$ .
    (6,7): |6-2|+|7-1|=10 $\rightarrow$ X

    Row 7:
    (7,0): |7-2|+|0-1|=6 $\rightarrow$ B
    (7,1): |7-2|+|1-1|=5 $\rightarrow$ M
    (7,2): |7-2|+|2-1|=6 $\rightarrow$ B
    (7,3): |7-2|+|3-1|=7 $\rightarrow$ G
    (7,4): |7-2|+|4-1|=8 $\rightarrow$ Y
    (7,5): |7-2|+|5-1|=9 $\rightarrow$ .
    (7,6): |7-2|+|6-1|=10 $\rightarrow$ X
    (7,7): |7-2|+|7-1|=11 $\rightarrow$ R

ANSWER:
A R A W M B G Y
R X R A W M B G
X . X R A W M B
R X R A W M B G
A R A W M B G Y
W A W M B G Y .
M W M B G Y . X
B M B G Y . X R

### Conceptual Digestion

1.  **CAST**:
    *   **Homes**: Blue (B) and Red (R).
    *   **Blue Owner's Pets**: Orange (O).
    *   **Red Owner's Pets**: Green (G).
    *   **Strays**: Any other colored pixels (though not present in the test input, they are ignored).
    *   **Field**: Black (.).

2.  **AFFORDANCES**:
    *   Pets move along the row or column they are currently in toward the home of their corresponding color.
    *   A pet "returns home" if it shares the same row or the same column as its home.
    *   The destination is the cell immediately adjacent to the home (up, down, left, or right).

3.  **TRANSFORMATION RULE**:
    *   **Identify Home Positions**: Find the coordinates of B and R.
    *   **Process Blue Pets (O)**: For every O at $(r, c)$:
        *   If $r = r_B$, it moves to $(r_B, c_B \pm 1)$ depending on whether $c < c_B$ or $c > c_B$.
        *   If $c = c_B$, it moves to $(r_B \pm 1, c_B)$ depending on whether $r < r_B$ or $r > r_B$.
        *   If neither, it is a stray and vanishes.
    *   **Process Red Pets (G)**: For every G at $(r, c)$:
        *   If $r = r_R$, it moves to $(r_R, c_R \pm 1)$ depending on whether $c < c_R$ or $c > c_R$.
        *   If $c = c_R$, it moves to $(r_R \pm 1, c_R)$ depending on whether $r < r_R$ or $r > r_R$.
        *   If neither, it is a stray and vanishes.
    *   **Final State**: Only the homes and the pets that successfully moved to the adjacent cells remain.

### Applying the Rule to Test Input

**1. Home Positions:**
*   Blue (B): Row 6, Col 5 (0-indexed)
*   Red (R): Row 3, Col 11 (0-indexed)

**2. Blue Pets (O):**
*   O at (6, 0): Same row as B (6). $0 < 5$, so it moves to (6, 4).
*   O at (6, 14): Same row as B (6). $14 > 5$, so it moves to (6, 6).

**3. Red Pets (G):**
*   G at (0, 11): Same column as R (11). $0 < 3$, so it moves to (2, 11).
*   G at (3, 0): Same row as R (3). $0 < 11$, so it moves to (3, 10).
*   G at (12, 11): Same column as R (11). $12 > 3$, so it moves to (4, 11).

**4. Final Grid Construction:**
*   B at (6, 5)
*   O at (6, 4) and (6, 6)
*   R at (3, 11)
*   G at (2, 11), (3, 10), and (4, 11)
*   All other cells are black (.).

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

### Conceptual Digestion

1.  **CAST**:
    *   **Nuclei**: Blue (B) and Red (R) pixels.
    *   **Organelles**: Orange (O) pixels associated with the Blue nucleus (B) and Green (G) pixels associated with the Red nucleus (R).
    *   **Stray Particles**: Any O or G pixels that do not share a row or column with their respective nucleus.
    *   **Cytoplasm**: Black (.) pixels.

2.  **AFFORDANCES**:
    *   Nuclei act as anchors.
    *   Organelles on the same row or column (orthogonal fibers) as their nucleus are "retracted" to the cell immediately adjacent to the nucleus in that direction.
    *   Stray particles (not on the axis) are deleted.

3.  **TRANSFORMATION RULE**:
    *   Locate Blue nucleus (B) at $(r_B, c_B)$ and Red nucleus (R) at $(r_R, c_R)$.
    *   For every Orange (O) pixel at $(r, c)$:
        *   If $r = r_B$ and $c < c_B$, move to $(r_B, c_B - 1)$.
        *   If $r = r_B$ and $c > c_B$, move to $(r_B, c_B + 1)$.
        *   If $c = c_B$ and $r < r_B$, move to $(r_B - 1, c_B)$.
        *   If $c = c_B$ and $r > r_B$, move to $(r_B + 1, c_B)$.
        *   Otherwise, delete.
    *   For every Green (G) pixel at $(r, c)$:
        *   If $r = r_R$ and $c < c_R$, move to $(r_R, c_R - 1)$.
        *   If $r = r_R$ and $c > c_R$, move to $(r_R, c_R + 1)$.
        *   If $c = c_R$ and $r < r_R$, move to $(r_R - 1, c_R)$.
        *   If $c = c_R$ and $r > r_R$, move to $(r_R + 1, c_R)$.
        *   Otherwise, delete.
    *   Clear all other pixels except B and R.

### Applying the Rule to Test Input

*   **Nuclei**: B is at (6, 5), R is at (3, 11). (0-indexed)
*   **Orange (O) pixels**:
    *   (6, 0): Same row as B (6=6), $0 < 5$. Retracts to (6, 4).
    *   (6, 14): Same row as B (6=6), $14 > 5$. Retracts to (6, 6).
*   **Green (G) pixels**:
    *   (0, 11): Same column as R (11=11), $0 < 3$. Retracts to (2, 11).
    *   (3, 0): Same row as R (3=3), $0 < 11$. Retracts to (3, 10).
    *   (12, 11): Same column as R (11=11), $12 > 3$. Retracts to (4, 11).

**Final Positions**:
*   B: (6, 5)
*   R: (3, 11)
*   O: (6, 4), (6, 6)
*   G: (2, 11), (3, 10), (4, 11)

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

### Conceptual Digestion

1.  **CAST**:
    *   **Vents**: Blue (B) and Red (R). These are the anchor points.
    *   **Mineral Deposits**: Orange (O) associated with Blue (B), and Green (G) associated with Red (R).
    *   **Sediment**: Any mineral deposit (O or G) that does not lie on the cardinal fracture lines (same row or same column) of its respective vent.
    *   **Fracture Lines**: The row and column intersecting at the vent's position.

2.  **AFFORDANCES**:
    *   Mineral deposits on a fracture line are "compressed" (moved) to the cell immediately adjacent to the vent along that specific axis.
    *   Mineral deposits not on a fracture line (sediment) are erased.

3.  **TRANSFORMATION RULE**:
    *   Locate vent **B** (Blue). Find all **O** (Orange) cells in the same row or column as B. For each cardinal direction (North, South, East, West), if at least one **O** exists in that direction, place one **O** adjacent to B in that direction. Erase all other **O** cells.
    *   Locate vent **R** (Red). Find all **G** (Green) cells in the same row or column as R. For each cardinal direction, if at least one **G** exists in that direction, place one **G** adjacent to R in that direction. Erase all other **G** cells.
    *   All other cells become black (.).

### Applying the Rule to Test Input

**1. Analyze Vent B (Blue) at (6, 5)** [Row 6, Col 5 - 0-indexed]:
*   **North (Col 5)**: No O.
*   **South (Col 5)**: No O.
*   **West (Row 6)**: O at (6, 0). Result: Place O at (6, 4).
*   **East (Row 6)**: O at (6, 14). Result: Place O at (6, 6).

**2. Analyze Vent R (Red) at (3, 11)** [Row 3, Col 11]:
*   **North (Col 11)**: G at (0, 11). Result: Place G at (2, 11).
*   **South (Col 11)**: G at (12, 11). Result: Place G at (4, 11).
*   **West (Row 3)**: G at (3, 0). Result: Place G at (3, 10).
*   **East (Row 3)**: No G.

**3. Final Grid Construction**:
*   B remains at (6, 5).
*   R remains at (3, 11).
*   O placed at (6, 4) and (6, 6).
*   G placed at (2, 11), (4, 11), and (3, 10).
*   All other cells are black (.).

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

### Conceptual Digestion

1.  **CAST**:
    *   **Conductors**: Blue (B) and Red (R). These are the anchor points.
    *   **Musicians**: Orange (O) and Green (G).
        *   Orange (O) musicians are associated with the Blue (B) conductor.
        *   Green (G) musicians are associated with the Red (R) conductor.
    *   **Stagehands**: Any other colored pixels not on the cardinal axes of their respective conductors (though in the examples, "stagehands" are simply musicians not aligned with their conductor).

2.  **AFFORDANCES**:
    *   Musicians (O, G) can "slide" along the cardinal axes (North, South, East, West) toward their respective conductor.
    *   The target position is the cell immediately adjacent to the conductor in the direction the musician was originally located.

3.  **TRANSFORMATION RULE**:
    *   Locate conductor **B** (Blue). Find all **O** (Orange) pixels that share the same row or column as B. Move each such O to the cell adjacent to B in that direction.
    *   Locate conductor **R** (Red). Find all **G** (Green) pixels that share the same row or column as R. Move each such G to the cell adjacent to R in that direction.
    *   Any O or G pixels that are **not** on the cardinal axes of their respective conductor are "stagehands" and are removed (set to black '.').

### Applying the Rule to Test Input

**1. Identify Conductors:**
*   B is at (6, 5) [row 6, col 5] (0-indexed).
*   R is at (3, 11) [row 3, col 11].

**2. Process Blue Conductor (B) at (6, 5):**
*   Look for Orange (O) on row 6 or col 5.
*   O at (6, 0): Same row. Direction: West. New position: (6, 4).
*   O at (6, 14): Same row. Direction: East. New position: (6, 6).
*   No O on col 5.

**3. Process Red Conductor (R) at (3, 11):**
*   Look for Green (G) on row 3 or col 11.
*   G at (3, 0): Same row. Direction: West. New position: (3, 10).
*   G at (0, 11): Same col. Direction: North. New position: (2, 11).
*   G at (12, 11): Same col. Direction: South. New position: (4, 11).

**4. Final Grid Construction:**
*   B remains at (6, 5).
*   R remains at (3, 11).
*   O moves to (6, 4) and (6, 6).
*   G moves to (3, 10), (2, 11), and (4, 11).
*   All other pixels are black (.).

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

### Conceptual Digestion

1.  **CAST**:
    *   **QBs (Quarterbacks)**: `B` (Blue QB) and `R` (Red QB). These are the target destinations.
    *   **Receivers**: `O` (Orange receivers associated with Blue QB) and `G` (Green receivers associated with Red QB).
    *   **Bystanders**: Any other colored elements (though none appear in the test input, the rules state they are removed).

2.  **AFFORDANCES**:
    *   Receivers move toward their respective QB along the cardinal axis (horizontal or vertical) they currently occupy relative to the QB.
    *   They stop exactly one cell (one yard) away from the QB.
    *   If a receiver is not on the same row or column as their QB, it is treated as a bystander and removed (based on the "straight lines" and "cardinal axes" description).

3.  **TRANSFORMATION RULE**:
    *   Locate `B` (Blue QB) and `R` (Red QB).
    *   For every `O` (Orange receiver):
        *   If `O` is in the same row as `B`, move it to the cell immediately to the left or right of `B` (depending on which side it started).
        *   If `O` is in the same column as `B`, move it to the cell immediately above or below `B`.
        *   Otherwise, remove it.
    *   For every `G` (Green receiver):
        *   If `G` is in the same row as `R`, move it to the cell immediately to the left or right of `R`.
        *   If `G` is in the same column as `R`, move it to the cell immediately above or below `R`.
        *   Otherwise, remove it.
    *   All other cells become `.` (black).

### Applying the Rule to Test Input

**1. Identify QBs:**
*   Blue QB (`B`): Row 6, Col 5 (0-indexed)
*   Red QB (`R`): Row 3, Col 11 (0-indexed)

**2. Process Orange Receivers (`O`):**
*   `O` at (6, 0): Same row as `B` (Row 6). It is to the left of `B`. New position: (6, 4).
*   `O` at (6, 14): Same row as `B` (Row 6). It is to the right of `B`. New position: (6, 6).

**3. Process Green Receivers (`G`):**
*   `G` at (0, 11): Same column as `R` (Col 11). It is above `R`. New position: (2, 11).
*   `G` at (3, 0): Same row as `R` (Row 3). It is to the left of `R`. New position: (3, 10).
*   `G` at (12, 11): Same column as `R` (Col 11). It is below `R`. New position: (4, 11).

**4. Final Grid Construction:**
*   (6, 5) = `B`
*   (6, 4) = `O`
*   (6, 6) = `O`
*   (3, 11) = `R`
*   (2, 11) = `G`
*   (3, 10) = `G`
*   (4, 11) = `G`
*   All others = `.`

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

### Conceptual Digestion

1.  **CAST**:
    *   **Monuments**: `B` (Blue) and `R` (Red). These are the anchor points.
    *   **Decorations**: `O` (Orange) and `G` (Green).
    *   **Roads**: The cardinal lines (horizontal and vertical) extending from the monuments.
    *   **Background**: `.` (Black).

2.  **AFFORDANCES**:
    *   `O` decorations are associated with monument `B`.
    *   `G` decorations are associated with monument `R`.
    *   Decorations can be "pulled" from their distant positions on the cardinal lines to the cell immediately adjacent to their respective monument.

3.  **TRANSFORMATION RULE**:
    *   Locate monument `B` and monument `R`.
    *   For every `O` located on the same row or column as `B`:
        *   Determine the direction (North, South, East, West) from `B`.
        *   Place an `O` at the position exactly 1 cell away from `B` in that direction.
    *   For every `G` located on the same row or column as `R`:
        *   Determine the direction (North, South, East, West) from `R`.
        *   Place a `G` at the position exactly 1 cell away from `R` in that direction.
    *   Remove all original decorations from their starting positions.
    *   Keep the monuments `B` and `R` in their original positions.
    *   All other cells become `.`.

### Applying the Rule to Test Input

1.  **Identify Monuments**:
    *   `B` is at (6, 5) [row 6, col 5] (0-indexed).
    *   `R` is at (3, 11) [row 3, col 11].

2.  **Process `O` (associated with `B` at 6, 5)**:
    *   `O` at (6, 0): Same row as `B`. Direction: West. New position: (6, 4).
    *   `O` at (6, 14): Same row as `B`. Direction: East. New position: (6, 6).

3.  **Process `G` (associated with `R` at 3, 11)**:
    *   `G` at (0, 11): Same column as `R`. Direction: North. New position: (2, 11).
    *   `G` at (3, 0): Same row as `R`. Direction: West. New position: (3, 10).
    *   `G` at (12, 11): Same column as `R`. Direction: South. New position: (4, 11).

4.  **Final Grid Construction**:
    *   (6, 5) = `B`
    *   (3, 11) = `R`
    *   (6, 4) = `O`
    *   (6, 6) = `O`
    *   (2, 11) = `G`
    *   (3, 10) = `G`
    *   (4, 11) = `G`
    *   All others = `.`

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

### Conceptual Digestion

1.  **CAST**:
    *   **Banners**: `B` (Blue) and `R` (Red). These are the anchor points (generals).
    *   **Scouts**: `O` (Orange) are scouts for `B`. `G` (Green) are scouts for `R`.
    *   **Unaffiliated**: Any other colored figures (though none appear in the test input besides B, R, O, G).
    *   **Background**: `.` (Black).

2.  **AFFORDANCES**:
    *   Scouts are positioned on the same row or column (axes) as their respective banner.
    *   The transformation "collapses" these scouts to the position immediately adjacent to the banner, maintaining the direction (North, South, East, West) they were originally in.

3.  **TRANSFORMATION RULE**:
    *   Locate `B` (Blue banner). Find all `O` (Orange scouts) on the same row or column. Move each `O` to the cell adjacent to `B` in that direction.
    *   Locate `R` (Red banner). Find all `G` (Green scouts) on the same row or column. Move each `G` to the cell adjacent to `R` in that direction.
    *   Remove any figures that are not `B`, `R`, or the correctly aligned scouts.

### Applying the Rule to Test Input

**1. Analyze Blue Banner (B) at (6, 5):** (Row 6, Col 5 - 0-indexed)
*   Scout `O` at (6, 0): Same row, to the left. Move to (6, 4).
*   Scout `O` at (6, 14): Same row, to the right. Move to (6, 6).
*   No `O` scouts on column 5.

**2. Analyze Red Banner (R) at (3, 11):** (Row 3, Col 11)
*   Scout `G` at (0, 11): Same column, above. Move to (2, 11).
*   Scout `G` at (3, 0): Same row, to the left. Move to (3, 10).
*   Scout `G` at (12, 11): Same column, below. Move to (4, 11).
*   No `G` scouts on row 3 to the right.

**3. Final Grid Construction:**
*   B at (6, 5)
*   O at (6, 4) and (6, 6)
*   R at (3, 11)
*   G at (2, 11), (3, 10), and (4, 11)
*   All other cells are `.`

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

### Conceptual Digestion

1.  **CAST**:
    *   **Trees**: `B` (Blue) and `R` (Red). These are the anchor points.
    *   **Growths**: `O` (Orange) and `G` (Green).
        *   `O` is associated with `B`.
        *   `G` is associated with `R`.
    *   **Weeds**: Any `O` or `G` that is not on the cardinal axes (North, South, East, West) of its respective tree.
    *   **Soil**: `.` (Black).

2.  **AFFORDANCES**:
    *   Growths on the same row or column as their tree are "pulled" toward the tree.
    *   The "pull" collapses the growth to the cell immediately adjacent (distance of 1) to the tree, maintaining the original direction (relative axis).
    *   Growths not on the cardinal axes of their tree are treated as weeds and removed.

3.  **TRANSFORMATION RULE**:
    *   Locate `B` (Blue tree) and `R` (Red tree).
    *   For every `O` (Orange growth):
        *   If `O` is in the same row as `B`, move it to the cell immediately left or right of `B` based on its original position.
        *   If `O` is in the same column as `B`, move it to the cell immediately above or below `B` based on its original position.
        *   Otherwise, remove it.
    *   For every `G` (Green growth):
        *   If `G` is in the same row as `R`, move it to the cell immediately left or right of `R` based on its original position.
        *   If `G` is in the same column as `R`, move it to the cell immediately above or below `R` based on its original position.
        *   Otherwise, remove it.
    *   All other cells remain `.`.

### Applying Rule to Test Input

*   **Tree B** is at (6, 5) [row 6, col 5].
*   **Tree R** is at (3, 11) [row 3, col 11].

**Processing Orange (O) for Tree B (6, 5):**
*   `O` at (6, 0): Same row. It is to the left of `B`. Move to (6, 4).
*   `O` at (6, 14): Same row. It is to the right of `B`. Move to (6, 6).

**Processing Green (G) for Tree R (3, 11):**
*   `G` at (0, 11): Same column. It is above `R`. Move to (2, 11).
*   `G` at (3, 0): Same row. It is to the left of `R`. Move to (3, 10).
*   `G` at (12, 11): Same column. It is below `R`. Move to (4, 11).

**Final Positions:**
*   `B` at (6, 5)
*   `O` at (6, 4)
*   `O` at (6, 6)
*   `R` at (3, 11)
*   `G` at (2, 11)
*   `G` at (3, 10)
*   `G` at (4, 11)

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

### Conceptual Digestion

1.  **CAST**:
    *   **Orange Lake (`O`)**: The background/environment. It remains unchanged.
    *   **Magenta Clay (`M`)**: Sculptures. These are connected components of `M` cells. They are the targets of the transformation.
    *   **Dye Droplets (`G`, `Y`, `B`, `A`, etc.)**: Single cells of colors other than `O` or `M`. These are the agents of change.
    *   **Other colors (`X`, etc.)**: Based on the description, these are treated as part of the environment or potential dyes.

2.  **AFFORDANCES**:
    *   A **Dye Droplet** can "touch" (be orthogonally adjacent to) a **Magenta Sculpture**.
    *   If a droplet touches a sculpture, the entire connected component of `M` cells is recolored to the droplet's color.
    *   The droplet itself always disappears (becomes `O`).
    *   If a droplet touches nothing (or only `O`), it disappears.
    *   If a sculpture is not touched by any droplet, it remains `M`.

3.  **TRANSFORMATION RULE**:
    1.  Identify all connected components of `M` (magenta sculptures).
    2.  Identify all single-cell "dye" droplets (any color that is not `O` or `M`).
    3.  For each dye droplet:
        *   Check its 4-neighbors (up, down, left, right).
        *   If a neighbor is part of a magenta sculpture, that entire sculpture changes to the dye's color.
        *   If a sculpture is touched by multiple different dyes, the rule doesn't explicitly state priority, but typically the first encountered or a specific order applies. However, in these puzzles, usually, one sculpture is touched by one dye.
    4.  Remove all dye droplets (replace with `O`).
    5.  The background `O` remains `O`.

---

### Applying the Rule to Test Input

**Input Grid:**
Row 0: Y O M O O O X
Row 1: M M M O O O O
Row 2: O O M O B O O
Row 3: O O O O M M O
Row 4: O O O O M M O
Row 5: M M A O O O O
Row 6: O M O O O O O

**Step 1: Identify Magenta Sculptures (Connected Components of M)**
*   **Sculpture 1**: (0,2), (1,0), (1,1), (1,2), (2,2)
*   **Sculpture 2**: (3,4), (3,5), (4,4), (4,5)
*   **Sculpture 3**: (5,0), (5,1), (6,1)

**Step 2: Identify Dye Droplets**
*   (0,0): Yellow (`Y`)
*   (0,6): Grey (`X`)
*   (2,4): Blue (`B`)
*   (5,2): Azure (`A`)

**Step 3: Check for Adjacency**
*   **Yellow (0,0)**: Neighbors are (0,1) `O` and (1,0) `M`. It touches **Sculpture 1**.
    *   Sculpture 1 becomes Yellow (`Y`).
*   **Grey (0,6)**: Neighbors are (0,5) `O` and (1,6) `O`. Touches nothing.
    *   Dissolves.
*   **Blue (2,4)**: Neighbors are (1,4) `O`, (3,4) `M`, (2,3) `O`, (2,5) `O`. It touches **Sculpture 2**.
    *   Sculpture 2 becomes Blue (`B`).
*   **Azure (5,2)**: Neighbors are (4,2) `O`, (6,2) `O`, (5,1) `M`, (5,3) `O`. It touches **Sculpture 3**.
    *   Sculpture 3 becomes Azure (`A`).

**Step 4: Final Assembly**
*   Background `O` stays `O`.
*   Sculpture 1 $\rightarrow$ `Y`
*   Sculpture 2 $\rightarrow$ `B`
*   Sculpture 3 $\rightarrow$ `A`
*   All dye droplets $\rightarrow$ `O`

**Mapping to Grid:**
(0,0): Y $\rightarrow$ O
(0,1): O $\rightarrow$ O
(0,2): M $\rightarrow$ Y
(0,3): O $\rightarrow$ O
(0,4): O $\rightarrow$ O
(0,5): O $\rightarrow$ O
(0,6): X $\rightarrow$ O

(1,0): M $\rightarrow$ Y
(1,1): M $\rightarrow$ Y
(1,2): M $\rightarrow$ Y
(1,3): O $\rightarrow$ O
(1,4): O $\rightarrow$ O
(1,5): O $\rightarrow$ O
(1,6): O $\rightarrow$ O

(2,0): O $\rightarrow$ O
(2,1): O $\rightarrow$ O
(2,2): M $\rightarrow$ Y
(2,3): O $\rightarrow$ O
(2,4): B $\rightarrow$ O
(2,5): O $\rightarrow$ O
(2,6): O $\rightarrow$ O

(3,0): O $\rightarrow$ O
(3,1): O $\rightarrow$ O
(3,2): O $\rightarrow$ O
(3,3): O $\rightarrow$ O
(3,4): M $\rightarrow$ B
(3,5): M $\rightarrow$ B
(3,6): O $\rightarrow$ O

(4,0): O $\rightarrow$ O
(4,1): O $\rightarrow$ O
(4,2): O $\rightarrow$ O
(4,3): O $\rightarrow$ O
(4,4): M $\rightarrow$ B
(4,5): M $\rightarrow$ B
(4,6): O $\rightarrow$ O

(5,0): M $\rightarrow$ A
(5,1): M $\rightarrow$ A
(5,2): A $\rightarrow$ O
(5,3): O $\rightarrow$ O
(5,4): O $\rightarrow$ O
(5,5): O $\rightarrow$ O
(5,6): O $\rightarrow$ O

(6,0): O $\rightarrow$ O
(6,1): M $\rightarrow$ A
(6,2): O $\rightarrow$ O
(6,3): O $\rightarrow$ O
(6,4): O $\rightarrow$ O
(6,5): O $\rightarrow$ O
(6,6): O $\rightarrow$ O

ANSWER:
O O Y O O O O
Y Y Y O O O O
O O Y O O O O
O O O O B B O
O O O O B B O
A A O O O O O
O A O O O O O

### Conceptual Digestion

**1. Identify the CAST:**
*   **Soil (Background):** The most frequent color in the grid (in the examples, `O` acts as the background/soil).
*   **Flower Clusters (Blooms):** Connected groups of `M` (magenta) cells.
*   **Bees (Pollen Carriers):** Single cells of colors other than `M` or the background color (`O`). In the test input, these are `Y` (yellow), `B` (blue), `A` (azure), and `X` (grey).

**2. Identify the AFFORDANCES:**
*   **Bees** can "touch" a flower cluster if they are orthogonally adjacent to any cell of that cluster.
*   **Flower Clusters** can be "pollinated," meaning all connected `M` cells change to the color of the bee that touched them.
*   **Bees** disappear after the transformation.
*   **Unpollinated Flowers** (those not touched by a bee) remain `M`.
*   **Bees** that don't touch a flower simply disappear.

**3. Identify the TRANSFORMATION RULE:**
1.  Identify all connected components of `M` (magenta).
2.  Identify all "bee" cells (any color that is not `M` or the background `O`).
3.  For each bee, check if it is adjacent (up, down, left, right) to any cell of a magenta cluster.
4.  If a bee is adjacent to a cluster, change all cells in that specific connected cluster to the bee's color.
5.  Remove all bee cells (change them to the background color `O`).
6.  Any `M` cells not adjacent to a bee remain `M`.

---

### Applying the Rule to Test Input

**Input Grid:**
Row 0: Y O M O O O X
Row 1: M M M O O O O
Row 2: O O M O B O O
Row 3: O O O O M M O
Row 4: O O O O M M O
Row 5: M M A O O O O
Row 6: O M O O O O O

**Step 1: Identify Magenta Clusters**
*   Cluster 1: (0,2), (1,0), (1,1), (1,2), (2,2)
*   Cluster 2: (3,4), (3,5), (4,4), (4,5)
*   Cluster 3: (5,0), (5,1), (6,1)

**Step 2: Identify Bees**
*   Bee 1: (0,0) - Yellow (Y)
*   Bee 2: (0,6) - Grey (X)
*   Bee 3: (2,4) - Blue (B)
*   Bee 4: (5,2) - Azure (A)

**Step 3: Check Adjacency and Pollinate**
*   **Bee 1 (Y) at (0,0):** Adjacent to (0,1) [O] and (1,0) [M]. (1,0) is part of Cluster 1.
    *   *Result:* Cluster 1 becomes Yellow (Y).
*   **Bee 2 (X) at (0,6):** Adjacent to (0,5) [O] and (1,6) [O].
    *   *Result:* No flower touched.
*   **Bee 3 (B) at (2,4):** Adjacent to (1,4) [O], (3,4) [M], (2,3) [O], (2,5) [O]. (3,4) is part of Cluster 2.
    *   *Result:* Cluster 2 becomes Blue (B).
*   **Bee 4 (A) at (5,2):** Adjacent to (4,2) [O], (6,2) [O], (5,1) [M], (5,3) [O]. (5,1) is part of Cluster 3.
    *   *Result:* Cluster 3 becomes Azure (A).

**Step 4: Final Grid Construction**
*   Background `O` remains `O`.
*   Bees (0,0), (0,6), (2,4), (5,2) become `O`.
*   Cluster 1 $\rightarrow$ Y: (0,2), (1,0), (1,1), (1,2), (2,2)
*   Cluster 2 $\rightarrow$ B: (3,4), (3,5), (4,4), (4,5)
*   Cluster 3 $\rightarrow$ A: (5,0), (5,1), (6,1)

**Final Grid Mapping:**
Row 0: O O Y O O O O
Row 1: Y Y Y O O O O
Row 2: O O Y O O O O
Row 3: O O O O B B O
Row 4: O O O O B B O
Row 5: A A O O O O O
Row 6: O A O O O O O

ANSWER:
O O Y O O O O
Y Y Y O O O O
O O Y O O O O
O O O O B B O
O O O O B B O
A A O O O O O
O A O O O O O