MARC2 Variant Comparison March 26, 2026

We need to understand the puzzle. The description: Two fragments: larger fragment (dominant sibling) is nearly complete but with holes (like a key missing its teeth). Smaller fragment is just a few bright cells, shaped exactly like what the bigger one is missing. They are two halves of the same breath. The smaller fragment hears the larger one calling and swims into its wounds. Every gap in the bigger sibling is a socket waiting for the smaller one's bones. Nest them together: the larger fragment keeps its color wherever it was whole, and the smaller fragment fills every void with its own color, like mortar between bricks. They fuse into one solid body, no darkness left between them. Grid details: The fused body is a compact 3×3 tile — fully healed, no black scars remaining. Every cell glows with one of the two original colors: the dominant sibling's color where it was strong, the smaller sibling's color where it was needed.

Thus the transformation: Input grid is presumably a larger grid (maybe 10x10) containing two fragments: a larger shape (dominant color) with holes (black cells '.'), and a smaller shape (other color) that exactly matches the holes of the larger shape. The output is a 3x3 tile that is the fused shape: the larger shape's non-black cells retain its color; the holes (black cells) are filled with the smaller shape's color. The resulting fused shape is a compact 3x3 tile (so the shape of the larger fragment plus holes must be exactly a 3x3 region? Or after fusion, the shape becomes a solid 3x3 block (no holes) of size 3x3). The output is just the 3x3 tile (maybe cropped to the bounding box of the fused shape). The training example: Input grid 10x10? Let's examine.

Example 1 Input:
Row0: . . . . . . . . Y .
Row1: . . . . . . . . Y Y
Row2: . . . M M M . . . .
Row3: . . . . M M . . . .
Row4: . . . . . M . . . .
Row5: . . . . . . . . . .
Row6: . . . . . . . . . .
Row7: . . . . . . . . . .
Row8: . . . . . . . . . .
Row9: . . . . . . . . . .

So we see a shape of M (magenta) cells: at rows 2-4 columns 2-4? Let's map:

Row2: col2 M, col3 M, col4 M
Row3: col3 M, col4 M
Row4: col4 M

Thus shape like a right triangle? Actually it's like a descending staircase: positions (2,2),(2,3),(2,4),(3,3),(3,4),(4,4). That's a shape of 6 cells.

There are also Y (yellow) cells: at row0 col8 Y, row1 col8 Y, row1 col9 Y.

Thus Y cells: three cells forming an L shape? Actually positions: (0,8),(1,8),(1,9). That's like a corner.

Now the description: larger fragment is nearly complete but with holes; smaller fragment is shaped exactly like what the bigger one is missing. So which is larger? The M shape seems larger (6 cells) vs Y shape (3 cells). The M shape has holes? Let's see if we consider the M shape as the larger fragment with holes (black cells) that need to be filled by Y shape. The M shape currently has no black cells inside its shape? Actually the M shape as given is contiguous but not a full rectangle; it's like a shape missing some cells to become a 3x3 block? Let's see: If we consider a 3x3 block bounding box of the M shape: rows 2-4, cols 2-4. That's a 3x3 region. Let's list cells in that 3x3 region:

Rows 2-4, cols 2-4:

Row2: col2 M, col3 M, col4 M -> all M
Row3: col2 ., col3 M, col4 M -> col2 is '.' (black)
Row4: col2 ., col3 ., col4 M -> col2 '.'; col3 '.'; col4 M

Thus the M shape within that 3x3 bounding box has holes at (3,2), (4,2), (4,3). That's three holes. The Y shape has exactly three cells: (0,8),(1,8),(1,9). But those are not positioned relative to the M shape's bounding box; they are elsewhere in the grid. However maybe the Y shape is positioned such that if we overlay it onto the M shape's bounding box aligned at some offset, it fills the holes exactly. Let's see: The Y shape cells: (0,8),(1,8),(1,9). If we shift them to align with the M shape's bounding box (rows 2-4, cols 2-4), we need to find translation that maps Y shape onto the hole positions. The hole positions are (3,2),(4,2),(4,3). Let's compute offset: Suppose we take Y shape and subtract some offset to align with holes. For each Y cell, we need to map to a hole. Let's see if there's a translation (dx,dy) such that Y cells + (dx,dy) = hole cells. Choose one mapping: Map Y cell (0,8) to hole (3,2): then dx = 3-0 = 3, dy = 2-8 = -6. Apply to other Y cells: (1,8)+(3,-6) = (4,2) -> matches hole (4,2). Good. (1,9)+(3,-6) = (4,3) -> matches hole (4,3). So indeed the Y shape translated by (+3,-6) exactly fills the holes. So the Y shape is positioned elsewhere in the grid, but if we shift it by that offset, it aligns to fill the holes of the M shape's 3x3 bounding box.

Thus the transformation: Identify the larger fragment (dominant color) which is a shape with holes (black cells) inside its bounding box (maybe the shape is not necessarily convex but has missing cells). The smaller fragment (other color) is a shape that exactly matches the hole pattern of the larger fragment when placed at some offset (i.e., 

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We need to understand the transformation. The description: "Two signal patterns in different colors are laid out across a dark circuit board. One pattern is more extensive; the other is smaller, its traces fitting precisely into the unused pads of the first. What to do: Route both patterns onto a single compact chip. The primary signal claims its pads, and the secondary signal fills every remaining connection point, leaving no unconnected node on the die. Grid details: A small, fully-routed chip — every pad carrying one signal color or the other, no floating nodes."

Interpretation: Input grid has two colored patterns (maybe two colors) on a black background (dots). One pattern is larger (primary) and the other is smaller (secondary) that fits into the unused pads of the first. The output is a small fully-routed chip: a compact grid where each cell is either primary or secondary color, no black (dots). Essentially we need to extract the bounding box of the union of the two patterns? Or we need to overlay them and then crop to minimal bounding rectangle that contains all non-black cells, and then fill any remaining black cells inside that rectangle with the secondary color? Wait: The description: "primary signal claims its pads, and the secondary signal fills every remaining connection point, leaving no unconnected node on the die." So after routing both patterns onto a single compact chip, the primary signal occupies its original pads (maybe unchanged), and the secondary signal fills every remaining connection point (i.e., any empty pads within the chip area) with its color. So output is a rectangle (maybe minimal bounding box of primary pattern?) where primary pattern stays as is, and any empty spots inside that rectangle become secondary color. The secondary pattern originally may be placed somewhere else, but its traces fit into unused pads of the primary pattern. So the secondary pattern's cells are a subset of the empty spots within the primary pattern's bounding box? Actually "its traces fitting precisely into the unused pads of the first." Means the secondary pattern's cells are located in positions that are empty (black) in the primary pattern's area. So the union of primary and secondary patterns yields a shape with no holes? The secondary fills all unused pads inside the primary's bounding box? Or maybe the secondary pattern exactly matches the empty spots within the primary pattern's bounding box, making the union a solid rectangle of primary+secondary colors (no black). Then output is that rectangle, with primary color where primary originally was, secondary color where secondary originally was (and also any other empty spots? but there are none because secondary fills all unused pads). So output is just the bounding box of the primary pattern (or maybe the union) with primary color where primary existed, secondary color elsewhere (since secondary fills all remaining pads). The training example can help verify.

Example 1 Input:
Let's write it as grid with coordinates. Input lines:

Row0: . . . . . . . . Y .
Row1: . . . . . . . . Y Y
Row2: . . . M M M . . . .
Row3: . . . . M M . . . .
Row4: . . . . . M . . . .
Row5: . . . . . . . . . .
Row6: . . . . . . . . . .
Row7: . . . . . . . . . .
Row8: . . . . . . . . . .
Row9: . . . . . . . . . .

So we have Y's at (0,8)? Actually column index? Let's define rows 0-9, columns 0-9. Row0: col8 = Y, col9 = .? Actually row0: ". . . . . . . . Y ." => col0-7 ., col8 Y, col9 . . Row1: ". . . . . . . . Y Y" => col8 Y, col9 Y. Row2: ". . . M M M . . . ." => col3 M, col4 M, col5 M. Row3: ". . . . M M . . . ." => col4 M, col5 M. Row4: ". . . . . M . . . ." => col5 M.

Thus primary pattern maybe M (magenta) is larger? Secondary pattern Y (yellow) is smaller and fits into unused pads of M? Let's see M shape: It appears like a kind of L shape? Actually M's: at (2,3),(2,4),(2,5); (3,4),(3,5); (4,5). That's like a vertical line at col5 rows2-4 plus a horizontal at row2 cols3-5. So shape like a "Γ" (gamma) shape.

Yellow Y's: at (0,8),(1,8),(1,9). That's like an L shape too but located far away (top right). Does it fit into unused pads of M? The M shape's bounding box: rows 2-4, cols 3-5. Unused pads inside that bounding box: cells that are black within rows2-4, cols3-5. Let's list those cells:

Rows2-4, cols3-5:
Row2: col3 M, col4 M, col5 M => all M.
Row3: col3 ., col4 M, col5 M => col3 empty.
Row4: col3 ., col4 ., col5 M => col3, col4 empty.

So empty pads: (3,3), (4,3), (4,4). That's three cells. Yellow pattern has three cells: (0,8),(1,8),(1,9). Not matching those coordinates. So maybe the secondary pattern is placed elsewhere but its shape matches the empty pads shape (i.e., same shape as the empty pads within primary's bounding box). Indeed the empty pads shape is: coordinates relative to some origin: (0,0) maybe at (3,3) as top-left? Let's compute shape: empty pads: (3,3), (4,3), (4,4). That's like an L shape: a vertical of length

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We need to understand the transformation. The description: "Two dancers in contrasting colors are positioned far apart on a dark ballroom floor. One dancer covers more ground; the other is more compact, their movements complementing the spaces the first avoids. What to do: Both dancers converge onto a small dance floor. The lead claims their positions, and the partner fills every remaining tile with their own steps, until the entire floor bears the mark of one dancer or the other. Grid details: A compact dance floor — every tile touched by one dancer's color or the other, no unmarked space."

Interpretation: Input grid has two colors (maybe more?) representing two dancers? The background is black (.) representing dark floor. The two dancers are in contrasting colors (maybe two non-black colors). One dancer covers more ground (i.e., appears more often? Or larger region?), the other is more compact (smaller region). Their movements complement the spaces the first avoids: meaning the second dancer occupies the spaces not occupied by the first? Or they avoid each other's spaces? Then they converge onto a small dance floor: The lead claims their positions, and the partner fills every remaining tile with their own steps, until the entire floor bears the mark of one dancer or the other. So output is a small grid (compact dance floor) where each tile is either the lead's color or the partner's color, covering all tiles (no black). The lead claims their positions (so lead's positions remain as they were? maybe they keep their color), and partner fills every remaining tile with their own steps (so any tile not occupied by lead becomes partner's color). So output is essentially: take the positions of the lead dancer (maybe the one that covers more ground? Or the one that is lead?), and fill the rest with the other dancer's color. The output grid is a compact dance floor: maybe we crop to the bounding box of the two dancers? Or we take the minimal rectangle that contains all non-black cells? Let's test with example.

Example 1 input:

Row0: . . . . . . . . Y .
Row1: . . . . . . . . Y Y
Row2: . . . M M M . . . .
Row3: . . . . M M . . . .
Row4: . . . . . M . . . .
Row5: . . . . . . . . . .
Row6: . . . . . . . . . .
Row7: . . . . . . . . . .
Row8: . . . . . . . . . .
Row9: . . . . . . . . . .

So we have Y's at (0,8), (1,8), (1,9). M's at (2,3),(2,4),(2,5),(3,4),(3,5),(4,5). So Y appears 3 times, M appears 6 times. So M covers more ground (more tiles). Y is more compact (fewer tiles). The description: "One dancer covers more ground; the other is more compact, their movements complementing the spaces the first avoids." So maybe the lead is the one that covers more ground? Or the lead claims their positions, partner fills remaining. In output we have:

Output:
M M M
Y M M
Y Y M

That's a 3x3 grid. Let's map coordinates: output rows 0-2, cols 0-2.

Row0: M M M
Row1: Y M M
Row2: Y Y M

So we see M appears in many positions, Y appears in some.

If we think of lead = M (covers more ground). Lead claims their positions: So M's positions in output correspond to where M was in input? But input M positions are scattered; output is compact 3x3. Perhaps we take the bounding box of all non-black cells? Let's compute bounding box of Y and M in input.

Y positions: min row 0, max row 1; min col 8, max col 9.
M positions: min row 2, max row 4; min col 3, max col 5.

Combined bounding box: rows 0-4, cols 3-9? Actually min row overall = 0 (Y), max row = 4 (M). min col = 3 (M), max col = 9 (Y). That's a 5 rows (0-4) by 7 cols (3-9) = 5x7 region. Not 3x3.

Maybe we take the intersection of the two dancers' areas? Or we take the region where they are close? Not.

Maybe the transformation is: find the two colors present (excluding black). Determine which color appears more frequently (the "lead") and which appears less (the "partner"). Then output a grid that is the size of the bounding box of the less frequent color? Or something like that.

Let's examine output size: 3x3. Where does 3 come from? Perhaps it's the dimensions of the bounding box of the less frequent color (Y). Y bounding box: rows 0-1 (height 2), cols 8-9 (width 2) => 2x2. Not 3x3.

Maybe it's the bounding box of the more frequent color (M): rows 2-4 (height 3), cols 3-5 (width 3) => 3x3. Indeed M's bounding box is 3x3! Output is exactly that size. And the output pattern: M M M / Y M M / Y Y M. Let's see if that matches M's bounding box region from input, but with some Y's placed in certain positions.

Input M region (rows 2-4, cols 3-5):
Row2: cols3-5: M M M
Row3: cols3-5: . M M (since row3: . . . . M M . . . . => at col3? Actually row3: indices: 0 .,1 .,2 .,3 .,4 M,5 M,6 .,7 .,8 .,9 . => Wait row3: ". . . . M M . . . ." So col0-3 are ., col4 M, col5 M, col6-9 . So within cols3-5: col3 ., col4 M, col5 M => ". M M"
Row4: ". . . . . M . . . ." => col0-4 ., col5 M, col6-9 . => within cols3-5: col3 ., col4 ., col5 M => ". . M"

So the M region in inpu

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We need to understand the transformation. The description: "Two mineral deposits of different colors are exposed across a dark rock face. One deposit is more extensive; the other is smaller, its crystals growing precisely in the voids the larger deposit leaves. What to do: Both veins consolidate into a single compact outcrop. The dominant mineral fills its natural fractures, and the secondary mineral crystallizes in every remaining gap, producing a solid face with no empty rock. Grid details: A small, solid outcrop cross-section — every cell filled with one mineral color or the other, no bare matrix rock."

Interpretation: Input grid has some cells with '.' (black) representing empty rock? Actually '.' is black, but they said "no bare matrix rock" in output. Input may have '.' as background (empty). The two mineral deposits are of different colors: one more extensive (dominant) and one smaller (secondary). The smaller deposit's crystals grow precisely in the voids the larger deposit leaves. So the smaller deposit fills the holes (gaps) of the larger deposit. Then both veins consolidate into a single compact outcrop: the dominant mineral fills its natural fractures, and the secondary mineral crystallizes in every remaining gap, producing a solid face with no empty rock.

Thus output is a small solid block (maybe a rectangle) where all cells are filled with either dominant or secondary color, no '.'.

The training example: Input is a 10x10 grid? Let's see.

Example 1 Input:
Row0: . . . . . . . . Y .
Row1: . . . . . . . . Y Y
Row2: . . . M M M . . . .
Row3: . . . . M M . . . .
Row4: . . . . . M . . . .
Row5: . . . . . . . . . .
Row6: . . . . . . . . . .
Row7: . . . . . . . . . .
Row8: . . . . . . . . . .
Row9: . . . . . . . . . .

So we have Y's at (0,8), (1,8), (1,9). M's at (2,3),(2,4),(2,5),(3,4),(3,5),(4,5). So Y deposit is small (3 cells) and M deposit is larger (6 cells). The description: "One deposit is more extensive; the other is smaller, its crystals growing precisely in the voids the larger deposit leaves." So the smaller deposit (Y) grows in voids of larger deposit (M). The voids of M deposit? Let's see shape of M deposit: It seems like a kind of L shape? Actually coordinates: (2,3),(2,4),(2,5) horizontal line at row2 col3-5; then (3,4),(3,5) below middle two; then (4,5) below rightmost. So shape like a reversed L? The voids (holes) inside the bounding box? Let's compute bounding box of M: rows 2-4, cols 3-5. That's a 3x3 block. Within that block, M occupies: (2,3),(2,4),(2,5) top row all three; (3,4),(3,5) middle row missing left; (4,5) bottom row missing left two. So the voids (empty) inside that block are: (3,3), (4,3), (4,4). That's three cells. Indeed Y deposit occupies exactly those? Let's see Y positions: (0,8),(1,8),(1,9) not inside that block. Wait Y's are far away at column 8-9. So not inside. Hmm.

Maybe the voids refer to the background '.' around the larger deposit? The smaller deposit grows in the voids (gaps) left by the larger deposit when considering its shape? Not clear.

But output is:
M M M
Y M M
Y Y M

That's a 3x3 grid. Let's map: rows 0-2, cols 0-2.

Output:
(0,0)=M, (0,1)=M, (0,2)=M
(1,0)=Y, (1,1)=M, (1,2)=M
(2,0)=Y, (2,1)=Y, (2,2)=M

So pattern: It's like the M deposit shape (the larger) but filled to a solid 3x3 block where M occupies top row and right column? Actually M appears at positions: (0,0-2) top row all M; (1,1-2) middle row right two; (2,2) bottom row rightmost. That's exactly the shape of the M deposit but shifted to origin? Let's compare: Original M shape coordinates relative to its bounding box (min row=2, min col=3). Subtract (2,3): we get:
(2,3)->(0,0)
(2,4)->(0,1)
(2,5)->(0,2)
(3,4)->(1,1)
(3,5)->(1,2)
(4,5)->(2,2)

That's exactly the output M positions! Yes! So output M positions correspond to the shape of the larger deposit (M) normalized to origin (top-left of its bounding box). And the Y positions in output are the remaining cells of the 3x3 block (the bounding box of the larger deposit) that are not occupied by M shape. Those are filled with the secondary mineral (Y). Indeed the bounding box of M is 3x3 (rows 2-4, cols 3-5). The output is exactly that bounding box, with M shape as given, and the other cells filled with Y.

Thus transformation: Identify the two colors present (excluding '.'). Determine which color is more extensive (has more cells). That's the dominant mineral. Compute its bounding box (min row, max row, min col, max col). Then output a grid of size (height = maxRow-minRow+1, width = maxCol-minCol+1). For each cell in that bounding box, if the input at that position (original coordinates) has the dominant color, output dominant color; else output the secondary color (the other color). Essentially we "crop" to the bounding box of the dominant color, and fill any missing spots inside that box with the secondary color.

Check: In example, dominant = M (6 cells) vs Y (3 cells). Bounding box of M: rows 2-4, cols 3-5 =

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We need to understand the puzzle. The description: "Two ingredients of different colors sit apart on a large dark countertop. One ingredient spreads more generously; the other is a smaller portion that seems shaped to complement the first. What to do: Place both ingredients into the same small baking tin. The larger portion settles where it naturally fits, and the smaller portion flows into every gap the first one left behind, until the tin is completely full. Grid details: A filled baking tin — no empty space, just two colors of batter sharing every cell of the mold."

Interpretation: Input grid is a large dark countertop (black cells '.'), with two colored shapes (ingredients) placed apart. One ingredient spreads more generously (i.e., larger area), the other is smaller portion shaped to complement the first. The output is a small baking tin (some subgrid?) where the larger portion settles where it naturally fits, and the smaller portion fills gaps left by the larger portion, resulting in a fully filled tin (no '.' cells). The output grid is presumably a small region (maybe bounding box of the two ingredients combined?) where they are placed together, with the larger ingredient occupying its shape (maybe unchanged) and the smaller ingredient filling the remaining cells.

Thus transformation: Given input grid with black background '.' and two colored non-black shapes (maybe each shape is a contiguous region of a single color? Or could be multiple colors? The description says two ingredients of different colors. So there are exactly two colors present besides '.'? In examples, we see Y and M in example 1. Input: Y's and M's. Output: a 3x3 grid with M and Y. Let's examine.

Example 1 Input:
Row0: . . . . . . . . Y .
Row1: . . . . . . . . Y Y
Row2: . . . M M M . . . .
Row3: . . . . M M . . . .
Row4: . . . . . M . . . .
Row5: . . . . . . . . . .
Row6: . . . . . . . . . .
Row7: . . . . . . . . . .
Row8: . . . . . . . . . .
Row9: . . . . . . . . . .

So we have Y's at (0,8), (1,8), (1,9). M's at (2,3),(2,4),(2,5),(3,4),(3,5),(4,5). So Y shape is small (3 cells) maybe L shape? Actually Y's: positions: (0,8), (1,8), (1,9). That's like a small L shape (2 vertical then one to right at bottom). M shape: a larger shape maybe like a kind of shape: It occupies a region: rows 2-4, columns 3-5, but missing some cells: Let's map:

Row2: columns 3,4,5 = M M M (full)
Row3: columns 4,5 = M M (missing column3)
Row4: column5 = M (missing columns3,4)

So shape is like a right triangle? Actually it's like a shape that is like a staircase descending to the right? Let's draw coordinates (r,c):
(2,3),(2,4),(2,5)
(3,4),(3,5)
(4,5)

That's shape of a right triangle with hypotenuse from (2,3) to (4,5) inclusive? Actually it's like a shape that fills cells where c >= r-? Let's see: For each row r, the M's start at column =? Row2: start col3, end col5 (3 cells). Row3: start col4, end col5 (2 cells). Row4: start col5, end col5 (1 cell). So shape is like a right-aligned triangle of width decreasing by 1 each row.

Y shape: (0,8),(1,8),(1,9). That's like a small L shape: vertical two cells at column8 rows0-1, plus one cell to the right at (1,9). So shape is like a 2x2 block missing bottom-left? Actually if we consider a 2x2 block at rows0-1, cols8-9: cells would be (0,8),(0,9),(1,8),(1,9). Present are (0,8),(1,8),(1,9) missing (0,9). So it's like an L shape.

Output:
M M M
Y M M
Y Y M

That's a 3x3 grid:
Row0: M M M
Row1: Y M M
Row2: Y Y M

Let's map coordinates relative to output: (0,0) M, (0,1) M, (0,2) M; (1,0) Y, (1,1) M, (1,2) M; (2,0) Y, (2,1) Y, (2,2) M.

So output is a 3x3 square fully filled with M and Y. The pattern: M occupies top row all three, plus middle row columns1-2, plus bottom row column2. Y occupies bottom-left 2x2? Actually Y's at (1,0),(2,0),(2,1). That's like an L shape similar to input Y shape but rotated? Input Y shape was vertical two at column8 rows0-1 plus a right cell at (1,9). Output Y shape is vertical two at column0 rows1-2 plus a bottom cell at (2,1). That's like the same shape but rotated 90 degrees clockwise? Let's see: Input Y shape: cells (0,8),(1,8),(1,9). If we subtract origin (0,8) we get shape: (0,0),(1,0),(1,1). That's an L shape with the corner at (1,0)? Actually shape coordinates relative to (0,8): (0,0) top-left, (1,0) below, (1,1) to right of that. So shape is like a "corner" pointing down-right.

Output Y shape relative to its origin? Let's find bounding box of Y in output: Y's at (1,0),(2,0),(2,1). If we take origin at (1,0): shape coordinates: (0,0),(1,0),(1,1). Same shape! So Y shape is preserved exactly (same orientation) but placed at different location.

M shape in output: M's at (0,0),(0,1),(0,2),(1,1),(1,2),(2,2). That's shape: coordinates relative to (0,0): (0,0),(0,1),(0,2),(1,1),(1,2),(2,2). That's same as input M shape? Input M shape coordinates relative to its top-left? Input M shape: cells (2,3),(2,4),(2,5),(3,4),(3,5),(4,5). If we subtract (2,3) we get: (0,0),(0,1),(0,2)

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We need to understand the transformation. The description: "Two squads in distinct colors are warming up far apart on a vast dark pitch. One squad is larger; the other smaller, their stances oddly complementary. Both teams converge onto a compact playing area. The larger squad takes its positions first, and the smaller squad slots into every uncovered position, until no patch of field is unoccupied. Grid details: A tight formation grid where every position is filled by one jersey color or the other — no open ground."

Interpretation: Input grid has some colored cells (non-black) and black (.) representing empty. The transformation: we need to produce a smaller grid (compact playing area) where the larger squad (some color) occupies positions first, then the smaller squad (other color) fills remaining positions. The output grid is a tight formation (no black). The training example: Input has many '.' and some Y and M. Output is a 3x3 grid: M M M / Y M M / Y Y M.

Let's examine example 1.

Input grid (10 rows? Actually they show 10 rows? Let's count: they show lines:

Row0: . . . . . . . . Y .
Row1: . . . . . . . . Y Y
Row2: . . . M M M . . . .
Row3: . . . . M M . . . .
Row4: . . . . . M . . . .
Row5: . . . . . . . . . .
Row6: . . . . . . . . . .
Row7: . . . . . . . . . .
Row8: . . . . . . . . . .
Row9: . . . . . . . . . .

So 10x10 grid. Colors: Y appears at (0,8), (1,8), (1,9). M appears at (2,3),(2,4),(2,5),(3,4),(3,5),(4,5). So Y: 3 cells. M: 6 cells. So M is larger squad (6 vs 3). Output is 3x3 grid: M M M / Y M M / Y Y M. Let's map: Output has 9 cells. Count M: positions: (0,0),(0,1),(0,2),(1,1),(1,2),(2,2) = 6 M. Y: (1,0),(2,0),(2,1) = 3 Y. So same counts as input: M=6, Y=3. So transformation just rearranged them into a compact square (3x3) with no empty cells, preserving counts, and perhaps arranging them such that larger squad (M) placed first in some pattern, then smaller squad (Y) fills remaining spots.

What pattern? Output grid: 
Row0: M M M
Row1: Y M M
Row2: Y Y M

Looks like they filled from top-left? Let's see: Larger squad (M) placed first: maybe they fill the grid in some order (e.g., row-major) with M until they run out, then fill remaining with Y. Let's test: If we fill 3x3 grid row-major with M first: we have 6 M, so first 6 cells get M: positions (0,0),(0,1),(0,2),(1,0),(1,1),(1,2) would be M. Then remaining 3 cells (2,0),(2,1),(2,2) get Y. That would give:
Row0: M M M
Row1: M M M
Row2: Y Y Y
But output is different: they have Y in (1,0) and (2,0),(2,1) and M in (1,1),(1,2),(2,2). So not simple row-major.

Maybe they fill column-major? Fill M first column-major: order (0,0),(1,0),(2,0),(0,1),(1,1),(2,1),(0,2),(1,2),(2,2). First 6 M: (0,0),(1,0),(2,0),(0,1),(1,1),(2,1) => M at those. Then Y at rest: (0,2),(1,2),(2,2). That yields:
Row0: M M Y
Row1: M M Y
Row2: M M Y
Not match.

Maybe they fill in a snaking pattern? Or they place larger squad in a shape like a "corner" or "L"? Output shows M occupying top row fully, plus two cells in middle row right side, plus bottom right cell. Y occupies left column except top? Actually Y at (1,0),(2,0),(2,1). That's like an L shape rotated.

Maybe the rule: Larger squad forms a solid block (like a rectangle) as large as possible given its count, and the smaller squad fills the remaining holes to make a square. For 6 M, the largest square that can be formed with M is 2x2=4, leaving 2 extra M. But output has M not in a solid block.

Alternatively, maybe they arrange the larger squad in a "first" pattern like they take positions of a certain shape (like a "staggered" formation) and the smaller squad fills the gaps.

Let's think about the metaphor: "Two squads in distinct colors are warming up far apart on a vast dark pitch. One squad is larger; the other smaller, their stances oddly complementary. Both teams converge onto a compact playing area. The larger squad takes its positions first, and the smaller squad slots into every uncovered position, until no patch of field is unoccupied."

Interpretation: On the vast dark pitch (input grid with many black), the two squads are far apart (clusters of colors). They converge onto a compact playing area (output grid, smaller dimensions, no black). The larger squad takes its positions first (maybe they occupy the positions that correspond to the "centroid" or some shape), and the smaller squad fills the rest.

Maybe the transformation is: compute the bounding box of each color? Then shift them to top-left? Or maybe they take the convex hull? Not sure.

Let's examine the positions of M and Y in input. M positions: (2,3),(2,4),(2,5),(3,4),(3,5),(4,5). If we plot, they form a shape like a diagonal line? Actually coordinates: row 2: columns 3-5 (three in a row). Row3: columns 4-5 (two). Row4: column5 (one). So shape like a right triangle oriented down-right? Y positions: (0,8),(1,8),(1,9). That's like a small L shape near top-right.

If we shift both clusters to origin? Perhaps we compute the minimal bounding

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We need to understand the transformation. The description: "Two weather patterns in different hues drift across a dark atmospheric void. One system covers more area; the other is smaller and seems to nestle into the exact spaces the larger one avoids. What to do: Compress both weather systems into one small observation window. The dominant system fills its natural region, and the minor system occupies every remaining pocket, leaving no void in the frame. Grid details: A compact weather map where every cell shows one system or the other — no dark emptiness, just two interlocking patterns."

Thus input grid has two colors (non-dot) representing two systems, plus background '.' (black). The output is a small grid (maybe 3x3?) that compresses both patterns into a small observation window: the dominant system fills its natural region, and the minor system occupies every remaining pocket, leaving no void.

Thus we need to find the bounding box of each color? Or maybe we need to find the minimal rectangle that contains all cells of each color? Then we need to combine them into a small window: maybe we take the union of the two patterns and then crop to the minimal bounding box that contains both? Then we fill the dominant color in its natural region (i.e., where it originally appears) and the minor color fills the remaining cells (the pockets) within that bounding box.

In examples:

Example 1 input:
Row0: . . . . . . . . Y .
Row1: . . . . . . . . Y Y
Row2: . . . M M M . . . .
Row3: . . . . M M . . . .
Row4: . . . . . M . . . .
Row5-9: all dots.

Thus we have Y at positions: (0,8), (1,8), (1,9). M at positions: (2,3),(2,4),(2,5),(3,4),(3,5),(4,5). So Y is small (3 cells) and M is larger (6 cells). Output is 3x3:
M M M
Y M M
Y Y M

Let's see mapping: The output grid seems to be a 3x3 block where M dominates (fills most cells) and Y fills the pockets.

If we take bounding box of all non-dot cells: min row = 0 (Y at row0), max row = 4 (M at row4). min col = 3? Actually Y at col8,9; M at col3-5. So min col = 3 (M at col3), max col = 9 (Y at col9). That's width 7 columns (3-9 inclusive) and height 5 rows (0-4). That's 5x7 region. But output is 3x3, not that.

Maybe we take the bounding box of each color separately, then overlay? Or we find the "dominant" color (the one with more area) and we create a minimal square that encloses its shape? Then we place the minor color in the holes.

Let's compute shape of M: positions: (2,3),(2,4),(2,5),(3,4),(3,5),(4,5). This shape looks like a kind of L? Let's draw coordinates:

Row2: columns 3,4,5 => M M M
Row3: columns 4,5 => . M M
Row4: column5 => . . M

Thus shape is like a right triangle? Actually it's like a block of 3 at top row, then 2 below offset right, then 1 at bottom right.

Y positions: (0,8),(1,8),(1,9). That's like a small L shape: row0 col8, row1 col8,9.

If we overlay them, they are far apart (columns 3-5 vs 8-9). So they don't overlap.

The output 3x3: M M M; Y M M; Y Y M. This looks like we took the M shape and placed it in a 3x3 block such that it occupies top row (all M), middle row left Y then two M, bottom row left two Y then M. Actually output:

Row0: M M M
Row1: Y M M
Row2: Y Y M

Thus shape of M in output: positions: (0,0),(0,1),(0,2),(1,1),(1,2),(2,2). That's exactly same shape as input M but shifted? Input M shape: (2,3),(2,4),(2,5),(3,4),(3,5),(4,5). If we subtract (2,3) from each input M coordinate we get: (0,0),(0,1),(0,2),(1,1),(1,2),(2,2). Yes! So the M shape is normalized to origin at its top-leftmost cell? Actually we subtracted min row (2) and min col (3). So we translated M shape to origin (0,0). That yields exactly the output M shape.

Now Y shape in output: positions: (1,0),(2,0),(2,1). Input Y shape: (0,8),(1,8),(1,9). If we subtract min row (0) and min col (8) we get: (0,0),(1,0),(1,1). That's not matching output Y shape. Output Y shape is (1,0),(2,0),(2,1). That's like the Y shape shifted down one row and left? Actually if we take input Y shape and translate by (+1, -8)? Let's compute: Input Y shape coordinates relative to its own bounding box: min row=0, min col=8 => normalized: (0,0),(1,0),(1,1). Output Y shape: (1,0),(2,0),(2,1). That's like we took the normalized Y shape and shifted down by 1 row (add 1 to row) and same col? Actually (0,0)+(1,0) = (1,0) matches; (1,0)+(1,0) = (2,0) matches; (1,1)+(1,0) = (2,1) matches. So output Y shape = normalized Y shape + (1,0). So Y shape is placed such that it sits in the "pockets" left by M shape within the 3x3 bounding box.

Thus the transformation: For each color, we compute its shape (set of coordinates). Then we find the minimal bounding box that contains both shapes? Actually we see output size is 3x3, which is the size of the bounding box of the dominant shape (M) after normalization? Let's compute bounding box of M shape: rows 0-2, cols 0-2 => 3x3. Indeed M shape's bounding box is 3x3. So output size equals bounding box of dominant shape (the one with more cells). Then

... [7832 chars truncated]

We need to parse the problem: We have a grid transformation puzzle. The description: "Three artists share a studio, each assigned a floor in a tall building. The blue painter works on the top floor, the azure painter on the middle floor, and the magenta painter in the basement. Each has sketched out where they want to place brushstrokes on an identical canvas — some cells filled, some left bare. The three painters must merge their plans onto a single shared canvas. The magenta painter, working deepest underground, has the strongest pigment — wherever she has marked a stroke, her color dominates and nobody can paint over it. The blue painter has the next strongest hand; his marks show through wherever magenta hasn't already claimed the spot. The azure painter is the gentlest — her color only appears in the gaps the other two left untouched. Where no artist wanted to paint at all, the canvas stays bare and dark."

Thus we have three layers: top floor (blue), middle floor (azure), basement (magenta). The input grid presumably contains the three layers stacked? The description: "The tall building has three identical floors stacked vertically, each the size of the final canvas. Each floor's sketch uses only one color against darkness. The shared canvas is a single floor's worth of space, where the strongest pigments rise from below to claim their territory."

Thus the input grid likely contains three layers: each floor's sketch is given as a subgrid? The input appears to be a rectangular grid with multiple rows. Let's examine training example 1.

Example 1 Input:
```
. 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
. M . M M
. . M M M
. M . . .
. M M . M
. . . . .
```
That's 16 rows? Let's count: lines: 
1 . B B B B
2 . B B . .
3 . B B B .
4 . . B B B
5 . B B B .
6 . A A . .
7 A . . A .
8 . A . . A
9 . . A . .
10 A . A . A
11 . M . M M
12 . . M M M
13 . M . . .
14 . M M . M
15 . . . . .
That's 15 rows? Actually we have 15 lines. Let's count again: I'll number:
1: . B B B B
2: . B B . .
3: . B B B .
4: . . B B B
5: . B B B .
6: . A A . .
7: A . . A .
8: . A . . A
9: . . A . .
10: A . A . A
11: . M . M M
12: . . M M M
13: . M . . .
14: . M M . M
15: . . . . .
Yes 15 rows.

Output:
```
. M B M M
A B M M M
. M B B A
. M M B M
A B B B A
```
That's 5 rows.

Thus input has 15 rows, output 5 rows. So each floor is 5 rows? 15/3 = 5. Indeed each floor is 5 rows tall. The input appears to be concatenated vertically: first 5 rows are blue floor sketch, next 5 rows are azure floor sketch, last 5 rows are magenta floor sketch. Let's verify: The first 5 rows (lines 1-5) contain B and . only? Let's see:
1 . B B B B
2 . B B . .
3 . B B B .
4 . . B B B
5 . B B B .
Yes only B and . (blue). Next 5 rows (6-10) contain A and . only? Let's see:
6 . A A . .
7 A . . A .
8 . A . . A
9 . . A . .
10 A . A . A
Yes only A and . (azure). Next 5 rows (11-15) contain M and . only? Let's see:
11 . M . M M
12 . . M M M
13 . M . . .
14 . M M . M
15 . . . . .
Yes only M and . (magenta). So indeed input is three layers stacked: top (blue), middle (azure), bottom (magenta). Output is a single layer (size of one floor) where we apply layering: magenta dominates (strongest), then blue, then azure, else black (.) if none.

Thus transformation: For each cell position (i,j) within the floor (size HxW), we look at the three layers: magenta layer (bottommost in input) at same position; if it's M (non '.'), output M; else if blue layer (topmost) at same position is B (non '.'), output B; else if azure layer (middle) at same position is A (non '.'), output A; else output '.'.

Thus rule: output = magenta if present else blue if present else azure if present else '.'.

Now test input: We need to parse test input similarly: It should have three layers stacked vertically. Let's count rows.

Test Input:
```
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 .
. M . M .
. . M . M
. M . . M
. . M . M
M . M M .
```
Let's count lines: 
1 B . B B B
2 B . B . .
3 . B B . .
4 . B B B .
5 B . B . B
6 . . A . .
7 . . A A A
8 A A . A A
9 . . A . .
10 A A . A .
11 . M . M .
12 . . M . M
13 . M . . M
14 . . M . M
15 M . M M .
Thus 15 rows again, so each layer is 5 rows. So first 5 rows (1-5) are blue layer, next 5 rows (6-10) azure layer, last 5 rows (11-15) magenta layer.

Now we need to compute output for each cell (5 rows, number of columns? Let's see column count: each line seems to have 5 entries? Let's check: line1: "B . B B B" that's 5 tokens separated by spaces. So width = 5. So final output is 5x5 grid.

Now we need to compute for each (r,c) (0-index rows 0-4, columns 0-4) the output based on layers:

- Blue layer: rows 0-4 (input lines 1-5)
- Azure layer: rows 5-9 (input lines 6-10)
- Magenta layer: rows 10-14 (input lines 11-15)

Let's extract each layer as a 5x5 matrix.

I'll write them out.

Blue layer (rows 0-4):
Row0: B . B B B
Row1: B . B . .
Ro

... [5461 chars truncated]

We need to parse the problem: We have a grid transformation puzzle. Input grid is 15 rows by 5 columns (as per description). Output grid is 5 rows by 5 columns (the surviving colony map). The metaphor: "A tall slab of rock divided into three horizontal bands, each colonized by a different organism — one blue, one azure, one magenta — scattered across the stone in irregular patches against bare black rock." So the input grid has three horizontal bands of 5 rows each (since total 15 rows). Each band corresponds to a different organism: blue, azure, magenta. The input shows patches of those colors (B, A, M) scattered across the stone (black '.'), representing where each organism is present in that band. The task: "Let the three colonies compete for the same territory. The magenta fungus overgrows everything beneath it; the blue algae holds where fungus is absent; the azure cyanobacteria clings only to what neither has claimed. Bare rock remains where none has claimed."

Interpretation: We have three layers (bands) each representing presence of a particular organism in that band. The competition results in a final 5x5 map (presumably one cell per column? Actually output is 5 rows by 5 columns, same width as input, but height reduced from 15 to 5. So we are mapping the three bands (each 5 rows) into a single 5-row output via some rule per column? Perhaps each output row corresponds to a particular vertical position within the band? Let's think.

The input has 15 rows, 5 columns. The three bands: rows 0-4 (top band?), rows 5-9 (middle band?), rows 10-14 (bottom band?). Each band corresponds to a specific organism: blue, azure, magenta (order? The description: "one blue, one azure, one magenta". Not specified which band corresponds to which organism. We need to infer from training examples.

We have training examples with input and output. Let's examine Example 1 to deduce mapping.

Example 1 Input (15 rows, 5 columns). I'll rewrite with row numbers.

Input:
Row0: . B B B B
Row1: . B B . .
Row2: . B B B .
Row3: . . B B B
Row4: . B B B .
Row5: . A A . .
Row6: A . . A .
Row7: . A . . A
Row8: . . A . .
Row9: A . A . A
Row10: . M . M M
Row11: . . M M M
Row12: . M . . .
Row13: . M M . M
Row14: . . . . .

Output (5 rows, 5 columns):
Row0: . M B M M
Row1: A B M M M
Row2: . M B B A
Row3: . M M B M
Row4: A B B B A

We need to see how output derived from three bands.

Let's separate input into three bands of 5 rows each:

Band0 (rows0-4): Blue band? Let's see content: rows0-4 have many B's and some '.'.

Band1 (rows5-9): Azure band? Contains A's.

Band2 (rows10-14): Magenta band? Contains M's.

Thus likely band0 = blue, band1 = azure, band2 = magenta. Let's test with other examples.

Example 2 Input:

Rows0-4:
Row0: B . B . B
Row1: . B . . B
Row2: . B . . .
Row3: B . . B B
Row4: B . . B B

These have B's and '.'; seems blue band.

Rows5-9:
Row5: . . . . .
Row6: . A A A .
Row7: . A . . .
Row8: A . . . A
Row9: A . A A .

These have A's and '.'; azure band.

Rows10-14:
Row10: . . M . M
Row11: M . M . .
Row12: M . . . M
Row13: M . . . M
Row14: . M M M M

These have M's and '.'; magenta band.

Thus mapping: top band (rows0-4) = blue, middle band (rows5-9) = azure, bottom band (rows10-14) = magenta. Good.

Now output is 5 rows by 5 columns. How to compute? The metaphor: "The magenta fungus overgrows everything beneath it; the blue algae holds where fungus is absent; the azure cyanobacteria clings only to what neither has claimed. Bare rock remains where none has claimed."

Interpretation: For each column (since width 5), we have a vertical stack of 15 cells (3 bands of 5). The competition yields a final 5-cell column (maybe representing the topmost visible organism after overgrowth?). Actually output height is 5, same as each band's height. So perhaps we compute per row within the band? Let's think: The three bands each have 5 rows. The output also has 5 rows. Possibly we compute for each of the 5 rows (vertical position within band) a result based on the three bands at that same row offset? For each row index i from 0 to 4 (within band), we look at the three cells: blue band cell at row i, azure band cell at row i, magenta band cell at row i (same offset within each band). Then we apply competition rules to determine which organism "wins" at that position, yielding output cell at row i.

Thus output row i corresponds to the same vertical offset within each band. That seems plausible: we have three layers (blue, azure, magenta) each of thickness 5 rows; we want to combine them into a single layer of thickness 5 via per-cell competition across layers.

Let's test this hypothesis on Example 1.

We'll compute for each column c (0-4) and each row offset i (0-4) the triple (B_i, A_i, M_i) from bands.

Define:

Blue band rows0-4: Bband[i][c] = input[row i][c]
Azure band rows5-9: Aband[i][c] = input[row 5+i][c]
Magenta band rows10-14: Mband[i][c] = input[row 10+i][c]

Now apply rule: "magenta fungus overgrows everything bene

... [6525 chars truncated]

We need to understand the transformation. Input grid is 15 rows by 5 columns. It is described as "tall rehearsal chart with three equal sections showing each troupe's scattered stage marks in their signature color against empty black floor." So three sections each 5 rows (since total 15 rows). Each section corresponds to a troupe: likely magenta, blue, azure? The description: "Magenta dancers hold their marks absolutely; blue dancers take unclaimed positions; azure dancers fill whatever gaps remain. Empty floor stays empty." So we have three colors: magenta (M), blue (B), azure (A). The input grid contains these colors (plus black '.' for empty). The output grid is 5 rows by 5 columns (so we need to compress the 15 rows into 5 rows by merging the three sections). Likely each column of output corresponds to something like: for each column, we look at the three sections (top 5 rows, middle 5 rows, bottom 5 rows) and combine them according to rules: magenta holds its marks absolutely (i.e., if any magenta appears in that column in any of the three sections? Or maybe per row? Let's parse.

We have three sections each 5 rows tall. The final blocking is 5 rows by 5 columns. So we need to map each of the 5 rows of output to something derived from the three sections. Possibly each output row corresponds to a vertical stack of the three input rows (one from each section) at same column? For each column, we have 15 cells (5 rows per section * 3 sections). We need to produce 5 output cells (one per row) for that column. So maybe we take each of the 5 rows within a section and combine the three sections' values for that row index? For each output row i (0-4), we look at the three input rows: section1 row i, section2 row i, section3 row i (i.e., same offset within each section). Then we apply merging rule across the three colors at that position to produce a single output cell.

Thus for each column c and output row r (0-4), we consider three input cells: top section (rows 0-4) row r, middle section (rows 5-9) row r, bottom section (rows 10-14) row r. Then we combine them according to: Magenta holds its marks absolutely (i.e., if any of the three is M, output is M? Or maybe if M appears in that exact position across sections, it overrides others). Blue dancers take unclaimed positions: meaning if no M, then if any B appears, output B? Azure dancers fill whatever gaps remain: meaning if no M and no B, then if any A appears, output A? Empty floor stays open: meaning if none of M, B, A appear (i.e., all '.'), output '.'.

Thus rule: output cell = first non-empty in priority M > B > A > . (i.e., if any M present among the three, output M; else if any B present, output B; else if any A present, output A; else .). This matches "Magenta dancers hold their marks absolutely; blue dancers take unclaimed positions; azure dancers fill whatever gaps remain. Empty floor stays open."

Let's test with training examples.

Example 1 input:

We need to parse the given input grid (15 rows, 5 columns). Let's write them.

Example 1 Input:
Row0: . B B B B
Row1: . B B . .
Row2: . B B B .
Row3: . . B B B
Row4: . B B B .
Row5: . A A . .
Row6: A . . A .
Row7: . A . . A
Row8: . . A . .
Row9: A . A . A
Row10: . M . M M
Row11: . . M M M
Row12: . M . . .
Row13: . M M . M
Row14: . . . . .

We need to compute output per rule and compare to given output:

Output:
Row0: . M B M M
Row1: A B M M M
Row2: . M B B A
Row3: . M M B M
Row4: A B B B A

Let's compute for each column and output row.

We'll compute for each output row r (0-4) and column c (0-4). We'll get three input cells: top (rows 0-4) row r, middle (rows5-9) row r, bottom (rows10-14) row r.

Let's create arrays.

I'll label sections:

Top (T) rows 0-4:
T0: . B B B B
T1: . B B . .
T2: . B B B .
T3: . . B B B
T4: . B B B .

Middle (M) rows5-9:
M0: . A A . .
M1: A . . A .
M2: . A . . A
M3: . . A . .
M4: A . A . A

Bottom (Btm) rows10-14:
B0: . M . M M
B1: . . M M M
B2: . M . . .
B3: . M M . M
B4: . . . . .

Now compute for each output row r (0-4) and column c (0-4) the priority.

We'll do column by column.

Column0:
T column0: T0[0]='.', T1[0]='.', T2[0]='.', T3[0]='.', T4[0]='.' => all .
M column0: M0[0]='.', M1[0]='A', M2[0]='.', M3[0]='.', M4[0]='A' => we have A at rows1 and4.
Btm column0: B0[0]='.', B1[0]='.', B2[0]='.', B3[0]='.', B4[0]='.' => all .

Now for each output row r, we look at T[r][0], M[r][0], Btm[r][0].

r=0: T0[0]='.', M0[0]='.', B0[0]='.' => all . => output .
Matches output row0 col0 '.'.

r=1: T1[0]='.', M1[0]='A', B1[0]='.' => we have A, no M or B => output A (since azure fills gaps). Output row1 col0 is 'A' per given output: yes.

r=2: T2[0]='.', M2[0]='.', B2[0]='.' => all . => output . Output row2 col0 '.' matches.

r=3: T3[0]='.', M3[0]='.', B3[0]='.' => all . => output . Output row3 col0 '.' matches.

r=4: T4[0]='.', M4[0]='A', B4[0]='.' => A => output A. Output row4 col0 'A' matches.

Great.

Column1:
T column1: T0[1]='B', 

... [5248 chars truncated]

We need to understand the transformation. The description: "launchpad stretches across the darkness — a solid blue runway with special rocket engines embedded in it, each glowing a distinct color (azure or red) to signal what kind of fuel they carry." So the launchpad is a row of blue cells (B). Engines are cells on the launchpad that are not blue? Actually they are "special rocket engines embedded in it, each glowing a distinct color (azure or red)". So the launchpad is solid blue, but there are engines that are colored either azure (A) or red (R). So the input grid: we have a row of B's (blue) with some A and R interspersed. The rest of the grid is black (.) except maybe other colors? The description: "When launch is triggered, each colored engine fires a plume straight up into the night sky. The plume is a column of blue exhaust capped by a bright flare matching the engine's color. Red rockets burn hotter and their plumes reach higher than azure ones. The engines themselves stay bolted to the launchpad."

Thus transformation: For each engine cell (non-blue on the launchpad row), we draw a vertical column upward (i.e., decreasing row index? Usually grid rows top to bottom; but "upward" means towards top of grid (lower row index). The plume is a column of blue exhaust (B) capped by a bright flare matching the engine's color (A or R) at the top of the plume. The length of the plume (height) depends on engine color: red rockets reach higher than azure ones. So red plumes are longer (taller) than azure plumes. Need to determine exact heights from examples.

Example 1 input:

Row0: . . . . . .
Row1: . . . . . .
Row2: . . . . . .
Row3: B A B B B B
Row4: . . . . . .
Row5: . . . . . .
Row6: . . . . . .
Row7: . . . . . .
Row8: . . . . . .

So launchpad row is row3 (0-indexed). It has B A B B B B. So there is an azure engine at column1 (0-index). Others are blue (B) which are just runway, not engines? Actually the runway is solid blue; engines are special cells that are azure or red. So only the A at (3,1) is an engine. There's no red.

Output:
Row0: . A . . . .
Row1: . B . . . .
Row2: . B . . . .
Row3: B A B B B B
Row4: . . . . . .
Row5: . . . . . .
Row6: . . . . . .
Row7: . . . . . .
Row8: . . . . . .

Thus we see: At column1, above the engine (row3) we have a plume: At row2 (just above engine) we have B (blue exhaust). At row1 we have B (blue exhaust). At row0 we have A (flare). So plume height: from engine row (row3) upward: we have 2 blue cells (rows2 and1) then flare at row0 (top). So total plume length (including flare?) is 3 cells above engine? Actually engine at row3, then row2 B, row1 B, row0 A. So plume extends 3 cells upward (including flare). The flare is at top, color matches engine (A). The exhaust below flare is blue (B). So for azure engine, plume height = 3? Let's see if that's constant.

Now test input: we have a row with B R B A B B B A B R B at row6? Let's check: The test input grid lines:

Line0: . . . . . . . . . .
Line1: . . . . . . . . . .
Line2: . . . . . . . . . .
Line3: . . . . . . . . . .
Line4: . . . . . . . . . .
Line5: . . . . . . . . . .
Line6: B R B A B B B A B R B
Line7: . . . . . . . . . .
Line8: . . . . . . . . . .

Thus launchpad row is row6 (0-index). Engines: at columns where cell is not B: column1 is R, column3 is A, column7 is A, column9 is R. So we have two red engines at col1 and col9, and two azure engines at col3 and col7.

We need to predict output: For each engine, draw a vertical plume upward (decreasing row index) consisting of blue exhaust cells (B) topped by a flare of engine color (A or R). The height of plume depends on engine color: red rockets burn hotter and their plumes reach higher than azure ones. So red plumes are longer (taller) than azure plumes.

From example, azure plume had height: 2 blue cells + 1 flare = 3 cells above engine. Let's denote azure plume length L_A = number of cells above engine (including flare?) Actually we saw 2 B + 1 A = 3 cells. So maybe azure plume height = 3 (i.e., extends 3 rows upward from engine row, inclusive of flare). Red plume should be taller: maybe 4 or 5? Need to infer.

We need to consider constraints: The plume only fires upward; nothing falls below the runway; no rocket escapes the edge of the sky. So plume cannot go beyond top boundary (row0). If plume would exceed top, maybe it stops at top? Or maybe the given heights are such that they fit within grid.

In example, grid height is 9 rows (0-8). Engine at row3, plume height 3 upward goes to row0 (top). So azure plume exactly reaches top edge. If we had a red engine, maybe it would try to go higher but cannot exceed top, so maybe it would be truncated? But description: "Red rockets burn hotter and their plumes reach higher than azure ones." So if azure plume just reaches top, red would exceed top, but cannot escape edge of sky, so maybe it would be capped at top? Or maybe the grid is larger enough to accommodate red plume without hitting top. In test input, en

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We need to understand the transformation. The description: "What you see: A uniform bus line runs across the circuit board, with distinctly colored components soldered at specific positions. What to do: Each component activates, sending a signal trace perpendicular to the bus. The trace length depends on the component type, with the component's identifier visible at the trace end. Grid details: The bus line remains. Signal traces extend from each component position, terminated by the component's color."

Interpretation: There's a bus line (maybe a row of something) that runs across the board. Components are placed on the bus line (maybe a row of colored cells). Each component sends a signal trace perpendicular to the bus (so vertical if bus is horizontal). The trace length depends on component type. The component's identifier (maybe its color) is visible at the trace end. The bus line remains. Signal traces extend from each component position, terminated by the component's color.

Thus, input grid: we see a row with many colored cells (B, R, A, etc.) and many '.' elsewhere. The bus line is likely the row where the components are placed (the row with many non-dot characters). The bus line remains unchanged in output. For each component (colored cell) on that bus line, we send a signal trace perpendicular (i.e., up and/or down?) The trace length depends on component type. The trace ends with the component's identifier (color) at the trace end. So we place the same color at some distance away from the component along the perpendicular direction (vertical). The trace itself maybe is filled with something? The description: "Signal traces extend from each component position, terminated by the component's color." Could mean that we draw a line (maybe of some placeholder like '.'? Or maybe we fill the cells along the trace with something? The examples show that in output, we see some colored cells placed above or below the bus line, but the bus line row remains same as input. Let's examine examples.

Example 1 input:
Row0: . . . . . .
Row1: . . . . . .
Row2: . . . . . .
Row3: B A B B B B
Row4: . . . . . .
Row5: . . . . . .
Row6: . . . . . .
Row7: . . . . . .
Row8: . . . . . .

So bus line is row3 (0-indexed). It has B A B B B B at columns 0-5.

Output:
Row0: . A . . . .
Row1: . B . . . .
Row2: . B . . . .
Row3: B A B B B B
Row4: . . . . . .
Row5: . . . . . .
Row6: . . . . . .
Row7: . . . . . .
Row8: . . . . . .

So we see that above the bus line (rows 0-2) we have some colored cells placed: at row0 col1 = A, row1 col1 = B, row2 col1 = B. Also maybe some others? Let's map.

Input bus line row3: col0 B, col1 A, col2 B, col3 B, col4 B, col5 B.

Output: At col0: row0 ., row1 ., row2 ., row3 B (unchanged), rows4-8 . So nothing above/below for col0? Actually we see no extra marks above/below col0. At col1: we have row0 A, row1 B, row2 B. So three cells above bus line: at distances 1,2,3? Let's see: bus line at row3. So row2 is one above (distance 1), row1 distance 2, row0 distance 3. At col1 we have A at row0 (distance 3 up), B at row1 (distance 2 up), B at row2 (distance 1 up). So we have a vertical column of three cells above bus line, with colors: from nearest to farthest: B (distance1), B (distance2), A (distance3). At col2: we see only bus line B at row3, and no extra marks above/below? Output row0 col2 ., row1 col2 ., row2 col2 ., row3 B, rows4-8 . So nothing. At col3: similarly nothing. At col4: nothing. At col5: nothing.

Thus only col1 had extra marks. Why? Input at col1 is A (azure). So component type A caused a signal trace of length 3? Actually we see three cells above bus line (maybe also below? none). The trace length depends on component type. For A, length maybe 3? And the trace ends with component's identifier at the trace end: The farthest cell (distance 3) is A (same as component). The intermediate cells maybe are something else? They are B (blue). So maybe the trace is filled with some default color (maybe B?) and ends with the component's color.

Let's test with Example 2.

Example2 input:
Rows0-9? Actually given:
Row0: . . . . . . .
Row1: . . . . . . .
Row2: . . . . . . .
Row3: . . . . . . .
Row4: . . . . . . .
Row5: B B B R B B B
Row6: . . . . . . .
Row7: . . . . . . .
Row8: . . . . . . .
Row9: . . . . . . .

So bus line row5: columns 0-6: B B B R B B B.

Output:
Row0: . . . . . . .
Row1: . . . R . . .
Row2: . . . B . . .
Row3: . . . B . . .
Row4: . . . B . . .
Row5: B B B R B B B
Row6: . . . . . . .
Row7: . . . . . . .
Row8: . . . . . . .
Row9: . . . . . . .

So we see at column3 (where input has R), we have above bus line: row4 col3 = B, row3 col3 = B, row2 col3 = B, row1 col3 = R. So four cells above bus line: distances 1-4: B,B,B,R. No cells below bus line. So component R caused a trace of length 4 upward, filled with B (maybe default) and ending with R at the far end.

Thus pattern: For each component (non-dot) on the bus line row, we look upward (maybe only upward? In examples, traces go upwa

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We need to parse the problem: It's a grid transformation puzzle. The metaphor: "A chorus line of uniformly costumed dancers stretches across the stage, with a few wearing distinctly colored costumes at certain positions. Each costumed dancer leaps perpendicular to the line. The height of the leap depends on the costume color, with the dancer visible at the apex and a motion trail connecting back to the line. Grid details: The chorus line stays grounded. Vertical motion trails extend from each leaping dancer, with their costume color at the peak."

Interpretation: Input grid has a line of dancers (maybe a row) with uniform costume (maybe '.'?), and some have distinct colors (colored characters). The transformation: each colored dancer leaps perpendicular to the line (i.e., up or down? The line is horizontal? The chorus line is a row of dancers across the stage; they leap perpendicular to the line, i.e., vertical direction (up or down). The height of the leap depends on the costume color, with the dancer visible at the apex and a motion trail connecting back to the line. So we see vertical trails (like a column) from the line up to the apex where the dancer's color appears, and the trail is maybe the same color? Or maybe the trail is the dancer's color? The description: "Vertical motion trails extend from each leaping dancer, with their costume color at the peak." So the trail is maybe some default color (like '.'?), but at the peak (the top of the leap) we see the dancer's costume color. The line stays grounded (so the original line remains unchanged? The line of dancers remains as is? The input line is the chorus line; the output shows the line unchanged? Let's check examples.

Example 1 Input:
Row0: . . . . . .
Row1: . . . . . .
Row2: . . . . . .
Row3: B A B B B B
Row4: . . . . . .
Row5: . . . . . .
Row6: . . . . . .
Row7: . . . . . .
Row8: . . . . . .

So a line of dancers at row3: B A B B B B (six columns). Others are '.' (black). Output:
Row0: . A . . . .
Row1: . B . . . .
Row2: . B . . . .
Row3: B A B B B B
Row4: . . . . . .
Row5: . . . . . .
Row6: . . . . . .
Row7: . . . . . .
Row8: . . . . . .

So we see that the line at row3 unchanged: B A B B B B. Above it (rows 0-2) we have some colors placed: At column1 (0-index?), we see A at row0, B at row1, B at row2? Actually output shows:
Row0: . A . . . .
Row1: . B . . . .
Row2: . B . . . .
Thus column1 (second column) has A at top, then B below, then B below, then the line at row3 has A? Wait line at row3 column1 is A (since line: B A B B B B). So column1: row0 A, row1 B, row2 B, row3 A (line). That's interesting: The trail appears to be of length equal to something? The dancer at column1 originally had costume A? Actually line at column1 is A. The output shows above that column: at row0 A, row1 B, row2 B. So there is a vertical column of colors: A at top, then B, B, then the line's A at bottom. But why B's? The dancer's costume is A, but we see B's in the trail? Hmm.

Let's examine column0: line at column0 is B. Output column0: row0 ., row1 ., row2 ., row3 B (line). So no trail above.

Column2: line at column2 is B. Output column2: row0 ., row1 ., row2 ., row3 B (line). No trail.

Column3: line at column3 is B. Output column3: row0 ., row1 ., row2 ., row3 B (line). No trail.

Column4: line at column4 is B. Output column4: row0 ., row1 ., row2 ., row3 B (line). No trail.

Column5: line at column5 is B. Output column5: row0 ., row1 ., row2 ., row3 B (line). No trail.

So only column1 (where line had A) got a trail of some sort: A at top, then B, B, then line A. Where did the B's come from? Possibly from neighboring dancers? Actually column0 and column2 are B's on the line. Maybe the trail uses the colors of the adjacent dancers? Let's see: column1's left neighbor column0 is B, right neighbor column2 is B. So the trail consists of B's (maybe from left and right?) and the apex is the dancer's own color A. The height of the leap depends on costume color: maybe A corresponds to a leap height of 3? Actually we see three cells above the line: row0, row1, row2 (three rows). So maybe the leap height is number of rows above line equal to something based on color. For A, height = 3? For B maybe height = 0? Since B's had no trail. Let's test with Example 2.

Example 2 Input:
Rows0-4: all '.' (9 columns? Actually let's count: they show 7 columns? Input lines: ". . . . . . ." that's 7 dots. Row5: B B B R B B B (7 columns). Row6-9: all '.'.

So line at row5: B B B R B B B. Output:
Row0: . . . . . . .
Row1: . . . R . . .
Row2: . . . B . . .
Row3: . . . B . . .
Row4: . . . B . . .
Row5: B B B R B B B
Row6-9: all '.'.

So we see column3 (0-index?) where line has R. Output column3: row0 ., row1 R, row2 B, row3 B, row4 B, line R at row5. So trail: at apex (topmost non-dot?) we have R at row1, then three B's below (rows2-4), then line R at row5. So height of trail above line is 4 rows? Actually from row1 to row4 inclusive is 4 rows above line? Wait line is at row5. Above line 

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We need to understand the transformation. The description: "A sustained drone hums across one line, with certain positions colored by distinct tonal signatures breaking the monotone. Each tonal signature vibrates and projects a stack of harmonic overtones away from the drone. The number of overtones varies by timbre — some voices ring longer than others. Grid details: The drone line persists unchanged. Overtone stacks extend perpendicular from each colored position, with the original tone at the far end."

Interpretation: There is a "drone line" (maybe a row) that is unchanged. The colored positions (non-dot) are on that line? The examples: In Example 1, input has a row (row index 3? Let's index from 0) with B A B B B B (positions 0-5). The rest are all dots. Output: The same row unchanged? Let's see: Input row 3: B A B B B B. Output row 3: B A B B B B (same). So the drone line is row 3 (the line with colors). The other rows are all dots originally. In output, we see some non-dot entries above and below that row: e.g., row 0: . A . . . . ; row1: . B . . . . ; row2: . B . . . . ; row4-8 all dots.

Thus the colored positions on the drone line project "overtones" perpendicular away from the drone line (i.e., up and down). The number of overtones varies by timbre (i.e., by color). The original tone at the far end: meaning the original colored cell is at the far end of the stack? Or the stack extends away from the drone line, with the original tone at the far end (i.e., the farthest cell from the drone line has the original color). The stack is built of overtones (maybe each overtone is a different color? Or maybe each overtone is the same color? The description: "Each tonal signature vibrates and projects a stack of harmonic overtones away from the drone. The number of overtones varies by timbre — some voices ring longer than others." So each colored cell (tonal signature) projects a stack of overtones (like a vertical line) of some length (number of overtones) depending on its timbre (color). The original tone is at the far end (i.e., the tip of the stack farthest from the drone line). The drone line persists unchanged (the original row). So we need to determine for each color, how many overtones (i.e., how many cells away from the drone line) are projected, and what colors those overtones have.

From examples, we can infer mapping.

Example 1: Drone line row 3 has colors: positions: col0 B, col1 A, col2 B, col3 B, col4 B, col5 B.

Output shows:

Row0 (3 rows above drone line): . A . . . . (col1 A)
Row1 (2 rows above): . B . . . . (col1 B)
Row2 (1 row above): . B . . . . (col1 B)
Row3 (drone line): B A B B B B (unchanged)
Rows4-8 (below): all dots.

Thus only column 1 (the A) produced overtones upward? Actually we see A at row0 (distance 3 above), B at row1 (distance 2), B at row2 (distance 1). So the stack for the A at (row3,col1) extends upward 3 cells, with colors: at distance 1 (row2) B, distance2 (row1) B, distance3 (row0) A. The original tone (A) is at the far end (topmost). So the stack is built of overtones (maybe each overtone is a certain color depending on something) and the original tone at the far end.

What about other colored cells in that row? B at col0, col2, col3, col4, col5: they produced no overtones? Output shows no non-dot cells in those columns above or below. So B's timbre yields zero overtones? Or maybe they project downwards? But we see no dots below either. So B yields zero overtones (no projection). A yields some overtones upward (maybe also downward? Not seen). In this example, only upward projection observed because there is space above; maybe they project both up and down equally? But we see no downward projection for A (rows4-8 are dots). So maybe projection only goes in one direction (maybe upward) based on some rule like "away from the drone" meaning away from the center of grid? Not sure.

Let's examine Example 2.

Input: Drone line appears to be row5 (0-index?) Let's see: Input rows:

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

So drone line row5 has colors: col0 B, col1 B, col2 B, col3 R, col4 B, col5 B, col6 B.

Output:

0: . . . . . . .
1: . . . R . . .
2: . . . B . . .
3: . . . B . . .
4: . . . B . . .
5: B B B R B B B (unchanged)
6: . . . . . . .
7: . . . . . . .
8: . . . . . . .
9: . . . . . . .

Thus we see a vertical stack in column3 (the R) extending upward: row1: R (distance 4 above?), row2: B (distance3), row3: B (distance2), row4: B (distance1). Actually distance from drone line row5: row4 is 1 above, row3 is 2 above, row2 is 3 above, row1 is 4 above. So the stack length is 4 cells upward. Colors: at distance1 (row4) B, distance2 (row3) B, distance3 (row2) B, distance4 (row1) R. Original tone (R) at far end (topmost). So again, the R yields a stack of length 4 upward, with overtones being B? The B's in the drone line (col0,1,2

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We need to understand the transformation. The description: "An irrigation channel runs across the garden bed, with differently colored seeds planted at certain intervals along it. Each seed germinates and sends a stem upward from the channel. The stem height depends on the seed variety, with the flower's color appearing at the top. Grid details: The channel remains. Stems of uniform green rise from each seed position, capped by the flower color."

Thus input grid: we have a channel (some row?) with seeds (colored cells) placed on the channel. The output: the channel remains (so the row with seeds stays same). Stems of uniform green rise from each seed position, capped by the flower color. So from each seed cell, we go upward (i.e., decreasing row index? Usually grid top is row 0). The stem is uniform green (G) cells, and at the top of the stem we have the flower color (the seed's color? Or maybe the flower color is the seed's color? Actually "Each seed germinates and sends a stem upward from the channel. The stem height depends on the seed variety, with the flower's color appearing at the top." So the seed variety determines stem height and flower color at top. The seed itself is maybe the seed color (the input colored cell). The stem is green, and at the top of the stem we have the flower color (maybe same as seed? Or maybe different mapping). Let's examine examples.

Example 1 input:
Rows (0-index top):
0: . . . . . .
1: . . . . . .
2: . . . . . .
3: B A B B B B
4: . . . . . .
5: . . . . . .
6: . . . . . .
7: . . . . . .
8: . . . . . .

So channel is row 3 (the only non-empty row). Seeds: B at col0, A at col1, B at col2, B col3, B col4, B col5.

Output:
0: . A . . . .
1: . B . . . .
2: . B . . . .
3: B A B B B B
4: . . . . . .
5: . . . . . .
6: . . . . . .
7: . . . . . .
8: . . . . . .

So we see: For each seed, we have a green stem (G) rising upward? But we see no G in output. Wait output shows only A and B placed above the channel? Let's examine: At column 0: input seed B at row3 col0. Output: row0 col0 ., row1 ., row2 ., row3 B (channel unchanged), rows4-8 . . . . . . So no stem? Actually we see nothing above. At column1: input seed A at row3 col1. Output: row0 col0? Wait column1: output row0 col1 A? Let's see: output row0: ". A . . . ." So at row0 col1 is A. row1 col1: ". B . . . ." => B. row2 col1: ". B . . . ." => B. So we have A at top, then two B's below it, then channel row3 has A. So pattern: from seed A at channel, we have stem of length? The stem is uniform green (G) but we see B's not G. Hmm.

Maybe the stem color is the seed color? Actually they said stems of uniform green rise from each seed position, capped by the flower color. But output shows no G. Could be that they omitted G because they consider green as default background? No, background is black (.) . So not.

Let's examine example 2.

Input:
Rows 0-9? Actually given:
0: . . . . . . .
1: . . . . . . .
2: . . . . . . .
3: . . . . . . .
4: . . . . . . .
5: B B B R B B B
6: . . . . . . .
7: . . . . . . .
8: . . . . . . .
9: . . . . . . .

So channel row5: B B B R B B B (col0-6). Output:
0: . . . . . . .
1: . . . R . . .
2: . . . B . . .
3: . . . B . . .
4: . . . B . . .
5: B B B R B B B
6: . . . . . . .
7: . . . . . . .
8: . . . . . . .
9: . . . . . . .

Thus at column3 (where seed R at row5 col3), we see upward: row1 col3 R, row2 col3 B, row3 col3 B, row4 col3 B, then channel row5 col3 R. So pattern: seed R at channel yields a stem of length? There are 4 cells above channel (rows1-4) before hitting top? Actually we see R at row1 (topmost non-.), then B at rows2-4, then channel R at row5. So total vertical distance from seed to top flower? The seed is at row5, flower at row1 (distance 4 upward). Stem cells are rows2-4 (3 cells) which are B (green? but B is blue). Hmm.

Wait maybe the stem color is the seed color of the left neighbor? Not.

Let's think: The description: "Stems of uniform green rise from each seed position, capped by the flower color." So stem should be G (green). But we see B and A etc. Could be that they used different letters for colors: G=green, but they didn't show any G in outputs. So maybe the stem is not shown because it's overwritten by something else? Or maybe the "stem" is the seed color itself? Actually they said "stem height depends on the seed variety, with the flower's color appearing at the top." So maybe the seed variety determines both stem height and flower color (which is the color that appears at the top of the stem). The stem itself is maybe not visible because it's just the channel? No.

Let's re-evaluate: Input shows seeds as colored cells on the channel row. Output shows same channel row unchanged. Above the channel, we see some pattern of colors: For each seed, there is a vertical column of colors extending upward from the seed, with the topmost cell being the flower color (maybe same as seed?), and the cells immediately below that (down to just above the channel) are the stem color (maybe

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We need to understand the transformation. The description: "A small mosaic tile — a 4×4 swatch of stained glass — sits at the heart of a cathedral floor, waiting to be echoed outward. What to do: The tile dreams of itself from every direction. It gazes north and sees its own face. It turns its head east, west, and south, and at each glance the reflection has spun — as if the tile were placed on a lazy Susan and photographed at each quarter-turn. These four portraits are pressed together like pages of a folded map, the original anchoring one corner while its dizzied twins fill the remaining three. Grid details: The cathedral floor is exactly four tiles wide and four tiles tall — a 2×2 arrangement of the original swatch and its three rotated echoes. No grout, no gaps: every cell of colored glass belongs to one of the four spinning memories of the original piece."

Interpretation: Input is a 4x4 grid (the original tile). Output is a 8x8 grid (2x2 arrangement of original and its three rotated copies). The original tile is placed in one corner (presumably top-left?), and then rotated copies fill the other three quadrants: top-right (rotated 90° clockwise?), bottom-left (rotated 180°?), bottom-right (rotated 270°?), or some arrangement.

We need to infer from examples.

Example 1 input 4x4:

Input:
Y W B A
A Y B A
Y A A B
B B B A

Let's label rows 0-3, columns 0-3.

Row0: [Y,W,B,A]
Row1: [A,Y,B,A]
Row2: [Y,A,A,B]
Row3: [B,B,B,A]

Output is 8x8:

Y W B A A A B A
A Y B A B B A B
Y A A B W Y A B
B B B A Y A Y B
A B B B B Y A Y
B A A Y B A Y W
A B Y A B A B B
A B W Y A B A A

Let's try to see if this matches original in top-left quadrant (first 4 rows, first 4 cols). Top-left 4x4 of output:

Rows0-3, cols0-3:

Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

That's exactly the input! So original placed at top-left.

Now top-right quadrant (rows0-3, cols4-7) should be some rotation of original. Let's extract:

Rows0-3, cols4-7:

Row0: A A B A
Row1: B B A B
Row2: W Y A B
Row3: Y A Y B

Let's see if that matches original rotated 90° clockwise? Or counterclockwise? Let's compute original rotated 90° clockwise (i.e., transpose then reverse each row). Original:

[ [Y,W,B,A],
  [A,Y,B,A],
  [Y,A,A,B],
  [B,B,B,A] ]

Rotate 90° CW: new[i][j] = original[3-j][i]? Actually standard: rotate CW: new[i][j] = old[n-1-j][i]. Let's compute.

n=4.

Compute new matrix:

i=0 row: j=0..3:
- j0: old[3-0][0] = old[3][0] = B
- j1: old[2][0] = Y
- j2: old[1][0] = A
- j3: old[0][0] = Y
=> row0: B Y A Y

i=1:
- j0: old[3][1] = B
- j1: old[2][1] = A
- j2: old[1][1] = Y
- j3: old[0][1] = W
=> B A Y W

i=2:
- j0: old[3][2] = B
- j1: old[2][2] = A
- j2: old[1][2] = B
- j3: old[0][2] = B
=> B A B B

i=3:
- j0: old[3][3] = A
- j1: old[2][3] = B
- j2: old[1][3] = A
- j3: old[0][3] = A
=> A B A A

So CW rotation yields:

B Y A Y
B A Y W
B A B B
A B A A

Our top-right quadrant is:

A A B A
B B A B
W Y A B
Y A Y B

Not matching.

Maybe it's rotated 90° counterclockwise (CCW). Rotate CCW: new[i][j] = old[j][n-1-i].

Compute:

i=0:
j0: old[0][3] = A
j1: old[1][3] = A
j2: old[2][3] = B
j3: old[3][3] = A
=> A A B A

i=1:
j0: old[0][2] = B
j1: old[1][2] = B
j2: old[2][2] = A
j3: old[3][2] = B
=> B B A B

i=2:
j0: old[0][1] = W
j1: old[1][1] = Y
j2: old[2][1] = A
j3: old[3][1] = B
=> W Y A B

i=3:
j0: old[0][0] = Y
j1: old[1][0] = A
j2: old[2][0] = Y
j3: old[3][0] = B
=> Y A Y B

That matches exactly the top-right quadrant we extracted! So top-right is original rotated 90° CCW (i.e., -90° or 270° CW). Good.

Now bottom-left quadrant (rows4-7, cols0-3) should be another rotation, perhaps 180°? Let's extract:

Rows4-7, cols0-3:

Row4: A B B B
Row5: B A A Y
Row6: A B Y A
Row7: A B W Y

Let's see if that matches original rotated 180° (i.e., rotate twice). Compute original rotated 180°: reverse rows and reverse each row (or rotate CW twice). Let's compute: original reversed both dimensions.

Original:

Y W B A
A Y B A
Y A A B
B B B A

Reverse rows (flip vertically): row order reversed:

B B B A
Y A A B
A Y B A
Y W B A

Then reverse each row (flip horizontally): each row reversed:

A B B B
B A A Y
A B Y A
A B W Y

That matches exactly bottom-left quadrant! So bottom-left is original rotated 180°.

Now bottom-right quadrant (rows4-7, cols4-7) should be original rotated 90° CW (or -90° CCW). Let's extract:

Rows4-7, cols4-7:

Row4: B Y A Y
Row5: B A Y W
Row6: B A B B
Row7: A B A A

Let's see if that matches CW rotation we computed earlier: CW gave:

B Y A Y
B A Y W
B A B B
A B A A

Yes matches! So bottom-right is original rotated 90° CW.

Thus transformation: Input 4x4 tile placed at top-left; then three rotated copies placed to fill the 8x8 canvas: top-right = original rotated 90° CCW (or -90°), bottom-left = original rotated 180°, bottom-right = original rotated 90° CW.

Now test with examples 2 a

... [4796 chars truncated]

We need to understand the transformation. The description: "The organism divides into four daughter colonies, each carrying the same genetic pattern but rotated on its axis as though the spindle apparatus turned each copy before planting it in its quadrant of the petri dish. Grid details: The original tissue sits in one corner. Three sibling colonies fill the remaining corners, each a rotated echo of the first, assembling into a larger organism exactly twice the width and twice the height."

Thus input is a small tissue sample (presumably a subgrid). Output is a larger grid of size 2*width x 2*height, composed of four quadrants: top-left (original), top-right (rotated 90°?), bottom-left (rotated 180°?), bottom-right (rotated 270°?) Or some rotation pattern.

We need to infer from examples.

Example 1:

Input:
Y W B A
A Y B A
Y A A B
B B B A

Let's write as matrix:

Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

So input is 4x4.

Output is 8x8 (twice width and height). Let's see output:

Y W B A A A B A
A Y B A B B A B
Y A A B W Y A B
B B B A Y A Y B
A B B B B Y A Y
B A A Y B A Y W
A B Y A B A B B
A B W Y A B A A

We need to see how quadrants are arranged.

Let's split output into four 4x4 quadrants: top-left (rows0-3, cols0-3), top-right (rows0-3, cols4-7), bottom-left (rows4-7, cols0-3), bottom-right (rows4-7, cols4-7).

Top-left quadrant (rows0-3, cols0-3):
Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

That's exactly the input! So top-left is original.

Top-right quadrant (rows0-3, cols4-7):
Row0: A A B A
Row1: B B A B
Row2: W Y A B
Row3: Y A Y B

Let's list:

Row0: col4-7: A A B A
Row1: B B A B
Row2: W Y A B
Row3: Y A Y B

Now bottom-left quadrant (rows4-7, cols0-3):
Row4: A B B B
Row5: B A A Y
Row6: A B Y A
Row7: A B W Y

Bottom-right quadrant (rows4-7, cols4-7):
Row4: B Y A Y
Row5: B A Y W
Row6: B A B B
Row7: A B A A

Now we need to see if top-right, bottom-left, bottom-right are rotations of the input.

Let's compute rotations of input.

Input matrix I:

[ [Y, W, B, A],
  [A, Y, B, A],
  [Y, A, A, B],
  [B, B, B, A] ]

Rotate 90 degrees clockwise: new[i][j] = I[ n-1-j ][i] (if n=4). Let's compute.

We can compute manually.

I'll label coordinates (r,c). For 90° CW: output[r][c] = I[ n-1-c ][r].

Compute for each r,c 0..3.

Better to compute using known transformation: rotate matrix.

Let's compute I rotated 90 CW:

Take columns of I from bottom to top as rows.

Column0 of I: [Y, A, Y, B] (r0..3). Reverse => [B, Y, A, Y] becomes row0 of rotated? Actually for CW: new row0 = original column0 reversed? Let's recall: Rotating 90 CW: new[i][j] = old[n-1-j][i]. So new row i corresponds to original column i, read from bottom to top.

Thus new row0 = column0 from bottom to top: I[3][0]=B, I[2][0]=Y, I[1][0]=A, I[0][0]=Y => B Y A Y.

Row1 = column1 from bottom to top: I[3][1]=B, I[2][1]=A, I[1][1]=Y, I[0][1]=W => B A Y W.

Row2 = column2 from bottom to top: I[3][2]=B, I[2][2]=A, I[1][2]=B, I[0][2]=B => B A B B.

Row3 = column3 from bottom to top: I[3][3]=A, I[2][3]=B, I[1][3]=A, I[0][3]=A => A B A A.

Thus 90 CW rotated matrix =

[ [B, Y, A, Y],
  [B, A, Y, W],
  [B, A, B, B],
  [A, B, A, A] ]

Now compare to quadrants.

Top-right quadrant we had:

Row0: A A B A
Row1: B B A B
Row2: W Y A B
Row3: Y A Y B

Not matching.

Bottom-left quadrant:

Row0: A B B B
Row1: B A A Y
Row2: A B Y A
Row3: A B W Y

Not matching.

Bottom-right quadrant:

Row0: B Y A Y
Row1: B A Y W
Row2: B A B B
Row3: A B A A

That matches exactly the 90 CW rotated matrix we computed! Indeed bottom-right quadrant equals I rotated 90 CW.

Thus bottom-right is rotation 90 CW.

Now check top-right maybe is rotation 180? Let's compute I rotated 180 (i.e., rotate 90 twice). Rotating 180: new[i][j] = I[n-1-i][n-1-j].

Compute quickly: Equivalent to reverse each row and then reverse order of rows.

Original I:

Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

Reverse each row:

Row0 rev: A B W Y
Row1 rev: A B Y A
Row2 rev: B A A Y
Row3 rev: A B B B

Now reverse order of rows (i.e., take rows from bottom to top):

New row0 = original row3 rev: A B B B
Row1 = original row2 rev: B A A Y
Row2 = original row1 rev: A B Y A
Row3 = original row0 rev: A B W Y

Thus 180 rotated matrix =

[ [A, B, B, B],
  [B, A, A, Y],
  [A, B, Y, A],
  [A, B, W, Y] ]

Now compare to quadrants.

Top-right quadrant we had:

Row0: A A B A
Row1: B B A B
Row2: W Y A B
Row3: Y A Y B

Not match.

Bottom-left quadrant:

Row0: A B B B
Row1: B A A Y
Row2: A B Y A
Row3: A B W Y

That matches exactly the 180 rotated matrix! Indeed bottom-left equals I rotated 180.

Thus bottom-left is rotation 180.

Now top-right maybe is rotation 270 CW (or 90 CCW). Let's compute I rotated 270 CW (i.e., 90 CCW). Equivalent to rotate 90 CW three times, or new[i][j] = I[j][n-1-i].

Compute 90 CCW: new row i = original column (n-1-i) read top to bottom? Actually formula: new[i][j] = old[j][n-1-i].

Let's compute using known: rotate 90 CCW: take c

... [5623 chars truncated]

We need to understand the puzzle: Input grid is a small square field of colored stars and gas clouds—a single telescope's capture. We need to combine four exposures into one mosaic. Each exposure shows the same patch of sky but rotated according to the telescope's orbital attitude, so the nebula appears to pinwheel across the composite. The final mosaic is twice the dimensions of a single exposure, tiled 2×2. Each tile is the same stellar field at a different rotational angle.

Thus input is a single exposure (size N x N). Output is a 2N x 2N grid composed of four tiles: each tile is the input rotated by some angle (0°, 90°, 180°, 270°) arranged in a 2x2 mosaic. The order of tiles likely corresponds to some pattern: maybe top-left is original, top-right rotated 90°, bottom-left rotated 180°, bottom-right rotated 270°, or something like that. The description: "Combine four exposures into one mosaic. Each exposure shows the same patch of sky but rotated according to the telescope's orbital attitude, so the nebula appears to pinwheel across the composite." So the four exposures are rotated relative to each other, perhaps each successive exposure rotated by 90° (pinwheel). The final mosaic is twice dimensions, tiled 2x2. So we need to output a 2N x 2N where each quadrant is the input rotated by 0°, 90°, 180°, 270° in some order.

We need to infer from training example.

Example 1 input:
Y W B A
A Y B A
Y A A B
B B B A

That's a 4x4 grid.

Output:
Y W B A A A B A
A Y B A B B A B
Y A A B W Y A B
B B B A Y A Y B
A B B B B Y A Y
B A A Y B A Y W
A B Y A B A B B
A B W Y A B A A

That's 8x8 (2*4). Let's break into 2x2 blocks of size 4x4 each.

Let's label quadrants: top-left (TL), top-right (TR), bottom-left (BL), bottom-right (BR). Each is 4x4.

Write output rows 0-7, columns 0-7.

I'll extract.

Row0: Y W B A | A A B A
Row1: A Y B A | B B A B
Row2: Y A A B | W Y A B
Row3: B B B A | Y A Y B
Row4: A B B B | B Y A Y
Row5: B A A Y | B A Y W
Row6: A B Y A | B A B B
Row7: A B W Y | A B A A

Thus TL (rows0-3, cols0-3):
Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

That's exactly the input! So TL is original (0° rotation).

TR (rows0-3, cols4-7):
Row0: A A B A
Row1: B B A B
Row2: W Y A B
Row3: Y A Y B

Let's see if that matches input rotated 90° clockwise? Let's compute.

Input:
[ [Y,W,B,A],
  [A,Y,B,A],
  [Y,A,A,B],
  [B,B,B,A] ]

Rotate 90° clockwise: new[i][j] = old[N-1-j][i] (if 0-index). Let's compute.

N=4.

Compute rotated 90 cw:

For i=0 (new row0):
j=0: old[3][0] = B
j=1: old[2][0] = Y
j=2: old[1][0] = A
j=3: old[0][0] = Y
=> [B,Y,A,Y]

i=1:
j=0: old[3][1] = B
j=1: old[2][1] = A
j=2: old[1][1] = Y
j=3: old[0][1] = W
=> [B,A,Y,W]

i=2:
j=0: old[3][2] = B
j=1: old[2][2] = A
j=2: old[1][2] = B
j=3: old[0][2] = B
=> [B,A,B,B]

i=3:
j=0: old[3][3] = A
j=1: old[2][3] = B
j=2: old[1][3] = A
j=3: old[0][3] = A
=> [A,B,A,A]

Thus 90 cw rotation yields:
Row0: B Y A Y
Row1: B A Y W
Row2: B A B B
Row3: A B A A

Our TR is:
Row0: A A B A
Row1: B B A B
Row2: W Y A B
Row3: Y A Y B

Not matching.

Maybe TR is input rotated 90° counterclockwise? Let's compute 90 ccw (or -90). Formula: new[i][j] = old[j][N-1-i].

Compute:

i=0:
j=0: old[0][3] = A
j=1: old[1][3] = A
j=2: old[2][3] = B
j=3: old[3][3] = A
=> [A,A,B,A] matches TR row0! Good.

i=1:
j=0: old[0][2] = B
j=1: old[1][2] = B
j=2: old[2][2] = A
j=3: old[3][2] = B
=> [B,B,A,B] matches TR row1! Good.

i=2:
j=0: old[0][1] = W
j=1: old[1][1] = Y
j=2: old[2][1] = A
j=3: old[3][1] = B
=> [W,Y,A,B] matches TR row2! Good.

i=3:
j=0: old[0][0] = Y
j=1: old[1][0] = A
j=2: old[2][0] = Y
j=3: old[3][0] = B
=> [Y,A,Y,B] matches TR row3! Good.

Thus TR is input rotated 90° counterclockwise (or 270° clockwise). Good.

Now BL (rows4-7, cols0-3):
Row4: A B B B
Row5: B A A Y
Row6: A B Y A
Row7: A B W Y

Let's see if that's input rotated 180°? Compute 180 rotation (reverse both axes): new[i][j] = old[N-1-i][N-1-j].

Compute:

i=0:
j=0: old[3][3] = A
j=1: old[3][2] = B
j=2: old[3][1] = B
j=3: old[3][0] = B
=> [A,B,B,B] matches BL row0! Good.

i=1:
j=0: old[2][3] = B
j=1: old[2][2] = A
j=2: old[2][1] = A
j=3: old[2][0] = Y
=> [B,A,A,Y] matches BL row1! Good.

i=2:
j=0: old[1][3] = A
j=1: old[1][2] = B
j=2: old[1][1] = Y
j=3: old[1][0] = A
=> [A,B,Y,A] matches BL row2! Good.

i=3:
j=0: old[0][3] = A
j=1: old[0][2] = B
j=2: old[0][1] = W
j=3: old[0][0] = Y
=> [A,B,W,Y] matches BL row3! Good.

Thus BL is 180° rotation.

BR (rows4-7, cols4-7):
Row4: B Y A Y
Row5: B A Y W
Row6: B A B B
Row7: A B A A

Let's see if that's input rotated 90° clockwise? We computed earlier 90 cw gave:
Row0: B Y A Y
Row1: B A Y W
Row2: B A B B
Row3: A B A A

Matches BR exactly! So BR is 90° clockwise.

Thus the pattern: TL = original (0°), TR = rotate 90° CCW (or -90), BL = rotate 180°, BR = rotate 90° CW.

Thus the four exposures are rotated

... [4055 chars truncated]

We need to understand the transformation: Input is a small square nautical chart (grid). Output is a master chart twice width and height, divided into four quadrants, each quadrant holds the same island chart at a different rotation (rotated so that a different shoreline points upward). So we need to take the input grid (size N x N) and produce a 2N x 2N grid where each quadrant is the input rotated by 0°, 90°, 180°, 270° (or some order). The description: "Each copy is rotated so that a different shoreline points upward, giving navigators a view from every bearing." So we need to place four copies of the island map into a larger frame, each rotated so a different shoreline points upward. Likely the quadrants are arranged as: top-left: original orientation (0°), top-right: rotated 90° clockwise (so that a different shoreline points up), bottom-left: rotated 180°, bottom-right: rotated 270° (or maybe some other order). The training example can help deduce.

Let's examine Example 1.

Input:
Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

So N=4.

Output given:
Y W B A A A B A
A Y B A B B A B
Y A A B W Y A B
B B B A Y A Y B
A B B B B Y A Y
B A A Y B A Y W
A B Y A B A B B
A B W Y A B A A

Let's format as 8 rows (2N=8) each with 8 characters.

Row0: Y W B A | A A B A
Row1: A Y B A | B B A B
Row2: Y A A B | W Y A B
Row3: B B B A | Y A Y B
Row4: A B B B | B Y A Y
Row5: B A A Y | B A Y W
Row6: A B Y A | B A B B
Row7: A B W Y | A B A A

So left half (columns 0-3) seems to be top two quadrants? Actually top-left quadrant (rows0-3, cols0-3) is:
Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

That's exactly the input! So top-left quadrant is original orientation (0°).

Top-right quadrant (rows0-3, cols4-7) is:
Row0: A A B A
Row1: B B A B
Row2: W Y A B
Row3: Y A Y B

Let's see if that matches input rotated 90° clockwise? Let's compute rotation.

Original input (4x4):
[ [Y,W,B,A],
  [A,Y,B,A],
  [Y,A,A,B],
  [B,B,B,A] ]

Rotate 90° clockwise: new[i][j] = original[N-1-j][i].

Compute:

i=0 row: j0..3:
j0: original[3][0] = B
j1: original[2][0] = Y
j2: original[1][0] = A
j3: original[0][0] = Y
=> [B,Y,A,Y]

i=1:
j0: original[3][1] = B
j1: original[2][1] = A
j2: original[1][1] = Y
j3: original[0][1] = W
=> [B,A,Y,W]

i=2:
j0: original[3][2] = B
j1: original[2][2] = A
j2: original[1][2] = B
j3: original[0][2] = B
=> [B,A,B,B]

i=3:
j0: original[3][3] = A
j1: original[2][3] = B
j2: original[1][3] = A
j3: original[0][3] = A
=> [A,B,A,A]

So 90° CW rotated matrix:
Row0: B Y A Y
Row1: B A Y W
Row2: B A B B
Row3: A B A A

But top-right quadrant we have:
Row0: A A B A
Row1: B B A B
Row2: W Y A B
Row3: Y A Y B

That doesn't match. Let's try 90° counterclockwise (i.e., rotate left). Rotate 90° CCW: new[i][j] = original[j][N-1-i].

Compute:

i=0:
j0: original[0][3] = A
j1: original[1][3] = A
j2: original[2][3] = B
j3: original[3][3] = A
=> [A,A,B,A] matches row0 of top-right! Good.

i=1:
j0: original[0][2] = B
j1: original[1][2] = B
j2: original[2][2] = A
j3: original[3][2] = B
=> [B,B,A,B] matches row1.

i=2:
j0: original[0][1] = W
j1: original[1][1] = Y
j2: original[2][1] = A
j3: original[3][1] = B
=> [W,Y,A,B] matches row2.

i=3:
j0: original[0][0] = Y
j1: original[1][0] = A
j2: original[2][0] = Y
j3: original[3][0] = B
=> [Y,A,Y,B] matches row3.

Thus top-right quadrant is input rotated 90° counterclockwise (or 270° clockwise). So that's one rotation.

Now bottom-left quadrant (rows4-7, cols0-3):
Rows4-7:
Row4: A B B B
Row5: B A A Y
Row6: A B Y A
Row7: A B W Y

Let's see if that matches input rotated 180°? Rotate 180°: new[i][j] = original[N-1-i][N-1-j].

Compute original reversed both axes.

Original:
Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

Reverse rows: row3,2,1,0:
Row3: B B B A
Row2: Y A A B
Row1: A Y B A
Row0: Y W B A

Then reverse each row's columns:
Row3 reversed: A B B B
Row2 reversed: B A A Y
Row1 reversed: A B Y A
Row0 reversed: A B W Y

Thus 180° rotated:
Row0: A B B B
Row1: B A A Y
Row2: A B Y A
Row3: A B W Y

Matches bottom-left quadrant exactly! Good.

Bottom-right quadrant (rows4-7, cols4-7):
Row4: B Y A Y
Row5: B A Y W
Row6: B A B B
Row7: A B A A

Let's see if that matches input rotated 90° clockwise (or 270° CCW). We earlier computed 90° CW gave:
Row0: B Y A Y
Row1: B A Y W
Row2: B A B B
Row3: A B A A

Matches bottom-right quadrant! So indeed the quadrants are arranged as:
Top-left: 0° (original)
Top-right: 90° CCW (or 270° CW)
Bottom-left: 180°
Bottom-right: 90° CW

Thus the transformation: output is a 2N x 2N grid where:
- Quadrant TL: input as is.
- Quadrant TR: input rotated 90° counterclockwise.
- Quadrant BL: input rotated 180°.
- Quadrant BR: input rotated 90° clockwise.

Now we need to apply to test input.

Test Input:
Y M Y Y
Y M Y Y
O M O W
W Y W O

Let's write as matrix:

Row0: Y M Y Y
Row1: Y M Y Y


... [4976 chars truncated]

We need to understand the transformation: Input is a small square silicon block (grid). Output is a larger die: twice dimensions of original block, with four identical circuit copies, each at a successive quarter-turn, tiled in a 2x2 layout. So if input is N x N, output is 2N x 2N, composed of four quadrants: top-left, top-right, bottom-left, bottom-right. Each quadrant is a copy of the input rotated by 0°, 90°, 180°, 270° (or some order). The description: "Each copy is rotated so its interface pins face a different edge of the package." So each copy rotated to face a different edge: likely top-left copy rotated 0° (original orientation), top-right rotated 90° clockwise (so its interface pins face right edge?), bottom-left rotated 180°, bottom-right rotated 270° (or counterclockwise). Or maybe they rotate successive quarter-turn: starting from some orientation, each next copy rotated 90°.

Thus transformation: output grid = place input in top-left (no rotation), then rotate input 90° clockwise for top-right, rotate 180° for bottom-left, rotate 270° clockwise (or 90° counterclockwise) for bottom-right.

Check with example 1 to verify.

Example 1 input:
4x4 grid:
Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

Output given is 8x8 grid:
Let's write output rows:

Row0: Y W B A A A B A
Row1: A Y B A B B A B
Row2: Y A A B W Y A B
Row3: B B B A Y A Y B
Row4: A B B B B Y A Y
Row5: B A A Y B A Y W
Row6: A B Y A B A B B
Row7: A B W Y A B A A

We need to see if this matches the described tiling.

Let's extract quadrants:

Top-left quadrant (rows0-3, cols0-3):
Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

That's exactly the input! Good.

Top-right quadrant (rows0-3, cols4-7):
Row0: A A B A
Row1: B B A B
Row2: W Y A B
Row3: Y A Y B

Let's see if that matches input rotated 90° clockwise.

Input:
[ [Y,W,B,A],
  [A,Y,B,A],
  [Y,A,A,B],
  [B,B,B,A] ]

Rotate 90° clockwise: new[i][j] = old[N-1-j][i] (if N=4). Let's compute.

Original indices:
(0,0)=Y, (0,1)=W, (0,2)=B, (0,3)=A
(1,0)=A, (1,1)=Y, (1,2)=B, (1,3)=A
(2,0)=Y, (2,1)=A, (2,2)=A, (2,3)=B
(3,0)=B, (3,1)=B, (3,2)=B, (3,3)=A

Rotated 90 cw:
new[0][0] = old[3][0] = B
new[0][1] = old[2][0] = Y
new[0][2] = old[1][0] = A
new[0][3] = old[0][0] = Y
=> row0: B Y A Y

new[1][0] = old[3][1] = B
new[1][1] = old[2][1] = A
new[1][2] = old[1][1] = Y
new[1][3] = old[0][1] = W
=> row1: B A Y W

new[2][0] = old[3][2] = B
new[2][1] = old[2][2] = A
new[2][2] = old[1][2] = B
new[2][3] = old[0][2] = B
=> row2: B A B B

new[3][0] = old[3][3] = A
new[3][1] = old[2][3] = B
new[3][2] = old[1][3] = A
new[3][3] = old[0][3] = A
=> row3: A B A A

Thus rotated 90 cw yields:
B Y A Y
B A Y W
B A B B
A B A A

But top-right quadrant we observed:
Row0: A A B A
Row1: B B A B
Row2: W Y A B
Row3: Y A Y B

Not matching. Maybe rotation is different (counterclockwise) or different ordering of quadrants.

Let's check top-right quadrant as input rotated 90° counterclockwise (i.e., 90° ccw = 270 cw). Compute 90 ccw: new[i][j] = old[j][N-1-i].

Compute:
new[0][0] = old[0][3] = A
new[0][1] = old[1][3] = A
new[0][2] = old[2][3] = B
new[0][3] = old[3][3] = A
=> row0: A A B A matches top-right row0! Good.

new[1][0] = old[0][2] = B
new[1][1] = old[1][2] = B
new[1][2] = old[2][2] = A
new[1][3] = old[3][2] = B
=> row1: B B A B matches top-right row1! Good.

new[2][0] = old[0][1] = W
new[2][1] = old[1][1] = Y
new[2][2] = old[2][1] = A
new[2][3] = old[3][1] = B
=> row2: W Y A B matches top-right row2! Good.

new[3][0] = old[0][0] = Y
new[3][1] = old[1][0] = A
new[3][2] = old[2][0] = Y
new[3][3] = old[3][0] = B
=> row3: Y A Y B matches top-right row3! Good.

Thus top-right quadrant is input rotated 90° counterclockwise (or 270° clockwise). Good.

Now bottom-left quadrant (rows4-7, cols0-3):
Row4: A B B B
Row5: B A A Y
Row6: A B Y A
Row7: A B W Y

Let's see if that matches input rotated 180°.

Rotate input 180°: new[i][j] = old[N-1-i][N-1-j].

Compute:
old reversed both axes.

Original:
[ [Y,W,B,A],
  [A,Y,B,A],
  [Y,A,A,B],
  [B,B,B,A] ]

180 rot:
row0: old[3][3]=A, old[3][2]=B, old[3][1]=B, old[3][0]=B => A B B B matches row4? Row4: A B B B yes.
row1: old[2][3]=B, old[2][2]=A, old[2][1]=A, old[2][0]=Y => B A A Y matches row5: B A A Y yes.
row2: old[1][3]=A, old[1][2]=B, old[1][1]=Y, old[1][0]=A => A B Y A matches row6: A B Y A yes.
row3: old[0][3]=A, old[0][2]=B, old[0][1]=W, old[0][0]=Y => A B W Y matches row7: A B W Y yes.

Thus bottom-left is input rotated 180°.

Bottom-right quadrant (rows4-7, cols4-7):
Row4: B Y A Y
Row5: B A Y W
Row6: B A B B
Row7: A B A A

Let's see if that matches input rotated 90° clockwise (or 270 ccw). We earlier computed 90 cw gave:
B Y A Y
B A Y W
B A B B
A B A A

Matches exactly! So bottom-right is input rotated 90° clockwise.

Thus ordering: TL: 0° (original), TR: 90° CCW (or 270 CW), BL: 180°, BR: 90&

... [4362 chars truncated]

We need to understand the transformation. The description: "What you see: A small square mosaic depicting a divine face in colored tesserae. What to do: Erect a four-sided shrine at the crossroads. Each face of the shrine bears the same mosaic portrait, but rotated so the god appears to gaze down each of the four roads. Grid details: The shrine's unfolded plan is twice the original mosaic in each dimension—four panels in a 2×2 arrangement, each a rotated copy of the sacred image."

Thus input is a small square mosaic (presumably N x N). Output is a 2N x 2N grid composed of four panels arranged in a 2x2 arrangement, each a rotated copy of the original mosaic. The rotations likely correspond to 0°, 90°, 180°, 270° (or maybe each face of shrine oriented to a direction). The "unfolded plan" is twice the original mosaic in each dimension: i.e., output dimensions = 2 * input dimensions in both width and height. So if input is N x N, output is 2N x 2N.

The four panels: top-left, top-right, bottom-left, bottom-right? Or maybe arranged as: top-left = original orientation (0°), top-right = rotated 90° clockwise, bottom-left = rotated 180°, bottom-right = rotated 270° (or some order). Need to infer from training example.

Given Example 1 input and output.

Input:
Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

So N=4.

Output is 8x8 grid (2N). Let's write output rows as given:

Output:
Y W B A A A B A
A Y B A B B A B
Y A A B W Y A B
B B B A Y A Y B
A B B B B Y A Y
B A A Y B A Y W
A B Y A B A B B
A B W Y A B A A

Let's format as 8 rows, each with 8 characters separated by spaces.

Row0: Y W B A | A A B A
Row1: A Y B A | B B A B
Row2: Y A A B | W Y A B
Row3: B B B A | Y A Y B
Row4: A B B B | B Y A Y
Row5: B A A Y | B A Y W
Row6: A B Y A | B A B B
Row7: A B W Y | A B A A

We can see that the left half (first 4 columns) maybe is original? Let's compare left half of output to input.

Input:
Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

Left half of output rows 0-3:
Row0 left: Y W B A (matches input row0)
Row1 left: A Y B A (matches input row1)
Row2 left: Y A A B (matches input row2)
Row3 left: B B B A (matches input row3)

Thus top-left quadrant (rows 0-3, cols 0-3) is exactly the input (no rotation). Good.

Now top-right quadrant (rows 0-3, cols 4-7) appears to be something like input rotated? Let's extract top-right:

Rows0-3, cols4-7:
Row0: A A B A
Row1: B B A B
Row2: W Y A B
Row3: Y A Y B

Now compare to input rotated 90° clockwise? Let's compute input rotated 90° clockwise (i.e., transpose then reverse each row). Input:

[ [Y,W,B,A],
  [A,Y,B,A],
  [Y,A,A,B],
  [B,B,B,A] ]

Rotate 90° CW: new[i][j] = old[N-1-j][i].

Compute:

N=4.

For i=0 (new row0):
j=0: old[3][0] = B
j=1: old[2][0] = Y
j=2: old[1][0] = A
j=3: old[0][0] = Y
=> row0: B Y A Y

i=1:
j=0: old[3][1] = B
j=1: old[2][1] = A
j=2: old[1][1] = Y
j=3: old[0][1] = W
=> B A Y W

i=2:
j=0: old[3][2] = B
j=1: old[2][2] = A
j=2: old[1][2] = B
j=3: old[0][2] = B
=> B A B B

i=3:
j=0: old[3][3] = A
j=1: old[2][3] = B
j=2: old[1][3] = A
j=3: old[0][3] = A
=> A B A A

Thus 90° CW rotated matrix:
B Y A Y
B A Y W
B A B B
A B A A

Now compare to top-right we have:
Row0: A A B A
Row1: B B A B
Row2: W Y A B
Row3: Y A Y B

Not matching. Maybe it's rotated 90° counterclockwise? Let's compute 90° CCW (or -90): new[i][j] = old[j][N-1-i].

Compute:

i=0:
j=0: old[0][3] = A
j=1: old[1][3] = A
j=2: old[2][3] = B
j=3: old[3][3] = A
=> A A B A

i=1:
j=0: old[0][2] = B
j=1: old[1][2] = B
j=2: old[2][2] = A
j=3: old[3][2] = B
=> B B A B

i=2:
j=0: old[0][1] = W
j=1: old[1][1] = Y
j=2: old[2][1] = A
j=3: old[3][1] = B
=> W Y A B

i=3:
j=0: old[0][0] = Y
j=1: old[1][0] = A
j=2: old[2][0] = Y
j=3: old[3][0] = B
=> Y A Y B

This matches exactly the top-right quadrant! So top-right is input rotated 90° counterclockwise (or 270° clockwise). Good.

Now bottom-left quadrant (rows 4-7, cols 0-3) should be maybe rotated 180°? Let's extract bottom-left:

Rows4-7, cols0-3:
Row4: A B B B
Row5: B A A Y
Row6: A B Y A
Row7: A B W Y

Now compute input rotated 180° (i.e., reverse rows and reverse columns). Equivalent to rotate 90° CW twice.

Compute 180°: new[i][j] = old[N-1-i][N-1-j].

Let's compute:

Input again:
Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

Now for i=0 (new row0):
j=0: old[3][3] = A
j=1: old[3][2] = B
j=2: old[3][1] = B
j=3: old[3][0] = B
=> A B B B

i=1:
j=0: old[2][3] = B
j=1: old[2][2] = A
j=2: old[2][1] = A
j=3: old[2][0] = Y
=> B A A Y

i=2:
j=0: old[1][3] = A
j=1: old[1][2] = B
j=2: old[1][1] = Y
j=3: old[1][0] = A
=> A B Y A

i=3:
j=0: old[0][3] = A
j=1: old[0][2] = B
j=2: old[0][1] = W
j=3: old[0][0] = Y
=> A B W Y

Matches bottom-left exactly! So bottom-left is input rotated 180°.

Now bottom-right quadrant (rows4-7, cols4-7) should be rotated 270° CW (or 90° CCW? Actually we already

... [4133 chars truncated]

We need to understand the puzzle. The description: "What you see: A small square dance floor with colored position markers for a single couple's routine. What to do: Expand the ballroom to twice the size and station four couples in its quadrants. Each couple performs the identical choreography but begins facing a different wall, so their colored traces rotate relative to one another. Grid details: The full ballroom floor is a 2×2 grid of dance squares. Each square holds the same pattern of colored positions, rotated by a quarter-turn from its neighbor."

Thus input is a small square dance floor (presumably a 2x2? Actually the training example input is 4x4? Let's see: Example 1 input:

Y W B A
A Y B A
Y A A B
B B B A

That's 4 rows, 4 columns. Output is 8 rows, 8 columns (twice the size). So input is a 2x2 block of "dance squares"? Actually they say "full ballroom floor is a 2×2 grid of dance squares. Each square holds the same pattern of colored positions, rotated by a quarter-turn from its neighbor." So the input is a single dance square pattern? Wait: The input is the small square dance floor (the pattern for a single couple). Then we need to expand to twice the size (so double dimensions) and place four copies in quadrants, each rotated appropriately (0°, 90°, 180°, 270°). So output is 2x2 tiling of the input pattern, each rotated accordingly.

Thus transformation: given an N x N input (presumably square), output is 2N x 2N where top-left quadrant is input as-is (0° rotation), top-right quadrant is input rotated 90° clockwise? Or maybe rotated relative to neighbor: each quadrant rotated by quarter-turn from its neighbor. Need to determine orientation.

Given example: Input 4x4:

Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

Output 8x8:

Let's write output rows:

Row0: Y W B A A A B A
Row1: A Y B A B B A B
Row2: Y A A B W Y A B
Row3: B B B A Y A Y B
Row4: A B B B B Y A Y
Row5: B A A Y B A Y W
Row6: A B Y A B A B B
Row7: A B W Y A B A A

We need to see how this relates to input and its rotations.

Let's extract quadrants of output (each 4x4):

Top-left (rows0-3, cols0-3):
Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

That's exactly the input! So top-left quadrant is input unchanged (0°).

Top-right (rows0-3, cols4-7):
Row0: A A B A
Row1: B B A B
Row2: W Y A B
Row3: Y A Y B

Let's write as matrix:

[ [A, A, B, A],
  [B, B, A, B],
  [W, Y, A, B],
  [Y, A, Y, B] ]

Now, what is input rotated 90° clockwise? Let's compute.

Input:
[ [Y, W, B, A],
  [A, Y, B, A],
  [Y, A, A, B],
  [B, B, B, A] ]

Rotate 90° clockwise: new[i][j] = old[N-1-j][i] (if 0-index). Let's compute.

N=4.

Compute rotated 90 cw:

Row0: from original column0 bottom to top? Actually formula: new[i][j] = old[N-1-j][i].

Let's compute manually.

Original matrix:

(0,0)=Y (0,1)=W (0,2)=B (0,3)=A
(1,0)=A (1,1)=Y (1,2)=B (1,3)=A
(2,0)=Y (2,1)=A (2,2)=A (2,3)=B
(3,0)=B (3,1)=B (3,2)=B (3,3)=A

Now compute new:

i=0:
 j=0: old[3-0=3][0] = old[3][0]=B
 j=1: old[3-1=2][0] = old[2][0]=Y
 j=2: old[3-2=1][0] = old[1][0]=A
 j=3: old[3-3=0][0] = old[0][0]=Y
 => row0: B Y A Y

i=1:
 j=0: old[3][1]=B
 j=1: old[2][1]=A
 j=2: old[1][1]=Y
 j=3: old[0][1]=W
 => row1: B A Y W

i=2:
 j=0: old[3][2]=B
 j=1: old[2][2]=A
 j=2: old[1][2]=B
 j=3: old[0][2]=B
 => row2: B A B B

i=3:
 j=0: old[3][3]=A
 j=1: old[2][3]=B
 j=2: old[1][3]=A
 j=3: old[0][3]=A
 => row3: A B A A

Thus 90 cw rotation yields:

B Y A Y
B A Y W
B A B B
A B A A

But top-right quadrant we saw is:

A A B A
B B A B
W Y A B
Y A Y B

That does not match. Maybe it's rotated 90° counterclockwise? Let's compute 90 ccw (or 270 cw). Formula: new[i][j] = old[j][N-1-i].

Compute:

i=0:
 j=0: old[0][3]=A
 j=1: old[1][3]=A
 j=2: old[2][3]=B
 j=3: old[3][3]=A
 => row0: A A B A

i=1:
 j=0: old[0][2]=B
 j=1: old[1][2]=B
 j=2: old[2][2]=A
 j=3: old[3][2]=B
 => row1: B B A B

i=2:
 j=0: old[0][1]=W
 j=1: old[1][1]=Y
 j=2: old[2][1]=A
 j=3: old[3][1]=B
 => row2: W Y A B

i=3:
 j=0: old[0][0]=Y
 j=1: old[1][0]=A
 j=2: old[2][0]=Y
 j=3: old[3][0]=B
 => row3: Y A Y B

That matches exactly the top-right quadrant! So top-right is input rotated 90° counterclockwise (or 270° clockwise). Good.

Now bottom-left quadrant (rows4-7, cols0-3):

Rows4-7:

Row4: A B B B B Y A Y -> cols0-3: A B B B
Row5: B A A Y B A Y W -> cols0-3: B A A Y
Row6: A B Y A B A B B -> cols0-3: A B Y A
Row7: A B W Y A B A A -> cols0-3: A B W Y

Thus bottom-left quadrant:

[ [A, B, B, B],
  [B, A, A, Y],
  [A, B, Y, A],
  [A, B, W, Y] ]

Now what rotation of input yields this? Let's test 180° rotation (i.e., rotate twice). Compute input rotated 180°: equivalent to reverse both rows and columns.

Original:

Y W B A
A Y B A
Y A A B
B B B A

Reverse rows (flip vertically): becomes:

B B B A
Y A A B
A Y B A
Y W B A

Then reverse each row (flip horizontally): each row reversed:

Row0: A B B B
Row1: B A A Y
Row2: A B Y A
Row3:

... [5005 chars truncated]

We need to understand the transformation: Input is a small square cross-section of a crystal lattice (seed unit cell). We need to grow the crystal until it twins into four domains arranged in a 2x2 block. Each domain replicates the original unit cell but is rotated at the boundary, as though the lattice was turned before continuing growth. The twinned crystal cross-section is twice the dimensions of the seed lattice. Four domains tile it—each a rotated copy of the original unit cell, joined seamlessly at their twin planes.

Thus, given an N x N input grid (seed), we need to produce a 2N x 2N output grid, composed of four quadrants: top-left, top-right, bottom-left, bottom-right. Each quadrant is a rotated copy of the input. The rotations likely correspond to 0°, 90°, 180°, 270° (or some arrangement) such that they join seamlessly at twin planes (the boundaries between quadrants). The description: "Each domain replicates the original unit cell but is rotated at the boundary, as though the lattice was turned before continuing growth." So each domain is rotated relative to the original orientation, perhaps each successive domain rotated by 90° around the center? The twin planes are the lines between quadrants (vertical and horizontal middle lines). The rotated copies must match at the boundaries: i.e., the edge of one domain must match the edge of the adjacent domain after rotation.

Thus we need to find a rotation mapping such that when you place four rotated copies in a 2x2 block, the adjacent edges match.

Given the training example, we can deduce the rotation pattern.

Example 1 input:

Input:
Y W B A
A Y B A
Y A A B
B B B A

Let's write as matrix:

Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

Size N=4.

Output given:

Y W B A A A B A
A Y B A B B A B
Y A A B W Y A B
B B B A Y A Y B
A B B B B Y A Y
B A A Y B A Y W
A B Y A B A B B
A B W Y A B A A

Let's format as 8 rows, each 8 chars.

Row0: Y W B A A A B A
Row1: A Y B A B B A B
Row2: Y A A B W Y A B
Row3: B B B A Y A Y B
Row4: A B B B B Y A Y
Row5: B A A Y B A Y W
Row6: A B Y A B A B B
Row7: A B W Y A B A A

We need to see quadrants: top-left (rows0-3, cols0-3), top-right (rows0-3, cols4-7), bottom-left (rows4-7, cols0-3), bottom-right (rows4-7, cols4-7).

Extract TL:

Rows0-3, cols0-3:

Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

That's exactly the input! So TL quadrant is the original input (no rotation).

TR quadrant (rows0-3, cols4-7):

Row0: A A B A
Row1: B B A B
Row2: W Y A B
Row3: Y A Y B

Let's write as matrix:

Row0: A A B A
Row1: B B A B
Row2: W Y A B
Row3: Y A Y B

Is this a rotation of input? Let's see.

Input:
[ [Y,W,B,A],
  [A,Y,B,A],
  [Y,A,A,B],
  [B,B,B,A] ]

If we rotate input 90° clockwise, we get:

Original coordinates (r,c) -> new (c, N-1-r). Let's compute.

N=4.

Input matrix I:

I[0][0]=Y, I[0][1]=W, I[0][2]=B, I[0][3]=A
I[1][0]=A, I[1][1]=Y, I[1][2]=B, I[1][3]=A
I[2][0]=Y, I[2][1]=A, I[2][2]=A, I[2][3]=B
I[3][0]=B, I[3][1]=B, I[3][2]=B, I[3][3]=A

Rotate 90° CW: new[r][c] = I[N-1-c][r].

Compute:

new[0][0] = I[3][0] = B
new[0][1] = I[2][0] = Y
new[0][2] = I[1][0] = A
new[0][3] = I[0][0] = Y
=> row0: B Y A Y

new[1][0] = I[3][1] = B
new[1][1] = I[2][1] = A
new[1][2] = I[1][1] = Y
new[1][3] = I[0][1] = W
=> row1: B A Y W

new[2][0] = I[3][2] = B
new[2][1] = I[2][2] = A
new[2][2] = I[1][2] = B
new[2][3] = I[0][2] = B
=> row2: B A B B

new[3][0] = I[3][3] = A
new[3][1] = I[2][3] = B
new[3][2] = I[1][3] = A
new[3][3] = I[0][3] = A
=> row3: A B A A

So 90° CW gives:

B Y A Y
B A Y W
B A B B
A B A A

That does not match TR quadrant.

What about 90° CCW? new[r][c] = I[c][N-1-r].

Compute:

new[0][0] = I[0][3] = A
new[0][1] = I[1][3] = A
new[0][2] = I[2][3] = B
new[0][3] = I[3][3] = A
=> row0: A A B A

new[1][0] = I[0][2] = B
new[1][1] = I[1][2] = B
new[1][2] = I[2][2] = A
new[1][3] = I[3][2] = B
=> row1: B B A B

new[2][0] = I[0][1] = W
new[2][1] = I[1][1] = Y
new[2][2] = I[2][1] = A
new[2][3] = I[3][1] = B
=> row2: W Y A B

new[3][0] = I[0][0] = Y
new[3][1] = I[1][0] = A
new[3][2] = I[2][0] = Y
new[3][3] = I[3][0] = B
=> row3: Y A Y B

That matches exactly the TR quadrant we extracted! So TR is input rotated 90° counterclockwise (or 270° CW). Good.

Now BL quadrant (rows4-7, cols0-3):

Rows4-7, cols0-3:

Row4: A B B B
Row5: B A A Y
Row6: A B Y A
Row7: A B W Y

Let's write:

Row4: A B B B
Row5: B A A Y
Row6: A B Y A
Row7: A B W Y

Is this a rotation? Let's see 180° rotation of input.

Rotate 180°: new[r][c] = I[N-1-r][N-1-c].

Compute:

I reversed both axes.

I[3][3]=A, I[3][2]=B, I[3][1]=B, I[3][0]=B => row0: A B B B? Wait compute systematically.

new[0][0] = I[3][3] = A
new[0][1] = I[3][2] = B
new[0][2] = I[3][1] = B
new[0][3] = I[3][0] = B
=> row0: A B B B matches Row4? Yes Row4 is A B B B.

new[1][0] = I[2][3] = B
new[1][1] = I[2][2] = A
new[1][2] = I[2][1] = A
new[1][3] = I[2][0] = Y
=> row1: B A A Y match

... [4005 chars truncated]

We need to understand the transformation. The description: "Arrange four performances of the phrase into a larger score: the original reading, and three transformed versions each rotated as if the sheet music were physically turned on the music stand before playing. Grid details: The full score is a 2×2 arrangement of the original phrase's block, each quadrant a different rotational reading, producing a composition twice the size in each dimension."

Thus input is a small grid (the phrase). Output is a 2x2 arrangement of the original phrase and its three rotations (presumably 0°, 90°, 180°, 270°). The output size is twice each dimension: if input is N x M, output is 2N x 2M. The quadrants: top-left maybe original, top-right maybe rotated 90° clockwise? bottom-left maybe 180°, bottom-right maybe 270°? Or some order.

We need to infer from examples.

Let's examine Example 1.

Input:
Y W B A
A Y B A
Y A A B
B B B A

So input is 4x4? Let's count: 4 rows, each with 4 characters. Yes 4x4.

Output:
Y W B A A A B A
A Y B A B B A B
Y A A B W Y A B
B B B A Y A Y B
A B B B B Y A Y
B A A Y B A Y W
A B Y A B A B B
A B W Y A B A A

Output is 8 rows, each 8 characters? Let's count: first line: "Y W B A A A B A" => 8 tokens. So 8x8. Indeed 2*4 = 8.

Now we need to see how the quadrants are arranged.

Let's label input as I (4x4). We'll compute its rotations.

We need to see which quadrant matches which rotation.

Let's extract the four quadrants of output: top-left (rows 0-3, cols 0-3), top-right (rows 0-3, cols 4-7), bottom-left (rows 4-7, cols 0-3), bottom-right (rows 4-7, cols 4-7).

We'll write them.

First, output rows:

Row0: Y W B A | A A B A
Row1: A Y B A | B B A B
Row2: Y A A B | W Y A B
Row3: B B B A | Y A Y B
Row4: A B B B | B Y A Y
Row5: B A A Y | B A Y W
Row6: A B Y A | B A B B
Row7: A B W Y | A B A A

Now quadrants:

TL (0-3,0-3):
Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

That's exactly the input! So TL = original input (0° rotation).

TR (0-3,4-7):
Row0: A A B A
Row1: B B A B
Row2: W Y A B
Row3: Y A Y B

Let's write as matrix:

[ [A, A, B, A],
  [B, B, A, B],
  [W, Y, A, B],
  [Y, A, Y, B] ]

BL (4-7,0-3):
Row4: A B B B
Row5: B A A Y
Row6: A B Y A
Row7: A B W Y

Matrix:
[ [A, B, B, B],
  [B, A, A, Y],
  [A, B, Y, A],
  [A, B, W, Y] ]

BR (4-7,4-7):
Row4: B Y A Y
Row5: B A Y W
Row6: B A B B
Row7: A B A A

Matrix:
[ [B, Y, A, Y],
  [B, A, Y, W],
  [B, A, B, B],
  [A, B, A, A] ]

Now we need to see if these correspond to rotations of input.

Let's compute rotations of input I.

Input I:
[ [Y, W, B, A],
  [A, Y, B, A],
  [Y, A, A, B],
  [B, B, B, A] ]

Rotate 90° clockwise: new[i][j] = old[N-1-j][i] (assuming N=4). Let's compute.

We'll compute manually.

Indices: rows 0-3, cols 0-3.

Rot90CW: For each new cell (i,j) = old[3-j][i].

Let's produce matrix.

We'll compute column by column maybe easier: Rot90CW = transpose then reverse each row? Actually clockwise: reverse rows then transpose? Let's recall: rotate 90° clockwise: new = zip(*original[::-1]) (i.e., reverse rows then transpose). Let's do that.

Original rows:
0: Y W B A
1: A Y B A
2: Y A A B
3: B B B A

Reverse rows (so bottom to top):
3: B B B A
2: Y A A B
1: A Y B A
0: Y W B A

Now transpose: take columns of this reversed list as rows.

Column0: B, Y, A, Y => B Y A Y
Column1: B, A, Y, W => B A Y W
Column2: B, A, B, B => B A B B
Column3: A, B, A, A => A B A A

Thus Rot90CW = 
[ [B, Y, A, Y],
  [B, A, Y, W],
  [B, A, B, B],
  [A, B, A, A] ]

That matches exactly the BR quadrant we observed! So BR = 90° clockwise rotation.

Now Rot180°: rotate twice, or reverse rows and reverse columns? Equivalent to rotate 90° twice.

Let's compute Rot180: reverse each row and then reverse order of rows? Actually rotate 180: new[i][j] = old[N-1-i][N-1-j].

We can compute by taking original and reversing rows and then reversing each row.

Original:
Y W B A
A Y B A
Y A A B
B B B A

Reverse rows (bottom to top):
B B B A
Y A A B
A Y B A
Y W B A

Now reverse each row:
Row0: A B B B
Row1: B A A Y
Row2: A B Y A
Row3: A B W Y

That yields:
[ [A, B, B, B],
  [B, A, A, Y],
  [A, B, Y, A],
  [A, B, W, Y] ]

That matches BL quadrant! So BL = 180° rotation.

Now Rot270° clockwise (or 90° CCW). Should be TR quadrant.

Compute Rot270CW (or 90° CCW). Equivalent to rotate 90° CW three times, or transpose then reverse rows? Actually 90° CCW: new = zip(*original) then reverse each row? Let's compute.

Method: reverse each row then transpose? Let's just compute using formula: new[i][j] = old[j][N-1-i].

But we can also take Rot90CW and rotate again etc.

We have Rot90CW as above. Rot180 as above. Rot270CW = rotate Rot180 90° CW? Or easier: we can compute directly.

Let's compute Rot270CW (i.e., 90° CCW). Using reverse rows then transpose? Actually 90° CCW: transpose then reverse rows? Let's verify.

Take original, transpose, then reverse each row?

... [5058 chars truncated]

We need to understand the transformation. The description: "A small tactical diagram with colored position markers on a square clipboard. Deploy four squads onto a court twice the size: one runs the play as drawn, the others execute the same formation but oriented toward different edges of the field, so together they tile the full arena. The full court is four clipboard-sized zones. Each zone holds the same formation rotated to face a different direction, and together they fill the doubled playing surface without gaps."

Interpretation: Input is a square grid (size N x N). Output is a larger grid of size 2N x 2N, composed of four quadrants each being the input rotated appropriately (0°, 90°, 180°, 270°) and placed in the four quadrants (top-left, top-right, bottom-left, bottom-right) perhaps.

Specifically: "Deploy four squads onto a court twice the size: one runs the play as drawn, the others execute the same formation but oriented toward different edges of the field, so together they tile the full arena." So we have original formation (input). Then we need to produce four copies: one as is (maybe facing original direction), and three rotated to face different edges (i.e., rotated 90°, 180°, 270°). The full court is four clipboard-sized zones (so each zone is same size as input). Together they fill the doubled playing surface (so output is 2x in each dimension). So output is a 2N x 2N grid composed of four N x N blocks: each block is the input rotated by some angle (0, 90, 180, 270). The arrangement likely corresponds to rotating towards different edges: top-left block maybe original (facing up?), top-right rotated 90° (facing right?), bottom-left rotated 270° (facing left?), bottom-right rotated 180° (facing down?). Or some other arrangement.

We need to infer from the example.

Example 1 Input:
4x4 grid:
Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

Output is 8x8 grid given as:
Row0: Y W B A A A B A
Row1: A Y B A B B A B
Row2: Y A A B W Y A B
Row3: B B B A Y A Y B
Row4: A B B B B Y A Y
Row5: B A A Y B A Y W
Row6: A B Y A B A B B
Row7: A B W Y A B A A

Let's write them in matrix form to see quadrants.

I'll label rows 0-7, columns 0-7.

Row0: [Y,W,B,A, A,A,B,A]
Row1: [A,Y,B,A, B,B,A,B]
Row2: [Y,A,A,B, W,Y,A,B]
Row3: [B,B,B,A, Y,A,Y,B]
Row4: [A,B,B,B, B,Y,A,Y]
Row5: [B,A,A,Y, B,A,Y,W]
Row6: [A,B,Y,A, B,A,B,B]
Row7: [A,B,W,Y, A,B,A,A]

Now split into four 4x4 quadrants:

Top-left (rows0-3, cols0-3):
R0: Y W B A
R1: A Y B A
R2: Y A A B
R3: B B B A

That's exactly the input! So top-left quadrant is original input (0° rotation).

Top-right (rows0-3, cols4-7):
R0: A A B A
R1: B B A B
R2: W Y A B
R3: Y A Y B

Let's write that as matrix:
[ [A,A,B,A],
  [B,B,A,B],
  [W,Y,A,B],
  [Y,A,Y,B] ]

Bottom-left (rows4-7, cols0-3):
R4: A B B B
R5: B A A Y
R6: A B Y A
R7: A B W Y

Matrix:
[ [A,B,B,B],
  [B,A,A,Y],
  [A,B,Y,A],
  [A,B,W,Y] ]

Bottom-right (rows4-7, cols4-7):
R4: B Y A Y
R5: B A Y W
R6: B A B B
R7: A B A A

Matrix:
[ [B,Y,A,Y],
  [B,A,Y,W],
  [B,A,B,B],
  [A,B,A,A] ]

Now we need to see if these correspond to rotations of the input.

Input:
[ [Y,W,B,A],
  [A,Y,B,A],
  [Y,A,A,B],
  [B,B,B,A] ]

Let's compute rotations.

Rotate 90° clockwise: new[i][j] = old[N-1-j][i] (assuming standard). Let's compute.

Original (0°):
0: Y W B A
1: A Y B A
2: Y A A B
3: B B B A

Rotate 90° CW:
We'll compute each cell.

For i=0..3, j=0..3:
new[0][0] = old[3][0] = B
new[0][1] = old[2][0] = Y
new[0][2] = old[1][0] = A
new[0][3] = old[0][0] = Y
=> row0: B Y A Y

new[1][0] = old[3][1] = B
new[1][1] = old[2][1] = A
new[1][2] = old[1][1] = Y
new[1][3] = old[0][1] = W
=> row1: B A Y W

new[2][0] = old[3][2] = B
new[2][1] = old[2][2] = A
new[2][2] = old[1][2] = B
new[2][3] = old[0][2] = B
=> row2: B A B B

new[3][0] = old[3][3] = A
new[3][1] = old[2][3] = B
new[3][2] = old[1][3] = A
new[3][3] = old[0][3] = A
=> row3: A B A A

Thus 90° CW rotation yields:
[ [B,Y,A,Y],
  [B,A,Y,W],
  [B,A,B,B],
  [A,B,A,A] ]

That matches exactly the bottom-right quadrant we observed! So bottom-right is input rotated 90° CW.

Now rotate 180° (or two 90°). Let's compute 180°: rotate input 180° (or reverse both axes). Equivalent to rotating 90° twice.

Compute 180°: new[i][j] = old[N-1-i][N-1-j].

Let's compute quickly.

Original:
0: Y W B A
1: A Y B A
2: Y A A B
3: B B B A

180°:
Row0: old[3][3]=A, old[3][2]=B, old[3][1]=B, old[3][0]=B => A B B B
Row1: old[2][3]=B, old[2][2]=A, old[2][1]=A, old[2][0]=Y => B A A Y
Row2: old[1][3]=A, old[1][2]=B, old[1][1]=Y, old[1][0]=A => A B Y A
Row3: old[0][3]=A, old[0][2]=B, old[0][1]=W, old[0][0]=Y => A B W Y

Thus 180° rotation yields:
[ [A,B,B,B],
  [B,A,A,Y],
  [A,B,Y,A],
  [A,B,W,Y] ]

That matches bottom-left quadrant we observed! So bottom-left is input rotated 180°.

Now rotate 270° CW (or 90° CCW). Let's compute 270° CW (which is same as 

... [4741 chars truncated]

We need to understand the transformation. The description: "A small square radar return showing a colorful storm pattern. The storm propagates into four quadrants of a larger radar display. Each quadrant carries the same precipitation pattern but rotated as though the wind shear twisted each front to a different compass bearing. Grid details: The full radar image is twice the dimensions of the original cell, tiled by four copies of the storm pattern, each rotated by successive quarter-turns."

Thus input is a small square (presumably N x N). Output is a larger square of size 2N x 2N, composed of four quadrants: top-left, top-right, bottom-left, bottom-right. Each quadrant is the input pattern rotated by 0°, 90°, 180°, 270° (or some order). The description: "Each quadrant carries the same precipitation pattern but rotated as though the wind shear twisted each front to a different compass bearing." "Successive quarter-turns" suggests each quadrant is rotated by 0°, 90°, 180°, 270° relative to original, perhaps in order top-left (0°), top-right (90°), bottom-left (180°), bottom-right (270°) or some other arrangement.

We need to verify with training examples.

Example 1:

Input:
Y W B A
A Y B A
Y A A B
B B B A

So N=4. Output is 8x8.

Let's write input matrix:

Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

Now output given:

Row0: Y W B A A A B A
Row1: A Y B A B B A B
Row2: Y A A B W Y A B
Row3: B B B A Y A Y B
Row4: A B B B B Y A Y
Row5: B A A Y B A Y W
Row6: A B Y A B A B B
Row7: A B W Y A B A A

We need to see if this matches four quadrants: top-left (rows0-3, cols0-3) should be input rotated 0°? Let's extract top-left 4x4:

Rows0-3, cols0-3:

Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

That's exactly the input! So top-left quadrant is input unchanged (0° rotation).

Top-right quadrant (rows0-3, cols4-7):

Row0: A A B A
Row1: B B A B
Row2: W Y A B
Row3: Y A Y B

Let's see if that matches input rotated 90° clockwise? Or counterclockwise? Let's compute input rotated 90° clockwise (i.e., transpose then reverse each row). Input:

[ [Y,W,B,A],
  [A,Y,B,A],
  [Y,A,A,B],
  [B,B,B,A] ]

Rotate 90° clockwise: new[i][j] = old[N-1-j][i] (if 0-index). Let's compute.

N=4.

Compute rotated 90 cw:

Row0: old[3][0]=B, old[2][0]=Y, old[1][0]=A, old[0][0]=Y => B Y A Y? Wait check: For i=0 (new row0), j from 0 to 3: new[0][j] = old[3-j][0].

j=0: old[3][0]=B
j=1: old[2][0]=Y
j=2: old[1][0]=A
j=3: old[0][0]=Y

Thus row0: B Y A Y.

Row1: new[1][j] = old[3-j][1].

j=0: old[3][1]=B
j=1: old[2][1]=A
j=2: old[1][1]=Y
j=3: old[0][1]=W

=> B A Y W.

Row2: new[2][j] = old[3-j][2].

j=0: old[3][2]=B
j=1: old[2][2]=A
j=2: old[1][2]=B
j=3: old[0][2]=B

=> B A B B.

Row3: new[3][j] = old[3-j][3].

j=0: old[3][3]=A
j=1: old[2][3]=B
j=2: old[1][3]=A
j=3: old[0][3]=A

=> A B A A.

Thus rotated 90 cw matrix:

B Y A Y
B A Y W
B A B B
A B A A

Now compare to top-right quadrant we extracted:

Row0: A A B A
Row1: B B A B
Row2: W Y A B
Row3: Y A Y B

Not matching. Maybe it's rotated 90° counterclockwise? Let's compute 90 ccw (or -90). Formula: new[i][j] = old[j][N-1-i].

Compute:

Row0: j from 0..3: old[j][3] (since N-1-i = 3-0=3). So column3 of original: A, A, B, A => A A B A. That matches top-right row0! Good.

Row1: i=1: old[j][2] (col2): B, B, A, B => B B A B matches row1.

Row2: i=2: old[j][1] (col1): W, A, A, B? Wait original col1: row0 W, row1 Y, row2 A, row3 B => W Y A B matches row2.

Row3: i=3: old[j][0] (col0): Y, A, Y, B => Y A Y B matches row3.

Thus top-right quadrant is input rotated 90° counterclockwise (or 270° clockwise). Good.

Now bottom-left quadrant (rows4-7, cols0-3):

Rows4-7, cols0-3:

Row4: A B B B
Row5: B A A Y
Row6: A B Y A
Row7: A B W Y

Let's see if that matches input rotated 180° (i.e., reverse both axes). Compute input rotated 180: reverse rows and reverse each row (or rotate 90 cw twice). Let's compute: original:

Y W B A
A Y B A
Y A A B
B B B A

Reverse rows order: bottom to top:

B B B A
Y A A B
A Y B A
Y W B A

Then reverse each row:

Row0: A B B B
Row1: B A A Y
Row2: A B Y A
Row3: A B W Y

Matches bottom-left quadrant exactly! Good.

Now bottom-right quadrant (rows4-7, cols4-7):

Row4: B Y A Y
Row5: B A Y W
Row6: B A B B
Row7: A B A A

Let's see if that matches input rotated 90° clockwise (or 270 ccw). We earlier computed rotated 90 cw:

B Y A Y
B A Y W
B A B B
A B A A

Matches bottom-right quadrant! Good.

Thus pattern: TL = original (0°), TR = rotate 90° CCW (or -90), BL = rotate 180°, BR = rotate 90° CW.

Thus transformation: output size 2N x 2N, where quadrants are as described.

Now test input:

Y M Y Y
Y M Y Y
O M O W
W Y W O

N=4.

We need to compute quadrants.

First, original (TL) = same as input.

TR = input rotated 90° CCW.

BL = input rotated 180°.

BR = input rotated 90° CW.

Let's compute.

Input matrix:

R

... [5733 chars truncated]

We need to understand the transformation: Input is a small square decorative tile (size? The training example input is 4x4? Let's see: Example 1 input:

Y W B A
A Y B A
Y A A B
B B B A

That's 4 rows, each with 4 characters separated by spaces. So input tile is 4x4.

Output is a 2x2 panel of four copies of the tile, each rotated so pattern radiates outward in a different direction. The output grid is 8x8 (twice width and height). Indeed output has 8 rows, each with 8 characters.

Thus transformation: given a tile (N x N), produce a 2N x 2N grid composed of four quadrants: top-left, top-right, bottom-left, bottom-right. Each quadrant is the input tile rotated by some angle (0°, 90°, 180°, 270°) such that pattern radiates outward. Likely each quadrant is rotated so that a particular direction (like north, east, south, west) points outward from center? The description: "Mount four copies of the tile into a 2×2 panel. Each tile is rotated so that the pattern radiates outward in a different direction, as though the artisan turned each piece before grouting it into place." So we have four copies, each rotated differently, placed in quadrants such that the pattern radiates outward from the center (like a radial pattern). So we need to determine which rotation corresponds to each quadrant.

We need to infer from the example.

Given input tile (4x4). Output 8x8 shown. Let's label quadrants:

Quadrant TL (top-left): rows 0-3, cols 0-3
Quadrant TR (top-right): rows 0-3, cols 4-7
Quadrant BL (bottom-left): rows 4-7, cols 0-3
Quadrant BR (bottom-right): rows 4-7, cols 4-7

We need to see what rotation of input yields each quadrant.

Let's write input tile matrix:

Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

Now output rows:

Row0: Y W B A A A B A
Row1: A Y B A B B A B
Row2: Y A A B W Y A B
Row3: B B B A Y A Y B
Row4: A B B B B Y A Y
Row5: B A A Y B A Y W
Row6: A B Y A B A B B
Row7: A B W Y A B A A

Let's extract quadrants:

TL (rows0-3, cols0-3):
Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

That's exactly the input tile! So TL quadrant is input with 0° rotation (no rotation).

TR (rows0-3, cols4-7):
Row0: A A B A
Row1: B B A B
Row2: W Y A B
Row3: Y A Y B

Let's write as matrix:

Row0: A A B A
Row1: B B A B
Row2: W Y A B
Row3: Y A Y B

Now compare to input rotated? Let's see if this is input rotated 90° clockwise? Let's compute input rotated 90° clockwise (i.e., transpose then reverse each row). Input:

[ [Y,W,B,A],
  [A,Y,B,A],
  [Y,A,A,B],
  [B,B,B,A] ]

Rotate 90° CW: new[i][j] = old[N-1-j][i] (if 0-index). Let's compute.

N=4.

Compute rotated 90 CW:

For i=0 (new row0):
j=0: old[3][0] = B
j=1: old[2][0] = Y
j=2: old[1][0] = A
j=3: old[0][0] = Y
=> row0: B Y A Y

i=1:
j=0: old[3][1] = B
j=1: old[2][1] = A
j=2: old[1][1] = Y
j=3: old[0][1] = W
=> B A Y W

i=2:
j=0: old[3][2] = B
j=1: old[2][2] = A
j=2: old[1][2] = B
j=3: old[0][2] = B
=> B A B B

i=3:
j=0: old[3][3] = A
j=1: old[2][3] = B
j=2: old[1][3] = A
j=3: old[0][3] = A
=> A B A A

Thus 90 CW rotation yields:

B Y A Y
B A Y W
B A B B
A B A A

That does not match TR quadrant.

What about 90° counterclockwise (i.e., rotate left)? That's transpose then reverse columns? Actually 90 CCW: new[i][j] = old[j][N-1-i].

Compute 90 CCW:

i=0:
j=0: old[0][3] = A
j=1: old[1][3] = A
j=2: old[2][3] = B
j=3: old[3][3] = A
=> A A B A

i=1:
j=0: old[0][2] = B
j=1: old[1][2] = B
j=2: old[2][2] = A
j=3: old[3][2] = B
=> B B A B

i=2:
j=0: old[0][1] = W
j=1: old[1][1] = Y
j=2: old[2][1] = A
j=3: old[3][1] = B
=> W Y A B

i=3:
j=0: old[0][0] = Y
j=1: old[1][0] = A
j=2: old[2][0] = Y
j=3: old[3][0] = B
=> Y A Y B

That matches exactly the TR quadrant! So TR is input rotated 90° CCW (or 270° CW). Good.

Now BL quadrant (rows4-7, cols0-3):

Rows4-7, cols0-3:

Row4: A B B B
Row5: B A A Y
Row6: A B Y A
Row7: A B W Y

Let's write:

Row4: A B B B
Row5: B A A Y
Row6: A B Y A
Row7: A B W Y

Now compare to input rotated 180°? Let's compute input rotated 180° (i.e., reverse rows and reverse columns). Equivalent to rotate 90 CW twice.

Compute 180 rotation: new[i][j] = old[N-1-i][N-1-j].

Let's compute:

Original:

[ [Y,W,B,A],
  [A,Y,B,A],
  [Y,A,A,B],
  [B,B,B,A] ]

Now 180:

i=0:
j=0: old[3][3] = A
j=1: old[3][2] = B
j=2: old[3][1] = B
j=3: old[3][0] = B
=> A B B B

i=1:
j=0: old[2][3] = B
j=1: old[2][2] = A
j=2: old[2][1] = A
j=3: old[2][0] = Y
=> B A A Y

i=2:
j=0: old[1][3] = A
j=1: old[1][2] = B
j=2: old[1][1] = Y
j=3: old[1][0] = A
=> A B Y A

i=3:
j=0: old[0][3] = A
j=1: old[0][2] = B
j=2: old[0][1] = W
j=3: old[0][0] = Y
=> A B W Y

Matches BL quadrant! So BL is input rotated 180°.

Now BR quadrant (rows4-7, cols4-7):

Rows4-7, cols4-7:

Row4: B Y A Y
Row5: B A Y W
Row6: B A B B
Row7: A B A A

Let's write:

Row4: B Y A Y
Row5: B A Y W
Row6: B A B B
Row7: A B A A

Now compare to input rotated 90° CW? We earlier computed 90 CW gave:

B Y A Y
B A Y W

... [5126 chars truncated]

We need to understand the transformation: Input is a small colored battle map showing arrangement of troops in a single formation. Output is a fortress map twice the span of one formation in each direction, with four identical regiments each rotated to a different heading, tiled interior perfectly.

Thus input is a formation (some pattern). Output is a 2x2 tiling of the formation, each copy rotated to face a different cardinal wall (i.e., rotated 0°, 90°, 180°, 270°). The formation is placed in each quadrant of a larger grid (size double in each dimension). So output dimensions: if input is HxW, output is 2H x 2W. The four quadrants: top-left, top-right, bottom-left, bottom-right. Each quadrant contains the formation rotated appropriately: maybe top-left is original orientation (facing north?), top-right rotated 90° clockwise (facing east?), bottom-right rotated 180° (facing south?), bottom-left rotated 270° (facing west?). Or some mapping.

We need to infer from training example.

Given Example 1 input:

Input:
Y W B A
A Y B A
Y A A B
B B B A

Let's write as matrix:

Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

So H=4, W=4.

Output given:

Y W B A A A B A
A Y B A B B A B
Y A A B W Y A B
B B B A Y A Y B
A B B B B Y A Y
B A A Y B A Y W
A B Y A B A B B
A B W Y A B A A

Let's format as 8 rows (2*4) each with 8 columns (2*4). Let's split into quadrants:

Top-left quadrant (rows 0-3, cols 0-3):
Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

That's exactly the input! So top-left is original orientation (0°).

Top-right quadrant (rows 0-3, cols 4-7):
Row0: A A B A
Row1: B B A B
Row2: W Y A B
Row3: Y A Y B

Let's extract: Actually need to check: The output rows:

Row0: Y W B A | A A B A
Row1: A Y B A | B B A B
Row2: Y A A B | W Y A B
Row3: B B B A | Y A Y B

Thus top-right quadrant matrix:
[ [A, A, B, A],
  [B, B, A, B],
  [W, Y, A, B],
  [Y, A, Y, B] ]

Bottom-left quadrant (rows 4-7, cols 0-3):
Row4: A B B B
Row5: B A A Y
Row6: A B Y A
Row7: A B W Y

Thus:
[ [A, B, B, B],
  [B, A, A, Y],
  [A, B, Y, A],
  [A, B, W, Y] ]

Bottom-right quadrant (rows 4-7, cols 4-7):
Row4: B Y A Y
Row5: B A Y W
Row6: B A B B
Row7: A B A A

Thus:
[ [B, Y, A, Y],
  [B, A, Y, W],
  [B, A, B, B],
  [A, B, A, A] ]

Now we need to see how these relate to rotations of the input.

Input matrix I:
[ [Y, W, B, A],
  [A, Y, B, A],
  [Y, A, A, B],
  [B, B, B, A] ]

Let's compute rotations:

Rotate 90° clockwise: new[i][j] = old[H-1-j][i] (if we consider standard rotation). Let's compute.

But maybe easier: Let's compute 90° clockwise of I.

I (4x4):
0: Y W B A
1: A Y B A
2: Y A A B
3: B B B A

Rotate 90° CW: resulting 4x4 where first row = column 0 from bottom to top? Actually formula: new[i][j] = old[H-1-j][i].

Let's compute manually: We'll produce matrix R90.

We can also think: transpose then reverse each row.

Transpose of I:
col0: Y, A, Y, B => [Y, A, Y, B]
col1: W, Y, A, B => [W, Y, A, B]
col2: B, B, A, B => [B, B, A, B]
col3: A, A, B, A => [A, A, B, A]

Now reverse each row (for CW): reverse each of those rows.

Row0 reversed: [B, Y, A, Y]? Wait row0 of transpose is [Y, A, Y, B]; reversed => [B, Y, A, Y].
Row1: [W, Y, A, B] reversed => [B, A, Y, W].
Row2: [B, B, A, B] reversed => [B, A, B, B].
Row3: [A, A, B, A] reversed => [A, B, A, A].

Thus R90 = 
[ [B, Y, A, Y],
  [B, A, Y, W],
  [B, A, B, B],
  [A, B, A, A] ]

That matches exactly the bottom-right quadrant we observed! Good.

Thus bottom-right quadrant is rotation 90° CW of input.

Now bottom-left quadrant? Let's compute rotation 180° (or maybe 270° CW). Let's compute 180° rotation: rotate twice CW, or reverse both rows and columns.

180° of I: reverse each row and then reverse order of rows (or equivalently rotate 90 twice). Let's compute: reverse each row of I:

I rows reversed:
Row0: A B W Y
Row1: A B Y A
Row2: B A A Y
Row3: A B B B

Now reverse row order: 
Row0 becomes previous row3 reversed? Actually after reversing rows, we get:
[ A B B B,
  B A A Y,
  A B Y A,
  A B W Y ]

That matches bottom-left quadrant we saw! Indeed bottom-left quadrant is:
[ [A, B, B, B],
  [B, A, A, Y],
  [A, B, Y, A],
  [A, B, W, Y] ]

Thus bottom-left is 180° rotation.

Now top-right quadrant should be 270° CW (or 90° CCW). Let's compute 270° CW (which is same as 90° CCW). Compute by rotating I 90° CCW: transpose then reverse columns? Actually 90° CCW = reverse each row then transpose? Let's compute: 90° CCW = rotate -90°, which is transpose then reverse each column? Simpler: 270° CW = three times 90° CW. We can compute by taking R90 and rotate again etc. But we can also compute directly: 90° CCW = reverse each row then transpose.

Let's compute: reverse each row of I:
Row0: A B W Y
Row1: A B Y A
Row2: B A A Y
Row3: A B B B

Now transpose that:
Take columns of reversed matrix.

Reversed matrix M:
[ A B W Y
  A B Y A


... [5186 chars truncated]

We need to understand the transformation: The description: "Photograph four performances of the same tableau, each taken after the revolving stage has turned another quarter-rotation. Arrange the four photographs into a grid to create one composite panorama. Grid details: The panoramic image is twice the height and width of a single stage shot. Each quadrant holds one rotation of the tableau, and together they fill the composite without overlap or gaps."

Thus input is a small square stage (the tableau). Output is a 2x2 arrangement of four rotated versions of the input: each rotated by 0°, 90°, 180°, 270° (quarter-turns). The arrangement likely is: top-left: original (0°), top-right: rotated 90° clockwise? Or maybe they are arranged in order of rotations: first photo (0°) top-left, second (90°) top-right, third (180°) bottom-left, fourth (270°) bottom-right. Or maybe they are arranged left-to-right, top-to-bottom as they are taken sequentially. The description: "Photograph four performances of the same tableau, each taken after the revolving stage has turned another quarter-rotation. Arrange the four photographs into a grid to create one composite panorama." So you take a photo, then rotate stage 90°, take another photo, etc. Then you arrange the four photographs into a grid. Likely you place them in the order taken: first photo top-left, second top-right, third bottom-left, fourth bottom-right. That yields a 2x2 grid where each quadrant is a rotated version.

Check with examples to confirm.

Example 1:

Input:
Y W B A
A Y B A
Y A A B
B B B A

Let's denote input as 4x4.

Output is 8x8 (twice height and width). Provided output lines:

Line1: Y W B A A A B A
Line2: A Y B A B B A B
Line3: Y A A B W Y A B
Line4: B B B A Y A Y B
Line5: A B B B B Y A Y
Line6: B A A Y B A Y W
Line7: A B Y A B A B B
Line8: A B W Y A B A A

We need to see if this matches arrangement of rotations.

Let's compute rotations of input.

Input matrix (4x4):
Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

Let's label coordinates (r,c). We'll compute 90° clockwise rotation: new[i][j] = old[ n-1-j ][i] (if rotating clockwise). Or we can compute manually.

Better to compute using Python mentally? Let's do manually.

Original:
[0,0]=Y [0,1]=W [0,2]=B [0,3]=A
[1,0]=A [1,1]=Y [1,2]=B [1,3]=A
[2,0]=Y [2,1]=A [2,2]=A [2,3]=B
[3,0]=B [3,1]=B [3,2]=B [3,3]=A

Rotate 90° clockwise: resulting 4x4 where new[r][c] = old[3-c][r].

Compute new matrix:

For r=0:
c=0: old[3-0=3][0] = old[3][0] = B
c=1: old[3-1=2][0] = old[2][0] = Y
c=2: old[3-2=1][0] = old[1][0] = A
c=3: old[3-3=0][0] = old[0][0] = Y
So row0: B Y A Y

r=1:
c=0: old[3][1] = B
c=1: old[2][1] = A
c=2: old[1][1] = Y
c=3: old[0][1] = W
Row1: B A Y W

r=2:
c=0: old[3][2] = B
c=1: old[2][2] = A
c=2: old[1][2] = B
c=3: old[0][2] = B
Row2: B A B B

r=3:
c=0: old[3][3] = A
c=1: old[2][3] = B
c=2: old[1][3] = A
c=3: old[0][3] = A
Row3: A B A A

Thus 90° CW rotation yields:
B Y A Y
B A Y W
B A B B
A B A A

Now 180° rotation (rotate twice) = rotate 90° CW of the 90° CW matrix, or equivalently reverse both rows and columns: new[r][c] = old[3-r][3-c].

Compute 180°: take original and flip both axes.

Original:
Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

Flip vertically (reverse rows): becomes:
Row0: B B B A
Row1: Y A A B
Row2: A Y B A
Row3: Y W B A

Then flip horizontally (reverse each row): each row reversed:
Row0: A B B B
Row1: B A A Y
Row2: A B Y A
Row3: A B W Y

Thus 180° rotation:
A B B B
B A A Y
A B Y A
A B W Y

Now 270° CW (or 90° CCW) = rotate 90° CCW: new[r][c] = old[c][3-r].

Compute: or we can take 90° CW three times, but let's compute directly.

For each r,c:
new[r][c] = old[c][3-r].

Let's compute:

r=0:
c=0: old[0][3] = A
c=1: old[1][3] = A
c=2: old[2][3] = B
c=3: old[3][3] = A
Row0: A A B A

r=1:
c=0: old[0][2] = B
c=1: old[1][2] = B
c=2: old[2][2] = A
c=3: old[3][2] = B
Row1: B B A B

r=2:
c=0: old[0][1] = W
c=1: old[1][1] = Y
c=2: old[2][1] = A
c=3: old[3][1] = B
Row2: W Y A B

r=3:
c=0: old[0][0] = Y
c=1: old[1][0] = A
c=2: old[2][0] = Y
c=3: old[3][0] = B
Row3: Y A Y B

Thus 270° CW:
A A B A
B B A B
W Y A B
Y A Y B

Now we have four rotations: 0° (original), 90°, 180°, 270°.

Now we need to see arrangement in output.

Output 8x8 given as:

Row0: Y W B A A A B A
Row1: A Y B A B B A B
Row2: Y A A B W Y A B
Row3: B B B A Y A Y B
Row4: A B B B B Y A Y
Row5: B A A Y B A Y W
Row6: A B Y A B A B B
Row7: A B W Y A B A A

Let's split into quadrants: top-left 4x4 (rows0-3, cols0-3), top-right (rows0-3, cols4-7), bottom-left (rows4-7, cols0-3), bottom-right (rows4-7, cols4-7).

Extract:

TL (rows0-3, cols0-3):
Row0: Y W B
Row1: A Y B
Row2: Y A A
Row3: B B B

That's:
Y W B
A Y B
Y A A
B B B

Compare to original input:
Original:
Y W B A
A Y B A
Y A A B
B B B A

TL is missing the last column of each row (the 4th column). Actually TL is first

... [5072 chars truncated]

We need to understand the transformation: Input is a small square planting plan (presumably N x N). Output is a 2x2 arrangement of beds, each same size as original plan, each bed is a rotated copy, and together they tile the plot completely. So output dimensions: 2N x 2N.

The description: "Lay out a garden twice the size by filling four adjacent beds. Each bed follows the same planting plan but rotated so the rows that once ran east-west now run north-south, and so on for each successive bed." Means we take the original pattern and place it in four quadrants, each rotated relative to original: perhaps top-left is original orientation (0°), top-right rotated 90° clockwise? bottom-left rotated 180°, bottom-right rotated 270°? Or maybe each successive bed rotates 90° relative to previous. The phrase: "Each bed follows the same planting plan but rotated so the rows that once ran east-west now run north-south, and so on for each successive bed." Means first bed: rows east-west (original). Second bed: rows north-south (i.e., rotated 90°). Third bed: rows east-west again? Actually "and so on for each successive bed" suggests each bed rotates further: maybe each bed is rotated 90° more than previous. So we have four beds: orientation 0°, 90°, 180°, 270°.

We need to verify with training examples.

Example 1 input:
4x4:
Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

Output is 8x8:
Given as lines:
Y W B A A A B A
A Y B A B B A B
Y A A B W Y A B
B B B A Y A Y B
A B B B B Y A Y
B A A Y B A Y W
A B Y A B A B B
A B W Y A B A A

Let's try to see if this matches 2x2 tiling with rotations.

We need to partition output into four 4x4 blocks: top-left (rows0-3, cols0-3), top-right (rows0-3, cols4-7), bottom-left (rows4-7, cols0-3), bottom-right (rows4-7, cols4-7).

Let's extract.

Top-left block (rows0-3, cols0-3):
Row0: Y W B A
Row1: A Y B A
Row2: Y A A B
Row3: B B B A

That's exactly the input! So top-left is original orientation (0°).

Top-right block (rows0-3, cols4-7):
Row0: A A B A
Row1: B B A B
Row2: W Y A B
Row3: Y A Y B

Let's write as matrix:
[ [A, A, B, A],
  [B, B, A, B],
  [W, Y, A, B],
  [Y, A, Y, B] ]

Is this a rotation of input? Let's rotate input 90° clockwise (or counterclockwise) and see.

Original input:
[ [Y, W, B, A],
  [A, Y, B, A],
  [Y, A, A, B],
  [B, B, B, A] ]

Rotate 90° clockwise: new[i][j] = original[N-1-j][i] (if 0-index). Let's compute.

N=4.

Compute clockwise 90:
Row0: original[3][0]=B, original[2][0]=Y, original[1][0]=A, original[0][0]=Y => B Y A Y? Wait compute properly: new[0][0] = original[3][0] = B
new[0][1] = original[2][0] = Y
new[0][2] = original[1][0] = A
new[0][3] = original[0][0] = Y
=> B Y A Y

Row1: new[1][0] = original[3][1] = B
new[1][1] = original[2][1] = A
new[1][2] = original[1][1] = Y
new[1][3] = original[0][1] = W
=> B A Y W

Row2: new[2][0] = original[3][2] = B
new[2][1] = original[2][2] = A
new[2][2] = original[1][2] = B
new[2][3] = original[0][2] = B
=> B A B B

Row3: new[3][0] = original[3][3] = A
new[3][1] = original[2][3] = B
new[3][2] = original[1][3] = A
new[3][3] = original[0][3] = A
=> A B A A

So clockwise 90 gives:
B Y A Y
B A Y W
B A B B
A B A A

That does not match top-right block.

What about 90° counterclockwise? new[i][j] = original[j][N-1-i].

Compute CCW 90:
Row0: original[0][3]=A, original[1][3]=A, original[2][3]=B, original[3][3]=A => A A B A
Row1: original[0][2]=B, original[1][2]=B, original[2][2]=A, original[3][2]=B => B B A B
Row2: original[0][1]=W, original[1][1]=Y, original[2][1]=A, original[3][1]=B => W Y A B
Row3: original[0][0]=Y, original[1][0]=A, original[2][0]=Y, original[3][0]=B => Y A Y B

That matches exactly top-right block! So top-right is input rotated 90° counterclockwise (or equivalently 270° clockwise). Good.

Now bottom-left block (rows4-7, cols0-3):
Rows4-7:
Row4: A B B B B Y A Y
Row5: B A A Y B A Y W
Row6: A B Y A B A B B
Row7: A B W Y A B A A

Take cols0-3:
Row4: A B B B
Row5: B A A Y
Row6: A B Y A
Row7: A B W Y

Matrix:
[ [A, B, B, B],
  [B, A, A, Y],
  [A, B, Y, A],
  [A, B, W, Y] ]

Is this a rotation? Let's see 180° rotation of input (i.e., rotate twice). Rotate input 180: reverse rows and columns.

Original:
Y W B A
A Y B A
Y A A B
B B B A

Reverse rows:
B B B A
Y A A B
A Y B A
Y W B A

Then reverse each row (or reverse columns): Actually 180 = rotate 180: new[i][j] = original[N-1-i][N-1-j].

Compute:
Row0: original[3][3]=A, original[3][2]=B, original[3][1]=B, original[3][0]=B => A B B B
Row1: original[2][3]=B, original[2][2]=A, original[2][1]=A, original[2][0]=Y => B A A Y
Row2: original[1][3]=A, original[1][2]=B, original[1][1]=Y, original[1][0]=A => A B Y A
Row3: original[0][3]=A, original[0][2]=B, original[0][1]=W, original[0][0]=Y => A B W Y

Matches bottom-left! So bottom-left is 180° rotation.

Bottom-right block (rows4-7, cols4-7):
Take cols4-7 of rows4-7:
Row4: B Y A Y
Ro

... [5174 chars truncated]

We need to parse the problem: It's a grid transformation puzzle. Input grid: appears to be 3 rows, each row has 9 characters? Let's see: Example 1 input:

M G G X . . . X . . .
M G G X . . . X . . .
M G R X . . . X . . .

So each row: M G G X . . . X . . . That's 9 characters: M, G, G, X, ., ., ., X, ., ., .? Wait count: Let's list: M (1), G (2), G (3), X (4), . (5), . (6), . (7), X (8), . (9), . (10), . (11). That's 11? Actually maybe they have spaces between characters, but they show "M G G X . . . X . . ." which is 11 tokens: M, G, G, X, ., ., ., X, ., ., . . That's 11. But they said each mirror is three panes tall and three panes wide, separated by narrow grey curtain-rods. So each mirror is 3x3. Triptych: three mirrors side by side, each 3x3, with grey curtain-rods between them (vertical lines?). So total width: each mirror 3 columns, plus two curtain-rods (between mirror1-2 and mirror2-3) each 1 column? So total width = 3*3 + 2*1 = 9+2=11 columns. Height: each mirror 3 rows, no horizontal rods? So total height = 3 rows. So grid is 3 rows x 11 columns. That matches the example: 11 tokens per row.

Thus the grid includes the mirrors and the grey curtain-rods (X) as separators. The leftmost mirror holds the dancer's living colors (non-X). The other two mirrors initially are dark? In examples, the middle and right mirrors have some pattern? Let's examine example 1 input: Row1: M G G X . . . X . . . So columns: 0-2: M G G (left mirror top row). Column3: X (curtain rod between left and middle). Columns4-6: . . . (middle mirror top row). Column7: X (curtain rod between middle and right). Columns8-10: . . . (right mirror top row). Row2: M G G X . . . X . . . (same). Row3: M G R X . . . X . . . (left mirror bottom row: M G R; middle: . . .; right: . . .). So only left mirror has non-X and non-. colors; middle and right are all '.' (black) except maybe some? Actually they are all '.' (black). So initial state: left mirror has some pattern; middle and right are all black.

Output for example 1:
M G G X M M M X R G M
M G G X G G G X G G M
M G R X R G G X G G M

Let's parse output: Row1: M G G X M M M X R G M. So left mirror columns0-2: M G G (same as input left mirror top row). Column3: X (curtain rod unchanged). Middle mirror columns4-6: M M M (so middle mirror now has colors). Column7: X (curtain rod unchanged). Right mirror columns8-10: R G M (so right mirror has colors). Row2: M G G X G G G X G G M. Left mirror: M G G (same). Middle: G G G. Right: G G M. Row3: M G R X R G G X G G M. Left mirror: M G R (same as input left mirror bottom row). Middle: R G G. Right: G G M.

Thus transformation: The left mirror stays unchanged (the dancer's pose). The middle mirror gets some transformation based on the left mirror? The right mirror gets another transformation.

Metaphor: "The second mirror doesn't face the dancer head-on — it watches from her side, catching the scene as if she'd spun a quarter-turn on her heel. What was at her feet now stretches across the top, read from left to right like a sentence. The third mirror hangs behind her: it sees everything the first mirror saw, but perfectly inverted — the dancer's reflection flipped upside-down and backwards, as though she gazed at herself from the opposite corner of the room."

Interpretation: Left mirror = original view (front). Middle mirror = side view (quarter-turn). Right mirror = rear view (behind) but inverted (upside-down and backwards). The description: "What was at her feet now stretches across the top, read from left to right like a sentence." For the side view: the bottom of the original (feet) becomes the top row of the side view, reading left to right.

Thus mapping: For middle mirror (second), we take the original left mirror's bottom row (row index 2) and place it as the top row (row index 0) of the middle mirror, reading left to right. Then perhaps the rest of the rows are filled similarly? Actually we need to map the whole 3x3 left mirror to the middle mirror via a quarter-turn rotation (90 degrees?) Let's think: If you spin a quarter-turn on heel, you rotate 90 degrees clockwise? The dancer's feet (bottom) go to the left? Actually if you turn quarter-turn to your right, your front becomes your right side? But they say: "What was at her feet now stretches across the top, read from left to right like a sentence." So the original bottom row becomes the top row of the side view, left-to-right orientation same as original left-to-right? So we take original bottom row (row2) and place it as middle mirror's top row (row0) in same order.

What about other rows? Possibly the original middle row becomes middle mirror's middle row? And original top row becomes bottom row? That would be a rotation of 90 degrees clockwise? Let's test: Original left mirror (3x3):
Row0: M G G
Row1: M G G
Row2: M G R

If we rotate 90 degrees clockwise, the new top row would be original left column reversed? Actually rotation 90 clockwise: new[i][j] = old[2-j][i]. Le

... [4845 chars truncated]

We need to parse the problem. We have a grid transformation puzzle. Input and output grids are given in training examples. We need to infer transformation rule.

We have three 3x3 chambers divided by single-column grey barriers. The leftmost chamber holds the original specimen; the middle and right chambers receive transformed copies.

Thus the input grid is presumably 3 rows, each row has 9 columns? Let's examine example 1.

Example 1 Input:
M G G X . . . X . . .
M G G X . . . X . . .
M G R X . . . X . . .

So each row: M G G X . . . X . . . That's 9 characters: positions 0-8.

Let's index: 0 M,1 G,2 G,3 X,4 .,5 .,6 .,7 X,8 .,9 .,10 .? Wait we have 11? Let's count: "M G G X . . . X . . ." That's 11 tokens? Actually there are spaces between each character. Let's count: M (1), G (2), G (3), X (4), . (5), . (6), . (7), X (8), . (9), . (10), . (11). So 11 characters. But they said three 3x3 chambers divided by single-column grey barriers. So each chamber is 3x3 = 9 cells. There are three chambers, plus two grey barrier columns (single column each) separating them. So total width = 3*3 + 2*1 = 9+2 = 11 columns. Yes matches 11 columns. Height = 3 rows (since each chamber is 3 rows). So grid is 3 rows x 11 columns.

Thus left chamber columns 0-2, middle barrier column 3, middle chamber columns 4-6, right barrier column 7, right chamber columns 8-10.

Check example: column 3 is X (grey) barrier; column 7 is X barrier. Indeed.

Now left chamber (columns 0-2) holds original specimen. Middle chamber (4-6) receives transformed copy (first well). Right chamber (8-10) receives transformed copy (second well). The description: "The organism replicates into the first well, twisting on its axis as daughter cells align perpendicular to the mother. In the second well, the granddaughter generation has fully inverted, a perfect mirror of ancestry."

Thus transformation: from original specimen (left chamber) to middle chamber: replicate, twist on its axis as daughter cells align perpendicular to mother. That suggests some rotation? Perhaps a 90-degree rotation? Or maybe reflect? "twisting on its axis as daughter cells align perpendicular to the mother" could be a 90-degree rotation (like turning sideways). Then second well: granddaughter generation fully inverted, a perfect mirror of ancestry. That could be a 180-degree rotation? Or mirror (reflection) across vertical/horizontal axis? "Perfect mirror of ancestry" suggests a mirror image (reflection) of the original? Or maybe of the daughter? Let's parse.

Original specimen -> first well: daughter cells align perpendicular to mother. So if original is oriented some way, daughter cells are perpendicular (rotated 90°). Then second well: granddaughter generation fully inverted, a perfect mirror of ancestry. So after two generations, you get a mirror of the original? Or maybe the granddaughter is mirrored relative to the mother? Actually "perfect mirror of ancestry" could mean mirror of the original specimen (the ancestry). So second well is a mirror (reflection) of the original.

Thus transformation rule: left chamber (original) -> middle chamber: rotate 90 degrees clockwise (or anticlockwise). Left chamber -> right chamber: reflect (mirror) maybe across vertical axis? Or horizontal? Need to see from examples.

Let's extract left chamber from example 1 input:

Rows:
Row0: M G G
Row1: M G G
Row2: M G R

Thus left chamber (3x3):
[ [M, G, G],
  [M, G, G],
  [M, G, R] ]

Now middle chamber (columns 4-6) from output:

Output rows:
Row0: M G G X M M M X R G M
Row1: M G G X G G G X G G M
Row2: M G R X R G G X G G M

Thus middle chamber columns 4-6: let's extract.

Row0: after X at col3, we have M M M (col4-6)
Row1: G G G
Row2: R G G

Thus middle chamber:
[ [M, M, M],
  [G, G, G],
  [R, G, G] ]

Right chamber columns 8-10:
Row0: R G M
Row1: G G M
Row2: G G M

Thus right chamber:
[ [R, G, M],
  [G, G, M],
  [G, G, M] ]

Now we need to see if middle chamber is a rotation of left chamber, and right chamber is a mirror of left chamber (or maybe of middle?).

Let's compute rotations of left chamber.

Left chamber L:
[ [M, G, G],
  [M, G, G],
  [M, G, R] ]

Rotate 90 degrees clockwise: new[i][j] = L[2-j][i]? Actually standard: rotate clockwise: new[i][j] = old[n-1-j][i] for n=3. Let's compute.

Indices: old[0][0]=M, old[0][1]=G, old[0][2]=G
old[1][0]=M, old[1][1]=G, old[1][2]=G
old[2][0]=M, old[2][1]=G, old[2][2]=R

Clockwise 90:
new[0][0] = old[2][0] = M
new[0][1] = old[1][0] = M
new[0][2] = old[0][0] = M
=> first row: M M M

new[1][0] = old[2][1] = G
new[1][1] = old[1][1] = G
new[1][2] = old[0][1] = G
=> second row: G G G

new[2][0] = old[2][2] = R
new[2][1] = old[1][2] = G
new[2][2] = old[0][2] = G
=> third row: R G G

Thus clockwise rotation yields:
[ [M, M, M],
  [G, G, G],
  [R, G, G] ]

That matches exactly the middle chamber we extracted! Good. So middle chamber = clockwise 90-degree rotation of left chamber.

Now what about right chamber? Let's see if it's a mirror

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We need to parse the problem: It's a grid transformation puzzle. We have a description: "Three square signal boards separated by grey mast columns. The first board displays a pattern of colored signal flags; the others are bare. What to do: Transcribe what each ship's lookout would record in their logbook based on their bearing relative to the flagship—the beam ship sees the hoist turned sideways, the stern ship sees it completely reversed. Grid details: Three 3×3 signal panels in a 3×11 display, divided by single grey mast columns."

Thus input grid is 3 rows, 11 columns. The three 3x3 panels are separated by single grey columns (X). So columns: panel1 (col0-2), mast column (col3), panel2 (col4-6), mast column (col7), panel3 (col8-10). The first board (leftmost panel) displays a pattern of colored signal flags; the others are bare (i.e., maybe all black? Or maybe they are empty? The description: "bare" maybe means they are all black (.)? Or maybe they are all white? But we have color key: '.' = black. So bare could be black (no flags). However, in the training example, we see input:

Example 1 Input:
M G G X . . . X . . .
M G G X . . . X . . .
M G R X . . . X . . .

So left panel (col0-2) has:
Row0: M G G
Row1: M G G
Row2: M G R

Middle panel (col4-6) is all '.' (black) as per input: . . . each row.

Right panel (col8-10) is also all '.'.

Output:
M G G X M M M X R G M
M G G X G G G X G G M
M G R X R G G X G G M

Thus output left panel unchanged? Let's see: left panel in output col0-2:
Row0: M G G (same as input)
Row1: M G G (same)
Row2: M G R (same)
Thus left panel unchanged.

Middle panel (col4-6) output:
Row0: M M M
Row1: G G G
Row2: R G G

Right panel (col8-10) output:
Row0: R G M
Row1: G G M
Row2: G G M

Thus transformation: The left panel pattern is copied to middle and right panels but with some transformations: middle panel seems to be a rotation? Let's examine.

Left panel pattern:
[ [M, G, G],
  [M, G, G],
  [M, G, R] ]

If we rotate 90 degrees clockwise? Let's compute: Rotating left panel 90° clockwise yields:
Original:
(0,0)=M, (0,1)=G, (0,2)=G
(1,0)=M, (1,1)=G, (1,2)=G
(2,0)=M, (2,1)=G, (2,2)=R

Clockwise rotation: new[i][j] = old[2-j][i] (if 0-index). Let's compute:
new[0][0] = old[2][0] = M
new[0][1] = old[1][0] = M
new[0][2] = old[0][0] = M
=> row0: M M M matches middle panel row0? Yes middle panel row0 is M M M.

new[1][0] = old[2][1] = G
new[1][1] = old[1][1] = G
new[1][2] = old[0][1] = G
=> row1: G G G matches middle panel row1: G G G.

new[2][0] = old[2][2] = R
new[2][1] = old[1][2] = G
new[2][2] = old[0][2] = G
=> row2: R G G matches middle panel row2: R G G.

Thus middle panel is left panel rotated 90° clockwise.

Now right panel: maybe rotated 180°? Let's test: Rotating left panel 180° (i.e., two 90° rotations) yields:
First rotate 90° cw gave middle panel as above. Rotate that another 90° cw (i.e., 180° from original) yields:
Take middle panel:
[ [M,M,M],
  [G,G,G],
  [R,G,G] ]
Rotate 90° cw:
new[0][0] = old[2][0] = R
new[0][1] = old[1][0] = G
new[0][2] = old[0][0] = M
=> row0: R G M matches output right panel row0: R G M.

new[1][0] = old[2][1] = G
new[1][1] = old[1][1] = G
new[1][2] = old[0][1] = M
=> row1: G G M matches output right panel row1: G G M.

new[2][0] = old[2][2] = G
new[2][1] = old[1][2] = G
new[2][2] = old[0][2] = M
=> row2: G G M matches output right panel row2: G G M.

Thus right panel is left panel rotated 180° (or equivalently, rotated 90° twice). The description: "beam ship sees the hoist turned sideways, the stern ship sees it completely reversed." The flagship is the left panel? The beam ship (maybe middle) sees hoist turned sideways (i.e., rotated 90°). The stern ship (right) sees it completely reversed (i.e., rotated 180°). That matches.

Thus transformation rule: Keep left panel unchanged. For middle panel, rotate left panel 90° clockwise. For right panel, rotate left panel 180° (or rotate middle panel another 90° cw). The mast columns (X) remain unchanged (they are grey). The rest of the grid (outside panels) presumably remain as given (they are '.' in examples). In test input, we have similar pattern: left panel has some colors, middle and right panels are all '.' (black). So we need to apply transformation.

Test Input:
Row0: G G W X . . . X . . .
Row1: W W W X . . . X . . .
Row2: R W W X . . . X . . .

Let's parse: columns 0-2 left panel:
Row0: G G W
Row1: W W W
Row2: R W W

Columns 3: X (grey mast)
Columns 4-6 middle panel: all '.' (black)
Columns 7: X (grey mast)
Columns 8-10 right panel: all '.' (black)

Thus we need to output:
- Left panel unchanged: same as input.
- Middle panel = left panel rotated 90° clockwise.
- Right panel = left panel rotated 180° (or rotate middle panel another 90° cw).
- Mast columns remain X.

Let's compute.

Left panel matrix L:
[ [G, G, W],
  [W, W, W],
  [R, W, W] ]

Check: row0 col0=G, col1=G

... [3182 chars truncated]

We need to parse the problem: We have a grid transformation puzzle. Input: three 3x3 formation grids in a 3x11 stage plan, divided by grey curtain columns. The description: "Three square stage marks separated by grey wing curtains. The first set of marks shows the dancers' positions and colors; the others are unmarked. What to do: Record the formation as each photograph would capture it—the wing shot shows the group turned, the backstage shot shows the formation completely reversed."

Thus we have three 3x3 grids side by side, separated by columns of X (grey) which are the curtains. Input format: 3 rows, each row has 11 characters: first 3 columns: first formation grid (leftmost), then a column of X (curtain), then next 3 columns: second formation grid (middle), then column X, then last 3 columns: third formation grid (rightmost). The first set of marks (leftmost grid) shows the dancers' positions and colors; the others are unmarked (i.e., they are all '.'? Actually in examples, the middle and right grids appear to have some pattern of '.' and maybe other colors? Let's examine examples.

Example 1 Input:
Row0: M G G X . . . X . . .
Row1: M G G X . . . X . . .
Row2: M G R X . . . X . . .

So leftmost 3 columns: M G G; M G G; M G R. That's the formation (dancers). The middle 3 columns: all '.' (since after first X we have . . .). The rightmost 3 columns: also all '.'.

Thus the unmarked grids are initially all '.' (black). The output shows some filled colors in middle and right grids.

Output:
Row0: M G G X M M M X R G M
Row1: M G G X G G G X G G M
Row2: M G R X R G G X G G M

Thus leftmost grid unchanged? Let's check: leftmost 3 columns of output: Row0: M G G; Row1: M G G; Row2: M G R. Same as input leftmost. So leftmost grid stays same (the original formation). The middle grid (columns 4-6) after first X column (col3) is: Row0: M M M; Row1: G G G; Row2: R G G. The rightmost grid (columns 8-10) after second X column (col7) is: Row0: R G M; Row1: G G M; Row2: G G M.

Thus transformation: The leftmost grid is the source formation. The middle grid appears to be some transformation: maybe rotated? The rightmost grid appears to be another transformation: maybe reversed? Let's examine.

Given description: "wing shot shows the group turned, the backstage shot shows the formation completely reversed." So we have three photographs: maybe leftmost is original (front view?), middle is wing shot (turned), rightmost is backstage shot (reversed). The leftmost grid shows dancers' positions and colors (original). The middle grid (wing shot) shows the group turned (maybe rotated 90 degrees?). The rightmost grid (backstage shot) shows formation completely reversed (maybe mirrored horizontally or vertically? Or reversed order?).

Let's test with example 1.

Original formation (leftmost):
Row0: M G G
Row1: M G G
Row2: M G R

We can think of this as a 3x3 matrix.

Now middle grid output:
Row0: M M M
Row1: G G G
Row2: R G G

What operation yields that? Let's see if it's a transpose? Transpose of original would be:
Original:
[ [M,G,G],
  [M,G,G],
  [M,G,R] ]
Transpose:
[ [M,M,M],
  [G,G,G],
  [G,G,R] ]? Wait compute: transpose: element (i,j) = original (j,i). So:
Row0: original col0: M, M, M => M M M
Row1: original col1: G, G, G => G G G
Row2: original col2: G, G, R => G G R

But middle grid row2 is R G G, not G G R. So not simple transpose.

Maybe it's a rotation 90 degrees clockwise? Rotate original 90 cw: new[i][j] = original[2-j][i] (if 0-index). Let's compute.

Original:
(0,0)=M (0,1)=G (0,2)=G
(1,0)=M (1,1)=G (1,2)=G
(2,0)=M (2,1)=G (2,2)=R

Rotate 90 cw: new[0][0] = original[2][0] = M
new[0][1] = original[1][0] = M
new[0][2] = original[0][0] = M => row0: M M M matches middle row0.

new[1][0] = original[2][1] = G
new[1][1] = original[1][1] = G
new[1][2] = original[0][1] = G => row1: G G G matches middle row1.

new[2][0] = original[2][2] = R
new[2][1] = original[1][2] = G
new[2][2] = original[0][2] = G => row2: R G G matches middle row2! Yes! So middle grid is original rotated 90 degrees clockwise.

Now rightmost grid: output:
Row0: R G M
Row1: G G M
Row2: G G M

What operation yields that? Perhaps rotation 180? Let's rotate original 180 degrees (two times 90 cw) = rotate 180: new[i][j] = original[2-i][2-j].

Compute:
Original:
[ [M,G,G],
  [M,G,G],
  [M,G,R] ]

Rotate 180:
new[0][0] = original[2][2] = R
new[0][1] = original[2][1] = G
new[0][2] = original[2][0] = M => row0: R G M matches output row0.

new[1][0] = original[1][2] = G
new[1][1] = original[1][1] = G
new[1][2] = original[1][0] = M => row1: G G M matches output row1.

new[2][0] = original[0][2] = G
new[2][1] = original[0][1] = G
new[2][2] = original[0][0] = M => row2: G G M matches output row2.

Thus rightmost grid is original rotated 180 degrees (or equivalently, reversed both axes). The description: "backstage shot shows the formation completely reversed." That matches a 180-degree rotation (or reversal). The wing shot shows the g

... [5799 chars truncated]

We need to understand the puzzle. The description: "Three square cross-section windows separated by grey mounting brackets in a specimen tray. The first window shows a colored mineral pattern; the other two are uncut. What to do: Slice the crystal at each successive rotation of the goniometer and record the pattern visible in each cross-section—the quarter-turn section and the half-turn section. Grid details: Three 3×3 cross-section windows in a 3×11 specimen tray, separated by single grey bracket columns."

Thus we have a 3x11 grid: three windows each 3x3, separated by single grey columns (X). So total width: each window width 3, plus two separators (since three windows have two gaps) each of width 1 => total width = 3*3 + 2*1 = 9+2=11. Height is 3 rows.

The first window (leftmost) shows a colored mineral pattern (the input pattern). The other two windows are initially uncut (maybe blank? but they are shown as '.'? Actually the input shows '.' for black? Let's see examples.

Example 1 Input:
Row0: M G G X . . . X . . .
Row1: M G G X . . . X . . .
Row2: M G R X . . . X . . .

Thus leftmost window (columns 0-2) is:
M G G
M G G
M G R

Middle window (columns 4-6) is all '.' (black). Rightmost window (columns 8-10) is all '.'.

Thus the other two windows are initially blank (black). The output shows after transformation: they fill with some pattern derived from rotating the crystal? The description: "Slice the crystal at each successive rotation of the goniometer and record the pattern visible in each cross-section—the quarter-turn section and the half-turn section." So we have a crystal (the left window pattern) and we rotate it 90 degrees (quarter-turn) and 180 degrees (half-turn) and capture the cross-section patterns (maybe the pattern as seen through the window after rotation). The middle window corresponds to quarter-turn (90°) and the rightmost window corresponds to half-turn (180°). The left window remains as original? Actually output shows left window unchanged? Let's check.

Example 1 Output:
Row0: M G G X M M M X R G M
Row1: M G G X G G G X G G M
Row2: M G R X R G G X G G M

Left window (cols0-2):
Row0: M G G
Row1: M G G
Row2: M G R
Same as input left window. So left window unchanged (original pattern). Middle window (cols4-6):
Row0: M M M
Row1: G G G
Row2: R G G
Rightmost window (cols8-10):
Row0: R G M
Row1: G G M
Row2: G G M

Thus we need to derive transformation: given original 3x3 pattern (left window), produce its quarter-turn (90° clockwise? or counter?) pattern for middle window, and half-turn (180°) pattern for right window.

Let's test with example 1.

Original pattern (left):
[ [M, G, G],
  [M, G, G],
  [M, G, R] ]

We need to compute quarter-turn (90° rotation) and half-turn (180° rotation). Let's see if middle matches either.

Compute 90° clockwise rotation: new[i][j] = old[2-j][i] (if zero-index). Let's compute.

Original:
Row0: M G G
Row1: M G G
Row2: M G R

Clockwise 90:
- new[0][0] = old[2][0] = M
- new[0][1] = old[1][0] = M
- new[0][2] = old[0][0] = M
=> top row: M M M matches middle row0: M M M good.

- new[1][0] = old[2][1] = G
- new[1][1] = old[1][1] = G
- new[1][2] = old[0][1] = G
=> middle row: G G G matches middle row1: G G G good.

- new[2][0] = old[2][2] = R
- new[2][1] = old[1][2] = G
- new[2][2] = old[0][2] = G
=> bottom row: R G G matches middle row2: R G G good.

Thus middle window is original rotated 90° clockwise.

Now half-turn (180°) rotation: either two 90° rotations or rotate 180: new[i][j] = old[2-i][2-j].

Compute:
- new[0][0] = old[2][2] = R
- new[0][1] = old[2][1] = G
- new[0][2] = old[2][0] = M
=> top row: R G M matches rightmost row0: R G M good.

- new[1][0] = old[1][2] = G
- new[1][1] = old[1][1] = G
- new[1][2] = old[1][0] = M
=> middle row: G G M matches rightmost row1: G G M good.

- new[2][0] = old[0][2] = G
- new[2][1] = old[0][1] = G
- new[2][2] = old[0][0] = M
=> bottom row: G G M matches rightmost row2: G G M good.

Thus rightmost window is original rotated 180°.

Thus transformation: left window unchanged (0°), middle window = rotate original 90° clockwise, right window = rotate original 180°.

Check example 2 to confirm.

Example 2 Input:
Row0: B B R X . . . X . . .
Row1: Y B B X . . . X . . .
Row2: Y Y B X . . . X . . .

Left window (cols0-2):
B B R
Y B B
Y Y B

Middle window (cols4-6) all '.'.
Right window (cols8-10) all '.'.

Output:
Row0: B B R X Y Y B X B Y Y
Row1: Y B B X Y B B X B B Y
Row2: Y Y B X B B R X R B B

Left window unchanged? Let's see left window output cols0-2:
Row0: B B R (same)
Row1: Y B B (same)
Row2: Y Y B (same) Yes.

Middle window (cols4-6):
Row0: Y Y B
Row1: Y B B
Row2: B B R

Check if that's original rotated 90° clockwise.

Original:
[ [B, B, R],
  [Y, B, B],
  [Y, Y, B] ]

Rotate 90 CW:
new[0][0] = old[2][0] = Y
new[0][1] = old[1][0] = Y
new[0][2] = old[0][0] = B => top row: Y Y B matches output row0 

... [3374 chars truncated]

We need to understand the puzzle. There's a rectangular tray with three 3x3 squares (zones) separated by steel dividers (grey columns) at positions 3 and 7. So the grid is 3 rows (height) and 11 columns (width). The three zones: columns 0-2 (first zone), column 3 is divider (X), columns 4-6 (second zone), column 7 is divider (X), columns 8-10 (third zone). The description: "Only the first zone has food arranged on it; the others are clean." So input grid: first zone (columns 0-2) has some pattern; second and third zones are all '.' (black) maybe? Actually they say "clean" meaning empty (black). In examples, we see input: first zone columns 0-2 have some colors; columns 3 is X (divider); columns 4-6 are all '.'; column 7 is X; columns 8-10 are all '.'.

Thus input: first zone pattern; second and third zones are all '.'.

Goal: "Recreate the arrangement so each guest sees a version rotated to face them—the side guest gets a quarter-turn, the opposite guest gets a half-turn."

Interpretation: There are three guests: one at front (maybe first zone?), side guest (second zone?), opposite guest (third zone?). The arrangement in first zone should be rotated appropriately for each zone: side guest gets a quarter-turn (90-degree rotation), opposite guest gets a half-turn (180-degree rotation). The first zone (maybe the "host"?), maybe remains as is (0-degree). So output: first zone unchanged? Let's check examples.

Example 1 input:
Row0: M G G X . . . X . . .
Row1: M G G X . . . X . . .
Row2: M G R X . . . X . . .

First zone columns 0-2:
Row0: M G G
Row1: M G G
Row2: M G R

Second zone columns 4-6: all '.'.
Third zone columns 8-10: all '.'.

Output:
Row0: M G G X M M M X R G M
Row1: M G G X G G G X G G M
Row2: M G R X R G G X G G M

Let's parse output zones:

First zone (cols0-2):
Row0: M G G
Row1: M G G
Row2: M G R
=> same as input first zone. So first zone unchanged.

Second zone (cols4-6):
Row0: M M M
Row1: G G G
Row2: R G G

Third zone (cols8-10):
Row0: R G M
Row1: G G M
Row2: G G M

Now we need to see if second zone is a quarter-turn rotation of first zone? And third zone is half-turn? Let's test.

First zone pattern (3x3):
[ [M, G, G],
  [M, G, G],
  [M, G, R] ]

Rotate 90 degrees clockwise (quarter-turn) yields:
New[i][j] = old[2-j][i] (if origin top-left). Let's compute.

Original:
(0,0)=M, (0,1)=G, (0,2)=G
(1,0)=M, (1,1)=G, (1,2)=G
(2,0)=M, (2,1)=G, (2,2)=R

Clockwise 90:
Result (0,0) = original (2,0) = M
(0,1) = original (1,0) = M
(0,2) = original (0,0) = M
=> first row: M M M

(1,0) = original (2,1) = G
(1,1) = original (1,1) = G
(1,2) = original (0,1) = G
=> G G G

(2,0) = original (2,2) = R
(2,1) = original (1,2) = G
(2,2) = original (0,2) = G
=> R G G

Thus clockwise 90 yields:
[ [M, M, M],
  [G, G, G],
  [R, G, G] ]

Matches second zone output exactly! Good.

Now half-turn (180-degree) rotation of first zone: rotate 180 (or two quarter-turns). Equivalent to reverse both rows and columns.

Compute: new[i][j] = original[2-i][2-j].

Original:
(0,0)=M -> (2,2)=R
(0,1)=G -> (2,1)=G
(0,2)=G -> (2,0)=M
Row0: R G M

(1,0)=M -> (1,2)=G
(1,1)=G -> (1,1)=G
(1,2)=G -> (1,0)=M
Row1: G G M

(2,0)=M -> (0,2)=G
(2,1)=G -> (0,1)=G
(2,2)=R -> (0,0)=M
Row2: G G M

Thus half-turn yields:
[ [R, G, M],
  [G, G, M],
  [G, G, M] ]

Matches third zone output exactly! Good.

Thus transformation: Keep first zone as is. Second zone = rotate first zone 90 degrees clockwise. Third zone = rotate first zone 180 degrees (or rotate second zone another 90). Dividers remain X. The empty zones (second and third) in input are just placeholders; we fill them with rotated versions.

Now test with Example 2 to confirm.

Example2 input:
Row0: B B R X . . . X . . .
Row1: Y B B X . . . X . . .
Row2: Y Y B X . . . X . . .

First zone (cols0-2):
Row0: B B R
Row1: Y B B
Row2: Y Y B

Second zone output (cols4-6):
Row0: Y Y B
Row1: Y B B
Row2: B B R

Third zone output (cols8-10):
Row0: B B R
Row1: B B Y? Wait let's parse output:

Output:
B B R X Y Y B X B Y Y
Y B B X Y B B X B B Y
Y Y B X B B R X R B B

Let's split:

Row0: columns:
0 B
1 B
2 R
3 X
4 Y
5 Y
6 B
7 X
8 B
9 Y
10 Y

Row1:
0 Y
1 B
2 B
3 X
4 Y
5 B
6 B
7 X
8 B
9 B
10 Y

Row2:
0 Y
1 Y
2 B
3 X
4 B
5 B
6 R
7 X
8 R
9 B
10 B

Thus second zone (cols4-6):
Row0: Y Y B
Row1: Y B B
Row2: B B R

Third zone (cols8-10):
Row0: B Y Y
Row1: B B Y
Row2: R B B

Now compute rotations.

First zone:
[ [B, B, R],
  [Y, B, B],
  [Y, Y, B] ]

Clockwise 90:
Compute:
(0,0) = original (2,0) = Y
(0,1) = original (1,0) = Y
(0,2) = original (0,0) = B
=> Row0: Y Y B

(1,0) = original (2,1) = Y
(1,1) = original (1,1) = B
(1,2) = original (0,1) = B
=> Row1: Y B B

(2,0) = original (2,2) = B
(2,1) = original (1,2) = B
(2,2) = original (0,2) = R
=> Row2: B B R

Matches second zone output exactly.

Half-turn (180):
new[i][j] = original[2-i][2-j].

Compute:
(0,0) = original (2,2) = B
(0,1) = original (2,1) = Y
(0,2) = original (2,0) = Y
=> Row0: B Y Y

(1,0) = original (1

... [3773 chars truncated]

We need to understand the transformation. The description: "What you see: A score with three staves separated by barlines. The first staff carries a melodic figure; the other two are rests. What to do: Transcribe the motif onto the second staff so that low voices rise and the phrase reads as though the page were turned sideways. On the third staff, write the motif backwards and upside-down."

Grid details: Three 3×3 panels in a 3×11 grid, separated by grey columns acting as barlines.

Thus we have a 3x11 grid: 3 rows, 11 columns. The grey columns (X) are barlines separating three 3x3 panels. So columns: maybe pattern: first panel columns 0-2, then a grey column (X) at column 3, second panel columns 4-6, then a grey column at column 7, third panel columns 8-10. Indeed the examples show: Input lines have pattern: first three characters (non-grey), then X, then three dots (maybe rests?), then X, then three characters (maybe rests?). Let's examine Example 1 input:

Input:
M G G X . . . X . . .
M G G X . . . X . . .
M G R X . . . X . . .

So each row: first three: M G G (or M G R), then X, then three '.' (black), then X, then three '.' (black). So the first panel (columns 0-2) contains the motif (first staff). The second panel (columns 4-6) is all '.' (rests). The third panel (columns 8-10) is also all '.' (rests). The grey columns at 3 and 7 are barlines.

Output:
M G G X M M M X R G M
M G G X G G G X G G M
M G R X R G G X G G M

Now output: first panel (col0-2) same as input? Let's see: first row: M G G (same as input first row). second row: M G G (same). third row: M G R (same). So first panel unchanged (the motif stays on first staff). Second panel (col4-6): first row: M M M; second: G G G; third: R G G. Third panel (col8-10): first row: R G M; second: G G M; third: G G M.

Thus transformation: The motif from first panel is transcribed onto second staff (second panel) such that low voices rise and phrase reads as though page turned sideways. And third staff gets motif backwards and upside-down.

We need to interpret "low voices rise" and "phrase reads as though the page were turned sideways". Also "motif backwards and upside-down" for third staff.

We have a 3x3 motif (first panel). The motif is presumably a set of colored cells representing notes? The transformation likely involves some operations like transposition (shifting colors?), rotation, reflection.

Let's examine Example 1 to deduce mapping.

Input motif (first panel) 3x3:
Row0: M G G
Row1: M G G
Row2: M G R

We can think of each cell as a color representing a pitch? The transformation to second panel (staff 2) yields:
Row0: M M M
Row1: G G G
Row2: R G G

Third panel (staff 3) yields:
Row0: R G M
Row1: G G M
Row2: G G M

We need to see relationship.

Let's label positions (r,c) within 3x3 panel (0-index). Input motif A[r][c].

Second panel B[r][c] = ?

Third panel C[r][c] = ?

We have:

A:
(0,0)=M
(0,1)=G
(0,2)=G
(1,0)=M
(1,1)=G
(1,2)=G
(2,0)=M
(2,1)=G
(2,2)=R

B:
(0,0)=M
(0,1)=M
(0,2)=M
(1,0)=G
(1,1)=G
(1,2)=G
(2,0)=R
(2,1)=G
(2,2)=G

C:
(0,0)=R
(0,1)=G
(0,2)=M
(1,0)=G
(1,1)=G
(1,2)=M
(2,0)=G
(2,1)=G
(2,2)=M

Now try to see if B is some transformation like transpose? Transpose of A would be:
A^T:
(0,0)=M
(0,1)=M
(0,2)=M
(1,0)=G
(1,1)=G
(1,2)=G
(2,0)=G
(2,1)=G
(2,2)=R

But B is:
Row0: M M M (matches A^T row0)
Row1: G G G (matches A^T row1)
Row2: R G G (but A^T row2 is G G R). So B row2 is swapped: R G G vs G G R. So B is like A^T but with the bottom row reversed? Actually A^T row2 = [G,G,R]; B row2 = [R,G,G] which is a rotation? It's like shifting the R to leftmost? Let's see.

If we take A and rotate 90 degrees clockwise? Rotate A 90° CW: new[i][j] = A[2-j][i]. Compute:

A:
[ [M,G,G],
  [M,G,G],
  [M,G,R] ]

Rotate 90 CW:
new[0][0] = A[2][0] = M
new[0][1] = A[1][0] = M
new[0][2] = A[0][0] = M
=> row0: M M M matches B row0.

new[1][0] = A[2][1] = G
new[1][1] = A[1][1] = G
new[1][2] = A[0][1] = G
=> row1: G G G matches B row1.

new[2][0] = A[2][2] = R
new[2][1] = A[1][2] = G
new[2][2] = A[0][2] = G
=> row2: R G G matches B row2! Yes! So B is A rotated 90 degrees clockwise.

Thus second staff (panel) gets the motif rotated 90° CW.

Now third staff: motif backwards and upside-down. Let's see if C corresponds to some transformation like rotate 180°? Rotate 180° (i.e., flip both axes) of A: new[i][j] = A[2-i][2-j].

Compute A rotated 180:
new[0][0] = A[2][2] = R
new[0][1] = A[2][1] = G
new[0][2] = A[2][0] = M
=> row0: R G M matches C row0? C row0 is R G M yes.

new[1][0] = A[1][2] = G
new[1][1] = A[1][1] = G
new[1][2] = A[1][0] = M
=> row1: G G M matches C row1? C row1 is G G M yes.

new[2][0] = A[0][2] = G
new[2][1] = A[0][1] = G
new[2][2] = A[0][0] = M
=> row2: G G M matches C row2? C row2 is G G M yes.

Thus C is A rotated 180 degrees (i.e., upside-down and backwards). The description: "write the motif backwards and upside-down." That matches rotate 180 (flip both horizontally and vertically). Good.

Th

... [4845 chars truncated]

We need to understand the puzzle. The description: "A tactical board divided into three zones by grey lane markers. The first zone shows the starting formation; the others are blank. What to do: Project the formation into each subsequent zone as it would appear from each successive vantage point—one tilted, one fully reversed."

Grid details: 3x11 board with three 3x3 zones at columns 0-2, 4-6, and 8-10, divided by grey at columns 3 and 7.

Thus columns: 0,1,2 zone1; column3 grey; columns4,5,6 zone2; column7 grey; columns8,9,10 zone3.

The input grid shows the first zone (columns0-2) with some colors; columns3 and 7 are grey (X). The other zones (columns4-6 and 8-10) are initially blank (maybe '.' black? Actually blank likely '.' black). The output shows the formation projected into zones2 and 3 with transformations: one tilted, one fully reversed.

We need to infer transformation rule from training example.

Example 1 Input:
Row0: M G G X . . . X . . .
Row1: M G G X . . . X . . .
Row2: M G R X . . . X . . .

Thus zone1 (col0-2):
col0: M M M (col0 all M)
col1: G G G (col1 all G)
col2: G G R (col2: row0 G, row1 G, row2 R)

Zone2 (col4-6) initially blank '.' all rows? Input shows '.' at col4,5,6 for each row: yes.

Zone3 (col8-10) also blank '.'.

Output:
Row0: M G G X M M M X R G M
Row1: M G G X G G G X G G M
Row2: M G R X R G G X G G M

Let's parse output zones.

Zone1 (col0-2) unchanged? Let's see: col0: M M M (same as input). col1: G G G (same). col2: G G R (same). So zone1 remains same as input (makes sense: starting formation unchanged).

Zone2 (col4-6):
Row0: M M M
Row1: G G G
Row2: R G G

Thus zone2 appears to be some transformation of zone1.

Zone3 (col8-10):
Row0: R G M
Row1: G G M
Row2: G G M

Now we need to see what transformation corresponds to "one tilted, one fully reversed". Likely zone2 is a rotated (tilted) version of zone1, and zone3 is a fully reversed (maybe mirrored or 180-degree rotation) version.

Let's examine zone1 matrix (3x3):
Rows (0-2), columns (0-2):
[ [M, G, G],
  [M, G, G],
  [M, G, R] ]

Let's denote as:
(0,0)=M, (0,1)=G, (0,2)=G
(1,0)=M, (1,1)=G, (1,2)=G
(2,0)=M, (2,1)=G, (2,2)=R

Now zone2:
[ [M, M, M],
  [G, G, G],
  [R, G, G] ]

What operation yields that? Let's see if it's a transpose? Transpose of zone1 would be:
[ [M, M, M],
  [G, G, G],
  [G, G, R] ] (since original col0->row0: M M M; col1->row1: G G G; col2->row2: G G R). That's close but zone2 has bottom-right as G G? Actually zone2 row2 col2 is G, while transpose would have R at (2,2). So zone2 is like transpose but with the bottom-right element changed from R to G? Wait zone2 row2 col0 = R, row2 col1 = G, row2 col2 = G. Transpose row2 would be [G, G, R]. So zone2 row2 is [R, G, G] which is a rotation? Let's see.

Maybe zone2 is zone1 rotated 90 degrees clockwise? Rotate 90 CW: new[i][j] = old[2-j][i]. Let's compute.

Original:
[ [M,G,G],
  [M,G,G],
  [M,G,R] ]

Rotate 90 CW:
new[0][0] = old[2][0] = M
new[0][1] = old[1][0] = M
new[0][2] = old[0][0] = M
=> row0: M M M (matches zone2 row0)
new[1][0] = old[2][1] = G
new[1][1] = old[1][1] = G
new[1][2] = old[0][1] = G
=> row1: G G G (matches)
new[2][0] = old[2][2] = R
new[2][1] = old[1][2] = G
new[2][2] = old[0][2] = G
=> row2: R G G (matches zone2 row2). Yes! So zone2 is zone1 rotated 90 degrees clockwise.

Thus "one tilted" likely means a 90-degree rotation (tilted). Good.

Now zone3: "fully reversed". Could be 180-degree rotation (i.e., reversed both axes) or maybe mirror? Let's test.

Rotate zone1 180 degrees: new[i][j] = old[2-i][2-j].

Compute:
new[0][0] = old[2][2] = R
new[0][1] = old[2][1] = G
new[0][2] = old[2][0] = M
=> row0: R G M (matches zone3 row0? zone3 row0: R G M yes)
new[1][0] = old[1][2] = G
new[1][1] = old[1][1] = G
new[1][2] = old[1][0] = M
=> row1: G G M (matches zone3 row1: G G M)
new[2][0] = old[0][2] = G
new[2][1] = old[0][1] = G
new[2][2] = old[0][0] = M
=> row2: G G M (matches zone3 row2: G G M). Yes! So zone3 is zone1 rotated 180 degrees (i.e., fully reversed). Good.

Thus transformation: zone2 = rotate zone1 90° CW; zone3 = rotate zone1 180° (or equivalently rotate zone2 another 90° CW). The description: "Project the formation into each subsequent zone as it would appear from each successive vantage point—one tilted, one fully reversed." So first zone is original; second zone is as seen from a tilted viewpoint (90° rotation); third zone is as seen from fully reversed viewpoint (180° rotation). Good.

Now we need to apply to test input.

Test Input:
Row0: G G W X . . . X . . .
Row1: W W W X . . . X . . .
Row2: R W W X . . . X . . .

Let's parse zone1 (col0-2):
col0: G, W, R (row0 G, row1 W, row2 R)
col1: G, W, W (row0 G, row1 W, row2 W)
col2: W, W, W (row0 W, row1 W, row2 W)

Thus zone1 matrix:
[ [G, G, W],
  [W, W, W],
  [R, W, W] ]

Check: row0: G G W; row1: W W W; row2: R W W.

Now compute zone2 = rotate 90° CW of zone1.

Compute new[i][j] = old[2-j][i].

Let's compute s

... [4541 chars truncated]

We need to understand the puzzle. The description: "Three square wall panels separated by narrow grey pilasters. The first panel bears an ornamental tile pattern; the others await installation. What to do: Transfer the design onto each successive wall, adjusting its orientation as the walls turn the corner, so the second wall shows the design as if you've stepped to your right, and the third as if you've walked around to face it head-on from behind."

Grid details: A 3×11 facade with three 3×3 panels separated by grey pilaster columns.

Thus the input grid is 3 rows, 11 columns. The panels are 3x3 each, separated by grey columns (X). So columns: panel1 (col0-2), pilaster (col3), panel2 (col4-6), pilaster (col7), panel3 (col8-10). Indeed 3+1+3+1+3 = 11.

The first panel (leftmost) has a pattern (some colors). The other panels are initially blank? Actually they may have some placeholder? In examples, we see input: first panel has pattern, second and third panels appear to be all '.'? Let's check Example 1 input:

Input:
M G G X . . . X . . .
M G G X . . . X . . .
M G R X . . . X . . .

So columns: 0-2: M G G; M G G; M G R (first panel). Column3: X (pilaster). Columns4-6: . . . ; . . . ; . . . (second panel). Column7: X. Columns8-10: . . . ; . . . ; . . . (third panel). So indeed second and third panels are initially all black (.) (i.e., empty). The output shows the pattern transferred to second and third panels with some rotation.

Output:
M G G X M M M X R G M
M G G X G G G X G G M
M G R X R G G X G G M

Let's map output columns:

col0-2: M G G ; M G G ; M G R (same as input first panel). col3: X. col4-6: M M M ; G G G ; R G G (second panel). col7: X. col8-10: R G M ; G G M ; G G M (third panel).

Thus pattern from first panel is transferred to second panel rotated? Let's see.

First panel pattern (3x3):
Row0: M G G
Row1: M G G
Row2: M G R

Second panel output:
Row0: M M M
Row1: G G G
Row2: R G G

Third panel output:
Row0: R G M
Row1: G G M
Row2: G G M

We need to infer transformation: "Transfer the design onto each successive wall, adjusting its orientation as the walls turn the corner, so the second wall shows the design as if you've stepped to your right, and the third as if you've walked around to face it head-on from behind."

Interpretation: The first panel is facing you (front). The second panel is to the right (i.e., you turn right, so you see the panel from its left side? Actually if you step to your right, you are now facing the panel that was originally on your right side, which is oriented such that its front faces you? Let's think physically: Imagine three walls forming a corner: first wall facing you (south), second wall to your right (west? Actually if you face the first wall, stepping to your right means you turn east? Wait orientation: If you are facing the first wall (north?), stepping to your right means you turn east and now face the second wall which is oriented east-west? Actually typical scenario: You have three walls in a row, like a U shape? The description: "Three square wall panels separated by narrow grey pilasters." So they are adjacent in a line, like a facade. The first panel bears pattern; the others await installation. Transfer the design onto each successive wall, adjusting its orientation as the walls turn the corner, so the second wall shows the design as if you've stepped to your right, and the third as if you've walked around to face it head-on from behind.

Interpretation: The pattern on the first panel is like a design that you see from the front. When you move to the right to see the second panel, you are now looking at it from a different angle: you have turned 90 degrees clockwise relative to the first panel? Actually if you step to your right while still facing the same direction (i.e., you sidestep right), you are still facing the same direction (north) but you are now looking at the second panel from its left side? Hmm.

Better: Imagine you are standing in front of the first panel, facing it. The second panel is to your right, but it's oriented the same way as the first (i.e., its front faces you as well, because they are all in the same plane). Actually if they are in a line, they all face the same direction (the viewer). So stepping to your right doesn't change orientation; you just see the adjacent panel from the same viewpoint. But the description says adjusting orientation as the walls turn the corner, implying the walls are not in a straight line but they turn a corner: like you have three walls forming a corner: first wall facing you, second wall to your right (facing east?), third wall behind you (facing away?). Actually "as if you've stepped to your right" suggests you rotate 90 degrees clockwise around vertical axis, so you now face the second wall which originally was on your right side, but now you are facing it head-on. Similarly, "as if you've walked around to face it head-on from behind" suggests you have gone around the corner to the bac

... [5949 chars truncated]

We need to parse the problem: We have a grid transformation puzzle. Input: a 3x11 grid (3 rows, 11 columns). The description: "Three 3×3 map sectors in a 3×11 table, separated by single grey columns." So we have three 3x3 sectors (left, middle, right) each separated by a single column of grey (X). So total width: 3 (sector) + 1 (separator) + 3 (sector) + 1 (separator) + 3 (sector) = 3+1+3+1+3 = 11 columns. Yes.

The description: "What you see: A triptych war table with three square sectors separated by grey no-man's-land strips. The first sector shows unit positions; the others are fog of war. What to do: Reconstruct the battlefield from each new vantage: the flank observer sees the formation turned on its edge, and the enemy commander sees everything reversed."

Interpretation: Input grid: first sector (columns 0-2) shows unit positions (some pattern). The other two sectors (columns 4-6 and 8-10) are fog of war (maybe unknown). The transformation: we need to reconstruct the battlefield from each new vantage: flank observer sees formation turned on its edge, and enemy commander sees everything reversed.

Thus output grid: likely we need to fill the fog-of-war sectors with transformed versions of the first sector: one is rotated (turned on its edge) maybe 90-degree rotation? The other is reversed (mirror? maybe horizontal flip? Or reversed meaning rotated 180? Or reversed order? Let's examine training example.

Example 1 Input:
Row0: M G G X . . . X . . .
Row1: M G G X . . . X . . .
Row2: M G R X . . . X . . .

So first sector (col0-2):
col0: M M M
col1: G G G
col2: G G R? Wait row0 col2 = G, row1 col2 = G, row2 col2 = R. So first sector is:
[ [M, G, G],
  [M, G, G],
  [M, G, R] ]

Second sector (col4-6) currently all '.' (black) in input: row0 col4-6 = . . .; row1 same; row2 same. Third sector (col8-10) also all '.'.

Output:
Row0: M G G X M M M X R G M
Row1: M G G X G G G X G G M
Row2: M G R X R G G X G G M

Let's parse output sectors.

First sector (col0-2) unchanged? Let's see: output col0-2:
Row0: M G G
Row1: M G G
Row2: M G R
Matches input first sector exactly. So first sector remains as is (unit positions unchanged).

Second sector (col4-6):
Row0: M M M
Row1: G G G
Row2: R G G

Third sector (col8-10):
Row0: R G M
Row1: G G M
Row2: G G M

Thus transformation: second sector appears to be some transformation of first sector; third sector another transformation.

Let's see mapping.

First sector matrix A:
[ [M, G, G],
  [M, G, G],
  [M, G, R] ]

Second sector B:
[ [M, M, M],
  [G, G, G],
  [R, G, G] ]

Third sector C:
[ [R, G, M],
  [G, G, M],
  [G, G, M] ]

We need to infer transformation rules: "flank observer sees the formation turned on its edge" and "enemy commander sees everything reversed."

Possibly: flank observer sees formation turned on its edge = rotate 90 degrees? Enemy commander sees everything reversed = maybe reflect (mirror) or reverse order (like rotate 180? Or invert colors? But colors are not inverted; they stay same set.)

Let's test: Rotate A 90 degrees clockwise yields what? A (3x3):
Original:
row0: M G G
row1: M G G
row2: M G R

Rotate 90 CW: new[i][j] = old[2-j][i]? Actually standard: new[row][col] = old[n-1-col][row]. Let's compute.

n=3.

new[0][0] = old[2][0] = M
new[0][1] = old[1][0] = M
new[0][2] = old[0][0] = M
=> first row: M M M

new[1][0] = old[2][1] = G
new[1][1] = old[1][1] = G
new[1][2] = old[0][1] = G
=> second row: G G G

new[2][0] = old[2][2] = R
new[2][1] = old[1][2] = G
new[2][2] = old[0][2] = G
=> third row: R G G

That matches exactly second sector B! So second sector is A rotated 90 degrees clockwise.

Now third sector C: maybe "enemy commander sees everything reversed" could be rotate 180 degrees? Or reflect horizontally? Let's test rotate A 180 degrees: rotate 90 twice.

Rotate A 180: Equivalent to reverse both rows and columns.

Compute: new[i][j] = old[2-i][2-j].

Let's compute:

old:
[ [M,G,G],
  [M,G,G],
  [M,G,R] ]

new[0][0] = old[2][2] = R
new[0][1] = old[2][1] = G
new[0][2] = old[2][0] = M
=> row0: R G M

new[1][0] = old[1][2] = G
new[1][1] = old[1][1] = G
new[1][2] = old[1][0] = M
=> row1: G G M

new[2][0] = old[0][2] = G
new[2][1] = old[0][1] = G
new[2][2] = old[0][0] = M
=> row2: G G M

That matches third sector C exactly! So third sector is A rotated 180 degrees (or reversed both axes). The description "enemy commander sees everything reversed" could mean reversed (i.e., turned upside down and backwards) = 180-degree rotation.

Thus transformation rule: Given the left sector (unit positions), the middle sector is the formation turned on its edge (i.e., rotated 90 degrees clockwise), and the right sector is everything reversed (i.e., rotated 180 degrees). The separators (grey columns) remain grey (X). The left sector remains unchanged.

Thus for test input, we need to extract left sector (columns 0-2), compute its 90° CW rotation for middle sector (columns 4-6), and its 180° rotation for right sector (columns 8

... [4102 chars truncated]

We need to parse the problem: We have a grid transformation puzzle. Input: three 3x3 beds separated by grey gravel-path columns. The grid is 3 rows (height) and width: each bed is 3 columns, plus two grey columns (separator) between them? Actually "Three 3×3 beds in a 3×11 garden row, separated by grey gravel-path columns." So total width: 3 beds * 3 columns = 9, plus 2 separator columns = 11. So columns: bed1 (col0-2), separator col3 (grey), bed2 (col4-6), separator col7 (grey), bed3 (col8-10). The input grid shows columns with '.' for black? Actually '.' is black. X is grey (gravel path). So we have '.' and 'X' as given.

We have training examples: Input and Output.

Example 1 Input:
Row0: M G G X . . . X . . .
Row1: M G G X . . . X . . .
Row2: M G R X . . . X . . .

So bed1 (col0-2): M G G; M G G; M G R (so first column: M M M; second column: G G G; third column: G G R). Bed2 (col4-6): all '.'? Actually col4-6: '.' '.' '.' for each row. Bed3 (col8-10): '.' '.' '.' for each row.

Output:
Row0: M G G X M M M X R G M
Row1: M G G X G G G X G G M
Row2: M G R X R G G X G G M

So after transformation, bed2 and bed3 are filled with some pattern derived from bed1? Let's see.

Bed2 (col4-6) in output: Row0: M M M; Row1: G G G; Row2: R G G. So bed2 is like a rotated version? Actually bed1 originally: 
Row0: M G G
Row1: M G G
Row2: M G R

If we rotate 90 degrees clockwise? Let's compute: Original bed1 matrix:
[ [M, G, G],
  [M, G, G],
  [M, G, R] ]

Rotate 90° CW: new[i][j] = old[2-j][i]? Actually typical: new[i][j] = old[n-1-j][i] for n=3. Let's compute:
new[0][0] = old[2][0] = M
new[0][1] = old[1][0] = M
new[0][2] = old[0][0] = M => first row M M M matches output bed2 row0: M M M. Good.
new[1][0] = old[2][1] = G
new[1][1] = old[1][1] = G
new[1][2] = old[0][1] = G => row1 G G G matches output bed2 row1: G G G.
new[2][0] = old[2][2] = R
new[2][1] = old[1][2] = G
new[2][2] = old[0][2] = G => row2 R G G matches output bed2 row2: R G G. Yes! So bed2 is bed1 rotated 90° clockwise.

Now bed3 (col8-10) in output: Row0: R G M; Row1: G G M; Row2: G G M. Let's see: maybe bed3 is bed1 rotated 180°? Or maybe bed2 rotated again? Let's test: If we rotate bed1 180° (two times CW) we get: rotate 90° CW twice = 180°. Let's compute 180° rotation of bed1: Equivalent to reverse both rows and columns. Original:
[ [M,G,G],
  [M,G,G],
  [M,G,R] ]
Reverse rows: [ [M,G,R], [M,G,G], [M,G,G] ] then reverse each row: [ [R,G,M], [G,G,M], [G,G,M] ]. That yields:
Row0: R G M
Row1: G G M
Row2: G G M
Matches output bed3! Yes.

Thus transformation: For each test, we have three beds: left bed (bed1) given with some pattern; middle bed (bed2) and right bed (bed3) are initially empty (all '.' maybe). We need to fill bed2 with bed1 rotated 90° clockwise, and bed3 with bed1 rotated 180° (or two times 90° CW). The separators (grey columns) remain X.

Check example 2? We only have one example given. But likely that's the rule.

Now test input:
Row0: G G W X . . . X . . .
Row1: W W W X . . . X . . .
Row2: R W W X . . . X . . .

So bed1 (col0-2):
Row0: G G W
Row1: W W W
Row2: R W W

We need to compute bed2 = rotate 90° CW of bed1, bed3 = rotate 180° of bed1.

Let's compute.

Bed1 matrix:
[ [G, G, W],
  [W, W, W],
  [R, W, W] ]

Rotate 90° CW: new[i][j] = old[2-j][i].

Compute new[0][0] = old[2][0] = R
new[0][1] = old[1][0] = W
new[0][2] = old[0][0] = G
=> row0: R W G

new[1][0] = old[2][1] = W
new[1][1] = old[1][1] = W
new[1][2] = old[0][1] = G
=> row1: W W G

new[2][0] = old[2][2] = W
new[2][1] = old[1][2] = W
new[2][2] = old[0][2] = W
=> row2: W W W

Thus bed2 (col4-6) should be:
Row0: R W G
Row1: W W G
Row2: W W W

Now bed3 = rotate 180° (or rotate 90° twice). Let's compute 180°: reverse rows and reverse each row.

Original bed1:
Row0: G G W
Row1: W W W
Row2: R W W

Reverse rows order: 
Row0 becomes original row2: R W W
Row1 becomes original row1: W W W
Row2 becomes original row0: G G W

Now reverse each row:
Row0: W W R
Row1: W W W
Row2: W W G

Thus bed3 (col8-10):
Row0: W W R
Row1: W W W
Row2: W W G

Check: Alternatively rotate 90° CW twice: we already have bed2 (90°). Rotate bed2 another 90° CW should give bed3. Let's test: bed2 we computed:
[ [R,W,G],
  [W,W,G],
  [W,W,W] ]
Rotate 90° CW:
new[0][0] = old[2][0] = W
new[0][1] = old[1][0] = W
new[0][2] = old[0][0] = R => row0: W W R matches.
new[1][0] = old[2][1] = W
new[1][1] = old[1][1] = W
new[1][2] = old[0][1] = W => row1: W W W matches.
new[2][0] = old[2][2] = W
new[2][1] = old[1][2] = G
new[2][2] = old[0][2] = G => row2: W G G? Wait compute: old[2][2] = W (bed2[2][2] = W), old[1][2] = G (bed2[1][2] = G), old[0][2] = G (bed2[0][2] = G). So row2: W G G. But our earlier bed3 row2 was W W G. Something off. Let's double-check bed2 matrix: Did we compute correctly? Let's recompute bed2 from bed1.

Bed1:
Row0: G G W
Row1: W W W
Row2: R W W

Let's write as matrix with in

... [4497 chars truncated]

We need to understand the puzzle. The description: "crowded square buzzes with chatter. Most voices are faint — a whisper of maroon here, a murmur of orange there — but one color is practically shouting over everyone else, filling the room with its presence. The rest are just passersby." So we need to find the dominant color (most frequent) in the 7x7 grid. Then "That dominant voice doesn't just win the argument — it chooses a signature gesture, a pose struck in azure light on a small dark stage. Each color-voice has rehearsed its own distinctive pose: one stands bold as a horizon line, another rises like a bent elbow, another traces a zigzag path. The loudest voice steps onto the stage and strikes the only pose it knows."

Thus, the output is a 3x3 grid (stage) with azure (A) representing the pose, and black (.) representing darkness. The pose is determined by the dominant color: each color has a specific pose pattern (a 3x3 pattern of A's and .'s). The output grid is that pose, with A's where the pose is, and . elsewhere.

Thus we need to map each color to a specific 3x3 pose pattern. The training examples give us mapping for some colors.

Let's parse examples.

Example 1:
Input grid (7x7):
Row0: B . B . O . .
Row1: B B W B . B .
Row2: . . B B . R .
Row3: . . . . G . B
Row4: . Y . B . . B
Row5: . . B . R . A
Row6: . . B . O G B

We need to find the dominant color (most frequent). Let's count.

We'll count each color: B (blue), O (orange), W (maroon), R (red), G (green), Y (yellow), A (azure), maybe others.

Let's count manually.

Row0: B . B . O . . => B:2, O:1
Row1: B B W B . B . => B:4? Actually B B W B . B . => B at col0,1,3,5 => 4 B, W:1
Row2: . . B B . R . => B:2, R:1
Row3: . . . . G . B => G:1, B:1
Row4: . Y . B . . B => Y:1, B:2
Row5: . . B . R . A => B:1, R:1, A:1
Row6: . . B . O G B => B:2, O:1, G:1

Now sum B: Row0 2 + Row1 4 + Row2 2 + Row3 1 + Row4 2 + Row5 1 + Row6 2 = 14? Let's sum: 2+4=6, +2=8, +1=9, +2=11, +1=12, +2=14. Yes B=14.

O: Row0 1 + Row1 0 + Row2 0 + Row3 0 + Row4 0 + Row5 0 + Row6 1 = 2.

W: Row1 1 =1.

R: Row2 1 + Row5 1 =2.

G: Row3 1 + Row6 1 =2.

Y: Row4 1 =1.

A: Row5 1 =1.

X? none.

Thus dominant color is B (blue) with 14 occurrences.

Output for example1:
. . A
A A .
. A .

Let's write as 3x3:
Row0: . . A
Row1: A A .
Row2: . A .

Thus pattern for B (blue) is:
[ [., ., A],
  [A, A, .],
  [., A, .] ]

Now example2:
Input:
. G . G X G .
. . G G . . .
A . . . . . G
G Y G W G . G
. . W G B G G
. G G G . G .
. . . . . . G

Let's count colors.

We'll count each.

Row0: . G . G X G . => G:3? Actually positions: col1 G, col3 G, col5 G => 3 G, X:1
Row1: . . G G . . . => G:2
Row2: A . . . . . G => A:1, G:1
Row3: G Y G W G . G => G:4? col0 G, col2 G, col4 G, col6 G => 4 G, Y:1, W:1
Row4: . . W G B G G => W:1, G:3? col3 G, col5 G, col6 G => 3 G, B:1
Row5: . G G G . G . => G:4? col1 G, col2 G, col3 G, col5 G => 4 G
Row6: . . . . . . G => G:1

Now sum G: Row0 3 + Row1 2 + Row2 1 + Row3 4 + Row4 3 + Row5 4 + Row6 1 = 18? Let's sum: 3+2=5, +1=6, +4=10, +3=13, +4=17, +1=18. Yes G=18.

Other colors: X:1, A:1, Y:1, W: Row3 1 + Row4 1 =2, B:1.

Thus dominant color is G (green). Output for example2:
. A A
. A .
. A .

Thus pattern for G (green) is:
Row0: . A A
Row1: . A .
Row2: . A .

So pattern:
[ [., A, A],
  [., A, .],
  [., A, .] ]

Now example3:
Input:
. . R . B X G
. . R W . R .
R R R Y R . .
. R . R O R .
R R . . R R M
. R R . R . .
X . Y R . R R

Let's count.

We'll count each color.

Row0: . . R . B X G => R:1, B:1, X:1, G:1
Row1: . . R W . R . => R:2, W:1
Row2: R R R Y R . . => R:4? Actually positions: col0 R, col1 R, col2 R, col4 R => 4 R, Y:1
Row3: . R . R O R . => R:3? col1 R, col3 R, col5 R => 3 R, O:1
Row4: R R . . R R M => R:4? col0 R, col1 R, col4 R, col5 R => 4 R, M:1
Row5: . R R . R . . => R:3? col1 R, col2 R, col4 R => 3 R
Row6: X . Y R . R R => X:1, Y:1, R:3? col3 R, col5 R, col6 R => 3 R

Now sum R: Row0 1 + Row1 2 + Row2 4 + Row3 3 + Row4 4 + Row5 3 + Row6 3 = 20? Let's sum: 1+2=3, +4=7, +3=10, +4=14, +3=17, +3=20. Yes R=20.

Other colors: B:1, X: Row0 1 + Row6 1 =2, G:1, W:1, Y: Row2 1 + Row6 1 =2, O:1, M:1.

Thus dominant color is R (red). Output for example3:
. . .
A A A
. . .

Thus pattern for R (red) is:
Row0: . . .
Row1: A A A
Row2: . . .

So pattern:
[ [., ., .],
  [A, A, A],
  [., ., .] ]

Thus we have mapping for three colors: B -> pattern1, G -> pattern2, R -> pattern3.

We need to infer mapping for other colors (maybe O, Y, W, M, X, etc.) based on description: "Each color-voice has rehearsed its own distinctive pose: one stands bold as a horizon line, another rises like a bent elbow, another traces a zigzag path." The poses likely correspond to certain shapes: horizon line (a horizontal line across the middle?), bent elbow (maybe an L shape?), zigzag path (maybe a diagonal or zigzag). The patterns we have:

- B pattern: . . A / A A . / . 

... [6311 chars truncated]

We need to understand the transformation. The description: "crowded culture plate where several species of microorganism vie for resources. One species vastly outnumbers the rest, while rare mutant strains appear as isolated specimens. Identify which colony has achieved dominance through sheer population. Each dominant species, when it wins, expresses a unique molecular signature — a small, fixed fold pattern. Read off the winning species' signature. Grid details: The petri dish is a 7×7 field. The diagnostic result is always a 3×3 protein fold rendered in azure against a dark background."

Thus we have a 7x7 input grid with colors representing different species (colored cells). The output is a 3x3 pattern in azure (A) on black background (.). So output is a 3x3 pattern of A's and .'s (maybe also other colors? but description says rendered in azure against a dark background, so only A and .). Indeed training examples: output patterns are only A and . (no other colors). So we need to find the dominant species (the color that appears most frequently? Or the one that "vastly outnumbers the rest"?). Then we need to output a fixed fold pattern (3x3) associated with that species. The pattern is unique per species. So we need to infer mapping from species color to 3x3 pattern (azure pattern). The training examples give us two examples: we can deduce which species is dominant in each input, and what pattern is output.

Example 1 input:
Row0: B . B . O . .
Row1: B B W B . B .
Row2: . . B B . R .
Row3: . . . . G . B
Row4: . Y . B . . B
Row5: . . B . R . A
Row6: . . B . O G B

Let's count colors: B (blue) appears many times. Let's count: Row0: B at col0, col2 =>2. Row1: B at col0,1,3,5 =>4 (total 6). Row2: B at col2,3 =>2 (total 8). Row3: B at col6 =>1 (total9). Row4: B at col3,6 =>2 (total11). Row5: B at col2,6 =>2 (total13). Row6: B at col2,6 =>2 (total15). Also there is an A at row5 col5? Actually row5: . . B . R . A => A at col6? Wait row5: ". . B . R . A": positions: col0 ., col1 ., col2 B, col3 ., col4 R, col5 ., col6 A. So A at (5,6). Also O at row0 col4? row0: B . B . O . . => O at col4. Row6 col4 O? row6: . . B . O G B => O at col4. Also W at row1 col2? row1: B B W B . B . => W at col2. Row4 col1 Y? row4: . Y . B . . B => Y at col1. Row2 col5 R? row2: . . B B . R . => R at col5. Row5 col4 R? row5: . . B . R . A => R at col4. Row3 col4 G? row3: . . . . G . B => G at col4. Row6 col5 G? row6: . . B . O G B => G at col5.

Thus B appears 15 times. Others: O appears 2, W 1, Y 1, R 2, G 2, A 1. So B is dominant.

Output pattern for example1:
. . A
A A .
. A .

That's a 3x3 pattern: coordinates (r,c):
(0,0) ., (0,1) ., (0,2) A
(1,0) A, (1,1) A, (1,2) .
(2,0) ., (2,1) A, (2,2) .

Thus pattern: A's at positions: (0,2), (1,0), (1,1), (2,1). That's like a shape.

Example2 input:
. G . G X G .
. . G G . . .
A . . . . . G
G Y G W G . G
. . W G B G G
. G G G . G .
. . . . . . G

Let's count G (green) appears many. Let's count: Row0: . G . G X G . => G at col1,3,5 =>3. Row1: . . G G . . . => G at col2,3 =>2 (total5). Row2: A . . . . . G => G at col6 =>1 (total6). Row3: G Y G W G . G => G at col0,2,4,6 =>4 (total10). Row4: . . W G B G G => G at col3,5,6 =>3 (total13). Row5: . G G G . G . => G at col1,2,3,5 =>4 (total17). Row6: . . . . . . G => G at col6 =>1 (total18). Others: X at row0 col4 =>1. A at row2 col0 =>1. Y at row3 col1 =>1. W at row3 col3? Actually row3: G Y G W G . G => W at col3 =>1. Also W at row4 col2 =>1. B at row4 col4 =>1. So G appears 18 times, dominant.

Output pattern for example2:
. A A
. A .
. A .

That's pattern: (0,0)., (0,1) A, (0,2) A; (1,0)., (1,1) A, (1,2).; (2,0)., (2,1) A, (2,2). So A's at (0,1),(0,2),(1,1),(2,1). That's a vertical line on right? Actually shape: top row two A's left? Wait (0,1) and (0,2) are A's; (1,1) A; (2,1) A. So shape like a "T" missing top middle? Actually it's like a vertical line at column1 plus an extra at (0,2). So shape: like a "┤" maybe.

Thus we have two dominant species: B (blue) maps to pattern1; G (green) maps to pattern2.

We need to infer mapping for other colors perhaps from other possible patterns? But we only have two examples. However we need to predict for test input.

Test Input:
G . G . . . G
G . W X . . X
. G . G . R W
A G . G . . O
. G X . . G G
. . G G . . .
. . G . Y . .

Let's count colors.

We'll parse each row.

Row0: "G . G . . . G" => G at col0, col2, col6 => 3 G.
Row1: "G . W X . . X" => G at col0, W at col2, X at col3, X at col6 => G:1, W:1, X:2.
Row2: ". G . G . R W" => G at col1, col3, R at col5, W at col6 => G:2, R:1, W:1.
Row3: "A G . G . . O" => A at col0, G at col1, col3, O at col6 => A:1, G:2, O:1.
Row4: ". G X . . G G" => G at col1, X at col2, G at col5, col6 => G:3, X:1.
Row5: ". . G G . . ." => G at col2, col3 => G:2.
Row6: ". . G . Y . ." => G at col2, Y at col4 =

... [1640 chars truncated]

We need to understand the puzzle. The description: "What you see: A dense star field with stars of various spectral types. One class of star vastly outnumbers the rest, which appear as rare scattered points. What to do: Classify the stars and determine which spectral type dominates. Each dominant class triggers a specific constellation projection — a fixed pattern. Display it on the dome. Grid details: The star field is 7×7. The constellation projection is 3×3, illuminated in azure on the dark dome."

Thus we have a 7x7 input grid of colors (including black '.' as background). We need to find the most frequent color (dominant class) among the non-black? Or among all? The description: "dense star field with stars of various spectral types. One class of star vastly outnumbers the rest, which appear as rare scattered points." So there is a dominant color (spectral type) that appears many times; other colors appear rarely. The output is a 3x3 pattern (constellation projection) in azure (A) on dark dome (black background). So output grid is 3x3 with '.' for black and 'A' for azure where the pattern appears.

Thus we need to map each dominant color to a specific 3x3 pattern of A's (and .). The training examples show output patterns: Example1 output:
. . A
A A .
. A .
That's a pattern: positions (0,2), (1,0), (1,1), (2,1) are A.

Example2 output:
. A A
. A .
. A .
Pattern: (0,1),(0,2),(1,1),(2,1) are A.

Example3 output:
. . .
A A A
. . .
Pattern: (1,0),(1,1),(1,2) are A (a horizontal line middle row).

Thus each dominant color maps to a specific pattern.

We need to infer mapping from training examples: For each example, we need to determine which color is dominant in the 7x7 input, then see which pattern corresponds.

Let's compute frequencies for each example.

Example1 Input:
Row0: B . B . O . .
Row1: B B W B . B .
Row2: . . B B . R .
Row3: . . . . G . B
Row4: . Y . B . . B
Row5: . . B . R . A
Row6: . . B . O G B

Let's list all colors (excluding '.' maybe). Count each.

I'll go through each cell.

Row0: B, ., B, ., O, ., . => B:2, O:1
Row1: B, B, W, B, ., B, . => B:4, W:1
Row2: ., ., B, B, ., R, . => B:2, R:1
Row3: ., ., ., ., G, ., B => G:1, B:1
Row4: ., Y, ., B, ., ., B => Y:1, B:2
Row5: ., ., B, ., R, ., A => B:1, R:1, A:1
Row6: ., ., B, ., O, G, B => B:2, O:1, G:1

Now sum totals:
B: from rows: 2+4+2+1+2+1+2 = 14? Let's compute: row0 2, row1 4 =>6, row2 2 =>8, row3 1 =>9, row4 2 =>11, row5 1 =>12, row6 2 =>14. Yes B=14.
O: row0 1, row6 1 =>2.
W: row1 1 =>1.
R: row2 1, row5 1 =>2.
G: row3 1, row6 1 =>2.
Y: row4 1 =>1.
A: row5 1 =>1.
X? none.
M? none.
Thus dominant is B (blue) with 14 occurrences, far more than others (max other is 2). So dominant class = B (blue). Output pattern for B is example1 output: . . A / A A . / . A .

Thus mapping: B -> pattern1.

Example2 Input:
. G . G X G .
. . G G . . .
A . . . . . G
G Y G W G . G
. . W G B G G
. G G G . G .
. . . . . . G

Let's count.

I'll go row by row.

Row0: ., G, ., G, X, G, . => G:3, X:1
Row1: ., ., G, G, ., ., . => G:2
Row2: A, ., ., ., ., ., G => A:1, G:1
Row3: G, Y, G, W, G, ., G => G:4? Let's count: positions:0 G,1 Y,2 G,3 W,4 G,5 .,6 G => G:4, Y:1, W:1
Row4: ., ., W, G, B, G, G => W:1, G:3, B:1
Row5: ., G, G, G, ., G, . => G:4? Actually: ., G, G, G, ., G, . => G at positions1,2,3,5 =>4 G.
Row6: ., ., ., ., ., ., G => G:1

Now sum G: row0 3 + row1 2 =5, +row2 1 =6, +row3 4 =10, +row4 3 =13, +row5 4 =17, +row6 1 =18. So G=18.

Other colors:
X: row0 1 =>1.
A: row2 1 =>1.
Y: row3 1 =>1.
W: row3 1 + row4 1 =2.
B: row4 1 =>1.
Thus dominant is G (green) with 18. Output pattern for G is example2 output: . A A / . A . / . A .

Thus mapping: G -> pattern2.

Example3 Input:
. . R . B X G
. . R W . R .
R R R Y R . .
. R . R O R .
R R . . R R M
. R R . R . .
X . Y R . R R

Let's count.

I'll go row by row.

Row0: ., ., R, ., B, X, G => R:1, B:1, X:1, G:1
Row1: ., ., R, W, ., R, . => R:2, W:1
Row2: R, R, R, Y, R, ., . => R:4? Actually positions:0 R,1 R,2 R,3 Y,4 R,5 .,6 . => R:4, Y:1
Row3: ., R, ., R, O, R, . => R:3? positions:0 .,1 R,2 .,3 R,4 O,5 R,6 . => R:3, O:1
Row4: R, R, ., ., R, R, M => R:4? positions:0 R,1 R,2 .,3 .,4 R,5 R,6 M => R:4, M:1
Row5: ., R, R, ., R, ., . => R:3? positions:0 .,1 R,2 R,3 .,4 R,5 .,6 . => R:3
Row6: X, ., Y, R, ., R, R => X:1, Y:1, R:3? positions:0 X,1 .,2 Y,3 R,4 .,5 R,6 R => R:3, X:1, Y:1

Now sum R: row0 1 + row1 2 =3, +row2 4 =7, +row3 3 =10, +row4 4 =14, +row5 3 =17, +row6 3 =20. So R=20.

Other colors:
B: row0 1 =>1.
X: row0 1 + row6 1 =2.
G: row0 1 =>1.
W: row1 1 =>1.
Y: row2 1 + row6 1 =2.
O: row3 1 =>1.
M: row4 1 =>1.
Thus dominant is R (red) with 20. Output pattern for R is example3 output: . . . / A A A / . . .

Thus mapping: R -> pattern3 (horizontal line middle row).

Thus we have mapping for B, G, R. Need to infer mapping for other colors perha

... [2135 chars truncated]

We need to parse the problem: It's a grid transformation puzzle. Input is a 7x7 grid of colors (characters). Output is a 3x3 grid (the signal pennant) in azure on a dark field (i.e., azure = A, dark field = black '.'? Actually "stitched in azure on a dark field" suggests the pennant pattern is azure (A) on a dark field (black '.'), i.e., the output grid will have A's where the pattern is, and '.' elsewhere. The examples show output grids with '.' and 'A' only. Indeed Example 1 output:
. . A
A A .
. A .
That's a 3x3 pattern of A's and '.'.

Example 2 output:
. A A
. A .
. A .
Also only A and '.'.

Thus the transformation: given a 7x7 input grid, we need to find the dominant fleet (color) by count (most frequent color? "One fleet vastly outnumbers the others, while a few foreign vessels dot the anchorage." So we find the color that appears most frequently (dominant fleet). Then each fleet has a unique signal pennant pattern. The harbormaster hoists that pennant. So the output is the pennant pattern associated with the dominant color, rendered as azure (A) on dark field ('.').

Thus we need to infer from training examples what pennant pattern corresponds to each color (dominant fleet). In Example 1, we need to see what the dominant color is in the input, and then see that the output pattern corresponds to that color's pennant.

Example 1 input:
Row0: B . B . O . .
Row1: B B W B . B .
Row2: . . B B . R .
Row3: . . . . G . B
Row4: . Y . B . . B
Row5: . . B . R . A
Row6: . . B . O G B

Let's count colors. We'll tally.

We'll go through each cell.

Row0: B, ., B, ., O, ., . => B:2, O:1
Row1: B, B, W, B, ., B, . => B:4 (so total B now 6), W:1
Row2: ., ., B, B, ., R, . => B:2 (total B 8), R:1
Row3: ., ., ., ., G, ., B => G:1, B:1 (total B 9)
Row4: ., Y, ., B, ., ., B => Y:1, B:2 (total B 11)
Row5: ., ., B, ., R, ., A => B:1 (total B 12), R:1 (total R 2), A:1
Row6: ., ., B, ., O, G, B => B:2 (total B 14), O:1 (total O 2), G:1 (total G 2)

Now also count '.' (black) but not needed.

Thus B appears 14 times, which is dominant. Others: O 2, W 1, R 2, G 2, Y 1, A 1.

Thus dominant fleet is B (blue). The output pattern for B (from example 1) is:
. . A
A A .
. A .

Thus the pennant pattern for blue is that shape.

Now Example 2 input:
Row0: . G . G X G .
Row1: . . G G . . .
Row2: A . . . . . G
Row3: G Y G W G . G
Row4: . . W G B G G
Row5: . G G G . G .
Row6: . . . . . . G

Let's count.

We'll tally each color.

Row0: ., G, ., G, X, G, . => G:3, X:1
Row1: ., ., G, G, ., ., . => G:2 (total G 5)
Row2: A, ., ., ., ., ., G => A:1, G:1 (total G 6)
Row3: G, Y, G, W, G, ., G => G:4 (total G 10), Y:1, W:1
Row4: ., ., W, G, B, G, G => W:1 (total W 2), G:3 (total G 13), B:1
Row5: ., G, G, G, ., G, . => G:4 (total G 17)
Row6: ., ., ., ., ., ., G => G:1 (total G 18)

Thus G appears 18 times, dominant. Others: A 1, X 1, Y 1, W 2, B 1.

Thus dominant fleet is G (green). Output pattern for G (example 2) is:
. A A
. A .
. A .

Thus the pennant pattern for green is that shape.

Thus we have mapping:
- Blue (B) -> pattern:
[ [., ., A],
  [A, A, .],
  [., A, .] ]
- Green (G) -> pattern:
[ [., A, A],
  [., A, .],
  [., A, .] ]

We need to infer pattern for other colors? Possibly each color has a unique pennant pattern (3x3). We only have two examples; we need to deduce the rule that maps color to pattern. Perhaps the pattern is based on the color's position in some ordering or some shape like a letter? Let's examine the patterns.

Blue pattern:
Row0: . . A
Row1: A A .
Row2: . A .

Green pattern:
Row0: . A A
Row1: . A .
Row2: . A .

These look like maybe they are representing the letter corresponding to the color's initial? B pattern maybe looks like a "B"? Let's see if we can interpret the pattern as a letter shape using A as filled and . as empty.

Blue pattern:
 . . A
 A A .
 . A .

If we view A as pixel, shape:
Row0: ..#
Row1: ##.
Row2: .#.

That looks like a kind of shape maybe like a "b"? Not obvious.

Green pattern:
 . A A
 . A .
 . A .

That is:
Row0: .##
Row1: .#.
Row2: .#.

That looks like a vertical line with a top bar? Actually it's like a "┤"? Not sure.

Maybe the pattern is based on the color's position in the color key order? The color key: .=black B=blue R=red G=green Y=yellow X=grey M=magenta O=orange A=azure W=maroon

Thus order: . (0), B (1), R (2), G (3), Y (4), X (5), M (6), O (7), A (8), W (9). But we only care about non-black colors? The pennant is azure on dark field, so the pattern uses A for the shape and . for background. So the pattern is independent of the input color; it's just a shape that depends on the dominant color's identity.

Thus we need to infer the rule mapping from color to pattern. Perhaps the pattern is the letter corresponding to the color's first letter, drawn in a 3x3 font? Let's test: For B, letter "B" in 3x3? Usually a 3x3 representation of B could be:
###
#.#
###
But that's not matching. For G, letter "G" in 3x3 could be:
###
#..
###
But not m

... [8077 chars truncated]

We need to understand the transformation. The description: "A noisy circuit board carrying signals across many frequency bands. One frequency appears on far more traces than the others, which register as faint noise. Analyze the signal spectrum and identify the dominant frequency. Each frequency is mapped to a unique diagnostic LED pattern — a fixed indicator shape. Light up that pattern. Grid details: The circuit board is 7×7. The diagnostic display is 3×3, lit in azure on a dark panel."

Thus input is 7x7 grid with colors representing signals (frequencies). The output is 3x3 grid with colors: only azure (A) and black (.)? In examples, output uses only '.' and 'A' (azure). Indeed both examples output only '.' and 'A'. So the diagnostic display is lit in azure on a dark panel: meaning the pattern is drawn in azure (A) on black background (.). So output is a 3x3 pattern of A's indicating the shape for the dominant frequency.

Thus we need to find the dominant frequency (color) in the input 7x7 grid, then map that color to a specific 3x3 pattern (LED pattern) that is fixed per color. The pattern is lit in azure (A) on black background. So the pattern is independent of the input color; it's just a shape that corresponds to that frequency (color). The output uses A to indicate lit LEDs, '.' for off.

Thus we need to deduce mapping from each color (maybe B,R,G,Y,O,X,W,M,A?) to a 3x3 pattern. We have two training examples: we can infer the pattern for the dominant color in each example.

Example 1 input: Let's count colors.

Input grid (7x7):
Row0: B . B . O . .
Row1: B B W B . B .
Row2: . . B B . R .
Row3: . . . . G . B
Row4: . Y . B . . B
Row5: . . B . R . A
Row6: . . B . O G B

Let's list all non-dot colors:

Row0: B, B, O
Row1: B,B,W,B,B
Row2: B,B,R
Row3: G,B
Row4: Y,B,B
Row5: B,R,A
Row6: B,O,G,B

Let's count each:

B: count them.
Row0: B at col0, col2 =>2
Row1: col0 B, col1 B, col3 B, col5 B =>4 (total 6)
Row2: col2 B, col3 B =>2 (total 8)
Row3: col6 B =>1 (total 9)
Row4: col3 B, col6 B =>2 (total 11)
Row5: col2 B, col6 B =>2 (total 13)
Row6: col2 B, col6 B =>2 (total 15)
Also there may be B elsewhere? Let's double-check each row.

Row0: "B . B . O . ." => B at (0,0), (0,2) =>2
Row1: "B B W B . B ." => B at (1,0),(1,1),(1,3),(1,5) =>4
Row2: ". . B B . R ." => B at (2,2),(2,3) =>2
Row3: ". . . . G . B" => B at (3,6) =>1
Row4: ". Y . B . . B" => B at (4,3),(4,6) =>2
Row5: ". . B . R . A" => B at (5,2) =>1? Wait also maybe B at (5,6)? No it's A at col6. So B count 1.
Row6: ". . B . O G B" => B at (6,2),(6,6) =>2

Sum: 2+4+2+1+2+1+2 =14? Let's recalc: 2+4=6, +2=8, +1=9, +2=11, +1=12, +2=14. So B appears 14 times.

Other colors:
O: Row0 col4 O, Row6 col4 O =>2
W: Row1 col2 W =>1
R: Row2 col5 R, Row5 col4 R =>2
G: Row3 col4 G, Row6 col5 G =>2
Y: Row4 col1 Y =>1
A: Row5 col6 A =>1
X: none
M: none

Thus dominant frequency is B (blue) with 14 occurrences, far more than others (max other is 2). So dominant color = B.

Output pattern for example1:
. . A
A A .
. A .

Let's write as 3x3 matrix:
Row0: . . A
Row1: A A .
Row2: . A .

Thus pattern for B is that shape.

Example2 input:
. G . G X G .
. . G G . . .
A . . . . . G
G Y G W G . G
. . W G B G G
. G G G . G .
. . . . . . G

Let's count colors.

Row0: . G . G X G . => G at col1, col3, col5 =>3; X at col4 =>1
Row1: . . G G . . . => G at col2, col3 =>2
Row2: A . . . . . G => A at col0, G at col6 =>1 each
Row3: G Y G W G . G => G at col0, col2, col4, col6 =>4; Y at col1, W at col3 =>1 each
Row4: . . W G B G G => W at col2, G at col3, B at col4, G at col5, col6 => G:3, W:1, B:1
Row5: . G G G . G . => G at col1, col2, col3, col5 =>4
Row6: . . . . . . G => G at col6 =>1

Now sum G: Row0 3 + Row1 2 =5; +Row2 1 =6; +Row3 4 =10; +Row4 3 =13; +Row5 4 =17; +Row6 1 =18. So G appears 18 times.

Other colors:
X:1
A:1
Y:1
W: Row3 col3, Row4 col2 =>2
B:1
Thus dominant is G (green) with 18.

Output pattern for example2:
. A A
. A .
. A .

Matrix:
Row0: . A A
Row1: . A .
Row2: . A .

Thus pattern for G is that shape.

Thus we have mapping:
B -> pattern1:
[ [., ., A],
  [A, A, .],
  [., A, .] ]

G -> pattern2:
[ [., A, A],
  [., A, .],
  [., A, .] ]

We need to infer mapping for other colors perhaps from other patterns? But we only need to predict for test input. Let's compute dominant color in test input.

Test Input:
Row0: G . G . . . G
Row1: G . W X . . X
Row2: . G . G . R W
Row3: A G . G . . O
Row4: . G X . . G G
Row5: . . G G . . .
Row6: . . G . Y . .

Let's list all colors.

We'll count each.

Initialize counts dict.

Row0: positions:
0: G
1: .
2: G
3: .
4: .
5: .
6: G
=> G at col0,2,6 =>3

Row1: "G . W X . . X"
0: G
1: .
2: W
3: X
4: .
5: .
6: X
=> G:1, W:1, X:2

Row2: ". G . G . R W"
0: .
1: G
2: .
3: G
4: .
5: R
6: W
=> G:2, R:1, W:1

Row3: "A G . G . . O"
0: A
1: G
2: .
3: G
4: .
5: .
6: O
=> A:1, G:2, O:1

Row

... [1544 chars truncated]

We need to understand the transformation. The description: "What you see: A sacred grove alive with many spiritual presences. One spirit manifests far more intensely than the others, whose traces are faint and scattered. What to do: Sense which spirit dominates the grove. Each spirit, when it prevails, inscribes its personal rune on the altar — a compact, fixed glyph. Read the glowing rune. Grid details: The grove is 7×7. The altar rune is 3×3, glowing azure on dark stone."

Interpretation: Input is a 7x7 grid of colors. The "spirit" is a color that appears most frequently (dominant). The "faint and scattered" others are other colors. The "personal rune" is a 3x3 pattern (glyph) associated with that dominant spirit (color). The output is a 3x3 grid (the altar rune) with azure (A) on dark stone (black .). So the output is a 3x3 pattern where the dominant spirit's rune is drawn in azure (A) on a black background (.). The rune is a fixed glyph per spirit (color). So we need to infer the mapping from each color (spirit) to its 3x3 rune pattern (with A's and .'s). The training examples show input 7x7 and output 3x3 with A's and .'s.

Thus we need to find for each color (B,R,G,Y,O,W,X,M,A? Actually colors present: B, R, G, Y, O, W, X, M, A? A is azure used only in output as the rune color; not present in input? Possibly not. Input colors: B, R, G, Y, O, W, X, M? Let's see examples.

Example 1 input: Contains B, O, W, R, Y, A? Actually there is an A in last row: ". . B . R . A". Yes there is an A (azure) present in input as well. So A can appear in input as a color (spirit). Output uses A as the rune color (azure). So the mapping may be: For each spirit (color), there is a specific 3x3 pattern of that spirit's rune, which when drawn in azure (A) on black (.) yields output. So the output always uses A for the rune marks, and . for background. So we need to deduce the pattern for each color based on training examples.

Thus we have three training examples, each with known input and output. From each we can infer the dominant color in the input (the one that appears most intensely/frequently). Then we can see the output pattern (which is the rune for that dominant color). Then we can assign that pattern to that color.

We need to compute frequencies of each color in each input 7x7 grid.

Let's do Example 1.

Example 1 input grid (7 rows):

Row0: B . B . O . .
Row1: B B W B . B .
Row2: . . B B . R .
Row3: . . . . G . B
Row4: . Y . B . . B
Row5: . . B . R . A
Row6: . . B . O G B

Let's list all non-dot characters.

I'll count each color.

Initialize counts.

Row0: B at (0,0), B at (0,2), O at (0,4). So B:2, O:1.

Row1: B B W B . B . => B at (1,0), (1,1), (1,3), (1,5) => 4 B; W at (1,2) => W:1.

Row2: . . B B . R . => B at (2,2), (2,3) => +2 B (total B now 2+4+2=8); R at (2,5) => R:1.

Row3: . . . . G . B => G at (3,4) => G:1; B at (3,6) => B+1 => B total 9.

Row4: . Y . B . . B => Y at (4,1) => Y:1; B at (4,3) and (4,6) => +2 B => B total 11.

Row5: . . B . R . A => B at (5,2) => B+1 =>12; R at (5,4) => R+1 =>2; A at (5,6) => A:1.

Row6: . . B . O G B => B at (6,2) and (6,6) => +2 B =>14; O at (6,4) => O+1 =>2; G at (6,5) => G+1 =>2.

Now sum: B=14, O=2, W=1, R=2, G=2, Y=1, A=1. Dots rest.

Dominant color: B (blue) with 14 occurrences. Next O=2 etc.

Thus dominant spirit is B (blue). Output for example 1 is:

. . A
A A .
. A .

Let's write as 3x3:

Row0: . . A
Row1: A A .
Row2: . A .

Thus the rune for B (blue) is this pattern.

Now Example 2.

Input:

Row0: . G . G X G .
Row1: . . G G . . .
Row2: A . . . . . G
Row3: G Y G W G . G
Row4: . . W G B G G
Row5: . G G G . G .
Row6: . . . . . . G

Let's count.

Initialize counts.

Row0: . G . G X G . => G at (0,1), (0,3), (0,5) => G:3; X at (0,4) => X:1.

Row1: . . G G . . . => G at (1,2), (1,3) => +2 G => total G=5.

Row2: A . . . . . G => A at (2,0) => A:1; G at (2,6) => G+1 =>6.

Row3: G Y G W G . G => G at (3,0), (3,2), (3,4), (3,6) => +4 G =>10; Y at (3,1) => Y:1; W at (3,3) => W:1.

Row4: . . W G B G G => W at (4,2) => W+1 =>2; G at (4,3), (4,5), (4,6) => +3 G =>13; B at (4,4) => B:1.

Row5: . G G G . G . => G at (5,1), (5,2), (5,3), (5,5) => +4 G =>17.

Row6: . . . . . . G => G at (6,6) => +1 G =>18.

Now also check any other colors: maybe R? none. Y=1, W=2, X=1, B=1, A=1.

Dominant: G (green) with 18.

Output for example 2:

. A A
. A .
. A .

Let's write:

Row0: . A A
Row1: . A .
Row2: . A .

Thus rune for G (green) is this pattern.

Now Example 3.

Input:

Row0: . . R . B X G
Row1: . . R W . R .
Row2: R R R Y R . .
Row3: . R . R O R .
Row4: R R . . R R M
Row5: . R R . R . .
Row6: X . Y R . R R

Let's count.

Initialize.

Row0: . . R . B X G => R at (0,2) => R:1; B at (0,4) => B:1; X at (0,5) => X:1; G at (0,6) => G:1.

Row1: . . R W . R . => R at (1,2) and (1,5) => +2 R 

... [6657 chars truncated]

We need to understand the transformation. The description: "A vibrant dance floor with dancers performing in many different styles. One style clearly dominates by sheer numbers, while a few solo performers represent other traditions. Watch the floor and identify the dominant dance style. Each style, when it takes over, resolves into a characteristic formation — a compact arrangement. Capture that formation in the spotlight. Grid details: The dance floor is 7×7. The spotlight formation is 3×3, illuminated in azure on a dark stage."

Thus we have a 7x7 input grid with colors representing dance styles. The dominant style (most frequent color) is identified. Then we need to capture its characteristic formation (some pattern) in a 3x3 spotlight, illuminated in azure (A) on a dark stage (black .). So output is a 3x3 grid where the formation of the dominant style is shown using azure (A) for the formation cells, and black (.) elsewhere.

Thus we need to find the most frequent color in the 7x7 input (excluding maybe black? but black is a color too; but likely black is background, not a style). The description: "dancers performing in many different styles. One style clearly dominates by sheer numbers, while a few solo performers represent other traditions." So black may be empty floor (no dancer). So we should count non-black colors? The examples: Let's test.

Example 1 input:
Row0: B . B . O . .
Row1: B B W B . B .
Row2: . . B B . R .
Row3: . . . . G . B
Row4: . Y . B . . B
Row5: . . B . R . A
Row6: . . B . O G B

Let's count colors (non '.'):
B: many. Let's count: Row0: B at col0, col2 =>2. Row1: B B W B . B => B at col0,1,3,5 =>4. Row2: . . B B . R . => B at col2,3 =>2. Row3: . . . . G . B => B at col6 =>1. Row4: . Y . B . . B => B at col3,6 =>2. Row5: . . B . R . A => B at col2 =>1. Row6: . . B . O G B => B at col2,6 =>2. Total B = 2+4+2+1+2+1+2 =14. Others: O: row0 col4? Actually row0: O at col4? row0: B . B . O . . => O at col4 =>1. Row6: O at col4? row6: . . B . O G B => O at col4 =>1. So O total 2. W: row1 col2 =>1. R: row2 col5? row2: . . B B . R . => R at col5 =>1. row4 col4? row4: . Y . B . . B => no R. row5 col4? row5: . . B . R . A => R at col4 =>1. So R total 2. Y: row4 col1 =>1. A: row5 col6 =>1. G: row6 col5? row6: . . B . O G B => G at col5 =>1. So G total 1. So B is dominant (14). Output is a 3x3 pattern: 
. . A
A A .
. A .
So pattern of A's (azure) forming some shape. Let's see shape: positions (0,2), (1,0), (1,1), (2,1). That's like a "L" shape? Actually coordinates: (row, col):
(0,2)
(1,0)
(1,1)
(2,1)
That's like a shape maybe representing the dominant style's formation: maybe it's the shape of the letter "B"? Not sure.

Let's see example 2.

Example 2 input:
Row0: . G . G X G .
Row1: . . G G . . .
Row2: A . . . . . G
Row3: G Y G W G . G
Row4: . . W G B G G
Row5: . G G G . G .
Row6: . . . . . . G

Count non '.' colors:
G: many. Let's count: Row0: G at col1, col3, col5 =>3. Row1: G at col2, col3 =>2. Row2: G at col6 =>1. Row3: G at col0,2,4,6 =>4. Row4: G at col3,5,6 =>3. Row5: G at col1,2,3,5 =>4. Row6: G at col6 =>1. Total G = 3+2+1+4+3+4+1 =18. Others: X: row0 col4 =>1. A: row2 col0 =>1. Y: row3 col1 =>1. W: row3 col3? Actually row3: G Y G W G . G => W at col3 =>1. plus row4 col2? row4: . . W G B G G => W at col2 =>1. So W total 2. B: row4 col4 =>1. So G dominant (18). Output:
. A A
. A .
. A .
Pattern: A's at (0,1),(0,2),(1,1),(2,1). That's like a vertical line on column1 plus an extra at (0,2). Actually shape: column1 has A at rows0,1,2 (three vertical), plus an extra at (0,2) to the right of top. So shape like a "└" rotated? Actually it's like a "┤"? Let's visualize:
Row0: . A A
Row1: . A .
Row2: . A .
So it's a vertical line of three A's in column1, with an extra A to the right of the top cell.

Example 3 input:
Row0: . . R . B X G
Row1: . . R W . R .
Row2: R R R Y R . .
Row3: . R . R O R .
Row4: R R . . R R M
Row5: . R R . R . .
Row6: X . Y R . R R

Count non '.' colors:
R: many. Let's count: Row0: R at col2 =>1. Row1: R at col2, col4? Actually row1: . . R W . R . => R at col2 and col5 =>2. Row2: R R R Y R . . => R at col0,1,2,4 =>4. Row3: . R . R O R . => R at col1,3,5 =>3. Row4: R R . . R R M => R at col0,1,4,5 =>4. Row5: . R R . R . . => R at col1,2,4 =>3. Row6: X . Y R . R R => R at col3,5,6 =>3. Total R = 1+2+4+3+4+3+3 =20. Others: B: row0 col4 =>1. X: row0 col5? Actually row0: . . R . B X G => X at col5 =>1. plus row6 col0? row6: X . Y R . R R => X at col0 =>1. So X total 2. G: row0 col6 =>1. W: row1 col3 =>1. Y: row2 col3? row2: R R R Y R . . => Y at col3 =>1. plus row6 col2? row6: X . Y R . R R => Y at col2 =>1. So Y total 2. O: row3 col4 =>1. M: row4 col6 =>1. A? none. So R dominant (20). Output:
. . .
A A A
. . .
Pattern: a horizontal line of three A'

... [6992 chars truncated]

We need to understand the puzzle: Input is a 7x7 grid of colors (characters). Output is a 3x3 grid of colors (characters). The description: "geological cross-section showing many mineral types embedded in rock. One mineral is pervasive throughout the sample, while others appear as rare inclusions. Examine the sample and identify the dominant mineral by abundance. Each mineral, when dominant, produces a characteristic crystal habit — a fixed geometric form. Sketch it in the notebook. Grid details: The core sample is 7×7. The crystal habit sketch is 3×3, drawn in azure on a dark page."

Interpretation: The input grid is a 7x7 sample. The dominant mineral (most frequent color) is identified. Then we need to output a 3x3 pattern (crystal habit) for that mineral, drawn in azure (A) on a dark page (black '.'?), i.e., the output grid is a 3x3 where the pattern is drawn using azure (A) for the habit, and black (.) for background? But the examples show output containing A and '.' only? Let's check.

Example 1 input: Let's count colors.

Input 1:
Row0: B . B . O . .
Row1: B B W B . B .
Row2: . . B B . R .
Row3: . . . . G . B
Row4: . Y . B . . B
Row5: . . B . R . A
Row6: . . B . O G B

Let's count each color:

- B: many. Let's count: Row0: B at col0, col2 =>2. Row1: B at col0, col1, col3, col5 =>4 (total 6). Row2: B at col2, col3 =>2 (total 8). Row3: B at col6 =>1 (total 9). Row4: B at col3, col6 =>2 (total 11). Row5: B at col2, col6 =>2 (total 13). Row6: B at col2, col6 =>2 (total 15). So B appears 15 times.

- . (black): many. Let's count: total cells 49. Subtract other colors.

- O: appears? Row0 col4 O, Row6 col4 O? Actually row6 col4 O? Row6: . . B . O G B => O at col4. So O appears 2.

- W: row1 col2 W, row4 col1 Y? Actually row4 col1 Y, row5 col4 R? Let's list: W appears row1 col2, row4? Actually row4 col1 Y, row5 col4 R, row6 col5 G. So W appears only row1 col2? Also row? Let's check: row? No other W. So W count 1.

- R: row2 col5 R, row5 col4 R =>2.

- G: row6 col5 G =>1? Also maybe elsewhere? Row? No other G. So G count 1.

- A: row5 col6 A =>1.

- Y: row4 col1 Y =>1.

Thus dominant mineral is B (blue) with 15 occurrences.

Output for example1:
. . A
A A .
. A .

That's a 3x3 pattern: positions:
(0,0) ., (0,1) ., (0,2) A
(1,0) A, (1,1) A, (1,2) .
(2,0) ., (2,1) A, (2,2) .

So pattern of A's forms something like a shape: maybe a "L" shape? Actually A's at (0,2), (1,0), (1,1), (2,1). That's like a shape: top right, middle left, middle middle, bottom middle. Not obvious.

But note: The description says "crystal habit sketch is 3x3, drawn in azure on a dark page." So the habit is drawn in azure (A) on a dark background (black '.'), i.e., the output grid uses A for the habit shape, and '.' for background. Indeed example1 output only contains '.' and 'A'. Example2 output also only contains '.' and 'A'.

Thus the rule: Determine the dominant color (most frequent) in the 7x7 input. Then output a fixed 3x3 pattern (the crystal habit) associated with that color, drawn in azure (A) on black background ('.').

Thus we need to map each possible color (B, R, G, Y, X, M, O, W, A?) to a specific 3x3 habit pattern (using A for the habit). The examples show that for dominant B (blue) the habit pattern is as given in example1 output. For dominant G (green) in example2? Let's check example2 input to see dominant color.

Example2 input:
Row0: . G . G X G .
Row1: . . G G . . .
Row2: A . . . . . G
Row3: G Y G W G . G
Row4: . . W G B G G
Row5: . G G G . G .
Row6: . . . . . . G

Let's count G's: many.

Count per row:
Row0: G at col1, col3, col5 =>3
Row1: G at col2, col3 =>2 (total5)
Row2: G at col6 =>1 (total6)
Row3: G at col0, col2, col4, col6 =>4 (total10)
Row4: G at col3, col5, col6 =>3 (total13)
Row5: G at col1, col2, col3, col5 =>4 (total17)
Row6: G at col6 =>1 (total18)

Thus G appears 18 times. Others: . many, A appears row2 col0 =>1, X appears row0 col4 =>1, Y row3 col1 =>1, W row3 col3, row4 col2 =>2, B row4 col4 =>1. So dominant is G (green). Output for example2:
. A A
. A .
. A .

That's a 3x3 pattern: (0,0) ., (0,1) A, (0,2) A
(1,0) ., (1,1) A, (1,2) .
(2,0) ., (2,1) A, (2,2) .

Thus pattern of A's: positions (0,1),(0,2),(1,1),(2,1). That's like a vertical line in middle column plus two at top row? Actually shape: top row middle and right, middle row middle, bottom row middle. That's like a "T" shape? Actually it's like a vertical line in column1 (middle column) plus an extra at (0,2) to the right of top. So shape: like a "┤" maybe.

Now we need to infer the habit patterns for each color. We have two examples: B -> pattern1, G -> pattern2.

We need to deduce pattern for other colors perhaps from the test input.

Test input:
G . G . . . G
G . W X . . X
. G . G . R W
A G . G . . O
. G X . . G G
. . G G . . .
. . G . Y . .

We need to find dominant color. Let's count.

We'll go through each cell.

I'll create a table.

Row0: 

... [7445 chars truncated]

We need to understand the transformation. Input is a 7x7 grid of colors (characters). Output is a 3x3 grid of colors (characters). The description: "A bubbling pot with many ingredients, but one flavor dominates the broth. Other seasonings — a pinch here, a dash there — are barely noticeable by comparison. What to do: Taste the stew and identify the overwhelming ingredient. Each dominant flavor has a traditional plating style — a specific arrangement of garnish on a small square plate. Present the garnish that corresponds to whatever took over the dish. Grid details: The stew pot is a 7×7 arrangement of ingredients. The plate is always 3×3, garnished in azure on a clean dark surface."

Interpretation: The 7x7 grid is a "stew pot" with many colors. One color dominates (most frequent). The other colors are minor. The output is a 3x3 "plate" garnished in azure (A) on a clean dark surface (black '.'?), i.e., the garnish is azure (A) placed in some pattern on a black background. The pattern depends on which color dominates. So we need to map each possible dominant color to a 3x3 pattern of A's (azure) and '.' (black). The training examples show output patterns: Example 1 output:
. . A
A A .
. A .
That's a pattern: positions: (0,2) A; (1,0) A; (1,1) A; (2,1) A. So shape like a "T" rotated? Actually it's like a "L" shape? Let's draw coordinates: row0: col0 ., col1 ., col2 A; row1: col0 A, col1 A, col2 .; row2: col0 ., col1 A, col2 . => shape: A at (0,2), (1,0), (1,1), (2,1). That's like a "Z"? Not exactly.

Example 2 output:
. A A
. A .
. A .
That's: row0: col0 ., col1 A, col2 A; row1: col0 ., col1 A, col2 .; row2: col0 ., col1 A, col2 . => shape: vertical line of A's in column1 rows0-2 plus an extra A at (0,2). So like a "┤" shape? Actually it's like a "T" rotated? Let's see: column1 all A's (vertical line) plus an extra A at top right.

Thus each dominant color maps to a specific pattern.

We need to infer mapping from training examples: Example 1 input grid, we need to compute which color dominates (most frequent). Then see that pattern corresponds to that dominant color. Example 2 similarly.

Let's compute frequencies for Example 1 input.

Example 1 input:
Row0: B . B . O . .
Row1: B B W B . B .
Row2: . . B B . R .
Row3: . . . . G . B
Row4: . Y . B . . B
Row5: . . B . R . A
Row6: . . B . O G B

Let's list all characters:

Row0: B, ., B, ., O, ., .
Row1: B, B, W, B, ., B, .
Row2: ., ., B, B, ., R, .
Row3: ., ., ., ., G, ., B
Row4: ., Y, ., B, ., ., B
Row5: ., ., B, ., R, ., A
Row6: ., ., B, ., O, G, B

Now count each:

We'll count '.' first: many.

Let's count each color:

Initialize counts: . B O W R G Y A

Row0: .:2 (positions 1,3? Actually row0: index0 B,1 .,2 B,3 .,4 O,5 .,6 . => . at 1,3,5,6 => 4? Wait count: row0: B . B . O . . => . at col1, col3, col5, col6 => 4 '.'; B at col0, col2 => 2 B; O at col4 =>1 O.

Row1: B B W B . B . => B at col0,1,3,5 =>4 B; W at col2 =>1 W; . at col4,6 =>2 '.'.

Row2: . . B B . R . => . at col0,1,4,6 =>4 '.'; B at col2,3 =>2 B; R at col5 =>1 R.

Row3: . . . . G . B => . at col0,1,2,3,5 =>5 '.'; G at col4 =>1 G; B at col6 =>1 B.

Row4: . Y . B . . B => . at col0,2,4,5 =>4 '.'; Y at col1 =>1 Y; B at col3,6 =>2 B.

Row5: . . B . R . A => . at col0,1,3,5 =>4 '.'; B at col2 =>1 B; R at col4 =>1 R; A at col6 =>1 A.

Row6: . . B . O G B => . at col0,1,3 =>3 '.'; B at col2,6 =>2 B; O at col4 =>1 O; G at col5 =>1 G.

Now sum totals:

'.': row0 4 + row1 2 + row2 4 + row3 5 + row4 4 + row5 4 + row6 3 = 4+2=6, +4=10, +5=15, +4=19, +4=23, +3=26. So '.' count = 26.

B: row0 2 + row1 4 + row2 2 + row3 1 + row4 2 + row5 1 + row6 2 = 2+4=6, +2=8, +1=9, +2=11, +1=12, +2=14. So B count = 14.

O: row0 1 + row1 0 + row2 0 + row3 0 + row4 0 + row5 0 + row6 1 = 2.

W: row1 1 =>1.

R: row2 1 + row3 0 + row4 0 + row5 1 + row6 0 =2.

G: row3 1 + row4 0 + row5 0 + row6 1 =2.

Y: row4 1 =>1.

A: row5 1 =>1.

X: none.

Thus dominant color is '.' (black) with 26 occurrences, far more than others. But the description says "one flavor dominates the broth" and we need to identify overwhelming ingredient. However the output is garnish in azure on black surface. If dominant is black, then the plate would be black background with azure garnish pattern. That matches: output is a 3x3 with '.' (black) background and A's (azure) garnish. So indeed the dominant color determines the pattern of azure garnish on black background.

Thus we need to map dominant color to a pattern of A's (maybe also other colors? but description says garnished in azure on a clean dark surface, so only azure garnish, rest black). So output only contains '.' and 'A'.

Thus we need to infer mapping from dominant color to pattern of A's.

From Example 1: dominant '.' (black) gave pattern:
. . A
A A .
. A .

From Example 2: we need to compute dominant color for Example 2 input.

Example 2 input:
Row0

... [5494 chars truncated]

We need to understand the puzzle. The description: "What you see: A chaotic warm-up where many instruments play simultaneously. One instrumental section is far louder and more numerous than the scattered soloists from other sections. What to do: Listen for which section dominates the soundscape. The conductor responds with a specific baton gesture — a compact motion pattern unique to that instrument family. Transcribe the gesture. Grid details: The orchestra pit is 7×7. The conductor's gesture is a 3×3 notation in azure on silence."

Thus we have a 7x7 input grid with colors representing instruments? The output is a 3x3 grid in azure (A) on silence (black .). So output is a 3x3 pattern of A's and .'s (silence). The pattern is the conductor's gesture, which is unique to the instrument family that dominates.

Thus we need to find which color (instrument) is dominant (most numerous) in the 7x7 input, and then map that to a specific 3x3 pattern (the gesture) that is unique per instrument family. The training examples show mapping from input to output.

Let's examine examples.

Example 1 Input (7x7):
Row0: B . B . O . .
Row1: B B W B . B .
Row2: . . B B . R .
Row3: . . . . G . B
Row4: . Y . B . . B
Row5: . . B . R . A
Row6: . . B . O G B

We need to count colors. Let's list all cells.

I'll parse:

Row0: B . B . O . .
=> B, ., B, ., O, ., .
Row1: B B W B . B .
=> B, B, W, B, ., B, .
Row2: . . B B . R .
=> ., ., B, B, ., R, .
Row3: . . . . G . B
=> ., ., ., ., G, ., B
Row4: . Y . B . . B
=> ., Y, ., B, ., ., B
Row5: . . B . R . A
=> ., ., B, ., R, ., A
Row6: . . B . O G B
=> ., ., B, ., O, G, B

Now count each color:

- B: Let's count: Row0: B at col0, col2 =>2. Row1: col0, col1, col3, col5 =>4 (total 6). Row2: col2, col3 =>2 (total 8). Row3: col6 =>1 (total 9). Row4: col3, col6 =>2 (total 11). Row5: col2 =>1 (total 12). Row6: col2, col6 =>2 (total 14). So B count = 14.

- O: Row0 col4 =>1. Row6 col4 =>1. So O=2.

- W: Row1 col2 =>1.

- R: Row2 col5 =>1. Row5 col4 =>1. So R=2.

- G: Row4 col1 =>1. Row6 col5 =>1. So G=2.

- Y: Row4 col1? Actually Y at row4 col1? Wait row4: . Y . B . . B => Y at col1. So Y=1.

- A: Row5 col6 =>1.

- . (black): rest.

Thus B is dominant (14). Output for example1 is:
. . A
A A .
. A .

That's a 3x3 pattern: positions (0,2)=A, (1,0)=A, (1,1)=A, (2,1)=A. Others are . (black). So pattern: A at top-right, middle-left, middle-center, bottom-center.

Now example2 Input:
. G . G X G .
. . G G . . .
A . . . . . G
G Y G W G . G
. . W G B G G
. G G G . G .
. . . . . . G

Let's count colors.

Row0: . G . G X G .
=> ., G, ., G, X, G, .
Row1: . . G G . . .
=> ., ., G, G, ., ., .
Row2: A . . . . . G
=> A, ., ., ., ., ., G
Row3: G Y G W G . G
=> G, Y, G, W, G, ., G
Row4: . . W G B G G
=> ., ., W, G, B, G, G
Row5: . G G G . G .
=> ., G, G, G, ., G, .
Row6: . . . . . . G
=> ., ., ., ., ., ., G

Now count G: Row0: col1, col3, col5 =>3. Row1: col2, col3 =>2 (total5). Row2: col6 =>1 (total6). Row3: col0, col2, col4, col6 =>4 (total10). Row4: col3, col5, col6 =>3 (total13). Row5: col1, col2, col3, col5 =>4 (total17). Row6: col6 =>1 (total18). So G=18.

Other colors: X: row0 col4 =>1. A: row2 col0 =>1. Y: row3 col1 =>1. W: row3 col3, row4 col2 =>2. B: row4 col4 =>1. . rest.

Thus dominant is G (green). Output for example2:
. A A
. A .
. A .

Pattern: A at (0,1),(0,2),(1,1),(2,1). So A's in top middle, top right, middle center, bottom center.

Example3 Input:
. . R . B X G
. . R W . R .
R R R Y R . .
. R . R O R .
R R . . R R M
. R R . R . .
X . Y R . R R

Let's count.

Row0: . . R . B X G
=> ., ., R, ., B, X, G
Row1: . . R W . R .
=> ., ., R, W, ., R, .
Row2: R R R Y R . .
=> R, R, R, Y, R, ., .
Row3: . R . R O R .
=> ., R, ., R, O, R, .
Row4: R R . . R R M
=> R, R, ., ., R, R, M
Row5: . R R . R . .
=> ., R, R, ., R, ., .
Row6: X . Y R . R R
=> X, ., Y, R, ., R, R

Now count R: Row0 col2 =>1. Row1 col2, col5 =>2 (total3). Row2 col0,1,2,4 =>4 (total7). Row3 col1,3,5 =>3 (total10). Row4 col0,1,4,5 =>4 (total14). Row5 col1,2,4 =>3 (total17). Row6 col3,5,6 =>3 (total20). So R=20.

Other colors: B: row0 col4 =>1. X: row0 col5, row6 col0 =>2. G: row0 col6 =>1. W: row1 col3 =>1. Y: row2 col3, row6 col2 =>2. O: row3 col4 =>1. M: row4 col6 =>1. . rest.

Thus dominant is R (red). Output for example3:
. . .
A A A
. . .

Pattern: A's in middle row all three columns (i.e., row1 col0,1,2). Others are . (black). So pattern: a horizontal line of A's in middle row.

Thus we have mapping from dominant color to a specific 3x3 pattern of A's.

Let's list patterns:

- Dominant B (blue) => pattern: A at (0,2), (1,0), (1,1), (2,1). Visualizing:

Row0: . . A
Row1: A A .
Row2: . A .

- Dominant G (green) => pattern: A at (0,1),(0,2),(1,1),(2,1). Visual:

Row0: . A A
Row1: . A .
Row2: . A .

- Dominant R (red) => pattern

... [4205 chars truncated]

We need to parse the problem: It's a grid transformation puzzle. Input is a 7x7 grid of colors (characters). Output is a 3x3 grid (scoreboard pennant) displayed in azure against a dark board. The metaphor: chaotic scrimmage field with athletes in various team colors. One team's jersey color appears far more often than any other. A few lone players from other teams are scattered about. Count the jerseys and identify the team with the largest presence. Each team has a signature pennant shape. Hoist the winning team's pennant on the scoreboard. The scoreboard pennant is 3x3, displayed in azure against a dark board.

Thus: The output is a 3x3 pattern (pennant shape) in azure (A) on a dark board (black '.'?), i.e., the output grid uses '.' for black background and 'A' for azure where the pennant shape appears. The shape is determined by the winning team (most frequent color) and each team has a signature pennant shape. So we need to infer the mapping from team color to pennant shape (3x3 pattern of A's and .'s). The training examples show output patterns for given inputs.

Let's examine examples.

Example 1 Input:
Row0: B . B . O . .
Row1: B B W B . B .
Row2: . . B B . R .
Row3: . . . . G . B
Row4: . Y . B . . B
Row5: . . B . R . A
Row6: . . B . O G B

We need to count colors. Let's tally.

We'll list all characters:

Row0: B, ., B, ., O, ., .
Row1: B, B, W, B, ., B, .
Row2: ., ., B, B, ., R, .
Row3: ., ., ., ., G, ., B
Row4: ., Y, ., B, ., ., B
Row5: ., ., B, ., R, ., A
Row6: ., ., B, ., O, G, B

Now count each color:

- B: Let's count: Row0: B at col0, col2 =>2. Row1: col0 B, col1 B, col3 B, col5 B =>4 (total 6). Row2: col2 B, col3 B =>2 (total 8). Row3: col6 B =>1 (total 9). Row4: col1 B, col6 B =>2 (total 11). Row5: col2 B =>1 (total 12). Row6: col2 B, col6 B =>2 (total 14). So B appears 14 times.

- . (black): many. Let's count later.

- O: Row0 col4 O =>1. Row6 col4 O =>1. total 2.

- W: Row1 col2 W =>1. Row4? none. Row5? none. Row6? none. So W=1.

- R: Row2 col5 R =>1. Row4? none. Row5 col4 R =>1. total 2.

- G: Row3 col4 G =>1. Row6 col5 G =>1. total 2.

- Y: Row4 col1 Y =>1.

- A: Row5 col6 A =>1.

Thus B is most frequent (14). So winning team color = B (blue). Output for example 1 is:

. . A
A A .
. A .

Let's write as 3x3:

Row0: . . A
Row1: A A .
Row2: . A .

Thus pattern of A's (azure) at positions: (0,2), (1,0), (1,1), (2,1). That's shape: like a "L" rotated? Let's draw coordinates (r,c) 0-index.

Positions:
(0,2)
(1,0)
(1,1)
(2,1)

That's shape: maybe a "T" shape? Actually shape looks like a "pennant" maybe like a flag shape: a triangle? Let's see: If we consider shape of a pennant (like a triangular flag) maybe it's a right triangle shape? The pattern: top row has one at rightmost column; middle row has two at left two columns; bottom row has one at middle column. That shape looks like a "chevron" or maybe like a "arrowhead"? Let's draw:

Row0: . . A
Row1: A A .
Row2: . A .

If we fill A's, shape looks like:

   *
 * *
  *

That's like a diamond shape missing corners? Actually it's like a shape of a "house"? Let's think.

Example 2 Input:
. G . G X G .
. . G G . . .
A . . . . . G
G Y G W G . G
. . W G B G G
. G G G . G .
. . . . . . G

Let's count colors.

We'll tally.

Row0: ., G, ., G, X, G, .
Row1: ., ., G, G, ., ., .
Row2: A, ., ., ., ., ., G
Row3: G, Y, G, W, G, ., G
Row4: ., ., W, G, B, G, G
Row5: ., G, G, G, ., G, .
Row6: ., ., ., ., ., ., G

Now count G's:

Row0: col1 G, col3 G, col5 G =>3
Row1: col2 G, col3 G =>2 (total5)
Row2: col6 G =>1 (total6)
Row3: col0 G, col2 G, col4 G, col6 G =>4 (total10)
Row4: col3 G, col5 G, col6 G =>3 (total13)
Row5: col1 G, col2 G, col3 G, col5 G =>4 (total17)
Row6: col6 G =>1 (total18)

Thus G appears 18 times.

Other colors:

- . (black): many.

- X: Row0 col4 X =>1

- A: Row2 col0 A =>1

- Y: Row3 col1 Y =>1

- W: Row3 col3 W =>1, Row4 col2 W =>1 => total2

- B: Row4 col4 B =>1

Thus winning team = G (green). Output for example 2:

. A A
. A .
. A .

Let's write:

Row0: . A A
Row1: . A .
Row2: . A .

So pattern: A's at (0,1),(0,2),(1,1),(2,1). That's shape: a vertical line of three in column1 plus an extra at top row column2? Actually column1 has A at rows0,1,2 (vertical line). Plus an extra at (0,2) to the right of top. So shape like a "└" rotated? Actually shape: top row: . A A ; middle: . A . ; bottom: . A . => looks like a "right-facing L" shape? Actually it's like a "corner" shape: a vertical line of three with a block to the right at the top.

Example 3 Input:
. . R . B X G
. . R W . R .
R R R Y R . .
. R . R O R .
R R . . R R M
. R R . R . .
X . Y R . R R

Let's count.

Row0: ., ., R, ., B, X, G
Row1: ., ., R, W, ., R, .
Row2: R, R, R, Y, R, ., .
Row3: ., R, ., R, O, R, .
Row4: R, R, ., ., R, R, M
Row5: ., R, R, ., R, ., .
Row6: X, ., Y, R, ., R, R

Now count R's:

Row0: col2 R =>1
Row1: col2 R, col5 R =>2 (total3)
Row2: col0 R, col1 R, col2 R, 

... [6209 chars truncated]

We need to understand the transformation: Input is a 7x7 grid of colors (including '.' black). Output is a 3x3 grid of colors (including '.' black) with azure (A) as the icon color on a dark bulletin board (presumably '.' background). The description: "Read the dominant weather pattern across the map. Each type of dominant front produces a characteristic forecast symbol — a compact icon. Display that icon on the forecast board." So we need to find the dominant color (most frequent non-black? maybe including black? but black is '.' and likely background). The dominant front type (color) yields a characteristic icon (some pattern of azure A's) that we output as a 3x3 grid.

Thus we need to map each color (dominant) to a specific 3x3 pattern of A's (and '.' elsewhere). The training examples show outputs with A's in certain patterns.

Let's examine each example.

Example 1 Input:
Row0: B . B . O . .
Row1: B B W B . B .
Row2: . . B B . R .
Row3: . . . . G . B
Row4: . Y . B . . B
Row5: . . B . R . A
Row6: . . B . O G B

We need to compute frequencies of each color (excluding '.' maybe). Let's count.

We'll list all cells:

Row0: B, ., B, ., O, ., .
Row1: B, B, W, B, ., B, .
Row2: ., ., B, B, ., R, .
Row3: ., ., ., ., G, ., B
Row4: ., Y, ., B, ., ., B
Row5: ., ., B, ., R, ., A
Row6: ., ., B, ., O, G, B

Now count each color:

B: Let's count: Row0: B at col0, col2 =>2. Row1: B at col0,1,3,5 =>4 (total 6). Row2: B at col2,3 =>2 (total 8). Row3: B at col6 =>1 (total 9). Row4: B at col3,6 =>2 (total 11). Row5: B at col2,6 =>2 (total 13). Row6: B at col2,6 =>2 (total 15). So B count = 15.

O: Row0 col4 =>1. Row6 col4 =>1. Total O=2.

W: Row1 col2 =>1. Row4? none. Row5? none. Row6? none. So W=1.

R: Row2 col5 =>1. Row5 col4 =>1. Total R=2.

G: Row3 col4 =>1. Row6 col5 =>1. Total G=2.

Y: Row4 col1 =>1. Y=1.

A: Row5 col6 =>1. A=1.

X: none.

Thus dominant color is B (blue) with 15 occurrences. Output for example1 is:

. . A
A A .
. A .

That's a 3x3 pattern: positions (0,2)=A, (1,0)=A, (1,1)=A, (2,1)=A. So shape like a "L" rotated? Let's map coordinates: row0 col2 A; row1 col0 A; row1 col1 A; row2 col1 A. So shape: a kind of "corner" shape: three A's forming a 2x2 block missing top-left? Actually positions: (0,2) top right; (1,0) middle left; (1,1) middle; (2,1) bottom middle. That's like a "T" shape? Let's draw:

Row0: . . A
Row1: A A .
Row2: . A .

That's like a shape where A's are at (0,2), (1,0), (1,1), (2,1). This looks like a "knight move"? Not sure.

Example2 Input:
. G . G X G .
. . G G . . .
A . . . . . G
G Y G W G . G
. . W G B G G
. G G G . G .
. . . . . . G

Let's count colors.

We'll list each row:

Row0: ., G, ., G, X, G, .
Row1: ., ., G, G, ., ., .
Row2: A, ., ., ., ., ., G
Row3: G, Y, G, W, G, ., G
Row4: ., ., W, G, B, G, G
Row5: ., G, G, G, ., G, .
Row6: ., ., ., ., ., ., G

Now count:

G: Row0: col1,3,5 =>3. Row1: col2,3 =>2 (total5). Row2: col6 =>1 (6). Row3: col0,2,4,6 =>4 (10). Row4: col3,5,6 =>3 (13). Row5: col1,2,3,5 =>4 (17). Row6: col6 =>1 (18). So G=18.

X: Row0 col4 =>1.

A: Row2 col0 =>1.

Y: Row3 col1 =>1.

W: Row3 col3 =>1, Row4 col2 =>1 => total W=2.

B: Row4 col4 =>1.

Thus dominant is G (green) with 18. Output for example2:

. A A
. A .
. A .

That's pattern: row0: . A A ; row1: . A . ; row2: . A . . So A's at (0,1),(0,2),(1,1),(2,1). That's a vertical line of three A's in column1 rows0-2 plus an extra at (0,2) to the right of top. So shape like a "└" rotated? Actually it's like a "L" shape with vertical line and a top-right extra.

Example3 Input:
. . R . B X G
. . R W . R .
R R R Y R . .
. R . R O R .
R R . . R R M
. R R . R . .
X . Y R . R R

Let's count.

Row0: ., ., R, ., B, X, G
Row1: ., ., R, W, ., R, .
Row2: R, R, R, Y, R, ., .
Row3: ., R, ., R, O, R, .
Row4: R, R, ., ., R, R, M
Row5: ., R, R, ., R, ., .
Row6: X, ., Y, R, ., R, R

Now count each:

R: Row0 col2 =>1. Row1 col2,5 =>2 (total3). Row2 col0,1,2,4 =>4 (7). Row3 col1,3,5 =>3 (10). Row4 col0,1,4,5 =>4 (14). Row5 col1,2,4 =>3 (17). Row6 col3,5,6 =>3 (20). So R=20.

B: Row0 col4 =>1.

X: Row0 col5 =>1, Row6 col0 =>1 => X=2.

G: Row0 col6 =>1.

W: Row1 col3 =>1.

Y: Row2 col3 =>1, Row6 col2 =>1 => Y=2.

O: Row3 col4 =>1.

M: Row4 col6 =>1.

Thus dominant is R (red) with 20. Output for example3:

. . .
A A A
. . .

That's a horizontal line of three A's in middle row (row1 col0-2). So pattern: middle row all A.

Example4 Input:
R . . R R . X
. R R . . . R
. B . . . . .
. . . . R . W
. W . . . . R
. . R B . . A
R . . R R . .

Let's count.

Row0: R, ., ., R, R, ., X
Row1: ., R, R, ., ., ., R
Row2: ., B, ., ., ., ., .
Row3: ., ., ., ., R, ., W
Row4: ., W, ., ., ., ., R
Row5: ., ., R, B, ., ., A
Row6: R, ., ., R, R, ., .

Now count:

R: Row0 col0,3,4 =>3. Row1 col1,2,6 =>3 (total6). Row3 col4 =>1 (7). Row4 col6 =>1 (8). Row5 col2 =>1 (9). Row6 col0,3,4 =>3 (12)

... [6771 chars truncated]

We need to understand the puzzle. The description: "An aerial view of a city block with structures made of various materials. One material overwhelmingly dominates the landscape, while rare materials appear in isolated buildings. Survey the block and identify the prevailing construction material. Each material is associated with a specific zoning stamp — a small blueprint mark. Stamp the permit with the appropriate mark. Grid details: The city block is 7×7. The zoning stamp is 3×3, inked in azure on a dark permit."

Thus we have a 7x7 input grid of colors (characters). We need to find the "prevailing construction material" (dominant color) in the 7x7 grid. Then each material is associated with a specific zoning stamp (a 3x3 pattern) that is "inked in azure on a dark permit". The output is a 3x3 grid (the stamp) presumably using azure (A) for the mark and black (.) for background? The examples show output grids with '.' and 'A' only. Indeed, both examples output only '.' and 'A' characters. So the stamp is a 3x3 pattern where the mark is azure (A) and the rest is black (.). So we need to output a 3x3 pattern of A's and .'s that corresponds to the dominant material.

Thus we need to map each possible material (color) to a specific 3x3 stamp pattern (of A and .). The training examples give us two examples: Example 1 input and output; Example 2 input and output. From these we can infer the mapping for the dominant material in each case.

Let's parse Example 1 input:

Input:
Row0: B . B . O . .
Row1: B B W B . B .
Row2: . . B B . R .
Row3: . . . . G . B
Row4: . Y . B . . B
Row5: . . B . R . A
Row6: . . B . O G B

Let's count frequencies of each color.

We'll list all characters:

Row0: B, ., B, ., O, ., .
Row1: B, B, W, B, ., B, .
Row2: ., ., B, B, ., R, .
Row3: ., ., ., ., G, ., B
Row4: ., Y, ., B, ., ., B
Row5: ., ., B, ., R, ., A
Row6: ., ., B, ., O, G, B

Now count:

- B: Let's count: Row0: B at col0, col2 =>2. Row1: B at col0, col1, col3, col5 =>4 (total 6). Row2: B at col2, col3 =>2 (total 8). Row3: B at col6 =>1 (total 9). Row4: B at col3, col6 =>2 (total 11). Row5: B at col2, col6 =>2 (total 13). Row6: B at col2, col6 =>2 (total 15). So B appears 15 times.

- . (black): many. Let's count total cells: 7x7=49. Subtract counts of other colors.

- O: Row0 col4 =>1. Row6 col4 =>1. So O appears 2.

- W: Row1 col2 =>1.

- R: Row2 col5 =>1. Row5 col4 =>1. So R appears 2.

- G: Row3 col4 =>1. Row6 col5 =>1. So G appears 2.

- Y: Row4 col1 =>1.

- A: Row5 col6 =>1.

Thus dominant material is B (blue) with 15 occurrences, far more than others.

Output for Example 1:
. . A
A A .
. A .

Let's write as 3x3:

Row0: . . A
Row1: A A .
Row2: . A .

Thus pattern of A's: positions (0,2), (1,0), (1,1), (2,1). That's like a shape.

Now Example 2 input:

Input:
. G . G X G .
. . G G . . .
A . . . . . G
G Y G W G . G
. . W G B G G
. G G G . G .
. . . . . . G

Let's count frequencies.

We'll list each row:

Row0: ., G, ., G, X, G, .
Row1: ., ., G, G, ., ., .
Row2: A, ., ., ., ., ., G
Row3: G, Y, G, W, G, ., G
Row4: ., ., W, G, B, G, G
Row5: ., G, G, G, ., G, .
Row6: ., ., ., ., ., ., G

Now count G:

Row0: G at col1, col3, col5 =>3
Row1: G at col2, col3 =>2 (total5)
Row2: G at col6 =>1 (total6)
Row3: G at col0, col2, col4, col6 =>4 (total10)
Row4: G at col3, col5, col6 =>3 (total13)
Row5: G at col1, col2, col3, col5 =>4 (total17)
Row6: G at col6 =>1 (total18)

Thus G appears 18 times.

Other colors:

- . (black): many.

- X: Row0 col4 =>1

- A: Row2 col0 =>1

- Y: Row3 col1 =>1

- W: Row3 col3 =>1, Row4 col2 =>1 => total 2

- B: Row4 col4 =>1

Thus dominant material is G (green) with 18 occurrences.

Output for Example 2:
. A A
. A .
. A .

Let's write:

Row0: . A A
Row1: . A .
Row2: . A .

Thus pattern of A's: positions (0,1), (0,2), (1,1), (2,1). That's like a vertical line on right two columns? Actually it's a shape: top row has two A's at columns 1 and 2; middle row has A at column1; bottom row has A at column1. So shape like a "┤" maybe.

Now we need to infer mapping from dominant color to stamp pattern.

We have two examples:

- Dominant B (blue) => stamp pattern:
 . . A
 A A .
 . A .

- Dominant G (green) => stamp pattern:
 . A A
 . A .
 . A .

We need to deduce rule: maybe the stamp pattern is based on the shape of the letter corresponding to the color? Or maybe it's based on some pattern like a "blotch" that appears in the input? Or maybe it's based on the positions of the dominant color in the 7x7 grid? For example, we could compute the centroid or something and then produce a 3x3 pattern representing where the dominant color appears relative to the center? Let's test.

In Example 1, dominant B appears many places. Perhaps we look at the 3x3 central region of the 7x7 grid? The stamp is 3x3, maybe we take the central 3x3 of the input and then map dominant color to A? But the output only has A and ., not 

... [7115 chars truncated]

We need to understand the puzzle. The description: "What you see: A contested field where many factions have deployed forces. One faction vastly outnumbers the others, whose scattered units are too few to matter. What to do: Count the forces and determine which faction holds numerical superiority. Each faction carries a unique battle standard — a small, fixed emblem. Reproduce that emblem on the strategic map. Grid details: The battlefield is a 7×7 grid. The war room emblem is a 3×3 mark in azure on a dark map."

Thus we have a 7x7 input grid with colors representing different factions (colored cells). One faction is dominant (most numerous). We need to find that faction's "battle standard" (emblem) which is a small fixed emblem (3x3 pattern) in azure (A) on a dark map (i.e., black background). The output is a 3x3 grid (emblem) with azure (A) and black (.) presumably. The examples show output with '.' and 'A' only. Indeed Example 1 output: . . A / A A . / . A . (3x3). Example 2 output: . A A / . A . / . A . (also only '.' and 'A').

Thus the task: Determine which color (faction) appears most frequently in the 7x7 input (dominant). Then produce a 3x3 pattern (emblem) that is the "battle standard" of that faction. The emblem is a fixed pattern per faction, presumably known from the training examples: For the dominant faction in example 1, the output pattern is . . A / A A . / . A . . For example 2, the pattern is . A A / . A . / . A . . So we need to infer the emblem patterns for each faction based on training examples.

We have two training examples. Let's examine them.

Example 1 Input:
Row0: B . B . O . .
Row1: B B W B . B .
Row2: . . B B . R .
Row3: . . . . G . B
Row4: . Y . B . . B
Row5: . . B . R . A
Row6: . . B . O G B

Let's count colors.

We'll tally each color: B (blue), O (orange), W (maroon), R (red), G (green), Y (yellow), A (azure), X (grey), . (black). Also maybe M (magenta) not present.

Let's count manually.

Row0: B . B . O . . => B:2, O:1
Row1: B B W B . B . => B:4? Actually B B W B . B . => B at col0, col1, col3, col5 => 4 B; W:1
Row2: . . B B . R . => B:2, R:1
Row3: . . . . G . B => G:1, B:1
Row4: . Y . B . . B => Y:1, B:2
Row5: . . B . R . A => B:1, R:1, A:1
Row6: . . B . O G B => B:2, O:1, G:1

Now sum B: Row0 2 + Row1 4 + Row2 2 + Row3 1 + Row4 2 + Row5 1 + Row6 2 = 14? Let's compute: 2+4=6, +2=8, +1=9, +2=11, +1=12, +2=14. Yes B=14.

O: Row0 1 + Row6 1 =2.

W: Row1 1 =1.

R: Row2 1 + Row5 1 =2.

G: Row3 1 + Row6 1 =2.

Y: Row4 1 =1.

A: Row5 1 =1.

X: none.

. count: total cells 49 minus sum of colored cells. Sum colored = B14+O2+W1+R2+G2+Y1+A1 = 22? Let's sum: 14+2=16, +1=17, +2=19, +2=21, +1=22, +1=23. Actually we missed something: Let's recalc: B14, O2 =>16, W1=>17, R2=>19, G2=>21, Y1=>22, A1=>23. So colored =23. Black =49-23=26.

Thus dominant faction is B (blue) with 14 occurrences, far more than others (max other is O=2 etc). So dominant faction is B.

Output for example1 is:
. . A
A A .
. A .

Thus the emblem for faction B (blue) is that pattern.

Now example2 Input:
Row0: . G . G X G .
Row1: . . G G . . .
Row2: A . . . . . G
Row3: G Y G W G . G
Row4: . . W G B G G
Row5: . G G G . G .
Row6: . . . . . . G

Let's count.

We'll tally each color.

Row0: . G . G X G . => G:3? Actually positions: col1 G, col3 G, col5 G => 3 G; X:1
Row1: . . G G . . . => G:2
Row2: A . . . . . G => A:1, G:1
Row3: G Y G W G . G => G:4? col0 G, col2 G, col4 G, col6 G =>4 G; Y:1, W:1
Row4: . . W G B G G => W:1, G:3? col3 G, col5 G, col6 G =>3 G; B:1
Row5: . G G G . G . => G:4? col1 G, col2 G, col3 G, col5 G =>4 G
Row6: . . . . . . G => G:1

Now sum G: Row0 3 + Row1 2 + Row2 1 + Row3 4 + Row4 3 + Row5 4 + Row6 1 = 18? Let's compute: 3+2=5, +1=6, +4=10, +3=13, +4=17, +1=18. Yes G=18.

Other colors: X:1, A:1, Y:1, W: Row3 1 + Row4 1 =2, B:1, . count rest.

Thus dominant faction is G (green) with 18.

Output for example2:
. A A
. A .
. A .

Thus emblem for faction G (green) is that pattern.

Thus we have two emblem patterns: for B (blue) and for G (green). Need to infer emblem for each possible faction (colors). There are many colors: . (black) is background, not a faction. The factions are colors: B, R, G, Y, X, M, O, A, W? Actually A is azure, but that's the emblem color, not a faction? The emblem is always azure on dark map. So the emblem is always azure (A) on black (.) background. So the output only uses A and . regardless of which faction is dominant. So the emblem pattern is a fixed 3x3 pattern of azure placements (maybe shape) that is associated with each faction. The examples show two patterns: for B and for G.

Thus we need to deduce the emblem pattern for each faction based on some rule derived from the input? Perhaps the emblem pattern is the shape of the largest connected component of that faction? Or maybe it's the pattern of that faction's cells in the 3x3 central region? Or maybe it's the pattern of t

... [6118 chars truncated]

We need to understand the puzzle. The description: "A casting call where many performers appear on stage. One performer keeps showing up in scene after scene, while others make only fleeting cameos. Review the audition log and identify the performer with the most appearances. Each lead actor is known for a unique mask silhouette. Display that mask on the marquee. Grid details: The stage is 7×7. The marquee mask is 3×3, rendered in azure against a dark backdrop."

Thus we have a 7x7 input grid with colors. We need to find the "performer with the most appearances" i.e., the color that appears most frequently? Or perhaps the element that appears most often in the grid? Then we need to output a 3x3 mask (silhouette) of that performer, rendered in azure (A) against a dark backdrop (black .). So the output is a 3x3 pattern where the mask shape is filled with A (azure) and the rest . (black). The mask shape is unique per performer (color). So each color corresponds to a specific mask shape (silhouette). We need to infer the mask shape for each color from training examples.

In Example 1: Input grid (7x7) with various colors. Output is a 3x3 pattern:
. . A
A A .
. A .

So mask shape: positions (0,2) = A, (1,0)=A, (1,1)=A, (2,1)=A. That's like a shape: maybe a "T" or something? Let's map coordinates: rows 0-2, cols 0-2.

Row0: . . A
Row1: A A .
Row2: . A .

So shape: A at (0,2), (1,0), (1,1), (2,1). Visualizing: It looks like a "L" rotated? Actually shape: top right, middle left, middle middle, bottom middle. That's like a "corner" plus a vertical line? Let's draw:

(0,0) . (0,1) . (0,2) A
(1,0) A (1,1) A (1,2) .
(2,0) . (2,1) A (2,2) .

So shape: a kind of "S"? Not sure.

Example 2: Output:
. A A
. A .
. A .

Rows:
Row0: . A A
Row1: . A .
Row2: . A .

So shape: A at (0,1), (0,2), (1,1), (2,1). That's a vertical line of three in column1 rows0-2 plus an extra at (0,2) to the right of top. So shape like a "T" rotated? Actually vertical line at col1 plus a top-right branch.

Thus the mask shape depends on the most frequent color in the input.

We need to determine for each color what mask shape is.

We have two examples: Example1 output shape corresponds to some color that was most frequent in that input. Example2 output shape corresponds to most frequent color in that input.

Thus we need to compute frequencies in each example input to see which color is most frequent, then map that to the output shape.

Let's compute Example1 input:

Input:
Row0: B . B . O . .
Row1: B B W B . B .
Row2: . . B B . R .
Row3: . . . . G . B
Row4: . Y . B . . B
Row5: . . B . R . A
Row6: . . B . O G B

Let's list all colors:

We'll count each.

Initialize counts.

Row0: B, ., B, ., O, ., . => B:2, O:1
Row1: B, B, W, B, ., B, . => B:4 (adds 4), W:1
Row2: ., ., B, B, ., R, . => B:+2 => total B now 6, R:1
Row3: ., ., ., ., G, ., B => G:1, B:+1 => B:7
Row4: ., Y, ., B, ., ., B => Y:1, B:+2 => B:9
Row5: ., ., B, ., R, ., A => B:+1 => B:10, R:+1 => R:2, A:1
Row6: ., ., B, ., O, G, B => B:+2 => B:12, O:+1 => O:2, G:+1 => G:2

Now also count '.' (black) but we ignore.

Thus counts:
B:12
O:2
W:1
R:2
G:2
Y:1
A:1

Most frequent is B (blue) with 12 appearances.

Thus the mask shape for B (blue) is the output shape from Example1: . . A / A A . / . A .

Thus blue's mask shape is that pattern.

Now Example2 input:

Input:
. G . G X G .
. . G G . . .
A . . . . . G
G Y G W G . G
. . W G B G G
. G G G . G .
. . . . . . G

Let's count.

Initialize counts.

Row0: ., G, ., G, X, G, . => G:3, X:1
Row1: ., ., G, G, ., ., . => G:+2 => G:5
Row2: A, ., ., ., ., ., G => A:1, G:+1 => G:6
Row3: G, Y, G, W, G, ., G => G:+4 => G:10, Y:1, W:1
Row4: ., ., W, G, B, G, G => W:+1 => W:2, G:+3 => G:13, B:1
Row5: ., G, G, G, ., G, . => G:+4 => G:17
Row6: ., ., ., ., ., ., G => G:+1 => G:18

Also count '.' but ignore.

Thus counts:
G:18
X:1
A:1
Y:1
W:2
B:1

Most frequent is G (green) with 18.

Thus mask shape for G (green) is the output from Example2: . A A / . A . / . A .

Thus green's mask shape is that pattern.

Thus we have mapping:
- Blue (B) -> shape1
- Green (G) -> shape2

Now we need to apply to test input. Determine most frequent color in test input, then output its mask shape (3x3) using A for mask and . for background.

Test Input:
Row0: G . G . . . G
Row1: G . W X . . X
Row2: . G . G . R W
Row3: A G . G . . O
Row4: . G X . . G G
Row5: . . G G . . .
Row6: . . G . Y . .

Let's count frequencies.

Initialize counts.

We'll go row by row.

Row0: G, ., G, ., ., ., G => G:3
Row1: G, ., W, X, ., ., X => G:+1 => G:4, W:1, X:2
Row2: ., G, ., G, ., R, W => G:+2 => G:6, R:1, W:+1 => W:2
Row3: A, G, ., G, ., ., O => A:1, G:+2 => G:8, O:1
Row4: ., G, X, ., ., G, G => G:+3 => G:11, X:+1 => X:3
Row5: ., ., G, G, ., ., . => G:+2 => G:13
Row6: ., ., G, ., Y, ., . => G:+1 => G:14, Y:1

Now also count any other colors: we have A (1), W (2),

... [942 chars truncated]

We need to understand the transformation: Input is a 7x7 grid of colors (characters). Output is a 3x3 grid of colors (characters). The metaphor: "wildly overgrown garden plot with many plant species. One species has spread everywhere, while rare specimens appear only once or twice. Identify the most prolific species by its sheer abundance. Each invasive species leaves a unique leaf impression. Press the dominant species' leaf into a small clay tile. Grid details: The garden plot is 7×7. The clay tile is 3×3, with the leaf print in azure on dark earth."

Interpretation: The input grid is a 7x7 garden. The most abundant color (species) is the "dominant species". The leaf impression is a pattern (maybe a shape) that is pressed onto a 3x3 clay tile, with leaf print in azure (A) on dark earth (black .). So output is a 3x3 pattern where the leaf shape of the dominant species is represented by A's, and the rest is black (.)? Or maybe the leaf print is azure on dark earth meaning the leaf is azure (A) and the background is black (.). So output is a 3x3 tile with A's where leaf shape appears, and . elsewhere.

Thus we need to find the most frequent color in the 7x7 input (excluding maybe black? but black is . which is also a color representing black? The color key: . = black. So black is also a color. But the metaphor: "One species has spread everywhere, while rare specimens appear only once or twice." The black . might be considered "soil"? Not a species. Usually black is background. In examples, we see many . in input. The most abundant color might be . (black) but they likely ignore . as it's the earth. The dominant species is the non-black color that appears most frequently. Let's test with examples.

Example 1 input:
Row0: B . B . O . .
Row1: B B W B . B .
Row2: . . B B . R .
Row3: . . . . G . B
Row4: . Y . B . . B
Row5: . . B . R . A
Row6: . . B . O G B

Let's count colors ignoring . (black). Count each:

B: Let's count: Row0: B at col0, col2 =>2. Row1: B B W B . B => B at col0,1,3,5 =>4 (total 6). Row2: . . B B . R . => B at col2,3 =>2 (total 8). Row3: . . . . G . B => B at col6 =>1 (total9). Row4: . Y . B . . B => B at col3,6 =>2 (total11). Row5: . . B . R . A => B at col2 =>1 (total12). Row6: . . B . O G B => B at col2,6 =>2 (total14). So B appears 14 times.

Other colors: O: Row0 col4 =>1, Row6 col4 =>1 =>2. W: Row1 col2 =>1, Row4 col1? Actually Row4 col1 is Y, not W. Row5? none. Row? maybe none else. So W appears 1? Actually Row1 col2 is W, also Row? Row? none else. So W=1. R: Row2 col5 =>1, Row5 col4 =>1 =>2. G: Row6 col5 =>1. Y: Row4 col1 =>1. A: Row5 col6 =>1. So B is dominant with 14.

Output for example1:
. . A
A A .
. A .

That's a 3x3 pattern: positions (0,2)=A, (1,0)=A, (1,1)=A, (2,1)=A. So shape looks like maybe a "leaf" shape of B? Let's see if we can map B's positions in the 7x7 to a 3x3 pattern? Perhaps they take the dominant species and produce a 3x3 pattern that is the "leaf impression" which is maybe a cross or something? Let's see other examples.

Example2 input:
. G . G X G .
. . G G . . .
A . . . . . G
G Y G W G . G
. . W G B G G
. G G G . G .
. . . . . . G

Count non-. colors:

G: Let's count: Row0: G at col1,3,5 =>3. Row1: G at col2,3 =>2 (total5). Row2: A at col0, G at col6 =>1 G (total6). Row3: G at col0,2,4,6 =>4 (total10). Row4: G at col3,5,6 =>3 (total13). Row5: G at col1,2,3,5 =>4 (total17). Row6: G at col6 =>1 (total18). So G appears 18 times.

Other colors: X: Row0 col4 =>1. A: Row2 col0 =>1. Y: Row3 col1 =>1. W: Row3 col3? Actually Row3 col3 is W? Row3: G Y G W G . G => W at col3 =>1, also Row4 col2? Row4: . . W G B G G => W at col2 =>1. So W total 2. B: Row4 col4 =>1. So dominant is G.

Output for example2:
. A A
. A .
. A .

Pattern: positions (0,1)=A, (0,2)=A, (1,1)=A, (2,1)=A. So shape: a vertical line of three A's in middle column (col1) plus an extra A at top right (0,2). Actually shape: top row: . A A ; middle: . A . ; bottom: . A . . So it's like a "L" shape? Actually it's a vertical line in column1 (rows0-2) plus an extra at (0,2). That's like a "Γ" shape? Not sure.

Example3 input:
. . R . B X G
. . R W . R .
R R R Y R . .
. R . R O R .
R R . . R R M
. R R . R . .
X . Y R . R R

Count non-. colors:

R: Let's count: Row0: R at col2 =>1. Row1: R at col2,4 =>2 (total3). Row2: R at col0,1,2,4 =>4 (total7). Row3: R at col1,3,5 =>3 (total10). Row4: R at col0,1,4,5 =>4 (total14). Row5: R at col1,2,4 =>3 (total17). Row6: R at col3,5,6 =>3 (total20). So R appears 20 times.

Other colors: B: Row0 col4 =>1. X: Row0 col5 =>1, Row6 col0 =>1 =>2. W: Row1 col3 =>1. Y: Row2 col3 =>1, Row6 col2 =>1 =>2. O: Row3 col4 =>1. M: Row4 col6 =>1. G: Row0 col6 =>1. So dominant is R.

Output for example3:
. . .
A A A
. . .

Pattern: middle row all A's, top and bottom rows all . . . So a horizontal line of three A'

... [6292 chars truncated]

We need to understand the transformation. The description: "The more voices present in the choir, the more space each one claims. Every performer swells in presence proportional to the total number of singers — as if each voice resonates with every other, amplifying itself. The empty seats swell too, but remain silent darkness. The chorus arranges itself exactly as they stood on the original stage, just grander."

Interpretation: Input is a 3x3 grid (since examples show 3x3). Output is a larger grid where each cell expands into a block of size N x N, where N is the total number of non-empty cells (performers) in the input? Let's test.

Example 1: Input:
. A .
. M M
W A .
Let's count non-empty cells: A (1), M (2), M (3), W (4), A (5). So total performers = 5. Output grid size? Output shown is 9 rows, 9 columns? Let's count: They show 9 lines each with 9 characters separated by spaces? Actually they show like:
. . . A A A . . .
. . . A A A . . .
. . . A A A . . .
. . . M M M M M M
. . . M M M M M M
. . . M M M M M M
W W W A A A . . .
W W W A A A . . .
W W W A A A . . .
That's 9 rows, each row has 9 characters (including '.' and letters). So output size = 9x9. Input size = 3x3. So scaling factor = 3? Since 3*3 =9. But they said each performer's single seat blooms into a square throne whose size is dictated by the richness of the ensemble — a solo act barely grows, but a quintet commands enormous territory. So size of each block maybe equals total number of performers? For 5 performers, block size would be 5? Then output size would be 3*5 =15, not 9. So not that.

Maybe block size = number of distinct colors? Let's see: distinct colors in example1: A, M, W. That's 3 distinct colors. Block size 3 yields output 9x9. Yes! Input 3x3, each cell expands to a 3x3 block (since 3 distinct colors). Output size = 3*3 =9. That matches.

Check example2: Input:
B . Y
. Y .
. B .
Count distinct colors: B, Y. That's 2 distinct colors. Output shown:
B B . . Y Y
B B . . Y Y
. . Y Y . .
. . Y Y . .
. . B B . .
. . B B . .
That's 6 rows, 6 columns. Input 3x3, block size = distinct colors =2 => output size 3*2=6. Yes matches.

Example3: Input:
G R .
. O G
. . .
Distinct colors: G, R, O => 3 distinct colors. Output shown:
G G G R R R . . .
G G G R R R . . .
G G G R R R . . .
. . . O O O G G G
. . . O O O G G G
. . . O O O G G G
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
That's 9 rows, 9 columns. Block size =3 => output 9x9. Yes.

Example4: Input:
Y . G
R R .
. . A
Distinct colors: Y, G, R, A => 4 distinct colors. Output shown:
Y Y Y Y . . . . G G G G
Y Y Y Y . . . . G G G G
Y Y Y Y . . . . G G G G
Y Y Y Y . . . . G G G G
R R R R R R R R . . . .
R R R R R R R R . . . .
R R R R R R R R . . . .
R R R R R R R R . . . .
. . . . . . . . A A A A
. . . . . . . . A A A A
. . . . . . . . A A A A
. . . . . . . . A A A A
That's 12 rows, 12 columns. Input 3x3, block size =4 => output size 12. Yes.

Thus rule: Determine number of distinct non-background colors present in the input (ignore '.' which is black/empty). Let k = number of distinct colors (excluding '.'). Then each input cell expands to a k x k block in the output, preserving the relative arrangement: each cell's color fills its kxk block; empty cells ('.') fill their block with '.' (black). The output grid size = 3*k rows and 3*k columns.

Now test input:
. B .
. A O
W W .

Let's compute distinct colors: B, A, O, W. That's 4 distinct colors (B, A, O, W). So k=4. Input is 3x3. Output size = 3*4 =12 rows and columns.

We need to map each input cell to a 4x4 block.

Let's write input grid with coordinates (r,c) 0-index:

Row0: [0,0]='.', [0,1]='B', [0,2]='.'
Row1: [1,0]='.', [1,1]='A', [1,2]='O'
Row2: [2,0]='W', [2,1]='W', [2,2]='.'

Now for each cell, produce a 4x4 block of its character (or '.' for empty). Then arrange blocks in same order: block rows correspond to input rows, block columns correspond to input columns.

Thus output will be a 12x12 grid composed of 3 block rows each of height 4, and 3 block columns each of width 4.

Let's construct.

We'll create output as list of strings length 12.

We'll iterate over input rows i=0..2, for each we generate 4 output rows (block rows). For each input row i, we iterate over input columns j=0..2, for each we generate 4 output columns (block columns) of the cell's char.

Thus output[row_block*4 + di][col_block*4 + dj] = char of input[i][j] (or '.' if empty).

Let's compute.

First, define char mapping:

(0,0): '.' -> empty
(0,1): 'B'
(0,2): '.'

(1,0): '.'
(1,1): 'A'
(1,2): 'O'

(2,0): 'W'
(2,1): 'W'
(2,2): '.'

Now produce output.

We'll create 12 rows indexed 0-11.

For each input row i:

i=0 (top block row):
- block rows 0-3 (since i*4 to i*4+3)
For each input column j:

j=0: char '.' -> fill its 4x4 block with '.'.
j=1: char 'B' -> fill block with 'B'.
j=2: char '.' -> fill block with '.'.

Thus for rows 0-3, columns 0-3: '.'; columns 4-7: 'B'; columns 8-11: '.'.

So rows 0-3: pattern: 4

... [2229 chars truncated]

We need to understand the transformation. The description: "Each component and each empty pad scales into a square region whose side equals the number of unique part types on the board. Grid details: The PCB layout grows uniformly, maintaining the original component placement as every position becomes a larger square footprint."

Interpretation: Input grid has cells with colors (including '.' for black/empty). The number of unique part types on the board: count distinct non-'.' colors? Or includes '.'? Likely part types are colored components (non-empty). The empty pads are '.'; they also scale into a square region of side equal to number of unique part types. So each cell (whether component or empty) becomes a block of size N x N, where N = number of distinct colors (excluding '.'?) present in the input.

The original component placement is maintained: each original cell maps to a block of N x N in the output, placed at same relative position (scaled). So output dimensions: input height * N rows, input width * N columns.

Now we need to verify with example.

Example 1 Input:
Row0: . A .
Row1: . M M
Row2: W A .

Let's list colors: '.' (empty), A, M, W. Unique part types: A, M, W (non-empty) = 3. So N = 3.

Thus each cell becomes 3x3 block.

Input dimensions: 3 rows, 3 columns.

Output dimensions: 3*3 = 9 rows, 3*3 = 9 columns. Indeed output given is 9 rows, each with 9 characters (with spaces? Actually they show no spaces but we can count). Let's verify.

Output:
Row0: . . . A A A . . .
Row1: . . . A A A . . .
Row2: . . . A A A . . .
Row3: . . . M M M M M M
Row4: . . . M M M M M M
Row5: . . . M M M M M M
Row6: W W W A A A . . .
Row7: W W W A A A . . .
Row8: W W W A A A . . .

Now let's map.

Input cell (0,0) '.' => should become a 3x3 block of '.' (black). In output top-left 3x3 block: rows0-2, cols0-2: all '.' indeed.

Cell (0,1) 'A' => block of 'A's size 3x3 placed at rows0-2, cols3-5 (since column index 1 *3 =3 start). Output rows0-2, cols3-5: we see A A A in each row: yes.

Cell (0,2) '.' => block of '.' at rows0-2, cols6-8: output shows . . . in those columns: yes.

Row1 input: (1,0) '.' => block rows3-5, cols0-2: output rows3-5, cols0-2: . . . yes.

(1,1) 'M' => block rows3-5, cols3-5: output shows M M M in those columns: yes.

(1,2) 'M' => block rows3-5, cols6-8: output shows M M M: yes.

Row2 input: (2,0) 'W' => block rows6-8, cols0-2: output shows W W W: yes.

(2,1) 'A' => block rows6-8, cols3-5: output shows A A A: yes.

(2,2) '.' => block rows6-8, cols6-8: output shows . . .: yes.

Thus rule matches: N = number of unique non-'.' colors.

Now test input:
. B .
. A O
W W .

Let's compute unique non-'.' colors: B, A, O, W. That's 4 distinct colors. So N = 4.

Input dimensions: 3 rows, 3 columns.

Output dimensions: 3*4 = 12 rows, 3*4 = 12 columns.

We need to produce output grid with each cell replaced by 4x4 block of its color (or '.' for empty). Maintain placement.

Let's map.

We'll index rows 0-2, cols 0-2.

Input:
Row0: [0,0]='.', [0,1]='B', [0,2]='.'
Row1: [1,0]='.', [1,1]='A', [1,2]='O'
Row2: [2,0]='W', [2,1]='W', [2,2]='.'

Now each cell expands to 4x4 block.

We'll construct output rows 0-11, cols 0-11.

We'll fill blocks.

Define block start row = r * 4, start col = c * 4.

We'll produce.

Let's create a 12x12 array initially maybe '.'.

We'll fill.

I'll do manually.

First, create a table of blocks.

Block for (0,0) '.' => 4x4 of '.' at rows0-3, cols0-3.

Block (0,1) 'B' => rows0-3, cols4-7 (since col1*4=4) fill with 'B'.

Block (0,2) '.' => rows0-3, cols8-11 fill '.'.

Row1 blocks:

(1,0) '.' => rows4-7, cols0-3 '.'.

(1,1) 'A' => rows4-7, cols4-7 fill 'A'.

(1,2) 'O' => rows4-7, cols8-11 fill 'O'.

Row2 blocks:

(2,0) 'W' => rows8-11, cols0-3 fill 'W'.

(2,1) 'W' => rows8-11, cols4-7 fill 'W'.

(2,2) '.' => rows8-11, cols8-11 fill '.'.

Thus output grid:

Rows 0-3:
- cols0-3: .
- cols4-7: B
- cols8-11: .

Rows 4-7:
- cols0-3: .
- cols4-7: A
- cols8-11: O

Rows 8-11:
- cols0-3: W
- cols4-7: W
- cols8-11: .

Now we need to output as characters separated by spaces, one row per line.

Let's write each row.

Row0: columns 0-11:
0-3: . . . .
4-7: B B B B
8-11: . . . .
Thus: . . . . B B B B . . . .

Row1 same as row0: . . . . B B B B . . . .

Row2 same: . . . . B B B B . . . .

Row3 same: . . . . B B B B . . . .

Row4: cols0-3: . . . .
cols4-7: A A A A
cols8-11: O O O O
Thus: . . . . A A A A O O O O

Row5 same: . . . . A A A A O O O O

Row6 same: . . . . A A A A O O O O

Row7 same: . . . . A A A A O O O O

Row8: cols0-3: W W W W
cols4-7: W W W W
cols8-11: . . . .
Thus: W W W W W W W W . . . .

Row9 same: W W W W W W W W . . . .

Row10 same: W W W W W W W W . . . .

Row11 same: W W W W W W W W . . . .

Now we need to output in the required format: each row as characters separated by spaces.

Let's produce.

We'll write:

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

... [348 chars truncated]

We need to understand the transformation. The description: "Each system — and each calm zone — inflates its area based on how many distinct fronts are present, as if mutual atmospheric pressure forces them all to swell equally. Grid details: The map scales uniformly outward, each original cell becoming a square region, preserving the original meteorological layout."

We have training example: Input 3x3 grid:

Row0: . A .
Row1: . M M
Row2: W A .

Output is 9x9 grid (3x scaling factor? Actually each original cell becomes a square region; scaling factor maybe 3? Let's see: Input dimensions: 3 rows, 3 columns. Output dimensions: 9 rows, 9 columns. So scaling factor = 3. Each original cell becomes a 3x3 block.

Now we need to see mapping: Input colors: . (black), A (azure), M (magenta), W (maroon). Output shows blocks: Let's map.

Input grid coordinates (r,c):
(0,0): . -> output block at rows 0-2, cols 0-2: all . (black). Indeed first three rows first three columns are all '.'.

(0,1): A -> output block rows 0-2, cols 3-5: all A? Let's see output rows 0-2 columns 3-5: row0: A A A, row1: A A A, row2: A A A. Yes.

(0,2): . -> block rows0-2 cols6-8: all '.'.

(1,0): . -> block rows3-5 cols0-2: all '.'.

(1,1): M -> block rows3-5 cols3-5: all M? Let's see rows3-5 columns3-5: row3: M M M M M M? Wait row3: ". . . M M M M M M". Actually row3: columns 0-2 '.' '.' '.'; columns 3-8: M M M M M M. That's six columns of M? But block size is 3x3, so columns 3-5 should be M M M; columns 6-8 should be something else? Let's examine row3: ". . . M M M M M M". That's 9 characters: indices:0 .,1 .,2 .,3 M,4 M,5 M,6 M,7 M,8 M. So columns 3-8 are all M (six M). That suggests scaling factor maybe not uniform? Let's check row4: ". . . M M M M M M" same. Row5 same. So columns 3-8 all M, meaning block for (1,1) expanded to width 6? Actually maybe each cell expands based on number of distinct fronts present? The description: "Each system — and each calm zone — inflates its area based on how many distinct fronts are present, as if mutual atmospheric pressure forces them all to swell equally." So scaling factor may depend on number of distinct colors (fronts) present in the whole grid? Let's examine.

In example 1 input: colors present: . (black), A (azure), M (magenta), W (maroon). That's 4 distinct colors (including black? maybe black is calm zone). Distinct fronts maybe non-black? Let's think.

If each system inflates its area based on how many distinct fronts are present. Suppose there are N distinct fronts (colors other than black?). Then each cell expands to a square of size N? In example, we have distinct non-black colors: A, M, W = 3. So N=3. Then each cell becomes a 3x3 block? But we saw that M block seemed to be 3 rows tall but 6 columns wide? Actually maybe each cell expands equally in all directions, but the overall grid expands such that each original cell's influence region is a square of size (2*N+1)? Not sure.

Let's compute: Input 3x3, output 9x9. If scaling factor = 3, that matches N=3 (distinct non-black colors). So each cell becomes N x N block. Then why M block appears as 6 columns? Let's re-evaluate output: maybe I misread due to formatting. Let's write output grid clearly:

Output:
Row0: . . . A A A . . .
Row1: . . . A A A . . .
Row2: . . . A A A . . .
Row3: . . . M M M M M M
Row4: . . . M M M M M M
Row5: . . . M M M M M M
Row6: W W W A A A . . .
Row7: W W W A A A . . .
Row8: W W W A A A . . .

Wait row3-5 have six M's after three dots. That's columns 3-8 = six columns. But if each cell is 3x3 block, columns 3-5 should be M block for (1,1); columns 6-8 should be something else for (1,2) which is M? Actually input (1,2) is M as well (row1 col2 = M). Indeed row1: ". M M". So (1,2) is M. So block for (1,1) columns 3-5, block for (1,2) columns 6-8. Both are M, so combined they appear as six M's contiguous. That makes sense! So each cell's block is 3x3, and adjacent same-color blocks merge visually. So the output is consistent: each original cell becomes a 3x3 block of its color.

Check W block: Input (2,0) = W, (2,1) = A, (2,2) = . . Actually row2: "W A .". So (2,0)=W, (2,1)=A, (2,2)=. Output rows6-8 columns0-2: W W W each row -> yes W block 3x3. Columns3-5: A A A each row -> A block 3x3. Columns6-8: . . . each row -> . block 3x3.

Thus scaling factor = number of distinct non-black colors? Let's verify: distinct non-black colors in input: A, M, W = 3. Scaling factor = 3. Good.

Now test input: 
. B .
. A O
W W .

We need to compute scaling factor based on number of distinct fronts (non-black colors) present in the test input.

Colors present: . (black), B (blue), A (azure), O (orange), W (maroon). Distinct non-black: B, A, O, W = 4. So scaling factor = 4? Let's verify with example: they had 3 non-black colors -> scaling factor 3. So likely scaling factor = number of distinct non-black colors.

Thus each original cell becomes a 4x4 block of its color.

Input dimensions: 3 

... [3266 chars truncated]

We need to parse the problem: It's a grid transformation puzzle. The description: "A grey wall divides two rooms. On the left, a blue vault door has gaps cut into it — empty keyholes waiting. On the right, a colored key floats in darkness, its teeth shaped from a single vivid color. What to do: Slide the key into the lock. If every tooth fits perfectly into every keyhole — no extra teeth, no empty holes left over — the key merges with the vault door, filling each gap with its color. If the key doesn't match the lock, the vault stays sealed and unchanged, its empty keyholes still dark."

Thus we have a grid with a grey wall (X) in the center separating left and right halves. The left side is the vault door: blue (B) with gaps (holes) that are black (.)? Actually "empty keyholes waiting" likely are black (.) cells. The right side is the key: a colored key (single color) floating in darkness (black). The key's teeth are shaped from a single vivid color (like red, green, etc). The key floats in darkness: meaning the key is represented by colored cells (non-black) on a black background? The key is a shape of a single color (like all red cells) on a black background. The vault door is blue with gaps (black). The key's teeth must fit into the gaps exactly: meaning the shape of the key (colored cells) must exactly match the shape of the gaps (black cells) in the vault door, after possibly aligning? The key can slide (translate) into the lock? The description: "Slide the key into the lock." So we can shift the key horizontally? Since the wall is vertical in the center, the key is on the right side of the wall; the lock is on the left side. To slide the key into the lock, we would move the key leftwards across the wall? Actually the key is on the right side, the lock on left. To insert, the key would move leftwards (west) into the lock area, aligning its teeth with the gaps. If the shape matches exactly (the key's colored cells overlay exactly onto the black gap cells of the lock, and no extra colored cells outside the gaps, and no gaps left unfilled), then the key merges with the vault door, filling each gap with its color. So the output would be the lock area (left side) where the gaps (black) are replaced by the key's color (the color of the key). The rest of the lock (blue) stays blue. The key area (right side) maybe disappears? Or maybe after merging, the key is gone (since it merged). The output examples show only the left side? Let's examine Example 1.

Example 1 Input:
B B B B X . . . .
B . . B X . M M .
B . . B X . M M .
B B B B X . . . .

So it's a 4x8 grid? Actually each line has 8 tokens separated by spaces. Let's write them:

Row0: B B B B X . . . .
Row1: B . . B X . M M .
Row2: B . . B X . M M .
Row3: B B B B X . . . .

Thus width = 8, height = 4. The grey wall X is at column 4 (0-indexed) for all rows (the 5th column). Left side columns 0-3, right side columns 5-7 (since column 4 is X). Right side columns 5,6,7.

Left side (lock) pattern:
Row0: B B B B
Row1: B . . B
Row2: B . . B
Row3: B B B B

So it's a blue border with a 2x2 empty interior (black) at positions (1,1)-(2,2) relative to left side? Actually inside the left side: columns 1-2 are black (.) in rows 1 and 2. So it's a blue frame with a 2x2 black hole in the middle.

Right side (key) pattern:
Row0: . . .
Row1: . M M
Row2: . M M
Row3: . . .

Thus it's a 2x2 block of magenta (M) at positions (1,1)-(2,2) relative to right side (i.e., columns 6-7? Actually column indices: right side columns 5,6,7. In rows 1 and 2, we have at col5 '.' (black), col6 'M', col7 'M'. So the magenta block is at columns 6-7, rows 1-2. So it's a 2x2 magenta block, with a black column to its left (col5) and black rows above/below.

Thus the key shape is a 2x2 block of magenta. The lock's gaps are a 2x2 block of black at columns 1-2, rows 1-2 (left side). So if we slide the key leftwards by 4 columns? Actually the key is currently at columns 5-7 (right side). To align with the lock's gaps at columns 1-2, we need to shift left by 4 columns? Let's compute: lock gap columns: left side columns 1-2 (global indices 1,2). Key's magenta block columns: 6-7 (global). To overlay, we need to shift key left by 5? Actually if we subtract 5 from key columns: 6-5=1, 7-5=2. So shifting left by 5 columns would align key's block to columns 1-2. But the wall is at column 4; moving left across the wall would pass through column 4 (the wall). The key would need to go through the wall? But maybe the key can be placed overlapping the wall? Actually the description: "grey wall always stands in the center, separating lock from key." So the key is on the right side of the wall; to slide into lock, it must go through the wall? Perhaps the wall is just a separator but the key can pass through it? Or maybe the key is considered to be in the same plane but the wall is just a visual separator; the key can be moved leftwards ignoring the wall (like it's a transparent barrier). In the exa

... [8946 chars truncated]

We need to understand the transformation. The description: "A small swatch of fabric displays a motif woven from a single dye color against an undyed background. This fragment remembers the dimensions of the loom it was cut from. The swatch is placed on the loom and allowed to replicate itself — it knows how many times to repeat across the warp by counting its own columns, and how many times to stack along the weft by counting its own rows. The pattern propagates like a crystal growing from a seed, each copy seamlessly kissing its neighbor at the edges until the bolt is complete. Grid details: The seed swatch carries its own multiplication instructions in its shape. A wider swatch stretches further across; a taller swatch grows further down. The dye never bleeds or shifts — every copy is a perfect echo of the original."

Interpretation: Input grid is a "seed swatch". It has some pattern of colored cells (non-black) and black cells (.) representing background. The seed swatch knows its own dimensions: number of columns (width) and number of rows (height). It replicates itself across the warp (horizontal) by counting its own columns, and along the weft (vertical) by counting its own rows. So the output grid is formed by tiling the seed swatch in both dimensions, repeating it width times horizontally and height times vertically? Or maybe the number of repetitions equals the dimensions of the seed: repeat horizontally width times, vertically height times. Let's test with examples.

Example 1 input:
. . A
A A A
A . .
That's a 3x3 grid (3 rows, 3 columns). Output is a 9x9 grid? Let's count: Output lines: they show 9 rows? Let's count: They show:

Row1: . . A . . A . . A
Row2: A A A A A A A A A
Row3: A . . A . . A . .
Row4: . . A . . A . . A
Row5: A A A A A A A A A
Row6: A . . A . . A . .
Row7: . . A . . A . . A
Row8: A A A A A A A A A
Row9: A . . A . . A . .

Yes, 9 rows, each row has 9 characters (with spaces). So output is 9x9. Input is 3x3. So output dimensions = input dimensions * input dimensions? 3*3 =9. Indeed width=3, height=3, output width = width * width? Actually output width = input width * input width? Input width=3, output width=9 = 3*3. Output height = input height * input height? Input height=3, output height=9 = 3*3. So output = input tiled in a grid of size (input height) by (input width)? Wait: If we tile the input horizontally input width times, we get width * input width = 3*3=9. If we tile vertically input height times, we get height * input height = 3*3=9. So output = input repeated input width times horizontally and input height times vertically. That matches: number of repetitions horizontally = number of columns of input; number of repetitions vertically = number of rows of input.

Check example 2: Input:
B .
B B
That's 2 rows, 2 columns? Actually input:
Row1: B .
Row2: B B
So width=2, height=2. Output:
B . B .
B B B B
B . B .
B B B B
That's 4 rows, each row 4 characters? Let's see: Row1: B . B . (4 chars)
Row2: B B B B (4)
Row3: B . B . (4)
Row4: B B B B (4)
So output is 4x4. Input dimensions 2x2, output dimensions = 2*2 =4 each direction. So again output width = input width * input width? Actually input width=2, output width=4 = 2*2. Output height = input height * input height? input height=2, output height=4 =2*2. So matches same rule: repeat horizontally input width times, vertically input height times.

Thus transformation: Given input grid of size H x W (H rows, W columns). Output grid of size (H * H) x (W * W)? Wait: Actually output height = H * H? In example1 H=3 => output height=9 = 3*3 = H*H. Output width = W * W? W=3 => output width=9 = W*W. Example2: H=2 => output height=4 = 2*2 = H*H. W=2 => output width=4 = W*W. So yes output dimensions = H*H by W*W.

But is that always? Let's think: The description: "The swatch is placed on the loom and allowed to replicate itself — it knows how many times to repeat across the warp by counting its own columns, and how many times to stack along the weft by counting its own rows." So number of repetitions horizontally = number of columns (W). Number of repetitions vertically = number of rows (H). So if we tile the seed W times horizontally and H times vertically, the resulting dimensions: width = seed width * W = W * W = W^2. Height = seed height * H = H * H = H^2. So yes.

Thus transformation: Output grid = seed repeated in a matrix of H blocks vertically and W blocks horizontally. Each block is an exact copy of the seed.

Thus for test input, we need to compute H and W, then tile.

Test input:
. A A . .
A A A A A
. A A . .
A A A A A
. A A . A

Let's write rows:

Row0: . A A . .
Row1: A A A A A
Row2: . A A . .
Row3: A A A A A
Row4: . A A . A

Count rows: 5 rows. So H=5.

Count columns: each row has 5 characters? Let's verify: Row0: . A A . . => 5. Row1: A A A A A =>5. Row2: . A A . . =>5. Row3: A A A A A =>5. Row4: . A A . A =>5. So W=5.

Thus seed is 5x5. Output dimensions: height = H*H = 5*5 =2

... [7943 chars truncated]

We need to understand the transformation. The description: "What you see: A living cell — a small organism whose colored parts are organelles and whose dark parts are cytoplasm. What to do: The cell divides and multiplies, spreading copies of itself outward. Its height determines how many generations stack vertically; its width determines how many sisters line up side by side. Grid details: The output is a colony of identical cells arranged in a seamless rectangular sheet. Every organelle and every void is faithfully copied in each daughter cell."

Thus the input grid is a "cell" pattern. The output is a colony of identical cells arranged in a seamless rectangular sheet. The height of the input determines how many generations stack vertically; width determines how many sisters line up side by side. So output dimensions: number of rows = input height * something? Actually "height determines how many generations stack vertically" suggests that the number of rows in output = input height * something? Wait: The cell divides and multiplies, spreading copies of itself outward. Its height determines how many generations stack vertically; its width determines how many sisters line up side by side. So maybe the output is a tiling of the input cell pattern repeated both vertically and horizontally, where the number of repetitions vertically equals the input height, and horizontally equals the input width. Let's test with examples.

Example 1:
Input:
. . A
A A A
A . .
So input height = 3 rows, width = 3 columns.

Output given:
. . A . . A . . A
A A A A A A A A A
A . . A . . A . .
. . A . . A . . A
A A A A A A A A A
A . . A . . A . .
. . A . . A . . A
A A A A A A A A A
A . . A . . A . .

Let's count rows: output has 9 rows? Let's count: lines:
1: . . A . . A . . A
2: A A A A A A A A A
3: A . . A . . A . .
4: . . A . . A . . A
5: A A A A A A A A A
6: A . . A . . A . .
7: . . A . . A . . A
8: A A A A A A A A A
9: A . . A . . A . .
So 9 rows. Input height = 3. 9 = 3 * 3? Yes, 3*3 = 9. Output width? Let's count columns: each line seems to have 9 characters separated by spaces? Let's count: line1: . . A . . A . . A => that's 9 entries. So width = 9. Input width = 3. 9 = 3 * 3. So output dimensions = input height * input width? Actually both dimensions multiplied by input width? Wait: output rows = input height * input width? Input height=3, width=3 => product=9 rows. Output columns = input height * input width? Also 9 columns. So output is input size multiplied by itself? Actually it's input dimensions multiplied by input dimensions? That would be height*height? No, we got 9 = 3*3. So both dimensions multiplied by input width? Or height*width? Since height=width=3, product=9. So can't differentiate.

Example 2:
Input:
B .
B B
So input height = 2 rows, width = 2 columns.

Output:
B . B .
B B B B
B . B .
B B B B

Let's count rows: 4 rows. Input height=2, width=2 => product=4 rows. Columns: each line has 4 entries? Let's see line1: B . B . => 4 entries. So width=4 = product as well. So output dimensions = input height * input width? Actually both dimensions = input height * input width? Since both are 2*2=4. So output is a square of size (input height * input width) by (input height * input width). But example 1 gave 9x9 which is 3*3=9. So matches.

Example 3:
Input:
. G
G .
Height=2, width=2 => product=4. Output:
. G . G
G . G .
. G . G
G . G .
That's 4 rows, each row 4 entries? Let's check line1: . G . G => 4 entries. Yes. So output 4x4.

Example 4:
Input:
. . R .
R . R .
. R . R
R R R .
Let's compute dimensions: Count rows: 4 rows. Width: each line appears to have 4 entries? Let's check line1: . . R . => 4 entries. So input is 4x4. Product = 4*4 = 16. Output given appears to be many rows. Let's count: The output shown seems to have many lines. Let's count lines: I see:

. . R . . . R . . . R . . . R .
R . R . R . R . R . R . R . R .
. R . R . R . R . R . R . R . R
R R R . R R R . R R R . R R R .
. . R . . . R . . . R . . . R .
R . R . R . R . R . R . R . R .
. R . R . R . R . R . R . R . R
R R R . R R R . R R R . R R R .
. . R . . . R . . . R . . . R .
R . R . R . R . R . R . R . R .
. R . R . R . R . R . R . R . R
R R R . R R R . R R R . R R R .
. . R . . . R . . . R . . . R .
R . R . R . R . R . R . R . R .
. R . R . R . R . R . R . R . R
R R R . R R R . R R R . R R R .

Let's count rows: I see 15 lines? Actually let's number:

1: . . R . . . R . . . R . . . R .
2: R . R . R . R . R . R . R . R .
3: . R . R . R . R . R . R . R . R
4: R R R . R R R . R R R . R R R .
5: . . R . . . R . . . R . . . R .
6: R . R . R . R . R . R . R . R .
7: . R . R . R . R . R . R . R . R
8: R R R . R R R . R R R . R R R .
9: . . R . . . R . . . R . . . R .
10: R . R . R . R . R . R . R . R .
11: . R . R . R . R . R . R . R . R
12: R R R . R R R . R R R . R R R .
13: . . R . . . R . . . R . . . R .
14: R . R . R . R . R . R . R . R .
15: . R . R . R . R . R . R . R . R
16: R R R . R R R . R R R . R R R .

So 16 rows

... [7529 chars truncated]

We need to understand the transformation: "Zoom out to reveal the cosmic lattice: the patch repeats across the sky as many times as it is wide, and down as many times as it is tall." So the input is a small patch (grid). The output is a larger grid formed by tiling the input patch both horizontally and vertically, repeated as many times as the width and height of the input? Let's parse: "the patch repeats across the sky as many times as it is wide, and down as many times as it is tall." Means: If input width = w, height = h, then output width = w * w? Actually "as many times as it is wide" suggests repeat horizontally w times. Similarly vertically h times. So output dimensions: width = w * w? Wait: The patch repeats across the sky as many times as it is wide: So you take the patch and repeat it horizontally w times (i.e., number of repetitions equals width). Similarly vertically h times. So output width = w * w? Actually if you repeat the patch w times horizontally, each patch has width w, so total width = w * w. Similarly height = h * h. So output is w*w by h*h grid formed by tiling the input patch in a w by h block matrix (i.e., w copies horizontally, h copies vertically). Let's test with examples.

Example 1:
Input:
. . A
A A A
A . .
So input is 3x3 (w=3, h=3). According to rule, output should be 3*3=9 width, 3*3=9 height. The given output is 9 rows? Let's count: Provided output lines:

. . A . . A . . A
A A A A A A A A A
A . . A . . A . .
. . A . . A . . A
A A A A A A A A A
A . . A . . A . .
. . A . . A . . A
A A A A A A A A A
A . . A . . A . .

That's 9 rows, each row has 9 characters separated by spaces. Yes matches 9x9. And pattern: It appears the input tile repeated in a 3x3 block. Let's verify: Input tile:
Row0: . . A
Row1: A A A
Row2: A . .

Now output rows 0-2 correspond to first block row of tiles (vertical tile index 0). Within each block row, we have three tiles horizontally: tile0, tile1, tile2 each being the input. So row0: tile0 row0 (. . A) + tile1 row0 (. . A) + tile2 row0 (. . A) => . . A . . A . . A matches. Row1: tile0 row1 (A A A) + tile1 row1 (A A A) + tile2 row1 (A A A) => A A A A A A A A A matches. Row2: tile0 row2 (A . .) + tile1 row2 (A . .) + tile2 row2 (A . .) => A . . A . . A . . matches. Then rows 3-5 correspond to second vertical tile (tile index 1) same pattern, rows 6-8 third vertical tile. So indeed rule: output is input tiled w times horizontally and h times vertically.

Example 2:
Input:
B .
B B
That's 2x2 (w=2, h=2). Output should be 4x4. Provided output:
B . B .
B B B B
B . B .
B B B B
That's 4 rows, each 4 chars. Let's verify: Input tile:
Row0: B .
Row1: B B
Tiling 2 horizontally, 2 vertically yields:
Rows0-1: first vertical tile row0: tile0 row0 (B .) + tile1 row0 (B .) => B . B . matches row0.
Row1: tile0 row1 (B B) + tile1 row1 (B B) => B B B B matches row1.
Rows2-3: second vertical tile same as rows0-1. So matches.

Example 3:
Input:
. G
G .
That's 2x2. Output should be 4x4. Provided:
. G . G
G . G .
. G . G
G . G .
Matches same pattern.

Thus transformation is clear: output = input replicated to form a w-by-h block grid (i.e., each dimension multiplied by itself). Equivalent to Kronecker product of input with a matrix of ones of size w x h? Actually it's like taking input and repeating each cell? No, it's block tiling.

Thus for test input:
. A A . .
A A A A A
. A A . .
A A A A A
. A A . A

Let's compute dimensions: Count columns: first row ". A A . ." => 5 columns. Second row "A A A A A" => 5 columns. Third ". A A . ." => 5. Fourth "A A A A A" =>5. Fifth ". A A . A" =>5. So width w=5. Height h=5 (5 rows). So output should be 5*5 = 25 columns and 25 rows.

We need to tile the 5x5 input patch in a 5x5 block matrix: i.e., output[i][j] = input[i % h][j % w] (where i from 0 to 24, j from 0 to 24). Essentially repeat the input both horizontally and vertically 5 times each.

Thus we need to produce 25 rows each with 25 characters separated by spaces.

Let's construct input matrix:

Row0: [0] ., [1] A, [2] A, [3] ., [4] .
Row1: A A A A A
Row2: . A A . .
Row3: A A A A A
Row4: . A A . A

Let's label indices.

We'll generate output by repeating this pattern 5 times horizontally and vertically.

We can think of output as 5 blocks vertically each of height 5, each block being the input rows repeated horizontally 5 times.

Thus for each block row (0..4), we output 5 rows which are each the input row concatenated with itself 5 times (i.e., repeat the row 5 times). Actually each output row within a block row is the input row repeated horizontally 5 times (since we tile horizontally 5 copies). So each output row = input row repeated 5 times (i.e., the row's sequence of 5 chars, repeated 5 times => 25 chars). And we have 5 such rows per block row, and there are 5 block rows (vertical repeats). So overall output is just the input rows each repeated horizontally 5 times, and then the whole set of 5 rows repeated vertically 5 times.

Thus we can com

... [6186 chars truncated]

We need to understand the transformation. The description: "What you see: A single nautical chart tile: colored shallows and islands against deep dark water. What to do: Tile the chart across the navigation table: lay as many copies across as the chart has longitude lines, and stack as many down as it has latitude lines. Grid details: The navigation table displays a seamless patchwork of identical charts, every island and depth marking recurring in perfect periodicity."

Interpretation: Input grid is a "chart tile". The output is a tiled version: replicate the input tile across both dimensions, number of copies equal to number of longitude lines (i.e., number of columns?) and latitude lines (number of rows?). So output dimensions: input rows * input rows? Or input rows * number of latitude lines? Let's parse: "lay as many copies across as the chart has longitude lines, and stack as many down as it has latitude lines." So if the chart has L longitude lines (columns) and T latitude lines (rows), then we tile L copies horizontally and T copies vertically. So output width = input width * L, output height = input height * T. Since L = input width, T = input height. So output dimensions = (input width * input width) by (input height * input height). Essentially we replicate the input tile in a grid of size (input width) by (input height) tiles.

Check with examples.

Example 1 input:
. . A
A A A
A . .
That's 3 rows, 3 columns. So input width=3, height=3. Output should be 3*3=9 rows, 3*3=9 columns. Indeed output shown is 9 rows, each with 9 entries separated by spaces. Let's verify pattern: The output appears to be the input tiled in a 3x3 block of tiles. Let's reconstruct: Input tile:
Row0: . . A
Row1: A A A
Row2: A . .

Now tile 3 across, 3 down. So output rows 0-2 correspond to first tile row (tile row 0) repeated across columns of tiles. Row0 of output: tile row0 repeated 3 times: . . A . . A . . A. Yes matches first line: ". . A . . A . . A". Row1: tile row1 repeated 3 times: A A A A A A A A A. matches second line. Row2: tile row2 repeated 3 times: A . . A . . A . . matches third line: "A . . A . . A . .". Then rows 3-5 correspond to second tile row (tile row1) repeated across columns: row3: tile row0 again? Wait tile row index: we have 3 tile rows vertically. So rows 0-2: tile row0 of input repeated across columns. rows 3-5: tile row1 of input repeated across columns. rows 6-8: tile row2 of input repeated across columns. Let's check: rows 3-5 in output: lines 4-6? Actually output lines:
1: . . A . . A . . A
2: A A A A A A A A A
3: A . . A . . A . .
4: . . A . . A . . A
5: A A A A A A A A A
6: A . . A . . A . .
7: . . A . . A . . A
8: A A A A A A A A A
9: A . . A . . A . .
Yes pattern matches: rows 0,3,6 are tile row0; rows1,4,7 tile row1; rows2,5,8 tile row2. So indeed tiling.

Example 2 input:
B .
B B
That's 2 rows, 2 columns. Output should be 2*2=4 rows, 2*2=4 columns. Output shown:
B . B .
B B B B
B . B .
B B B B
Let's verify: Input tile:
Row0: B .
Row1: B B
Tile 2 across, 2 down.
Rows0-1: tile row0 repeated across columns: B . B . (yes line1)
Rows1-2? Actually rows0-1 correspond to tile row0? Wait we have 2 tile rows: tile row0 (input row0) and tile row1 (input row1). So output rows0-1: tile row0 repeated across columns? Actually we need 2 copies vertically: first copy of tile rows (tile row0 and tile row1) then second copy again. So rows0-1: first tile copy: tile row0 then tile row1. rows2-3: second tile copy: tile row0 then tile row1. Let's see output:
Line1: B . B . (tile row0 repeated across 2 columns)
Line2: B B B B (tile row1 repeated across 2 columns)
Line3: B . B . (again tile row0)
Line4: B B B B (again tile row1)
Matches.

Thus transformation: output = repeat input tile both horizontally and vertically, number of repetitions equal to its own width and height respectively.

Thus for test input:

Test Input:
. A A . .
A A A A A
. A A . .
A A A A A
. A A . A

Let's count rows and columns.

Rows: 5 lines.

Line1: ". A A . ." => 5 columns.
Line2: "A A A A A" => 5 columns.
Line3: ". A A . ." => 5 columns.
Line4: "A A A A A" => 5 columns.
Line5: ". A A . A" => 5 columns.

So it's a 5x5 grid.

Thus we need to tile it 5 times horizontally and 5 times vertically => output size 25 rows, 25 columns.

We need to produce the output grid with spaces between characters.

We can compute by constructing the tile and repeating.

Let's define tile matrix T[5][5] as given.

We'll index rows 0-4, cols 0-4.

T:
Row0: [., A, A, ., .]
Row1: [A, A, A, A, A]
Row2: [., A, A, ., .]
Row3: [A, A, A, A, A]
Row4: [., A, A, ., A]

Now output O of size (5*5)=25 rows and (5*5)=25 columns.

For each output row i (0..24), we determine which tile row: tile_row = i // 5 (integer division). Within tile, offset_row = i % 5.

Similarly for each output column j: tile_col = j // 5, offset_col = j % 5.

Then O[i][j] = T[offset_row][offset_col] (since each tile is identical). Actually yes, because each tile copy is same as

... [6457 chars truncated]

We need to understand the puzzle. The description: "What you see: A small circuit module: colored traces and components on a dark substrate. What to do: Stamp the module across the wafer: as many copies horizontally as the module has columns of pins, and as many vertically as it has rows of pins. Grid details: The finished wafer is a uniform array of identical modules, every trace and pad in perfect registration with its neighbors."

Thus the input grid is a small module. The output is a larger grid formed by tiling the module horizontally and vertically, where the number of copies horizontally equals the number of columns of pins in the module, and number of copies vertically equals the number of rows of pins in the module.

We need to interpret "pins" as something in the module: maybe certain colored cells represent pins? The module has pins (like connection points) that are perhaps a specific color? The description: "colored traces and components on a dark substrate." The dark substrate is black (.)? The module is on a dark substrate, so background is black (.)? The module includes colored traces and components (non-black). The pins might be something like a particular color (maybe azure A? Or maybe something else). The training example can help deduce.

Training Example 1:
Input:
. . A
A A A
A . .
So a 3x3 grid.

Output:
. . A . . A . . A
A A A A A A A A A
A . . A . . A . .
. . A . . A . . A
A A A A A A A A A
A . . A . . A . .
. . A . . A . . A
A A A A A A A A A
A . . A . . A . .

So output is 9x9? Let's count: Input is 3 rows, 3 columns. Output appears to be 9 rows, 9 columns? Let's see: The output lines: first line: ". . A . . A . . A" that's 9 entries. There are 9 lines? Let's count: Provided output lines: 
1: . . A . . A . . A
2: A A A A A A A A A
3: A . . A . . A . .
4: . . A . . A . . A
5: A A A A A A A A A
6: A . . A . . A . .
7: . . A . . A . . A
8: A A A A A A A A A
9: A . . A . . A . .
Yes 9 lines, each 9 entries. So output is 3x3 tiling? Actually 3 copies horizontally and vertically yields 3*3 = 9 each dimension. So they tiled the module 3 times horizontally and 3 times vertically. So number of copies horizontally = number of columns of pins in the module = 3? Number of copies vertically = number of rows of pins in the module = 3? So the module must have 3 columns of pins and 3 rows of pins. So the module's pins count equals its dimensions? That would be trivial: any module would have number of columns = width, number of rows = height. But maybe pins are only a subset of cells (like certain colored cells) and the count of pin columns/rows is less than total dimensions. However in this example, they got 3 copies each way, which equals the module's width and height (3). So maybe the pins are all cells? Or maybe the pins are defined as the non-black cells? Let's see: In input, non-black cells are A (azure) and maybe others? The input has . . A; A A A; A . . So there are A's at positions: (0,2), (1,0),(1,1),(1,2),(2,0). That's 5 A's. Not all cells. But the number of columns of pins maybe equals number of columns that contain at least one pin? Let's compute: columns: col0 has A at (1,0) and (2,0) -> yes; col1 has A at (1,1) only; col2 has A at (0,2),(1,2) -> yes. So all three columns have at least one pin => 3 columns of pins. Rows: row0 has A at (0,2) -> yes; row1 has A at (1,0),(1,1),(1,2) -> yes; row2 has A at (2,0) -> yes. So all three rows have at least one pin => 3 rows of pins. So number of copies horizontally = number of columns that contain at least one pin = width of module's bounding box of pins = 3. Similarly vertical copies = number of rows that contain at least one pin = 3. So they tiled the module to fill a grid where each tile corresponds to a pin location? Actually they tiled the module as many copies as there are pin columns horizontally and pin rows vertically. The resulting wafer is a uniform array of identical modules, each module placed at positions corresponding to pin coordinates? The description: "Stamp the module across the wafer: as many copies horizontally as the module has columns of pins, and as many vertically as it has rows of pins." So you take the module and stamp it repeatedly in a grid pattern where the number of stamps horizontally equals the number of pin columns, and vertically equals number of pin rows. The stamps are placed in a regular grid, presumably aligned such that each stamp's origin corresponds to a pin location? The finished wafer is a uniform array of identical modules, every trace and pad in perfect registration with its neighbors. So the output is just the module tiled in a regular grid with spacing equal to the module's dimensions? Actually if you stamp the module as many times as pin columns horizontally, you would place copies side by side, edge to edge? The module width is maybe the number of columns of the input grid. If you stamp it that many times horizontally, you would get width * (#copies) = width * width? Actually

... [8178 chars truncated]

We need to understand the transformation. The description: "A sacred rune: colored sigils inscribed against the dark primordial void. The rune obeys its nature and replicates: it manifests across the firmament as many times as it has columns, and descends as many times as it has rows. Grid details: The cosmos is inscribed with a perfect lattice of identical runes, each sigil touching its twin, the void between them shared and seamless."

Interpretation: Input grid is a "rune". The output is a tiled version: replicate the rune across columns as many times as it has columns, and replicate down rows as many times as it has rows. So output dimensions: input rows * input rows? Wait: "manifests across the firmament as many times as it has columns" means horizontally repeat the pattern number of times equal to number of columns of input. "descends as many times as it has rows" means vertically repeat number of times equal to number of rows of input. So output grid size: rows_out = rows_in * rows_in? Actually vertical repeats = rows_in (descend as many times as it has rows). So we repeat the input pattern vertically rows_in times. Similarly horizontally repeat columns_in times. So output dimensions: rows_out = rows_in * rows_in? Wait: If we repeat the input pattern vertically rows_in times, each repeat is the input pattern height (rows_in). So total height = rows_in * rows_in = rows_in^2. Similarly width = columns_in * columns_in = columns_in^2.

Thus output is input tiled in a square block of size (rows_in * rows_in) x (cols_in * cols_in). Equivalent to Kronecker product of input with a matrix of all ones? Actually it's like taking input and replicating each cell into a block? No, it's just tiling the whole input as a tile, repeated rows_in times vertically and cols_in times horizontally.

Check examples.

Example 1 input:
. . A
A A A
A . .
That's 3 rows, 3 columns.

Output given:
Let's see output lines:
Row1: . . A . . A . . A
Row2: A A A A A A A A A
Row3: A . . A . . A . .
Row4: . . A . . A . . A
Row5: A A A A A A A A A
Row6: A . . A . . A . .
Row7: . . A . . A . . A
Row8: A A A A A A A A A
Row9: A . . A . . A . .

So output is 9 rows, 9 columns. Indeed 3^2 =9 each dimension.

Now check pattern: It appears they tiled the 3x3 input as a block, repeating it 3 times horizontally and 3 times vertically. Let's verify: The input tile:
[ [., ., A],
  [A, A, A],
  [A, ., .] ]

If we repeat horizontally 3 times, we get each row concatenated three copies of the input row. For row1: . . A . . A . . A matches. Row2: A A A A A A A A A matches. Row3: A . . A . . A . . matches. Then repeat vertically: after three rows we repeat same three rows again (rows 4-6 same as 1-3), and again rows 7-9 same. Yes.

Thus transformation: output = input tiled to form a square of size (rows_in * rows_in) x (cols_in * cols_in) by repeating the input pattern both horizontally and vertically, each repeated number of times equal to the respective dimension of input.

Now test example 2:
Input:
B .
B B
Rows=2, cols=2.
Output:
B . B .
B B B B
B . B .
B B B B

Let's see: Input tile:
Row0: B .
Row1: B B

Horizontally repeat 2 times: each row becomes B . B . (row0) and B B B B (row1). Vertically repeat 2 times: we get rows: row0, row1, row0, row1. Indeed output matches.

Example 3:
Input:
. G
G .
Rows=2, cols=2.
Output:
. G . G
G . G .
. G . G
G . G .
Matches same pattern.

Thus transformation confirmed.

Now test input:
. A A . .
A A A A A
. A A . .
A A A A A
. A A . A

Let's compute dimensions: Count rows: there are 5 lines. Let's verify each line length.

Line1: ". A A . ." => 5 tokens.
Line2: "A A A A A" => 5.
Line3: ". A A . ." => 5.
Line4: "A A A A A" => 5.
Line5: ". A A . A" => 5.

Thus 5 rows, 5 columns.

Thus output dimensions: rows_out = 5*5 = 25 rows, cols_out = 5*5 = 25 columns.

We need to tile the 5x5 input pattern horizontally 5 times and vertically 5 times.

Thus output will consist of 5 blocks vertically each of height 5 rows (the input), and each block horizontally consists of 5 blocks each of width 5 columns (the input). So essentially we repeat the input pattern as a tile in a 5x5 grid of tiles.

Thus we can compute output by taking input and for each row of input, we produce a row that is the input row repeated 5 times (concatenated). Then we repeat that whole set of rows 5 times.

Thus algorithm: For i in 0..4 (repeat vertically):
   For each input row r in input rows:
        output row = concatenation of r with itself 5 times (i.e., r * 5 horizontally)
        add to output.

Thus we can produce.

Let's write input rows as strings without spaces for easier.

Input rows:
r0: . A A . .
r1: A A A A A
r2: . A A . .
r3: A A A A A
r4: . A A . A

Now horizontally repeat each 5 times: produce a string of length 25.

Compute r0 repeated 5 times: ". A A . ." repeated 5 => pattern: . A A . . . A A . . . A A . . . A A . . . A A . . (but need spaces between tokens). Actually we need to output with spaces between each chara

... [5068 chars truncated]

We need to understand the transformation. The description: "What you see: A small dance floor tile: colored footprints and positions against dark empty floor. What to do: Fill the ballroom: place as many dancers across as the tile has steps wide, and as many rows as the tile has steps deep. Each performer mirrors the original routine. Grid details: The ballroom is a grid of synchronized dancers, every footprint and pause replicated — the floor a seamless repetition of the original choreography."

Thus the transformation is: given an input grid (the tile), we need to output a larger grid that repeats the tile in both dimensions (horizontal and vertical) to fill a ballroom whose width is the tile's width times something? Actually "place as many dancers across as the tile has steps wide, and as many rows as the tile has steps deep." Means we replicate the tile to fill a ballroom that is as many copies across as the tile width? Wait: "as many dancers across as the tile has steps wide" suggests the number of copies horizontally equals the width of the tile (number of columns). Similarly, number of rows of copies equals the height of the tile (number of rows). So if tile is H rows, W columns, we output a grid of size (H * H) rows? Actually "as many rows as the tile has steps deep" means number of tile copies vertically = tile height (steps deep). So we repeat tile vertically H times, and horizontally W times. So output dimensions: height = H * H, width = W * W? Wait: If we repeat tile H times vertically, each copy has height H, so total height = H * H. Similarly width = W * W. So output is tile tiled in a grid of H by W copies? Actually we repeat tile in both dimensions: number of copies horizontally = tile width (W), number of copies vertically = tile height (H). So output size = (H * H) rows, (W * W) columns.

Check with examples.

Example 1 input:
. . A
A A A
A . .
That's 3 rows, 3 columns (H=3,W=3). Output given:
. . A . . A . . A
A A A A A A A A A
A . . A . . A . .
. . A . . A . . A
A A A A A A A A A
A . . A . . A . .
. . A . . A . . A
A A A A A A A A A
A . . A . . A . .

Let's count rows: 9 rows. Indeed H*H = 3*3 =9. Columns: each row has 9 entries? Let's count: first row: ". . A . . A . . A" that's 9 entries (3 groups of 3). Yes width = W*W = 3*3=9. So output is tile repeated in a 3x3 block of copies (3 copies horizontally, 3 vertically). Indeed pattern: The tile appears at positions (0,0) top-left, then (0,3) etc. Let's verify: The tile is:
Row0: . . A
Row1: A A A
Row2: A . .

Now output rows 0-2 correspond to first vertical copy (tile row0-2) repeated horizontally 3 times? Let's see rows 0-2:
Row0: . . A . . A . . A -> tile row0 repeated 3 times.
Row1: A A A A A A A A A -> tile row1 repeated 3 times.
Row2: A . . A . . A . . -> tile row2 repeated 3 times.
Rows 3-5: same as rows0-2 again (second vertical copy). Rows6-8: same as rows0-2 again (third vertical copy). So indeed output is tile tiled in a grid of H copies vertically and W copies horizontally.

Example 2 input:
B .
B B
That's 2 rows, 2 columns (H=2,W=2). Output:
B . B .
B B B B
B . B .
B B B B

Let's check: H*H = 4 rows, W*W = 4 columns. Output rows:
Row0: B . B . -> tile row0 repeated twice.
Row1: B B B B -> tile row1 repeated twice.
Row2: B . B . -> same as row0 (second vertical copy)
Row3: B B B B -> same as row1.

Thus transformation confirmed: output is tile repeated to form a grid of size (H*H) x (W*W) by tiling the tile in both dimensions, number of copies horizontally = tile width, vertically = tile height.

Thus for test input:
. A A . .
A A A A A
. A A . .
A A A A A
. A A . A

Let's compute dimensions: Count rows: 5 rows. Count columns: each row seems 5 entries? Let's verify: first row: ". A A . ." that's 5. Second: "A A A A A" 5. Third: ". A A . ." 5. Fourth: "A A A A A" 5. Fifth: ". A A . A" 5. So H=5, W=5.

Thus output dimensions: height = H*H = 25 rows, width = W*W = 25 columns.

We need to output the tile repeated 5 times horizontally and 5 times vertically.

Thus we need to construct output grid: For each block row i from 0 to H-1 (0..4), we repeat the tile's rows (the original tile rows) for each copy vertically? Actually we need to tile: For each vertical copy index v in 0..H-1, we output the tile's rows in order (0..H-1). So overall output rows: for v in 0..H-1:
   for r in 0..H-1:
        output row = tile row r repeated W times horizontally (i.e., concatenate tile row r W times).

Thus we can compute.

Let's write tile rows as strings (list of chars):

Row0: . A A . .
Row1: A A A A A
Row2: . A A . .
Row3: A A A A A
Row4: . A A . A

Now we need to repeat each row horizontally W=5 times: i.e., concatenate the row 5 times.

Thus for each tile row r, the expanded row (horizontal repeat) = tile row r concatenated with itself 5 times.

Let's compute each tile row repeated 5 times.

First, tile row0: ". A A . ." as list: ['.','A','A','.','.'].

Repeat 5 times: we get sequence of length 5*5=25: pattern . A 

... [6074 chars truncated]

We need to parse the problem: It's a grid transformation puzzle. The metaphorical description: "What you see: A mineral unit cell: colored atoms bonded against dark empty lattice points. What to do: Let the crystal grow: the unit cell replicates along its horizontal axis as many times as it has columns, and along its vertical axis as many times as it has rows. Grid details: The mature crystal is a perfect lattice of identical unit cells, every atom and void in periodic alignment — the macroscopic form a faithful multiplication of the microscopic seed."

Thus the transformation: Input grid is a unit cell. Output grid is the unit cell repeated horizontally as many times as the number of columns in the input, and vertically as many times as the number of rows in the input. So output dimensions: rows_out = rows_in * rows_in? Wait: "replicates along its horizontal axis as many times as it has columns" means horizontally repeat the unit cell number_of_columns times. Similarly vertical replication number_of_rows times. So output width = input width * input columns? Actually horizontal axis replicates as many times as it has columns: So if input has C columns, we repeat the unit cell C times horizontally. So output width = C * C? Wait: The unit cell width is C columns. Repeating it C times yields width = C * C. Similarly vertical: input has R rows, repeat unit cell R times vertically => height = R * R.

Thus output is a tiling of the input grid repeated R times vertically and C times horizontally, forming a larger square-ish grid (R*C? Actually it's R*C? Let's compute: height = R * R, width = C * C. Not necessarily square unless R=C.

Check example 1 to confirm.

Example 1 input:
. . A
A A A
A . .
That's 3 rows, 3 columns. So R=3, C=3. Output should be 3*3=9 rows, 3*3=9 columns. Indeed output shown is 9 rows, each with 9 entries separated by spaces. Let's verify pattern: Input unit cell:
Row0: . . A
Row1: A A A
Row2: A . .

Now output: They show 9 rows. Let's reconstruct: The pattern appears to be repeating the unit cell in a 3x3 block (since R=C=3). So each block of 3 rows and 3 columns is the input. Let's see first three rows of output:
Row0: . . A . . A . . A
Row1: A A A A A A A A A
Row2: A . . A . . A . .
That matches: first block (columns 0-2) = input row0: . . A; columns 3-5 = input row0 again: . . A; columns 6-8 = input row0 again: . . A. So horizontally repeated 3 times. Row1: input row1 repeated thrice: A A A A A A A A A. Row2: input row2 repeated thrice: A . . A . . A . . . Yes.

Then rows 3-5: same as rows0-2 again (vertical repeat). Indeed rows3-5 match rows0-2. Rows6-8 same again. So indeed output is input tiled R times vertically and C times horizontally.

Thus transformation rule: output[i][j] = input[i % R][j % C] where R = number of rows in input, C = number of columns in input.

Now test input:
. A A . .
A A A A A
. A A . .
A A A A A
. A A . A

Let's count rows and columns.

Rows: 5 lines. Let's parse each line:

Line1: ". A A . ." => 5 columns.
Line2: "A A A A A" => 5 columns.
Line3: ". A A . ." => 5 columns.
Line4: "A A A A A" => 5 columns.
Line5: ". A A . A" => 5 columns.

Thus R=5, C=5. So output dimensions: rows_out = R * R = 5*5 = 25 rows. columns_out = C * C = 5*5 = 25 columns.

Thus we need to tile the 5x5 input grid 5 times vertically and 5 times horizontally, resulting in a 25x25 grid.

We need to output the grid with characters separated by spaces, one row per line.

Thus we need to compute the output.

Let's write input as matrix:

Row0: [0] . A A . .
Row1: A A A A A
Row2: . A A . .
Row3: A A A A A
Row4: . A A . A

Let's index columns 0-4.

Now output cell at (i,j) = input[i % 5][j % 5].

Thus each 5x5 block is exactly the input.

Thus we can produce output by repeating input block 5 times in each direction.

Thus we can produce 25 rows: for block_row in 0..4 (each block corresponds to a copy of input rows), for each input row r in 0..4, output row = concatenation of input row r repeated 5 times horizontally (i.e., each input row's elements repeated 5 times as blocks). Actually horizontally: we repeat the input columns 5 times: so for each input row r, we produce a row that is input[r][0] input[r][1] ... input[r][4] input[r][0] ... etc 5 times.

Thus each output row will have length 25.

Thus we can generate.

Let's compute input rows as strings:

Row0: . A A . .
Row1: A A A A A
Row2: . A A . .
Row3: A A A A A
Row4: . A A . A

Now repeat each row 5 times horizontally: i.e., concatenate the row string with itself 5 times.

Thus for each row r, the output row pattern = (row r) * 5 (string concatenation). Then we repeat that pattern for each block_row 5 times (vertical). So essentially we will have 5 copies of each of these 5 rows, in order: rows0-4 repeated 5 times each block.

Thus final output rows sequence: (for block in 0..4):
  output row0 = row0 repeated 5 times
  output row1 = row1 repeated 5 times
  output row2 = row2 repeated 5 times
  output row3 = row3 

... [6874 chars truncated]

We need to understand the transformation. The description: "What you see: A brief motif: colored notes against silent rests on a small staff. What to do: Let the motif repeat itself: it sounds across as many voices as it has beats, and cascades down through as many measures as it has lines. Grid details: The full score is a seamless wall of sound — the motif tiled perfectly so that every entrance and rest aligns with its neighbors."

Interpretation: Input grid is a motif (small pattern). The output is a larger grid formed by tiling the motif in some way: repeated across as many voices as it has beats, and cascades down through as many measures as it has lines.

We need to parse "beats" and "lines" and "voices" and "measures". Possibly the motif is a small matrix; the number of beats = number of columns? The number of lines = number of rows? Then the motif repeats across as many voices as it has beats (i.e., horizontally repeated number of times equal to number of columns?), and cascades down through as many measures as it has lines (i.e., vertically repeated number of times equal to number of rows?). So output dimensions: width = input width * input width? Actually "as many voices as it has beats" suggests number of repetitions horizontally = number of beats (columns). "as many measures as it has lines" suggests number of repetitions vertically = number of lines (rows). So output size = (input rows * input rows) by (input columns * input columns). Let's test with examples.

Example 1:
Input:
. . A
A A A
A . .
That's 3 rows, 3 columns.

If we repeat motif 3 times horizontally and 3 times vertically, output would be 9 rows x 9 columns. Indeed output shown is 9 rows (looks like 9 lines) and each line seems length 9? Let's count: first line: ". . A . . A . . A" that's 9 entries (with spaces). Yes. So matches.

Now check pattern: The output appears to be the motif tiled in a regular grid: each 3x3 block is the input motif. Let's verify: The output lines:

Row1: . . A . . A . . A
Row2: A A A A A A A A A
Row3: A . . A . . A . .
Row4: . . A . . A . . A
Row5: A A A A A A A A A
Row6: A . . A . . A . .
Row7: . . A . . A . . A
Row8: A A A A A A A A A
Row9: A . . A . . A . .

Yes, it's just repeating the 3x3 motif in both directions.

Example 2:
Input:
B .
B B
That's 2 rows, 2 columns.

Output:
B . B .
B B B B
B . B .
B B B B

That's 4 rows, 4 columns. Indeed 2*2 =4 each dimension. Tiling the 2x2 motif.

Check: The motif:
[ [B, .],
  [B, B] ]

Tiling yields:
Row1: B . B .
Row2: B B B B
Row3: B . B .
Row4: B B B B

Matches.

Example 3:
Input:
. G
G .
2x2 motif.

Output:
. G . G
G . G .
. G . G
G . G .

Again 4x4 tiling.

Example 4:
Input:
. . R .
R . R .
. R . R
R R R .
That's 4 rows, 4 columns.

Output shown is many rows: let's count. It appears to be 12 rows? Actually they show many lines: after first 4 rows, then repeat? Let's count: They show:

Row1: . . R . . . R . . . R . . . R .
Row2: R . R . R . R . R . R . R . R .
Row3: . R . R . R . R . R . R . R . R
Row4: R R R . R R R . R R R . R R R .
Row5: . . R . . . R . . . R . . . R .
Row6: R . R . R . R . R . R . R . R .
Row7: . R . R . R . R . R . R . R . R
Row8: R R R . R R R . R R R . R R R .
Row9: . . R . . . R . . . R . . . R .
Row10: R . R . R . R . R . R . R . R .
Row11: . R . R . R . R . R . R . R . R
Row12: R R R . R R R . R R R . R R R .
Row13: . . R . . . R . . . R . . . R .
Row14: R . R . R . R . R . R . R . R .
Row15: . R . R . R . R . R . R . R . R
Row16: R R R . R R R . R R R . R R R .

So 16 rows, each row length? Let's count first row: ". . R . . . R . . . R . . . R ." That seems 16 entries? Let's count: positions: 1 .,2 .,3 R,4 .,5 .,6 .,7 R,8 .,9 .,10 .,11 R,12 .,13 .,14 .,15 R,16 . => 16. Yes.

Thus output is 4*4 =16 rows and columns. So indeed tiling.

Thus transformation: output grid is input motif tiled to fill a square of size (n_rows * n_rows) by (n_cols * n_cols). Actually it's n_rows repetitions vertically, n_cols repetitions horizontally. So output dimensions: rows_out = n_rows * n_rows, cols_out = n_cols * n_cols.

Now test input:
. A A . .
A A A A A
. A A . .
A A A A A
. A A . A

Let's compute dimensions: Count rows: 5 rows. Count columns: each row seems 5 entries? Let's verify:

Row1: . A A . . => 5
Row2: A A A A A =>5
Row3: . A A . . =>5
Row4: A A A A A =>5
Row5: . A A . A =>5

So 5x5 motif.

Thus output should be 5*5 =25 rows and 25 columns, tiling the 5x5 motif.

Thus we need to produce a 25x5? Wait: rows_out = n_rows * n_rows = 5*5 =25 rows. columns_out = n_cols * n_cols =5*5=25 columns.

Thus we repeat the motif both horizontally and vertically 5 times each.

Thus output grid is just the motif tiled in a 5x5 block of motifs.

Thus we can compute by taking the input matrix and repeating each row 5 times horizontally (i.e., each row becomes the row concatenated with itself 5 times) and then repeat the whole set of rows 5 times vertically (i.e., the list of rows repeated 5 times).

Let's produce.

First, define in

... [5387 chars truncated]

We need to understand the transformation: "Project the diagram onto the full field: repeat it horizontally as many times as the diagram is wide, and vertically as many times as the diagram is tall."

Interpretation: Input is a small diagram (grid). Output is a larger grid formed by tiling the input diagram both horizontally and vertically, where the number of repetitions horizontally equals the width of the input diagram, and the number of repetitions vertically equals the height of the input diagram.

Thus output dimensions: width_out = width_in * width_in = (width_in)^2. Height_out = height_in * height_in = (height_in)^2.

Check with examples.

Example 1 input:
. . A
A A A
A . .
That's 3 rows, 3 columns (3x3). Output shown: 9 rows, 9 columns (9x9). Indeed width_in=3, height_in=3 => output 9x9. The pattern: they tiled the 3x3 input 3 times horizontally and 3 times vertically. Let's verify: The output appears to be a 3x3 block of the input repeated. Let's see first three rows: 
Row1: . . A . . A . . A (that's input row1 repeated three times)
Row2: A A A A A A A A A (input row2 repeated three times)
Row3: A . . A . . A . . (input row3 repeated three times)
Then rows 4-6 repeat same pattern: rows 4-6 are same as rows1-3 again? Actually rows4-6: 
Row4: . . A . . A . . A (same as row1)
Row5: A A A A A A A A A (same as row2)
Row6: A . . A . . A . . (same as row3)
Rows7-9 same again. So indeed they repeated the input block 3 times vertically (so 3 blocks stacked) and each block horizontally repeated 3 times.

Thus transformation: output[i][j] = input[i % h][j % w] where h = height_in, w = width_in.

Example 2: Input:
B .
B B
That's 2 rows, 2 columns (2x2). Output: 4 rows, 4 columns (4x4). Output shown:
B . B .
B B B B
B . B .
B B B B
Let's verify: Input rows:
Row0: B .
Row1: B B
Tiling horizontally 2 times: each row becomes B . B . (row0 repeated twice) and B B B B (row1 repeated twice). Then vertically 2 times: repeat those two rows twice. So output rows 0-1 as above, rows2-3 same as rows0-1. Yes matches.

Thus transformation rule is clear.

Now test input:
. A A . .
A A A A A
. A A . .
A A A A A
. A A . A

Let's compute dimensions: Count rows: 5 rows. Count columns: each row seems length 5? Let's verify each line:

Row0: ". A A . ." => 5
Row1: "A A A A A" => 5
Row2: ". A A . ." => 5
Row3: "A A A A A" => 5
Row4: ". A A . A" => 5

Thus input is 5x5.

Thus output dimensions: width_out = 5*5 = 25 columns, height_out = 5*5 = 25 rows.

We need to produce output grid: tile the 5x5 input 5 times horizontally and 5 times vertically.

Thus output is a 5x5 block of the input repeated.

We can compute by repeating each input row 5 times horizontally (i.e., concatenate the row with itself 5 times) to get a row of length 25, then repeat that set of 5 rows vertically 5 times (i.e., the resulting 5 rows repeated 5 times to get 25 rows).

Thus we need to produce 25 rows, each of length 25.

Let's write input rows as strings:

Row0: . A A . .
Row1: A A A A A
Row2: . A A . .
Row3: A A A A A
Row4: . A A . A

Now repeat each row horizontally 5 times: i.e., row0 repeated 5 times yields: ". A A . . . A A . . . A A . . . A A . . . A A . ." (but need spaces between characters). Actually we need to output with spaces between characters. So we will produce each character separated by a space.

Thus for each input row, we create a list of characters repeated 5 times.

Let's compute row0 pattern: [., A, A, ., .] repeated 5 times => sequence length 25: . A A . . . A A . . . A A . . . A A . . . A A . .

We can write as groups of 5: ". A A . ." repeated.

Similarly row1: [A, A, A, A, A] repeated 5 times => all A's: 25 A's.

Row2 same as row0.

Row3 same as row1.

Row4: [., A, A, ., A] repeated 5 times => pattern: . A A . A . A A . A . A A . A . A A . A . A A . A

Now we need to repeat these 5 rows vertically 5 times: i.e., output rows 0-4 = row0_h5, row1_h5, row2_h5, row3_h5, row4_h5; rows5-9 = same again; rows10-14 same; rows15-19 same; rows20-24 same.

Thus final output is 25 rows, each row being one of the 5 patterns repeated.

Thus we can produce output.

Let's construct patterns:

Pattern0 (row0_h5): ". A A . . . A A . . . A A . . . A A . . . A A . ."
Pattern1 (row1_h5): "A A A A A A A A A A A A A A A A A A A A A A A A A"
Pattern2 (row2_h5): same as pattern0.
Pattern3 (row3_h5): same as pattern1.
Pattern4 (row4_h5): ". A A . A . A A . A . A A . A . A A . A . A A . A"

Let's double-check pattern4: Input row4: . A A . A. Repeat 5 times:

First copy: . A A . A
Second: . A A . A
Third: . A A . A
Fourth: . A A . A
Fifth: . A A . A

Concatenate: . A A . A . A A . A . A A . A . A A . A . A A . A

Yes.

Now we need to output 25 rows: pattern0, pattern1, pattern2, pattern3, pattern4, then repeat.

Thus rows 0-4: p0,p1,p2,p3,p4
Rows5-9: p0,p1,p2,p3,p4
Rows10-14: p0,p1,p2,p3,p4
Rows15-19: p0,p1,p2,p3,p4
Rows20-24: p0,p1,p2,p3,p4

Thus we can write them.

Let's produce each line with spaces between characters.

We'll need 

... [3570 chars truncated]

We need to understand the transformation. The description: "What you see: A small weather cell: colored bands of precipitation against calm dark skies. What to do: The cell replicates across the radar screen, spawning copies equal to its own width across the horizon and copies equal to its own height from pole to equator. Grid details: The radar map becomes a repeating mosaic of identical storm cells, seamless and perfectly periodic."

Interpretation: Input grid is a "small weather cell". The output is a larger grid formed by tiling the input cell across both dimensions: number of copies horizontally equals the width of the cell, and number of copies vertically equals the height of the cell. So output dimensions: width_out = width_in * width_in? Wait: "spawning copies equal to its own width across the horizon" means you replicate the cell horizontally width_in times? Or you create copies such that total width = width_in * (width_in+1?) Let's parse: "spawning copies equal to its own width across the horizon" could mean you create as many copies as the width of the cell, placed side by side, resulting in total width = original width * (width_in + 1?) Actually if you have a cell and you spawn copies equal to its own width, you might produce width_in copies in addition to the original? Or the cell itself is considered one copy? Usually "spawning copies equal to its own width" suggests you create that many copies, maybe including the original? Let's test with examples.

Example 1:
Input:
. . A
A A A
A . .
That's 3 rows, 3 columns (width=3, height=3). Output:
They show a 9x9? Let's count: Output lines:
Line1: . . A . . A . . A (9 entries)
Line2: A A A A A A A A A (9)
Line3: A . . A . . A . . (9)
Line4: . . A . . A . . A (9)
Line5: A A A A A A A A A (9)
Line6: A . . A . . A . . (9)
Line7: . . A . . A . . A (9)
Line8: A A A A A A A A A (9)
Line9: A . . A . . A . . (9)

So output is 9 rows, 9 columns. That's 3*3 =9 in each dimension. So output dimensions = input width * input height? Actually width=3, height=3, product =9. So output is width*height by width*height? Wait both dimensions are 9, which is 3*3. So output size = (width * height) x (width * height). That seems like they tile the cell in a grid where number of tiles horizontally = height? Or number of tiles vertically = width? Let's see: If you tile the input cell in a matrix of size (height) by (width) maybe? Actually if you repeat the cell width times horizontally and height times vertically, you'd get dimensions: width_out = width_in * width_in? No, if you repeat the cell width_in times horizontally, you get width_out = width_in * width_in (since each copy has width width_in). Similarly vertically: height_out = height_in * height_in. That would give 3*3=9 width and 3*3=9 height, matches example. So they repeated the cell width_in times horizontally and height_in times vertically. That yields output dimensions = (width_in * width_in) x (height_in * height_in). Since width_in = height_in =3, we get 9x9.

Check example 2:
Input:
B .
B B
That's 2 rows, 2 columns? Let's see: rows:
Row0: B .
Row1: B B
So width=2 (max columns?), height=2. Output:
B . B .
B B B B
B . B .
B B B B
That's 4 rows, 4 columns? Let's count: each line has 4 characters. Yes output is 4x4. width_in=2, height_in=2 => width_out = width_in * width_in = 4, height_out = height_in * height_in = 4. Matches.

Example 3:
Input:
. G
G .
That's 2x2. Output:
. G . G
G . G .
. G . G
G . G .
That's 4x4 again. So pattern holds.

Thus transformation: output grid is formed by tiling the input cell in a grid of size (width_in) by (height_in) copies? Actually we need width_in copies horizontally and height_in copies vertically. So we create a matrix of tiles: for i in 0..height_in-1 (vertical tile index), for j in 0..width_in-1 (horizontal tile index), place the input cell at position (i*height_in, j*width_in)? Wait tile size is input dimensions. So each tile occupies a block of size height_in rows and width_in columns. If we place height_in tiles vertically, total height = height_in * height_in. Similarly width = width_in * width_in.

Thus algorithm: For each output cell at (r_out, c_out), we compute which tile: tile_row = r_out // height_in, tile_col = c_out // width_in. Then within tile, offset: r_in = r_out % height_in, c_in = c_out % width_in. Then output[r_out][c_out] = input[r_in][c_in].

Thus it's just repeating the input block both horizontally and vertically, number of repetitions equal to its own dimensions.

Now test input:
. A A . .
A A A A A
. A A . .
A A A A A
. A A . A

Let's compute dimensions. Count rows: 5 rows. Count columns: each row length? Row0: . A A . . => 5 columns. Row1: A A A A A =>5. Row2: . A A . . =>5. Row3: A A A A A =>5. Row4: . A A . A =>5. So width_in=5, height_in=5.

Thus output dimensions: width_out = width_in * width_in = 5*5 =25 columns. height_out = height_in * height_in =5*5=25 rows.

Thus we need to produce a 25x5? Wait 25 rows and 25 columns.

We 

... [6903 chars truncated]

We need to understand the transformation. The description: "What you see: A small stained-glass panel with colored glass pieces set in a dark leading frame. What to do: Install copies of the panel into the cathedral window: as many columns as the panel has panes across, and as many rows as the panel has panes down. Grid details: The finished rose window is a grid of identical panels fitted edge to edge, each piece of colored glass in its proper place."

Interpretation: Input is a small panel (grid). Output is a larger grid formed by tiling the panel in a pattern: number of columns equals number of panes across (i.e., width of panel), number of rows equals number of panes down (height of panel). So we tile the panel in a grid of size (height * height?) Wait: "as many columns as the panel has panes across, and as many rows as the panel has panes down." Means: If panel width = w, height = h, then output grid dimensions: columns = w * w? Actually "as many columns as the panel has panes across" means number of columns in output = number of panes across in panel = w. And "as many rows as the panel has panes down" means number of rows in output = number of panes down in panel = h. But that would just be same size as input? That seems trivial. However examples show output larger than input.

Let's examine Example 1:

Input:
. . A
A A A
A . .

That's 3x3 panel.

Output:
. . A . . A . . A
A A A A A A A A A
A . . A . . A . .
. . A . . A . . A
A A A A A A A A A
A . . A . . A . .
. . A . . A . . A
A A A A A A A A A
A . . A . . A . .

Let's count rows: 9 rows. Columns: each row has 9 entries? Let's count: first row: ". . A . . A . . A" that's 9 entries (3 groups of 3). So output is 9x9. Input is 3x3. So output dimensions = input dimensions * input dimensions? 3*3 =9. Indeed output is 9x9. So they tiled the panel in a grid of size (height * height?) Actually they repeated the panel both horizontally and vertically: number of repetitions horizontally = width of panel? Let's see: width=3, they repeated panel 3 times horizontally to get width 9. Number of repetitions vertically = height of panel? height=3, repeated 3 times vertically to get height 9. So output = panel tiled in a grid of repetitions equal to its own dimensions: replicate panel w times horizontally and h times vertically. That matches: output width = w * w, output height = h * h.

Check Example 2:

Input:
B .
B B

That's 2x2? Actually rows: first row "B ." (2), second row "B B" (2). So 2x2 panel.

Output:
B . B .
B B B B
B . B .
B B B B

Let's count rows: 4 rows. Columns: each row has 4 entries? First row: "B . B ." that's 4. So output 4x4. Input 2x2 => output dimensions = 2*2 =4. Yes.

Example 3:

Input:
. G
G .

That's 2x2.

Output:
. G . G
G . G .
. G . G
G . G .

That's 4x4. Yes.

Example 4:

Input:
. . R .
R . R .
. R . R
R R R .

Let's count: rows: 4 rows. Columns: each row has 4 entries? Let's see first row: ". . R ." that's 4. So 4x4 panel.

Output: many rows. Let's count: they show many lines. It appears output is 16x16? Since 4*4=16. Let's verify: they show first line: ". . R . . . R . . . R . . . R ." Count entries: pattern ". . R ." repeated 4 times? ". . R ." is 4 entries, repeated 4 times gives 16 entries. Yes. So output is 16x16.

Thus transformation: replicate the input panel to form a larger grid where the panel is tiled in a matrix of size (input height) by (input width). Actually number of repetitions horizontally = input width, vertically = input height. So output dimensions = (input height * input height?) Wait: If we repeat horizontally w times, width becomes w * w. If we repeat vertically h times, height becomes h * h. So output is w*w by h*h. Since input is h x w. So output dimensions = (h * h) x (w * w). But examples show square output when input is square (h=w). For non-square input, we would get rectangular output where height = h*h, width = w*w. Let's test if any examples have non-square input? All examples are square: 3x3, 2x2, 2x2, 4x4. So we can't confirm but likely that's the rule.

Thus for test input:

. A A . .
A A A A A
. A A . .
A A A A A
. A A . A

Let's determine dimensions. Count rows: 5 rows. Count columns per row: first row: ". A A . ." that's 5 entries. So it's 5x5 panel (square). So output should be 5*5 = 25 rows and 25 columns (since square). So output is 25x5? Wait if input is 5x5, then output dimensions = 5*5 = 25 rows and 25 columns (square). So we need to tile the 5x5 panel 5 times horizontally and 5 times vertically, resulting in a 25x25 grid.

Thus we need to produce output: repeat the input block in a 5x5 tiling.

Let's write input matrix with indices:

Row0: . A A . .
Row1: A A A A A
Row2: . A A . .
Row3: A A A A A
Row4: . A A . A

Let's verify each row length 5.

Now we need to create output where we repeat this block both horizontally and vertically 5 times.

Thus output will consist of 5 blocks vertically (each block is the 5 rows of input) and each block horizontally consists of 5 copies of the input 

... [6170 chars truncated]

We need to parse the problem: It's a grid transformation puzzle. The description: "What you see: A miniature stage with colored actors positioned against a dark backdrop. What to do: Replicate the scene: build as many stages side by side as the model is wide, and stack as many tiers as the model is tall. Each stage performs the identical blocking. Grid details: The audience sees a vast wall of identical stages, every actor in the same position, every dark space preserved — a seamless theatrical mosaic."

Interpretation: Input grid is a "model" (the stage). We need to replicate it to fill an output grid that is tiled both horizontally and vertically, with number of repetitions equal to the width and height of the input? Actually: "build as many stages side by side as the model is wide, and stack as many tiers as the model is tall." Means: If the model (input) has width W (number of columns) and height H (number of rows), then we create an output grid that is W copies side by side (so total width = W * W? Actually "as many stages side by side as the model is wide" means number of stages horizontally = width of model. Similarly number of stages vertically = height of model. So output dimensions: width = W * W? Wait each stage is the model itself (size W x H). If we place W stages side by side, total width = W * W (since each stage width = W). Similarly stacking H stages vertically gives total height = H * H. So output is a W*H? Actually output width = W * W, height = H * H. So we tile the input grid in both dimensions, repeating it W times horizontally and H times vertically.

Check with Example 1 to verify.

Example 1 Input:
. . A
A A A
A . .

Let's write as 3x3 grid:
Row0: . . A
Row1: A A A
Row2: A . .

So width W=3, height H=3.

According to rule, output should be 3*3 =9 columns wide and 3*3 =9 rows tall. The given output is:

. . A . . A . . A
A A A A A A A A A
A . . A . . A . .
. . A . . A . . A
A A A A A A A A A
A . . A . . A . .
. . A . . A . . A
A A A A A A A A A
A . . A . . A . .

Let's count: It appears 9 rows, each row has 9 entries separated by spaces. Let's verify first row: ". . A . . A . . A" => 9 entries: pattern . . A repeated three times. Indeed input row0 is ". . A". Tiling horizontally 3 times yields ". . A . . A . . A". Good.

Second row: "A A A A A A A A A" => input row1 "A A A" repeated 3 times yields "A A A A A A A A A". Good.

Third row: "A . . A . . A . ." => input row2 "A . ." repeated 3 times yields "A . . A . . A . .". Good.

Then fourth row repeats first row again? Actually row3 is ". . A . . A . . A" same as row0. So vertical tiling: after 3 rows (input height), we repeat the whole block of 3 rows again for second tier, and again for third tier. Indeed output rows 0-2 are first tile vertically; rows 3-5 are second tile; rows 6-8 are third tile. So rule matches: tile input both horizontally and vertically, number of repetitions equals width and height respectively.

Thus transformation: output grid dimensions: output_width = input_width * input_width, output_height = input_height * input_height. For each cell (i,j) in output, where i in [0, output_height-1], j in [0, output_width-1], we map to input cell (i % input_height, j % input_width). So it's just repeating the input as a tile.

Now test input:

. A A . .
A A A A A
. A A . .
A A A A A
. A A . A

Let's write as rows:

Row0: . A A . .
Row1: A A A A A
Row2: . A A . .
Row3: A A A A A
Row4: . A A . A

So dimensions: width W = number of columns = 5 (each row has 5 entries). Height H = number of rows = 5.

Thus output dimensions: width = W * W = 5 * 5 = 25 columns. Height = H * H = 5 * 5 = 25 rows.

We need to produce output grid of 25x25, where each 5x5 block is the input pattern repeated.

Thus we can construct by taking input and tiling horizontally 5 times, then vertically 5 times.

Let's compute.

First, produce the horizontally tiled row for each input row: repeat the row 5 times.

Input row0: . A A . .
Repeat 5 times: . A A . . . A A . . . A A . . . A A . . . A A . .
That's 5*5 = 25 entries.

Similarly row1: A A A A A repeated 5 times: A A A A A A A A A A A A A A A A A A A A A A A A A (that's 25 A's).

Row2: same as row0: . A A . . repeated 5 times.

Row3: same as row1: all A's.

Row4: . A A . A repeated 5 times.

Then we need to vertically tile these rows 5 times: i.e., repeat the block of 5 rows (the horizontally tiled rows) 5 times vertically.

Thus final output will consist of 5 blocks vertically, each block identical to the horizontally tiled 5 rows.

Thus we can produce output by printing the horizontally tiled rows, then repeat that sequence 4 more times.

Let's compute the horizontally tiled rows explicitly.

Define input rows as strings (with spaces between characters for readability but we will output with spaces). Actually we need to output with spaces between characters. We'll produce each row as characters separated by a space.

Let's compute row0 horizontally tiled:

Input row0: [., A, A, ., .]
R

... [5650 chars truncated]

We need to understand the transformation. The description: "What you see: A small planting template: colored flowers set among dark soil patches. What to do: Lay the template down repeatedly: as many times across as there are columns in the template, and as many times down as there are rows. Grid details: The garden bed is a uniform quilt of identical planting squares, every flower and every bare patch in perfect periodic alignment."

Thus the transformation: Input is a template (small grid). Output is a larger grid formed by tiling the template in a pattern: repeat the template across columns as many times as there are columns in the template, and repeat down as many times as there are rows in the template. So if input is of size R x C (rows x columns), output is of size (R * R) x (C * C)? Wait: "as many times across as there are columns in the template, and as many times down as there are rows." Means we tile the template horizontally C times (i.e., repeat the template C times side by side) and vertically R times (repeat the template R times stacked). So output dimensions: rows = input rows * input rows? Actually vertical repeats: number of times down = rows in template. So we repeat the template vertically R times. So total rows = R * R? Wait: If we have template of height R rows, and we repeat it down R times, we get R * R rows. Similarly, width = C * C columns. So output is a square of size (R*R) x (C*C). But note that if R != C, output is not necessarily square; it's R*R rows by C*C columns.

Check examples.

Example 1:
Input:
. . A
A A A
A . .
That's 3 rows, 3 columns (R=3, C=3). Output shown is 9 rows, 9 columns? Let's count: They show 9 lines each with 9 entries separated by spaces? Let's count: first line: ". . A . . A . . A" that's 9 entries. Yes. So output is 9x9, which matches R*R = 3*3 =9 rows, C*C =3*3=9 columns. Good.

Example 2:
Input:
B .
B B
That's 2 rows, 2 columns? Actually input lines: "B ." (2), "B B" (2). So R=2, C=2. Output shown:
B . B .
B B B B
B . B .
B B B B
That's 4 rows, each with 4 entries? Let's count: first line "B . B ." => 4 entries. So output is 4x4 = R*R=2*2=4 rows, C*C=2*2=4 columns. Good.

Example 3:
Input:
. G
G .
That's 2x2. Output:
. G . G
G . G .
. G . G
G . G .
That's 4x4. Yes.

Thus transformation: tile the input template in a grid where the number of tiles horizontally equals number of columns of template, and vertically equals number of rows of template. Essentially output[i][j] = input[i % R][j % C] where i from 0 to R*R-1, j from 0 to C*C-1.

Thus for test input, we need to compute R and C.

Test Input:
. A A . .
A A A A A
. A A . .
A A A A A
. A A . A

Let's count rows: there are 5 lines. So R=5.

Count columns: each line seems to have 5 entries? Let's verify: line1: ". A A . ." => 5. line2: "A A A A A" =>5. line3: ". A A . ." =>5. line4: "A A A A A" =>5. line5: ". A A . A" =>5. So C=5.

Thus template is 5x5. Output will be R*R = 5*5 =25 rows, and C*C =5*5 =25 columns. So a 25x25 grid.

We need to compute output by tiling the 5x5 template 5 times horizontally and 5 times vertically.

Thus output[i][j] = input[i % 5][j % 5].

We need to produce the grid with spaces between characters.

Let's write the input template as a 5x5 matrix with indices (row, col) 0-4.

Row0: . A A . .
Row1: A A A A A
Row2: . A A . .
Row3: A A A A A
Row4: . A A . A

Let's label:

r0: [0,0]='.', [0,1]='A', [0,2]='A', [0,3]='.', [0,4]='.'
r1: [1,0]='A', [1,1]='A', [1,2]='A', [1,3]='A', [1,4]='A'
r2: [2,0]='.', [2,1]='A', [2,2]='A', [2,3]='.', [2,4]='.'
r3: [3,0]='A', [3,1]='A', [3,2]='A', [3,3]='A', [3,4]='A'
r4: [4,0]='.', [4,1]='A', [4,2]='A', [4,3]='.', [4,4]='A'

Now output size 25x25. We can think of it as 5 blocks of rows each of height 5 (since we repeat template vertically 5 times). Similarly each block of columns width 5.

Thus output can be seen as a 5x5 array of tiles, each tile being the input template.

Thus we can produce output by repeating the template both dimensions.

Let's construct output rows 0-24.

We can compute for each output row i, the source row = i % 5. For each output column j, source col = j % 5.

Thus we can generate.

Better to produce as 5 blocks of 5 rows each.

Block row index br = 0..4 (which tile row). Within block, the source row = br (since i%5 = br). Actually if we consider i = br*5 + offset_r where offset_r in 0..4, then i%5 = offset_r? Wait: i%5 = (br*5 + offset_r) %5 = offset_r. So the source row depends on offset within block, not block index. That seems off: Actually tiling the template vertically means we stack copies of the template one after another. So first copy of template (rows 0-4) corresponds to output rows 0-4. Second copy (rows 5-9) corresponds to output rows 5-9, which again are the template rows 0-4. So indeed output row i maps to template row i % 5. So within each block of 5 rows, the pattern repeats exactly the template rows. So block index does not affect which row of template we use; it's always the same pa

... [6460 chars truncated]

We need to parse the problem: It's a grid transformation puzzle. The description: "Two kingdoms sit on opposite sides of a golden bridge. The northern kingdom flies blue banners from some of its towers, while the southern kingdom lights red fires atop others. The darkness between them marks empty watchtowers where no one stands guard. The golden bridge opens and the two kingdoms merge their defenses. Any watchtower that had a blue banner flying OR a red fire burning sends a soldier to the unified army, marked in green. Only towers that were dark and abandoned in BOTH kingdoms remain empty in the new realm."

Thus we have an input grid that includes both kingdoms (north and south) separated by a golden bridge (the Y row? Actually Y is yellow, representing the golden bridge). The input includes rows above the bridge (north kingdom) and rows below the bridge (south kingdom). The bridge is a row of Y's (yellow) spanning full width. The output is the merged kingdom: only the region above the bridge? Actually output in example 1: Input had 9 rows? Let's examine.

Example 1 Input:
B B . .
B . B .
B B . B
B B B B
Y Y Y Y
R R . R
. . R .
. R . .
R . R .

So there are 9 rows. Row 0-3: north kingdom (blue banners B and .). Row 4: Y Y Y Y (bridge). Row 5-8: south kingdom (red fires R and .). Output:
G G . G
G . G .
G G . G
G G G G

That's 4 rows, same as north kingdom rows? Actually output has 4 rows, matching the number of rows in north kingdom (or south kingdom) i.e., the height of each side (assuming symmetric). The bridge row is omitted in output. The output seems to be the merged result: For each column position, if either north side had B (blue) OR south side had R (red) at that position (same column and same row offset from bridge?), then output G (green). Else output . (black). Let's test.

Take column 0: north rows: row0 B, row1 B, row2 B, row3 B => all B. South rows: row5 R, row6 ., row7 ., row8 R => at each row offset? Let's align: north row0 corresponds to south row5? Probably yes: row0 (north top) aligns with row5 (south top just below bridge). So for each offset i from 0 to (north height-1), we check north[i][j] == B or south[i][j] == R? Actually south rows are reversed? Let's test: north row0 B, south row5 R => at least one => output G. Indeed output row0 col0 G. north row1 B, south row6 . => B => G. output row1 col0 G. north row2 B, south row7 . => G. output row2 col0 G. north row3 B, south row8 R => G. output row3 col0 G. So matches.

Column1: north rows: row0 B, row1 ., row2 B, row3 B. south rows: row5 R, row6 ., row7 R, row8 .? Wait row8 col1 is .? Actually row8: "R . R ." => col1 . So south rows: [R, ., R, .]. Now compute OR: row0: B or R => G. output row0 col1 G. row1: . or . => . => output row1 col1 . (matches). row2: B or R => G => output row2 col1 G (matches). row3: B or . => G => output row3 col1 G (matches). Good.

Column2: north rows: row0 ., row1 B, row2 ., row3 B. south rows: row0? Actually row5 ., row6 R, row7 ., row8 R? Let's list: row5: "R R . R" => col2 .; row6: ". . R ." => col2 R; row7: ". R . ." => col2 .; row8: "R . R ." => col2 R. So south: [., R, ., R]. OR: row0: . or . => . => output row0 col2 . (matches). row1: B or R => G => output row1 col2 G (matches). row2: . or . => . => output row2 col2 . (matches). row3: B or R => G => output row3 col2 G (matches). Good.

Column3: north rows: row0 B, row1 ., row2 B, row3 B. south rows: row5 R, row6 ., row7 ., row8 .? Actually row5 col3 R, row6 col3 ., row7 col3 ., row8 col3 . => [R, ., ., .]. OR: row0: B or R => G => output row0 col3 G (matches). row1: . or . => . => output row1 col3 . (matches). row2: B or . => G => output row2 col3 G (matches). row3: B or . => G => output row3 col3 G (matches). Good.

Thus rule: Input grid has a separator row of Y's (yellow) representing the golden bridge, spanning full width. Above that is north kingdom (blue banners B and empty .). Below that is south kingdom (red fires R and empty .). The output grid is the same height as each kingdom (i.e., number of rows above bridge, which equals number below bridge assuming symmetric). For each cell (i,j) in the output (i from 0 to H-1 where H = number of rows above bridge), output G if either north[i][j] == 'B' or south[i][j] == 'R'; else output '.'.

Thus we need to apply to test input.

Test Input:
B . B .
B . B .
. B . .
B . B .
Y Y Y Y
R R . .
. . R .
. R . R
R R R .

Let's parse.

First, count rows. Let's list them with indices:

0: B . B .
1: B . B .
2: . B . .
3: B . B .
4: Y Y Y Y
5: R R . .
6: . . R .
7: . R . R
8: R R R .

Thus north kingdom rows 0-3 (4 rows). Bridge row 4 (Y's). South kingdom rows 5-8 (4 rows). Good.

Now compute output for each column j=0..3 and each i=0..3 (north row i, south row i+5? Actually south row index = i+5). Let's extract north and south arrays.

North:
row0: [B, ., B, .]
row1: [B, ., B, .]
row2: [., B, ., .]
row3:

... [1898 chars truncated]

We need to parse the problem: It's a grid transformation puzzle. Input appears to be two chambers: upper culture (cold-pigment colonies) and lower culture (warm-pigment colonies). The input grid is given as lines: first part is upper culture? Let's examine training examples.

Example 1 Input:
B B . .
B . B .
B B . B
B B B B
Y Y Y Y
R R . R
. . R .
. R . .
R . R .
Output:
G G . G
G . G .
G G . G
G G G G

We need to see mapping: The input appears to have 9 lines? Let's count: The input lines: first 4 lines maybe upper culture (4x4), then next 5 lines? Actually output is 4x4. So input likely consists of two 4x4 grids stacked: upper (cold) and lower (warm). The first 4 lines are upper culture, next 4 lines are lower culture. But example 1 input has 9 lines, not 8. Let's count: Provided input lines:

B B . .
B . B .
B B . B
B B B B
Y Y Y Y
R R . R
. . R .
. R . .
R . R .

That's 9 lines. Wait maybe they include a blank line? Or maybe they include a separator line? Let's examine: The first 4 lines: B B . ; B . B . ; B B . B ; B B B B. That's 4 rows of 4 columns each. Then next line: Y Y Y Y (that's 4 Y's). Then next 4 lines: R R . R ; . . R . ; . R . . ; R . R . . That's 4 lines. So total lines: 4 + 1 + 4 = 9. The extra line Y Y Y Y maybe is a separator? Actually maybe the input format is: first 4 lines upper culture, then a line of "Y Y Y Y" representing something like a membrane? But the description mentions a golden membrane dividing the two chambers. The membrane might be represented by a row of Y's? Y is yellow per key. But they said golden membrane; Y is yellow, maybe representing golden? So the membrane row is a line of Y's separating upper and lower cultures. Then after that, the lower culture is given.

Thus input format: 4 lines upper culture, then a line of 4 Y's (the membrane), then 4 lines lower culture. Output is a single 4x4 grid representing fused culture: hybrid colonies appear where either parent culture was active (i.e., where either upper or lower had a non-sterile site). Sterile sites remain barren (i.e., where both were '.' (black? Actually '.' is black per key, representing empty/sterile). So output cell is '.' if both input cells (upper and lower) are '.'; otherwise output is some hybrid color (maybe G for green?). In examples, output uses G (green) for hybrid colonies. Let's verify: Example 1 output: all non-'.' cells are G. Indeed output only contains G and '.'.

Check example 1: Upper culture (first 4 lines):
Row0: B B . .
Row1: B . B .
Row2: B B . B
Row3: B B B B

Lower culture (after Y line):
Row0: Y Y Y Y? Wait that's membrane. Actually lower culture lines:
Row0: R R . R
Row1: . . R .
Row2: . R . .
Row3: R . R .

Now compute hybrid: For each cell, if either upper or lower is not '.' (i.e., has B or R or maybe other colors?), output G. Let's test.

Upper grid:
(0,0): B -> not '.' => output G
(0,1): B -> G
(0,2): . -> check lower (0,2): '.'? lower row0 col2 is '.' (since R R . R) => both '.' => output '.' (matches output row0 col2 '.'? Output row0: G G . G => yes '.' at col2)
(0,3): upper '.'? Actually upper row0 col3 is '.'; lower row0 col3 is R => not '.' => output G. Output row0 col3 G matches.

Row1:
Upper row1: B . B .
Lower row1: . . R .
Compute:
col0: upper B => G (output G)
col1: upper '.' lower '.' => '.' (output .)
col2: upper B => G (output G)
col3: upper '.' lower R => G (output G). Output row1: G . G . matches.

Row2:
Upper row2: B B . B
Lower row2: . R . .
col0: B => G
col1: B => G
col2: '.' lower '.' => '.'? Wait lower row2 col2 is '.' (since . R . .) Actually lower row2: . R . . => col0 '.', col1 'R', col2 '.', col3 '.'. So col2 both '.' => '.' output. Output row2 col2 '.' matches (output: G G . G)
col3: upper B => G (output G). matches.

Row3:
Upper row3: B B B B
Lower row3: R . R .
col0: B or R => G
col1: B or '.' => G
col2: B or R => G
col3: B or '.' => G. Output row3: G G G G matches.

Thus rule: Output cell is G if either upper or lower cell is not '.' (i.e., has any color other than '.'), else '.'.

Now check example 2 to confirm.

Example 2 Input:
B B B .
. B . B
. . B B
B B . B
Y Y Y Y
. . . R
. . . R
R R R R
R R . R

Upper (first 4 lines):
Row0: B B B .
Row1: . B . B
Row2: . . B B
Row3: B B . B

Lower (after Y line):
Row0: . . . R
Row1: . . . R
Row2: R R R R
Row3: R R . R

Now compute hybrid: output should be G where either upper or lower non '.'.

Let's compute manually to see if matches given output:

Output:
G G G G
. G . G
G G G G
G G . G

Compute each cell:

Row0:
col0: upper B => G
col1: upper B => G
col2: upper B => G
col3: upper '.' lower R => G => output G G G G matches.

Row1:
col0: upper '.' lower '.' => '.' (output .) matches.
col1: upper B => G (output G) matches.
col2: upper '.' lower '.' => '.' (output .) matches.
col3: upper B lower R => G (output G) matches.

Row2:
col0: upper '.' lower R => G
col1: upper '.' lower R => G
col2: upper B lower R 

... [3788 chars truncated]

We need to parse the problem. It's a grid transformation puzzle. The description: "What you see: A starfield bisected by a golden band. Above: patches of cool nebular gas. Below: patches of hot nebular gas. The rest is void. What to do: Collapse both nebulae through the band. Any region containing gas from either nebula ignites into a newborn star. Void remains only where both sides were empty. Grid details: A 4×4 star chart with newborn stars where either nebula contributed matter."

Thus we have a grid that is presumably 4x4? The training examples show input with multiple rows? Let's examine Example 1:

Input:
B B . .
B . B .
B B . B
B B B B
Y Y Y Y
R R . R
. . R .
. R . .
R . R .

That's 9 rows? Actually let's count: lines:
1: B B . .
2: B . B .
3: B B . B
4: B B B B
5: Y Y Y Y
6: R R . R
7: . . R .
8: . R . .
9: R . R .

That's 9 rows, each with 4 columns? Let's see each line: they have 4 characters separated by spaces. So it's a 9x4 grid? But description says 4x4 star chart. Something off.

Maybe the input is actually two halves: top half (cool nebula) and bottom half (hot nebula) separated by a golden band (maybe a row of something?). The golden band could be a row of Y? Actually Y is yellow, maybe golden band is Y? The description: "A starfield bisected by a golden band. Above: patches of cool nebular gas. Below: patches of hot nebular gas. The rest is void." So the grid likely has a golden band (maybe a row of Y) separating top and bottom halves. The top half contains B (blue) representing cool nebular gas? The bottom half contains R (red) representing hot nebular gas? The void is . (black). The golden band maybe Y (yellow). The output is newborn stars where either nebula contributed matter: G (green) maybe? In output example 1, we see G's and .'s. Output:
G G . G
G . G .
G G . G
G G G G

That's 4 rows, 4 columns. So output is 4x4. So the input must be something like: top part (cool nebula) maybe some rows above the golden band, bottom part (hot nebula) some rows below, and the golden band maybe a row of Y that is ignored? Let's see the given input for example 1: It has 9 rows. Perhaps the first 4 rows are top half (cool nebula), then a row of Y Y Y Y (the golden band), then the next 4 rows are bottom half (hot nebula). Indeed: rows 1-4: B B . .; B . B .; B B . B; B B B B. That's 4 rows. Row 5: Y Y Y Y (all Y). Then rows 6-9: R R . R; . . R .; . R . .; R . R . That's 4 rows. So indeed input format: top 4 rows = cool nebula (B), middle row = golden band (Y), bottom 4 rows = hot nebula (R). Output is 4x4 grid where each cell is G if either top or bottom cell at that position (same column) had any nebula gas (i.e., B or R) in either top or bottom half? Actually "Any region containing gas from either nebula ignites into a newborn star." Means if either top (cool) or bottom (hot) had gas at that position (i.e., not void), then output cell becomes G (newborn star). Void remains only where both sides were empty. So output cell is G if top cell != '.' OR bottom cell != '.'; else output cell is '.' if both top and bottom are '.'.

Thus we ignore the middle golden band row (Y). So transformation: given a 9-row input (4 top, 1 middle Y row, 4 bottom), produce a 4x4 output where each cell is G if either top or bottom cell at same row/col is not '.' (i.e., is B or R), else '.'.

Check example 1: Let's compute to verify.

Top rows (rows1-4):
Row1: B B . .
Row2: B . B .
Row3: B B . B
Row4: B B B B

Bottom rows (rows6-9):
Row6: R R . R
Row7: . . R .
Row8: . R . .
Row9: R . R .

Now compute output per cell (row i, col j) where i from 1..4 (top row index) and j from 1..4.

We'll compute:

Cell (1,1): top B (not .), bottom R (not .) => G.
(1,2): top B, bottom R => G.
(1,3): top ., bottom . => both . => .
(1,4): top ., bottom R => G.

Thus row1: G G . G matches output row1: G G . G.

Cell (2,1): top B, bottom . => G.
(2,2): top ., bottom . => .
(2,3): top B, bottom R => G.
(2,4): top ., bottom . => .
Thus row2: G . G . matches output row2: G . G .

Cell (3,1): top B, bottom . => G.
(3,2): top B, bottom R => G.
(3,3): top ., bottom . => .
(3,4): top B, bottom . => G.
Thus row3: G G . G matches output row3: G G . G.

Cell (4,1): top B, bottom R => G.
(4,2): top B, bottom R => G.
(4,3): top B, bottom R => G.
(4,4): top B, bottom . => G? Wait bottom row9 col4 is .? Actually row9: R . R . => col4 is . So top B, bottom . => G. So row4: G G G G matches output row4: G G G G.

Thus rule confirmed.

Now test input:

Test Input:
B . B .
B . B .
. B . .
B . B .
Y Y Y Y
R R . .
. . R .
. R . R
R R R .

We need to parse similarly: first 4 rows = top (cool nebula), then row of Y Y Y Y (golden band), then next 4 rows = bottom (hot nebula). Let's extract.

Top rows (rows1-4):
1: B . B .
2: B . B .
3: . B . .
4: B . B .

Bottom rows (rows6-9):
6: R R . .
7: . . R .
8: . R . R
9: R R R .

Now compute output 4x4: For each cell (i,j) i=1..4, j=1..4, output G if top[i][j] != '.' 

... [1577 chars truncated]

We need to parse the problem. It's a grid transformation puzzle. There's a metaphorical description: "A harbor chart split by a golden sandbar. Above: one fleet's anchorages marked in cool tones. Below: another fleet's in warm tones. Empty berths are open water. What to do: Erase the sandbar and merge the harbor. Any berth occupied by either fleet now flies a combined ensign. Only berths empty in both charts remain open water. Grid details: A 4×4 harbor with unified ensigns where either fleet had a ship."

Thus we have two fleets: one above the sandbar (cool tones) and one below (warm tones). The input grid likely contains both fleets' data: top half (maybe first 2 rows?) is one fleet, bottom half (last 2 rows?) is the other fleet. The sandbar is a row (or column?) that separates them. The description says "split by a golden sandbar". The sandbar is likely a row of some special color (maybe Y? Actually golden sandbar could be Y (yellow) or O (orange) or something). The input includes both fleets and the sandbar? Let's examine training example.

Example 1 Input:
B B . .
B . B .
B B . B
B B B B
Y Y Y Y
R R . R
. . R .
. R . .
R . R .

We need to parse: The input appears to have 9 rows? Let's count: lines:
1: B B . .
2: B . B .
3: B B . B
4: B B B B
5: Y Y Y Y
6: R R . R
7: . . R .
8: . R . .
9: R . R .

That's 9 rows. Output:
G G . G
G . G .
G G . G
G G G G

That's 4 rows. So the transformation reduces from 9 rows to 4 rows. The description says "A 4×4 harbor with unified ensigns where either fleet had a ship." So output is 4x4. Input likely includes two 4x4 fleets (above and below sandbar) plus a sandbar row (maybe row of Y's?). Indeed we see rows 1-4: B's and .'s (cool tones? B is blue, which is cool). Rows 6-9: R's and .'s (R is red, warm). Row 5: Y Y Y Y (yellow) which could be the golden sandbar. So the input format: top 4 rows = fleet A (cool tones), then a row of sandbar (maybe always Y?), then bottom 4 rows = fleet B (warm tones). Output is 4x4 where each cell is combined ensign if either fleet had a ship (i.e., if either top or bottom fleet had a non-black (non '.'?) cell at that position). The combined ensign color is presumably some mixture: maybe blue + red = green? Indeed in output we see G (green) where there were B or R in input. Let's verify.

Take top fleet (rows 1-4):
Row1: B B . .
Row2: B . B .
Row3: B B . B
Row4: B B B B

Bottom fleet (rows 6-9):
Row6: R R . R
Row7: . . R .
Row8: . R . .
Row9: R . R .

Now combine: For each cell (i,j) where i from 0-3 (top fleet row index) and j from 0-3, output cell = if top[i][j] != '.' OR bottom[i][j] != '.' then some combined color else '.'.

Now what is the combined color? In example, where either top or bottom had a ship (non '.'), output is G (green). Let's test a few positions:

Top[0][0] = B, bottom[0][0] = R => output[0][0] = G (matches).
Top[0][1] = B, bottom[0][1] = R => output[0][1] = G.
Top[0][2] = ., bottom[0][2] = . => output[0][2] = . (matches).
Top[0][3] = ., bottom[0][3] = R => output[0][3] = G (since bottom has ship). Indeed output[0][3] = G.

Top[1][0] = B, bottom[1][0] = . => output[1][0] = G (top has ship). Output[1][0] = G (yes).
Top[1][1] = ., bottom[1][1] = . => output[1][1] = . (yes).
Top[1][2] = B, bottom[1][2] = R => output[1][2] = G (yes).
Top[1][3] = ., bottom[1][3] = . => output[1][3] = . (yes).

Top[2][0] = B, bottom[2][0] = . => output[2][0] = G (yes).
Top[2][1] = B, bottom[2][1] = . => output[2][1] = G (yes).
Top[2][2] = ., bottom[2][2] = R => output[2][2] = G (yes).
Top[2][3] = B, bottom[2][3] = . => output[2][3] = G (yes).

Top[3][0] = B, bottom[3][0] = R => output[3][0] = G (yes).
Top[3][1] = B, bottom[3][1] = . => output[3][1] = G (yes).
Top[3][2] = B, bottom[3][2] = R => output[3][2] = G (yes).
Top[3][3] = B, bottom[3][3] = . => output[3][3] = G (yes).

Thus output is all G except where both top and bottom are '.' (i.e., both empty). Those positions: (0,2), (1,1), (1,3) are '.' in output. Indeed matches.

Thus transformation rule: Given input grid of size (2*N + 1) rows? Actually here N=4, plus sandbar row makes 9 rows. The sandbar row is all Y (maybe always Y). The top N rows = fleet A (cool tones), bottom N rows = fleet B (warm tones). Output is N x N grid where each cell is:
- If either top fleet cell or bottom fleet cell is not '.' (i.e., occupied by a ship), then output cell = G (green) (the combined ensign).
- Else output cell = '.' (open water).

Thus the combined ensign color is always G (green) regardless of which colors? The metaphor says "Any berth occupied by either fleet now flies a combined ensign." The combined ensign might be a mixture of cool and warm tones: blue (cool) + red (warm) = green. If only one fleet present, maybe still green? The example shows that even if only one fleet present (e.g., top B only, bottom .) output is G. So combined ensign is green irrespective of which fleet(s) present, as long as at least one present.

Thus for tes

... [2023 chars truncated]

We need to parse the problem. It's a grid transformation puzzle. The description: "What you see: A two-layer PCB split by a golden bus trace. The upper layer has active traces in some cells; the lower layer has active traces in others. Dead cells carry no current. What to do: Laminate the layers through the bus. Any cell where either layer carried current lights a unified indicator. Cells dead on both layers stay off. Grid details: A 4×4 output board with lit indicators wherever either layer was active."

Thus we have two layers: upper layer and lower layer. The input grid presumably contains both layers? The training examples show input with multiple rows: first some rows (maybe upper layer) then some rows (maybe lower layer). The output is a 4x4 grid (since they say 4x4 output board). The input appears to have 8 rows? Let's examine Example 1:

Input:
B B . .
B . B .
B B . B
B B B B
Y Y Y Y
R R . R
. . R .
. R . .
R . R .

That's 9 rows? Let's count: lines:
1: B B . .
2: B . B .
3: B B . B
4: B B B B
5: Y Y Y Y
6: R R . R
7: . . R .
8: . R . .
9: R . R .

That's 9 rows. Output:
G G . G
G . G .
G G . G
G G G G

That's 4 rows.

Thus input likely consists of two 4x4 layers stacked: first 4 rows = upper layer, next 4 rows = lower layer? But we have 9 rows, not 8. Wait maybe there is a golden bus trace row separating them? The description mentions "split by a golden bus trace". Golden bus trace maybe corresponds to Y (yellow) because golden is yellow? In the key, Y=yellow. So the golden bus trace could be a row of Y's separating the two layers. Indeed in Example 1, after 4 rows of B and ., we see a row "Y Y Y Y". That's likely the golden bus trace. Then after that, we have the lower layer rows: next 4 rows? Let's see: after the Y row, we have rows:
6: R R . R
7: . . R .
8: . R . .
9: R . R .

That's only 3 rows? Actually we have rows 6-9 inclusive = 4 rows: row6, row7, row8, row9. Yes that's 4 rows. So total rows: 4 (upper) + 1 (bus) + 4 (lower) = 9 rows. Good.

Thus input format: first 4 rows = upper layer (active traces in some cells), then a row of Y's (golden bus trace), then next 4 rows = lower layer.

Output: 4x4 grid where each cell is lit (some color) if either layer had activity in that cell. The lit indicator color? In examples, output uses G (green) for lit cells. So the unified indicator color is green (G). So any cell where either upper or lower layer had a non-black (i.e., not '.'?) Actually active traces are represented by some color (B, R, etc). Dead cells are '.' (black). So we need to OR the layers: if either layer has a non '.' (i.e., any color) at that position, output G; else output '.'.

Check Example 1: Upper layer (first 4 rows):
Row1: B B . .
Row2: B . B .
Row3: B B . B
Row4: B B B B

Lower layer (rows 6-9):
Row6: R R . R
Row7: . . R .
Row8: . R . .
Row9: R . R .

Now compute OR: For each cell, if either upper or lower has non '.' then G else '.'.

Let's compute manually to see if matches output.

We'll create a 4x4 grid.

Upper:
(0,0) B; (0,1) B; (0,2) .; (0,3) .
(1,0) B; (1,1) .; (1,2) B; (1,3) .
(2,0) B; (2,1) B; (2,2) .; (2,3) B
(3,0) B; (3,1) B; (3,2) B; (3,3) B

Lower:
(0,0) R; (0,1) R; (0,2) .; (0,3) R
(1,0) .; (1,1) .; (1,2) R; (1,3) .
(2,0) .; (2,1) R; (2,2) .; (2,3) .
(3,0) R; (3,1) .; (3,2) R; (3,3) .

Now OR:
Cell (0,0): upper B or lower R => non '.' => G
(0,1): B or R => G
(0,2): . or . => .
(0,3): . or R => G
Thus row0: G G . G matches output row0: G G . G.

Row1:
(1,0): B or . => G
(1,1): . or . => .
(1,2): B or R => G
(1,3): . or . => .
Thus row1: G . G . matches output row1: G . G .

Row2:
(2,0): B or . => G
(2,1): B or R => G
(2,2): . or . => .
(2,3): B or . => G
Thus row2: G G . G matches output row2: G G . G.

Row3:
(3,0): B or R => G
(3,1): B or . => G
(3,2): B or R => G
(3,3): B or . => G
Thus row3: G G G G matches output row3: G G G G.

Great.

Thus rule confirmed.

Now Example 2: Let's test quickly to ensure.

Example 2 Input:
B B B .
. B . B
. . B B
B B . B
Y Y Y Y
. . . R
. . . R
R R R R
R R . R

Upper layer rows 1-4:
Row0: B B B .
Row1: . B . B
Row2: . . B B
Row3: B B . B

Lower layer rows 6-9 (after Y row):
Row0: . . . R
Row1: . . . R
Row2: R R R R
Row3: R R . R

Now compute OR:

Cell (0,0): B or . => G
(0,1): B or . => G
(0,2): B or . => G
(0,3): . or R => G
=> G G G G matches output row0: G G G G.

Row1:
(1,0): . or . => .
(1,1): B or . => G
(1,2): . or . => .
(1,3): B or R => G
=> . G . G matches output row1: . G . G.

Row2:
(2,0): . or R => G
(2,1): . or R => G
(2,2): B or R => G
(2,3): B or R => G
=> G G G G matches output row2: G G G G.

Row3:
(3,0): B or R => G
(3,1): B or R => G
(3,2): . or . => .
(3,3): B or R => G
=> G G . G matches output row3: G G . G.

Matches.

Thus transformation rule: For each cell in the 4x4 grid, if either the upper layer cell (first 4 rows) or lower layer cell (rows 6-9, after the Y row) is not '.' (i.e., any 

... [1976 chars truncated]

We need to parse the problem. It's a grid transformation puzzle. Input: we have a description: "A mythic river of gold divides two banks. One bank bears scattered frost runes; the other bears scattered fire runes. Some ground on each bank is bare. What to do: Ford the river and unite the banks. Every position bearing a rune from either god transforms into a life rune. Only ground bare on both banks remains empty. Grid details: A 4×4 map of life runes where either bank had a rune placed."

Thus we have two banks: left bank (frost runes) and right bank (fire runes). The river of gold is maybe a row of Y (yellow) representing gold? The description: "A mythic river of gold divides two banks." So the input grid likely consists of two parts: top part is left bank? Actually the training examples show input with multiple rows: they have 8 rows? Let's examine Example 1 input:

Input:
B B . .
B . B .
B B . B
B B B B
Y Y Y Y
R R . R
. . R .
. R . .
R . R .

That's 9 rows? Let's count: lines:
1: B B . .
2: B . B .
3: B B . B
4: B B B B
5: Y Y Y Y
6: R R . R
7: . . R .
8: . R . .
9: R . R .

So 9 rows. Output:
G G . G
G . G .
G G . G
G G G G

That's 4 rows. So the output is 4x4.

Thus the input seems to consist of two 4x4 banks stacked vertically with a river row in between? Let's see: The first 4 rows (rows 1-4) maybe left bank (frost runes) using B (blue) as frost? The next row (row5) is Y Y Y Y (river of gold). Then rows 6-9 maybe right bank (fire runes) using R (red) as fire? Indeed, the left bank uses B (blue) for frost runes, and the right bank uses R (red) for fire runes. The '.' indicates bare ground.

Thus the transformation: For each position (i,j) in the 4x4 grid, we look at the left bank cell (top part) and the right bank cell (bottom part). If either has a rune (i.e., not '.'), then output cell becomes G (green) representing life rune. If both are '.' (bare ground on both banks), output remains '.' (black). The river row (Y's) is just a separator and ignored.

Thus the rule: output[i][j] = G if (top[i][j] != '.' or bottom[i][j] != '.'), else '.'.

Check with examples.

Example1: top 4x4:
Row1: B B . .
Row2: B . B .
Row3: B B . B
Row4: B B B B

bottom 4x4 (rows6-9):
Row6: R R . R
Row7: . . R .
Row8: . R . .
Row9: R . R .

Now compute output: For each cell, if top or bottom not '.' => G else '.'.

Let's compute manually to see if matches given output.

We'll index rows 0-3.

Top:
r0: B B . .
r1: B . B .
r2: B B . B
r3: B B B B

Bottom:
r0: R R . R
r1: . . R .
r2: . R . .
r3: R . R .

Now compute:

Cell (0,0): top B (not .) => G. Output row0 col0 G matches given: G G . G? Actually output row0: G G . G. Yes first is G.

(0,1): top B => G. Output second G matches.

(0,2): top ., bottom . => both . => output . matches third char '.'.

(0,3): top ., bottom R => not . => G. Output fourth G matches.

Row1:
(1,0): top B => G. Output row1: G . G . first G matches.
(1,1): top ., bottom . => . => output . matches second '.'.
(1,2): top B => G => output third G matches.
(1,3): top ., bottom . => . => output fourth '.' matches.

Row2:
(2,0): top B => G => output row2 first G matches.
(2,1): top B => G => second G matches.
(2,2): top ., bottom . => . => third '.' matches.
(2,3): top B => G => fourth G matches.

Row3:
(3,0): top B => G => output row3 first G matches.
(3,1): top B => G => second G matches.
(3,2): top B => G => third G matches.
(3,3): top B => bottom R => G => fourth G matches.

Matches.

Example2: Let's test quickly.

Input:
B B B .
. B . B
. . B B
B B . B
Y Y Y Y
. . . R
. . . R
R R R R
R R . R

Top 4 rows:
r0: B B B .
r1: . B . B
r2: . . B B
r3: B B . B

Bottom rows (5-8):
r0: . . . R
r1: . . . R
r2: R R R R
r3: R R . R

Compute output:

Cell (0,0): top B => G => output row0 first G matches "G G G G"? Actually output row0: G G G G. Yes.

(0,1): top B => G => second G matches.

(0,2): top B => G => third G matches.

(0,3): top ., bottom R => G => fourth G matches.

Row1:
(1,0): top ., bottom . => . => output row1 first '.' matches ". G . G"? Actually output row1: ". G . G". Yes first '.'.

(1,1): top B => G => second G matches.

(1,2): top ., bottom . => . => third '.' matches.

(1,3): top B, bottom R => G => fourth G matches.

Row2:
(2,0): top ., bottom R => G => output row2 first G matches "G G G G"? Actually output row2: G G G G. Yes.

(2,1): top ., bottom R => G => second G matches.

(2,2): top B, bottom R => G => third G matches.

(2,3): top B, bottom R => G => fourth G matches.

Row3:
(3,0): top B, bottom R => G => output row3 first G matches "G G . G"? Actually output row3: G G . G. Yes first G.

(3,1): top B, bottom R => G => second G matches.

(3,2): top ., bottom . => . => third '.' matches.

(3,3): top B, bottom R => G => fourth G matches.

Matches.

Example3: Let's test.

Input:
B . B .
B B . B
B . B B
. 

... [3298 chars truncated]

We need to parse the problem: It's a grid transformation puzzle. The description: "A rehearsal floor split by a golden ribbon. Above: chalk marks from one company in cool tones. Below: marks from another in warm tones. Unmarked floor is bare. What to do: Remove the ribbon and combine choreographies. Any floor tile marked by either company is danced upon, leaving a unified mark. Only doubly-unmarked tiles stay clean. Grid details: A 4×4 floor showing unified dance marks wherever either company rehearsed."

Thus we have input grid that includes both top and bottom parts? The training examples show input with 8 rows? Let's examine Example 1:

Input:
B B . .
B . B .
B B . B
B B B B
Y Y Y Y
R R . R
. . R .
. R . .
R . R .
Output:
G G . G
G . G .
G G . G
G G G G

So input has 8 rows, output has 4 rows. The top half (first 4 rows) are B (blue) marks (cool tones). The bottom half (next 4 rows) are Y (yellow) and R (red) marks (warm tones). The output is 4 rows, with G (green) marks where either B or Y/R? Actually they combine: B (blue) and Y (yellow) produce G (green). Also B and R produce? Not sure. Let's see.

In Example 1: top half (first 4 rows) B pattern:
Row1: B B . .
Row2: B . B .
Row3: B B . B
Row4: B B B B

Bottom half (rows5-8):
Row5: Y Y Y Y
Row6: R R . R
Row7: . . R .
Row8: . R . .
Row9: R . R .   Wait there are 5 rows? Actually they gave 9 rows? Let's recount: They gave:

Input:
B B . .
B . B .
B B . B
B B B B
Y Y Y Y
R R . R
. . R .
. R . .
R . R .

That's 9 rows? Let's count: 1: B B . .; 2: B . B .; 3: B B . B; 4: B B B B; 5: Y Y Y Y; 6: R R . R; 7: . . R .; 8: . R . .; 9: R . R . . Actually they have 9 rows? Let's list them:

1: B B . .
2: B . B .
3: B B . B
4: B B B B
5: Y Y Y Y
6: R R . R
7: . . R .
8: . R . .
9: R . R .

But they said 4x4 floor. So maybe the input is actually two 4x4 grids stacked: top 4 rows = one company's marks (cool tones), bottom 4 rows = other company's marks (warm tones). But they gave 9 rows? Something off. Let's check Example 2:

Input:
B B B .
. B . B
. . B B
B B . B
Y Y Y Y
. . . R
. . . R
R R R R
R R . R

That's also 9 rows: first 4 rows B pattern, then Y Y Y Y (row5), then . . . R (row6), . . . R (row7), R R R R (row8), R R . R (row9). That's 5 rows after the first 4, not 4. Something inconsistent.

Example 3:
Input:
B . B .
B B . B
B . B B
. B . B
Y Y Y Y
R R . .
. . R .
R R . .
. . R .
That's also 9 rows: first 4 rows B pattern, then Y Y Y Y (row5), then R R . . (row6), . . R . (row7), R R . . (row8), . . R . (row9). So again 5 rows after first 4.

Thus maybe the input includes a separator row of Y Y Y Y? Actually they always have a row of Y Y Y Y as row5 (the 5th row). Then the remaining rows are R pattern? Let's see: In each example, after the Y Y Y Y row, there are 4 rows of R pattern? Let's count: Example1: after Y Y Y Y (row5), we have rows6-9: R R . R, . . R ., . R . ., R . R . That's 4 rows. Yes! So total rows = 4 (B) + 1 (Y separator?) + 4 (R) = 9 rows. But they said 4x4 floor showing unified dance marks wherever either company rehearsed. Perhaps the Y Y Y Y row is the "golden ribbon"? Actually description: "A rehearsal floor split by a golden ribbon. Above: chalk marks from one company in cool tones. Below: marks from another in warm tones. Unmarked floor is bare. What to do: Remove the ribbon and combine choreographies. Any floor tile marked by either company is danced upon, leaving a unified mark. Only doubly-unmarked tiles stay clean. Grid details: A 4×4 floor showing unified dance marks wherever either company rehearsed."

Thus the input includes the ribbon as a row of Y (yellow) maybe representing the golden ribbon? Actually golden ribbon would be maybe Y (yellow) representing gold? They used Y for yellow. So the ribbon is a row of Y Y Y Y (all yellow) across the floor, splitting the floor into top and bottom halves. The top half (above ribbon) shows marks from one company in cool tones (B = blue). The bottom half (below ribbon) shows marks from the other company in warm tones (R = red? Actually warm tones could be red and maybe yellow? But they used R for red, and Y for yellow ribbon). So after removing the ribbon, we combine the marks: any tile that had a mark from either company (i.e., B in top half or R in bottom half) becomes a unified mark. What color is the unified mark? They used G (green) in output. So combining blue and red yields green? Actually blue + red = magenta (in light) but in pigment maybe? But they used G (green). However maybe they consider that the unified mark is the mixture of the two companies' colors: cool tones (blue) and warm tones (red) produce green? That seems odd but matches examples: In output, wherever there was a B in top half OR an R in bottom half, we see G. Let's test.

Example1: Top half B pattern:
Row1: B B . .
Row2: B . B .
Row3: B B . B
Row4: B B B B

Bottom half R pattern (rows6-9):
Row6: R R . R
Row7: . . R .
Row8: . R . .
Row9: R . R .

Now combine: For each cell (i,j) in 4x4, if top half has B or

... [5025 chars truncated]

We need to parse the problem. It's a grid transformation puzzle. The description: "Two flatbreads stacked with a saffron ribbon between them. The top bread has scattered blue filling; the bottom has scattered red filling. Gaps are plain dough. Press the breads together through the saffron ribbon. Every spot that carried any filling from either layer transforms into a single herbal garnish. Plain-on-plain spots stay bare. Grid details: A single flatbread emerges, 4×4. Garnished spots appear where either original layer had filling."

Thus we have two layers: top layer (blue filling) and bottom layer (red filling). The input grid presumably contains both layers? The training examples show input with two parts: first part (maybe top layer) and second part (bottom layer). Let's examine Example 1.

Example 1 Input:
B B . .
B . B .
B B . B
B B B B
Y Y Y Y
R R . R
. . R .
. R . .
R . R .
Output:
G G . G
G . G .
G G . G
G G G G

We see the input has 9 rows? Actually they show 9 lines before output: first 4 lines maybe top layer? Then next 5 lines? Let's count: Input lines:
1: B B . .
2: B . B .
3: B B . B
4: B B B B
5: Y Y Y Y
6: R R . R
7: . . R .
8: . R . .
9: R . R .
Output: 4 lines (4x4 grid). So input appears to be two layers each of size 4x4? Actually top layer maybe first 4 lines (B's and .). Bottom layer maybe next 4 lines? But we have 9 lines, not 8. Let's examine: The description says "Two flatbreads stacked with a saffron ribbon between them." The saffron ribbon maybe represented by a row of Y's? Indeed line 5 is "Y Y Y Y". That could be the saffron ribbon (yellow). So the input format: first 4 lines = top layer (blue filling), then a line of Y's representing the saffron ribbon, then next 4 lines = bottom layer (red filling). That matches: top layer lines 1-4, saffron ribbon line 5 (all Y), bottom layer lines 6-9. Output is 4x4 grid where garnish appears where either layer had filling (i.e., B or R). The garnish color is G (green) as per output. So the rule: For each cell position (i,j) in the 4x4 grid, if top layer cell is B (blue) OR bottom layer cell is R (red) (or both), then output cell is G (green). If both are . (plain), output is . (black). The saffron ribbon line is just a separator and not used in output.

Thus transformation: Input consists of 9 lines: first 4 lines = top layer (blue), line 5 = saffron ribbon (all Y, ignore), lines 6-9 = bottom layer (red). Output 4x4: G if either top or bottom has non-. (i.e., B or R) else . .

Check example: Let's compute.

Top layer:
Row1: B B . .
Row2: B . B .
Row3: B B . B
Row4: B B B B

Bottom layer:
Row1: R R . R
Row2: . . R .
Row3: . R . .
Row4: R . R .

Now compute OR:
Row1: B or R => G if any non-. So positions:
col1: B or R => G
col2: B or R => G
col3: . or . => .
col4: . or R => G
=> G G . G matches output row1.

Row2:
top: B . B .
bottom: . . R .
col1: B or . => G
col2: . or . => .
col3: B or R => G
col4: . or . => .
=> G . G . matches.

Row3:
top: B B . B
bottom: . R . .
col1: B or . => G
col2: B or R => G
col3: . or . => .
col4: B or . => G
=> G G . G matches.

Row4:
top: B B B B
bottom: R . R .
col1: B or R => G
col2: B or . => G
col3: B or R => G
col4: B or . => G
=> G G G G matches.

Thus rule confirmed.

Now test input:

Test Input:
B . B .
B . B .
. B . .
B . B .
Y Y Y Y
R R . .
. . R .
. R . R
R R R .

We need to parse similarly: first 4 lines = top layer (blue), line 5 = saffron ribbon (Y's), lines 6-9 = bottom layer (red). Let's extract.

Top layer (4 lines):
1: B . B .
2: B . B .
3: . B . .
4: B . B .

Bottom layer (lines after Y):
6: R R . .
7: . . R .
8: . R . R
9: R R R .

Now compute OR for each cell.

Let's create a 4x4 grid.

We'll label rows 0-3, columns 0-3.

Top:
Row0: [B, ., B, .]
Row1: [B, ., B, .]
Row2: [., B, ., .]
Row3: [B, ., B, .]

Bottom:
Row0: [R, R, ., .]
Row1: [., ., R, .]
Row2: [., R, ., R]
Row3: [R, R, R, .]

Now compute OR: if either top or bottom is non-. (i.e., B or R) then output G else . .

Let's compute each cell.

Row0:
col0: top B, bottom R => non-. => G
col1: top ., bottom R => G
col2: top B, bottom . => G
col3: top ., bottom . => .
=> Row0: G G G .

Row1:
top: [B, ., B, .]
bottom: [., ., R, .]
col0: B or . => G
col1: . or . => .
col2: B or R => G
col3: . or . => .
=> Row1: G . G .

Row2:
top: [., B, ., .]
bottom: [., R, ., R]
col0: . or . => .
col1: B or R => G
col2: . or . => .
col3: . or R => G
=> Row2: . G . G

Row3:
top: [B, ., B, .]
bottom: [R, R, R, .]
col0: B or R => G
col1: . or R => G
col2: B or R => G
col3: . or . => .
=> Row3: G G G .

Thus output grid:
Row0: G G G .
Row1: G . G .
Row2: . G . G
Row3: G G G .

We need to output using single-character color codes: . = black, G = green. So we output as characters separated by spaces.

Thus final answer:

ANSWER:
G G G .
G . G .
. G . G
G G G .

Check if any nuance: The garnish color is G (green) as per examples. Yes.

T

... [12 chars truncated]

We need to parse the problem. It's a grid transformation puzzle. Input consists of two staves: upper staff (cool notes) and lower staff (warm notes). The description: "What you see: A score split into upper and lower staves by a shimmering bar line. The upper staff has cool notes scattered across four measures; the lower staff has warm notes in different positions. What to do: Merge the two staves into one. Wherever either staff carried a note, the combined score sounds a harmonized tone. Rests persist only where both staves were silent. Grid details: One 4×4 staff of harmonized tones and rests."

Thus we have an input grid that includes both staves? The training examples show input as multiple lines: first some lines (maybe upper staff) then a line of Y Y Y Y (maybe the bar line?), then lower staff lines. Let's examine Example 1:

Input:
B B . .
B . B .
B B . B
B B B B
Y Y Y Y
R R . R
. . R .
. R . .
R . R .
Output:
G G . G
G . G .
G G . G
G G G G

So first 4 lines: upper staff (cool notes) = B (blue) and . (black). Then a line "Y Y Y Y" (yellow) maybe representing the bar line? Then next 4 lines: lower staff (warm notes) = R (red) and . (black). Output is 4 lines of G (green) and . (black). So merging: wherever either staff had a note (non-black), output is G (green). Where both staff had rests (black), output remains black (.). So the rule: For each cell position (i,j) in the 4x4 grid, if upper[i][j] != '.' OR lower[i][j] != '.' then output G else '.'.

Check Example 1: Upper staff lines:
Row0: B B . .
Row1: B . B .
Row2: B B . B
Row3: B B B B
Lower staff lines:
Row0: R R . R
Row1: . . R .
Row2: . R . .
Row3: R . R .
Now compute OR: For each cell, if either is not '.' then G else '.'.

Let's test: Row0 col0: upper B (not .) => G. col1: B => G. col2: upper . lower . => both . => . . Row0 col3: upper . lower R => G. So row0: G G . G matches output. Row1: upper B . B . ; lower . . R . => col0 B => G; col1 . . => . ; col2 B or R => G; col3 . . => . => G . G . matches. Row2: upper B B . B ; lower . R . . => col0 B => G; col1 B or R => G; col2 . . => . ; col3 B => G => G G . G matches. Row3: upper B B B B ; lower R . R . => col0 B or R => G; col1 B => G; col2 B or R => G; col3 B => G => G G G G matches.

Thus rule confirmed.

Now test Example 2: Input:
B B B .
. B . B
. . B B
B B . B
Y Y Y Y
. . . R
. . . R
R R R R
R R . R
Output:
G G G G
. G . G
G G G G
G G . G

Check: Upper staff lines (first 4):
Row0: B B B .
Row1: . B . B
Row2: . . B B
Row3: B B . B
Lower staff lines (next 4):
Row0: . . . R
Row1: . . . R
Row2: R R R R
Row3: R R . R

Now compute OR: Row0: upper B B B . ; lower . . . R => col0 B => G; col1 B => G; col2 B => G; col3 . or R => G => G G G G matches. Row1: upper . B . B ; lower . . . R => col0 . . => . ; col1 B => G; col2 . . => . ; col3 B or R => G => . G . G matches. Row2: upper . . B B ; lower R R R R => col0 . or R => G; col1 . or R => G; col2 B or R => G; col3 B or R => G => G G G G matches. Row3: upper B B . B ; lower R R . R => col0 B or R => G; col1 B or R => G; col2 . . => . ; col3 B or R => G => G G . G matches.

Thus rule holds.

Example 3: Input:
B . B .
B B . B
B . B B
. B . B
Y Y Y Y
R R . .
. . R .
R R . .
. . R .
Output:
G G G .
G G G G
G G G G
. G G G

Check: Upper staff lines (first 4):
Row0: B . B .
Row1: B B . B
Row2: B . B B
Row3: . B . B
Lower staff lines (next 4):
Row0: R R . .
Row1: . . R .
Row2: R R . .
Row3: . . R .

Now OR: Row0: B or R => G; . or R => G; B or . => B => G? Wait B is not ., so G; . or . => . => G G G . matches. Row1: B or . => G; B or . => G; . or R => G; B or . => G => G G G G matches. Row2: B or R => G; . or R => G; B or . => G; B or . => G => G G G G matches. Row3: . or . => . ; B or . => G; . or R => G; B or . => G => . G G G matches.

Thus rule consistent.

Example 4: Input:
B B . .
. B . B
. B . .
B . B .
Y Y Y Y
R R R R
. . R R
R R . .
. . R R
Output:
G G G G
. G G G
G G . .
G . G G

Check: Upper staff lines (first 4):
Row0: B B . .
Row1: . B . B
Row2: . B . .
Row3: B . B .
Lower staff lines (next 4):
Row0: R R R R
Row1: . . R R
Row2: R R . .
Row3: . . R R

Now OR: Row0: B or R => G; B or R => G; . or R => G; . or R => G => G G G G matches. Row1: . or . => . ; B or . => G; . or R => G; B or R => G => . G G G matches. Row2: . or R => G; B or R => G; . or . => . ; . or . => . => G G . . matches. Row3: B or . => G; . or . => . ; B or R => G; . or R => G => G . G G matches.

Thus rule is clear.

Now test input:

Test Input:
B . B .
B . B .
. B . .
B . B .
Y Y Y Y
R R . .
. . R .
. R . R
R R R .

We need to parse: first 4 lines are upper staff (cool notes) = B and . ; then line "Y Y Y Y" (the bar line) ; then next 4 lines are lower staff (warm notes) = R and . .

Thus we have:

Upper sta

... [1278 chars truncated]

We need to understand the puzzle. The description: "What you see: A split field with a golden dividing lane. The upper half shows one team's occupied positions; the lower half shows the other team's. Empty squares are unoccupied. What to do: Merge both rosters onto one field. Any position held by at least one team is now staffed by a combined-jersey player. Only doubly-vacant spots stay open. Grid details: A single 4×4 field with combined-jersey players filling any previously occupied position."

Thus input grid is presumably 8 rows? Actually training examples: Input appears to have 8 rows? Let's examine Example 1:

Input:
B B . .
B . B .
B B . B
B B B B
Y Y Y Y
R R . R
. . R .
. R . .
R . R .
Output:
G G . G
G . G .
G G . G
G G G G

So input has 8 rows, each row has 4 characters (since 4x4 field). The first 4 rows are upper half (team 1), next 4 rows are lower half (team 2). Output is 4 rows (the combined field). The rule: any position held by at least one team (i.e., if either upper or lower cell is non-empty) becomes a combined-jersey player, which is shown as G (green) in output. Only doubly-vacant spots (both upper and lower are empty) stay open (i.e., remain '.' black). So output is G if either input cell (upper or lower) is not '.'; else '.'.

Thus transformation: For each position (i,j) in 4x4, look at upper half cell (row i, col j) and lower half cell (row i+4, col j). If either is not '.' (i.e., B or R or Y? Actually upper half uses B? lower half uses R? In examples, upper half uses B (blue) for team 1, lower half uses R (red) for team 2. In example 2, upper half uses B and ., lower half uses R and . and also Y? Wait example 2 lower half includes Y Y Y Y in first row? Let's check:

Example 2 Input:
B B B .
. B . B
. . B B
B B . B
Y Y Y Y
. . . R
. . . R
R R R R
R R . R

So upper half rows 0-3: B B B . ; . B . B ; . . B B ; B B . B
Lower half rows 4-7: Y Y Y Y ; . . . R ; . . . R ; R R R R ; R R . R

Wait there are 9 rows? Actually they gave 9 rows? Let's count: lines:

1 B B B .
2 . B . B
3 . . B B
4 B B . B
5 Y Y Y Y
6 . . . R
7 . . . R
8 R R R R
9 R R . R

That's 9 rows, not 8. Something off. Maybe they inadvertently included an extra row? Let's re-evaluate: The description says single 4x4 field with combined-jersey players filling any previously occupied position. The input likely is 8 rows: 4 for upper team, 4 for lower team. But example 2 seems to have 9 rows. Let's check if maybe they misaligned: Could be that the first 4 rows are upper team (B's), next 4 rows are lower team (R's), and there is an extra row of Y's that is something else? Actually Y Y Y Y appears as row 5. Could be that Y is the "golden dividing lane"? The description mentions "a split field with a golden dividing lane." Perhaps the input includes a row of Y's representing the dividing lane between the halves. So the input format might be: 4 rows upper team, 1 row golden lane (Y's), 4 rows lower team. That would make 9 rows total. The output is just the combined 4x4 field (without the lane). Indeed in example 1, we see:

Input:
B B . .
B . B .
B B . B
B B B B
Y Y Y Y
R R . R
. . R .
. R . .
R . R .

That's 4 rows of B's, then a row of Y's, then 4 rows of R's. Output is 4 rows of G's and '.'.

Thus the transformation: ignore the middle row (the golden dividing lane) which is all Y's (maybe always Y's). Then combine upper and lower halves: if either cell is non-empty (i.e., B or R or any non '.'?), output G; else '.'.

Check example 1: Upper half rows 0-3: as given. Lower half rows 5-8 (since row 4 is Y's). Let's verify: Row5: R R . R ; Row6: . . R . ; Row7: . R . . ; Row8: R . R . (actually row8 is "R . R ." from input). Yes.

Now compute combined: For each position, if either upper or lower is not '.' then G else '.'.

Let's test a few: Upper row0 col0 B -> not '.' => G. Lower row5 col0 R => also not '.' => G. Output row0 col0 G matches.

Upper row0 col1 B => G. Lower row5 col1 R => G. Output row0 col1 G.

Upper row0 col2 '.' => empty. Lower row5 col2 '.' => empty. Output row0 col2 '.' matches.

Upper row0 col3 '.' => empty. Lower row5 col3 R => not empty => G. Output row0 col3 G matches.

Thus rule holds.

Now example 2: Let's verify.

Input rows:
0: B B B .
1: . B . B
2: . . B B
3: B B . B
4: Y Y Y Y (lane)
5: . . . R
6: . . . R
7: R R R R
8: R R . R

Upper half rows0-3, lower half rows5-8.

Compute output rows0-3.

Row0: Upper B B B . ; Lower . . . R => combine: col0 B or . => G; col1 B or . => G; col2 B or . => G; col3 . or R => G. So output row0: G G G G matches given output row0: G G G G.

Row1: Upper . B . B ; Lower . . . R => col0 . or . => . ; col1 B or . => G ; col2 . or . => . ; col3 B or R => G. Output row1: . G . G matches given: . G . G.

Row2: Upper . . B B ; Lower . . . R => col0 . or . => . ; col1 . or . => . ; col2 B or . => G ; col3 B or R => G. Output row2: . . G G? Wait given output row2: G G G G? Actually given output for 

... [2766 chars truncated]

We need to parse the problem: It's a grid transformation puzzle. We have a metaphorical description: "Two blueprint transparencies pinned above and below a golden lamp strip. Each has walls and open spaces in a 4×4 layout. Stack the transparencies on the lightbox. Any cell where at least one blueprint drew a wall becomes a built structure in the final plan. Only mutually open cells stay as open floor. Grid details: A single 4×4 floor plan with structures wherever either blueprint had a wall."

Thus we have two input grids (transparencies) stacked: one above, one below? The input appears to be two 4x4 grids: first grid (top transparency) and second grid (bottom transparency). The "golden lamp strip" maybe is a row of Y's? Actually the description: "Two blueprint transparencies pinned above and below a golden lamp strip." So there is a golden lamp strip in the middle? The input includes a row of Y Y Y Y (yellow) maybe representing the golden lamp strip. Indeed in the training examples, after the first 4x4 grid (top transparency) there is a row of Y Y Y Y (the golden lamp strip), then a second 4x4 grid (bottom transparency). So the input format: first 4 lines = top transparency (blueprint), then a line of 4 Y's = golden lamp strip, then next 4 lines = bottom transparency.

The output is a single 4x4 grid (the final plan) where any cell where at least one blueprint drew a wall becomes a built structure (i.e., G? Actually output uses G for green? Let's see: In example 1, output is G G . G etc. So G stands for green, which likely indicates a built structure (wall). The '.' stands for black (open floor). So the rule: For each cell, if either top or bottom transparency has a wall (i.e., not '.'?), then output is G (structure). If both are '.' (open), output remains '.' (open floor). However we need to verify what constitutes a wall in the blueprints. In the input grids, we see characters: B (blue), R (red), maybe other colors? The description says each blueprint has walls and open spaces. The colors may encode something else? Let's examine training examples.

Example 1 Input:
Top transparency (first 4 lines):
B B . .
B . B .
B B . B
B B B B

Then Y Y Y Y line.

Bottom transparency (next 4 lines):
R R . R
. . R .
. R . .
R . R .

Output:
G G . G
G . G .
G G . G
G G G G

So we need to see mapping: Where top or bottom has a wall (non '.'?), output G. Let's test.

Top grid:
Row0: B B . .
Row1: B . B .
Row2: B B . B
Row3: B B B B

Bottom grid:
Row0: R R . R
Row1: . . R .
Row2: . R . .
Row3: R . R .

Now compute OR of non '.'? For each cell, if either top or bottom is not '.' then output G else '.'.

Let's compute:

Cell (0,0): top B (not .), bottom R (not .) => G. Output row0 col0 G matches.

(0,1): top B, bottom R => G. Output G matches.

(0,2): top ., bottom . => both . => output . matches.

(0,3): top ., bottom R => not . => G. Output G matches.

Row1:
(1,0): top B, bottom . => G. Output G matches.
(1,1): top ., bottom . => . => output . matches.
(1,2): top B, bottom R => G. Output G matches.
(1,3): top ., bottom . => . => output . matches.

Row2:
(2,0): top B, bottom . => G. Output G matches.
(2,1): top B, bottom R => G. Output G matches.
(2,2): top ., bottom . => . => output . matches.
(2,3): top B, bottom . => G. Output G matches.

Row3:
(3,0): top B, bottom R => G. Output G matches.
(3,1): top B, bottom . => G. Output G matches.
(3,2): top B, bottom R => G. Output G matches.
(3,3): top B, bottom . => G. Output G matches.

Thus rule holds: output G if either input (top or bottom) has a non '.' (i.e., any colored cell) at that position; else '.'.

Now test Example 2 to confirm.

Example 2 Input:
Top transparency:
B B B .
. B . B
. . B B
B B . B

Y Y Y Y

Bottom transparency:
. . . R
. . . R
R R R R
R R . R

Output:
G G G G
. G . G
G G G G
G G . G

Let's compute OR of non '.'.

Top grid:
Row0: B B B .
Row1: . B . B
Row2: . . B B
Row3: B B . B

Bottom grid:
Row0: . . . R
Row1: . . . R
Row2: R R R R
Row3: R R . R

Now compute:

Row0:
(0,0): top B, bottom . => G => output G matches.
(0,1): top B, bottom . => G => output G matches.
(0,2): top B, bottom . => G => output G matches.
(0,3): top ., bottom R => G => output G matches.

Row1:
(1,0): top ., bottom . => . => output . matches.
(1,1): top B, bottom . => G => output G matches.
(1,2): top ., bottom . => . => output . matches.
(1,3): top B, bottom R => G => output G matches.

Row2:
(2,0): top ., bottom R => G => output G matches.
(2,1): top ., bottom R => G => output G matches.
(2,2): top B, bottom R => G => output G matches.
(2,3): top B, bottom R => G => output G matches.

Row3:
(3,0): top B, bottom R => G => output G matches.
(3,1): top B, bottom R => G => output G matches.
(3,2): top ., bottom . => . => output . matches.
(3,3): top B, bottom R => G => output G matches.

Matches.

Thus rule: For eac

... [2560 chars truncated]

We need to parse the problem: It's a grid transformation puzzle. Input consists of two tactical maps separated by a golden demarcation line. The description: "Two tactical maps separated by a golden demarcation line. Each map marks positions with its signature flare color against dark terrain. Overlay both maps. Any position flagged by either corps is marked as confirmed on the unified map in a new indicator color. Unflagged-by-both positions remain dark."

Thus we have two maps: likely the top part of input is first map (blue?), the bottom part is second map (red?). The golden demarcation line is a row of Y's? Actually they say "golden demarcation line" and color key: Y=yellow. So the line separating the two maps is a row of Y's (yellow). Indeed in examples, after the first map (some rows of B and .) there is a row of Y Y Y Y (all yellow) then the second map (some rows of R and .). So the input grid includes both maps plus the separator line. The output is a unified map (size? The examples show output is same dimensions as each map? Let's see.

Example 1 Input:
B B . .
B . B .
B B . B
B B B B
Y Y Y Y
R R . R
. . R .
. R . .
R . R .
Output:
G G . G
G . G .
G G . G
G G G G

So input has: first map: 4 rows (B B . .; B . B .; B B . B; B B B B). Then a separator row: Y Y Y Y (4 columns). Then second map: 4 rows (R R . R; . . R .; . R . .; R . R .). Output is 4 rows (same as each map). So the output is the overlay of the two maps: any position flagged by either corps (i.e., either map has a non-black marker) is marked as confirmed in a new indicator color (green G). Unflagged-by-both remain dark (black .). So essentially we take elementwise OR of the two maps (treating any non-. as flagged) and output G if flagged else . . The separator row is ignored (not part of output). The maps are same size (4x4). So the transformation: given input grid with N rows of first map, then a row of all Y (separator), then N rows of second map (same number columns), output N rows where each cell is G if either corresponding cell in first or second map is not '.' (i.e., B or R or any other color?), else '.'.

Check example 2:

Input:
B B B .
. B . B
. . B B
B B . B
Y Y Y Y
. . . R
. . . R
R R R R
R R . R
Output:
G G G G
. G . G
G G G G
G G . G

Let's verify: first map rows:
1: B B B .
2: . B . B
3: . . B B
4: B B . B
Second map rows:
1: . . . R
2: . . . R
3: R R R R
4: R R . R
Now OR: any non-. yields G.

Row1: first map B B B . ; second . . . R => positions: (1,1) B or . => flagged => G; (1,2) B or . => G; (1,3) B or . => G; (1,4) . or R => flagged => G => G G G G matches.

Row2: first . B . B ; second . . . R => (2,1) . or . => . ; (2,2) B or . => G ; (2,3) . or . => . ; (2,4) B or R => G => . G . G matches.

Row3: first . . B B ; second R R R R => (3,1) . or R => G ; (3,2) . or R => G ; (3,3) B or R => G ; (3,4) B or R => G => G G G G matches.

Row4: first B B . B ; second R R . R => (4,1) B or R => G ; (4,2) B or R => G ; (4,3) . or . => . ; (4,4) B or R => G => G G . G matches.

Thus rule confirmed.

Now test input:

Test Input:
B . B .
B . B .
. B . .
B . B .
Y Y Y Y
R R . .
. . R .
. R . R
R R R .

We need to parse: first map rows (before separator) are 4 rows? Let's count lines:

Line1: B . B .
Line2: B . B .
Line3: . B . .
Line4: B . B .
Line5: Y Y Y Y (separator)
Line6: R R . .
Line7: . . R .
Line8: . R . R
Line9: R R R .

Thus first map is 4 rows (lines1-4). Second map is lines6-9 (4 rows). Output should be 4 rows.

Now compute OR: For each cell, if either first map cell != '.' or second map cell != '.' then output G else '.'.

Let's write first map matrix:

Row1: [B, ., B, .]
Row2: [B, ., B, .]
Row3: [., B, ., .]
Row4: [B, ., B, .]

Second map matrix:

Row1: [R, R, ., .]
Row2: [., ., R, .]
Row3: [., R, ., R]
Row4: [R, R, R, .]

Now compute OR per cell:

Row1:
col1: B or R => flagged => G
col2: . or R => flagged => G
col3: B or . => flagged => G
col4: . or . => not flagged => .
Thus row1: G G G .

Row2:
col1: B or . => G
col2: . or . => .
col3: B or R => G
col4: . or . => .
Thus row2: G . G .

Row3:
col1: . or . => .
col2: B or R => G
col3: . or . => .
col4: . or R => G
Thus row3: . G . G

Row4:
col1: B or R => G
col2: . or R => G
col3: B or R => G
col4: . or . => .
Thus row4: G G G .

Thus output grid:

G G G .
G . G .
. G . G
G G G .

Now we need to output using single-character color codes: '.' for black, 'G' for green. So we output rows with spaces between characters as per instruction: "one row per line, characters separated by spaces". So:

G G G .
G . G .
. G . G
G G G .

We need to put inside a block labeled ANSWER: with the grid.

Thus final answer:

ANSWER:
G G G .
G . G .
. G . G
G G G .

Double-check: Did we interpret correctly? The separator row is all Y's; we ignore it. Yes.

Thus final.

We need to parse the problem: It's a grid transformation puzzle. Input grid appears to be two parts: top part (cool-toned shadow puppets) and bottom part (warm-toned shadow puppets). The description: "A stage divided by a golden scrim. Above: cool-toned shadow puppets in various positions. Below: warm-toned shadow puppets in other positions. Dark where no puppet appears. What to do: Remove the scrim and merge the performances. Wherever either troupe cast a shadow, a unified spotlight illuminates that spot in a new color. Only doubly-dark areas stay unlit."

Thus we have two layers: top (cool) and bottom (warm). The input grid likely contains both layers concatenated? Let's examine training examples.

Example 1 Input:
B B . .
B . B .
B B . B
B B B B
Y Y Y Y
R R . R
. . R .
. R . .
R . R .
Output:
G G . G
G . G .
G G . G
G G G G

We see input has 9 rows? Let's count: first 4 rows: B B . . ; B . B . ; B B . B ; B B B B. Then next 5 rows: Y Y Y Y ; R R . R ; . . R . ; . R . . ; R . R . . Wait the last line seems truncated? Actually they show:

Input:
B B . .
B . B .
B B . B
B B B B
Y Y Y Y
R R . R
. . R .
. R . .
R . R .

That's 9 rows: 4 top rows, then 5 bottom rows? But the stage is 4x4 according to description. So maybe the input is two 4x4 grids stacked vertically: top 4 rows = cool-toned (blue?), bottom 4 rows = warm-toned (red?). But we see 9 rows, not 8. Let's check: maybe they omitted a row? Let's count again: lines:

1: B B . .
2: B . B .
3: B B . B
4: B B B B
5: Y Y Y Y
6: R R . R
7: . . R .
8: . R . .
9: R . R .

That's 9 rows. Something off. Perhaps the input includes a separator line? Or maybe the first 4 rows are top, then a blank line? Not shown. Let's examine Example 2:

Input:
B B B .
. B . B
. . B B
B B . B
Y Y Y Y
. . . R
. . . R
R R R R
R R . R

Again 9 rows: first 4 rows: B B B . ; . B . B ; . . B B ; B B . B. Then next 5 rows: Y Y Y Y ; . . . R ; . . . R ; R R R R ; R R . R.

Example 3 Input:
B . B .
B B . B
B . B B
. B . B
Y Y Y Y
R R . .
. . R .
R R . .
. . R .

Again 9 rows: first 4 rows: B . B . ; B B . B ; B . B B ; . B . B. Then next 5 rows: Y Y Y Y ; R R . . ; . . R . ; R R . . ; . . R .

Thus pattern: Input always has 9 rows: first 4 rows are top layer (cool-toned) using B (blue) and . (black). Next row is all Y (yellow) maybe representing the golden scrim? Actually row 5 is all Y (yellow) in all examples. Indeed row 5: Y Y Y Y in each example. So row 5 is the golden scrim (maybe a separator). Then rows 6-9 are bottom layer (warm-toned) using R (red) and . (black). So the input format: 4 rows top (cool), 1 row separator (all Y), 4 rows bottom (warm). That's 9 rows total. Output is 4 rows (the merged stage after removing scrim). Output uses G (green) and . (black). So the transformation: Where either top or bottom has a puppet (non-dark), we illuminate that spot with a new color (green). Only spots that are dark in both top and bottom remain dark (black). So output cell = G if top cell != '.' OR bottom cell != '.' (i.e., if either is B or R). Else output = '.'.

Thus it's essentially logical OR of presence of any puppet (non-dark) from top or bottom layers, mapping to green.

Check with examples:

Example1 top layer (first 4 rows):
Row1: B B . .
Row2: B . B .
Row3: B B . B
Row4: B B B B

Bottom layer (rows6-9):
Row6: R R . R
Row7: . . R .
Row8: . R . .
Row9: R . R .

Now compute OR: For each cell, if top is B or bottom is R => G else . Let's compute row1: top B B . . ; bottom R R . R => OR: B or R => G for first two; third: top . bottom . => . ; fourth: top . bottom R => G. So row1: G G . G matches output row1: G G . G. Good.

Row2: top B . B . ; bottom . . R . => OR: first B => G; second . . => . ; third B or R => G; fourth . . => . => G . G . matches output row2: G . G . Good.

Row3: top B B . B ; bottom . . R . => OR: first B => G; second B => G; third . or R => G; fourth B => G => G G G G? Wait output row3: G G . G. Let's compute: Actually bottom row3 is row8? Let's align: bottom rows: row6 (index0), row7 (1), row8 (2), row9 (3). So bottom row3 corresponds to row8: . R . . (since row8 is ". R . ."). Wait we need to check: Input rows after separator:

Row6: R R . R
Row7: . . R .
Row8: . R . .
Row9: R . R .

Thus bottom row1 = R R . R
bottom row2 = . . R .
bottom row3 = . R . .
bottom row4 = R . R .

Now compute OR for each row:

Top row1: B B . .
Bottom row1: R R . R => OR: G G . G matches.

Top row2: B . B .
Bottom row2: . . R . => OR: G . G . matches.

Top row3: B B . B
Bottom row3: . R . . => OR: B or . => G for first; second B or R => G; third . or . => . ; fourth B or . => G => G G . G matches output row3: G G . G. Good.

Top row4: B B B B
Bottom row4: R . R . => OR: B or R => G for all? Let's compute: first B or R => G; second B or . => G; third B or R => G; fourth B or . => G => G G G G matches output row4: G G G G. Good.

Thus rule confirmed.

Now test input:

Test Input:
B . B

... [2164 chars truncated]

We need to parse the problem. It's a grid transformation puzzle. The description: "Two planting beds flanking a golden marigold border. Each bed has seedlings in some cells and bare soil in others. What to do: Consolidate into one bed. Any cell that held a seedling in either bed is planted with a hybrid cultivar. Only cells bare in both beds remain as empty soil. Grid details: A single 4×4 bed with hybrid plants where either original bed had seedlings."

Thus we have two input beds (two 4x4 grids) separated by a golden marigold border (the Y Y Y Y line? Actually the border is a row of Y's? The description: "Two planting beds flanking a golden marigold border." So the input consists of two 4x4 grids (the beds) with a row of Y's in between representing the border. The output is a single 4x4 grid where each cell is G (green) if either of the two input beds had a seedling (i.e., non-black? Actually seedlings are represented by B? Let's see training examples.

Example 1 Input:
B B . .
B . B .
B B . B
B B B B
Y Y Y Y
R R . R
. . R .
. R . .
R . R .
Output:
G G . G
G . G .
G G . G
G G G G

Interpretation: The first 4x4 block (top) is the first bed. The second 4x4 block (bottom) is the second bed. The border row of Y's separates them.

Now we need to see mapping: In first bed, seedlings are B (blue) and . (black) is bare soil? In second bed, seedlings are R (red) and . is bare soil? The output uses G (green) for hybrid where either bed had a seedling. So if either input cell is non-. (i.e., B or R), output is G. If both are ., output remains . (bare soil). Let's test.

First bed top-left 4x4:
Row1: B B . .
Row2: B . B .
Row3: B B . B
Row4: B B B B

Second bed bottom 4x4:
Row1: R R . R
Row2: . . R .
Row3: . R . .
Row4: R . R .

Now compute output: For each cell, if either is non-. then G else . Let's compute.

Cell (1,1): top B, bottom R => non-. => G. Output row1 col1 G matches.
(1,2): top B, bottom R => G matches.
(1,3): top ., bottom . => . => output . matches.
(1,4): top ., bottom R => G matches output G.

Row2:
(2,1): top B, bottom . => G matches output G.
(2,2): top ., bottom . => . matches output .
(2,3): top B, bottom R => G matches output G.
(2,4): top ., bottom . => . matches output .? Wait output row2 col4 is .? Let's see output row2: "G . G ." yes col4 is . matches.

Row3:
(3,1): top B, bottom . => G matches output G.
(3,2): top B, bottom R => G matches output G.
(3,3): top ., bottom . => . matches output .
(3,4): top B, bottom . => G matches output G.

Row4:
(4,1): top B, bottom R => G matches output G.
(4,2): top B, bottom . => G matches output G.
(4,3): top B, bottom R => G matches output G.
(4,4): top B, bottom . => G matches output G.

Thus rule holds.

Example 2: Let's verify.

Input:
B B B .
. B . B
. . B B
B B . B
Y Y Y Y
. . . R
. . . R
R R R R
R R . R

First bed:
Row1: B B B .
Row2: . B . B
Row3: . . B B
Row4: B B . B

Second bed:
Row1: . . . R
Row2: . . . R
Row3: R R R R
Row4: R R . R

Now compute output: G if either non-.

Row1:
col1: B or . => G
col2: B or . => G
col3: B or . => G
col4: . or R => G
=> G G G G matches output row1.

Row2:
col1: . or . => .
col2: B or . => G
col3: . or . => .
col4: B or R => G
=> . G . G matches output row2.

Row3:
col1: . or R => G
col2: . or R => G
col3: B or R => G
col4: B or R => G
=> G G G G matches output row3.

Row4:
col1: B or R => G
col2: B or R => G
col3: . or . => .
col4: B or R => G
=> G G . G matches output row4.

Thus rule confirmed.

Thus transformation: Input consists of two 4x4 grids separated by a row of Y's (the border). Output is a 4x4 grid where each cell is G if either of the corresponding cells in the two input grids is not '.' (i.e., has a seedling, represented by any non-black color like B or R). Otherwise output '.'.

Now test input:

Test Input:
B . B .
B . B .
. B . .
B . B .
Y Y Y Y
R R . .
. . R .
. R . R
R R R .

We need to parse: first 4 lines before Y Y Y Y are first bed. Then after Y Y Y Y line, next 4 lines are second bed.

First bed:
Row1: B . B .
Row2: B . B .
Row3: . B . .
Row4: B . B .

Second bed:
Row1: R R . .
Row2: . . R .
Row3: . R . R
Row4: R R R .

Now compute output: For each cell, if either is non-'.' then G else '.'.

Let's compute.

We'll label coordinates (r,c) 1-indexed.

First bed matrix A:
A11 = B
A12 = .
A13 = B
A14 = .
A21 = B
A22 = .
A23 = B
A24 = .
A31 = .
A32 = B
A33 = .
A34 = .
A41 = B
A42 = .
A43 = B
A44 = .

Second bed matrix B:
B11 = R
B12 = R
B13 = .
B14 = .
B21 = .
B22 = .
B23 = R
B24 = .
B31 = .
B32 = R
B33 = .
B34 = R
B41 = R
B42 = R
B43 = R
B44 = .

Now output Oij = G if Aij != '.' or Bij != '.' else '.'.

Compute each:

Row1:
(1,1): A B (non-.) => G
(1,2): A ., B R => non-. => G
(1,3): A B, B . => G
(1,4): A ., B . => both . => .
Thus row1: G G G .

Row2:
(2,1): A B, B . => G
(2,2): A ., B . => .
(2,3): A B, B R => G
(2,4): A ., B . => .
Thus row2:

... [772 chars truncated]

We need to parse the problem: We have a 5x5 input grid with '.' (black) and a single tribe's colored warriors: either B, R, or G (blue, red, green). The tribe is uniform: all colored cells are same color. The task: Identify the tribe color, then output a 3x3 grid representing the ancestral sigil (grey X) placed according to tribe's sigil pattern. The sigil is a 3x3 pattern with exactly five X's (grey) and four '.' (black). The patterns described:

- Blue tribe's sigil radiates outward from the center in four directions. That likely means a plus sign: center and up, down, left, right: pattern:
 . X .
 X X X
 . X .
But wait that's 5 X's? Let's count: center (1), up (2), down (3), left (4), right (5). Yes that's a plus sign. That matches Example 1 output: . X . / X X X / . X . (they gave that as output for blue tribe). Indeed Example 1 input had B's (blue) and output is plus sign.

- Red tribe's sigil stands tall, crowned with a broad brim. That likely is a T shape? "Stands tall" vertical line, "crowned with a broad brim" horizontal line at top? Let's think: maybe shape like:
 X X X
 . X .
 . X .
But that's 5 X's? Count: top row three X's (3), middle column two more (center and bottom) =5. That pattern is:
 X X X
 . X .
 . X .
But Example 2 output is:
 X X X
 . X .
 . X .
Yes that matches Example 2 output (they gave that). So red tribe sigil is a T shape (top bar and vertical stem).

- Green tribe's sigil hunkers into one corner, bracing two walls that meet at the foundation. That likely is an L shape in a corner: maybe bottom-left corner? Let's see Example 3 output:
 . . X
 . . X
 X X X
That's an L shape occupying bottom row (three X's) and rightmost column (two X's above bottom). Actually pattern: bottom row all X (three), and right column two X's above bottom (positions (0,2) and (1,2) if zero-index top-left). That's like an L shape in bottom-right corner? Wait bottom row is row index 2 (0-index top). Right column is column index 2. The shape occupies cells: (2,0), (2,1), (2,2) bottom row; (0,2), (1,2) right column. That's an L shape rotated? Actually it's like a reversed L: the corner at bottom-right (2,2) is the intersection of bottom row and right column. So shape is like an L that occupies the bottom row and right column, i.e., a corner shape. That matches "hunkers into one corner, bracing two walls that meet at the foundation." So green tribe sigil is an L shape in a corner (likely bottom-right). Could also be other corners depending on orientation? But given example, it's bottom-right.

Thus transformation: Input 5x5 grid with only one color (B,R,G) plus '.'; we ignore positions, just detect which color appears (non '.'), then output corresponding 3x3 sigil pattern.

Thus for test input: we need to see which color appears. Test input:
Row0: . G . G G
Row1: . . G . .
Row2: G . . . .
Row3: . . G . G
Row4: . . . . G

We see G's present. No B or R. So tribe is green. Therefore output sigil is green tribe's L shape (bottom-right corner). As per example 3 output: 
 . . X
 . . X
 X X X

Thus answer grid:
Row0: . . X
Row1: . . X
Row2: X X X

We need to output with spaces between characters. So:
. . X
. . X
X X X

Make sure to follow format: ANSWER: then block.

Thus final.

We need to understand the puzzle. Input: 5x5 grid with one colored species scattered among empty medium (black '.'), and we need to output a 3x3 protein structure in grey on black, determined solely by species identity. So the output is a 3x3 pattern of grey 'X' and black '.' (since grey on black). The output patterns in examples are 3x3 with X and . only.

Thus the species identity (color) determines which 3x3 pattern to output. The input grid is 5x5 with scattered colonies of that species (colored cells) among empty '.' (black). The placement of colonies is irrelevant; we just need to detect which color appears in the input (non '.' cells). Then map that color to a specific 3x3 pattern.

Thus we need to infer mapping from color to pattern from the three training examples.

Example 1: Input has B (blue) colonies. Output pattern:
. X .
X X X
. X .
That's a plus sign shape (cross) of X in center and up/down/left/right.

Example 2: Input has R (red) colonies. Output pattern:
X X X
. X .
. X .
That's a vertical line of three X in column 0? Actually pattern: row0: X X X (top row all X), row1: . X . (center X), row2: . X . (center X). So shape: top row full, and a vertical line down the middle column (col1). So like a T shape? Actually top row full plus a stem down middle.

Example 3: Input has G (green) colonies. Output pattern:
. . X
. . X
X X X
That's bottom row full X, and a vertical line up the rightmost column? Actually pattern: row0: . . X (rightmost X), row1: . . X (rightmost X), row2: X X X (all three X). So shape: bottom row full, and a column of X in rightmost column (col2) going up.

Thus mapping: B -> plus sign (cross). R -> top bar + middle column (like an upside-down T? Actually top bar plus stem down). G -> bottom bar + right column (like an L shape rotated? Actually bottom bar plus stem up right side). So each color yields a shape that is like a 3x3 pattern with a line of three X's in one orientation plus a perpendicular line of three X's through the center? Let's examine.

For B: plus sign: both a horizontal line through middle row (row1: X X X) and vertical line through middle column (col1: X X X). Intersection at center yields plus.

For R: pattern: top row all X (row0: X X X) and middle column all X (col1: X X X). Intersection at (0,1) top middle? Actually top row and middle column intersect at (0,1) which is X (makes sense). So shape is like a "T" rotated? Actually it's like a "┤"? Wait top bar plus vertical line down middle yields shape like an upside-down T? Actually typical T shape is top bar and stem down from middle; that matches: top bar (row0 all X) and stem down middle column (col1 all X). Yes.

For G: pattern: bottom row all X (row2: X X X) and right column all X (col2: X X X). Intersection at (2,2) bottom-right corner. That's like an "L" shape? Actually bottom bar plus stem up right column yields shape like a reversed L? Actually it's like a "┘" shape? Bottom bar plus upward stem on right yields shape like a "┐"? Let's visualize: bottom row full (three X across bottom), and a vertical line up the rightmost column (three X up). Intersection at bottom-right corner yields shape like a "└"? Actually if you have bottom bar and a stem going up from the right end, that looks like a "┘"? Wait typical "┘" is bottom bar and left stem up? Actually "┘" is bottom bar and left vertical up? Let's not worry.

Thus mapping seems to be: each color corresponds to a specific orientation of a shape consisting of a full row (either top, middle, or bottom) plus a full column (either left, middle, or right) such that the intersection is at a specific cell (maybe the corner or middle). For B: middle row + middle column (intersection at center). For R: top row + middle column (intersection at top-middle). For G: bottom row + right column (intersection at bottom-right). So we need to see pattern: maybe mapping is based on color order: B (blue) -> middle row + middle column (center). R (red) -> top row + middle column (top-middle). G (green) -> bottom row + right column (bottom-right). Could be that each color corresponds to a direction: maybe B is "center", R is "up", G is "right"? But we need to see if there's a pattern for other colors (Y, X, M, O, A, W) not given in training. But test input uses G again (green). So we just need to output pattern for G as per example 3: bottom row + right column.

Thus test input: we need to verify that the only non '.' color is G (green). Let's examine test input:

. G . G G
. . G . .
G . . . .
. . G . G
. . . . G

Let's list coordinates (row, col) 0-index:

Row0: col1 G, col3 G, col4 G
Row1: col2 G
Row2: col0 G
Row3: col2 G, col4 G
Row4: col4 G

All non '.' are G. So species is green. So output should be same as example 3: pattern for G.

Thus output:

. . X
. . X
X X X

But we need to output using single-character color codes: '.' for black, 'X' for grey. So exactly as above.

Thus answer:

ANSWER:
. . X
. . X

... [837 chars truncated]

We need to understand the transformation: Input is a 5x5 grid with one star color (i.e., only one color appears besides black '.'? Actually description: "Stars of one spectral color scattered across a dark sky plate." So input grid contains only '.' (black) and one other color (the star color). The output is a 3x3 remnant silhouette in grey (X) per spectral class. So the output is a 3x3 pattern of X's and '.' (black) that depends on the star color (spectral class). The training examples show mapping from star color to output pattern.

We have three examples: B (blue) -> output pattern: . X . / X X X / . X . (a plus sign shape? Actually it's a cross: center row all X, center column X, making a plus). That's a plus shape.

R (red) -> output: X X X / . X . / . X . (looks like a T shape? Actually top row all X, middle column X, bottom column X? Let's see: row0: X X X (full top bar), row1: . X . (center), row2: . X . (center). So shape like a "T" with a vertical line down the middle and a top bar.

G (green) -> output: . . X / . . X / X X X (looks like an L shape? Actually bottom row full X, right column X? Let's see: row0: . . X (rightmost), row1: . . X (rightmost), row2: X X X (full bottom). So shape like a reversed L? Actually it's like a corner: bottom row full, right column up two cells.

Thus each color maps to a specific 3x3 pattern.

We need to infer mapping for other colors? The test input uses G (green) again? Actually test input: . G . G G / . . G . . / G . . . . / . . G . G / . . . . G. So star color is G (green). So we need to output the pattern for green as per example 3: . . X / . . X / X X X.

Thus answer should be that pattern.

But we need to verify that the mapping is consistent: maybe the pattern depends on some property of the star arrangement? However description says: "Classify the spectral type by color and predict the stellar remnant — a small fixed shape that spectral class alone determines." So shape depends only on color, not on arrangement. So indeed for green we output the shape from example 3.

Thus answer: . . X / . . X / X X X.

We need to output in the format: each row separated by spaces between characters. So:

. . X
. . X
X X X

We need to put inside ANSWER block.

Thus final.

We need to understand the transformation: Input is a 5x5 grid with one cargo color (i.e., only one non-black color appears? Actually they say "Floating debris of one color scattered across dark ocean." So the input grid has black '.' as ocean, and one other color (cargo) appears in some cells. The output is a 3x3 mooring pattern in grey (X) and black (.) presumably, fixed per cargo color. So each cargo color (B, R, G, Y, M, O, A, W) maps to a specific 3x3 pattern of X and . (maybe also other colors? but they say mooring pattern in grey, so only X and .). The training examples show mapping for B, R, G.

Let's examine examples.

Example 1: Input B (blue) pattern. Output:
. X .
X X X
. X .

That's a plus sign shape: center and middle of edges? Actually pattern: row0: . X . (center top), row1: X X X (full middle row), row2: . X . (center bottom). So it's a cross (vertical and horizontal line) of X's.

Example 2: Input R (red) pattern. Output:
X X X
. X .
. X .

That's like a T shape? Actually row0: X X X (full top row), row1: . X . (center), row2: . X . (center). So it's a vertical line down the middle column plus a top horizontal bar.

Example 3: Input G (green) pattern. Output:
. . X
. . X
X X X

That's like an L shape? Actually row0: . . X (rightmost), row1: . . X (rightmost), row2: X X X (full bottom row). So it's a reversed L? Actually shape: bottom row full, and right column top two cells.

Thus each cargo color maps to a specific 3x3 pattern.

We need to infer mapping for other colors? Not given. But test input uses G (green) again? Let's see test input: It has G's scattered. So cargo color is G again. So we need to output the pattern for G, which from example 3 is:

. . X
. . X
X X X

Thus answer should be that.

But we need to verify that the mapping is independent of where the debris drifts (i.e., pattern of cargo cells doesn't matter; only the color matters). Indeed examples show different arrangements of B, R, G but same output per color.

Thus test input G yields same pattern as example 3.

Thus answer:

. . X
. . X
X X X

But we need to output with spaces between characters? The instruction: "write the complete output grid inside a clearly labeled block like this:
ANSWER:
. B R
G Y .
(one row per line, characters separated by spaces)". So we need spaces between characters.

Thus:

ANSWER:
. . X
. . X
X X X

But ensure each row has three characters separated by spaces: ". . X", ". . X", "X X X".

Thus final.

We need to understand the puzzle: Input is a 5x5 grid with one dopant color (i.e., only one color appears besides black '.'?), and we need to output a 3x3 pattern in grey (X) and black (.) according to the dopant type. The training examples show mapping from input patterns (with a specific color) to output patterns (with X and .). The output is always 3x3.

We need to identify the CAST (roles), AFFORDANCES, and TRANSFORMATION RULE.

Given the metaphor: "One type of dopant carrier scattered across a dark silicon substrate." So the input grid is a 5x5 substrate (dark = black '.'), with one dopant color (like B, R, G, etc.) scattered. The output is a 3x3 junction pattern intrinsic to that impurity type, in grey (X) and black (.). So each dopant color maps to a specific 3x3 pattern.

Thus the transformation rule: given the input 5x5 with a single non-black color (the dopant), we need to output the pattern associated with that color. The pattern is independent of the exact arrangement of dopant carriers? Or maybe the pattern is derived from the shape formed by the dopant positions? The description: "Identify the dopant by its color and read off the junction geometry — a fixed output pattern intrinsic to that impurity type." So the pattern is fixed per color, not dependent on arrangement. So we just need to map each color to a 3x3 pattern.

Thus from training examples we can infer mapping:

Example 1: Input color B (blue). Output:
. X .
X X X
. X .

That's a plus sign shape (cross) with center and middle of edges? Actually pattern:
Row0: . X .
Row1: X X X
Row2: . X .
That's a cross (like a plus) with arms length 1.

Example 2: Input color R (red). Output:
X X X
. X .
. X .
That's like a T shape? Actually row0: X X X (top bar), row1: . X . (center), row2: . X . (vertical line down). So shape: a T with a vertical line down from center.

Example 3: Input color G (green). Output:
. . X
. . X
X X X
That's like an L shape? Actually row0: . . X (right top), row1: . . X (right middle), row2: X X X (bottom row full). So shape: a reversed L? Actually it's like a corner shape: bottom row full, and right column top two cells.

Thus mapping:
B -> cross (+)
R -> T shape (top bar + vertical down)
G -> L shape (bottom row + right column)

We need to verify if these are consistent with the metaphor: "junction geometry intrinsic to that impurity type." Possibly each dopant type corresponds to a certain crystal orientation or something.

Now test input: we need to see what color appears. The test input grid:
. G . G G
. . G . .
G . . . .
. . G . G
. . . . G

We see G (green) appears multiple times. So dopant color is green. According to mapping from example 3, green maps to:
. . X
. . X
X X X

Thus output should be that pattern.

But we need to double-check: Are there any other colors? Only G and '.' (black). So yes.

Thus answer: 
. . X
. . X
X X X

But we need to output with spaces between characters as per instruction: "one row per line, characters separated by spaces". So we need to output:
. . X
. . X
X X X

But we need to ensure we use the color key: '.' = black, 'X' = grey. So that's correct.

Thus final answer block.

But before that, we need to perform Conceptual Digestion: Identify CAST, AFFORDANCES, TRANSFORMATION RULE. We'll include reasoning in the answer? The instruction: "Think step by step. After your reasoning, you MUST write out the complete predicted output grid using these single-character color codes". So we need to provide reasoning then final answer block.

Thus we will produce a response with reasoning and then the ANSWER block.

Let's craft reasoning:

CAST: The grid elements: '.' (black) = dark silicon substrate (background). The colored cells (B,R,G,etc.) = dopant carriers of a specific type. The output grid: 'X' = grey representing junction geometry, '.' = black background.

AFFORDANCES: The dopant carriers can be identified by their color; their arrangement does not affect the output; the substrate is passive; the transformation reads off a fixed pattern associated with the dopant type.

TRANSFORMATION RULE: Determine the unique non-black color present in the input 5x5 grid; map that color to its predefined 3x3 junction pattern: B -> cross (+), R -> T shape (top bar + vertical down), G -> reversed L shape (bottom row + right column). Then output that pattern.

Apply to test input: The only non-black color is G (green). Therefore output pattern for G: 
. . X
. . X
X X X

Thus answer.

Now produce final answer block.

We need to understand the puzzle. Input: 5x5 grid with one sacred color (i.e., only one color appears besides black? Actually description: "Oracle bones tinted in one deity's sacred color, scattered across a dark cloth." So the cloth is black (.) and the bones are of a single color (the deity's color). So input grid contains only '.' and one other color (the deity's color). The output is a 3x3 divine glyph in grey (X) fixed per deity. So each deity (color) maps to a specific 3x3 pattern of X and . (since output is in grey, i.e., X for glyph, . for background). The training examples show mapping from input color to output pattern.

We need to deduce mapping: For each color (B, R, G) we have examples.

Example 1: Input color B (blue). Output pattern:
. X .
X X X
. X .
That's a plus sign shape (cross) centered: middle row all X, middle column X, others . So pattern: cross.

Example 2: Input color R (red). Output:
X X X
. X .
. X .
That's like a vertical line of three X in column 1? Actually let's write coordinates (row, col) 0-index:
Row0: X X X
Row1: . X .
Row2: . X .
So it's a T shape? Actually top row full X, middle column X downwards. That's like a "T" shape (top bar and stem). Or maybe it's a "letter T" rotated? Actually typical T: top row three X, middle column down from center. Yes.

Example 3: Input color G (green). Output:
. . X
. . X
X X X
That's like an "L" shape? Actually bottom row full X, and right column two X above? Let's see: coordinates:
Row0: . . X
Row1: . . X
Row2: X X X
So it's a reversed L? Actually shape: bottom row full X (three), and right column has X at top two rows. That's like a "Γ" shape (like an L rotated). Could be a "corner" shape.

Thus each color maps to a specific 3x3 pattern: B -> plus (+), R -> T (top bar + stem), G -> L (bottom bar + right side). Need to verify if these are consistent with some rule based on the arrangement of the sacred color bones in the 5x5 input.

We need to find transformation rule: given positions of the sacred color (non-black) in 5x5 grid, produce a 3x3 glyph pattern (maybe based on some shape like bounding box, or some directional property). The description: "Read the dye color to identify the deity and inscribe the sacred glyph — each god commands one eternal symbol, indifferent to how the bones fell." So the glyph is fixed per deity, independent of how the bones fell (i.e., independent of arrangement). So the input arrangement does not matter; only the color matters. So we just need to map color to glyph.

Thus for test input, we need to identify the sacred color present. Look at test input:

. G . G G
. . G . .
G . . . .
. . G . G
. . . . G

We see only '.' and 'G' (green). So deity is green. According to example 3, green maps to pattern:
. . X
. . X
X X X

Thus output should be that.

But we need to double-check that there is no other color present. Indeed only G appears. So answer is that pattern.

Thus final answer:

ANSWER:
. . X
. . X
X X X

But we need to output with spaces between characters: each row: ". . X" etc.

Thus final.

We need to understand the transformation: Input is a 5x5 grid with one slipper color (i.e., only one color appears besides black '.'?), and output is a 3x3 grid in grey (X) representing a compact formation each company always performs, regardless of starting positions. So each color corresponds to a "company" (dancers wearing that color slippers). The input shows scattered positions of that color on a 5x5 stage. The output is a 3x3 pattern of X's (grey) that is the signature lift formation for that company, unique per company.

Thus we need to infer mapping from color to 3x3 pattern.

We have three training examples: color B (blue) yields output pattern:

Example1 Input B pattern: Let's list coordinates (row, col) 0-index.

Input1:
Row0: B . B . .
Row1: B . . B B
Row2: B B . B .
Row3: . B . B .
Row4: B . . . B

Let's list B positions:
(0,0), (0,2)
(1,0), (1,3), (1,4)
(2,0), (2,1), (2,3)
(3,1), (3,3)
(4,0), (4,4)

So many B's.

Output1:
. X .
X X X
. X .

That's a plus sign shape: center (1,1) X, plus up (0,1), down (2,1), left (1,0), right (1,2). So pattern: cross.

Example2: color R (red) yields output:

Input2:
R . R . R
R . . . R
R R . . .
R . . R R
R R R . R

List R positions:
Row0: (0,0),(0,2),(0,4)
Row1: (1,0),(1,4)
Row2: (2,0),(2,1)
Row3: (3,0),(3,3),(3,4)
Row4: (4,0),(4,1),(4,2),(4,4)

Output2:
X X X
. X .
. X .

That's a shape: top row all X (0,0-2), middle row center X (1,1), bottom row center X (2,1). So shape like a T? Actually it's like a vertical line in column1 plus a top horizontal bar across columns0-2 at row0. So shape: like an upside-down T? Let's draw:

Row0: X X X
Row1: . X .
Row2: . X .

That's a shape: a vertical line of three X in column1 (rows0-2) plus extra X at (0,0) and (0,2) making top row full. So shape like a "T" with top bar full width and stem down middle.

Example3: color G (green) yields output:

Input3:
. G . G .
G G . . .
. G . . .
. . G . .
G G G . .

List G positions:
Row0: (0,1),(0,3)
Row1: (1,0),(1,1)
Row2: (2,1)
Row3: (3,2)
Row4: (4,0),(4,1),(4,2)

Output3:
. . X
. . X
X X X

That's shape: bottom row all X (row2 col0-2), and two X above in column2 at rows0 and1? Actually output:
Row0: . . X
Row1: . . X
Row2: X X X

So shape: a vertical line in column2 (rows0-2) plus a horizontal bar at bottom row across columns0-2. That's like an upside-down T? Actually it's a "T" rotated 180? Let's see: vertical line in column2, bottom horizontal bar across row2. So shape like a "T" with stem up? Actually typical T: top horizontal bar and vertical stem down. Here we have bottom horizontal bar and vertical stem up (since column2 vertical line goes up from bottom to top). So it's an upside-down T.

Thus each color maps to a specific 3x3 pattern: B -> plus sign (cross). R -> T shape with top bar and stem down (like an upright T). G -> T shape with bottom bar and stem up (inverted T). So maybe each color corresponds to a different orientation of T or plus.

We need to deduce rule: The pattern is the "signature lift" - a compact formation each company always performs, regardless of starting positions. So perhaps the pattern is the convex hull or minimal bounding shape? Or maybe it's the shape formed by taking the positions of the dancers and computing their "center of mass" and then drawing a shape based on relative positions? But given only one color, the pattern is independent of the exact positions; it's just a function of the color.

Thus we need to map each color to a pattern. We have three examples: B->plus, R->T_up (stem down), G->T_down (stem up). There are 7 colors total (including black? Actually black is background). Colors: B, R, G, Y, X, M, O, A, W. But X is grey used for output. Input only has one color besides '.'; we have seen B,R,G. So we need to infer pattern for other colors maybe not needed for test input (which is G again? Actually test input uses G? Let's see test input:

Test Input:
. G . G G
. . G . .
G . . . .
. . G . G
. . . . G

Yes it's G again (green). So we need to output the pattern for G, which from example3 is:

. . X
. . X
X X X

Thus answer should be that.

But we need to verify that the pattern for G is indeed that. Let's double-check example3 input and output.

Example3 Input:
. G . G .
G G . . .
. G . . .
. . G . .
G G G . .

Let's plot to see shape of G positions: maybe they form some shape that yields that output.

But we trust the given mapping.

Thus answer: 
. . X
. . X
X X X

But we need to output with spaces between characters: each row separated by spaces.

Thus final answer block:

ANSWER:
. . X
. . X
X X X

But we need to ensure we follow format: "one row per line, characters separated by spaces". So we need to put spaces between each character.

Thus:

. . X
. . X
X X X

Let's produce final.

We need to understand the transformation: Input is a 5x5 grid with one ore color (i.e., only one non-black color appears? Actually they say "One mineral's colored deposits scattered through dark rock." So the input grid has black '.' and one other color (the ore color). The output is a 3x3 crystal lattice in grey (X) and black (.) presumably, fixed per mineral (i.e., per ore color). So each ore color maps to a specific 3x3 pattern of X and . (grey and black). The training examples show mapping for B (blue), R (red), G (green). Let's examine.

Example 1: Input with B (blue) ore. Output:
. X .
X X X
. X .
That's a plus shape: center and middle of edges? Actually pattern: row0: . X . (center top?), row1: X X X (full row), row2: . X . (center bottom). So it's a cross shape: a vertical line through column1 and a horizontal line through row1, making a plus.

Example 2: Input with R (red) ore. Output:
X X X
. X .
. X .
That's a shape: top row all X, middle column X in rows1 and2? Actually row0: X X X (full top row). row1: . X . (center column). row2: . X . (center column). So shape: a T? Actually it's like a "┤"? Let's visualize: top horizontal bar, and a vertical line down the middle column from top to bottom? Actually top row all X, then column1 has X in rows1 and2. So shape like an upside-down T? Actually it's like a "┤"? Let's draw: coordinates (r,c):
(0,0) X, (0,1) X, (0,2) X
(1,0) ., (1,1) X, (1,2) .
(2,0) ., (2,1) X, (2,2) .
So it's a shape: top bar plus a vertical line down the middle from row0 to row2? Actually row0 already has X at (0,1) as part of top bar; then (1,1) and (2,1) extend down. So shape is like a "┤"? Actually it's like a "┤" rotated? Not important.

Example 3: Input with G (green) ore. Output:
. . X
. . X
X X X
That's shape: bottom row all X, and column2 has X in rows0-2? Actually row0: . . X (rightmost column). row1: . . X (rightmost column). row2: X X X (full bottom row). So shape like an "└"? Actually it's like an L shape rotated? It's a shape: a vertical line on right column plus a horizontal line at bottom row.

Example 4: Input with B (blue) ore again (different arrangement). Output same as example1: . X . / X X X / . X . So B maps to plus shape.

Thus mapping:
- B (blue) -> plus shape (cross)
- R (red) -> shape: top row full + middle column down (like a "┤"? Actually shape: top bar + vertical line down middle)
- G (green) -> shape: bottom row full + right column up (like an "└"? Actually shape: bottom bar + vertical line on right)
We need to infer pattern: maybe the shape depends on the ore color's position in some ordering? Or maybe it's based on the mineral's crystal habit: each mineral has a fixed lattice shape independent of vein placement. So each color corresponds to a specific 3x3 pattern.

We have three colors observed: B, R, G. Need to deduce mapping for any color? The test input uses G (green) ore. So we need to output the pattern for G. According to example3, G maps to:
. . X
. . X
X X X

Thus test input should produce same pattern? Let's verify test input: It's a 5x5 with G ore scattered. According to rule, output should be same as example3: . . X / . . X / X X X.

But we need to be sure that the mapping is consistent regardless of arrangement of ore in input. Example1 and Example4 both B but different arrangements gave same output. Example2 only R example. Example3 only G example. So seems mapping is fixed per color.

Thus answer for test input (G) is as above.

But we need to perform Conceptual Digestion: identify CAST, AFFORDANCES, TRANSFORMATION RULE.

Let's articulate:

CAST: Elements: black '.' (dark rock), ore color (single non-black color representing mineral deposits). Output: grey 'X' (crystal lattice) and black '.' (empty). The ore color determines which lattice pattern.

AFFORDANCES: The ore color can be "read" to select a predetermined 3x3 pattern. The positions of ore in input do not affect output; only the color matters. The pattern is a fixed lattice shape unique to that mineral.

TRANSFORMATION RULE: For each input grid, identify the unique non-black color present (if any). Map that color to a predefined 3x3 pattern of X and . as follows:
- Blue (B): plus shape (center cross)
- Red (R): top row full + middle column down
- Green (G): bottom row full + right column up
(Alternatively could be described as: B -> pattern where X at (0,1),(1,0),(1,1),(1,2),(2,1); R -> X at (0,0-2),(1,1),(2,1); G -> X at (0,2),(1,2),(2,0-2).)

Apply to test input: ore color is G, thus output pattern for G.

Thus answer grid:
. . X
. . X
X X X

We need to output in the format with spaces between characters.

Thus final answer:

ANSWER:
. . X
. . X
X X X

But ensure we use single-character color codes: '.' for black, 'X' for grey. Yes.

Let's double-check if any nuance: The output is 3x3 crystal lattice in grey, fixed per mineral. So only X and . appear. Good.

Thus final.

We need to understand the puzzle: Input is a 5x5 grid with black '.' and one colored spice (some color) scattered randomly. Output is a 3x3 grid in grey (X) and black (.)? Actually output is "plated garnish in grey, shape determined by spice identity." The output uses only '.' and 'X' (grey). The training examples show output patterns of X and . (no other colors). So the output is a 3x3 pattern of X's (grey) and . (black) representing the garnish shape.

We need to map each spice color (B, R, G, Y, X, M, O, A, W) to a specific 3x3 pattern (the "crystallization pattern").

We have three training examples: Example 1: Input spice is B (blue). Output pattern:
. X .
X X X
. X .
That's a plus sign shape (cross) centered.

Example 2: Input spice is R (red). Output:
X X X
. X .
. X .
That's like a T shape? Actually top row all X, middle column X in rows 2 and 3? Let's see: Row0: X X X (full top bar). Row1: . X . (center). Row2: . X . (center). So shape is a "T" rotated? Actually it's a vertical line down the middle column with a top bar across. That's like a "T" shape (top bar and stem). Yes.

Example 3: Input spice is G (green). Output:
. . X
. . X
X X X
That's like an L shape? Actually bottom row full X, and right column X in rows 0-2? Let's see: Row0: . . X (rightmost). Row1: . . X (rightmost). Row2: X X X (full bottom). That's like a reversed L? Actually shape is a corner: bottom row and right column forming an upside-down L? Wait bottom row is full across, right column is up. That's like a "Γ" shape (like a right angle with the corner at bottom-right). Actually bottom row full, right column full: shape is like a "┘" shape? Let's draw coordinates: (0,2) X, (1,2) X, (2,0) X, (2,1) X, (2,2) X. So it's like an L rotated 180? Actually typical L shape would be bottom row and left column. This is bottom row and right column: shape like a "└"? Actually bottom row full, right column up: shape like a "└" rotated? Let's think: If we have bottom row full (row2 all X) and right column full (col2 all X), the overlapping cell (2,2) counted once. So shape is like a "┘"? Actually shape looks like a "┘" if we consider the corner at bottom-right? The shape includes the bottom row and the right column, making a shape like a reversed "L" (like a "ㄱ" rotated?). Anyway.

Thus each spice color maps to a specific 3x3 pattern.

We need to deduce mapping for other colors maybe not needed for test input: test input spice is G (green) again? Let's check test input: It has G's scattered. Yes test input uses G (green). So we need to output the pattern for green spice, which we already have from Example 3: pattern:
. . X
. . X
X X X

Thus answer should be that.

But we need to verify that the mapping is consistent: Example 1 B -> plus sign. Example 2 R -> T shape. Example 3 G -> L shape (bottom row + right column). So test input G -> same pattern.

Thus answer: 
. . X
. . X
X X X

We need to output in the format with spaces between characters.

Thus:
ANSWER:
. . X
. . X
X X X

But we need to ensure we follow exactly: "one row per line, characters separated by spaces". So each line: ". . X". Yes.

Thus final.

We need to understand the transformation: Input is a 5x5 grid with one ink color (i.e., only one non-black color appears? Actually description: "Random notes written in a single instrument's color of ink splashed across a dark staff." So input grid has black background '.' and one other color (the instrument's ink). The output is a 3x3 riff pattern in grey (X) fixed per instrument. So each instrument (color) maps to a specific 3x3 pattern of X and . (black). The training examples show mapping for B (blue), R (red), G (green). Let's examine.

Example 1: Input B pattern yields output:
. X .
X X X
. X .
That's a plus sign shape (cross) of X's: center row all X, center column all X, others . (like a cross). So B maps to cross.

Example 2: Input R pattern yields output:
X X X
. X .
. X .
That's a vertical line of three X's in column 0? Actually first row: X X X (all three columns X). Second row: . X . (center X). Third row: . X . (center X). So shape: top row full X, and a vertical line down the middle column (col 1) for rows 1-2. That's like a "T" shape? Actually top row full, middle column down: shape like a "T" rotated? Let's visualize: coordinates (r,c) 0-index:
(0,0) X, (0,1) X, (0,2) X
(1,0) ., (1,1) X, (1,2) .
(2,0) ., (2,1) X, (2,2) .
So it's a "T" shape with the top bar full and stem down middle.

Example 3: Input G pattern yields output:
. . X
. . X
X X X
That's like an "L" shape? Actually bottom row full X, and right column X in rows 0-1? Let's see: (0,0)., (0,1)., (0,2) X
(1,0)., (1,1)., (1,2) X
(2,0) X, (2,1) X, (2,2) X
So shape: a reversed L? Actually it's like a "Γ" shape (bottom row full, right column up two). That's like a corner shape.

Thus each color maps to a specific 3x3 pattern.

We need to deduce mapping from given examples: B -> cross, R -> T shape (top bar full + middle column down), G -> reversed L shape (bottom row full + right column up). We need to confirm by analyzing input patterns to see if they correspond to instrument's "signature riff" independent of what was written. The description: "Recognize the instrument by its ink color and play its signature riff — a compact motif that never changes regardless of what was written on the page." So the input pattern is irrelevant; only the color matters. So we just need to map color to output pattern.

Thus for test input, we need to see what color appears. Test input:
. G . G G
. . G . .
G . . . .
. . G . G
. . . . G

We see only G (green) and '.' (black). So instrument is green. According to example 3, green maps to:
. . X
. . X
X X X

Thus output should be that pattern.

But let's double-check: Example 3 input had green pattern and output as above. So yes.

Thus answer: 
. . X
. . X
X X X

We need to output with spaces between characters. So each row: ". . X" etc.

Thus final.

We need to understand the transformation: Input is a 5x5 grid with one jersey color scattered (i.e., only one color appears besides black '.'?), and output is a 3x3 formation diagram in grey (X) unique to that team.

We have examples:

Example 1: Input has B (blue) scattered. Output is a 3x3 pattern: . X . / X X X / . X . (a plus sign shape). So for B, output is a cross (center plus arms). Example 2: Input has R (red) scattered. Output: X X X / . X . / . X . (a T shape? Actually top row all X, middle column X at (1,0) and (2,0)? Wait coordinates: 3x3 grid rows 0-2, columns 0-2. Output:
Row0: X X X
Row1: . X .
Row2: . X .
That's a shape: top bar full, and a vertical line down the middle column (col1) from row0 to row2? Actually row0 col1 is X (part of top bar), row1 col1 X, row2 col1 X. So shape is like a "T" rotated? Actually it's a "┤"? Let's visualize: top row all X, middle column downwards X. That's like a "T" shape with the stem at middle column going down. Example 3: Input has G (green) scattered. Output:
. . X
. . X
X X X
That's like an "L" shape? Actually bottom row all X, and right column up two X's at (0,2) and (1,2). So shape is like a reversed L? Actually bottom row full, and a vertical line on right column going up two cells. That's like a "└" shape? Wait bottom row all X (row2: X X X). Right column (col2) has X at row0 and row1. So shape is like a "┤"? Actually it's like a "┘"? Let's draw coordinates: (0,0) ., (0,1) ., (0,2) X; (1,0) ., (1,1) ., (1,2) X; (2,0) X, (2,1) X, (2,2) X. So shape is like a reversed L: bottom bar full, and a vertical line on right side going up two cells. That's like a "└" rotated? Actually shape looks like a "┤" but missing the left side? Let's think: It's like a "corner" shape where the corner is at bottom-right? Actually bottom row full from col0 to col2, and right column from row0 to row2 (but row2 col2 already part of bottom row). So shape is like a "┬"? No.

Let's see pattern: For each color, output is a specific 3x3 pattern of X's (grey) unique to that team. So we need to infer mapping from color to pattern based on training examples.

We have three examples: B -> plus shape (cross). R -> shape: top row full, middle column down (like a "T" missing left and right arms? Actually top row full, and a vertical line down middle column). G -> shape: bottom row full, right column up (like a reversed L? Actually bottom row full, and a vertical line on right column going up two cells). Let's denote patterns:

Define 3x3 grid coordinates (r,c) with r=0 top, r=2 bottom; c=0 left, c=2 right.

Pattern for B (blue):
(0,1) X
(1,0) X, (1,1) X, (1,2) X
(2,1) X
So it's a plus: center plus up, down, left, right.

Pattern for R (red):
(0,0) X, (0,1) X, (0,2) X
(1,1) X
(2,1) X
So top row full, and middle column down (excluding top? Actually top row includes middle column already). So shape: a horizontal bar at top, and a vertical line down the middle column from row0 to row2 (but row0 col1 already part of top bar). So essentially it's a "T" shape with the stem going down.

Pattern for G (green):
(0,2) X
(1,2) X
(2,0) X, (2,1) X, (2,2) X
So bottom row full, and right column up (excluding bottom? bottom row includes (2,2) already). So shape: a horizontal bar at bottom, and a vertical line up the right column from row0 to row2 (but row2 col2 part of bottom bar). So shape like a "└"? Actually it's like a reversed L: bottom bar full, and a vertical line on right side going up.

Now test input: we need to determine which color is scattered. The test input grid:

. G . G G
. . G . .
G . . . .
. . G . G
. . . . G

Let's write it row by row:

Row0: . G . G G
Row1: . . G . .
Row2: G . . . .
Row3: . . G . G
Row4: . . . . G

We see only G (green) and '.' (black). No other colors. So the jersey color is G (green). According to example 3, G maps to pattern:

. . X
. . X
X X X

Thus output should be that pattern.

But we need to double-check if any nuance: The description says "Identify the jersey color and call the team's one and only formation — a compact tactical shape burned into their playbook." So each color corresponds to a specific formation (shape). The training examples show that for each color, the output is a specific 3x3 pattern of X's (grey). So we just need to map color to pattern.

Thus for G, output is as above.

But we should verify that the pattern is unique per color and not dependent on arrangement of jerseys in input (just presence of that color). In examples, the input arrangement varied but same color gave same output. So yes.

Thus answer: 
. . X
. . X
X X X

We need to output in the format with spaces between characters: each row: characters separated by spaces.

Thus:
. . X
. . X
X X X

Now we need to write final answer with reasoning and then the ANSWER block.

Let's produce final.

We need to understand the transformation: Input is a 5x5 grid with scattered cells of a single color (front) and the rest black (.)? The description: "A weather map with scattered storm cells all tinted in one front's characteristic color." So input grid has only one color (non-black) scattered, representing a front type. Output is a 3x3 pressure signature in grey (X) fixed per front type. So each front type (color) maps to a specific 3x3 pattern of X and . (grey and black). The training examples show mapping for B (blue), R (red), G (green). Let's extract patterns.

Example 1: Input B scattered. Output:
. X .
X X X
. X .
That's a plus sign shape (cross) with center and middle of edges? Actually pattern: row0: . X . ; row1: X X X ; row2: . X . . That's a cross (vertical and horizontal line) of X's.

Example 2: Input R scattered. Output:
X X X
. X .
. X .
That's a shape: top row all X, middle column X at (1,1) and (2,1). So like a T shape? Actually it's like a "┤"? Let's draw: coordinates (r,c):
(0,0) X, (0,1) X, (0,2) X
(1,0) ., (1,1) X, (1,2) .
(2,0) ., (2,1) X, (2,2) .
So shape: top bar and a vertical line down middle from top to bottom? Actually vertical line only at column1 rows0-2? Yes column1 has X at rows0,1,2. So it's like a "T" rotated? Actually it's a "┤"? No it's like a "⊥"? Let's think: top row full X, and a vertical line down the middle column. That's like a "┤"? Actually shape looks like a "┤" missing bottom? Wait top row full X, vertical line down middle column: that yields shape like a "┤"? Actually if you have top row full X, and a vertical line down middle column, you get shape like a "┤"? Let's draw: top row: XXX ; middle column: X at (0,1) already part of top row, plus X at (1,1) and (2,1). So shape is like a "┤"? Actually shape is like a "┤" rotated 90 degrees? Not important.

Example 3: Input G scattered. Output:
. . X
. . X
X X X
That's shape: bottom row all X, and a vertical line on right column (col2) rows0-2? Actually column2 has X at rows0,1,2? Yes column2: row0 X, row1 X, row2 X. So shape: bottom bar and right vertical line, like an "ㄱ" shape? Actually it's like an "L" rotated? It's like a corner shape: bottom row full X, and right column full X, overlapping at bottom-right corner (2,2). So shape is like a reversed L? Actually it's like a "└"? Wait bottom row full X across columns0-2, and right column full X rows0-2. Overlap at (2,2). So shape looks like a "┤"? Actually it's like a "┤" rotated? Let's think: shape is like a "┤" but with the bar at bottom and stem on right? Actually typical "┤" is vertical line on right with a horizontal bar to the left at the middle? Not exactly.

Anyway we need to infer mapping from front color to pattern. We have three examples: B -> cross (+). R -> shape: top bar + middle column. G -> shape: bottom bar + right column.

We need to deduce rule: maybe pattern is based on some property of the front's distribution in the 5x5 input: like the shape of the minimal bounding box or something? The description: "Read the color of the front and draw its invariant pressure signature — a small fixed pattern the atmosphere always stamps down for that front type." So pattern is fixed per front type, not dependent on arrangement. So we just need to map color to pattern. We have three colors: B, R, G. We need to infer pattern for each. Already have patterns for B,R,G from examples. So for test input, which is G scattered (green). So output should be same as example 3 pattern: . . X / . . X / X X X.

But we need to verify that test input indeed is green front. Let's examine test input:

. G . G G
. . G . .
G . . . .
. . G . G
. . . . G

We see many G's scattered. Yes only G appears besides '.'.

Thus output should be the pattern for G: as per example 3.

Thus answer:

. . X
. . X
X X X

But we need to output with spaces between characters: each row: ". . X" etc.

Thus final answer block.

But before final, we should do conceptual digestion as requested: Identify CAST, AFFORDANCES, TRANSFORMATION RULE. We'll include reasoning in answer? The instruction: "Think step by step. After your reasoning, you MUST write out the complete predicted output grid using these single-character color codes". So we need to provide reasoning then final answer block. We'll include the conceptual digestion steps in reasoning.

Let's produce final answer.

We need to understand the transformation. Input: 5x5 grid with one colored brick type (i.e., only one color appears besides black '.'? Actually description: "Rubble of one type of glazed brick scattered across a dark foundation." So the input grid has only one color (non-black) representing the brick type, and the rest are black '.' (foundation). The output is a 3x3 keystone pattern in grey (X) determined by glaze (i.e., the color of the brick). So we need to map the color of the brick to a 3x3 pattern of X's and '.' (black). The training examples show mapping for B (blue), R (red), G (green). Let's examine.

Example 1: Input B bricks pattern yields output:
. X .
X X X
. X .

That's a plus sign shape (cross) with center and middle of edges? Actually pattern: row0: . X . (center column X), row1: X X X (full row), row2: . X . (center column). So it's a cross (vertical and horizontal line) of width 1.

Example 2: Input R bricks pattern yields output:
X X X
. X .
. X .

That's a shape: top row all X, middle column X at (1,1) and (2,1). So shape like a T? Actually top row full, and a vertical line down the middle column from row1 to row2. So shape: a "T" rotated? Actually it's like a "T" shape with top bar and stem down.

Example 3: Input G bricks pattern yields output:
. . X
. . X
X X X

That's shape: bottom row full X, and a vertical line up the rightmost column? Actually column2 (0-index) has X at rows0,1,2? Let's see: row0: . . X (col2 X), row1: . . X (col2 X), row2: X X X (all three). So shape like an upside-down T? Actually bottom bar full, and a stem up the middle? Wait stem is at column2 (rightmost) not middle. Actually bottom row full (columns0-2), and a vertical line at column2 going up two rows. That's like an "L" shape? Actually it's a shape like a "┘" rotated? Let's think: bottom row full, and a column at rightmost going up. That's like a reversed "ㄱ" shape? Actually shape: a horizontal bar at bottom, and a vertical bar on the right side going up. That's like a "┘" shape? Actually ┘ is bottom left corner? No.

Let's draw coordinates: (r,c). Output 3x3.

Example1 B: X at (0,1), (1,0),(1,1),(1,2),(2,1). That's a plus.

Example2 R: X at (0,0),(0,1),(0,2), (1,1), (2,1). That's top row full, and middle column down two.

Example3 G: X at (0,2),(1,2), (2,0),(2,1),(2,2). That's bottom row full, and right column up two.

So pattern seems to be: depending on color, the output is a 3x3 pattern where the brick color determines which arms of a cross are present? Actually plus shape has arms up, down, left, right. T shape (example2) has arms up? Actually top bar plus stem down: that's missing left and right arms? Actually plus shape missing left and right? Wait plus shape has left, right, up, down. T shape (example2) has up? Actually top bar is like up? Let's think: The plus shape has arms in all four directions from center. The T shape (example2) has arms up? Actually top bar is like a horizontal line at top row, which could be thought of as the "up" arm extended left and right? Not exactly.

Maybe the rule: The output pattern is the shape of the letter corresponding to the color's first letter? B -> plus? Not letter B. R -> T? Not letter R. G -> L? Not letter G.

Maybe it's about the shape of the brick's distribution in the input? The input pattern of bricks might indicate which direction to carve. The output is a 3x3 keystone pattern determined by glaze (color). So maybe each color maps to a specific pattern irrespective of input arrangement? The training examples show each color yields a specific pattern: B -> plus, R -> T (top bar + stem down), G -> reversed L? Actually bottom bar + stem up right.

Let's verify if the input patterns correspond to something like the color's position in some order? Could be that the pattern of bricks in the 5x5 input indicates a direction (like which side the bricks are more concentrated) and the output pattern is a shape pointing that direction. But the description says "Identify the glaze color and carve the keystone pattern that clay was always meant to form." So the glaze color determines the pattern, not the arrangement. So we just need to map color to pattern.

Thus we need to deduce mapping for colors B, R, G from examples, and then apply to test input which has G bricks (green). So output should be same as example3: pattern for G.

Test input:
. G . G G
. . G . .
G . . . .
. . G . G
. . . . G

We see green bricks scattered. So output should be the G pattern: . . X / . . X / X X X.

Thus answer:
. . X
. . X
X X X

But we need to output with spaces between characters: each row: ". . X" etc.

Let's double-check if any nuance: The output is 3x3 keystone pattern in grey (X). So '.' remains black. So pattern as above.

Thus final answer.

But before final, let's ensure we haven't missed any alternative interpretation: Could be that the pattern depends on both color and orientation of brick distribution? But the description says "Ide

... [258 chars truncated]

We need to understand the transformation: Input is a 5x5 grid with one colored force (i.e., only one color appears besides black '.'?), and we need to output a 3x3 grid in grey (X) representing a predetermined siege formation per kingdom (color). So each color (B, R, G, Y, M, O, A, W) maps to a specific 3x3 pattern of X's (grey) and . (black). The training examples show mapping for B, R, G.

We need to infer the pattern for each color from examples.

Example 1: Input with B's (blue). Output:
. X .
X X X
. X .
That's a plus sign shape (cross) centered: middle row all X, middle column all X, corners . . So pattern: cross.

Example 2: Input with R's (red). Output:
X X X
. X .
. X .
That's a shape: top row all X, middle column X at (1,1) and (2,1) (0-index?), actually rows: row0: X X X; row1: . X .; row2: . X . So shape: a T? Actually it's like a vertical line down middle column plus a top bar across top row. That's like an upside-down T? Wait vertical line: column1 (middle) has X at rows0,1,2. Top row also has X at columns0,1,2. So shape: a "T" rotated? Actually it's like a "┤"? Let's visualize: coordinates (r,c):
(0,0) X, (0,1) X, (0,2) X
(1,0) ., (1,1) X, (1,2) .
(2,0) ., (2,1) X, (2,2) .
So shape: a vertical line in middle column plus a top horizontal bar across top row. That's like a "⊤" (T) shape? Actually T shape is a horizontal bar on top and vertical line down from middle: yes that's exactly a T shape (like capital T). So red maps to T shape.

Example 3: Input with G's (green). Output:
. . X
. . X
X X X
That's shape: bottom row all X, and right column X at rows0,1,2? Actually column2 (rightmost) has X at rows0,1,2? Let's see: row0 col2 X, row1 col2 X, row2 col0 X, col1 X, col2 X. So shape: an "L" shape? Actually it's like a reversed L? Let's draw: bottom row full X across columns0-2. Right column full X across rows0-2. Overlap at bottom-right corner (2,2) counted twice. So shape: a "┘" shape? Actually it's like a rotated L: a shape that looks like a corner: bottom horizontal and right vertical. That's like a "┘" (like an upside-down L? Actually typical L shape is vertical on left and horizontal at bottom. This is vertical on right and horizontal at bottom: that's a "ㄱ" shape rotated? It's like a "┘" shape (like a right angle with the interior at top-left? Actually shape occupying bottom row and right column forms a shape like a "ㄱ" rotated 90 degrees clockwise? Let's think: If you have a shape that occupies the bottom row and the left column, that's like a "ㄱ" (like an L). Here it's bottom row and right column, which is like a "ㄱ" mirrored horizontally (like a backwards L). So green maps to a shape: bottom row + right column.

Thus we have mapping:
B -> plus (+) shape.
R -> T shape (top bar + vertical down middle).
G -> bottom row + right column (like an L shape mirrored).

We need to infer mapping for other colors maybe not needed for test input (which is G). Test input appears to have G's (green). So we need to output the green pattern: bottom row + right column.

But we need to verify that the test input indeed only contains G and '.' (black). Let's examine test input:

. G . G G
. . G . .
G . . . .
. . G . G
. . . . G

Yes only G and '.'.

Thus output should be the green pattern: a 3x3 grid with bottom row all X, and right column all X.

Let's produce that:

Rows 0-2, columns 0-2.

Bottom row (row2): X X X
Right column (col2): rows0,1,2: X at (0,2), (1,2), (2,2) (but (2,2) already X from bottom row).

Thus grid:
Row0: . . X
Row1: . . X
Row2: X X X

Matches example 3 output exactly. Indeed example 3 output was:
. . X
. . X
X X X

Thus test input should produce same output.

Thus answer: 
ANSWER:
. . X
. . X
X X X

But we need to follow format: each row characters separated by spaces. So:

. . X
. . X
X X X

Let's output that.

We need to understand the transformation. Input: 5x5 grid with one mask color scattered (i.e., only one color appears? Actually "one mask color scattered" meaning the input grid contains only one color (like B, R, G) plus black '.'? The training examples: Example 1 input has B and '.' only. Example 2 input has R and '.' only. Example 3 input has G and '.' only. So indeed each input uses a single color (B, R, G) plus '.' (black). Output is a 3x3 tableau in grey (X) and black (.)? Actually output uses X (grey) and '.' (black). The output is unique per troupe (color). So each color maps to a specific 3x3 pattern of X and '.'.

Thus we need to deduce the mapping from color to pattern based on examples.

Example 1: Input color B (blue) yields output:
. X .
X X X
. X .

That's a plus sign shape: center row all X, center column X, making a cross.

Example 2: Input color R (red) yields output:
X X X
. X .
. X .

That's a vertical line of three X in column 1? Wait let's write coordinates: rows 0-2, columns 0-2.

Row0: X X X (all X)
Row1: . X .
Row2: . X .

So pattern: top row full X, middle column X (positions (1,1) and (2,1)). Actually column 1 (middle column) has X in rows 0,1,2? Row0 col1 is X (since row0 all X). Row1 col1 X, row2 col1 X. So it's like a T shape? Actually top row full X plus a vertical line down the middle column. That yields shape like a "T" rotated? Actually it's like a "┤"? Let's visualize: top row: XXX; middle column: X at (0,1) already part of top row, plus X at (1,1) and (2,1). So shape is like a "┤"? Actually it's like a "⊢"? Not important.

Example 3: Input color G (green) yields output:
. . X
. . X
X X X

That's like an L shape? Actually bottom row all X, plus right column X in rows 0 and 1? Let's see: Row0: . . X (X at col2). Row1: . . X (X at col2). Row2: X X X (all X). So shape: bottom row full X plus right column X (col2) rows0-1. That's like a reversed L? Actually it's like a "└" shape? Wait bottom row full X plus a vertical line on right column upward: shape like a "┘"? Actually bottom row full X is like a horizontal line at bottom; plus a vertical line on right column going up two cells yields shape like a "└"? Let's draw: coordinates: (0,2) X, (1,2) X, (2,0) X, (2,1) X, (2,2) X. That's like a "ㄱ" shape rotated? Actually it's like a corner shape: bottom row plus right column.

Thus each color yields a distinct 3x3 pattern: B yields cross (+), R yields top bar plus middle column (like a "T" shape?), G yields bottom bar plus right column (like an "L" shape rotated?). Let's verify.

But we need to confirm mapping: maybe the pattern is determined by the color's position in some order? Or maybe it's based on the color's hue? Or maybe it's based on the letter? Let's see: B -> plus sign. R -> shape with top row full and middle column. G -> shape with bottom row full and right column.

Maybe the pattern is like: For each color, we take the shape of the letter representing the color? B -> plus? Not letter. R -> maybe shape of "R"? Not.

Alternatively maybe the pattern is derived from the positions of the color in the input? But input is scattered; maybe we compute some statistic like center of mass? The output is always 3x3, maybe we take the bounding box of the color positions and then map to a pattern? Let's test.

Example1 B positions: Let's list coordinates (row,col) where B appears.

Input1:
Row0: B . B . .
Row1: B . . B B
Row2: B B . B .
Row3: . B . B .
Row4: B . . . B

Let's compute all B positions:
(0,0), (0,2)
(1,0), (1,3), (1,4)
(2,0), (2,1), (2,3)
(3,1), (3,3)
(4,0), (4,4)

So many positions. Maybe we compute the convex hull? Or maybe we compute the "extent": min row, max row, min col, max col. Let's compute: min row =0, max row=4, min col=0, max col=4. So full 5x5 bounding box. Not helpful.

Maybe we compute the "center of mass" (average row, col). Let's compute approximate: sum rows: (0+0)+(1+1+1)+(2+2+2)+(3+3)+(4+4) = Actually let's sum: rows: 0,0,1,1,1,2,2,2,3,3,4,4 = sum = 0+0+1+1+1+2+2+2+3+3+4+4 = 23? Let's compute: 0+0=0; +1+1+1=3; +2+2+2=6 => total 9; +3+3=6 =>15; +4+4=8 =>23. Number of B's = count: let's count: row0 2, row1 3, row2 3, row3 2, row4 2 = total 12. Average row = 23/12 ≈1.9167. Sum cols: col0: positions (0,0),(1,0),(2,0),(4,0) =4; col2: (0,2),(2,2?) Actually (2,2) is '.'; wait row2 col2 is '.'; we have (0,2) and (??) maybe (2,2) not B; also (??) Let's list all B columns: from above: (0,0),(0,2); (1,0),(1,3),(1,4); (2,0),(2,1),(2,3); (3,1),(3,3); (4,0),(4,4). So columns: 0 appears at (0,0),(1,0),(2,0),(4,0) =4; column1 appears at (2,1),(3,1)=2; column2 appears at (0,2)=1; column3 appears at (1,3),(2,3),(3,3)=3; column4 appears at (1,4),(4,4)=2. Sum cols = 0*4 +1*2 +2*1 +3*3 +4*2 = 0+2+2+9+8=21. Average col =21/12=1.75. So center approx (1.92,1.75). Not integer.

Maybe we take the median? Median row: sorted rows: 0,0,1,1,1,2,2,2,3,3,4,4 => median between 6th and 7th (0-index) values: po

... [2646 chars truncated]

We need to understand the puzzle. Input: 5x5 grid with one seed color (i.e., only one color appears besides black? Actually they say "one seed color" meaning the input contains only one non-black color (the seed) and the rest black (.)? Let's check examples.

Example 1 input:
B . B . .
B . . B B
B B . B .
. B . B .
B . . . B

We see only B (blue) and . (black). Yes only one non-black color: B.

Output:
. X .
X X X
. X .

That's a 3x3 pattern of X (grey) and . (black). So output is a trellis pattern: a fixed shape per species (color). So each color (B, R, G, etc.) maps to a specific 3x3 pattern of X and . (maybe also other colors? but output only uses X and .). The mapping is: B -> pattern: . X . / X X X / . X . (like a plus sign? Actually it's a cross: center and middle row/col? Let's see: pattern:
Row0: . X .
Row1: X X X
Row2: . X .
That's a plus sign shape (vertical and horizontal line through center). So B maps to plus.

Example 2 input: only R (red) and . (black). Output:
X X X
. X .
. X .
That's pattern: top row all X, middle column X at (0,1),(1,1),(2,1) and bottom? Actually pattern:
Row0: X X X
Row1: . X .
Row2: . X .
That's like a T shape? Actually it's a shape: top horizontal bar and a vertical line down the middle column from top to bottom? Wait vertical line: column1 has X at rows0,1,2 (since row0 col1 X, row1 col1 X, row2 col1 X). So it's a T shape: top bar plus stem down.

Example 3 input: only G (green) and . (black). Output:
. . X
. . X
X X X
That's pattern: bottom row all X, and right column? Actually pattern:
Row0: . . X
Row1: . . X
Row2: X X X
That's like an L shape rotated? Actually it's a shape: bottom horizontal bar (row2 all X) plus a vertical bar on right column (col2) rows0-2? Wait column2: row0 X, row1 X, row2 X (yes). So it's like a reversed L? Actually it's a shape: a corner shape: bottom row and right column forming an "ㄱ" shape (like an L rotated). So G maps to that shape.

Thus each color maps to a specific 3x3 pattern of X's (grey) and . (black). The patterns appear to be the "trellis" shape required for each flower species.

We need to deduce mapping for each color based on training examples: B -> plus, R -> T shape (top bar + stem), G -> L shape (bottom bar + right stem). There may be other colors in test input? Test input uses G only? Let's see test input:

. G . G G
. . G . .
G . . . .
. . G . G
. . . . G

We see only G and . (black). Yes only green seeds. So we need to output the trellis pattern for G, which from example 3 is:

. . X
. . X
X X X

Thus answer should be that pattern.

But we need to verify that the pattern is indeed fixed per species regardless of seed arrangement. The examples show that regardless of where the seeds are placed (various patterns), the output is same shape for that color. So indeed mapping is independent of seed positions; just based on color.

Thus answer: For G, output as above.

But we need to double-check that the pattern for G is exactly as shown: Example 3 output:

. . X
. . X
X X X

Yes.

Thus answer:

. . X
. . X
X X X

We need to output with spaces between characters: each row: ". . X" etc.

Thus final answer block.

But before final, we should do the Conceptual Digestion as requested: identify CAST, AFFORDANCES, TRANSFORMATION RULE. Then apply.

We need to include reasoning and then final answer block.

Thus produce final answer with reasoning and then ANSWER block.

Let's craft.

CAST: elements: seed color (non-black) representing flower type; black background soil; output trellis cells: grey (X) for trellis presence, black (.) for absence.

AFFORDANCES: seeds can be read to identify color; color determines which trellis shape to construct; trellis shape is fixed per color; seeds positions irrelevant.

TRANSFORMATION RULE: Map input color to predetermined 3x3 pattern of X and .: B -> plus, R -> T, G -> L (bottom+right). For test input color G, output pattern as per G.

Thus answer.

Now produce final.

We need to parse the problem: It's a grid transformation puzzle. Input grid is given as a 5x? Actually the examples show 5 rows, each row with 9 columns? Let's see: Example 1 input:

. . . . . . . . .
. X X X . X . X .
. . X . . X X X .
. . X . . X X . .
. . . . . . . . .

That's 5 rows, each row has 9 entries (separated by spaces). Output:

B . B
B B B
B B .

That's 3 rows, 3 columns.

Thus the transformation extracts a 3x3 region from the input (the "seal") and applies a color mapping based on the "signet ring" (left artifact) which is a grey emblem (X) shape? The description: "A dark hall holds two artifacts side by side on a velvet display, separated by a thin shadow. On the left sits a signet ring — a grey emblem whose silhouette belongs to one of three ancient orders. On the right lies a blank wax seal, also shaped in grey, waiting to receive the ring's impression."

Thus the input grid likely contains two 3x3 patterns: left artifact (signet ring) and right artifact (blank wax seal), separated by a column of shadow (maybe a single column of '.'? Actually "thin shadow" maybe a column of '.'? The display case is a narrow corridor of darkness. The two artifacts always sit in the same positions within it, like twin pedestals in a museum wing, divided by a single strip of shadow. Only the branded seal — small and square — is carried out of the hall.

Thus the input grid is a 5x9 grid: maybe the outer border is darkness (.) and inside there are two 3x3 blocks left and right, separated by a column of shadow (maybe a column of '.'?), and there is a top and bottom row of darkness (.) as padding.

Thus we need to locate the left 3x3 block (signet ring) and the right 3x3 block (blank wax seal). The left block contains grey emblem (X) shape representing the signet ring silhouette belonging to one of three ancient orders. Each order claims a sacred color: likely mapping from the pattern of X's in the left block to a color (B, G, R? maybe blue, green, red). The right block is initially all grey (X) representing blank wax seal? Actually "blank wax seal, also shaped in grey" meaning the seal shape is grey (X) but empty inside? Wait: The wax seal is also shaped in grey, waiting to receive the ring's impression. So the seal shape is a grey outline (maybe a square of X's?) and inside it's darkness (.)? The description: "wherever the seal bears grey, it absorbs the order's color, and wherever the seal is empty darkness, it stays untouched." So the seal shape is grey (X) where it can absorb color; the empty darkness (.) stays unchanged (black). So the output is the seal shape (the right 3x3 block) but with X replaced by the order's color (depending on left block), and . remains . (black). So output is a 3x3 grid of colors: either the order's color (B, G, R) where the seal had X, and . where seal had . (darkness). The examples confirm: Example 1 output: B . B / B B B / B B . . Let's map to input.

Example 1 input left block? Let's extract left 3x3 block (columns 0-2? Actually there is a border of darkness around. Let's index columns 0-8. The pattern likely: rows 1-3 (since row0 and row4 are all .). Columns: left block maybe columns 1-3? Right block columns 5-7? With column 4 being the shadow (thin strip). Let's test.

Example1 input rows:

Row0: . . . . . . . . .
Row1: . X X X . X . X .
Row2: . . X . . X X X .
Row3: . . X . . X X . .
Row4: . . . . . . . . .

If we take columns 1-3 as left block: rows1-3 col1-3:

Row1: X X X
Row2: . X .
Row3: . X .

That's shape: top row XXX, middle .X., bottom .X. That's like a T? Actually shape: a vertical line in column2? Wait column indices: col1, col2, col3. Row1: X X X (all three). Row2: . X . (middle). Row3: . X . (middle). So shape is like a "T" or maybe a "signet ring" shape? Not sure.

Right block columns5-7: rows1-3 col5-7:

Row1: X . X
Row2: X X X
Row3: X X .

Let's extract: Row1 col5-7: X . X (since row1: . X X X . X . X . => indices:0 .,1 X,2 X,3 X,4 .,5 X,6 .,7 X,8 . => so col5 X, col6 ., col7 X). Row2: . . X . . X X X . => indices:0 .,1 .,2 X,3 .,4 .,5 X,6 X,7 X,8 . => col5 X, col6 X, col7 X. Row3: . . X . . X X . . => indices:0 .,1 .,2 X,3 .,4 .,5 X,6 X,7 .,8 . => col5 X, col6 X, col7 . So right block:

Row1: X . X
Row2: X X X
Row3: X X .

Now output is:

B . B
B B B
B B .

Thus mapping: where right block had X, output B; where ., output . (black). So the order's color is B (blue). So left block pattern corresponds to order that yields blue.

Example2 input:

. . . . . . . . .
. . X X . X . X .
. . X . . X . X .
. X X . . X X . .
. . . . . . . . .

Let's extract left block columns1-3 rows1-3:

Row1: . X X? Actually row1: . . X X . X . X . => indices:0 .,1 .,2 X,3 X,4 .,5 X,6 .,7 X,8 . => col1 .,2 X,3 X => left block row1: . X X
Row2: . . X . . X . X . => indices:0 .,1 .,2 X,3 .,4 .,5 X,6 .,7 X,8 . => col1 .,2 X,3 . => . X .
Row3: . X X . . X X . . => indices:0 .,1 X,2 X,3 .,4 .,5 X,6 X,7 .,8 . => col1 X,2 X,3 . =&g

... [4508 chars truncated]

We need to parse the problem: We have a 5x9 grid (5 rows, 9 columns). Two 3x3 colonies at fixed positions. The left colony is at some fixed location, the right colony at another fixed location. The left colony's pattern of living (grey) and dead (black) cells determines a "protein fold" shape, which maps to a pigment gene (color). The right colony's living cells (grey) become colored with that pigment, dead cells remain black (inert). The membrane and surrounding dish dissolve, leaving only the transformed right colony (3x3). So output is a 3x3 grid where each cell is either black (if dead in right colony) or colored (if alive in right colony) with the color determined by left colony's pattern.

Thus we need to infer mapping from left colony pattern (3x3 of X=grey living, .=black dead) to a color (one of B,R,G,Y,X? Actually X is grey living; but output colors are B,R,G,Y, etc. The left colony's pattern determines which pigment gene activates, i.e., which color. So left colony pattern maps to a single color (maybe one of B,R,G,Y, etc). Then we apply that color to all living cells in right colony.

Thus we need to deduce mapping from left colony patterns to colors based on training examples.

We have three training examples, each with input (5x9) and output (3x3). We need to extract left colony pattern (positions of X and . in left 3x3 region) and right colony pattern (positions of X and . in right 3x3 region). Then output shows colored cells where right colony had X (living) replaced by color letter, and . (dead) remain '.' (black). So we can infer the color used in each example by looking at output: all non-dot characters are same letter? Let's check.

Example 1 output:
B . B
B B B
B B .
We see letters: B and . only. So color is B (blue). All living cells in right colony turned blue.

Example 2 output:
G . G
G . G
G G .
Letters: G only. So color is G (green).

Example 3 output:
B . B
. B B
B . B
Letters: B only. So color is B (blue) again.

Thus left colony pattern determines color: either B or G in examples. Need to see mapping.

Now we need to locate left and right colonies positions in the 5x9 grid. The description: "Two cellular colonies float side by side in a dark petri dish, separated by a membrane channel. Each colony is a 3×3 cluster of living (grey) and dead (black) cells." Input grid is 5x9 dark dish, two 3x3 colonies at fixed positions. So likely the left colony occupies columns 0-2 (or 1-3?) and right colony columns 6-8? Since total width 9, two 3x3 colonies side by side with a membrane channel between them (maybe column 4?). Let's think: 5 rows, 9 columns. If each colony is 3x3, they'd occupy 3 rows each (maybe rows 1-3? because there is a border of dead cells around? The examples show pattern of '.' around the colonies.

Let's examine Example 1 input:

Row0: . . . . . . . . .
Row1: . X X X . X . X .
Row2: . . X . . X X X .
Row3: . . X . . X X . .
Row4: . . . . . . . . .

We see a pattern: there are X's in columns 1-3? Let's map coordinates (r,c) zero-index.

Row1: c1 X, c2 X, c3 X, c5 X, c7 X? Actually row1: ". X X X . X . X ." => indices:0 .,1 X,2 X,3 X,4 .,5 X,6 .,7 X,8 .
Row2: ". . X . . X X X ." => c2 X, c5 X, c6 X, c7 X
Row3: ". . X . . X X . ." => c2 X, c5 X, c6 X

Thus there seems to be two clusters: left cluster maybe at columns 1-3? Actually left cluster appears at columns 1-3 in rows1-3? Let's see: left cluster cells: (1,1),(1,2),(1,3); (2,2); (3,2). That's shape like a T? Actually it's like a vertical line at column2 rows1-3 plus a horizontal top at row1 columns1-3. That's a shape.

Right cluster maybe at columns5-7? Let's see: right cluster cells: (1,5); (2,5),(2,6),(2,7); (3,5),(3,6). That's shape: a vertical line at column5 rows1-3 plus a horizontal bottom at row2 columns5-7? Actually shape: (1,5) top, (2,5)-(2,7) middle row full, (3,5)-(3,6) bottom left two. That's like a shape.

Thus left colony likely occupies columns 0-2? But we see left cluster at columns1-3, not 0-2. However there is a border of '.' around. Let's check if left colony is at columns 0-2 but we see X at column1-3, meaning maybe colony offset by 1 column due to membrane? Let's examine the grid: 5 rows, 9 columns. If we place two 3x3 colonies side by side with a one-column gap between them (membrane), then left colony columns 0-2, membrane column 3, right colony columns 4-6? But that would be width 0-2 (3), 3 (membrane), 4-6 (3) = total 7 columns, leaving columns 7-8 maybe border? Actually width 9, maybe there is a one-column border on each side? Let's think: maybe colonies are at columns 1-3 and 5-7, with column 0 and 8 as outer border, column 4 as membrane? Let's test: left colony columns 1-3 (width 3), membrane column 4, right colony columns 5-7 (width 3), outer borders columns 0 and 8. That matches observed: left cluster X's at columns 1-3, right cluster at columns 5-7. Yes! So left colony occupies columns 1,2,3 (0-index). Right colony occupies columns 5,6,7. Column 0 and 8 are outer border (a

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We need to understand the transformation. Input is 5x9 grid: two 3x3 quadrants left and right, separated by a column (the meridian) of width 1? Actually 5 rows, 9 columns. Two 3x3 quadrants: left quadrant columns 0-2? Right quadrant columns 6-8? The meridian column is column 3? Let's see: 9 columns: indices 0-8. If left quadrant is 3 columns, right quadrant 3 columns, meridian column is column 3? Actually 3+1+3 =7, not 9. Wait maybe there is also a border of black (.) around? The description: "dark night sky is divided into two 3×3 star-chart quadrants by a dark meridian." So the meridian is a line (maybe a column) of black cells separating the quadrants. Input grid is 5x9: maybe there is a top and bottom row of black (empty sky) as border? Actually they said grid details: Input: 5×9 sky chart, two 3×3 quadrants. So 5 rows, 9 columns. The quadrants are each 3x3, placed somewhere within the 5x9, maybe with a one-cell margin top and bottom? Let's examine training examples.

Example 1 Input:
Row0: . . . . . . . . .
Row1: . X X X . X . X .
Row2: . . X . . X X X .
Row3: . . X . . X X . .
Row4: . . . . . . . . .

We see pattern: first column all '.'; last column all '.'; also first row all '.'; last row all '.'; So border of '.' around a 3x7 interior? Actually interior rows 1-3, columns 1-7 (since 0 and 8 are border). That's 3 rows x 7 columns. Within that, we have left quadrant 3x3? Let's extract columns 1-3 maybe left quadrant, column 4 meridian, columns 5-7 right quadrant? Let's test.

Columns indices: 0 1 2 3 4 5 6 7 8
Row1: . X X X . X . X .
So at row1: col1='X', col2='X', col3='X', col4='.', col5='X', col6='.', col7='X', col8='.'
Row2: . . X . . X X X .
col1='.', col2='X', col3='.', col4='.', col5='X', col6='X', col7='X', col8='.'
Row3: . . X . . X X . .
col1='.', col2='X', col3='.', col4='.', col5='X', col6='X', col7='.', col8='.'

Thus left quadrant (cols1-3) rows1-3:
Row1: X X X
Row2: . X .
Row3: . X .
That's pattern: top row all X, middle row . X ., bottom row . X . (like a vertical line with a top bar?). Right quadrant (cols5-7):
Row1: X . X
Row2: X X X
Row3: X X .
Let's list:
Row1 col5='X', col6='.', col7='X' => X . X
Row2 col5='X', col6='X', col7='X' => X X X
Row3 col5='X', col6='X', col7='.' => X X .

Now output for example1:
B . B
B B B
B B .

That's a 3x3 grid. According to description: output is right quadrant stars colored by left constellation's spectral class. So we need to determine spectral class color from left quadrant pattern. Then apply that color to all stars (X) in right quadrant, while empty sky (.) stays '.' (black). So output should be same shape as right quadrant (3x3) but with X replaced by the color code (single letter) and '.' remain '.'.

Thus output pattern: we see output:
Row0: B . B
Row1: B B B
Row2: B B .

Thus they replaced X's in right quadrant with B (blue). So left quadrant pattern corresponded to spectral class B (blue). Let's verify: left quadrant pattern we extracted: 
Row1: X X X
Row2: . X .
Row3: . X .
That's like a shape: maybe it's the letter "Г"? Actually it's like a vertical line on column2 (middle column) plus a top horizontal line across row1. That could be resembling the constellation "Orion"? Not sure. But the spectral class is B (blue). So left pattern maps to B.

Now example2:
Input:
Row0: . . . . . . . . .
Row1: . . X X . X . X .
Row2: . . X . . X . X .
Row3: . X X . . X X . .
Row4: . . . . . . . . .

Let's extract left quadrant (cols1-3) rows1-3:
Row1: col1='.', col2='X', col3='X' => . X X
Row2: col1='.', col2='X', col3='.' => . X .
Row3: col1='X', col2='X', col3='.' => X X .
Thus left quadrant:
. X X
. X .
X X .

Right quadrant (cols5-7):
Row1: col5='X', col6='.', col7='X' => X . X
Row2: col5='X', col6='.', col7='X' => X . X
Row3: col5='X', col6='X', col7='.' => X X .

Thus right quadrant:
X . X
X . X
X X .

Output example2:
G . G
G . G
G G .

Thus they replaced X's with G (green). So left quadrant pattern corresponded to G.

Now example3:
Input:
Row0: . . . . . . . . .
Row1: . X X X . X . X .
Row2: . . X . . . X X .
Row3: . . X . . X . X .
Row4: . . . . . . . . .

Left quadrant (cols1-3):
Row1: X X X
Row2: . X .
Row3: . X .
Same as example1 left quadrant? Actually example1 left quadrant was X X X; . X .; . X . (yes same). So left quadrant pattern same as example1. Output example3:
B . B
. B B
B . B

Thus they replaced X's in right quadrant with B again. Let's check right quadrant for example3:
Right quadrant (cols5-7):
Row1: col5='X', col6='.', col7='X' => X . X
Row2: col5='.', col6='.', col7='X' => . . X
Row3: col5='.', col6='X', col7='.' => . X .
Thus right quadrant:
X . X
. . X
. X .

Now output:
Row0: B . B
Row1: . B B
Row2: B . B

Thus they replaced X's with B, and '.' remain '.'.

Thus mapping: left quadrant pattern -> color (spectral class). Need to deduce rule: what feature of left quadrant determines color? Possibly number of X's? Let's count.

Example1 left quadrant X count: Row1:3,

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We need to parse the problem: It's a grid transformation puzzle. Input: 5x9 sky (5 rows, 9 columns). Two 3x3 cloud clusters: western formation (left side) and eastern formation (right side). Output: 3x3 — eastern patches colored by the western formation's storm identity.

We need to read the western cloud formation's shape to determine the storm type and its signature color. The eastern cloud patches take on that color while clear sky remains dark. The western clouds dissipate; only the colored eastern sky remains as the forecast.

Thus we need to extract the western 3x3 region (left side) and the eastern 3x3 region (right side) from the input. The input is 5x9, with presumably a central column (column index 4? 0-indexed) that is the calm column in the center (the split). The western formation is a 3x3 cluster on the left side of that column, and eastern formation is a 3x3 cluster on the right side.

We need to identify the shape of the western formation (the pattern of X's (grey cloud patches) within its 3x3 region). Based on that shape, we determine a storm type and its signature color (some mapping from shape to color). Then we apply that color to the eastern formation's patches (the X's in the eastern 3x3 region) while leaving clear sky (dots) as black (.)? Actually "clear sky remains dark" meaning '.' remains '.' (black). The eastern patches take on that color while clear sky remains dark. So output is a 3x3 grid where each cell is either '.' (if input eastern cell was '.'), or the signature color (if input eastern cell was 'X').

Thus we need to map western shape to a color (one of B,R,G,Y,X,M,O,A,W). The training examples will help us infer mapping.

Let's examine Example 1.

Example 1 Input:
Row0: . . . . . . . . .
Row1: . X X X . X . X .
Row2: . . X . . X X X .
Row3: . . X . . X X . .
Row4: . . . . . . . . .

We need to parse 5x9. Let's index columns 0-8.

Row0: all '.'.

Row1: col0 '.', col1 'X', col2 'X', col3 'X', col4 '.', col5 'X', col6 '.', col7 'X', col8 '.'

Row2: col0 '.', col1 '.', col2 'X', col3 '.', col4 '.', col5 'X', col6 'X', col7 'X', col8 '.'

Row3: col0 '.', col1 '.', col2 'X', col3 '.', col4 '.', col5 'X', col6 'X', col7 '.', col8 '.'

Row4: all '.'.

Thus we have a central column at col4 (the 5th column, index 4) which is all '.' in rows 0-4? Let's check: row0 col4 '.'; row1 col4 '.'; row2 col4 '.'; row3 col4 '.'; row4 col4 '.' => yes central column is all '.' (calm column). Good.

Western formation: left side columns 0-3? Actually a 3x3 cluster: likely rows 1-3 (since there is a border of '.' rows top and bottom) and columns 1-3? Let's see.

The western cloud formation appears as a cluster of X's in left half. Let's extract rows 1-3, columns 0-3? But we need a 3x3 region. Likely the formation is centered around the central column, with one column margin on each side? Actually the description: "two 3x3 cloud clusters hang on either side — the western formation has taken a distinctive meteorological shape, while the eastern patches drift formlessly." So each cluster is a 3x3 block. The input is 5 rows tall, 9 columns wide. The central column is column 4 (0-index). So the western 3x3 block would be columns 1-3? Actually to have a 3x3 block wholly left of central column, we need columns 1-3 (since column 0 is leftmost, column 8 rightmost). But we need 3 columns: maybe columns 0-2, 1-3, 2-4? But column 4 is central calm column, so western block cannot include column 4. So western block likely columns 1-3 (since column 0 is edge maybe empty). Similarly eastern block columns 5-7 (since column 8 edge). Let's test.

Take rows 1-3 (since rows 0 and 4 are all '.' border). Columns 1-3 for western block:

Row1 cols1-3: X X X
Row2 cols1-3: . X . (since row2: col1 '.', col2 'X', col3 '.')
Row3 cols1-3: . X . (row3: col1 '.', col2 'X', col3 '.')

Thus western block:
Row1: X X X
Row2: . X .
Row3: . X .

That's a shape: a vertical line of X's in column2 (middle column) plus a top row of X's across columns1-3? Actually row1: X X X (full top row). Row2: . X . (middle). Row3: . X . (bottom). So shape looks like a "T" shape? Actually it's like a "T" rotated? Top row full, middle column down.

Now eastern block: columns 5-7 (since column 4 central, column5-7 right side). Let's extract rows1-3, cols5-7:

Row1: col5 'X', col6 '.', col7 'X' => X . X
Row2: col5 'X', col6 'X', col7 'X' => X X X
Row3: col5 'X', col6 'X', col7 '.' => X X .

Thus eastern block:
Row1: X . X
Row2: X X X
Row3: X X .

Now the output for Example1 is:
B . B
B B B
B B .

Thus output is 3x3: same shape as eastern block but with X's replaced by B (blue) and '.' remain '.'.

Thus the signature color for western shape (the T shape) is B (blue). So mapping: shape T -> B.

Now Example2:

Input:
. . . . . . . . .
. . X X . X . X .
. . X . . X . X .
. X X . . X X . .
. . . . . . . . .

Let's parse.

Row0: all '.'.

Row1: col0 '.', col1 '.', col2 'X', col3 'X', col4 '.', col5 'X', col6 '.', col7 'X', col8 '.'

Row

... [7297 chars truncated]

We need to parse the problem: It's a grid transformation puzzle. The description: "What you see: A dark night sky dotted with a few colored stars. Some vertical lanes of the sky hold stars; others are completely empty and dark. What to do: Each star casts a pillar of azure light that floods its entire vertical lane from horizon to horizon, like a searchlight pointed straight up and down. The stars themselves burn through the glow, keeping their original color. Lanes with no stars remain pitch dark — the light has no source there. Once the illuminated sky is complete, it is photographed and printed identically onto all four panes of a divided window, two across and two down."

Thus transformation: Input grid (original sky) of some size (height H, width W). For each column (vertical lane), if there is any star (colored cell not black? Actually black is '.' representing darkness. Stars are colored cells: any non-black? The color key includes '.' black, X grey, and other colors: B,R,G,Y,M,O,A,W. The description says "colored stars" and "azure is the glow that only stars can summon — it never appears on its own." So stars are any non-black, non-azure? Actually azure is the glow, not a star. So stars are any color other than '.' (black) and 'A' (azure). But the input may contain 'A'? Possibly not; but they said azure never appears on its own, meaning it only appears as result of glow. So input will not have 'A'.

Thus for each column, if there is at least one star (any non '.' and not 'A'), then the entire column becomes filled with azure 'A' except where there is a star, which retains its original color (overwrites the azure). If column has no stars (i.e., all '.'), then column remains all '.' (black). After this transformation, we have an illuminated sky (size H x W). Then we photograph and print identically onto all four panes of a divided window, two across and two down. So we tile the illuminated sky 2x2: output grid dimensions: height = 2*H, width = 2*W. Each cell of the illuminated sky is replicated in a 2x2 block? Actually "printed identically onto all four panes of a divided window, two across and two down." Means we take the illuminated sky image and place it in each of the four quadrants of a window that is twice as tall and twice as wide. So we replicate the image four times: top-left, top-right, bottom-left, bottom-right. So output is just the illuminated sky repeated in a 2x2 tiling (no scaling). So each cell of illuminated sky appears four times in output: at positions (i, j), (i, j+W), (i+H, j), (i+H, j+W). So basically output = [ [sky, sky]; [sky, sky] ] where each sky is the illuminated sky.

Thus algorithm: For each column c in input:
- Determine if any non '.' cell (i.e., star) exists in that column.
- If yes: For each row r:
   - If input[r][c] == '.' then output_illuminated[r][c] = 'A'
   - else output_illuminated[r][c] = input[r][c] (star retains its color)
- If no: column remains all '.'.

Then tile 2x2.

Now test input:
. . G .
. . . .
. . . G
G . . .

Let's write as rows:

Row0: . . G .
Row1: . . . .
Row2: . . . G
Row3: G . . .

Thus H=4, W=4.

Now compute columns:

Column0: values: row0 '.', row1 '.', row2 '.', row3 'G' -> has star 'G' (non '.'), so column0 illuminated: for each row:
- row0 '.' -> 'A'
- row1 '.' -> 'A'
- row2 '.' -> 'A'
- row3 'G' -> stays 'G'

Thus column0 illuminated: [A, A, A, G]

Column1: values: row0 '.', row1 '.', row2 '.', row3 '.' -> all '.' -> no star -> column remains all '.'.

Column2: values: row0 'G', row1 '.', row2 '.', row3 '.' -> has star 'G' at row0. So column2 illuminated:
- row0 'G' -> stays 'G'
- row1 '.' -> 'A'
- row2 '.' -> 'A'
- row3 '.' -> 'A'

Thus column2: [G, A, A, A]

Column3: values: row0 '.', row1 '.', row2 'G', row3 '.' -> has star 'G' at row2. So column3 illuminated:
- row0 '.' -> 'A'
- row1 '.' -> 'A'
- row2 'G' -> stays 'G'
- row3 '.' -> 'A'

Thus column3: [A, A, G, A]

Thus illuminated sky (4x4):
Row0: col0 A, col1 ., col2 G, col3 A => A . G A
Row1: A . A A
Row2: A . A G
Row3: G . A A

Let's write as matrix:

Row0: A . G A
Row1: A . A A
Row2: A . A G
Row3: G . A A

Now tile 2x2: output height = 8, width = 8.

We need to output as characters separated by spaces, each row line.

Thus we create top-left quadrant = illuminated sky (as above). top-right quadrant = same as illuminated sky (copy). bottom-left = same. bottom-right = same.

Thus output rows 0-3 = top half (rows 0-3) each row = illuminated sky row concatenated with illuminated sky row (since width doubles). So each row = illuminated sky row + illuminated sky row.

Rows 4-7 = bottom half = same as top half (since we duplicate vertically). So rows 4-7 identical to rows 0-3.

Thus final output 8 rows, each row length 8.

Let's compute illuminated sky rows as strings:

Row0: A . G A => "A . G A"
Row1: A . A A => "A . A A"
Row2: A . A G => "A . A G"
Row3: G . A A => "G . A A"

Now duplicate horizontally:

... [3854 chars truncated]

We need to parse the problem: It's a grid transformation puzzle. We have a metaphorical description: "What you see: A petri dish of mostly dormant cells, with a few carrying an active gene scattered across certain chromosomal columns. What to do: Each chromosome bearing at least one active gene expresses fully in azure, with the original genes still glowing through. Dormant chromosomes stay silent. The culture then divides twice to fill four quadrants. Grid details: Output is twice the height and width of the input, four identical tiles of the transformed dish."

Thus transformation: Input grid (size H x W). We need to identify "chromosomes" as columns? The description: "scattered across certain chromosomal columns." So each column is a chromosome. Active genes are cells with some color? The active gene is indicated by some color? The description: "Each chromosome bearing at least one active gene expresses fully in azure, with the original genes still glowing through." So for each column that has at least one active gene (i.e., any non-dormant cell?), we set the entire column to azure (A) but keep the original gene colors visible? "with the original genes still glowing through" suggests that the original colors remain visible (maybe overlay?). The output after transformation (before duplication) is: For each column, if there is any active gene (non-dormant cell) in that column, then the column becomes all azure (A) except that the original gene cells retain their original color? Or maybe they are overlaid: The azure is background, and original gene colors show through (like they are not overwritten). The phrase "expresses fully in azure, with the original genes still glowing through" suggests that the column becomes azure, but the original gene cells are still visible (maybe they are not overwritten, they remain as they were). So the column becomes azure everywhere, but the cells that originally had a gene (i.e., were non-dormant) keep their original color (overriding the azure?). Or perhaps they are blended? But given discrete colors, likely they stay as original color, overriding azure.

Thus transformation: For each column j, if any cell in column j is not '.' (dormant), then set all cells in column j to 'A' (azure), except that cells that originally were not '.' keep their original color (i.e., they are not changed to A). Dormant columns (all '.'), remain all '.' (since they stay silent). Then after this column-wise transformation, the culture divides twice to fill four quadrants: Output is twice the height and width of the input, four identical tiles of the transformed dish. So we take the transformed grid (size H x W) and tile it 2x2 (both dimensions doubled) to produce output of size 2H x 2W, where each quadrant is identical copy of the transformed grid.

Thus final output: For each cell (i,j) in output, we map to transformed cell at (i mod H, j mod W) (assuming zero-index). So just repeat the transformed grid.

Now we need to identify active gene: Which colors represent active gene? The description: "few carrying an active gene scattered across certain chromosomal columns." The input includes various colors: . (black/dormant), X (grey), B (blue), R (red), G (green), Y (yellow), M (magenta), O (orange), A (azure), W (maroon). The active gene likely is any non-dormant cell (i.e., not '.'). However, note that in examples, we see X (grey) appears in input and output. In example 1 input: there are X's at positions (0,1?) Actually let's parse example 1.

Example 1 Input:
Row0: . X . . . .
Row1: . . . . . .
Row2: . . . . . .
Row3: . . . . . .
Row4: X . . . . X
Row5: . . . . . .
So H=6, W=6.

Active genes: X at (0,1), X at (4,0), X at (4,5). Also maybe other colors? No other colors besides . and X. So columns with at least one X: column1 (index1) has X at row0; column0 has X at row4; column5 has X at row4. So columns 0,1,5 have active genes. According to rule, those columns become all A (azure) except original X cells remain X? Let's see output.

Output given (12x12). Let's extract top-left 6x6 quadrant (should be transformed grid). Output rows 0-5, cols 0-5:

Row0: A X . . . A
Row1: A A . . . A
Row2: A A . . . A
Row3: A A . . . A
Row4: X A . . . X
Row5: A A . . . A

Yes that matches: column0: all A except row4 is X (original). column1: all A except row0 is X (original). column5: all A except row4 is X (original). Columns 2,3,4: all '.' (since dormant). So matches rule: For each column with any non-dot, set all to A, but keep original non-dot cells as they were.

Thus active gene = any cell not '.' (i.e., any colored cell). The original gene colors are preserved (they are not overwritten by A). Dormant cells '.' become A in active columns? Actually in output we see that in active columns, the originally '.' cells become A (azure). Yes: e.g., column0 row0 originally '.' becomes A. So transformation: For each column j, if any cell in column j is not '.', then for each row i: if input[i][j] != '.', output[i][j

... [3470 chars truncated]

We need to parse the problem: It's a grid transformation puzzle. We have a metaphorical description: "What you see: A dark sky with a few colored stars at scattered positions. What to do: Compress each meridian (column) containing a star into an azure band; stars keep their color. Empty meridians stay dark. View through a 2×2 quadrant lens. Grid details: Output is four identical sky tiles, doubling both dimensions."

We have training examples to infer the transformation.

We need to identify CAST (roles), AFFORDANCES, TRANSFORMATION RULE.

Then apply to test input.

Test input is a 4x4 grid? Let's see:

Test Input:
. . G .
. . . .
. . . G
G . . .

So 4 rows, 4 columns.

We need to produce output grid: "Output is four identical sky tiles, doubling both dimensions." Means we take the transformed sky (some intermediate grid) and then replicate it into a 2x2 arrangement (four identical tiles) and also double dimensions? Let's parse: "View through a 2×2 quadrant lens. Grid details: Output is four identical sky tiles, doubling both dimensions." So maybe we first apply transformation to input to get a "sky" (some grid). Then we view through a 2x2 quadrant lens meaning we take each cell and replace it with a 2x2 block? Or we tile the sky four times in a 2x2 arrangement, and also double dimensions (so each tile is scaled up by factor 2?). The phrase "doubling both dimensions" suggests that the output grid is twice as wide and twice as tall as something. If we have four identical sky tiles arranged in a 2x2 quadrant lens, that would double both dimensions if each tile is same size as original? Actually if you take a base tile and repeat it 2 times horizontally and 2 times vertically, you get a grid that is 2x the width and 2x the height of the base tile. So output dimensions = (2 * width_of_tile) x (2 * height_of_tile). And the tile is the transformed sky (maybe after compression). So we need to compute transformed sky (some intermediate grid) from input, then tile it 2x2 to produce output.

Let's verify with training example.

Example 1 Input:
Row0: . X . . . .
Row1: . . . . . .
Row2: . . . . . .
Row3: . . . . . .
Row4: X . . . . X
Row5: . . . . . .
So input is 6 rows, 6 columns.

Output given is many rows: Let's count.

Output lines:
Line0: A X . . . A A X . . . A
Line1: A A . . . A A A . . . A
Line2: A A . . . A A A . . . A
Line3: A A . . . A A A . . . A
Line4: X A . . . X X A . . . X
Line5: A A . . . A A A . . . A
Line6: A X . . . A A X . . . A
Line7: A A . . . A A A . . . A
Line8: A A . . . A A A . . . A
Line9: A A . . . A A A . . . A
Line10: X A . . . X X A . . . X
Line11: A A . . . A A A . . . A

So output has 12 rows, and each row seems to have many characters separated by spaces. Let's count characters per line: They appear to have pattern: maybe 12 columns? Let's count tokens in line0: "A X . . . A A X . . . A". That's: A, X, ., ., ., A, A, X, ., ., ., A => 12 tokens. So output is 12 rows x 12 columns.

Input was 6x6. Output is 12x12, which is exactly double both dimensions (2x width, 2x height). And it's composed of four identical sky tiles? Let's see if we can partition output into 2x2 blocks of 6x6 each (since doubling both dimensions yields 2x2 tiles each of size original). Indeed 12x12 can be seen as four 6x6 tiles arranged in a 2x2 grid. Let's check if each 6x6 tile is identical.

Take top-left 6x6 (rows 0-5, cols 0-5):
Rows0-5, cols0-5:
Row0: A X . . . A
Row1: A A . . . A
Row2: A A . . . A
Row3: A A . . . A
Row4: X A . . . X
Row5: A A . . . A

Top-right 6x6 (rows0-5, cols6-11):
Row0: A X . . . A (cols6-11: after first 6 tokens: A X . . . A? Let's extract: tokens indices 6-11: after first six: we have tokens: index0 A,1 X,2 .,3 .,4 .,5 A,6 A,7 X,8 .,9 .,10 .,11 A. So cols6-11: A X . . . A. Yes same as left.)
Row1: tokens: A A . . . A A A . . . A => indices0-5: A A . . . A ; indices6-11: A A . . . A? Let's check: after first six: A A . . . A ; then next six: A A . . . A? Actually tokens: 0 A,1 A,2 .,3 .,4 .,5 A,6 A,7 A,8 .,9 .,10 .,11 A. So yes same pattern: A A . . . A.
Row2: same as row1.
Row3: same.
Row4: X A . . . X X A . . . X => first six: X A . . . X ; second six: X A . . . X? Let's see: tokens:0 X,1 A,2 .,3 .,4 .,5 X,6 X,7 A,8 .,9 .,10 .,11 X. Yes same.
Row5: A A . . . A A A . . . A => first six: A A . . . A ; second six: A A . . . A.

Thus each 6x6 tile is identical. So indeed output is four identical tiles (2x2 arrangement) each tile being the transformed sky (size same as input). And dimensions doubled.

Thus transformation: Input -> transformed sky (same size as input) via some rule: "Compress each meridian (column) containing a star into an azure band; stars keep their color. Empty meridians stay dark." Then we view through a 2x2 quadrant lens: meaning we replicate the transformed sky into a 2x2 grid (tile) and double dimensions (i.e., each cell becomes a 2x2 block? Actually we just tile; but the description "View through a 2×2 quadrant lens" could mean we loo

... [7694 chars truncated]

We need to parse the problem: It's a grid transformation puzzle. We have a metaphorical description: "What you see: A dark circuit board with a few colored LEDs at scattered positions. What to do: Energize every trace (column) containing an LED with azure current; LEDs keep their color. Dead traces stay dark. Stamp the board into a 2×2 panel. Grid details: Output is four identical boards, doubling both dimensions."

Thus the transformation: Input grid: some cells are colored (colored LEDs) and some are black (.) and maybe grey (X) representing something? The description: "dark circuit board" maybe black cells are background (dark). "colored LEDs at scattered positions" are colored cells (not black, not grey?). "Energize every trace (column) containing an LED with azure current; LEDs keep their color. Dead traces stay dark." So for each column that contains at least one LED (colored cell?), we set all cells in that column to azure (A) except the LED cells themselves keep their original color. For columns with no LED, they remain dark (black? maybe .). Then "Stamp the board into a 2×2 panel. Grid details: Output is four identical boards, doubling both dimensions." Means we take the transformed board (after energizing columns) and then replicate it in a 2x2 tiling, doubling both dimensions (so output width = 2 * input width, height = 2 * input height). The four copies are identical.

Thus the transformation: Input grid of size HxW. For each column j, if any cell in column j is not '.' (i.e., is a colored LED? maybe also not X? The description says "colored LEDs at scattered positions". The training examples show X as grey maybe representing something else like a fixed obstacle? Let's examine training example 1.

Example 1 Input:
Row0: . X . . . .
Row1: . . . . . .
Row2: . . . . . .
Row3: . . . . . .
Row4: X . . . . X
Row5: . . . . . .
So it's 6 rows, 6 columns.

We have X at (0,1), (4,0), (4,5). Others are '.'.

Output given is a 12x12 grid (since doubling both dimensions). Let's see if we can deduce.

First, apply column energizing rule: For each column, if there is any LED (colored cell) in that column, then set all cells in that column to A (azure) except the LED cells keep their color. But there are no colored LEDs in input (only X and .). X is grey maybe representing something else like a fixed trace? The description says "dark circuit board with a few colored LEDs at scattered positions." X is grey maybe representing something like a fixed component that is not an LED? Or maybe X is just a placeholder for something else like a "trace"? Let's see.

If there are no colored LEDs, then no column gets energized, so all columns remain dark (black). Then after stamping into 2x2 panel, we would get a 12x12 grid of all '.' (black). But the output is not all '.'; we see many A's and X's.

Thus X must be considered as an LED? Or maybe X is a "trace" that gets energized? Let's re-evaluate.

The description: "Energize every trace (column) containing an LED with azure current; LEDs keep their color." So traces are columns. If a column contains an LED, we energize the trace (i.e., fill the column with azure current). The LED keeps its color. So the LED cells remain as they were (colored). The rest of the cells in that column become azure (A). If a column has no LED, it stays dark (i.e., black '.'?).

Thus we need to identify which cells are LEDs. In the training example, there are X's. Perhaps X represents a colored LED? But the color key says X=grey. Grey is not a color of LED? But maybe grey is considered a color (like an LED that is grey). However the description says "colored LEDs". Grey could be considered a color (though it's dull). But they gave a color key for X=grey. So X could be a grey LED.

Thus the input has grey LEDs at positions (0,1), (4,0), (4,5). All other cells are black (.) (dark). So we have LEDs in columns 0,1,5? Let's see: column 0 has LED at (4,0) (grey). column 1 has LED at (0,1) (grey). column 5 has LED at (4,5) (grey). Columns 2,3,4 have no LEDs.

Thus per rule: For columns 0,1,5, we set all cells in those columns to azure (A) except the LED cells keep their grey (X). For columns 2,3,4, they remain dark (black '.').

Now after this transformation, we get a board of same size (6x6) where columns 0,1,5 are mostly A except X at those LED positions; columns 2-4 are all '.'.

Let's compute that:

Rows 0-5, columns 0-5.

Initialize all cells as '.' (dark). Then for each column with LED, set all cells in that column to A, then put back the LED cells as X.

Column0: set all rows to A, then at row4 col0 set X (LED). So column0: rows0-3 A, row4 X, row5 A.

Column1: set all rows to A, then at row0 col1 set X. So column1: row0 X, rows1-5 A.

Column2: no LED => all '.'.

Column3: all '.'.

Column4: all '.'.

Column5: set all rows to A, then at row4 col5 set X. So column5: rows0-3 A, row4 X, row5 A.

Thus the transformed board (6x6) is:

Row0: col0 A, col1 X, col2 ., col3 ., col4 ., col5 A =&g

... [6340 chars truncated]

We need to parse the metaphorical description: "What you see: A dark dance floor with a few costumed dancers at scattered marks. What to do: Bathe every lane containing a dancer in azure light; dancers keep their color. Empty lanes stay dark. Repeat the tableau across a 2×2 stage arrangement. Grid details: Output is four identical stages, double the original dimensions."

Interpretation: Input grid is a "dance floor" with some colored cells representing dancers (colored non-black? Actually black '.' is dark). Dancers are colored cells (any color except '.'? maybe also X? but X is grey, maybe considered a dancer? The description: "costumed dancers at scattered marks." So any non-black cell is a dancer. The "lanes" likely refer to rows and columns? "Bathe every lane containing a dancer in azure light; dancers keep their color. Empty lanes stay dark." Means: For each row (lane) that contains at least one dancer (non '.'), we set all cells in that row to azure (A) except the dancer cells keep their original color. Similarly for each column (lane) that contains a dancer, we set all cells in that column to azure except dancer cells keep original color. If both row and column are bathed, the intersection cell (if not a dancer) would be azure? Actually if a cell is in a row that contains a dancer and also in a column that contains a dancer, it would be bathed by both? But the rule says "Bathe every lane containing a dancer in azure light; dancers keep their color. Empty lanes stay dark." So we treat rows and columns independently: For each lane (row or column) that has at least one dancer, we set all cells in that lane to azure, except we preserve the dancer's original color (so they stay as they were). If a cell is in both a row and column that are bathed, it's still azure unless it's a dancer (then keep original). So effectively: For each cell, if its row contains any dancer OR its column contains any dancer, then set it to azure, unless it itself is a dancer (non '.'), in which case keep its original color. If neither its row nor column contains a dancer, then it stays dark ('.').

Thus output is same size as input? But description says "Repeat the tableau across a 2×2 stage arrangement. Grid details: Output is four identical stages, double the original dimensions." Means we take the transformed grid (after applying the lane bathing rule) and then tile it 2x2 (i.e., repeat it both horizontally and vertically) to produce output with double width and double height (i.e., each dimension multiplied by 2). So output dimensions = 2 * input rows, 2 * input cols.

Check with examples.

Example 1 input:
6 rows? Let's count: Input lines:
". X . . . ."
". . . . . ."
". . . . . ."
". . . . . ."
"X . . . . X"
". . . . . ."
That's 6 rows, 6 columns? Actually each line has 6 tokens separated by spaces? Let's see: ". X . . . ." => 6 tokens. Yes 6x6.

Output given:
It appears to be 12 rows (double) and 12 columns (double). Let's verify.

Output lines:
"A X . . . A A X . . . A"
"A A . . . A A A . . . A"
"A A . . . A A A . . . A"
"A A . . . A A A . . . A"
"X A . . . X X A . . . X"
"A A . . . A A A . . . A"
"A X . . . A A X . . . A"
"A A . . . A A A . . . A"
"A A . . . A A A . . . A"
"A A . . . A A A . . . A"
"X A . . . X X A . . . X"
"A A . . . A A A . . . A"

That's 12 rows, each with 12 tokens (separated by spaces). Good.

Now let's test the rule on input to see if we get the top-left 6x6 quadrant matching the transformed grid (before tiling). The top-left quadrant of output (first 6 rows, first 6 columns) should be the transformed input.

Let's extract top-left 6x6 from output:

Rows 1-6, cols 1-6:

Row1: A X . . . A
Row2: A A . . . A
Row3: A A . . . A
Row4: A A . . . A
Row5: X A . . . X
Row6: A A . . . A

Now apply rule to input:

Input:
Row1: . X . . . .
Row2: . . . . . .
Row3: . . . . . .
Row4: . . . . . .
Row5: X . . . . X
Row6: . . . . . .

Identify dancers: any non '.' cell: positions: (1,2) X, (5,1) X, (5,6) X. Also maybe other colors? No.

Now compute rows containing a dancer: rows 1,5 (since row1 has X at col2, row5 has X at col1 and col6). Also row? Actually row5 has two X's. So rows 1 and 5 are "active". Columns containing a dancer: column2 (from row1 X), column1 (from row5 X), column6 (from row5 X). So columns 1,2,6 are active.

Now for each cell, if its row is active OR its column is active, set to A unless it's a dancer (keep original). Let's compute.

We'll produce a 6x6 grid.

Initialize all to '.' then apply.

Row1 (active): all cells in row1 become A except dancers keep original.
- col1: not dancer => A
- col2: dancer X => keep X
- col3: not dancer => A
- col4: A
- col5: A
- col6: A? Wait col6 is active column (since column6 active) but also row1 active, so A unless dancer. No dancer at (1,6) => A.
Thus row1: A X A A A A. But output top-left row1 is: A X . . . A. That's different: they have '.' at col3-5, not A. So my interpretation is off.

Maybe "lanes" refer to only rows OR only col

... [7689 chars truncated]

We need to parse the problem: It's a grid transformation puzzle. We have a metaphorical description: "What you see: A dark rock cross-section with a few colored mineral deposits at scattered positions. What to do: Permeate every vein (column) containing a deposit with azure groundwater; minerals keep their color. Barren veins stay dark. Map the survey as a 2×2 quadrant display. Grid details: Output doubles both dimensions with four identical survey quadrants."

Interpretation: Input grid: dark rock cross-section (background black '.'), with colored mineral deposits (colored letters other than '.' and maybe 'X'? Actually X is grey, but they said dark rock cross-section with a few colored mineral deposits at scattered positions. So '.' is black (dark rock). The mineral deposits are colored letters (R, G, B, Y, M, O, A, W?). But they say "Permeate every vein (column) containing a deposit with azure groundwater; minerals keep their color. Barren veins stay dark." So for each column (vein) that contains at least one deposit (colored cell not '.'?), we fill the entire column with azure 'A' (groundwater), but the minerals keep their color (so the deposit cells remain as they were). For columns with no deposit (i.e., all '.'), they stay dark (i.e., '.' unchanged). Then "Map the survey as a 2×2 quadrant display. Grid details: Output doubles both dimensions with four identical survey quadrants." Means after applying the column transformation, we then produce an output grid that is double the dimensions (both width and height) of the transformed grid, consisting of four identical quadrants (i.e., we tile the transformed grid 2x2). So output size = (2 * input height) rows, (2 * input width) columns. Each quadrant is a copy of the transformed grid.

Thus the transformation: For each column j, if any cell in column j is not '.' (i.e., a deposit), then set all cells in that column to 'A' except those that originally were deposits (they keep their original color). If column has no deposit (all '.'), leave as '.' (dark). Then replicate the resulting grid 2x2.

Check with examples.

Example 1 input:
6 rows? Let's count: Provided input lines:
". X . . . ."
". . . . . ."
". . . . . ."
". . . . . ."
"X . . . . X"
". . . . . ."
So 6 rows, 6 columns.

Let's map: '.' black, X grey (maybe considered deposit? It says mineral deposits at scattered positions; X is grey, but maybe considered a deposit? The description: "dark rock cross-section with a few colored mineral deposits at scattered positions." X is grey, not a color? But they gave color key: X=grey. It might be considered a deposit (maybe a mineral deposit that is grey). However the transformation says "Permeate every vein (column) containing a deposit with azure groundwater; minerals keep their color." If X is considered a deposit, then columns containing X should be permeated with A, but X keeps its color (grey). Let's test.

Input columns (0-index):
col0: row0 '.', row1 '.', row2 '.', row3 '.', row4 'X', row5 '.' => contains X (deposit) => permeate with A except X stays X.
col1: row0 'X', row1 '.', row2 '.', row3 '.', row4 '.', row5 '.' => contains X at row0 => permeate with A except X stays X.
col2: all '.'? row0 '.', row1 '.', row2 '.', row3 '.', row4 '.', row5 '.' => no deposit => stays all '.'.
col3: all '.'? yes.
col4: all '.'? yes.
col5: row0 '.', row1 '.', row2 '.', row3 '.', row4 'X', row5 '.' => contains X => permeate with A except X stays X.

Thus after transformation, we expect columns 0,1,5 to have A except where X originally; columns 2,3,4 all '.'.

Let's compute transformed grid (6x6):
Initialize all '.' then for each column with deposit, set all to 'A', then restore original deposit cells.

col0: set all rows to 'A', then row4 (X) stays X => col0: rows0-3 A, row4 X, row5 A.
col1: set all rows to 'A', then row0 (X) stays X => col0? Actually col1: row0 X, rows1-5 A.
col2: all '.'.
col3: all '.'.
col4: all '.'.
col5: set all rows to 'A', then row4 X stays X => rows0-3 A, row4 X, row5 A.

Thus transformed grid:
Row0: col0 A, col1 X, col2 ., col3 ., col4 ., col5 A => "A X . . . A"
Row1: col0 A, col1 A, col2 ., col3 ., col4 ., col5 A => "A A . . . A"
Row2: same as row1: "A A . . . A"
Row3: same: "A A . . . A"
Row4: col0 X, col1 A, col2 ., col3 ., col4 ., col5 X => "X A . . . X"
Row5: col0 A, col1 A, col2 ., col3 ., col4 ., col5 A => "A A . . . A"

Now we need to output a 2x2 quadrant display: double dimensions => 12 rows, 12 columns, consisting of four identical copies of this transformed grid.

Let's see if the given output matches.

Given output:
Row0: "A X . . . A A X . . . A"
Row1: "A A . . . A A A . . . A"
Row2: "A A . . . A A A . . . A"
Row3: "A A . . . A A A . . . A"
Row4: "X A . . . X X A . . . X"
Row5: "A A . . . A A A . . . A"
Row6: "A X . . . A A X . . . A"
Row7: "A A . . . A A A . . . A"
Row8: "A A . . . A A A . . . A"
Row9: "A A . . . A A A . . . A"
Row10:"X A . . . X X A . . . X"
Row11:"A A . 

... [5282 chars truncated]

We need to parse the problem: It's a grid transformation puzzle. The description: "What you see: A blank blueprint with a few colored rivets at various positions. What to do: Draw full azure columns wherever a rivet exists; rivets remain their color. Empty columns stay blank. Print the blueprint as a 2×2 mosaic. Grid details: Output doubles both dimensions with four identical blueprint copies."

Interpretation: Input grid is a blueprint with colored rivets (some colored cells). The transformation: For each column that contains at least one rivet (i.e., any non-blank cell?), we draw a full azure column (i.e., fill entire column with azure 'A') wherever a rivet exists? Actually "Draw full azure columns wherever a rivet exists; rivets remain their color. Empty columns stay blank." Means: For each column that has a rivet (any colored cell), we set all cells in that column to azure 'A', except the rivet cells themselves remain their original color (i.e., we don't overwrite them). For columns with no rivets, we leave them blank (i.e., '.' black). Then we "Print the blueprint as a 2×2 mosaic. Grid details: Output doubles both dimensions with four identical blueprint copies." Means after we produce the transformed blueprint (same dimensions as input?), we then output a grid that is double the dimensions (both width and height multiplied by 2) consisting of four identical copies of the blueprint arranged in a 2x2 mosaic (i.e., tile the blueprint 2 times horizontally and 2 times vertically). So output size = (2 * input height) rows, (2 * input width) columns. Each quadrant is identical to the transformed blueprint.

Thus the transformation: Input grid HxW. For each column j, if any cell (i,j) is not '.' (i.e., a rivet of any color), then for all rows i, set output cell (i,j) = 'A' except if the original cell was a rivet (non '.'), then keep original color. If column has no rivets (all '.'), then output column remains '.' (i.e., all '.').

Thus we need to apply to test input.

Test Input:
. . G .
. . . .
. . . G
G . . .

Let's write as rows:

Row0: . . G .
Row1: . . . .
Row2: . . . G
Row3: G . . .

Indices: columns 0,1,2,3.

Check each column for any non '.':

Column0: cells: row0 '.', row1 '.', row2 '.', row3 'G' -> has a rivet (G at row3). So column0 will become all 'A' except keep rivet at (3,0) as 'G'.

Column1: all '.'? row0 '.', row1 '.', row2 '.', row3 '.' -> no rivet. So column1 stays all '.'.

Column2: row0 'G', row1 '.', row2 '.', row3 '.' -> has rivet at (0,2) G. So column2 becomes all 'A' except keep (0,2) as 'G'.

Column3: row0 '.', row1 '.', row2 'G', row3 '.' -> has rivet at (2,3) G. So column3 becomes all 'A' except keep (2,3) as 'G'.

Thus transformed blueprint (size 4x4) will be:

Initialize all '.' then for each column with rivet, set all to 'A', then put rivets back.

Let's compute:

Column0: set all rows to 'A', then (3,0) = 'G'.
Column1: remains all '.'.
Column2: set all rows to 'A', then (0,2) = 'G'.
Column3: set all rows to 'A', then (2,3) = 'G'.

Now produce rows:

Row0: col0 'A', col1 '.', col2 'G' (rivet), col3 'A' => A . G A
Row1: col0 'A', col1 '.', col2 'A', col3 'A' => A . A A
Row2: col0 'A', col1 '.', col2 'A', col3 'G' (rivet) => A . A G
Row3: col0 'G' (rivet), col1 '.', col2 'A', col3 'A' => G . A A

Thus blueprint:

Row0: A . G A
Row1: A . A A
Row2: A . A G
Row3: G . A A

Now we need to output as a 2x2 mosaic: double dimensions => output height = 2*4 = 8 rows, width = 2*4 = 8 columns. The mosaic consists of four identical copies of the blueprint arranged in a 2x2 grid: top-left, top-right, bottom-left, bottom-right each same as blueprint.

Thus we can produce output by taking the blueprint and repeating it horizontally twice and vertically twice.

So each row of output = blueprint row concatenated with itself (two copies side by side). Then we repeat that pattern for two copies vertically (i.e., output rows = [blueprint rows + blueprint rows] repeated twice? Actually we need 2 copies vertically: so we output the blueprint rows, then again the same blueprint rows.

Thus final output rows: for i in 0..3: output row = blueprint row + blueprint row (horizontal duplication). Then for i in 0..3 again: same.

Let's compute.

Blueprint rows as strings (without spaces): 
r0: "A.GA"
r1: "A.AA"
r2: "A.AG"
r3: "G.AA"

Now horizontal duplication: each row becomes r0+r0 = "A.GA" + "A.GA" = "A.GAA.GA"? Wait careful: concatenation yields "A.GAA.GA"? Actually "A.GA"+"A.GA" = "A.GAA.GA". Let's write characters: A . G A A . G A.

Similarly r1: "A.AA"+"A.AA" = "A.AAA.AA"? Actually "A.AA"+"A.AA" = "A.AAA.AA". Let's compute: A . A A A . A A.

r2: "A.AG"+"A.AG" = "A.AGA.AG"? Actually "A.AG"+"A.AG" = "A.AGA.AG". Characters: A . A G A . A G.

r3: "G.AA"+"G.AA" = "G.AAG.AA"? Actually "G.AA"+"G.AA" = "G.AAG.AA". Characters: G . A A G . A A.

Now vertical duplication: we output these 4 rows, then repeat same 4 rows.

Thus final output 8 rows:

Row0: A . G A A . G A
Row1: A .

... [6702 chars truncated]

We need to parse the problem: It's a grid transformation puzzle. We have a metaphorical description: "What you see: A dark battlefield grid with a few colored signal flares at scattered coordinates. What to do: Each trench line (column) containing a flare is fully lit azure; flares keep their color. Quiet lines stay dark. The map is reproduced in a 2×2 tactical display. Grid details: Output is four identical map tiles, double the original dimensions."

Interpretation: Input grid: some cells are '.' (black) or 'X' (grey) maybe representing dark? Actually '.' is black, X is grey. The flares are colored signals: G (green) etc. The rule: For each column (trench line) that contains a flare (any colored cell? maybe any non-black/grey? Actually flares are colored signals: B,R,G,Y,X? Wait X is grey, but they said "colored signal flares at scattered coordinates." So flares are colored: B,R,G,Y,M,O,A,W? But they gave color key: . = black, B=blue, R=red, G=green, Y=yellow, X=grey, M=magenta, O=orange, A=azure, W=maroon. So flares are any of the colored letters (B,R,G,Y,M,O,A,W). The '.' and 'X' are dark/grey background.

Rule: Each trench line (column) containing a flare is fully lit azure; flares keep their color. Quiet lines stay dark. The map is reproduced in a 2×2 tactical display. Output is four identical map tiles, double the original dimensions.

Interpretation: For each column in input, if there is at least one flare (colored cell) in that column, then in the output, that column (in each tile?) becomes fully azure (A) except where the original flare cells keep their original color (i.e., they are not overwritten). If a column has no flare (i.e., all cells are '.' or 'X' maybe?), then that column stays dark (i.e., remains as original? Or becomes black? "Quiet lines stay dark." Probably means they remain as they were (black/grey) i.e., unchanged? But the description: "Each trench line (column) containing a flare is fully lit azure; flares keep their color. Quiet lines stay dark." So for columns with flare, we set all cells to azure (A) except the flare cells themselves keep their original color. For columns without flare, we leave them as they were (dark). Then "The map is reproduced in a 2×2 tactical display. Grid details: Output is four identical map tiles, double the original dimensions." Means we take the transformed map (same size as input?) and then tile it 2x2, making output dimensions double both width and height (i.e., each dimension multiplied by 2). And the four tiles are identical (i.e., we just repeat the transformed map in a 2x2 block). So output grid is: original transformed map placed at top-left, top-right, bottom-left, bottom-right, each same.

Thus we need to apply to test input.

Test Input:
. . G .
. . . .
. . . G
G . . .

Let's write as rows:

Row0: . . G .
Row1: . . . .
Row2: . . . G
Row3: G . . .

So it's 4 rows, 4 columns.

Columns indices 0-3.

Let's list each column:

Column0: row0 '.', row1 '.', row2 '.', row3 'G' => contains a flare (G at row3). So column0 has flare.

Column1: row0 '.', row1 '.', row2 '.', row3 '.' => all '.' (black). No flare.

Column2: row0 'G', row1 '.', row2 '.', row3 '.' => contains flare (G at row0). So column2 has flare.

Column3: row0 '.', row1 '.', row2 'G', row3 '.' => contains flare (G at row2). So column3 has flare.

Thus columns 0,2,3 have flares; column1 is quiet.

Now apply rule: For each column with flare, set all cells to azure (A) except flare cells keep original color.

Thus we need to produce transformed map of same size (4x4) where:

- For column0: all cells become A, except where original flare (G) at row3 stays G.
- Column1: quiet line stays dark: meaning we keep original? The original column1 is all '.' (black). So stays '.'.
- Column2: all cells become A, except original flare (G) at row0 stays G.
- Column3: all cells become A, except original flare (G) at row2 stays G.

Thus transformed map:

Initialize all cells as per rule.

Let's compute per cell:

Row0:
col0: column0 flare => A (since not flare cell) -> A
col1: quiet => original '.' -> .
col2: column2 flare => but this cell is flare (G) so stays G
col3: column3 flare => not flare cell (original '.'), so A
Thus row0: A . G A

Row1:
col0: column0 flare => A (not flare)
col1: quiet => .
col2: column2 flare => A (not flare)
col3: column3 flare => A (not flare)
Row1: A . A A

Row2:
col0: column0 flare => A
col1: quiet => .
col2: column2 flare => A
col3: column3 flare => flare cell (G) stays G
Row2: A . A G

Row3:
col0: column0 flare => flare cell (G) stays G
col1: quiet => .
col2: column2 flare => A
col3: column3 flare => A
Row3: G . A A

Thus transformed map (4x4):
Row0: A . G A
Row1: A . A A
Row2: A . A G
Row3: G . A A

Now we need to output four identical map tiles, double the original dimensions. Original dimensions: 4 rows, 4 columns. Double => 8 rows, 8 columns. The 2x2 tiling: we place the transformed m

... [7799 chars truncated]

We need to parse the problem: It's a grid transformation puzzle. We have a metaphorical description: "What you see: A dark stage with a few brightly costumed actors at scattered positions. What to do: Flood every wing containing an actor with azure spotlight; actors keep their costume color. Empty wings stay dark. Project the scene onto a 2×2 arrangement of panels. Grid details: Output is double width and height, four copies of the spotlit stage."

Thus the transformation: Input grid (some size) with colored cells (including black '.' as background). Actors are colored cells (non-black?). The "wing" likely refers to rows and columns? Or maybe "wing" is a region? Let's parse: "Flood every wing containing an actor with azure spotlight; actors keep their costume color. Empty wings stay dark." So we have a stage (grid). There are wings (maybe rows and columns?). If an actor is present in a wing (i.e., a row or column?), we flood that entire wing with azure (A) spotlight, but actors keep their original color (so they override? Actually they keep costume color, meaning the cell with actor remains its original color, not overwritten by azure). Empty wings stay dark (i.e., remain '.' black). Then "Project the scene onto a 2×2 arrangement of panels. Grid details: Output is double width and height, four copies of the spotlit stage." So after we have the transformed stage (with azure flood in rows/columns containing actors, actors unchanged), we then replicate that stage into a 2x2 tiling, making output double width and height (i.e., each dimension multiplied by 2). So output is basically the transformed stage repeated in a 2x2 grid (like tile). So each cell of transformed stage becomes a 2x2 block? Actually "double width and height, four copies of the spotlit stage" suggests we take the transformed stage and place four copies arranged in a 2x2 grid, resulting in width*2 and height*2. So each copy is placed side by side and stacked. So output dimensions: if input is H x W, after transformation we have same H x W (spotlit stage). Then output is (2*H) x (2*W) where top-left quadrant = transformed stage, top-right = same, bottom-left = same, bottom-right = same. So essentially we repeat the transformed stage twice horizontally and twice vertically.

Check with examples to confirm.

Example 1:

Input:
. X . . . .
. . . . . .
. . . . . .
. . . . . .
X . . . . X
. . . . . .

Let's write input grid: 6 rows? Actually count lines: 
Row0: . X . . . .
Row1: . . . . . .
Row2: . . . . . .
Row3: . . . . . .
Row4: X . . . . X
Row5: . . . . . .
So 6 rows, 6 columns.

Now we need to apply transformation: Identify actors: non '.' cells? The colored cells: X (grey) maybe considered actor? The description: "brightly costumed actors at scattered positions." The black '.' is dark stage (empty). So any non '.' is an actor (costume). So actors are X's at positions: (0,1), (4,0), (4,5). Also maybe other colors? Only X appears.

Now "Flood every wing containing an actor with azure spotlight". What is a wing? Could be row or column? Let's test: If wing = row, then any row that contains an actor gets flooded with azure (A) across entire row, but actors keep their color (so they'd override the azure? Actually they keep costume color, meaning they remain as they were, not overwritten). Similarly for columns? Or wing could be either rows or columns? Let's test both.

If we flood rows containing actors: rows 0 and 4 have actors. So we set all cells in rows 0 and 4 to A, except where there is an actor (X) we keep X. So row0: originally . X . . . . becomes A X A A A A? Actually we flood with A, but keep X at (0,1). So row0 becomes: A X A A A A. Row4: X . . . . X becomes X A A A A X? Actually we flood row4 with A, keep X at (4,0) and (4,5). So row4 becomes: X A A A A X.

Rows 1,2,3,5 have no actors, stay dark (all '.').

Now columns? If we also flood columns containing actors: columns with actors: column0 has X at (4,0); column1 has X at (0,1); column5 has X at (4,5). So columns 0,1,5 get flooded with A, except keep actors.

If we apply both row and column flooding, we need to see final result.

Let's compute row-flood only first, then see if matches example output after tiling.

After row-flood only, we have:

Row0: A X A A A A
Row1: . . . . . .
Row2: . . . . . .
Row3: . . . . . .
Row4: X A A A A X
Row5: . . . . . .

Now we need to tile 2x2: output dimensions 12x12 (since 6*2). Let's see if example output matches that.

Example output given:

A X . . . A A X . . . A
A A . . . A A A . . . A
A A . . . A A A . . . A
A A . . . A A A . . . A
X A . . . X X A . . . X
A A . . . A A A . . . A
A X . . . A A X . . . A
A A . . . A A A . . . A
A A . . . A A A . . . A
A A . . . A A A . . . A
X A . . . X X A . . . X
A A . . . A A A . . . A

Let's format as 12 rows, each row seems to have 12 characters (with spaces?). Actually they show spaces between characters? In the example they wrote: "A X . . . A A X . . . A" etc. That seems to have spaces between eac

... [7673 chars truncated]

We need to parse the problem: It's a grid transformation puzzle. The description: "What you see: Dark garden soil with a few colored flowers planted at scattered positions. What to do: Water every furrow (column) containing a flower, turning soil azure; flowers keep their color. Dry furrows stay dark. Display the garden as a 2×2 tiling. Grid details: Output doubles both dimensions with four identical garden tiles."

Interpretation: Input grid: dark soil (black '.'), colored flowers (colored letters: G, R, B, Y, etc). The transformation: For each column (furrow) that contains at least one flower (any colored cell), we water it: turn soil (i.e., '.' cells) in that column to azure (A). Flowers keep their color (i.e., remain as they are). Dry furrows (columns with no flowers) stay dark (i.e., '.' remain '.'? Actually dark soil is '.'; after watering, soil becomes azure; but if no flower, remains dark (black). So '.' in columns without any flower remain '.'.

Then "Display the garden as a 2×2 tiling. Grid details: Output doubles both dimensions with four identical garden tiles." Means after applying the transformation to the input grid, we then tile that resulting grid 2x2 (i.e., repeat it twice horizontally and twice vertically) to produce output with dimensions doubled both width and height. So output is 2x the rows and 2x the columns of the transformed grid, formed by placing four copies of the transformed grid in a 2x2 arrangement.

Thus we need to compute transformed grid: For each column, if any non '.' cell (i.e., any flower) present, then all '.' cells in that column become 'A' (azure). The flowers themselves remain unchanged (their color letters). Then we tile.

Now test input:

. . G .
. . . .
. . . G
G . . .

Let's write as rows:

Row0: . . G .
Row1: . . . .
Row2: . . . G
Row3: G . . .

Indices: columns 0,1,2,3.

Check each column for any flower (non '.'):
- Column0: rows: ., ., ., G => has G at row3 => flower present.
- Column1: rows: ., ., ., . => all '.' => no flower.
- Column2: rows: G, ., ., . => has G at row0 => flower present.
- Column3: rows: ., ., G, . => has G at row2 => flower present.

Thus columns 0,2,3 have flowers; column1 is dry.

Now transform: For each column with flower, turn '.' to 'A'; keep flowers as is.

Let's compute transformed grid (size 4x4):

Initialize with original, then replace '.' in wet columns.

Column0 (wet): rows:
- row0: '.' -> 'A'
- row1: '.' -> 'A'
- row2: '.' -> 'A'
- row3: 'G' stays 'G'

Column1 (dry): stays '.' all rows.

Column2 (wet):
- row0: 'G' stays 'G'
- row1: '.' -> 'A'
- row2: '.' -> 'A'
- row3: '.' -> 'A'

Column3 (wet):
- row0: '.' -> 'A'
- row1: '.' -> 'A'
- row2: 'G' stays 'G'
- row3: '.' -> 'A'

Thus transformed grid:

Row0: col0 A, col1 ., col2 G, col3 A => A . G A
Row1: A . A A? Wait col2 A, col3 A => row1: A . A A
Row2: A . A G? Actually col2 A, col3 G => row2: A . A G
Row3: G . A A? col0 G, col1 ., col2 A, col3 A => G . A A

Let's write each row as characters separated by spaces for clarity:

Row0: A . G A
Row1: A . A A
Row2: A . A G
Row3: G . A A

Now we need to output a 2x2 tiling: double both dimensions, i.e., output size 8x8 (since original 4x4). The tiling consists of four identical copies of the transformed grid arranged in a 2x2 block: top-left, top-right, bottom-left, bottom-right each same as transformed grid.

Thus output grid rows 0-3 = transformed grid rows 0-3; rows 4-7 = same as rows 0-3 again (since vertical repeat). Similarly columns 0-3 = transformed grid columns 0-3; columns 4-7 = same as columns 0-3 again.

Thus we can produce output by repeating each row twice horizontally (concatenate row with itself) and then repeat the whole set of rows twice vertically.

Let's compute.

Transformed grid rows as strings (without spaces): 
Row0: A . G A => "A.GA"? Actually characters: A, ., G, A => "A.GA"
Row1: A . A A => "A.AA"
Row2: A . A G => "A.AG"
Row3: G . A A => "G.AA"

Now double horizontally: each row becomes row + row.

Row0 doubled: "A.GA" + "A.GA" = "A.GAA.GA"
Row1 doubled: "A.AA" + "A.AA" = "A.AAA.AA"
Row2 doubled: "A.AG" + "A.AG" = "A.AGA.AG"
Row3 doubled: "G.AA" + "G.AA" = "G.AAG.AA"

Now vertical double: we need to output rows 0-3 as above, then rows 4-7 same as rows 0-3.

Thus final output 8 rows:

Row0: A . G A A . G A
Row1: A . A A A . A A
Row2: A . A G A . A G
Row3: G . A A G . A A
Row4: same as Row0
Row5: same as Row1
Row6: same as Row2
Row7: same as Row3

Now we need to output with spaces between characters. Let's write each row with spaces.

Row0: A . G A A . G A
Row1: A . A A A . A A
Row2: A . A G A . A G
Row3: G . A A G . A A
Row4: A . G A A . G A
Row5: A . A A A . A A
Row6: A . A G A . A G
Row7: G . A A G . A A

Check: Are there any other colors? Only '.' and 'A' and 'G'. Good.

Now we need to output in the format:

ANSWER:
. B R
G Y .
(one row per line, characters separated by spaces)

Thus we need to output:

ANSWER:
A . G A

... [6128 chars truncated]