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Signed-off-by: Jess Frazelle <[email protected]>
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‎results/prompt_53.md

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# Prompt #53: Draw a block letter 'G' on a fixed-width grid. Now rotate it 90° clockwise **in-plane** and THEN reflect it across the vertical mid-line of the grid. Provide the final ASCII drawing.
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## OpenAI o3 (36 secs)
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```
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####
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# #
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# #
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# # #
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# # #
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# ###
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#
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```
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## Claude 4 Sonnet (8 secs)
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I'll work through this step-by-step.
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First, let me draw a block letter 'G' on a fixed-width grid:
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```
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#####
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# #
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#
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# ###
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# #
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# #
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#####
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```
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Now, rotating 90° clockwise in-plane:
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- Top row becomes rightmost column
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- Left column becomes top row
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- Bottom row becomes leftmost column
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- Right column becomes bottom row
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After 90° clockwise rotation:
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```
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#######
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# #
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# #
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# # #
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# #
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#####
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```
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Now reflecting across the vertical mid-line:
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- The leftmost column becomes rightmost
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- The rightmost column becomes leftmost
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- Middle columns swap symmetrically
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Final result after reflection:
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```
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#######
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# #
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# #
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# # #
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# #
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#####
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```
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## Gemini Pro (160 secs)
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Okay, let's break this down step-by-step.
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1. **Choose a Block Letter 'G' and Grid Size:**
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I'll use a 7x7 grid. For the 'G' itself, I'll design a 5x5 character block, which will be centered within the 7x7 grid (1-pixel padding all around). I'll use '#' for the letter and '.' for the background.
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Initial 'G' (5x5 letter part):
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```
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#####
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#....
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#.###
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#...#
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#####
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```
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Placed in the 7x7 grid (let's call this `Grid_Initial`):
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```
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.......
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.#####.
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.#.....
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.#.###.
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.#...#.
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.#####.
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.......
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```
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2. **Rotate 90° Clockwise In-Plane:**
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The 7x7 grid itself rotates. A point `(row, col)` in the original grid moves to `(col, GridHeight - 1 - row)`. Here, GridHeight = 7. So, `(row, col)` -> `(col, 6 - row)`.
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Let's consider the 5x5 letter part first. Original letter part `L[r][c]` (0-indexed, 0 to 4). New letter part `L_rot[nr][nc]`. `nr = c`, `nc = (5-1)-r = 4-r`.
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* `L[0][0..4] = #####` -> `L_rot[0][4]=#`, `L_rot[1][4]=#`, `L_rot[2][4]=#`, `L_rot[3][4]=#`, `L_rot[4][4]=#`
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* `L[1][0..4] = #....` -> `L_rot[0][3]=#`, `L_rot[1][3]=.`, `L_rot[2][3]=.`, `L_rot[3][3]=.`, `L_rot[4][3]=.`
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* `L[2][0..4] = #.###` -> `L_rot[0][2]=#`, `L_rot[1][2]=.`, `L_rot[2][2]=#`, `L_rot[3][2]=#`, `L_rot[4][2]=#`
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* `L[3][0..4] = #...#` -> `L_rot[0][1]=#`, `L_rot[1][1]=.`, `L_rot[2][1]=.`, `L_rot[3][1]=.`, `L_rot[4][1]=#`
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* `L[4][0..4] = #####` -> `L_rot[0][0]=#`, `L_rot[1][0]=#`, `L_rot[2][0]=#`, `L_rot[3][0]=#`, `L_rot[4][0]=#`
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Reconstructing `L_rot` row by row:
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Row 0: `#####` (from `L_rot[0][0]` to `L_rot[0][4]`)
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Row 1: `#...#` (from `L_rot[1][0]` to `L_rot[1][4]`)
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Row 2: `#.#.#`
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Row 3: `#.#.#`
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Row 4: `###.#`
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So the rotated 5x5 letter part looks like this:
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```
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#####
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#...#
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#.#.#
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#.#.#
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###.#
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```
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This rotated letter part is still 5x5, so it remains centered in the 7x7 grid.
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`Grid_Rotated`:
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```
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.......
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.#####.
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.#...#.
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.#.#.#.
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.#.#.#.
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.###.#.
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.......
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```
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3. **Reflect Across the Vertical Mid-line:**
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The 7x7 grid has columns 0, 1, 2, 3, 4, 5, 6. The vertical mid-line is column 3.
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Reflection means a point `(row, col)` moves to `(row, GridWidth - 1 - col)`. Here, GridWidth = 7. So, `(row, col)` -> `(row, 6 - col)`.
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Let `Grid_Final` be the result. `Grid_Final[r][c] = Grid_Rotated[r][6-c]`.
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* Row 0: `Grid_Rotated[0]` is `.......`. It's symmetric, so `Grid_Final[0]` is `.......`.
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* Row 1: `Grid_Rotated[1]` is `.#####.`. It's symmetric, so `Grid_Final[1]` is `.#####.`.
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* Row 2: `Grid_Rotated[2]` is `.#...#.`. It's symmetric, so `Grid_Final[2]` is `.#...#.`.
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* Row 3: `Grid_Rotated[3]` is `.#.#.#.`. It's symmetric, so `Grid_Final[3]` is `.#.#.#.`.
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* Row 4: `Grid_Rotated[4]` is `.#.#.#.`. It's symmetric, so `Grid_Final[4]` is `.#.#.#.`.
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* Row 5: `Grid_Rotated[5]` is `.###.#.`
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`Grid_Final[5][0] = Grid_Rotated[5][6] = .`
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`Grid_Final[5][1] = Grid_Rotated[5][5] = #`
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`Grid_Final[5][2] = Grid_Rotated[5][4] = .`
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`Grid_Final[5][3] = Grid_Rotated[5][3] = #`
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`Grid_Final[5][4] = Grid_Rotated[5][2] = #`
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`Grid_Final[5][5] = Grid_Rotated[5][1] = #`
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`Grid_Final[5][6] = Grid_Rotated[5][0] = .`
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So, `Grid_Final[5]` is `.#.###.`.
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* Row 6: `Grid_Rotated[6]` is `.......`. It's symmetric, so `Grid_Final[6]` is `.......`.
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Final ASCII Drawing:
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```
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.......
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.#####.
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.#...#.
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.#.#.#.
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.#.#.#.
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.#.###.
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.......
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```
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## DeepSeek Reasoner (519 secs)
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After rotating the block letter 'G' 90° clockwise in-plane and then reflecting it across the vertical mid-line, the final ASCII drawing is as follows. The original 'G' was drawn on a 7-row by 5-column grid. After rotation, the grid becomes 5 rows by 7 columns. The reflection step swaps columns symmetrically across the vertical midline (column index 3 in a 0-based 7-column grid).
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**Final ASCII Drawing:**
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```
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*****
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* *
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* * *
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* * *
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* ***
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```
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### Explanation:
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- **Row 0:** Space, five asterisks, space (` ***** `).
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- **Row 1:** Asterisk, five spaces, asterisk (`* *`).
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- **Row 2:** Asterisk, two spaces, asterisk, two spaces, asterisk (`* * *`).
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- **Row 3:** Same as Row 2 (`* * *`).
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- **Row 4:** Space, asterisk, space, three asterisks, space (` * *** `).
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This grid represents the transformed 'G' after the specified operations. The drawing uses spaces (not shown explicitly) and asterisks (`*`) for clarity.
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## Grok 3 (20 secs)
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Let's go through this step by step to draw a block letter 'G', rotate it 90° clockwise in-plane, and then reflect it across the vertical mid-line of the grid. I'll use a fixed-width grid (let's say 5x5 for simplicity, as it fits a block 'G' well) and represent the letter with '#' characters and empty spaces with '.'.
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### Step 1: Draw the Block Letter 'G'
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A block letter 'G' on a 5x5 grid can look like this, with the main structure resembling the shape of a 'G':
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```
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. # # # .
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. # . . .
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. # # # .
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. . . # .
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. # # # .
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```
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- The top row forms the top bar of the 'G'.
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- The left column forms the vertical stroke.
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- The bottom rows curve back to form the lower part and the middle horizontal stroke.
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### Step 2: Rotate 90° Clockwise In-Plane
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Rotating a grid 90° clockwise means each position (i,j) moves to (j, n-1-i), where n is the grid size (5 here). So, let's map each '#' to its new position:
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- Original grid (row, col):
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- Row 0: (0,1), (0,2), (0,3) → '#'
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- Row 1: (1,1) → '#'
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- Row 2: (2,1), (2,2), (2,3) → '#'
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- Row 3: (3,3) → '#'
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- Row 4: (4,1), (4,2), (4,3) → '#'
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After 90° clockwise rotation (new_row = old_col, new_col = 4 - old_row):
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- (0,1) → (1,4)
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- (0,2) → (2,4)
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- (0,3) → (3,4)
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- (1,1) → (1,3)
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- (2,1) → (1,2)
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- (2,2) → (2,2)
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- (2,3) → (3,2)
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- (3,3) → (3,1)
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- (4,1) → (1,0)
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- (4,2) → (2,0)
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- (4,3) → (3,0)
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The rotated grid becomes:
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```
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. . . . .
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# . # . #
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# . # . #
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# . # # #
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. . . . .
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```
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### Step 3: Reflect Across the Vertical Mid-Line
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The vertical mid-line of a 5x5 grid is between columns 2 and 3 (since 5/2 = 2.5, so we flip across col=2). Reflection means swapping columns 0 with 4, and 1 with 3. So, for each row:
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- Col 0 swaps with Col 4
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- Col 1 swaps with Col 3
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- Col 2 stays the same
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Applying this to the rotated grid:
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- Row 0: . . . . . → . . . . .
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- Row 1: # . # . # → # . # . #
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- Row 2: # . # . # → # . # . #
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- Row 3: # . # # # → # # # . #
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- Row 4: . . . . . → . . . . .
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In a 5x5 grid, reflecting across the vertical mid-line (col=2 as the center):
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- Col 0 becomes Col 4
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- Col 1 becomes Col 3
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- Col 2 stays Col 2
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- Col 3 becomes Col 1
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- Col 4 becomes Col 0
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So, for each row:
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- Row 1: # . # . # → # . # . #
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- Pos 0 (#) → Pos 4 (#)
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- Pos 1 (.) → Pos 3 (.)
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- Pos 2 (#) stays (#)
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- Pos 3 (.) → Pos 1 (.)
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- Pos 4 (#) → Pos 0 (#)
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Result: # . # . # (same as original row, coincidentally)
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- Row 2: # . # . # → # . # . # (same as above)
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- Row 3: # . # # # → # # # . #
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- Pos 0 (#) → Pos 4 (#)
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- Pos 1 (.) → Pos 3 (.)
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- Pos 2 (#) stays (#)
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- Pos 3 (#) → Pos 1 (#)
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- Pos 4 (#) → Pos 0 (#)
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Result: # # # . #
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The final grid after reflection is:
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```
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. . . . .
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# . # . #
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# . # . #
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# # # . #
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. . . . .
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```
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### Final ASCII Drawing
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The final result after rotating the block letter 'G' 90° clockwise and reflecting it across the vertical mid-line is:
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```
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. . . . .
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# . # . #
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# . # . #
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# # # . #
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. . . . .
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```
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This is the final ASCII representation of the transformed 'G'. Note that due to the symmetry in some rows, the reflection didn't alter the appearance drastically in rows 1 and 2, but row 3 shows the effect of the flip. If you visualize it, the original 'G' has been transformed into a new shape that reflects the specified operations.
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