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Signed-off-by: Jess Frazelle <[email protected]>
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‎results/prompt_56.md

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# Prompt #56: An ASCII 6×6 square has an 'O' hole at coordinate (1,4). Reflect the entire square (with hole) across the line y = x and show the result.
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## OpenAI o3 (30 secs)
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The reflection simply swaps the x- and y-coordinates of every point, so the hole that was at (1,4) moves to (4,1).
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Using “#” for the square’s edges, spaces for the interior, and “O” for the hole, the 6 × 6 square after the reflection is:
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```
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######
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# O#
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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 (14 secs)
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I need to reflect a 6×6 ASCII square with a hole at coordinate (1,4) across the line y = x.
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First, let me set up the original 6×6 square with the hole at (1,4). Using 0-based indexing where (0,0) is the top-left:
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- Row 1 (index 1), Column 4 (index 4) will have the hole 'O'
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Original square:
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```
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######
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####O#
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######
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######
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######
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######
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```
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When reflecting across the line y = x, each point (x,y) maps to (y,x). This means:
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- The hole at (1,4) will move to (4,1)
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- This is equivalent to transposing the matrix
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Reflected square:
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```
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######
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######
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######
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######
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#O####
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######
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```
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The hole 'O' is now at coordinate (4,1) - row 4, column 1 (using 0-based indexing).
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## Gemini Pro (39 secs)
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Okay, let's break this down.
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1. **The Grid and Coordinate System:**
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We have a 6x6 ASCII square. We'll assume a coordinate system where (row, column) starts from (0,0) at the top-left.
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So, the grid ranges from (0,0) to (5,5).
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The hole is at coordinate (1,4), meaning row 1, column 4.
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2. **Initial State:**
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Let's represent the square with '*' and the hole with 'O'.
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```
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(col) 0 1 2 3 4 5
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(row)
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0 * * * * * *
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1 * * * * O * <-- Hole at (1,4)
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2 * * * * * *
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3 * * * * * *
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4 * * * * * *
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5 * * * * * *
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```
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3. **Reflection Across the Line y = x:**
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When you reflect a point (x, y) across the line y = x, the new coordinates become (y, x).
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In our grid context, if we consider our coordinates as (row, col):
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* Let the original row be `r` and original column be `c`. So, the hole is at `(r, c) = (1, 4)`.
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* When reflecting across the line where the row index equals the column index (analogous to y=x), the new coordinates `(r', c')` will be `(c, r)`.
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* So, the new position of the hole will be `(4, 1)`.
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4. **Resulting State:**
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The 6x6 square itself remains a 6x6 square. Only the position of the hole changes.
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The hole moves from (1,4) to (4,1).
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```
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(col) 0 1 2 3 4 5
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(row)
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0 * * * * * *
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1 * * * * * *
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2 * * * * * *
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3 * * * * * *
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4 * O * * * * <-- Hole at (4,1)
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5 * * * * * *
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```
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**Initial Square (hole at (1,4)):**
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(Using 0-indexed rows and columns, (row, col))
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```
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* * * * * *
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* * * * O *
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* * * * * *
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* * * * * *
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* * * * * *
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* * * * * *
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```
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**Reflected Square (hole at (4,1)):**
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```
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* * * * * *
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* * * * * *
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* * * * * *
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* * * * * *
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* O * * * *
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* * * * * *
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```
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## DeepSeek Reasoner (234 secs)
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To reflect the 6×6 ASCII square with a hole at coordinate (1,4) across the line \(y = x\), the entire grid is transposed. This means that the row and column indices are swapped for each cell. Specifically, the hole at (row 1, column 4) in the original grid moves to (row 4, column 1) in the reflected grid.
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The original grid has all cells as '.' (representing solid) except for the hole at (1,4), which is 'O'. After reflection, the new grid has all cells as '.' except for the hole at (4,1), which is 'O'.
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### Original Grid (for reference):
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- Row 1: . . . O . . (→ "...O..")
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- Row 2: . . . . . . (→ "......")
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- Row 3: . . . . . . (→ "......")
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- Row 4: . . . . . . (→ "......")
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- Row 5: . . . . . . (→ "......")
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- Row 6: . . . . . . (→ "......")
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### Reflected Grid (result after reflection across \(y = x\)):
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- Row 1: . . . . . . (→ "......")
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- Row 2: . . . . . . (→ "......")
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- Row 3: . . . . . . (→ "......")
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- Row 4: O . . . . . (→ "O.....")
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- Row 5: . . . . . . (→ "......")
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- Row 6: . . . . . . (→ "......")
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The ASCII representation of the reflected grid is:
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```
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......
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......
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......
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O.....
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......
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......
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```
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In this output, the hole 'O' is at row 4, column 1 (i.e., the fourth line, first character). All other positions are filled with '.'.
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\boxed{\text{......}} \\
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\boxed{\text{......}} \\
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\boxed{\text{......}} \\
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\boxed{\text{O.....}} \\
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\boxed{\text{......}} \\
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\boxed{\text{......}}
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## Grok 3 (7 secs)
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To solve this problem, we will first represent a 6×6 ASCII square, place an 'O' hole at the specified coordinate (1,4), reflect the entire square across the line y = x, and then display the resulting grid. Let's break this down step by step.
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### Step 1: Understand the Initial Setup
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- A 6×6 grid has rows and columns indexed from 0 to 5 (since it's a 6×6 square).
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- Coordinates are typically given as (row, column). So, (1,4) refers to row 1, column 4 (0-based indexing).
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- We'll assume the grid is initially filled with a character like '#' to represent the square, and 'O' represents the hole at position (1,4).
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Initial grid (before reflection):
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- Rows: 0 to 5
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- Columns: 0 to 5
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- Hole at (1,4) means row 1, column 4.
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Here's the initial grid:
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```
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# # # # # #
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# # # # O #
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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 2: Understand Reflection Across y = x
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- Reflecting a point (x, y) across the line y = x swaps the coordinates to (y, x).
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- For a grid, this means that the element at position (i, j) moves to position (j, i).
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- This effectively transposes the grid (swaps rows and columns).
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### Step 3: Apply the Reflection
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- The hole at (1,4) will move to (4,1) after reflection (since row becomes column and vice versa).
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- All other positions remain filled with '#'.
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Reflected grid (after swapping rows and columns):
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- Original position (1,4) → New position (4,1)
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- The rest of the grid structure remains the same, just transposed.
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Here's the reflected 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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# O # # # #
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# # # # # #
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```
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### Step 4: Display the Result
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The final result after reflecting the 6×6 square with a hole at (1,4) across the line y = x 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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# O # # # #
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# # # # # #
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```
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This grid shows the hole 'O' now at position (4,1), which is the result of the reflection.
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