60

Signed-off-by: Jess Frazelle <[email protected]>

Author: jessfraz

results/prompt_60.md

@@ -0,0 +1,212 @@
+# Prompt #60: ASCII spiral: '@' at (0,0), '#' at (1,0), '$' at (1,1), '%' at (0,1). Unwind the spiral by rotating those four points 90° CCW about the square’s center. Output the new 2×2 grid.
+
+## OpenAI o3  (13 secs)
+%@
+$#
+
+## Claude 4 Sonnet  (7 secs)
+I need to rotate the 2×2 grid 90° counterclockwise about its center.
+
+Initial grid:
+```
+% $
+@ #
+```
+
+The center of a 2×2 grid is at (0.5, 0.5).
+
+For a 90° CCW rotation about the center:
+- (0,0) → (1,0)
+- (1,0) → (1,1)
+- (1,1) → (0,1)
+- (0,1) → (0,0)
+
+So:
+- '@' at (0,0) moves to (1,0)
+- '#' at (1,0) moves to (1,1)
+- '$' at (1,1) moves to (0,1)
+- '%' at (0,1) moves to (0,0)
+
+New grid:
+```
+$ #
+% @
+```
+
+## Gemini Pro  (18 secs)
+Okay, let's break this down.
+
+**1. Initial Grid & Coordinates:**
+
+We can represent the initial grid and the coordinates (x,y) of each character, assuming (0,0) is the top-left:
+
+*   `@` at (0,0)
+*   `#` at (1,0)
+*   `%` at (0,1)
+*   `$` at (1,1)
+
+Visually:
+```
+@ #
+% $
+```
+
+**2. Center of Rotation:**
+
+The square is defined by the points (0,0), (1,0), (0,1), and (1,1).
+The center of this square is at x = (0+1)/2 = 0.5, and y = (0+1)/2 = 0.5.
+So, the center of rotation P is (0.5, 0.5).
+
+**3. 90° CCW Rotation Formula:**
+
+To rotate a point (x,y) 90° CCW around a point (px,py) to get (x',y'):
+x' = px - (y - py)
+y' = py + (x - px)
+
+In our case, (px,py) = (0.5, 0.5):
+x' = 0.5 - (y - 0.5) = 0.5 - y + 0.5 = 1 - y
+y' = 0.5 + (x - 0.5) = 0.5 + x - 0.5 = x
+
+So, the new point (x', y') is (1-y, x).
+
+**4. Apply Rotation to Each Point:**
+
+*   **@ (0,0):**
+    x' = 1 - 0 = 1
+    y' = 0
+    New position for '@': (1,0)
+
+*   **# (1,0):**
+    x' = 1 - 0 = 1
+    y' = 1
+    New position for '#': (1,1)
+
+*   **$ (1,1):**
+    x' = 1 - 1 = 0
+    y' = 1
+    New position for '$': (0,1)
+
+*   **% (0,1):**
+    x' = 1 - 1 = 0
+    y' = 0
+    New position for '%': (0,0)
+
+**5. Construct the New Grid:**
+
+Mapping the characters to their new (x,y) coordinates:
+*   (0,0) is now occupied by '%'
+*   (1,0) is now occupied by '@'
+*   (0,1) is now occupied by '$'
+*   (1,1) is now occupied by '#'
+
+So the new 2x2 grid is:
+
+```
+% @
+$ #
+```
+
+## DeepSeek Reasoner  (246 secs)
+The task involves rotating the given 2×2 grid of characters 90 degrees counter-clockwise (CCW) about the center of the square. The original positions are:
+
+- '@' at (0,0)
+- '#' at (1,0)
+- '$' at (1,1)
+- '%' at (0,1)
+
+The center of the square is at (0.5, 0.5). To rotate a point (x, y) 90° CCW around (0.5, 0.5), use the transformation:
+- New x = 1 - y
+- New y = x
+
+Apply this transformation to each point:
+
+- '@' at (0,0):  
+  New position: (1 - 0, 0) = (1, 0)  
+  So, '@' moves to (1,0)
+
+- '#' at (1,0):  
+  New position: (1 - 0, 1) = (1, 1)  
+  So, '#' moves to (1,1)
+
+- '$' at (1,1):  
+  New position: (1 - 1, 1) = (0, 1)  
+  So, '$' moves to (0,1)
+
+- '%' at (0,1):  
+  New position: (1 - 1, 0) = (0, 0)  
+  So, '%' moves to (0,0)
+
+After rotation, the new grid assignments are:
+- (0,0): '%'
+- (1,0): '@'
+- (0,1): '$'
+- (1,1): '#'
+
+Output the grid row by row, with the top row first (y=0) and the bottom row second (y=1):
+- Top row (y=0): (0,0) and (1,0) → "%@"
+- Bottom row (y=1): (0,1) and (1,1) → "$#"
+
+Thus, the new 2×2 grid is:
+
+```
+%@
+$#
+```
+
+## Grok 3  (7 secs)
+Let's solve this step by step.
+
+1. First, we identify the initial positions of the symbols in the 2×2 grid:
+   - '@' at (0,0) — top-left
+   - '#' at (1,0) — top-right
+   - '$' at (1,1) — bottom-right
+   - '%' at (0,1) — bottom-left
+
+   This forms the initial grid:
+   ```
+   @ #
+   % $
+   ```
+
+2. Next, we need to rotate these points 90° counterclockwise (CCW) around the center of the 2×2 grid. The center of a 2×2 grid, with coordinates from (0,0) to (1,1), is at (0.5, 0.5).
+
+3. To rotate a point (x, y) 90° CCW around a center (cx, cy), we use the transformation:
+   - New x = cx - (y - cy)
+   - New y = cy + (x - cx)
+
+   Applying this to each point with cx = 0.5 and cy = 0.5:
+   - For '@' at (0,0):
+     - New x = 0.5 - (0 - 0.5) = 0.5 - (-0.5) = 1
+     - New y = 0.5 + (0 - 0.5) = 0.5 - 0.5 = 0
+     - New position: (1,0)
+   - For '#' at (1,0):
+     - New x = 0.5 - (0 - 0.5) = 0.5 - (-0.5) = 1
+     - New y = 0.5 + (1 - 0.5) = 0.5 + 0.5 = 1
+     - New position: (1,1)
+   - For '$' at (1,1):
+     - New x = 0.5 - (1 - 0.5) = 0.5 - 0.5 = 0
+     - New y = 0.5 + (1 - 0.5) = 0.5 + 0.5 = 1
+     - New position: (0,1)
+   - For '%' at (0,1):
+     - New x = 0.5 - (1 - 0.5) = 0.5 - 0.5 = 0
+     - New y = 0.5 + (0 - 0.5) = 0.5 - 0.5 = 0
+     - New position: (0,0)
+
+4. Now, we place the symbols in their new positions:
+   - (0,0): '%'
+   - (1,0): '@'
+   - (0,1): '$'
+   - (1,1): '#'
+
+   This forms the new grid:
+   ```
+   % @
+   $ #
+   ```
+
+**Output:**
+```
+% @
+$ #
+```
+