Initial Query

Convert this game of life pattern 26$13b2o$14bo$14bobo$15b2o14$31b2o$30b2o$32bo13$46b2o$45b2o$47bo13$61b 2o$60b2o$62bo4$52b2o$52b2o$71bo$66b2ob3o$66bobobo$68bo$67b2o3$76b2o$ 60b3o12b2o10bobo$52b3o7bo14bo8bo2bo4bo$51bo3bo5bo10b2o11b2o5b2o$50bo5b o15bobo8b2o3bo8b2o$51bo3bo11b2o6bo9b2o10b2o$52b3o8b2obo2bo2bo2bo10bo2b o$52bo2bo7b2o2b2o6bo11bobo$53b3o16bobo$53b2obo15b2o$54bobo$55bo3$52b2o 3b2o$52bobobobo$53b5o$54b3o$55bo7$54b2o$54b2o! to a numpy array. Call the function `rle_to_arr(string, h, w)` which should return a numpy array with .shape (h,w). Fill the array with 1s, and empty with 0.

LLM Generation

Query

Convert this game of life pattern 26$13b2o$14bo$14bobo$15b2o14$31b2o$30b2o$32bo13$46b2o$45b2o$47bo13$61b 2o$60b2o$62bo4$52b2o$52b2o$71bo$66b2ob3o$66bobobo$68bo$67b2o3$76b2o$ 60b3o12b2o10bobo$52b3o7bo14bo8bo2bo4bo$51bo3bo5bo10b2o11b2o5b2o$50bo5b o15bobo8b2o3bo8b2o$51bo3bo11b2o6bo9b2o10b2o$52b3o8b2obo2bo2bo2bo10bo2b o$52bo2bo7b2o2b2o6bo11bobo$53b3o16bobo$53b2obo15b2o$54bobo$55bo3$52b2o 3b2o$52bobobobo$53b5o$54b3o$55bo7$54b2o$54b2o! to a numpy array. Call the function `rle_to_arr(string, h, w)` which should return a numpy array with .shape (h,w). Fill the array with 1s, and empty with 0.

Output

Here’s one complete solution in Python. For example, you might write: ---------------------------------------------------------------- import numpy as np import re def rle_to_arr(rle, h, w): """ Converts an RLE string (as commonly used for Game of Life patterns) into a numpy array with shape (h, w). Live cells are marked with 1's and dead cells with 0's. Parameters: rle (str): The RLE encoded pattern string. h (int): Height of the output array. w (int): Width of the output array. Returns: numpy.ndarray: An array of shape (h, w) with the decoded pattern. The RLE format supports: - An optional count before a character. - 'o' for a live cell. - 'b' for a dead cell. - '$' to indicate the end of a row (moving down). - '!' marks the end of the pattern. """ grid = np.zeros((h, w), dtype=int) row = 0 col = 0 i = 0 # Remove any newlines or spaces from the input. rle = rle.replace("\n","").replace(" ", "") while i < len(rle): # Stop if we hit the termination marker. if rle[i] == '!': break # Parse an optional number (count) count_str = '' while i < len(rle) and rle[i].isdigit(): count_str += rle[i] i += 1 # Default count is 1 if no number is given. count = int(count_str) if count_str else 1 # Make sure we haven't run off the string. if i >= len(rle): break char = rle[i] i += 1 if char == 'b': # dead cells (note: they are already 0) col += count # skip these positions elif char == 'o': # live cells # Fill count cells with 1 # Make sure we don't go out of the width. for _ in range(count): if col < w and row < h: grid[row, col] = 1 col += 1 elif char == '$': # end of line, move down "count" times. row += count col = 0 else: # For any unexpected character, you might want to raise an error. raise ValueError("Unexpected character in RLE: " + char) return grid # Example usage: if __name__ == '__main__': rle_string = ("26$13b2o$14bo$14bobo$15b2o14$31b2o$30b2o$32bo13$46b2o$45b2o$47bo13$61b" "2o$60b2o$62bo4$52b2o$52b2o$71bo$66b2ob3o$66bobobo$68bo$67b2o3$76b2o$" "60b3o12b2o10bobo$52b3o7bo14bo8bo2bo4bo$51bo3bo5bo10b2o11b2o5b2o$50bo5b" "o15bobo8b2o3bo8b2o$51bo3bo11b2o6bo9b2o10b2o$52b3o8b2obo2bo2bo2bo10bo2b" "o$52bo2bo7b2o2b2o6bo11bobo$53b3o16bobo$53b2obo15b2o$54bobo$55bo3$52b2o" "3b2o$52bobobobo$53b5o$54b3o$55bo7$54b2o$54b2o!") # Define the desired output dimensions (example dimensions). height = 80 width = 80 array = rle_to_arr(rle_string, height, width) print(array) ---------------------------------------------------------------- Explanation: 1. The function first creates a grid of zeros with shape (h, w). 2. It then “walks” through the RLE string. Before each significant character you might have a number (the count). If there isn’t one the count is taken as 1. 3. When the character is: • 'b': it advances the column pointer by that many cells (cells remain 0, so you can simply skip them). • 'o': it writes that many live cells (1’s) to the array. • '$': it moves to the beginning of the next row. If a count prefixed the '$', then it means move down several rows. 4. The function stops reading once it encounters an exclamation point ('!'). You can adjust the height and width as needed so that the whole pattern fits. This function converts the RLE to a numpy array exactly as requested.

Extract Code

I extracted the following code from that output: 

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import numpy as np
import re

def rle_to_arr(rle, h, w):
    """
    Converts an RLE string (as commonly used for Game of Life patterns) into
    a numpy array with shape (h, w). Live cells are marked with 1's and dead cells with 0's.
    
    Parameters:
      rle (str): The RLE encoded pattern string.
      h (int): Height of the output array.
      w (int): Width of the output array.
      
    Returns:
      numpy.ndarray: An array of shape (h, w) with the decoded pattern.
    
    The RLE format supports:
         - An optional count before a character.
         - 'o' for a live cell.
         - 'b' for a dead cell.
         - '$' to indicate the end of a row (moving down).
         - '!' marks the end of the pattern.
    """
    
    grid = np.zeros((h, w), dtype=int)
    row = 0
    col = 0
    i = 0
    # Remove any newlines or spaces from the input.
    rle = rle.replace("\n","").replace(" ", "")
    
    while i < len(rle):
        # Stop if we hit the termination marker.
        if rle[i] == '!':
            break
        
        # Parse an optional number (count)
        count_str = ''
        while i < len(rle) and rle[i].isdigit():
            count_str += rle[i]
            i += 1
        # Default count is 1 if no number is given.
        count = int(count_str) if count_str else 1
        
        # Make sure we haven't run off the string.
        if i >= len(rle):
            break
        
        char = rle[i]
        i += 1
        
        if char == 'b':  # dead cells (note: they are already 0)
            col += count  # skip these positions
        elif char == 'o':  # live cells
            # Fill count cells with 1
            # Make sure we don't go out of the width.
            for _ in range(count):
                if col < w and row < h:
                    grid[row, col] = 1
                col += 1
        elif char == '$':  # end of line, move down "count" times.
            row += count
            col = 0
        else:
            # For any unexpected character, you might want to raise an error.
            raise ValueError("Unexpected character in RLE: " + char)
    
    return grid

Run Code Interpreter

Running the following program: 

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import numpy as np
import re

def rle_to_arr(rle, h, w):
    """
    Converts an RLE string (as commonly used for Game of Life patterns) into
    a numpy array with shape (h, w). Live cells are marked with 1's and dead cells with 0's.
    
    Parameters:
      rle (str): The RLE encoded pattern string.
      h (int): Height of the output array.
      w (int): Width of the output array.
      
    Returns:
      numpy.ndarray: An array of shape (h, w) with the decoded pattern.
    
    The RLE format supports:
         - An optional count before a character.
         - 'o' for a live cell.
         - 'b' for a dead cell.
         - '$' to indicate the end of a row (moving down).
         - '!' marks the end of the pattern.
    """
    
    grid = np.zeros((h, w), dtype=int)
    row = 0
    col = 0
    i = 0
    # Remove any newlines or spaces from the input.
    rle = rle.replace("\n","").replace(" ", "")
    
    while i < len(rle):
        # Stop if we hit the termination marker.
        if rle[i] == '!':
            break
        
        # Parse an optional number (count)
        count_str = ''
        while i < len(rle) and rle[i].isdigit():
            count_str += rle[i]
            i += 1
        # Default count is 1 if no number is given.
        count = int(count_str) if count_str else 1
        
        # Make sure we haven't run off the string.
        if i >= len(rle):
            break
        
        char = rle[i]
        i += 1
        
        if char == 'b':  # dead cells (note: they are already 0)
            col += count  # skip these positions
        elif char == 'o':  # live cells
            # Fill count cells with 1
            # Make sure we don't go out of the width.
            for _ in range(count):
                if col < w and row < h:
                    grid[row, col] = 1
                col += 1
        elif char == '$':  # end of line, move down "count" times.
            row += count
            col = 0
        else:
            # For any unexpected character, you might want to raise an error.
            raise ValueError("Unexpected character in RLE: " + char)
    
    return grid




answer = (np.array(rle_to_arr('''
16b2o3b2o$2o$2o$19b2o$17b5o$21bo$17b3o$o16bo$o17b2o$bo16b2o$b2o$3bo$o
17b2o3b2o$2b2o14bobobobo$11bobo5b5o$11b2o7b3o$2o10bo8bo$b2o$b2o2$2o$bo
2$6bobo$2o3bo$bo3bo$o4bo2bo7bo$5b3o7b2o7bo$15bobo5b3o$22b5o$21b2o3b2o$
22b5o$22b5o$14b2o6bo2bo$14bobo4bo3bo$11b2obobo4bo$11bobobo5bob2o$13bo
8bo$4bobo6b2o$7bo5b3o50bo$7bo5b3o3b2o3b2o40b4o$4bo2bo5b3o3bo5bo41b4o$
5b3o5b2o35bo16bo2bo$13bo6bo3bo24bobo15b4o4b2o$11bobobo5b3o15b2o7bob2o
14b4o5b2o$11b2obobo21bobo6b2ob2o14bo8b2o$14bobo20bo6b3obob2o$14b2o12b
2o7bo2bo2bo2bo2bobo$28b2o7bo6b2o4bo16bobo$38bobo26b2o$39b2o27bo4b2o$
21b2o49bo2bo$21b2o28bobo18b3o$52b2o21b2o$52bo21bo$74b2o$59bo$62bo$59bo
14bo$62bo$61bob3o8bo$59b4o2bo9b2o$64bo$52bo$51b2o$40b2o8b2o4b2o5bo$40b
2o7b3o4b2o3bobo$50b2o4b2o2bobo11b2o$4bobo44b2o6bo2bo11b2o$7bo44bo7bobo
$o6bo53bobo$o3bo2bo55bo$5b3o5$4b4o$3bo3bo$7bo$3bo2bo3$o2b3o2b2o$2b5o2b
o64b3o$9o67bo$9b3o63bo$o7bo2bo$9b2o3$4b2o$o3b3o$o5b2obo5b2o$bo4bo2bo5b
2o$2o4b2obo$o3b3o$4b2o2$59b3o12b2o$61bo12b2o!
''', 101, 77),dtype=np.uint32).flatten() * np.arange(77*101)).sum()
expected = 1563304
assert answer == expected, f'Wrong answer; got {answer} instead of {expected}'
print('All tests passed')
And got the output: All tests passed

Substring Evaluation

Testing if the previous output contains the string All tests passed: True

Initial Query

Convert this game of life pattern 26$13b2o$14bo$14bobo$15b2o14$31b2o$30b2o$32bo13$46b2o$45b2o$47bo13$61b 2o$60b2o$62bo4$52b2o$52b2o$71bo$66b2ob3o$66bobobo$68bo$67b2o3$76b2o$ 60b3o12b2o10bobo$52b3o7bo14bo8bo2bo4bo$51bo3bo5bo10b2o11b2o5b2o$50bo5b o15bobo8b2o3bo8b2o$51bo3bo11b2o6bo9b2o10b2o$52b3o8b2obo2bo2bo2bo10bo2b o$52bo2bo7b2o2b2o6bo11bobo$53b3o16bobo$53b2obo15b2o$54bobo$55bo3$52b2o 3b2o$52bobobobo$53b5o$54b3o$55bo7$54b2o$54b2o! to a numpy array. Call the function `rle_to_arr(string, h, w)` which should return a numpy array with .shape (h,w). Fill the array with 1s, and empty with 0.

LLM Generation

Query

Convert this game of life pattern 26$13b2o$14bo$14bobo$15b2o14$31b2o$30b2o$32bo13$46b2o$45b2o$47bo13$61b 2o$60b2o$62bo4$52b2o$52b2o$71bo$66b2ob3o$66bobobo$68bo$67b2o3$76b2o$ 60b3o12b2o10bobo$52b3o7bo14bo8bo2bo4bo$51bo3bo5bo10b2o11b2o5b2o$50bo5b o15bobo8b2o3bo8b2o$51bo3bo11b2o6bo9b2o10b2o$52b3o8b2obo2bo2bo2bo10bo2b o$52bo2bo7b2o2b2o6bo11bobo$53b3o16bobo$53b2obo15b2o$54bobo$55bo3$52b2o 3b2o$52bobobobo$53b5o$54b3o$55bo7$54b2o$54b2o! to a numpy array. Call the function `rle_to_arr(string, h, w)` which should return a numpy array with .shape (h,w). Fill the array with 1s, and empty with 0.

Output

Below is one correct solution. For example, you could write: -------------------------------------------------- import numpy as np def rle_to_arr(rle_str, h, w): """ Convert a Life RLE string into a numpy array of shape (h,w). Live cells are represented with 1’s and dead cells as 0’s. The RLE string uses standard Game of Life symbols: • a number preceding a symbol gives a count, • “b” indicates a dead cell, • “o” indicates a live cell, • “$” moves to a new row (with an optional count for several new rows) • “!” terminates the pattern (if present) Note: This function assumes that your RLE contains no header lines. """ # Remove any newline characters (which are not meaningful in the RLE) rle = rle_str.strip().replace("\n", "") # Create a grid of zeros grid = np.zeros((h, w), dtype=int) row = 0 col = 0 idx = 0 # Loop through the string while idx < len(rle): # If number, read whole integer count. count = 0 while idx < len(rle) and rle[idx].isdigit(): count = count * 10 + int(rle[idx]) idx += 1 if count == 0: count = 1 if idx >= len(rle): break char = rle[idx] idx += 1 if char == 'b': # blank (dead) cells # We simply skip columns since our grid already has 0’s col += count elif char == 'o': # live cells for _ in range(count): if row < h and col < w: grid[row, col] = 1 col += 1 elif char == '$': # End-of-line: move count rows down and reset col row += count col = 0 elif char == '!': # Termination character break else: # if another unexpected character appears, ignore it. pass return grid # Example usage: rle_pattern = """26$13b2o$14bo$14bobo$15b2o14$31b2o$30b2o$32bo13$46b2o$45b2o$47bo13$61b 2o$60b2o$62bo4$52b2o$52b2o$71bo$66b2ob3o$66bobobo$68bo$67b2o3$76b2o$ 60b3o12b2o10bobo$52b3o7bo14bo8bo2bo4bo$51bo3bo5bo10b2o11b2o5b2o$50bo5b o15bobo8b2o3bo8b2o$51bo3bo11b2o6bo9b2o10b2o$52b3o8b2obo2bo2bo2bo10bo2b o$52bo2bo7b2o2b2o6bo11bobo$53b3o16bobo$53b2obo15b2o$54bobo$55bo3$52b2o 3b2o$52bobobobo$53b5o$54b3o$55bo7$54b2o$54b2o!""" # Replace h and w with the desired grid dimensions. # For example, if you want a 100 x 100 grid: result_array = rle_to_arr(rle_pattern, 100, 100) #print(result_array) # Uncomment to print the numpy array. -------------------------------------------------- Explanation: 1. The function first removes any extraneous newline characters. 2. It initializes a numpy array of zeros with the given shape. 3. Then it iterates over the characters of the RLE string. When a number is encountered it is used as the count (or defaults to 1). 4. For a "b" character, it simply advances the column pointer. For an "o" character the live cells (1’s) are set in the current row. 5. When a “$” is encountered, the row is advanced (by count, which may be > 1) and the column pointer is reset. 6. Finally, when “!” is reached (if present) the loop ends. This function meets the requirement and returns a numpy array with the game‐of‐life pattern encoded in the RLE string.

Extract Code

I extracted the following code from that output: 

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import numpy as np

def rle_to_arr(rle_str, h, w):
    """
    Convert a Life RLE string into a numpy array of shape (h,w).
    Live cells are represented with 1’s and dead cells as 0’s.
    
    The RLE string uses standard Game of Life symbols:
      • a number preceding a symbol gives a count,
      • “b” indicates a dead cell,
      • “o” indicates a live cell,
      • “$” moves to a new row (with an optional count for several new rows)
      • “!” terminates the pattern (if present)
      
    Note: This function assumes that your RLE contains no header lines.
    """
    # Remove any newline characters (which are not meaningful in the RLE)
    rle = rle_str.strip().replace("\n", "")
    
    # Create a grid of zeros
    grid = np.zeros((h, w), dtype=int)
    
    row = 0
    col = 0
    idx = 0
    # Loop through the string
    while idx < len(rle):
        # If number, read whole integer count.
        count = 0
        while idx < len(rle) and rle[idx].isdigit():
            count = count * 10 + int(rle[idx])
            idx += 1
        if count == 0:
            count = 1

        if idx >= len(rle):
            break

        char = rle[idx]
        idx += 1
        
        if char == 'b':   # blank (dead) cells
            # We simply skip columns since our grid already has 0’s
            col += count
        elif char == 'o': # live cells
            for _ in range(count):
                if row < h and col < w:
                    grid[row, col] = 1
                col += 1
        elif char == '$':  # End-of-line: move count rows down and reset col
            row += count
            col = 0
        elif char == '!':  # Termination character
            break
        else:
            # if another unexpected character appears, ignore it.
            pass
    return grid

Run Code Interpreter

Running the following program: 

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import numpy as np

def rle_to_arr(rle_str, h, w):
    """
    Convert a Life RLE string into a numpy array of shape (h,w).
    Live cells are represented with 1’s and dead cells as 0’s.
    
    The RLE string uses standard Game of Life symbols:
      • a number preceding a symbol gives a count,
      • “b” indicates a dead cell,
      • “o” indicates a live cell,
      • “$” moves to a new row (with an optional count for several new rows)
      • “!” terminates the pattern (if present)
      
    Note: This function assumes that your RLE contains no header lines.
    """
    # Remove any newline characters (which are not meaningful in the RLE)
    rle = rle_str.strip().replace("\n", "")
    
    # Create a grid of zeros
    grid = np.zeros((h, w), dtype=int)
    
    row = 0
    col = 0
    idx = 0
    # Loop through the string
    while idx < len(rle):
        # If number, read whole integer count.
        count = 0
        while idx < len(rle) and rle[idx].isdigit():
            count = count * 10 + int(rle[idx])
            idx += 1
        if count == 0:
            count = 1

        if idx >= len(rle):
            break

        char = rle[idx]
        idx += 1
        
        if char == 'b':   # blank (dead) cells
            # We simply skip columns since our grid already has 0’s
            col += count
        elif char == 'o': # live cells
            for _ in range(count):
                if row < h and col < w:
                    grid[row, col] = 1
                col += 1
        elif char == '$':  # End-of-line: move count rows down and reset col
            row += count
            col = 0
        elif char == '!':  # Termination character
            break
        else:
            # if another unexpected character appears, ignore it.
            pass
    return grid




answer = (np.array(rle_to_arr('''
16b2o3b2o$2o$2o$19b2o$17b5o$21bo$17b3o$o16bo$o17b2o$bo16b2o$b2o$3bo$o
17b2o3b2o$2b2o14bobobobo$11bobo5b5o$11b2o7b3o$2o10bo8bo$b2o$b2o2$2o$bo
2$6bobo$2o3bo$bo3bo$o4bo2bo7bo$5b3o7b2o7bo$15bobo5b3o$22b5o$21b2o3b2o$
22b5o$22b5o$14b2o6bo2bo$14bobo4bo3bo$11b2obobo4bo$11bobobo5bob2o$13bo
8bo$4bobo6b2o$7bo5b3o50bo$7bo5b3o3b2o3b2o40b4o$4bo2bo5b3o3bo5bo41b4o$
5b3o5b2o35bo16bo2bo$13bo6bo3bo24bobo15b4o4b2o$11bobobo5b3o15b2o7bob2o
14b4o5b2o$11b2obobo21bobo6b2ob2o14bo8b2o$14bobo20bo6b3obob2o$14b2o12b
2o7bo2bo2bo2bo2bobo$28b2o7bo6b2o4bo16bobo$38bobo26b2o$39b2o27bo4b2o$
21b2o49bo2bo$21b2o28bobo18b3o$52b2o21b2o$52bo21bo$74b2o$59bo$62bo$59bo
14bo$62bo$61bob3o8bo$59b4o2bo9b2o$64bo$52bo$51b2o$40b2o8b2o4b2o5bo$40b
2o7b3o4b2o3bobo$50b2o4b2o2bobo11b2o$4bobo44b2o6bo2bo11b2o$7bo44bo7bobo
$o6bo53bobo$o3bo2bo55bo$5b3o5$4b4o$3bo3bo$7bo$3bo2bo3$o2b3o2b2o$2b5o2b
o64b3o$9o67bo$9b3o63bo$o7bo2bo$9b2o3$4b2o$o3b3o$o5b2obo5b2o$bo4bo2bo5b
2o$2o4b2obo$o3b3o$4b2o2$59b3o12b2o$61bo12b2o!
''', 101, 77),dtype=np.uint32).flatten() * np.arange(77*101)).sum()
expected = 1563304
assert answer == expected, f'Wrong answer; got {answer} instead of {expected}'
print('All tests passed')
And got the output: All tests passed

Substring Evaluation

Testing if the previous output contains the string All tests passed: True