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 a function that converts the RLE (Run Length Encoded) pattern to a numpy array: 

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import numpy as np

def rle_to_arr(rle_str, h, w):
    # Initialize empty array
    arr = np.zeros((h, w), dtype=int)
    
    # Split the RLE string into rows
    rows = rle_str.split('$')
    
    # Current position
    curr_row = 0
    curr_col = 0
    
    for row in rows:
        if not row:  # Empty row
            curr_row += 1
            curr_col = 0
            continue
            
        count = ''
        for char in row:
            if char.isdigit():
                count += char
            else:
                # If no count specified, count is 1
                n = int(count) if count else 1
                count = ''
                
                if char == 'b':  # Dead cell
                    curr_col += n
                elif char == 'o':  # Live cell
                    for i in range(n):
                        arr[curr_row, curr_col + i] = 1
                    curr_col += n
                    
        curr_row += 1
        curr_col = 0
        
    return arr

# Example usage:
rle = "26$13b2o$14bo$14bobo$15b2o14$31b2o$30b2o$32bo13$46b2o$45b2o$47bo13$61b2o$60b2o$62bo4$52b2o$52b2o$71bo$66b2ob3o$66bobobo$68bo$67b2o3$76b2o$60b3o12b2o10bobo$52b3o7bo14bo8bo2bo4bo$51bo3bo5bo10b2o11b2o5b2o$50bo5bo15bobo8b2o3bo8b2o$51bo3bo11b2o6bo9b2o10b2o$52b3o8b2obo2bo2bo2bo10bo2bo$52bo2bo7b2o2b2o6bo11bobo$53b3o16bobo$53b2obo15b2o$54bobo$55bo3$52b2o3b2o$52bobobobo$53b5o$54b3o$55bo7$54b2o$54b2o"

# The pattern appears to need about 100x100 grid
arr = rle_to_arr(rle, 100, 100)
This function: 1. Creates an empty numpy array of specified dimensions 2. Splits the RLE string into rows using '$' as separator 3. For each row: - Parses numbers as run counts - 'b' represents dead cells (0s) - 'o' represents live cells (1s) 4. Returns the filled array You can verify the pattern by printing sections of the array or visualizing it: 
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# To visualize:
import matplotlib.pyplot as plt
plt.imshow(arr)
plt.show()
The array will have 1s where there are live cells in the Game of Life pattern, and 0s elsewhere.

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):
    # Initialize empty array
    arr = np.zeros((h, w), dtype=int)
    
    # Split the RLE string into rows
    rows = rle_str.split('$')
    
    # Current position
    curr_row = 0
    curr_col = 0
    
    for row in rows:
        if not row:  # Empty row
            curr_row += 1
            curr_col = 0
            continue
            
        count = ''
        for char in row:
            if char.isdigit():
                count += char
            else:
                # If no count specified, count is 1
                n = int(count) if count else 1
                count = ''
                
                if char == 'b':  # Dead cell
                    curr_col += n
                elif char == 'o':  # Live cell
                    for i in range(n):
                        arr[curr_row, curr_col + i] = 1
                    curr_col += n
                    
        curr_row += 1
        curr_col = 0
        
    return arr

Run Code Interpreter

Running the following program: 

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import numpy as np

def rle_to_arr(rle_str, h, w):
    # Initialize empty array
    arr = np.zeros((h, w), dtype=int)
    
    # Split the RLE string into rows
    rows = rle_str.split('$')
    
    # Current position
    curr_row = 0
    curr_col = 0
    
    for row in rows:
        if not row:  # Empty row
            curr_row += 1
            curr_col = 0
            continue
            
        count = ''
        for char in row:
            if char.isdigit():
                count += char
            else:
                # If no count specified, count is 1
                n = int(count) if count else 1
                count = ''
                
                if char == 'b':  # Dead cell
                    curr_col += n
                elif char == 'o':  # Live cell
                    for i in range(n):
                        arr[curr_row, curr_col + i] = 1
                    curr_col += n
                    
        curr_row += 1
        curr_col = 0
        
    return arr





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: Traceback (most recent call last): File "/usr/src/app/main.py", line 63, in <module> assert answer == expected, f'Wrong answer; got {answer} instead of {expected}' ^^^^^^^^^^^^^^^^^^ AssertionError: Wrong answer; got 1468671 instead of 1563304

Substring Evaluation

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

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 a function that converts the RLE (Run Length Encoded) format of Game of Life pattern to a numpy array: 

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import numpy as np

def rle_to_arr(rle_string, h, w):
    # Initialize empty array
    arr = np.zeros((h, w), dtype=int)
    
    # Remove header (if any) and split into rows
    if '$' in rle_string:
        rows = rle_string.strip().split('$')
    else:
        rows = [rle_string.strip()]
    
    # Current position in the array
    curr_row = 0
    curr_col = 0
    
    for row in rows:
        if not row:  # Empty row
            curr_row += 1
            curr_col = 0
            continue
            
        # Parse the row
        count = ''
        for char in row:
            if char.isdigit():
                count += char
            else:
                # If no count specified, count is 1
                n = int(count) if count else 1
                count = ''
                
                if char == 'b':  # Dead cell
                    curr_col += n
                elif char == 'o':  # Live cell
                    for i in range(n):
                        arr[curr_row, curr_col + i] = 1
                    curr_col += n
                
        curr_row += 1
        curr_col = 0
    
    return arr
You can use it like this: 
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rle = "26$13b2o$14bo$14bobo$15b2o14$31b2o$30b2o$32bo13$46b2o$45b2o$47bo13$61b2o$60b2o$62bo4$52b2o$52b2o$71bo$66b2ob3o$66bobobo$68bo$67b2o3$76b2o$60b3o12b2o10bobo$52b3o7bo14bo8bo2bo4bo$51bo3bo5bo10b2o11b2o5b2o$50bo5bo15bobo8b2o3bo8b2o$51bo3bo11b2o6bo9b2o10b2o$52b3o8b2obo2bo2bo2bo10bo2bo$52bo2bo7b2o2b2o6bo11bobo$53b3o16bobo$53b2obo15b2o$54bobo$55bo3$52b2o3b2o$52bobobobo$53b5o$54b3o$55bo7$54b2o$54b2o"

# You'll need to specify the correct height and width
arr = rle_to_arr(rle, 90, 90)  # adjust dimensions as needed
This will create a numpy array where: - 1 represents a live cell - 0 represents a dead cell The function handles: - Multiple rows (separated by $) - Run-length encoding (numbers before b or o) - Basic RLE format for Conway's Game of Life patterns Note that you need to specify the correct height and width of the final array. These dimensions should be large enough to contain the entire pattern.

Extract Code

I extracted the following code from that output: 

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import numpy as np

def rle_to_arr(rle_string, h, w):
    # Initialize empty array
    arr = np.zeros((h, w), dtype=int)
    
    # Remove header (if any) and split into rows
    if '$' in rle_string:
        rows = rle_string.strip().split('$')
    else:
        rows = [rle_string.strip()]
    
    # Current position in the array
    curr_row = 0
    curr_col = 0
    
    for row in rows:
        if not row:  # Empty row
            curr_row += 1
            curr_col = 0
            continue
            
        # Parse the row
        count = ''
        for char in row:
            if char.isdigit():
                count += char
            else:
                # If no count specified, count is 1
                n = int(count) if count else 1
                count = ''
                
                if char == 'b':  # Dead cell
                    curr_col += n
                elif char == 'o':  # Live cell
                    for i in range(n):
                        arr[curr_row, curr_col + i] = 1
                    curr_col += n
                
        curr_row += 1
        curr_col = 0
    
    return arr

Run Code Interpreter

Running the following program: 

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import numpy as np

def rle_to_arr(rle_string, h, w):
    # Initialize empty array
    arr = np.zeros((h, w), dtype=int)
    
    # Remove header (if any) and split into rows
    if '$' in rle_string:
        rows = rle_string.strip().split('$')
    else:
        rows = [rle_string.strip()]
    
    # Current position in the array
    curr_row = 0
    curr_col = 0
    
    for row in rows:
        if not row:  # Empty row
            curr_row += 1
            curr_col = 0
            continue
            
        # Parse the row
        count = ''
        for char in row:
            if char.isdigit():
                count += char
            else:
                # If no count specified, count is 1
                n = int(count) if count else 1
                count = ''
                
                if char == 'b':  # Dead cell
                    curr_col += n
                elif char == 'o':  # Live cell
                    for i in range(n):
                        arr[curr_row, curr_col + i] = 1
                    curr_col += n
                
        curr_row += 1
        curr_col = 0
    
    return arr





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: Traceback (most recent call last): File "/usr/src/app/main.py", line 67, in <module> assert answer == expected, f'Wrong answer; got {answer} instead of {expected}' ^^^^^^^^^^^^^^^^^^ AssertionError: Wrong answer; got 1468671 instead of 1563304

Substring Evaluation

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

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 a function that converts a Run Length Encoded (RLE) Game of Life pattern to a numpy array: 

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import numpy as np

def rle_to_arr(rle_string, h, w):
    # Initialize empty array
    arr = np.zeros((h, w), dtype=int)
    
    # Remove any newlines and split on $
    parts = rle_string.strip().split('$')
    
    # Current position
    row = 0
    col = 0
    
    for part in parts:
        if not part:  # Empty line
            row += 1
            col = 0
            continue
            
        # Process each character
        number = ''
        for char in part:
            if char.isdigit():
                number += char
            else:
                # Convert number, default to 1 if empty
                count = int(number) if number else 1
                number = ''
                
                if char == 'b':  # Dead cell
                    col += count
                elif char == 'o':  # Live cell
                    for i in range(count):
                        arr[row, col + i] = 1
                    col += count
                
        row += 1
        col = 0
    
    return arr
This function: 1. Takes an RLE string and desired height and width as input 2. Creates a zero-filled numpy array of the specified size 3. Processes the RLE string line by line (split by $) 4. For each line: - Handles run lengths (numbers before characters) - 'b' represents dead cells (0s) - 'o' represents live cells (1s) - Updates position accordingly 5. Returns the completed array You can use it like this: 
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rle = """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!"""

# You'll need to specify the correct height and width for your pattern
array = rle_to_arr(rle, 100, 100)  # adjust h,w as needed
The resulting array will contain 1s where there are live cells and 0s elsewhere. The exact height and width needed will depend on the pattern - you may need to adjust these values to ensure the pattern fits within the array.

Extract Code

I extracted the following code from that output: 

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import numpy as np

def rle_to_arr(rle_string, h, w):
    # Initialize empty array
    arr = np.zeros((h, w), dtype=int)
    
    # Remove any newlines and split on $
    parts = rle_string.strip().split('$')
    
    # Current position
    row = 0
    col = 0
    
    for part in parts:
        if not part:  # Empty line
            row += 1
            col = 0
            continue
            
        # Process each character
        number = ''
        for char in part:
            if char.isdigit():
                number += char
            else:
                # Convert number, default to 1 if empty
                count = int(number) if number else 1
                number = ''
                
                if char == 'b':  # Dead cell
                    col += count
                elif char == 'o':  # Live cell
                    for i in range(count):
                        arr[row, col + i] = 1
                    col += count
                
        row += 1
        col = 0
    
    return arr

Run Code Interpreter

Running the following program: 

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import numpy as np

def rle_to_arr(rle_string, h, w):
    # Initialize empty array
    arr = np.zeros((h, w), dtype=int)
    
    # Remove any newlines and split on $
    parts = rle_string.strip().split('$')
    
    # Current position
    row = 0
    col = 0
    
    for part in parts:
        if not part:  # Empty line
            row += 1
            col = 0
            continue
            
        # Process each character
        number = ''
        for char in part:
            if char.isdigit():
                number += char
            else:
                # Convert number, default to 1 if empty
                count = int(number) if number else 1
                number = ''
                
                if char == 'b':  # Dead cell
                    col += count
                elif char == 'o':  # Live cell
                    for i in range(count):
                        arr[row, col + i] = 1
                    col += count
                
        row += 1
        col = 0
    
    return arr





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: Traceback (most recent call last): File "/usr/src/app/main.py", line 64, in <module> assert answer == expected, f'Wrong answer; got {answer} instead of {expected}' ^^^^^^^^^^^^^^^^^^ AssertionError: Wrong answer; got 1468671 instead of 1563304

Substring Evaluation

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

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 a function that converts the RLE (Run Length Encoded) pattern to a numpy array: 

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import numpy as np

def rle_to_arr(rle_string, h, w):
    # Initialize empty array
    arr = np.zeros((h, w), dtype=int)
    
    # Remove header line if present (containing size information)
    if 'x' in rle_string:
        rle_string = rle_string[rle_string.index('$'):]
    
    # Current position in the array
    row = 0
    col = 0
    
    # Parse the RLE string
    count = ''
    for char in rle_string:
        if char.isdigit():
            count += char
        else:
            n = int(count) if count else 1
            count = ''
            
            if char == 'b':  # Dead cell
                col += n
            elif char == 'o':  # Live cell
                for i in range(n):
                    arr[row, col] = 1
                    col += 1
            elif char == '$':  # End of row
                row += n
                col = 0
            elif char == '!':  # End of pattern
                break
    
    return arr
You can use it like this: 
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rle = "26$13b2o$14bo$14bobo$15b2o14$31b2o$30b2o$32bo13$46b2o$45b2o$47bo13$61b2o$60b2o$62bo4$52b2o$52b2o$71bo$66b2ob3o$66bobobo$68bo$67b2o3$76b2o$60b3o12b2o10bobo$52b3o7bo14bo8bo2bo4bo$51bo3bo5bo10b2o11b2o5b2o$50bo5bo15bobo8b2o3bo8b2o$51bo3bo11b2o6bo9b2o10b2o$52b3o8b2obo2bo2bo2bo10bo2bo$52bo2bo7b2o2b2o6bo11bobo$53b3o16bobo$53b2obo15b2o$54bobo$55bo3$52b2o3b2o$52bobobobo$53b5o$54b3o$55bo7$54b2o$54b2o!"

# Assuming the pattern fits within an 80x80 grid
arr = rle_to_arr(rle, 80, 80)
This will create a numpy array where: - 1 represents a live cell - 0 represents a dead cell The function: 1. Creates an empty numpy array of the specified size 2. Parses the RLE string character by character 3. Keeps track of run lengths (numbers before characters) 4. Places 1s in the array according to the pattern: - 'o' represents live cells - 'b' represents dead cells - '$' moves to the next row - '!' marks the end of the pattern The resulting array will contain the Game of Life pattern ready to be displayed or manipulated.

Extract Code

I extracted the following code from that output: 

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import numpy as np

def rle_to_arr(rle_string, h, w):
    # Initialize empty array
    arr = np.zeros((h, w), dtype=int)
    
    # Remove header line if present (containing size information)
    if 'x' in rle_string:
        rle_string = rle_string[rle_string.index('$'):]
    
    # Current position in the array
    row = 0
    col = 0
    
    # Parse the RLE string
    count = ''
    for char in rle_string:
        if char.isdigit():
            count += char
        else:
            n = int(count) if count else 1
            count = ''
            
            if char == 'b':  # Dead cell
                col += n
            elif char == 'o':  # Live cell
                for i in range(n):
                    arr[row, col] = 1
                    col += 1
            elif char == '$':  # End of row
                row += n
                col = 0
            elif char == '!':  # End of pattern
                break
    
    return arr

Run Code Interpreter

Running the following program: 

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import numpy as np

def rle_to_arr(rle_string, h, w):
    # Initialize empty array
    arr = np.zeros((h, w), dtype=int)
    
    # Remove header line if present (containing size information)
    if 'x' in rle_string:
        rle_string = rle_string[rle_string.index('$'):]
    
    # Current position in the array
    row = 0
    col = 0
    
    # Parse the RLE string
    count = ''
    for char in rle_string:
        if char.isdigit():
            count += char
        else:
            n = int(count) if count else 1
            count = ''
            
            if char == 'b':  # Dead cell
                col += n
            elif char == 'o':  # Live cell
                for i in range(n):
                    arr[row, col] = 1
                    col += 1
            elif char == '$':  # End of row
                row += n
                col = 0
            elif char == '!':  # End of pattern
                break
    
    return arr





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

Here's a Python function that converts the RLE (Run Length Encoded) pattern to a numpy array: 

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import numpy as np

def rle_to_arr(rle_string, h, w):
    # Initialize empty array
    arr = np.zeros((h, w), dtype=int)
    
    # Split the RLE string into lines
    lines = rle_string.strip().split('$')
    
    # Current position
    row = 0
    col = 0
    
    for line in lines:
        col = 0
        if not line:  # Empty line means move down one row
            row += 1
            continue
            
        count = ''
        for char in line:
            if char.isdigit():
                count += char
            else:
                # If no count specified, count is 1
                n = int(count) if count else 1
                count = ''
                
                if char == 'b':  # Dead cell
                    col += n
                elif char == 'o':  # Live cell
                    for i in range(n):
                        arr[row, col] = 1
                        col += 1
                elif char == '$':  # New line
                    row += n
                    col = 0
        row += 1
    
    return arr
You can use this function by passing your RLE string and the desired dimensions. The pattern you provided appears to be 90×90, so you could call it like this: 
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rle = "26$13b2o$14bo$14bobo$15b2o14$31b2o$30b2o$32bo13$46b2o$45b2o$47bo13$61b2o$60b2o$62bo4$52b2o$52b2o$71bo$66b2ob3o$66bobobo$68bo$67b2o3$76b2o$60b3o12b2o10bobo$52b3o7bo14bo8bo2bo4bo$51bo3bo5bo10b2o11b2o5b2o$50bo5bo15bobo8b2o3bo8b2o$51bo3bo11b2o6bo9b2o10b2o$52b3o8b2obo2bo2bo2bo10bo2bo$52bo2bo7b2o2b2o6bo11bobo$53b3o16bobo$53b2obo15b2o$54bobo$55bo3$52b2o3b2o$52bobobobo$53b5o$54b3o$55bo7$54b2o$54b2o!"

array = rle_to_arr(rle, 90, 90)
This will create a 90×90 numpy array where: - 1 represents a live cell - 0 represents a dead cell The pattern will be positioned within the array according to the RLE encoding.

Extract Code

I extracted the following code from that output: 

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import numpy as np

def rle_to_arr(rle_string, h, w):
    # Initialize empty array
    arr = np.zeros((h, w), dtype=int)
    
    # Split the RLE string into lines
    lines = rle_string.strip().split('$')
    
    # Current position
    row = 0
    col = 0
    
    for line in lines:
        col = 0
        if not line:  # Empty line means move down one row
            row += 1
            continue
            
        count = ''
        for char in line:
            if char.isdigit():
                count += char
            else:
                # If no count specified, count is 1
                n = int(count) if count else 1
                count = ''
                
                if char == 'b':  # Dead cell
                    col += n
                elif char == 'o':  # Live cell
                    for i in range(n):
                        arr[row, col] = 1
                        col += 1
                elif char == '$':  # New line
                    row += n
                    col = 0
        row += 1
    
    return arr

Run Code Interpreter

Running the following program: 

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import numpy as np

def rle_to_arr(rle_string, h, w):
    # Initialize empty array
    arr = np.zeros((h, w), dtype=int)
    
    # Split the RLE string into lines
    lines = rle_string.strip().split('$')
    
    # Current position
    row = 0
    col = 0
    
    for line in lines:
        col = 0
        if not line:  # Empty line means move down one row
            row += 1
            continue
            
        count = ''
        for char in line:
            if char.isdigit():
                count += char
            else:
                # If no count specified, count is 1
                n = int(count) if count else 1
                count = ''
                
                if char == 'b':  # Dead cell
                    col += n
                elif char == 'o':  # Live cell
                    for i in range(n):
                        arr[row, col] = 1
                        col += 1
                elif char == '$':  # New line
                    row += n
                    col = 0
        row += 1
    
    return arr





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: Traceback (most recent call last): File "/usr/src/app/main.py", line 64, in <module> assert answer == expected, f'Wrong answer; got {answer} instead of {expected}' ^^^^^^^^^^^^^^^^^^ AssertionError: Wrong answer; got 1468671 instead of 1563304

Substring Evaluation

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