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 to convert the Run-Length Encoding (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):
    # Create an empty array of zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Remove whitespace and line breaks
    rle_string = rle_string.replace('\n', '').replace('$', '')
    
    # Parse the RLE string
    y, x = 0, 0
    i = 0
    while i < len(rle_string):
        # Check for run length
        run_length = 0
        while i < len(rle_string) and rle_string[i].isdigit():
            run_length = run_length * 10 + int(rle_string[i])
            i += 1
        
        # If no explicit run length, default to 1
        if run_length == 0:
            run_length = 1
        
        # Get cell type
        if rle_string[i] == 'o':
            # Fill run with 1s
            for j in range(run_length):
                if x < w and y < h:
                    arr[y, x] = 1
                    x += 1
        elif rle_string[i] == 'b':
            # Skip cells (already zeros)
            x += run_length
        
        # Reset x and increment y if needed
        if x >= w:
            x = 0
            y += 1
        
        i += 1
        
        if y >= h:
            break
    
    return arr
This function does the following: 1. Create a zero-filled NumPy array of the specified height and width. 2. Remove whitespace and line breaks from the RLE string. 3. Parse the RLE string: - Extract run lengths (optional numeric prefix) - If no run length specified, default to 1 - For 'o' (live cell), fill with 1s - For 'b' (dead/empty cell), skip cells (already zeros) 4. Track x and y positions while filling the array 5. Stop if array bounds are exceeded Example usage: 
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rle_string = "26$13b2o$14bo$14bobo$15b2o14$31b2o$30b2o$32bo13$46b2o$45b2o$47bo..."
h, w = 100, 100  # Adjust height and width as needed
arr = rle_to_arr(rle_string, h, w)

# Optional: visualize the array
import matplotlib.pyplot as plt
plt.imshow(arr, cmap='binary')
plt.show()
Note: You may need to adjust the height and width to fully capture the pattern. The function will stop filling if the specified height or width is reached.

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):
    # Create an empty array of zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Remove whitespace and line breaks
    rle_string = rle_string.replace('\n', '').replace('$', '')
    
    # Parse the RLE string
    y, x = 0, 0
    i = 0
    while i < len(rle_string):
        # Check for run length
        run_length = 0
        while i < len(rle_string) and rle_string[i].isdigit():
            run_length = run_length * 10 + int(rle_string[i])
            i += 1
        
        # If no explicit run length, default to 1
        if run_length == 0:
            run_length = 1
        
        # Get cell type
        if rle_string[i] == 'o':
            # Fill run with 1s
            for j in range(run_length):
                if x < w and y < h:
                    arr[y, x] = 1
                    x += 1
        elif rle_string[i] == 'b':
            # Skip cells (already zeros)
            x += run_length
        
        # Reset x and increment y if needed
        if x >= w:
            x = 0
            y += 1
        
        i += 1
        
        if y >= h:
            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):
    # Create an empty array of zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Remove whitespace and line breaks
    rle_string = rle_string.replace('\n', '').replace('$', '')
    
    # Parse the RLE string
    y, x = 0, 0
    i = 0
    while i < len(rle_string):
        # Check for run length
        run_length = 0
        while i < len(rle_string) and rle_string[i].isdigit():
            run_length = run_length * 10 + int(rle_string[i])
            i += 1
        
        # If no explicit run length, default to 1
        if run_length == 0:
            run_length = 1
        
        # Get cell type
        if rle_string[i] == 'o':
            # Fill run with 1s
            for j in range(run_length):
                if x < w and y < h:
                    arr[y, x] = 1
                    x += 1
        elif rle_string[i] == 'b':
            # Skip cells (already zeros)
            x += run_length
        
        # Reset x and increment y if needed
        if x >= w:
            x = 0
            y += 1
        
        i += 1
        
        if y >= h:
            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: Traceback (most recent call last): File "/usr/src/app/main.py", line 69, in <module> assert answer == expected, f'Wrong answer; got {answer} instead of {expected}' ^^^^^^^^^^^^^^^^^^ AssertionError: Wrong answer; got 558423 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 Python function to convert the RLE (Run Length Encoding) Game of Life pattern to a NumPy array: 

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import numpy as np

def rle_to_arr(rle_string, h, w):
    # Create an empty array of zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Remove newline characters and split the string
    rle_string = rle_string.replace('\n', '')
    
    # Remove the final '!' if present
    if rle_string.endswith('!'):
        rle_string = rle_string[:-1]
    
    # Track current position
    row, col = 0, 0
    
    # Parse the RLE string
    i = 0
    while i < len(rle_string):
        # Check if current character is a number
        num_str = ''
        while i < len(rle_string) and rle_string[i].isdigit():
            num_str += rle_string[i]
            i += 1
        
        # Default count is 1 if no number specified
        count = int(num_str) if num_str else 1
        
        # Get the current character
        if i < len(rle_string):
            char = rle_string[i]
            
            # If '$' is encountered, move to next row
            if char == '$':
                row += count
                col = 0
            # If 'b' or 'o' is encountered, fill cells
            elif char == 'b':
                col += count
            elif char == 'o':
                # Fill count number of cells with 1
                arr[row, col:col+count] = 1
                col += count
            
            i += 1
    
    return arr
To use this function, you would call it like: 
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# Example usage
rle_pattern = "26$13b2o$14bo$14bobo$15b2o"
pattern_height = 40  # Adjust based on the pattern
pattern_width = 80   # Adjust based on the pattern

game_of_life_array = rle_to_arr(rle_pattern, pattern_height, pattern_width)
print(game_of_life_array)
Key points about the function: 1. It creates an empty NumPy array of specified height and width 2. Parses the RLE string character by character 3. Supports run-length encoding (e.g., '2o' means two consecutive live cells) 4. Uses '$' to move to a new row 5. Fills cells with 1 for 'o' and skips cells for 'b' Make sure to adjust the height and width parameters based on the specific pattern you're converting. Would you like me to explain any part of the implementation in more detail?

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):
    # Create an empty array of zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Remove newline characters and split the string
    rle_string = rle_string.replace('\n', '')
    
    # Remove the final '!' if present
    if rle_string.endswith('!'):
        rle_string = rle_string[:-1]
    
    # Track current position
    row, col = 0, 0
    
    # Parse the RLE string
    i = 0
    while i < len(rle_string):
        # Check if current character is a number
        num_str = ''
        while i < len(rle_string) and rle_string[i].isdigit():
            num_str += rle_string[i]
            i += 1
        
        # Default count is 1 if no number specified
        count = int(num_str) if num_str else 1
        
        # Get the current character
        if i < len(rle_string):
            char = rle_string[i]
            
            # If '$' is encountered, move to next row
            if char == '$':
                row += count
                col = 0
            # If 'b' or 'o' is encountered, fill cells
            elif char == 'b':
                col += count
            elif char == 'o':
                # Fill count number of cells with 1
                arr[row, col:col+count] = 1
                col += count
            
            i += 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):
    # Create an empty array of zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Remove newline characters and split the string
    rle_string = rle_string.replace('\n', '')
    
    # Remove the final '!' if present
    if rle_string.endswith('!'):
        rle_string = rle_string[:-1]
    
    # Track current position
    row, col = 0, 0
    
    # Parse the RLE string
    i = 0
    while i < len(rle_string):
        # Check if current character is a number
        num_str = ''
        while i < len(rle_string) and rle_string[i].isdigit():
            num_str += rle_string[i]
            i += 1
        
        # Default count is 1 if no number specified
        count = int(num_str) if num_str else 1
        
        # Get the current character
        if i < len(rle_string):
            char = rle_string[i]
            
            # If '$' is encountered, move to next row
            if char == '$':
                row += count
                col = 0
            # If 'b' or 'o' is encountered, fill cells
            elif char == 'b':
                col += count
            elif char == 'o':
                # Fill count number of cells with 1
                arr[row, col:col+count] = 1
                col += count
            
            i += 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: 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 given Run-Length Encoding (RLE) pattern to a NumPy array: 

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import numpy as np

def rle_to_arr(rle_string, h, w):
    # Initialize an empty array of zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Remove newline characters and split the RLE string
    rle_string = rle_string.replace('\n', '')
    
    # Split the RLE string into rows
    rows = rle_string.split('$')
    
    # Track current position
    current_row = 0
    
    for row_str in rows:
        # If row is empty or just a number, skip
        if not row_str:
            current_row += 1
            continue
        
        # Parse the row
        col = 0
        i = 0
        while i < len(row_str):
            # Check if current character is a number
            num_str = ''
            while i < len(row_str) and row_str[i].isdigit():
                num_str += row_str[i]
                i += 1
            
            # Determine repeat count
            repeat = int(num_str) if num_str else 1
            
            # Check if current character is live cell
            if i < len(row_str) and row_str[i] in ['o', 'b']:
                if row_str[i] == 'o':
                    # Fill with 1s
                    arr[current_row, col:col+repeat] = 1
                col += repeat
                i += 1
        
        current_row += 1
    
    return arr

# Example usage
rle_string = "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!"

# Specify height and width of the array
h, w = 100, 100

# Convert RLE to array
result = rle_to_arr(rle_string, h, w)

# Optional: Visualize or print parts of the array
print(result)
The function works as follows: 1. Create a zero-filled NumPy array of the specified height and width. 2. Parse the RLE string row by row. 3. For each row: - Handle run-length encoding (numbers before 'o' or 'b') - Fill '1's for live cells ('o') - Move to the next column/row 4. Return the resulting array. You can adjust the height and width (`h` and `w`) as needed to capture the entire pattern. In this example, I used 100x100, but you might want to adjust based on the specific pattern's size.

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 an empty array of zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Remove newline characters and split the RLE string
    rle_string = rle_string.replace('\n', '')
    
    # Split the RLE string into rows
    rows = rle_string.split('$')
    
    # Track current position
    current_row = 0
    
    for row_str in rows:
        # If row is empty or just a number, skip
        if not row_str:
            current_row += 1
            continue
        
        # Parse the row
        col = 0
        i = 0
        while i < len(row_str):
            # Check if current character is a number
            num_str = ''
            while i < len(row_str) and row_str[i].isdigit():
                num_str += row_str[i]
                i += 1
            
            # Determine repeat count
            repeat = int(num_str) if num_str else 1
            
            # Check if current character is live cell
            if i < len(row_str) and row_str[i] in ['o', 'b']:
                if row_str[i] == 'o':
                    # Fill with 1s
                    arr[current_row, col:col+repeat] = 1
                col += repeat
                i += 1
        
        current_row += 1
    
    return arr

# Example usage
rle_string = "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!"

# Specify height and width of the array
h, w = 100, 100

# Convert RLE to array
result = rle_to_arr(rle_string, h, w)

# Optional: Visualize or print parts of the array
print(result)

Run Code Interpreter

Running the following program: 

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import numpy as np

def rle_to_arr(rle_string, h, w):
    # Initialize an empty array of zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Remove newline characters and split the RLE string
    rle_string = rle_string.replace('\n', '')
    
    # Split the RLE string into rows
    rows = rle_string.split('$')
    
    # Track current position
    current_row = 0
    
    for row_str in rows:
        # If row is empty or just a number, skip
        if not row_str:
            current_row += 1
            continue
        
        # Parse the row
        col = 0
        i = 0
        while i < len(row_str):
            # Check if current character is a number
            num_str = ''
            while i < len(row_str) and row_str[i].isdigit():
                num_str += row_str[i]
                i += 1
            
            # Determine repeat count
            repeat = int(num_str) if num_str else 1
            
            # Check if current character is live cell
            if i < len(row_str) and row_str[i] in ['o', 'b']:
                if row_str[i] == 'o':
                    # Fill with 1s
                    arr[current_row, col:col+repeat] = 1
                col += repeat
                i += 1
        
        current_row += 1
    
    return arr

# Example usage
rle_string = "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!"

# Specify height and width of the array
h, w = 100, 100

# Convert RLE to array
result = rle_to_arr(rle_string, h, w)

# Optional: Visualize or print parts of the array
print(result)





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: Timeout: function took too long to complete

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 Python function to convert the Run-Length Encoding (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 an empty array of zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Remove newline characters and $ signs
    rle_string = rle_string.replace('$', '')
    rle_string = rle_string.replace('!', '')
    
    # Initialize row and column trackers
    row, col = 0, 0
    
    # Parse the RLE string
    i = 0
    while i < len(rle_string):
        # Check if current character is a digit
        run_length = 0
        while i < len(rle_string) and rle_string[i].isdigit():
            run_length = run_length * 10 + int(rle_string[i])
            i += 1
        
        # If no run length specified, assume 1
        if run_length == 0:
            run_length = 1
        
        # Get the cell type (o for live, b for dead)
        cell_type = rle_string[i]
        
        # Fill the array
        if cell_type == 'o':
            arr[row, col:col+run_length] = 1
        
        # Update column and row
        col += run_length
        
        # Move to next row if column exceeds width
        if col >= w:
            row += 1
            col = 0
        
        i += 1
    
    return arr
You can use this function like this: 
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# Example usage
rle_string = "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!"

# Determine appropriate height and width based on pattern
h, w = 100, 100  # Adjust these values as needed

# Convert RLE to array
pattern_arr = rle_to_arr(rle_string, h, w)

# Optionally, you can visualize or print the array
print(pattern_arr)
Key points about the function: - It takes the RLE string, height, and width as inputs - Converts runs of cells (specified by numbers) to 1s or 0s - `o` represents live cells (1s) - `b` or implicit empty spaces represent dead cells (0s) - Handles multi-digit run lengths - Automatically moves to next row when column exceeds width Make sure to adjust the height and width parameters to match the actual pattern size. You may need to experiment to find the right dimensions that fully capture the 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 an empty array of zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Remove newline characters and $ signs
    rle_string = rle_string.replace('$', '')
    rle_string = rle_string.replace('!', '')
    
    # Initialize row and column trackers
    row, col = 0, 0
    
    # Parse the RLE string
    i = 0
    while i < len(rle_string):
        # Check if current character is a digit
        run_length = 0
        while i < len(rle_string) and rle_string[i].isdigit():
            run_length = run_length * 10 + int(rle_string[i])
            i += 1
        
        # If no run length specified, assume 1
        if run_length == 0:
            run_length = 1
        
        # Get the cell type (o for live, b for dead)
        cell_type = rle_string[i]
        
        # Fill the array
        if cell_type == 'o':
            arr[row, col:col+run_length] = 1
        
        # Update column and row
        col += run_length
        
        # Move to next row if column exceeds width
        if col >= w:
            row += 1
            col = 0
        
        i += 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 an empty array of zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Remove newline characters and $ signs
    rle_string = rle_string.replace('$', '')
    rle_string = rle_string.replace('!', '')
    
    # Initialize row and column trackers
    row, col = 0, 0
    
    # Parse the RLE string
    i = 0
    while i < len(rle_string):
        # Check if current character is a digit
        run_length = 0
        while i < len(rle_string) and rle_string[i].isdigit():
            run_length = run_length * 10 + int(rle_string[i])
            i += 1
        
        # If no run length specified, assume 1
        if run_length == 0:
            run_length = 1
        
        # Get the cell type (o for live, b for dead)
        cell_type = rle_string[i]
        
        # Fill the array
        if cell_type == 'o':
            arr[row, col:col+run_length] = 1
        
        # Update column and row
        col += run_length
        
        # Move to next row if column exceeds width
        if col >= w:
            row += 1
            col = 0
        
        i += 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 68, in <module> assert answer == expected, f'Wrong answer; got {answer} instead of {expected}' ^^^^^^^^^^^^^^^^^^ AssertionError: Wrong answer; got 568507 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 Python function that converts the 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 an empty array
    arr = np.zeros((h, w), dtype=int)
    
    # Remove whitespace from the RLE string
    rle_string = rle_string.replace('\n', '').replace(' ', '')
    
    # Parse the RLE string
    x, y = 0, 0
    i = 0
    while i < len(rle_string):
        # Check for run length
        run_length = 0
        while i < len(rle_string) and rle_string[i].isdigit():
            run_length = run_length * 10 + int(rle_string[i])
            i += 1
        
        # If no run length specified, default to 1
        if run_length == 0:
            run_length = 1
        
        # Check the cell type
        if rle_string[i] == 'b':
            # Skip (blank/dead cells)
            x += run_length
        elif rle_string[i] == 'o':
            # Fill live cells
            for j in range(run_length):
                if x < w and y < h:
                    arr[y, x] = 1
                x += 1
        elif rle_string[i] == '$':
            # Move to next row
            y += run_length
            x = 0
        
        i += 1
    
    return arr

# Example usage
rle_pattern = "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!"

# Create an appropriately sized array (adjust h and w as needed)
h, w = 100, 100
result = rle_to_arr(rle_pattern, h, w)

# Optional: visualize the result
import matplotlib.pyplot as plt
plt.imshow(result, cmap='binary')
plt.show()
Key points: - The function takes the RLE string, height, and width as inputs - It parses the RLE encoding rules: - Numbers before 'b' or 'o' indicate run length - 'b' represents blank/dead cells - 'o' represents live cells - '$' moves to the next row - It populates a NumPy array with 1s for live cells and 0s for dead cells - The array is created with the specified height and width You can adjust the height and width as needed to ensure the entire pattern fits. In the example, I used 100x100, but you might want to adjust based on the specific pattern. The visualization with matplotlib is optional but can help you verify the 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 an empty array
    arr = np.zeros((h, w), dtype=int)
    
    # Remove whitespace from the RLE string
    rle_string = rle_string.replace('\n', '').replace(' ', '')
    
    # Parse the RLE string
    x, y = 0, 0
    i = 0
    while i < len(rle_string):
        # Check for run length
        run_length = 0
        while i < len(rle_string) and rle_string[i].isdigit():
            run_length = run_length * 10 + int(rle_string[i])
            i += 1
        
        # If no run length specified, default to 1
        if run_length == 0:
            run_length = 1
        
        # Check the cell type
        if rle_string[i] == 'b':
            # Skip (blank/dead cells)
            x += run_length
        elif rle_string[i] == 'o':
            # Fill live cells
            for j in range(run_length):
                if x < w and y < h:
                    arr[y, x] = 1
                x += 1
        elif rle_string[i] == '$':
            # Move to next row
            y += run_length
            x = 0
        
        i += 1
    
    return arr

# Example usage
rle_pattern = "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!"

# Create an appropriately sized array (adjust h and w as needed)
h, w = 100, 100
result = rle_to_arr(rle_pattern, h, w)

# Optional: visualize the result
import matplotlib.pyplot as plt
plt.imshow(result, cmap='binary')
plt.show()

Run Code Interpreter

Running the following program: 

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import numpy as np

def rle_to_arr(rle_string, h, w):
    # Initialize an empty array
    arr = np.zeros((h, w), dtype=int)
    
    # Remove whitespace from the RLE string
    rle_string = rle_string.replace('\n', '').replace(' ', '')
    
    # Parse the RLE string
    x, y = 0, 0
    i = 0
    while i < len(rle_string):
        # Check for run length
        run_length = 0
        while i < len(rle_string) and rle_string[i].isdigit():
            run_length = run_length * 10 + int(rle_string[i])
            i += 1
        
        # If no run length specified, default to 1
        if run_length == 0:
            run_length = 1
        
        # Check the cell type
        if rle_string[i] == 'b':
            # Skip (blank/dead cells)
            x += run_length
        elif rle_string[i] == 'o':
            # Fill live cells
            for j in range(run_length):
                if x < w and y < h:
                    arr[y, x] = 1
                x += 1
        elif rle_string[i] == '$':
            # Move to next row
            y += run_length
            x = 0
        
        i += 1
    
    return arr

# Example usage
rle_pattern = "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!"

# Create an appropriately sized array (adjust h and w as needed)
h, w = 100, 100
result = rle_to_arr(rle_pattern, h, w)

# Optional: visualize the result
import matplotlib.pyplot as plt
plt.imshow(result, cmap='binary')
plt.show()





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 52, in <module> import matplotlib.pyplot as plt ModuleNotFoundError: No module named 'matplotlib'

Substring Evaluation

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