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 Game of Life pattern (in Run Length Encoded format) to a numpy array: 

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import numpy as np

def rle_to_arr(string, h, w):
    # Initialize an empty array filled with zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Remove the header (everything before the first $)
    pattern = string.split('$', 1)[1]
    
    # Split the pattern into rows
    rows = pattern.split('$')
    
    x, y = 0, 0
    for row in rows:
        x = 0
        i = 0
        while i < len(row):
            # Check for run count
            run_count = ''
            while i < len(row) and row[i].isdigit():
                run_count += row[i]
                i += 1
            
            if run_count == '':
                run_count = 1
            else:
                run_count = int(run_count)
            
            # Check for cell state
            if i < len(row):
                if row[i] == 'o':
                    arr[y, x:x+run_count] = 1
                x += run_count
            
            i += 1
        y += 1
        if y >= h:
            break
    
    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!"

# Determine the height and width of the pattern
h = len(rle_string.split('$'))
w = max(len(row) for row in rle_string.split('$'))

result = rle_to_arr(rle_string, h, w)
print(result)
print(result.shape)
This function does the following: 1. It initializes a numpy array filled with zeros. 2. It parses the RLE string, interpreting run counts and cell states. 3. It fills the array with 1s where living cells are indicated in the RLE string. 4. It returns the resulting numpy array. The function handles the RLE format, including run counts for both living ('o') and dead ('b') cells, as well as newline characters ('$'). It also correctly interprets the implied '1' when no run count is specified. You can call this function with your RLE string and the desired height and width of the array. The function will return a numpy array representing the Game of Life pattern.

Extract Code

I extracted the following code from that output: 

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import numpy as np

def rle_to_arr(string, h, w):
    # Initialize an empty array filled with zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Remove the header (everything before the first $)
    pattern = string.split('$', 1)[1]
    
    # Split the pattern into rows
    rows = pattern.split('$')
    
    x, y = 0, 0
    for row in rows:
        x = 0
        i = 0
        while i < len(row):
            # Check for run count
            run_count = ''
            while i < len(row) and row[i].isdigit():
                run_count += row[i]
                i += 1
            
            if run_count == '':
                run_count = 1
            else:
                run_count = int(run_count)
            
            # Check for cell state
            if i < len(row):
                if row[i] == 'o':
                    arr[y, x:x+run_count] = 1
                x += run_count
            
            i += 1
        y += 1
        if y >= h:
            break
    
    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!"

# Determine the height and width of the pattern
h = len(rle_string.split('$'))
w = max(len(row) for row in rle_string.split('$'))

result = rle_to_arr(rle_string, h, w)
print(result)
print(result.shape)

Run Code Interpreter

Running the following program: 

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import numpy as np

def rle_to_arr(string, h, w):
    # Initialize an empty array filled with zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Remove the header (everything before the first $)
    pattern = string.split('$', 1)[1]
    
    # Split the pattern into rows
    rows = pattern.split('$')
    
    x, y = 0, 0
    for row in rows:
        x = 0
        i = 0
        while i < len(row):
            # Check for run count
            run_count = ''
            while i < len(row) and row[i].isdigit():
                run_count += row[i]
                i += 1
            
            if run_count == '':
                run_count = 1
            else:
                run_count = int(run_count)
            
            # Check for cell state
            if i < len(row):
                if row[i] == 'o':
                    arr[y, x:x+run_count] = 1
                x += run_count
            
            i += 1
        y += 1
        if y >= h:
            break
    
    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!"

# Determine the height and width of the pattern
h = len(rle_string.split('$'))
w = max(len(row) for row in rle_string.split('$'))

result = rle_to_arr(rle_string, h, w)
print(result)
print(result.shape)





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: [[0 0 0 ... 0 0 0] [0 0 0 ... 0 0 0] [0 0 0 ... 0 0 0] ... [0 0 0 ... 0 0 0] [0 0 0 ... 0 0 0] [0 0 0 ... 0 0 0]] (40, 27) Traceback (most recent call last): File "/usr/src/app/main.py", line 75, in <module> assert answer == expected, f'Wrong answer; got {answer} instead of {expected}' AssertionError: Wrong answer; got 1428376 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 given Game of Life pattern (in Run Length Encoded format) to a numpy array: 

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import numpy as np

def rle_to_arr(string, h, w):
    # Initialize an empty array filled with zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Remove the header (everything before the first $)
    pattern = string.split('$', 1)[1]
    
    # Split the pattern into rows
    rows = pattern.split('$')
    
    x, y = 0, 0
    for row in rows:
        x = 0
        i = 0
        while i < len(row):
            # Check for run count
            run_count = ''
            while i < len(row) and row[i].isdigit():
                run_count += row[i]
                i += 1
            
            if run_count == '':
                run_count = 1
            else:
                run_count = int(run_count)
            
            # Check for the cell state
            if i < len(row):
                if row[i] == 'o':
                    arr[y, x:x+run_count] = 1
                x += run_count
            
            i += 1
        y += 1
        if y >= h:
            break
    
    return arr

# Example usage:
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!"""

# Determine the height and width from the pattern
h = rle_string.count('$') + 1
w = max(len(row) for row in rle_string.split('$'))

result = rle_to_arr(rle_string, h, w)
print(result)
This function does the following: 1. It initializes a numpy array of zeros with the specified height and width. 2. It removes the header from the RLE string (everything before the first $). 3. It splits the pattern into rows. 4. For each row, it decodes the RLE format: - It reads run counts (numbers before letters). - It sets cells to 1 for 'o' characters, leaving 'b' as 0. 5. It returns the resulting numpy array. You can call this function with your RLE string and the desired height and width. The function will return a numpy array representing the Game of Life pattern, where live cells are represented by 1s and dead cells by 0s.

Extract Code

I extracted the following code from that output: 

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import numpy as np

def rle_to_arr(string, h, w):
    # Initialize an empty array filled with zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Remove the header (everything before the first $)
    pattern = string.split('$', 1)[1]
    
    # Split the pattern into rows
    rows = pattern.split('$')
    
    x, y = 0, 0
    for row in rows:
        x = 0
        i = 0
        while i < len(row):
            # Check for run count
            run_count = ''
            while i < len(row) and row[i].isdigit():
                run_count += row[i]
                i += 1
            
            if run_count == '':
                run_count = 1
            else:
                run_count = int(run_count)
            
            # Check for the cell state
            if i < len(row):
                if row[i] == 'o':
                    arr[y, x:x+run_count] = 1
                x += run_count
            
            i += 1
        y += 1
        if y >= h:
            break
    
    return arr

# Example usage:
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!"""

# Determine the height and width from the pattern
h = rle_string.count('$') + 1
w = max(len(row) for row in rle_string.split('$'))

result = rle_to_arr(rle_string, h, w)
print(result)

Run Code Interpreter

Running the following program: 

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import numpy as np

def rle_to_arr(string, h, w):
    # Initialize an empty array filled with zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Remove the header (everything before the first $)
    pattern = string.split('$', 1)[1]
    
    # Split the pattern into rows
    rows = pattern.split('$')
    
    x, y = 0, 0
    for row in rows:
        x = 0
        i = 0
        while i < len(row):
            # Check for run count
            run_count = ''
            while i < len(row) and row[i].isdigit():
                run_count += row[i]
                i += 1
            
            if run_count == '':
                run_count = 1
            else:
                run_count = int(run_count)
            
            # Check for the cell state
            if i < len(row):
                if row[i] == 'o':
                    arr[y, x:x+run_count] = 1
                x += run_count
            
            i += 1
        y += 1
        if y >= h:
            break
    
    return arr

# Example usage:
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!"""

# Determine the height and width from the pattern
h = rle_string.count('$') + 1
w = max(len(row) for row in rle_string.split('$'))

result = rle_to_arr(rle_string, h, w)
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: [[0 0 0 ... 0 0 0] [0 0 0 ... 0 0 0] [0 0 0 ... 0 0 0] ... [0 0 0 ... 0 0 0] [0 0 0 ... 0 0 0] [0 0 0 ... 0 0 0]] Traceback (most recent call last): File "/usr/src/app/main.py", line 79, in <module> assert answer == expected, f'Wrong answer; got {answer} instead of {expected}' AssertionError: Wrong answer; got 1428376 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 given Game of Life pattern (in Run Length Encoded format) to a numpy array: 

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import numpy as np

def rle_to_arr(string, h, w):
    # Initialize an empty array
    arr = np.zeros((h, w), dtype=int)
    
    # Remove the header (everything before the first $)
    pattern = string.split('$', 1)[1]
    
    # Split the pattern into rows
    rows = pattern.split('$')
    
    x, y = 0, 0
    for row in rows:
        x = 0
        i = 0
        while i < len(row):
            # Check for run count
            run_count = ''
            while i < len(row) and row[i].isdigit():
                run_count += row[i]
                i += 1
            
            if run_count == '':
                run_count = 1
            else:
                run_count = int(run_count)
            
            # Check for cell state
            if i < len(row):
                if row[i] == 'o':
                    arr[y, x:x+run_count] = 1
                x += run_count
            
            i += 1
        y += 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!"

# Assuming the dimensions are 80x80 (adjust as needed)
result = rle_to_arr(rle_string, 80, 80)

# Print the result
print(result)
This function does the following: 1. It initializes a numpy array of zeros with the specified dimensions. 2. It removes the header from the RLE string (everything before the first $). 3. It splits the pattern into rows. 4. For each row, it parses the run counts and cell states. 5. It fills the array with 1s where there are live cells (represented by 'o' in the RLE). You can adjust the dimensions (80x80 in this example) as needed to fit your specific pattern. The function will fill the array up to the specified dimensions, ignoring any pattern that goes beyond these limits.

Extract Code

I extracted the following code from that output: 

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import numpy as np

def rle_to_arr(string, h, w):
    # Initialize an empty array
    arr = np.zeros((h, w), dtype=int)
    
    # Remove the header (everything before the first $)
    pattern = string.split('$', 1)[1]
    
    # Split the pattern into rows
    rows = pattern.split('$')
    
    x, y = 0, 0
    for row in rows:
        x = 0
        i = 0
        while i < len(row):
            # Check for run count
            run_count = ''
            while i < len(row) and row[i].isdigit():
                run_count += row[i]
                i += 1
            
            if run_count == '':
                run_count = 1
            else:
                run_count = int(run_count)
            
            # Check for cell state
            if i < len(row):
                if row[i] == 'o':
                    arr[y, x:x+run_count] = 1
                x += run_count
            
            i += 1
        y += 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!"

# Assuming the dimensions are 80x80 (adjust as needed)
result = rle_to_arr(rle_string, 80, 80)

# Print the result
print(result)

Run Code Interpreter

Running the following program: 

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import numpy as np

def rle_to_arr(string, h, w):
    # Initialize an empty array
    arr = np.zeros((h, w), dtype=int)
    
    # Remove the header (everything before the first $)
    pattern = string.split('$', 1)[1]
    
    # Split the pattern into rows
    rows = pattern.split('$')
    
    x, y = 0, 0
    for row in rows:
        x = 0
        i = 0
        while i < len(row):
            # Check for run count
            run_count = ''
            while i < len(row) and row[i].isdigit():
                run_count += row[i]
                i += 1
            
            if run_count == '':
                run_count = 1
            else:
                run_count = int(run_count)
            
            # Check for cell state
            if i < len(row):
                if row[i] == 'o':
                    arr[y, x:x+run_count] = 1
                x += run_count
            
            i += 1
        y += 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!"

# Assuming the dimensions are 80x80 (adjust as needed)
result = rle_to_arr(rle_string, 80, 80)

# Print the result
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: [[0 0 0 ... 0 0 0] [0 0 0 ... 0 0 0] [0 0 0 ... 0 0 0] ... [0 0 0 ... 0 0 0] [0 0 0 ... 0 0 0] [0 0 0 ... 0 0 0]] Traceback (most recent call last): File "/usr/src/app/main.py", line 71, in <module> assert answer == expected, f'Wrong answer; got {answer} instead of {expected}' AssertionError: Wrong answer; got 1428376 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 given Game of Life pattern in Run-Length Encoded (RLE) format to a numpy array: 

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import numpy as np

def rle_to_arr(string, h, w):
    # Initialize an empty array filled with zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Remove the header (everything before the first $)
    pattern = string.split('$', 1)[1]
    
    # Split the pattern into rows
    rows = pattern.split('$')
    
    x, y = 0, 0
    for row in rows:
        x = 0
        i = 0
        while i < len(row):
            # Get the run count (if any)
            run_count = ''
            while i < len(row) and row[i].isdigit():
                run_count += row[i]
                i += 1
            run_count = int(run_count) if run_count else 1
            
            # Process the character
            if i < len(row):
                if row[i] == 'o':
                    arr[y, x:x+run_count] = 1
                x += run_count
            i += 1
        y += 1
        if y >= h:
            break
    
    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!"

# Assuming the dimensions of the pattern
height, width = 100, 100

result = rle_to_arr(rle_string, height, width)
print(result)
This function does the following: 1. It initializes a numpy array of zeros with the specified height and width. 2. It removes the header from the RLE string (everything before the first $). 3. It splits the pattern into rows. 4. For each row, it processes the RLE encoding: - It handles run counts (e.g., '3o' means three consecutive 'o's). - It sets 1s in the array for 'o' characters, leaving 'b' (empty) as 0. 5. It returns the resulting numpy array. Note that this function assumes that 'o' represents live cells (1) and 'b' or no character represents dead cells (0). It also assumes that the pattern fits within the specified dimensions. If the pattern is larger than the specified dimensions, it will be truncated. You can adjust the `height` and `width` variables to match the actual dimensions of your pattern. The function will fill the array up to these dimensions, padding with zeros if the pattern is smaller.

Extract Code

I extracted the following code from that output: 

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import numpy as np

def rle_to_arr(string, h, w):
    # Initialize an empty array filled with zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Remove the header (everything before the first $)
    pattern = string.split('$', 1)[1]
    
    # Split the pattern into rows
    rows = pattern.split('$')
    
    x, y = 0, 0
    for row in rows:
        x = 0
        i = 0
        while i < len(row):
            # Get the run count (if any)
            run_count = ''
            while i < len(row) and row[i].isdigit():
                run_count += row[i]
                i += 1
            run_count = int(run_count) if run_count else 1
            
            # Process the character
            if i < len(row):
                if row[i] == 'o':
                    arr[y, x:x+run_count] = 1
                x += run_count
            i += 1
        y += 1
        if y >= h:
            break
    
    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!"

# Assuming the dimensions of the pattern
height, width = 100, 100

result = rle_to_arr(rle_string, height, width)
print(result)

Run Code Interpreter

Running the following program: 

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import numpy as np

def rle_to_arr(string, h, w):
    # Initialize an empty array filled with zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Remove the header (everything before the first $)
    pattern = string.split('$', 1)[1]
    
    # Split the pattern into rows
    rows = pattern.split('$')
    
    x, y = 0, 0
    for row in rows:
        x = 0
        i = 0
        while i < len(row):
            # Get the run count (if any)
            run_count = ''
            while i < len(row) and row[i].isdigit():
                run_count += row[i]
                i += 1
            run_count = int(run_count) if run_count else 1
            
            # Process the character
            if i < len(row):
                if row[i] == 'o':
                    arr[y, x:x+run_count] = 1
                x += run_count
            i += 1
        y += 1
        if y >= h:
            break
    
    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!"

# Assuming the dimensions of the pattern
height, width = 100, 100

result = rle_to_arr(rle_string, height, width)
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: [[0 0 0 ... 0 0 0] [0 0 0 ... 0 0 0] [0 0 0 ... 0 0 0] ... [0 0 0 ... 0 0 0] [0 0 0 ... 0 0 0] [0 0 0 ... 0 0 0]] 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 1428376 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

To convert the given Game of Life pattern to a numpy array, we can create a function `rle_to_arr(string, h, w)` that parses the RLE (Run-Length Encoded) string and creates a 2D numpy array. Here's the implementation: 

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import numpy as np

def rle_to_arr(string, h, w):
    # Initialize the array with zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Remove the header (everything before the first $)
    pattern = string.split('$', 1)[1]
    
    # Split the pattern into rows
    rows = pattern.split('$')
    
    x, y = 0, 0
    for row in rows:
        x = 0
        i = 0
        while i < len(row):
            # Check for run count
            run_count = ''
            while i < len(row) and row[i].isdigit():
                run_count += row[i]
                i += 1
            
            if run_count == '':
                run_count = 1
            else:
                run_count = int(run_count)
            
            # Check for cell state
            if i < len(row):
                if row[i] == 'o' or row[i] == 'b':
                    state = 1 if row[i] == 'o' else 0
                    arr[y, x:x+run_count] = state
                    x += run_count
                elif row[i] == '!':
                    break
            i += 1
        y += 1
        if y >= h:
            break
    
    return arr

# Example usage:
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!"""

# Assuming the pattern fits within a 100x100 grid
result = rle_to_arr(rle_string, 100, 100)
print(result)
This function does the following: 1. Initializes a numpy array of zeros with the specified dimensions. 2. Removes the header from the RLE string. 3. Splits the pattern into rows. 4. Iterates through each row, parsing run counts and cell states. 5. Fills the array with 1s for live cells and 0s for dead cells. You can adjust the dimensions (100x100 in this example) to fit your specific needs. The function will fill the array up to the specified dimensions or until it reaches the end of the pattern, whichever comes first.

Extract Code

I extracted the following code from that output: 

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import numpy as np

def rle_to_arr(string, h, w):
    # Initialize the array with zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Remove the header (everything before the first $)
    pattern = string.split('$', 1)[1]
    
    # Split the pattern into rows
    rows = pattern.split('$')
    
    x, y = 0, 0
    for row in rows:
        x = 0
        i = 0
        while i < len(row):
            # Check for run count
            run_count = ''
            while i < len(row) and row[i].isdigit():
                run_count += row[i]
                i += 1
            
            if run_count == '':
                run_count = 1
            else:
                run_count = int(run_count)
            
            # Check for cell state
            if i < len(row):
                if row[i] == 'o' or row[i] == 'b':
                    state = 1 if row[i] == 'o' else 0
                    arr[y, x:x+run_count] = state
                    x += run_count
                elif row[i] == '!':
                    break
            i += 1
        y += 1
        if y >= h:
            break
    
    return arr

# Example usage:
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!"""

# Assuming the pattern fits within a 100x100 grid
result = rle_to_arr(rle_string, 100, 100)
print(result)

Run Code Interpreter

Running the following program: 

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import numpy as np

def rle_to_arr(string, h, w):
    # Initialize the array with zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Remove the header (everything before the first $)
    pattern = string.split('$', 1)[1]
    
    # Split the pattern into rows
    rows = pattern.split('$')
    
    x, y = 0, 0
    for row in rows:
        x = 0
        i = 0
        while i < len(row):
            # Check for run count
            run_count = ''
            while i < len(row) and row[i].isdigit():
                run_count += row[i]
                i += 1
            
            if run_count == '':
                run_count = 1
            else:
                run_count = int(run_count)
            
            # Check for cell state
            if i < len(row):
                if row[i] == 'o' or row[i] == 'b':
                    state = 1 if row[i] == 'o' else 0
                    arr[y, x:x+run_count] = state
                    x += run_count
                elif row[i] == '!':
                    break
            i += 1
        y += 1
        if y >= h:
            break
    
    return arr

# Example usage:
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!"""

# Assuming the pattern fits within a 100x100 grid
result = rle_to_arr(rle_string, 100, 100)
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: [[0 0 0 ... 0 0 0] [0 0 0 ... 0 0 0] [0 0 0 ... 0 0 0] ... [0 0 0 ... 0 0 0] [0 0 0 ... 0 0 0] [0 0 0 ... 0 0 0]] Traceback (most recent call last): File "/usr/src/app/main.py", line 78, in <module> assert answer == expected, f'Wrong answer; got {answer} instead of {expected}' AssertionError: Wrong answer; got 1437410 instead of 1563304

Substring Evaluation

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