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 RLE (Run Length Encoded) pattern from Conway's Game of Life to a numpy array, we need to parse the RLE string and then fill in a numpy array accordingly. Below is the implementation of the `rle_to_arr` function: 

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import numpy as np
import re

def rle_to_arr(string, h, w):
    # Initialize the numpy array with zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Split the string by '$' which represents the end of a row
    rows = string.split('$')
    
    # Pattern to match numbers and letters
    pattern = re.compile(r'(\d*)([bo])')
    
    row_index = 0
    for row in rows:
        if not row:
            row_index += 1
            continue
        
        col_index = 0
        for match in pattern.finditer(row):
            count = int(match.group(1)) if match.group(1) else 1
            char = match.group(2)
            
            if char == 'b':
                col_index += count
            elif char == 'o':
                arr[row_index, col_index:col_index+count] = 1
                col_index += count
        
        row_index += 1
    
    return arr

# Sample 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$54b2o!"
h, w = 80, 80  # Example dimensions
arr = rle_to_arr(rle_string, h, w)
print(arr)
Here's how the function works: 1. It initializes a numpy array of shape `(h, w)` with zeros. 2. It splits the RLE string by the `$` character, which signifies the end of a row. 3. It uses a regular expression to match sequences of digits followed by either `b` (for blank cells) or `o` (for filled cells). 4. It parses each row and fills in the array with `1`s for `o` and leaves cells as `0`s for `b`. 5. It handles the case where a digit is not present by defaulting to `1`. This function should correctly interpret the RLE string and convert it to a numpy array representation of the Game of Life pattern.

Extract Code

I extracted the following code from that output: 

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import numpy as np
import re

def rle_to_arr(string, h, w):
    # Initialize the numpy array with zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Split the string by '$' which represents the end of a row
    rows = string.split('$')
    
    # Pattern to match numbers and letters
    pattern = re.compile(r'(\d*)([bo])')
    
    row_index = 0
    for row in rows:
        if not row:
            row_index += 1
            continue
        
        col_index = 0
        for match in pattern.finditer(row):
            count = int(match.group(1)) if match.group(1) else 1
            char = match.group(2)
            
            if char == 'b':
                col_index += count
            elif char == 'o':
                arr[row_index, col_index:col_index+count] = 1
                col_index += count
        
        row_index += 1
    
    return arr

# Sample 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!"
h, w = 80, 80  # Example dimensions
arr = rle_to_arr(rle_string, h, w)
print(arr)

Run Code Interpreter

Running the following program: 

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import numpy as np
import re

def rle_to_arr(string, h, w):
    # Initialize the numpy array with zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Split the string by '$' which represents the end of a row
    rows = string.split('$')
    
    # Pattern to match numbers and letters
    pattern = re.compile(r'(\d*)([bo])')
    
    row_index = 0
    for row in rows:
        if not row:
            row_index += 1
            continue
        
        col_index = 0
        for match in pattern.finditer(row):
            count = int(match.group(1)) if match.group(1) else 1
            char = match.group(2)
            
            if char == 'b':
                col_index += count
            elif char == 'o':
                arr[row_index, col_index:col_index+count] = 1
                col_index += count
        
        row_index += 1
    
    return arr

# Sample 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!"
h, w = 80, 80  # Example dimensions
arr = rle_to_arr(rle_string, h, w)
print(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: [[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 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

To convert the Game of Life pattern from the RLE (Run Length Encoded) format to a NumPy array, we need to parse the RLE string and then populate the NumPy array accordingly. Here is the implementation of the function `rle_to_arr`: 

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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)
    
    # Split the RLE string into lines and remove any empty lines
    lines = rle_string.split('$')
    lines = [line for line in lines if line]
    
    # Variables to keep track of the current position in the array
    row, col = 0, 0
    
    for line in lines:
        i = 0
        while i < len(line):
            if line[i].isdigit():
                # If the character is a digit, we need to find the complete number
                num_str = ''
                while i < len(line) and line[i].isdigit():
                    num_str += line[i]
                    i += 1
                num = int(num_str)
                
                # Determine whether the number represents 'b' (empty space) or 'o' (filled space)
                if i < len(line) and line[i] == 'b':
                    col += num
                elif i < len(line) and line[i] == 'o':
                    arr[row, col:col+num] = 1
                    col += num
                i += 1
            else:
                if line[i] == 'b':
                    col += 1
                elif line[i] == 'o':
                    arr[row, col] = 1
                    col += 1
                i += 1
        
        # Move to the next row and reset column index
        row += 1
        col = 0
    
    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!"
h, w = 80, 80  # Set height and width of the array
arr = rle_to_arr(rle_string, h, w)
print(arr)
This function will convert the given RLE string into a NumPy array of shape `(h, w)`, where `1` represents a filled cell (alive) and `0` represents an empty cell (dead). Note that the function assumes the RLE string uses 'b' for empty cells and 'o' for filled cells, as per the standard Game of Life RLE format.

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)
    
    # Split the RLE string into lines and remove any empty lines
    lines = rle_string.split('$')
    lines = [line for line in lines if line]
    
    # Variables to keep track of the current position in the array
    row, col = 0, 0
    
    for line in lines:
        i = 0
        while i < len(line):
            if line[i].isdigit():
                # If the character is a digit, we need to find the complete number
                num_str = ''
                while i < len(line) and line[i].isdigit():
                    num_str += line[i]
                    i += 1
                num = int(num_str)
                
                # Determine whether the number represents 'b' (empty space) or 'o' (filled space)
                if i < len(line) and line[i] == 'b':
                    col += num
                elif i < len(line) and line[i] == 'o':
                    arr[row, col:col+num] = 1
                    col += num
                i += 1
            else:
                if line[i] == 'b':
                    col += 1
                elif line[i] == 'o':
                    arr[row, col] = 1
                    col += 1
                i += 1
        
        # Move to the next row and reset column index
        row += 1
        col = 0
    
    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!"
h, w = 80, 80  # Set height and width of the array
arr = rle_to_arr(rle_string, h, w)
print(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
    arr = np.zeros((h, w), dtype=int)
    
    # Split the RLE string into lines and remove any empty lines
    lines = rle_string.split('$')
    lines = [line for line in lines if line]
    
    # Variables to keep track of the current position in the array
    row, col = 0, 0
    
    for line in lines:
        i = 0
        while i < len(line):
            if line[i].isdigit():
                # If the character is a digit, we need to find the complete number
                num_str = ''
                while i < len(line) and line[i].isdigit():
                    num_str += line[i]
                    i += 1
                num = int(num_str)
                
                # Determine whether the number represents 'b' (empty space) or 'o' (filled space)
                if i < len(line) and line[i] == 'b':
                    col += num
                elif i < len(line) and line[i] == 'o':
                    arr[row, col:col+num] = 1
                    col += num
                i += 1
            else:
                if line[i] == 'b':
                    col += 1
                elif line[i] == 'o':
                    arr[row, col] = 1
                    col += 1
                i += 1
        
        # Move to the next row and reset column index
        row += 1
        col = 0
    
    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!"
h, w = 80, 80  # Set height and width of the array
arr = rle_to_arr(rle_string, h, w)
print(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 50, in <module> arr = rle_to_arr(rle_string, h, w) File "/usr/src/app/main.py", line 37, in rle_to_arr arr[row, col] = 1 IndexError: index 87 is out of bounds for axis 1 with size 80

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 RLE (Run Length Encoded) pattern to a numpy array, we need to parse the RLE string and translate it into a 2D numpy array. Here’s a function that does just that: 

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import numpy as np

def rle_to_arr(string, h, w):
    # Initialize an empty numpy array of the given dimensions
    arr = np.zeros((h, w), dtype=int)
    
    rows = string.split('$')
    x, y = 0, 0
    
    for row in rows:
        i = 0
        while i < len(row):
            if row[i].isdigit():
                num = 0
                while i < len(row) and row[i].isdigit():
                    num = num * 10 + int(row[i])
                    i += 1
                if row[i] == 'b':
                    x += num
                elif row[i] == 'o':
                    arr[y, x:x+num] = 1
                    x += num
            elif row[i] == 'b':
                x += 1
            elif row[i] == 'o':
                arr[y, x] = 1
                x += 1
            i += 1
        y += 1
        x = 0
        
    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!"
height = 80
width = 80

arr = rle_to_arr(rle_string, height, width)
print(arr)
Here's a breakdown: 1. **Initialize the Array**: Create an empty numpy array with the specified height and width. 2. **Parse the RLE String**: Split the RLE string by `$`, which separates the rows. 3. **Process Each Row**: For each row, parse through the characters. 4. **Decode Runs**: Handle digits followed by `b` or `o` to determine the number of cells to fill. 5. **Advance to Next Row**: After processing a row, move to the next row and reset the column index. This script converts the provided RLE string into a numpy array, where filled cells are represented by 1s and empty 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 numpy array of the given dimensions
    arr = np.zeros((h, w), dtype=int)
    
    rows = string.split('$')
    x, y = 0, 0
    
    for row in rows:
        i = 0
        while i < len(row):
            if row[i].isdigit():
                num = 0
                while i < len(row) and row[i].isdigit():
                    num = num * 10 + int(row[i])
                    i += 1
                if row[i] == 'b':
                    x += num
                elif row[i] == 'o':
                    arr[y, x:x+num] = 1
                    x += num
            elif row[i] == 'b':
                x += 1
            elif row[i] == 'o':
                arr[y, x] = 1
                x += 1
            i += 1
        y += 1
        x = 0
        
    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!"
height = 80
width = 80

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

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 numpy array of the given dimensions
    arr = np.zeros((h, w), dtype=int)
    
    rows = string.split('$')
    x, y = 0, 0
    
    for row in rows:
        i = 0
        while i < len(row):
            if row[i].isdigit():
                num = 0
                while i < len(row) and row[i].isdigit():
                    num = num * 10 + int(row[i])
                    i += 1
                if row[i] == 'b':
                    x += num
                elif row[i] == 'o':
                    arr[y, x:x+num] = 1
                    x += num
            elif row[i] == 'b':
                x += 1
            elif row[i] == 'o':
                arr[y, x] = 1
                x += 1
            i += 1
        y += 1
        x = 0
        
    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!"
height = 80
width = 80

arr = rle_to_arr(rle_string, height, width)
print(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 40, in <module> arr = rle_to_arr(rle_string, height, width) File "/usr/src/app/main.py", line 19, in rle_to_arr if row[i] == 'b': IndexError: string index out of range

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 an RLE (Run Length Encoded) pattern from Conway's Game of Life to a numpy array, we need to decode the RLE string and populate the numpy array accordingly. The provided RLE pattern needs to be parsed, and then we can fill the array based on the parsed information. Here's the function `rle_to_arr` which takes the RLE string, height `h`, and width `w` as inputs and returns the corresponding numpy array: 

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import numpy as np

def rle_to_arr(string, h, w):
    # Initialize the numpy array with zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Split the RLE string into lines and process each line
    lines = string.split('$')
    
    row = 0
    col = 0
    
    for line in lines:
        count = ''
        for char in line:
            if char.isdigit():
                count += char
            else:
                if count:
                    num = int(count)
                else:
                    num = 1
                
                if char == 'o':
                    arr[row, col:col+num] = 1
                    col += num
                elif char == 'b':
                    col += num
                
                count = ''
        
        row += 1
        col = 0
    
    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!"
height = 80
width = 80

arr = rle_to_arr(rle_string, height, width)
print(arr)
This function decodes the RLE string, creating a numpy array with the specified height and width. The array is filled with 1s where the pattern's live cells are and 0s otherwise. Let's break down the steps of the function: 1. Initialize the numpy array with zeros. 2. Split the input string by the `$` character, which separates each line in the RLE pattern. 3. Loop through each line, and for each character, decode the number of cells to be filled with 1s (`o`) or skipped (`b`). 4. Update the row and column indices accordingly. 5. Finally, return the populated numpy array. Make sure that the dimensions `h` and `w` are sufficient to accommodate the pattern described by the RLE string.

Extract Code

I extracted the following code from that output: 

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import numpy as np

def rle_to_arr(string, h, w):
    # Initialize the numpy array with zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Split the RLE string into lines and process each line
    lines = string.split('$')
    
    row = 0
    col = 0
    
    for line in lines:
        count = ''
        for char in line:
            if char.isdigit():
                count += char
            else:
                if count:
                    num = int(count)
                else:
                    num = 1
                
                if char == 'o':
                    arr[row, col:col+num] = 1
                    col += num
                elif char == 'b':
                    col += num
                
                count = ''
        
        row += 1
        col = 0
    
    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!"
height = 80
width = 80

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

Run Code Interpreter

Running the following program: 

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import numpy as np

def rle_to_arr(string, h, w):
    # Initialize the numpy array with zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Split the RLE string into lines and process each line
    lines = string.split('$')
    
    row = 0
    col = 0
    
    for line in lines:
        count = ''
        for char in line:
            if char.isdigit():
                count += char
            else:
                if count:
                    num = int(count)
                else:
                    num = 1
                
                if char == 'o':
                    arr[row, col:col+num] = 1
                    col += num
                elif char == 'b':
                    col += num
                
                count = ''
        
        row += 1
        col = 0
    
    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!"
height = 80
width = 80

arr = rle_to_arr(rle_string, height, width)
print(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: [[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 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

To convert an RLE (Run Length Encoded) pattern from Conway's Game of Life into a NumPy array, we need to parse the RLE string and then fill the array accordingly. Here is the function `rle_to_arr` that accomplishes this: 

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import numpy as np
import re

def rle_to_arr(string, h, w):
    # Initialize an array filled with zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Split the RLE string into rows and process each row
    rows = string.strip().split('$')
    
    x, y = 0, 0  # Start at the top-left corner of the array
    for row in rows:
        count = 0
        for char in row:
            if char.isdigit():
                count = count * 10 + int(char)
            else:
                if count == 0:
                    count = 1
                if char == 'b':
                    x += count
                elif char == 'o':
                    arr[y, x:x+count] = 1
                    x += count
                count = 0
        y += 1
        x = 0  # Reset to the beginning of the next row
    
    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!"

h = 100  # height of the array
w = 100  # width of the array

arr = rle_to_arr(rle_string, h, w)
print(arr)
This function will parse the RLE string and return a NumPy array of the specified shape, filled with 1s for live cells and 0s for dead cells. Note that the height (`h`) and width (`w`) should be large enough to contain the entire pattern described by the RLE string.

Extract Code

I extracted the following code from that output: 

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import numpy as np
import re

def rle_to_arr(string, h, w):
    # Initialize an array filled with zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Split the RLE string into rows and process each row
    rows = string.strip().split('$')
    
    x, y = 0, 0  # Start at the top-left corner of the array
    for row in rows:
        count = 0
        for char in row:
            if char.isdigit():
                count = count * 10 + int(char)
            else:
                if count == 0:
                    count = 1
                if char == 'b':
                    x += count
                elif char == 'o':
                    arr[y, x:x+count] = 1
                    x += count
                count = 0
        y += 1
        x = 0  # Reset to the beginning of the next row
    
    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!"

h = 100  # height of the array
w = 100  # width of the array

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

Run Code Interpreter

Running the following program: 

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import numpy as np
import re

def rle_to_arr(string, h, w):
    # Initialize an array filled with zeros
    arr = np.zeros((h, w), dtype=int)
    
    # Split the RLE string into rows and process each row
    rows = string.strip().split('$')
    
    x, y = 0, 0  # Start at the top-left corner of the array
    for row in rows:
        count = 0
        for char in row:
            if char.isdigit():
                count = count * 10 + int(char)
            else:
                if count == 0:
                    count = 1
                if char == 'b':
                    x += count
                elif char == 'o':
                    arr[y, x:x+count] = 1
                    x += count
                count = 0
        y += 1
        x = 0  # Reset to the beginning of the next row
    
    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!"

h = 100  # height of the array
w = 100  # width of the array

arr = rle_to_arr(rle_string, h, w)
print(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: [[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 62, 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