Initial Query

Here's a python program. Please rewrite it in c with the type signature float solve_dp(float* arr, int size, int lines_left) def solve_dp(graph, lines_left): n = len(graph) dp = [[[1e9 for _ in range(10)] for _ in range(lines_left + 1)] for _ in range(n + 1)] # Initialize how as -1 indicating no decision made how = [[[-1 for _ in range(10)] for _ in range(lines_left + 1)] for _ in range(n + 1)] for i in range(n, -1, -1): for j in range(lines_left + 1): for k in range(10): if i == n and j == 0: dp[i][j][k] = 0 else: # Option 1: don't take the line if i < n and k < 9: if dp[i + 1][j][k + 1] < dp[i][j][k]: dp[i][j][k] = dp[i + 1][j][k + 1] how[i][j][k] = k + 1 # Representing choosing not to take the line # Option 2: take the line if i < n and j > 0: cost = graph[i] + (k - 8)**2 if cost + dp[i + 1][j - 1][0] < dp[i][j][k]: dp[i][j][k] = cost + dp[i + 1][j - 1][0] how[i][j][k] = 0 # Representing choosing to take the line # Reconstruct the solution i, j, k = 0, lines_left, 6 taken_lines = [] while i < n: if how[i][j][k] == 0: # Chose to take the line taken_lines.append(n - i) i += 1 j -= 1 k = 0 else: # Chose not to take the line i += 1 k += 1 return dp[0][lines_left][6]

LLM Generation

Query

Here's a python program. Please rewrite it in c with the type signature float solve_dp(float* arr, int size, int lines_left) def solve_dp(graph, lines_left): n = len(graph) dp = [[[1e9 for _ in range(10)] for _ in range(lines_left + 1)] for _ in range(n + 1)] # Initialize how as -1 indicating no decision made how = [[[-1 for _ in range(10)] for _ in range(lines_left + 1)] for _ in range(n + 1)] for i in range(n, -1, -1): for j in range(lines_left + 1): for k in range(10): if i == n and j == 0: dp[i][j][k] = 0 else: # Option 1: don't take the line if i < n and k < 9: if dp[i + 1][j][k + 1] < dp[i][j][k]: dp[i][j][k] = dp[i + 1][j][k + 1] how[i][j][k] = k + 1 # Representing choosing not to take the line # Option 2: take the line if i < n and j > 0: cost = graph[i] + (k - 8)**2 if cost + dp[i + 1][j - 1][0] < dp[i][j][k]: dp[i][j][k] = cost + dp[i + 1][j - 1][0] how[i][j][k] = 0 # Representing choosing to take the line # Reconstruct the solution i, j, k = 0, lines_left, 6 taken_lines = [] while i < n: if how[i][j][k] == 0: # Chose to take the line taken_lines.append(n - i) i += 1 j -= 1 k = 0 else: # Chose not to take the line i += 1 k += 1 return dp[0][lines_left][6]

Output

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#include <stdio.h>
#include <stdlib.h>
#include <math.h>

float solve_dp(float* graph, int size, int lines_left) {
    float dp[size + 1][lines_left + 1][10];
    int how[size + 1][lines_left + 1][10];

    for (int i = size; i >= 0; i--) {
        for (int j = lines_left; j >= 0; j--) {
            for (int k = 0; k < 10; k++) {
                if (i == size && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    float option1 = dp[i + 1][j][k + 1];
                    float option2 = dp[i + 1][j - 1][0];
                    if (i < size && k < 9 && option1 < option2) {
                        dp[i][j][k] = option1;
                        how[i][j][k] = k + 1;
                    } else if (i < size && j > 0 && option2 < option1) {
                        dp[i][j][k] = graph[i] + pow(k - 8, 2) + option2;
                        how[i][j][k] = 0;
                    } else {
                        dp[i][j][k] = option1;
                        how[i][j][k] = k + 1;
                    }
                }
            }
        }
    }

    int taken_lines[size];
    int index = 0;
    int i = 0;
    int j = lines_left;
    int k = 6;
    while (i < size) {
        if (how[i][j][k] == 0) {
            taken_lines[index] = size - i;
            index++;
            i++;
            j--;
            k = 0;
        } else {
            i++;
            k++;
        }
    }

    return dp[0][lines_left][6];
}

Extract Code

I extracted the following code from that output: 

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#include <stdio.h>
#include <stdlib.h>
#include <math.h>

float solve_dp(float* graph, int size, int lines_left) {
    float dp[size + 1][lines_left + 1][10];
    int how[size + 1][lines_left + 1][10];

    for (int i = size; i >= 0; i--) {
        for (int j = lines_left; j >= 0; j--) {
            for (int k = 0; k < 10; k++) {
                if (i == size && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    float option1 = dp[i + 1][j][k + 1];
                    float option2 = dp[i + 1][j - 1][0];
                    if (i < size && k < 9 && option1 < option2) {
                        dp[i][j][k] = option1;
                        how[i][j][k] = k + 1;
                    } else if (i < size && j > 0 && option2 < option1) {
                        dp[i][j][k] = graph[i] + pow(k - 8, 2) + option2;
                        how[i][j][k] = 0;
                    } else {
                        dp[i][j][k] = option1;
                        how[i][j][k] = k + 1;
                    }
                }
            }
        }
    }

    int taken_lines[size];
    int index = 0;
    int i = 0;
    int j = lines_left;
    int k = 6;
    while (i < size) {
        if (how[i][j][k] == 0) {
            taken_lines[index] = size - i;
            index++;
            i++;
            j--;
            k = 0;
        } else {
            i++;
            k++;
        }
    }

    return dp[0][lines_left][6];
}

Run Code Interpreter

Running the following program: 

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#include <stdio.h>
#include <stdlib.h>
#include <math.h>

float solve_dp(float* graph, int size, int lines_left) {
    float dp[size + 1][lines_left + 1][10];
    int how[size + 1][lines_left + 1][10];

    for (int i = size; i >= 0; i--) {
        for (int j = lines_left; j >= 0; j--) {
            for (int k = 0; k < 10; k++) {
                if (i == size && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    float option1 = dp[i + 1][j][k + 1];
                    float option2 = dp[i + 1][j - 1][0];
                    if (i < size && k < 9 && option1 < option2) {
                        dp[i][j][k] = option1;
                        how[i][j][k] = k + 1;
                    } else if (i < size && j > 0 && option2 < option1) {
                        dp[i][j][k] = graph[i] + pow(k - 8, 2) + option2;
                        how[i][j][k] = 0;
                    } else {
                        dp[i][j][k] = option1;
                        how[i][j][k] = k + 1;
                    }
                }
            }
        }
    }

    int taken_lines[size];
    int index = 0;
    int i = 0;
    int j = lines_left;
    int k = 6;
    while (i < size) {
        if (how[i][j][k] == 0) {
            taken_lines[index] = size - i;
            index++;
            i++;
            j--;
            k = 0;
        } else {
            i++;
            k++;
        }
    }

    return dp[0][lines_left][6];
}



#include<stdio.h>
#include<stdlib.h>
int main() {
float arr[] = {71, 89, 34, 63, 19, 94, 54, 61, 88, 20, 66, 46, 26, 87, 55, 81, 6, 2, 72, 75, 98, 78, 24, 95, 73, 7, 56, 48, 14, 99, 64, 51, 69, 77, 28, 47, 8, 22, 49, 3, 62, 32, 10, 82, 35, 18, 85, 60, 83, 23, 5, 40, 41, 68, 53, 52, 44, 45, 65, 84, 93, 25, 13, 1, 31, 11, 12, 97, 38, 0, 43, 90, 36, 70, 33, 17, 21, 30, 16, 15, 74, 67, 58, 37, 39, 96, 79, 29, 27, 92, 86, 9, 80, 42, 57, 91, 59, 4, 76, 50};

int answer = solve_dp(arr, 100, 100);
int expected = 11290;
if (answer != expected) {
    printf("Wrong answer; got %d instead of %d.\n", answer, expected);
    exit(1);
}
printf("All tests passed\n");
}
And got the output: Wrong answer; got -2147483648 instead of 11290.

Substring Evaluation

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

Initial Query

Here's a python program. Please rewrite it in c with the type signature float solve_dp(float* arr, int size, int lines_left) def solve_dp(graph, lines_left): n = len(graph) dp = [[[1e9 for _ in range(10)] for _ in range(lines_left + 1)] for _ in range(n + 1)] # Initialize how as -1 indicating no decision made how = [[[-1 for _ in range(10)] for _ in range(lines_left + 1)] for _ in range(n + 1)] for i in range(n, -1, -1): for j in range(lines_left + 1): for k in range(10): if i == n and j == 0: dp[i][j][k] = 0 else: # Option 1: don't take the line if i < n and k < 9: if dp[i + 1][j][k + 1] < dp[i][j][k]: dp[i][j][k] = dp[i + 1][j][k + 1] how[i][j][k] = k + 1 # Representing choosing not to take the line # Option 2: take the line if i < n and j > 0: cost = graph[i] + (k - 8)**2 if cost + dp[i + 1][j - 1][0] < dp[i][j][k]: dp[i][j][k] = cost + dp[i + 1][j - 1][0] how[i][j][k] = 0 # Representing choosing to take the line # Reconstruct the solution i, j, k = 0, lines_left, 6 taken_lines = [] while i < n: if how[i][j][k] == 0: # Chose to take the line taken_lines.append(n - i) i += 1 j -= 1 k = 0 else: # Chose not to take the line i += 1 k += 1 return dp[0][lines_left][6]

LLM Generation

Query

Here's a python program. Please rewrite it in c with the type signature float solve_dp(float* arr, int size, int lines_left) def solve_dp(graph, lines_left): n = len(graph) dp = [[[1e9 for _ in range(10)] for _ in range(lines_left + 1)] for _ in range(n + 1)] # Initialize how as -1 indicating no decision made how = [[[-1 for _ in range(10)] for _ in range(lines_left + 1)] for _ in range(n + 1)] for i in range(n, -1, -1): for j in range(lines_left + 1): for k in range(10): if i == n and j == 0: dp[i][j][k] = 0 else: # Option 1: don't take the line if i < n and k < 9: if dp[i + 1][j][k + 1] < dp[i][j][k]: dp[i][j][k] = dp[i + 1][j][k + 1] how[i][j][k] = k + 1 # Representing choosing not to take the line # Option 2: take the line if i < n and j > 0: cost = graph[i] + (k - 8)**2 if cost + dp[i + 1][j - 1][0] < dp[i][j][k]: dp[i][j][k] = cost + dp[i + 1][j - 1][0] how[i][j][k] = 0 # Representing choosing to take the line # Reconstruct the solution i, j, k = 0, lines_left, 6 taken_lines = [] while i < n: if how[i][j][k] == 0: # Chose to take the line taken_lines.append(n - i) i += 1 j -= 1 k = 0 else: # Chose not to take the line i += 1 k += 1 return dp[0][lines_left][6]

Output

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#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

float solve_dp(float* arr, int size, int lines_left) {
    float dp[size + 1][lines_left + 1][10];
    int how[size + 1][lines_left + 1][10];

    for (int i = 0; i <= size; i++) {
        for (int j = 0; j <= lines_left; j++) {
            for (int k = 0; k < 10; k++) {
                dp[i][j][k] = FLT_MAX;
                how[i][j][k] = -1;
            }
        }
    }

    dp[size][0][6] = 0;

    for (int i = size - 1; i >= 0; i--) {
        for (int j = 0; j <= lines_left; j++) {
            for (int k = 0; k < 10; k++) {
                if (i == size && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    // Option 1: don't take the line
                    if (i < size && k < 9) {
                        if (dp[i + 1][j][k + 1] < dp[i][j][k]) {
                            dp[i][j][k] = dp[i + 1][j][k + 1];
                            how[i][j][k] = k + 1;  // Representing choosing not to take the line
                        }
                    }

                    // Option 2: take the line
                    if (i < size && j > 0) {
                        float cost = arr[i] + (k - 8) * (k - 8);
                        if (cost + dp[i + 1][j - 1][0] < dp[i][j][k]) {
                            dp[i][j][k] = cost + dp[i + 1][j - 1][0];
                            how[i][j][k] = 0;  // Representing choosing to take the line
                        }
                    }
                }
            }
        }
    }

    return dp[0][lines_left][6];
}

Extract Code

I extracted the following code from that output: 

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#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

float solve_dp(float* arr, int size, int lines_left) {
    float dp[size + 1][lines_left + 1][10];
    int how[size + 1][lines_left + 1][10];

    for (int i = 0; i <= size; i++) {
        for (int j = 0; j <= lines_left; j++) {
            for (int k = 0; k < 10; k++) {
                dp[i][j][k] = FLT_MAX;
                how[i][j][k] = -1;
            }
        }
    }

    dp[size][0][6] = 0;

    for (int i = size - 1; i >= 0; i--) {
        for (int j = 0; j <= lines_left; j++) {
            for (int k = 0; k < 10; k++) {
                if (i == size && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    // Option 1: don't take the line
                    if (i < size && k < 9) {
                        if (dp[i + 1][j][k + 1] < dp[i][j][k]) {
                            dp[i][j][k] = dp[i + 1][j][k + 1];
                            how[i][j][k] = k + 1;  // Representing choosing not to take the line
                        }
                    }

                    // Option 2: take the line
                    if (i < size && j > 0) {
                        float cost = arr[i] + (k - 8) * (k - 8);
                        if (cost + dp[i + 1][j - 1][0] < dp[i][j][k]) {
                            dp[i][j][k] = cost + dp[i + 1][j - 1][0];
                            how[i][j][k] = 0;  // Representing choosing to take the line
                        }
                    }
                }
            }
        }
    }

    return dp[0][lines_left][6];
}

Run Code Interpreter

Running the following program: 

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#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

float solve_dp(float* arr, int size, int lines_left) {
    float dp[size + 1][lines_left + 1][10];
    int how[size + 1][lines_left + 1][10];

    for (int i = 0; i <= size; i++) {
        for (int j = 0; j <= lines_left; j++) {
            for (int k = 0; k < 10; k++) {
                dp[i][j][k] = FLT_MAX;
                how[i][j][k] = -1;
            }
        }
    }

    dp[size][0][6] = 0;

    for (int i = size - 1; i >= 0; i--) {
        for (int j = 0; j <= lines_left; j++) {
            for (int k = 0; k < 10; k++) {
                if (i == size && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    // Option 1: don't take the line
                    if (i < size && k < 9) {
                        if (dp[i + 1][j][k + 1] < dp[i][j][k]) {
                            dp[i][j][k] = dp[i + 1][j][k + 1];
                            how[i][j][k] = k + 1;  // Representing choosing not to take the line
                        }
                    }

                    // Option 2: take the line
                    if (i < size && j > 0) {
                        float cost = arr[i] + (k - 8) * (k - 8);
                        if (cost + dp[i + 1][j - 1][0] < dp[i][j][k]) {
                            dp[i][j][k] = cost + dp[i + 1][j - 1][0];
                            how[i][j][k] = 0;  // Representing choosing to take the line
                        }
                    }
                }
            }
        }
    }

    return dp[0][lines_left][6];
}



#include<stdio.h>
#include<stdlib.h>
int main() {
float arr[] = {71, 89, 34, 63, 19, 94, 54, 61, 88, 20, 66, 46, 26, 87, 55, 81, 6, 2, 72, 75, 98, 78, 24, 95, 73, 7, 56, 48, 14, 99, 64, 51, 69, 77, 28, 47, 8, 22, 49, 3, 62, 32, 10, 82, 35, 18, 85, 60, 83, 23, 5, 40, 41, 68, 53, 52, 44, 45, 65, 84, 93, 25, 13, 1, 31, 11, 12, 97, 38, 0, 43, 90, 36, 70, 33, 17, 21, 30, 16, 15, 74, 67, 58, 37, 39, 96, 79, 29, 27, 92, 86, 9, 80, 42, 57, 91, 59, 4, 76, 50};

int answer = solve_dp(arr, 100, 100);
int expected = 11290;
if (answer != expected) {
    printf("Wrong answer; got %d instead of %d.\n", answer, expected);
    exit(1);
}
printf("All tests passed\n");
}
And got the output: main.c: In function 'solve_dp': main.c:13:31: error: 'FLT_MAX' undeclared (first use in this function) 13 | dp[i][j][k] = FLT_MAX; | ^~~~~~~ main.c:5:1: note: 'FLT_MAX' is defined in header '<float.h>'; did you forget to '#include <float.h>'? 4 | #include <limits.h> +++ |+#include <float.h> 5 | main.c:13:31: note: each undeclared identifier is reported only once for each function it appears in 13 | dp[i][j][k] = FLT_MAX; | ^~~~~~~ main.sh: line 2: ./a.out: No such file or directory

Substring Evaluation

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

Initial Query

Here's a python program. Please rewrite it in c with the type signature float solve_dp(float* arr, int size, int lines_left) def solve_dp(graph, lines_left): n = len(graph) dp = [[[1e9 for _ in range(10)] for _ in range(lines_left + 1)] for _ in range(n + 1)] # Initialize how as -1 indicating no decision made how = [[[-1 for _ in range(10)] for _ in range(lines_left + 1)] for _ in range(n + 1)] for i in range(n, -1, -1): for j in range(lines_left + 1): for k in range(10): if i == n and j == 0: dp[i][j][k] = 0 else: # Option 1: don't take the line if i < n and k < 9: if dp[i + 1][j][k + 1] < dp[i][j][k]: dp[i][j][k] = dp[i + 1][j][k + 1] how[i][j][k] = k + 1 # Representing choosing not to take the line # Option 2: take the line if i < n and j > 0: cost = graph[i] + (k - 8)**2 if cost + dp[i + 1][j - 1][0] < dp[i][j][k]: dp[i][j][k] = cost + dp[i + 1][j - 1][0] how[i][j][k] = 0 # Representing choosing to take the line # Reconstruct the solution i, j, k = 0, lines_left, 6 taken_lines = [] while i < n: if how[i][j][k] == 0: # Chose to take the line taken_lines.append(n - i) i += 1 j -= 1 k = 0 else: # Chose not to take the line i += 1 k += 1 return dp[0][lines_left][6]

LLM Generation

Query

Here's a python program. Please rewrite it in c with the type signature float solve_dp(float* arr, int size, int lines_left) def solve_dp(graph, lines_left): n = len(graph) dp = [[[1e9 for _ in range(10)] for _ in range(lines_left + 1)] for _ in range(n + 1)] # Initialize how as -1 indicating no decision made how = [[[-1 for _ in range(10)] for _ in range(lines_left + 1)] for _ in range(n + 1)] for i in range(n, -1, -1): for j in range(lines_left + 1): for k in range(10): if i == n and j == 0: dp[i][j][k] = 0 else: # Option 1: don't take the line if i < n and k < 9: if dp[i + 1][j][k + 1] < dp[i][j][k]: dp[i][j][k] = dp[i + 1][j][k + 1] how[i][j][k] = k + 1 # Representing choosing not to take the line # Option 2: take the line if i < n and j > 0: cost = graph[i] + (k - 8)**2 if cost + dp[i + 1][j - 1][0] < dp[i][j][k]: dp[i][j][k] = cost + dp[i + 1][j - 1][0] how[i][j][k] = 0 # Representing choosing to take the line # Reconstruct the solution i, j, k = 0, lines_left, 6 taken_lines = [] while i < n: if how[i][j][k] == 0: # Chose to take the line taken_lines.append(n - i) i += 1 j -= 1 k = 0 else: # Chose not to take the line i += 1 k += 1 return dp[0][lines_left][6]

Output

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#include <stdio.h>
#include <stdlib.h>
#include <math.h>

float solve_dp(float* arr, int size, int lines_left) {
    float dp[size + 1][lines_left + 1][10];
    int how[size + 1][lines_left + 1][10];

    for (int i = size; i >= 0; i--) {
        for (int j = lines_left; j >= 0; j--) {
            for (int k = 0; k < 10; k++) {
                if (i == size && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    // Option 1: don't take the line
                    if (i < size && k < 9) {
                        if (dp[i + 1][j][k + 1] < dp[i][j][k]) {
                            dp[i][j][k] = dp[i + 1][j][k + 1];
                            how[i][j][k] = k + 1; // Representing choosing not to take the line
                        }
                    }

                    // Option 2: take the line
                    if (i < size && j > 0) {
                        float cost = arr[i] + pow(k - 8, 2);
                        if (cost + dp[i + 1][j - 1][0] < dp[i][j][k]) {
                            dp[i][j][k] = cost + dp[i + 1][j - 1][0];
                            how[i][j][k] = 0; // Representing choosing to take the line
                        }
                    }
                }
            }
        }
    }

    return dp[0][lines_left][6];
}

Extract Code

I extracted the following code from that output: 

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#include <stdio.h>
#include <stdlib.h>
#include <math.h>

float solve_dp(float* arr, int size, int lines_left) {
    float dp[size + 1][lines_left + 1][10];
    int how[size + 1][lines_left + 1][10];

    for (int i = size; i >= 0; i--) {
        for (int j = lines_left; j >= 0; j--) {
            for (int k = 0; k < 10; k++) {
                if (i == size && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    // Option 1: don't take the line
                    if (i < size && k < 9) {
                        if (dp[i + 1][j][k + 1] < dp[i][j][k]) {
                            dp[i][j][k] = dp[i + 1][j][k + 1];
                            how[i][j][k] = k + 1; // Representing choosing not to take the line
                        }
                    }

                    // Option 2: take the line
                    if (i < size && j > 0) {
                        float cost = arr[i] + pow(k - 8, 2);
                        if (cost + dp[i + 1][j - 1][0] < dp[i][j][k]) {
                            dp[i][j][k] = cost + dp[i + 1][j - 1][0];
                            how[i][j][k] = 0; // Representing choosing to take the line
                        }
                    }
                }
            }
        }
    }

    return dp[0][lines_left][6];
}

Run Code Interpreter

Running the following program: 

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#include <stdio.h>
#include <stdlib.h>
#include <math.h>

float solve_dp(float* arr, int size, int lines_left) {
    float dp[size + 1][lines_left + 1][10];
    int how[size + 1][lines_left + 1][10];

    for (int i = size; i >= 0; i--) {
        for (int j = lines_left; j >= 0; j--) {
            for (int k = 0; k < 10; k++) {
                if (i == size && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    // Option 1: don't take the line
                    if (i < size && k < 9) {
                        if (dp[i + 1][j][k + 1] < dp[i][j][k]) {
                            dp[i][j][k] = dp[i + 1][j][k + 1];
                            how[i][j][k] = k + 1; // Representing choosing not to take the line
                        }
                    }

                    // Option 2: take the line
                    if (i < size && j > 0) {
                        float cost = arr[i] + pow(k - 8, 2);
                        if (cost + dp[i + 1][j - 1][0] < dp[i][j][k]) {
                            dp[i][j][k] = cost + dp[i + 1][j - 1][0];
                            how[i][j][k] = 0; // Representing choosing to take the line
                        }
                    }
                }
            }
        }
    }

    return dp[0][lines_left][6];
}



#include<stdio.h>
#include<stdlib.h>
int main() {
float arr[] = {71, 89, 34, 63, 19, 94, 54, 61, 88, 20, 66, 46, 26, 87, 55, 81, 6, 2, 72, 75, 98, 78, 24, 95, 73, 7, 56, 48, 14, 99, 64, 51, 69, 77, 28, 47, 8, 22, 49, 3, 62, 32, 10, 82, 35, 18, 85, 60, 83, 23, 5, 40, 41, 68, 53, 52, 44, 45, 65, 84, 93, 25, 13, 1, 31, 11, 12, 97, 38, 0, 43, 90, 36, 70, 33, 17, 21, 30, 16, 15, 74, 67, 58, 37, 39, 96, 79, 29, 27, 92, 86, 9, 80, 42, 57, 91, 59, 4, 76, 50};

int answer = solve_dp(arr, 100, 100);
int expected = 11290;
if (answer != expected) {
    printf("Wrong answer; got %d instead of %d.\n", answer, expected);
    exit(1);
}
printf("All tests passed\n");
}
And got the output: Wrong answer; got -3565 instead of 11290.

Substring Evaluation

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

Initial Query

Here's a python program. Please rewrite it in c with the type signature float solve_dp(float* arr, int size, int lines_left) def solve_dp(graph, lines_left): n = len(graph) dp = [[[1e9 for _ in range(10)] for _ in range(lines_left + 1)] for _ in range(n + 1)] # Initialize how as -1 indicating no decision made how = [[[-1 for _ in range(10)] for _ in range(lines_left + 1)] for _ in range(n + 1)] for i in range(n, -1, -1): for j in range(lines_left + 1): for k in range(10): if i == n and j == 0: dp[i][j][k] = 0 else: # Option 1: don't take the line if i < n and k < 9: if dp[i + 1][j][k + 1] < dp[i][j][k]: dp[i][j][k] = dp[i + 1][j][k + 1] how[i][j][k] = k + 1 # Representing choosing not to take the line # Option 2: take the line if i < n and j > 0: cost = graph[i] + (k - 8)**2 if cost + dp[i + 1][j - 1][0] < dp[i][j][k]: dp[i][j][k] = cost + dp[i + 1][j - 1][0] how[i][j][k] = 0 # Representing choosing to take the line # Reconstruct the solution i, j, k = 0, lines_left, 6 taken_lines = [] while i < n: if how[i][j][k] == 0: # Chose to take the line taken_lines.append(n - i) i += 1 j -= 1 k = 0 else: # Chose not to take the line i += 1 k += 1 return dp[0][lines_left][6]

LLM Generation

Query

Here's a python program. Please rewrite it in c with the type signature float solve_dp(float* arr, int size, int lines_left) def solve_dp(graph, lines_left): n = len(graph) dp = [[[1e9 for _ in range(10)] for _ in range(lines_left + 1)] for _ in range(n + 1)] # Initialize how as -1 indicating no decision made how = [[[-1 for _ in range(10)] for _ in range(lines_left + 1)] for _ in range(n + 1)] for i in range(n, -1, -1): for j in range(lines_left + 1): for k in range(10): if i == n and j == 0: dp[i][j][k] = 0 else: # Option 1: don't take the line if i < n and k < 9: if dp[i + 1][j][k + 1] < dp[i][j][k]: dp[i][j][k] = dp[i + 1][j][k + 1] how[i][j][k] = k + 1 # Representing choosing not to take the line # Option 2: take the line if i < n and j > 0: cost = graph[i] + (k - 8)**2 if cost + dp[i + 1][j - 1][0] < dp[i][j][k]: dp[i][j][k] = cost + dp[i + 1][j - 1][0] how[i][j][k] = 0 # Representing choosing to take the line # Reconstruct the solution i, j, k = 0, lines_left, 6 taken_lines = [] while i < n: if how[i][j][k] == 0: # Chose to take the line taken_lines.append(n - i) i += 1 j -= 1 k = 0 else: # Chose not to take the line i += 1 k += 1 return dp[0][lines_left][6]

Output

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#include <stdio.h>
#include <stdlib.h>
#include <math.h>

float solve_dp(float* arr, int size, int lines_left) {
    float dp[size + 1][lines_left + 1][10];
    int how[size + 1][lines_left + 1][10];

    for (int i = size; i >= 0; i--) {
        for (int j = lines_left; j >= 0; j--) {
            for (int k = 0; k < 10; k++) {
                if (i == size && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    // Option 1: don't take the line
                    if (i < size && k < 9) {
                        if (dp[i + 1][j][k + 1] < dp[i][j][k]) {
                            dp[i][j][k] = dp[i + 1][j][k + 1];
                            how[i][j][k] = k + 1;  // Representing choosing not to take the line
                        }
                    }

                    // Option 2: take the line
                    if (i < size && j > 0) {
                        float cost = arr[i] + pow(k - 8, 2);
                        if (cost + dp[i + 1][j - 1][0] < dp[i][j][k]) {
                            dp[i][j][k] = cost + dp[i + 1][j - 1][0];
                            how[i][j][k] = 0;  // Representing choosing to take the line
                        }
                    }
                }
            }
        }
    }

    return dp[0][lines_left][6];
}

Extract Code

I extracted the following code from that output: 

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#include <stdio.h>
#include <stdlib.h>
#include <math.h>

float solve_dp(float* arr, int size, int lines_left) {
    float dp[size + 1][lines_left + 1][10];
    int how[size + 1][lines_left + 1][10];

    for (int i = size; i >= 0; i--) {
        for (int j = lines_left; j >= 0; j--) {
            for (int k = 0; k < 10; k++) {
                if (i == size && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    // Option 1: don't take the line
                    if (i < size && k < 9) {
                        if (dp[i + 1][j][k + 1] < dp[i][j][k]) {
                            dp[i][j][k] = dp[i + 1][j][k + 1];
                            how[i][j][k] = k + 1;  // Representing choosing not to take the line
                        }
                    }

                    // Option 2: take the line
                    if (i < size && j > 0) {
                        float cost = arr[i] + pow(k - 8, 2);
                        if (cost + dp[i + 1][j - 1][0] < dp[i][j][k]) {
                            dp[i][j][k] = cost + dp[i + 1][j - 1][0];
                            how[i][j][k] = 0;  // Representing choosing to take the line
                        }
                    }
                }
            }
        }
    }

    return dp[0][lines_left][6];
}

Run Code Interpreter

Running the following program: 

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#include <stdio.h>
#include <stdlib.h>
#include <math.h>

float solve_dp(float* arr, int size, int lines_left) {
    float dp[size + 1][lines_left + 1][10];
    int how[size + 1][lines_left + 1][10];

    for (int i = size; i >= 0; i--) {
        for (int j = lines_left; j >= 0; j--) {
            for (int k = 0; k < 10; k++) {
                if (i == size && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    // Option 1: don't take the line
                    if (i < size && k < 9) {
                        if (dp[i + 1][j][k + 1] < dp[i][j][k]) {
                            dp[i][j][k] = dp[i + 1][j][k + 1];
                            how[i][j][k] = k + 1;  // Representing choosing not to take the line
                        }
                    }

                    // Option 2: take the line
                    if (i < size && j > 0) {
                        float cost = arr[i] + pow(k - 8, 2);
                        if (cost + dp[i + 1][j - 1][0] < dp[i][j][k]) {
                            dp[i][j][k] = cost + dp[i + 1][j - 1][0];
                            how[i][j][k] = 0;  // Representing choosing to take the line
                        }
                    }
                }
            }
        }
    }

    return dp[0][lines_left][6];
}



#include<stdio.h>
#include<stdlib.h>
int main() {
float arr[] = {71, 89, 34, 63, 19, 94, 54, 61, 88, 20, 66, 46, 26, 87, 55, 81, 6, 2, 72, 75, 98, 78, 24, 95, 73, 7, 56, 48, 14, 99, 64, 51, 69, 77, 28, 47, 8, 22, 49, 3, 62, 32, 10, 82, 35, 18, 85, 60, 83, 23, 5, 40, 41, 68, 53, 52, 44, 45, 65, 84, 93, 25, 13, 1, 31, 11, 12, 97, 38, 0, 43, 90, 36, 70, 33, 17, 21, 30, 16, 15, 74, 67, 58, 37, 39, 96, 79, 29, 27, 92, 86, 9, 80, 42, 57, 91, 59, 4, 76, 50};

int answer = solve_dp(arr, 100, 100);
int expected = 11290;
if (answer != expected) {
    printf("Wrong answer; got %d instead of %d.\n", answer, expected);
    exit(1);
}
printf("All tests passed\n");
}
And got the output: Wrong answer; got -2147483648 instead of 11290.

Substring Evaluation

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

Initial Query

Here's a python program. Please rewrite it in c with the type signature float solve_dp(float* arr, int size, int lines_left) def solve_dp(graph, lines_left): n = len(graph) dp = [[[1e9 for _ in range(10)] for _ in range(lines_left + 1)] for _ in range(n + 1)] # Initialize how as -1 indicating no decision made how = [[[-1 for _ in range(10)] for _ in range(lines_left + 1)] for _ in range(n + 1)] for i in range(n, -1, -1): for j in range(lines_left + 1): for k in range(10): if i == n and j == 0: dp[i][j][k] = 0 else: # Option 1: don't take the line if i < n and k < 9: if dp[i + 1][j][k + 1] < dp[i][j][k]: dp[i][j][k] = dp[i + 1][j][k + 1] how[i][j][k] = k + 1 # Representing choosing not to take the line # Option 2: take the line if i < n and j > 0: cost = graph[i] + (k - 8)**2 if cost + dp[i + 1][j - 1][0] < dp[i][j][k]: dp[i][j][k] = cost + dp[i + 1][j - 1][0] how[i][j][k] = 0 # Representing choosing to take the line # Reconstruct the solution i, j, k = 0, lines_left, 6 taken_lines = [] while i < n: if how[i][j][k] == 0: # Chose to take the line taken_lines.append(n - i) i += 1 j -= 1 k = 0 else: # Chose not to take the line i += 1 k += 1 return dp[0][lines_left][6]

LLM Generation

Query

Here's a python program. Please rewrite it in c with the type signature float solve_dp(float* arr, int size, int lines_left) def solve_dp(graph, lines_left): n = len(graph) dp = [[[1e9 for _ in range(10)] for _ in range(lines_left + 1)] for _ in range(n + 1)] # Initialize how as -1 indicating no decision made how = [[[-1 for _ in range(10)] for _ in range(lines_left + 1)] for _ in range(n + 1)] for i in range(n, -1, -1): for j in range(lines_left + 1): for k in range(10): if i == n and j == 0: dp[i][j][k] = 0 else: # Option 1: don't take the line if i < n and k < 9: if dp[i + 1][j][k + 1] < dp[i][j][k]: dp[i][j][k] = dp[i + 1][j][k + 1] how[i][j][k] = k + 1 # Representing choosing not to take the line # Option 2: take the line if i < n and j > 0: cost = graph[i] + (k - 8)**2 if cost + dp[i + 1][j - 1][0] < dp[i][j][k]: dp[i][j][k] = cost + dp[i + 1][j - 1][0] how[i][j][k] = 0 # Representing choosing to take the line # Reconstruct the solution i, j, k = 0, lines_left, 6 taken_lines = [] while i < n: if how[i][j][k] == 0: # Chose to take the line taken_lines.append(n - i) i += 1 j -= 1 k = 0 else: # Chose not to take the line i += 1 k += 1 return dp[0][lines_left][6]

Output

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#include <stdio.h>
#include <stdlib.h>
#include <math.h>

float solve_dp(float* arr, int size, int lines_left) {
    float*** dp = malloc((size + 1) * sizeof(float**));
    int*** how = malloc((size + 1) * sizeof(int**));

    for (int i = 0; i <= size; i++) {
        dp[i] = malloc((lines_left + 1) * sizeof(float*));
        how[i] = malloc((lines_left + 1) * sizeof(int*));

        for (int j = 0; j <= lines_left; j++) {
            dp[i][j] = malloc(10 * sizeof(float));
            how[i][j] = malloc(10 * sizeof(int));

            for (int k = 0; k < 10; k++) {
                dp[i][j][k] = 1e9;
                how[i][j][k] = -1;
            }
        }
    }

    dp[size][0][6] = 0;

    for (int i = size - 1; i >= 0; i--) {
        for (int j = 0; j <= lines_left; j++) {
            for (int k = 0; k < 10; k++) {
                // Option 1: don't take the line
                if (i < size && k < 9) {
                    if (dp[i + 1][j][k + 1] < dp[i][j][k]) {
                        dp[i][j][k] = dp[i + 1][j][k + 1];
                        how[i][j][k] = k + 1;
                    }
                }

                // Option 2: take the line
                if (i < size && j > 0) {
                    float cost = arr[i] + pow(k - 8, 2);
                    if (cost + dp[i + 1][j - 1][0] < dp[i][j][k]) {
                        dp[i][j][k] = cost + dp[i + 1][j - 1][0];
                        how[i][j][k] = 0;
                    }
                }
            }
        }
    }

    int i = 0, j = lines_left, k = 6;
    int* taken_lines = malloc(size * sizeof(int));
    int taken_lines_count = 0;

    while (i < size) {
        if (how[i][j][k] == 0) {
            taken_lines[taken_lines_count++] = size - i;
            i++;
            j--;
            k = 0;
        } else {
            i++;
            k++;
        }
    }

    float result = dp[0][lines_left][6];

    for (int i = 0; i < size + 1; i++) {
        for (int j = 0; j <= lines_left; j++) {
            free(dp[i][j]);
            free(how[i][j]);
        }

        free(dp[i]);
        free(how[i]);
    }

    free(dp);
    free(how);
    free(taken_lines);

    return result;
}

Extract Code

I extracted the following code from that output: 

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#include <stdio.h>
#include <stdlib.h>
#include <math.h>

float solve_dp(float* arr, int size, int lines_left) {
    float*** dp = malloc((size + 1) * sizeof(float**));
    int*** how = malloc((size + 1) * sizeof(int**));

    for (int i = 0; i <= size; i++) {
        dp[i] = malloc((lines_left + 1) * sizeof(float*));
        how[i] = malloc((lines_left + 1) * sizeof(int*));

        for (int j = 0; j <= lines_left; j++) {
            dp[i][j] = malloc(10 * sizeof(float));
            how[i][j] = malloc(10 * sizeof(int));

            for (int k = 0; k < 10; k++) {
                dp[i][j][k] = 1e9;
                how[i][j][k] = -1;
            }
        }
    }

    dp[size][0][6] = 0;

    for (int i = size - 1; i >= 0; i--) {
        for (int j = 0; j <= lines_left; j++) {
            for (int k = 0; k < 10; k++) {
                // Option 1: don't take the line
                if (i < size && k < 9) {
                    if (dp[i + 1][j][k + 1] < dp[i][j][k]) {
                        dp[i][j][k] = dp[i + 1][j][k + 1];
                        how[i][j][k] = k + 1;
                    }
                }

                // Option 2: take the line
                if (i < size && j > 0) {
                    float cost = arr[i] + pow(k - 8, 2);
                    if (cost + dp[i + 1][j - 1][0] < dp[i][j][k]) {
                        dp[i][j][k] = cost + dp[i + 1][j - 1][0];
                        how[i][j][k] = 0;
                    }
                }
            }
        }
    }

    int i = 0, j = lines_left, k = 6;
    int* taken_lines = malloc(size * sizeof(int));
    int taken_lines_count = 0;

    while (i < size) {
        if (how[i][j][k] == 0) {
            taken_lines[taken_lines_count++] = size - i;
            i++;
            j--;
            k = 0;
        } else {
            i++;
            k++;
        }
    }

    float result = dp[0][lines_left][6];

    for (int i = 0; i < size + 1; i++) {
        for (int j = 0; j <= lines_left; j++) {
            free(dp[i][j]);
            free(how[i][j]);
        }

        free(dp[i]);
        free(how[i]);
    }

    free(dp);
    free(how);
    free(taken_lines);

    return result;
}

Run Code Interpreter

Running the following program: 

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#include <stdio.h>
#include <stdlib.h>
#include <math.h>

float solve_dp(float* arr, int size, int lines_left) {
    float*** dp = malloc((size + 1) * sizeof(float**));
    int*** how = malloc((size + 1) * sizeof(int**));

    for (int i = 0; i <= size; i++) {
        dp[i] = malloc((lines_left + 1) * sizeof(float*));
        how[i] = malloc((lines_left + 1) * sizeof(int*));

        for (int j = 0; j <= lines_left; j++) {
            dp[i][j] = malloc(10 * sizeof(float));
            how[i][j] = malloc(10 * sizeof(int));

            for (int k = 0; k < 10; k++) {
                dp[i][j][k] = 1e9;
                how[i][j][k] = -1;
            }
        }
    }

    dp[size][0][6] = 0;

    for (int i = size - 1; i >= 0; i--) {
        for (int j = 0; j <= lines_left; j++) {
            for (int k = 0; k < 10; k++) {
                // Option 1: don't take the line
                if (i < size && k < 9) {
                    if (dp[i + 1][j][k + 1] < dp[i][j][k]) {
                        dp[i][j][k] = dp[i + 1][j][k + 1];
                        how[i][j][k] = k + 1;
                    }
                }

                // Option 2: take the line
                if (i < size && j > 0) {
                    float cost = arr[i] + pow(k - 8, 2);
                    if (cost + dp[i + 1][j - 1][0] < dp[i][j][k]) {
                        dp[i][j][k] = cost + dp[i + 1][j - 1][0];
                        how[i][j][k] = 0;
                    }
                }
            }
        }
    }

    int i = 0, j = lines_left, k = 6;
    int* taken_lines = malloc(size * sizeof(int));
    int taken_lines_count = 0;

    while (i < size) {
        if (how[i][j][k] == 0) {
            taken_lines[taken_lines_count++] = size - i;
            i++;
            j--;
            k = 0;
        } else {
            i++;
            k++;
        }
    }

    float result = dp[0][lines_left][6];

    for (int i = 0; i < size + 1; i++) {
        for (int j = 0; j <= lines_left; j++) {
            free(dp[i][j]);
            free(how[i][j]);
        }

        free(dp[i]);
        free(how[i]);
    }

    free(dp);
    free(how);
    free(taken_lines);

    return result;
}



#include<stdio.h>
#include<stdlib.h>
int main() {
float arr[] = {71, 89, 34, 63, 19, 94, 54, 61, 88, 20, 66, 46, 26, 87, 55, 81, 6, 2, 72, 75, 98, 78, 24, 95, 73, 7, 56, 48, 14, 99, 64, 51, 69, 77, 28, 47, 8, 22, 49, 3, 62, 32, 10, 82, 35, 18, 85, 60, 83, 23, 5, 40, 41, 68, 53, 52, 44, 45, 65, 84, 93, 25, 13, 1, 31, 11, 12, 97, 38, 0, 43, 90, 36, 70, 33, 17, 21, 30, 16, 15, 74, 67, 58, 37, 39, 96, 79, 29, 27, 92, 86, 9, 80, 42, 57, 91, 59, 4, 76, 50};

int answer = solve_dp(arr, 100, 100);
int expected = 11290;
if (answer != expected) {
    printf("Wrong answer; got %d instead of %d.\n", answer, expected);
    exit(1);
}
printf("All tests passed\n");
}
And got the output: Wrong answer; got 1000000000 instead of 11290.

Substring Evaluation

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