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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float solve_dp(float* arr, int size, int lines_left) {
    int n = size;

    float dp[n + 1][lines_left + 1][10];
    int how[n + 1][lines_left + 1][10];

    for (int i = n; i >= 0; i--) {
        for (int j = 0; j <= lines_left; j++) {
            for (int k = 0; k < 10; k++) {
                if (i == n && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    // Option 1: don't take the line
                    if (i < n && 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 < n && 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;
    float taken_lines[n];
    int idx = 0;
    while (i < n) {
        if (how[i][j][k] == 0) {
            taken_lines[idx++] = n - i;
            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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float solve_dp(float* arr, int size, int lines_left) {
    int n = size;

    float dp[n + 1][lines_left + 1][10];
    int how[n + 1][lines_left + 1][10];

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

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

Run Code Interpreter

Running the following program: 

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float solve_dp(float* arr, int size, int lines_left) {
    int n = size;

    float dp[n + 1][lines_left + 1][10];
    int how[n + 1][lines_left + 1][10];

    for (int i = n; i >= 0; i--) {
        for (int j = 0; j <= lines_left; j++) {
            for (int k = 0; k < 10; k++) {
                if (i == n && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    // Option 1: don't take the line
                    if (i < n && 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 < n && 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;
    float taken_lines[n];
    int idx = 0;
    while (i < n) {
        if (how[i][j][k] == 0) {
            taken_lines[idx++] = n - i;
            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. main.c: In function 'solve_dp': main.c:24:47: warning: implicit declaration of function 'pow' [-Wimplicit-function-declaration] 24 | float cost = arr[i] + pow((k - 8), 2); | ^~~ main.c:1:1: note: include '<math.h>' or provide a declaration of 'pow' +++ |+#include <math.h> 1 | main.c:24:47: warning: incompatible implicit declaration of built-in function 'pow' [-Wbuiltin-declaration-mismatch] 24 | float cost = arr[i] + pow((k - 8), 2); | ^~~ main.c:24:47: note: include '<math.h>' or provide a declaration of 'pow'

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>

float solve_dp(float* arr, int size, int lines_left) {
    int n = size;

    float dp[n + 1][lines_left + 1][10];
    int how[n + 1][lines_left + 1][10];

    for (int i = n; i >= 0; i--) {
        for (int j = 0; j <= lines_left; j++) {
            for (int k = 0; k < 10; k++) {
                if (i == n && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    if (i < n && 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;
                        }
                    }
                    if (i < n && 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;
                        }
                    }
                }
            }
        }
    }

    int i = 0, j = lines_left, k = 6;
    float taken_lines[n];
    int idx = 0;
    while (i < n) {
        if (how[i][j][k] == 0) {
            taken_lines[idx++] = n - i;
            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>

float solve_dp(float* arr, int size, int lines_left) {
    int n = size;

    float dp[n + 1][lines_left + 1][10];
    int how[n + 1][lines_left + 1][10];

    for (int i = n; i >= 0; i--) {
        for (int j = 0; j <= lines_left; j++) {
            for (int k = 0; k < 10; k++) {
                if (i == n && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    if (i < n && 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;
                        }
                    }
                    if (i < n && 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;
                        }
                    }
                }
            }
        }
    }

    int i = 0, j = lines_left, k = 6;
    float taken_lines[n];
    int idx = 0;
    while (i < n) {
        if (how[i][j][k] == 0) {
            taken_lines[idx++] = n - i;
            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>

float solve_dp(float* arr, int size, int lines_left) {
    int n = size;

    float dp[n + 1][lines_left + 1][10];
    int how[n + 1][lines_left + 1][10];

    for (int i = n; i >= 0; i--) {
        for (int j = 0; j <= lines_left; j++) {
            for (int k = 0; k < 10; k++) {
                if (i == n && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    if (i < n && 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;
                        }
                    }
                    if (i < n && 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;
                        }
                    }
                }
            }
        }
    }

    int i = 0, j = lines_left, k = 6;
    float taken_lines[n];
    int idx = 0;
    while (i < n) {
        if (how[i][j][k] == 0) {
            taken_lines[idx++] = n - i;
            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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float solve_dp(float* graph, int size, int lines_left) {
    int n = size;

    float dp[n + 1][lines_left + 1][10];
    int how[n + 1][lines_left + 1][10];

    for(int i = n; i >= 0; i--) {
        for(int j = 0; j <= lines_left; j++) {
            for(int k = 0; k < 10; k++) {
                if (i == n && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    // Option 1: don't take the line
                    if (i < n && 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 < n && j > 0) {
                        float cost = graph[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;
    float taken_lines[n];
    int index = 0;
    while (i < n) {
        if (how[i][j][k] == 0) {
            taken_lines[index++] = n - i;
            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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float solve_dp(float* graph, int size, int lines_left) {
    int n = size;

    float dp[n + 1][lines_left + 1][10];
    int how[n + 1][lines_left + 1][10];

    for(int i = n; i >= 0; i--) {
        for(int j = 0; j <= lines_left; j++) {
            for(int k = 0; k < 10; k++) {
                if (i == n && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    // Option 1: don't take the line
                    if (i < n && 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 < n && j > 0) {
                        float cost = graph[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;
    float taken_lines[n];
    int index = 0;
    while (i < n) {
        if (how[i][j][k] == 0) {
            taken_lines[index++] = n - i;
            i++;
            j--;
            k = 0;
        } else {
            i++;
            k++;
        }
    }

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

Run Code Interpreter

Running the following program: 

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float solve_dp(float* graph, int size, int lines_left) {
    int n = size;

    float dp[n + 1][lines_left + 1][10];
    int how[n + 1][lines_left + 1][10];

    for(int i = n; i >= 0; i--) {
        for(int j = 0; j <= lines_left; j++) {
            for(int k = 0; k < 10; k++) {
                if (i == n && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    // Option 1: don't take the line
                    if (i < n && 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 < n && j > 0) {
                        float cost = graph[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;
    float taken_lines[n];
    int index = 0;
    while (i < n) {
        if (how[i][j][k] == 0) {
            taken_lines[index++] = n - i;
            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. main.c: In function 'solve_dp': main.c:24:49: warning: implicit declaration of function 'pow' [-Wimplicit-function-declaration] 24 | float cost = graph[i] + pow((k - 8), 2); | ^~~ main.c:1:1: note: include '<math.h>' or provide a declaration of 'pow' +++ |+#include <math.h> 1 | main.c:24:49: warning: incompatible implicit declaration of built-in function 'pow' [-Wbuiltin-declaration-mismatch] 24 | float cost = graph[i] + pow((k - 8), 2); | ^~~ main.c:24:49: note: include '<math.h>' or provide a declaration of 'pow'

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>

float solve_dp(float* arr, int size, int lines_left) {
    int n = size;

    float dp[n+1][lines_left+1][10];
    int how[n+1][lines_left+1][10];

    for(int i = n; i >= 0; i--) {
        for(int j = 0; j <= lines_left; j++) {
            for(int k = 0; k < 10; k++) {
                if(i == n && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    if(i < n && 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;
                        }
                    }
                    if(i < n && 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;
                        }
                    }
                }
            }
        }
    }

    int i = 0, j = lines_left, k = 6;
    while(i < n) {
        if(how[i][j][k] == 0) {
            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>

float solve_dp(float* arr, int size, int lines_left) {
    int n = size;

    float dp[n+1][lines_left+1][10];
    int how[n+1][lines_left+1][10];

    for(int i = n; i >= 0; i--) {
        for(int j = 0; j <= lines_left; j++) {
            for(int k = 0; k < 10; k++) {
                if(i == n && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    if(i < n && 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;
                        }
                    }
                    if(i < n && 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;
                        }
                    }
                }
            }
        }
    }

    int i = 0, j = lines_left, k = 6;
    while(i < n) {
        if(how[i][j][k] == 0) {
            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>

float solve_dp(float* arr, int size, int lines_left) {
    int n = size;

    float dp[n+1][lines_left+1][10];
    int how[n+1][lines_left+1][10];

    for(int i = n; i >= 0; i--) {
        for(int j = 0; j <= lines_left; j++) {
            for(int k = 0; k < 10; k++) {
                if(i == n && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    if(i < n && 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;
                        }
                    }
                    if(i < n && 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;
                        }
                    }
                }
            }
        }
    }

    int i = 0, j = lines_left, k = 6;
    while(i < n) {
        if(how[i][j][k] == 0) {
            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 0 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>

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

    for (int i = n; i >= 0; i--) {
        for (int j = 0; j <= lines_left; j++) {
            for (int k = 0; k < 10; k++) {
                if (i == n && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    if (i < n && 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;
                        }
                    }

                    if (i < n && j > 0) {
                        float cost = graph[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;
    float result = dp[0][lines_left][6];
    return result;
}

int main() {
    float graph[] = {1.0, 2.0, 3.0, 4.0, 5.0}; // example graph data
    int n = 5; // size of the graph
    int lines_left = 3; // example number of lines left

    float result = solve_dp(graph, n, lines_left);
    printf("Result: %f\n", result);

    return 0;
}

1

Extract Code

I extracted the following code from that output: 

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

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

    for (int i = n; i >= 0; i--) {
        for (int j = 0; j <= lines_left; j++) {
            for (int k = 0; k < 10; k++) {
                if (i == n && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    if (i < n && 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;
                        }
                    }

                    if (i < n && j > 0) {
                        float cost = graph[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;
    float result = dp[0][lines_left][6];
    return result;
}

Run Code Interpreter

Running the following program: 

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

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

    for (int i = n; i >= 0; i--) {
        for (int j = 0; j <= lines_left; j++) {
            for (int k = 0; k < 10; k++) {
                if (i == n && j == 0) {
                    dp[i][j][k] = 0;
                } else {
                    if (i < n && 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;
                        }
                    }

                    if (i < n && j > 0) {
                        float cost = graph[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;
    float result = dp[0][lines_left][6];
    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 -2147483648 instead of 11290. main.c: In function 'solve_dp': main.c:22:49: warning: implicit declaration of function 'pow' [-Wimplicit-function-declaration] 22 | float cost = graph[i] + pow((k - 8), 2); | ^~~ main.c:3:1: note: include '<math.h>' or provide a declaration of 'pow' 2 | #include <stdlib.h> +++ |+#include <math.h> 3 | main.c:22:49: warning: incompatible implicit declaration of built-in function 'pow' [-Wbuiltin-declaration-mismatch] 22 | float cost = graph[i] + pow((k - 8), 2); | ^~~ main.c:22:49: note: include '<math.h>' or provide a declaration of 'pow'

Substring Evaluation

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