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

Okay, here's the C implementation of the provided Python dynamic programming code. **Key Changes and Considerations:** 1. **Memory Allocation:** Python lists grow dynamically. In C, we need to allocate memory explicitly, especially for the `dp` and `how` tables whose sizes depend on runtime inputs (`size`, `lines_left`). We'll use `malloc` for heap allocation to avoid potential stack overflow with large inputs. 2. **3D Array Simulation:** C doesn't have built-in dynamic multi-dimensional arrays like Python lists of lists. We'll allocate a single flat block of memory for `dp` and `how` and calculate the 3D index manually or using helper macros. This is generally more efficient than allocating arrays of pointers. 3. **Indexing:** Python uses 0-based indexing. C also uses 0-based indexing, so the logic translates directly. `graph[i]` becomes `arr[i]`. 4. **Types:** Python's numbers are flexible. We explicitly use `float` for costs (`dp` table and `arr`) and `int` for indices, counts, and the `how` table. 5. **Constants:** `1e9` in Python becomes `1e9f` in C to denote a float literal. 6. **Math:** `(k - 8)**2` becomes `(k - 8.0f) * (k - 8.0f)` or `powf(k - 8.0f, 2.0f)` (requires `<math.h>`). Direct multiplication is often preferred for simple squares. 7. **Return Value:** The function signature requires returning the final float value `dp[0][lines_left][6]`. The reconstruction part from the Python code is not needed for the return value but is included in the calculation logic via the `how` array as it was part of the original program's computation. 8. **Error Handling:** Basic checks for `malloc` failure are included. 9. **Includes:** Necessary headers like `<stdlib.h>` (for `malloc`/`free`), `<stdio.h>` (for potential error messages), and `<float.h>` (useful for `FLT_MAX` if needed, though `1e9f` is used here) are included. 

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#include <stdio.h>
#include <stdlib.h>
#include <math.h> // Can be used for powf, though direct multiplication is used here
#include <float.h> // Can be used for FLT_MAX

// Define a large float value to represent infinity, similar to 1e9 in Python
#define INF_COST 1e9f

// Helper macro for calculating flat array index for 3D simulation
// Assumes dimensions are dim1, dim2, dim3
#define INDEX(i, j, k, dim2, dim3) ((i) * (dim2) * (dim3) + (j) * (dim3) + (k))

/**
 * @brief Solves the dynamic programming problem described in the Python code.
 *
 * @param arr Pointer to the array of floats representing the graph costs.
 * @param size The number of elements in the arr (n in Python).
 * @param lines_left The initial number of lines allowed to be taken.
 * @return The minimum cost calculated by the DP, or -1.0f on memory allocation error.
 */
float solve_dp(float* arr, int size, int lines_left) {
    int n = size; // Use 'n' internally for closer resemblance to Python version

    // Dimensions for the DP and how tables
    int dim1 = n + 1;
    int dim2 = lines_left + 1;
    int dim3 = 10; // The k dimension size

    // Allocate memory for dp and how tables on the heap
    // Using a single flat array for efficiency and easier memory management
    float* dp = (float*)malloc((size_t)dim1 * dim2 * dim3 * sizeof(float));
    int* how = (int*)malloc((size_t)dim1 * dim2 * dim3 * sizeof(int));

    // Check if allocation was successful
    if (dp == NULL || how == NULL) {
        fprintf(stderr, "Error: Memory allocation failed for DP tables.\n");
        free(dp); // free(NULL) is safe
        free(how);
        return -1.0f; // Indicate error
    }

    // Initialize dp table with infinity and how table with -1
    for (int i = 0; i < dim1; ++i) {
        for (int j = 0; j < dim2; ++j) {
            for (int k = 0; k < dim3; ++k) {
                dp[INDEX(i, j, k, dim2, dim3)] = INF_COST;
                how[INDEX(i, j, k, dim2, dim3)] = -1;
            }
        }
    }

    // Base case: dp[n][0][k] = 0 for all k
    // If we are past the last element (i == n) and have used exactly the allowed lines (j == 0),
    // the cost is 0.
    int base_i = n;
    int base_j = 0;
    if (base_j < dim2) { // Ensure j=0 is within bounds (it always should be if lines_left >= 0)
       for (int k = 0; k < dim3; ++k) {
           dp[INDEX(base_i, base_j, k, dim2, dim3)] = 0.0f;
       }
    }


    // Fill the DP table using bottom-up approach (iterating backwards on i)
    for (int i = n; i >= 0; --i) {
        for (int j = 0; j <= lines_left; ++j) {
            for (int k = 0; k < 10; ++k) {

                // Skip the base case we already initialized
                if (i == n && j == 0) {
                    continue;
                }

                // --- Transitions ---
                float current_min_cost = dp[INDEX(i, j, k, dim2, dim3)]; // Read current value
                int current_choice = how[INDEX(i, j, k, dim2, dim3)];

                // Option 1: Don't take the line arr[i]
                // Condition: i < n (still have lines to consider) and k < 9 (consecutive skip limit)
                if (i < n && k < 9) {
                    float cost_option1 = dp[INDEX(i + 1, j, k + 1, dim2, dim3)];
                    if (cost_option1 < current_min_cost) {
                        current_min_cost = cost_option1;
                        current_choice = k + 1; // Store next k state as choice marker
                    }
                }

                // Option 2: Take the line arr[i]
                // Condition: i < n (still have lines to consider) and j > 0 (have budget left)
                if (i < n && j > 0) {
                    // Calculate cost: graph[i] + (k - 8)^2
                    float k_penalty = (float)(k - 8); // Cast k for float calculation
                    float cost_of_taking = arr[i] + k_penalty * k_penalty;
                    float cost_option2 = cost_of_taking + dp[INDEX(i + 1, j - 1, 0, dim2, dim3)];

                    if (cost_option2 < current_min_cost) {
                        current_min_cost = cost_option2;
                        current_choice = 0; // Store 0 as choice marker for taking the line
                    }
                }

                // Update the DP table with the minimum cost found and the choice made
                dp[INDEX(i, j, k, dim2, dim3)] = current_min_cost;
                how[INDEX(i, j, k, dim2, dim3)] = current_choice;
            }
        }
    }

    // The final result is stored at dp[0][lines_left][6]
    float final_result = dp[INDEX(0, lines_left, 6, dim2, dim3)];

    // Free the allocated memory before returning
    free(dp);
    free(how);

    // Check if the result is still the initial infinity value, meaning no solution found
    // (This might happen if lines_left is too small or other constraints prevent reaching the base case)
    // Note: Comparing floats for exact equality can be tricky, but comparing against the initial INF_COST is usually okay here.
    if (final_result >= INF_COST) {
         // Depending on problem constraints, maybe return a specific value indicating impossibility
         // For now, just return the large value itself, or potentially 0 or -1 if specified.
         // Let's return the calculated value as is, even if it's INF_COST.
    }


    return final_result;
}

// Example Usage (optional, for testing)
int main() {
    // Example matching the structure, values are arbitrary
    float graph_data[] = {10.0f, 2.0f, 5.0f, 8.0f, 3.0f, 7.0f};
    int n = sizeof(graph_data) / sizeof(graph_data[0]);
    int k = 3; // Example: max lines to take

    printf("Graph data (n=%d):\n", n);
    for(int i=0; i<n; ++i) {
        printf("%.2f ", graph_data[i]);
    }
    printf("\nLines left to take: %d\n", k);

    float min_cost = solve_dp(graph_data, n, k);

    if (min_cost == -1.0f) {
         printf("Function failed due to memory allocation error.\n");
    } else if (min_cost >= INF_COST) {
         printf("Minimum cost: No valid solution found (Cost is effectively infinity).\n");
    }
    else {
        printf("Minimum cost: %.4f\n", min_cost);
    }

    return 0;
}
**Explanation of the C Code:** 1. **Includes:** Basic I/O, memory allocation, and potentially math functions. 2. **`INF_COST`:** A macro for the large float value used for initialization. 3. **`INDEX` Macro:** A helper to convert 3D indices `(i, j, k)` into a single index for the flat `dp` and `how` arrays. This makes the code cleaner than writing the formula repeatedly. 4. **Function Signature:** Matches the requested `float solve_dp(float* arr, int size, int lines_left)`. 5. **Memory Allocation:** `malloc` is used to request space for `dp` (float) and `how` (int) arrays. `size_t` cast is good practice for multiplication results passed to `malloc`. Error checking is performed. 6. **Initialization:** Nested loops set initial `dp` values to `INF_COST` and `how` values to `-1`. 7. **Base Case:** `dp[n][0][k]` is set to `0.0f`. 8. **DP Calculation:** The nested loops iterate through the states `(i, j, k)` exactly like the Python version (with `i` going backward). 9. **Transitions:** The logic for "Option 1: Don't take" and "Option 2: Take" directly mirrors the Python code, using the `INDEX` macro for array access and C arithmetic/comparison operators. 10. **Result:** The value at `dp[INDEX(0, lines_left, 6, dim2, dim3)]` is retrieved. 11. **Memory Deallocation:** `free(dp)` and `free(how)` are called to release the allocated memory, preventing memory leaks. 12. **Return:** The calculated minimum cost is returned. 13. **`main` Function (Example):** Demonstrates how to call `solve_dp` with sample data and print the result.

Extract Code

I extracted the following code from that output: 

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#include <stdio.h>
#include <stdlib.h>
#include <math.h> // Can be used for powf, though direct multiplication is used here
#include <float.h> // Can be used for FLT_MAX

// Define a large float value to represent infinity, similar to 1e9 in Python
#define INF_COST 1e9f

// Helper macro for calculating flat array index for 3D simulation
// Assumes dimensions are dim1, dim2, dim3
#define INDEX(i, j, k, dim2, dim3) ((i) * (dim2) * (dim3) + (j) * (dim3) + (k))

/**
 * @brief Solves the dynamic programming problem described in the Python code.
 *
 * @param arr Pointer to the array of floats representing the graph costs.
 * @param size The number of elements in the arr (n in Python).
 * @param lines_left The initial number of lines allowed to be taken.
 * @return The minimum cost calculated by the DP, or -1.0f on memory allocation error.
 */
float solve_dp(float* arr, int size, int lines_left) {
    int n = size; // Use 'n' internally for closer resemblance to Python version

    // Dimensions for the DP and how tables
    int dim1 = n + 1;
    int dim2 = lines_left + 1;
    int dim3 = 10; // The k dimension size

    // Allocate memory for dp and how tables on the heap
    // Using a single flat array for efficiency and easier memory management
    float* dp = (float*)malloc((size_t)dim1 * dim2 * dim3 * sizeof(float));
    int* how = (int*)malloc((size_t)dim1 * dim2 * dim3 * sizeof(int));

    // Check if allocation was successful
    if (dp == NULL || how == NULL) {
        fprintf(stderr, "Error: Memory allocation failed for DP tables.\n");
        free(dp); // free(NULL) is safe
        free(how);
        return -1.0f; // Indicate error
    }

    // Initialize dp table with infinity and how table with -1
    for (int i = 0; i < dim1; ++i) {
        for (int j = 0; j < dim2; ++j) {
            for (int k = 0; k < dim3; ++k) {
                dp[INDEX(i, j, k, dim2, dim3)] = INF_COST;
                how[INDEX(i, j, k, dim2, dim3)] = -1;
            }
        }
    }

    // Base case: dp[n][0][k] = 0 for all k
    // If we are past the last element (i == n) and have used exactly the allowed lines (j == 0),
    // the cost is 0.
    int base_i = n;
    int base_j = 0;
    if (base_j < dim2) { // Ensure j=0 is within bounds (it always should be if lines_left >= 0)
       for (int k = 0; k < dim3; ++k) {
           dp[INDEX(base_i, base_j, k, dim2, dim3)] = 0.0f;
       }
    }


    // Fill the DP table using bottom-up approach (iterating backwards on i)
    for (int i = n; i >= 0; --i) {
        for (int j = 0; j <= lines_left; ++j) {
            for (int k = 0; k < 10; ++k) {

                // Skip the base case we already initialized
                if (i == n && j == 0) {
                    continue;
                }

                // --- Transitions ---
                float current_min_cost = dp[INDEX(i, j, k, dim2, dim3)]; // Read current value
                int current_choice = how[INDEX(i, j, k, dim2, dim3)];

                // Option 1: Don't take the line arr[i]
                // Condition: i < n (still have lines to consider) and k < 9 (consecutive skip limit)
                if (i < n && k < 9) {
                    float cost_option1 = dp[INDEX(i + 1, j, k + 1, dim2, dim3)];
                    if (cost_option1 < current_min_cost) {
                        current_min_cost = cost_option1;
                        current_choice = k + 1; // Store next k state as choice marker
                    }
                }

                // Option 2: Take the line arr[i]
                // Condition: i < n (still have lines to consider) and j > 0 (have budget left)
                if (i < n && j > 0) {
                    // Calculate cost: graph[i] + (k - 8)^2
                    float k_penalty = (float)(k - 8); // Cast k for float calculation
                    float cost_of_taking = arr[i] + k_penalty * k_penalty;
                    float cost_option2 = cost_of_taking + dp[INDEX(i + 1, j - 1, 0, dim2, dim3)];

                    if (cost_option2 < current_min_cost) {
                        current_min_cost = cost_option2;
                        current_choice = 0; // Store 0 as choice marker for taking the line
                    }
                }

                // Update the DP table with the minimum cost found and the choice made
                dp[INDEX(i, j, k, dim2, dim3)] = current_min_cost;
                how[INDEX(i, j, k, dim2, dim3)] = current_choice;
            }
        }
    }

    // The final result is stored at dp[0][lines_left][6]
    float final_result = dp[INDEX(0, lines_left, 6, dim2, dim3)];

    // Free the allocated memory before returning
    free(dp);
    free(how);

    // Check if the result is still the initial infinity value, meaning no solution found
    // (This might happen if lines_left is too small or other constraints prevent reaching the base case)
    // Note: Comparing floats for exact equality can be tricky, but comparing against the initial INF_COST is usually okay here.
    if (final_result >= INF_COST) {
         // Depending on problem constraints, maybe return a specific value indicating impossibility
         // For now, just return the large value itself, or potentially 0 or -1 if specified.
         // Let's return the calculated value as is, even if it's INF_COST.
    }


    return final_result;
}

// Example Usage (optional, for testing)
int main() {
    // Example matching the structure, values are arbitrary
    float graph_data[] = {10.0f, 2.0f, 5.0f, 8.0f, 3.0f, 7.0f};
    int n = sizeof(graph_data) / sizeof(graph_data[0]);
    int k = 3; // Example: max lines to take

    printf("Graph data (n=%d):\n", n);
    for(int i=0; i<n; ++i) {
        printf("%.2f ", graph_data[i]);
    }
    printf("\nLines left to take: %d\n", k);

    float min_cost = solve_dp(graph_data, n, k);

    if (min_cost == -1.0f) {
         printf("Function failed due to memory allocation error.\n");
    } else if (min_cost >= INF_COST) {
         printf("Minimum cost: No valid solution found (Cost is effectively infinity).\n");
    }
    else {
        printf("Minimum cost: %.4f\n", min_cost);
    }

    return 0;
}

Run Code Interpreter

Running the following program: 

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#include <stdio.h>
#include <stdlib.h>
#include <math.h> // Can be used for powf, though direct multiplication is used here
#include <float.h> // Can be used for FLT_MAX

// Define a large float value to represent infinity, similar to 1e9 in Python
#define INF_COST 1e9f

// Helper macro for calculating flat array index for 3D simulation
// Assumes dimensions are dim1, dim2, dim3
#define INDEX(i, j, k, dim2, dim3) ((i) * (dim2) * (dim3) + (j) * (dim3) + (k))

/**
 * @brief Solves the dynamic programming problem described in the Python code.
 *
 * @param arr Pointer to the array of floats representing the graph costs.
 * @param size The number of elements in the arr (n in Python).
 * @param lines_left The initial number of lines allowed to be taken.
 * @return The minimum cost calculated by the DP, or -1.0f on memory allocation error.
 */
float solve_dp(float* arr, int size, int lines_left) {
    int n = size; // Use 'n' internally for closer resemblance to Python version

    // Dimensions for the DP and how tables
    int dim1 = n + 1;
    int dim2 = lines_left + 1;
    int dim3 = 10; // The k dimension size

    // Allocate memory for dp and how tables on the heap
    // Using a single flat array for efficiency and easier memory management
    float* dp = (float*)malloc((size_t)dim1 * dim2 * dim3 * sizeof(float));
    int* how = (int*)malloc((size_t)dim1 * dim2 * dim3 * sizeof(int));

    // Check if allocation was successful
    if (dp == NULL || how == NULL) {
        fprintf(stderr, "Error: Memory allocation failed for DP tables.\n");
        free(dp); // free(NULL) is safe
        free(how);
        return -1.0f; // Indicate error
    }

    // Initialize dp table with infinity and how table with -1
    for (int i = 0; i < dim1; ++i) {
        for (int j = 0; j < dim2; ++j) {
            for (int k = 0; k < dim3; ++k) {
                dp[INDEX(i, j, k, dim2, dim3)] = INF_COST;
                how[INDEX(i, j, k, dim2, dim3)] = -1;
            }
        }
    }

    // Base case: dp[n][0][k] = 0 for all k
    // If we are past the last element (i == n) and have used exactly the allowed lines (j == 0),
    // the cost is 0.
    int base_i = n;
    int base_j = 0;
    if (base_j < dim2) { // Ensure j=0 is within bounds (it always should be if lines_left >= 0)
       for (int k = 0; k < dim3; ++k) {
           dp[INDEX(base_i, base_j, k, dim2, dim3)] = 0.0f;
       }
    }


    // Fill the DP table using bottom-up approach (iterating backwards on i)
    for (int i = n; i >= 0; --i) {
        for (int j = 0; j <= lines_left; ++j) {
            for (int k = 0; k < 10; ++k) {

                // Skip the base case we already initialized
                if (i == n && j == 0) {
                    continue;
                }

                // --- Transitions ---
                float current_min_cost = dp[INDEX(i, j, k, dim2, dim3)]; // Read current value
                int current_choice = how[INDEX(i, j, k, dim2, dim3)];

                // Option 1: Don't take the line arr[i]
                // Condition: i < n (still have lines to consider) and k < 9 (consecutive skip limit)
                if (i < n && k < 9) {
                    float cost_option1 = dp[INDEX(i + 1, j, k + 1, dim2, dim3)];
                    if (cost_option1 < current_min_cost) {
                        current_min_cost = cost_option1;
                        current_choice = k + 1; // Store next k state as choice marker
                    }
                }

                // Option 2: Take the line arr[i]
                // Condition: i < n (still have lines to consider) and j > 0 (have budget left)
                if (i < n && j > 0) {
                    // Calculate cost: graph[i] + (k - 8)^2
                    float k_penalty = (float)(k - 8); // Cast k for float calculation
                    float cost_of_taking = arr[i] + k_penalty * k_penalty;
                    float cost_option2 = cost_of_taking + dp[INDEX(i + 1, j - 1, 0, dim2, dim3)];

                    if (cost_option2 < current_min_cost) {
                        current_min_cost = cost_option2;
                        current_choice = 0; // Store 0 as choice marker for taking the line
                    }
                }

                // Update the DP table with the minimum cost found and the choice made
                dp[INDEX(i, j, k, dim2, dim3)] = current_min_cost;
                how[INDEX(i, j, k, dim2, dim3)] = current_choice;
            }
        }
    }

    // The final result is stored at dp[0][lines_left][6]
    float final_result = dp[INDEX(0, lines_left, 6, dim2, dim3)];

    // Free the allocated memory before returning
    free(dp);
    free(how);

    // Check if the result is still the initial infinity value, meaning no solution found
    // (This might happen if lines_left is too small or other constraints prevent reaching the base case)
    // Note: Comparing floats for exact equality can be tricky, but comparing against the initial INF_COST is usually okay here.
    if (final_result >= INF_COST) {
         // Depending on problem constraints, maybe return a specific value indicating impossibility
         // For now, just return the large value itself, or potentially 0 or -1 if specified.
         // Let's return the calculated value as is, even if it's INF_COST.
    }


    return final_result;
}

// Example Usage (optional, for testing)
int __delete_this__main() {
    // Example matching the structure, values are arbitrary
    float graph_data[] = {10.0f, 2.0f, 5.0f, 8.0f, 3.0f, 7.0f};
    int n = sizeof(graph_data) / sizeof(graph_data[0]);
    int k = 3; // Example: max lines to take

    printf("Graph data (n=%d):\n", n);
    for(int i=0; i<n; ++i) {
        printf("%.2f ", graph_data[i]);
    }
    printf("\nLines left to take: %d\n", k);

    float min_cost = solve_dp(graph_data, n, k);

    if (min_cost == -1.0f) {
         printf("Function failed due to memory allocation error.\n");
    } else if (min_cost >= INF_COST) {
         printf("Minimum cost: No valid solution found (Cost is effectively infinity).\n");
    }
    else {
        printf("Minimum cost: %.4f\n", min_cost);
    }

    return 0;
}



#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: All tests passed

Substring Evaluation

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