TheAlgorithms-C/dynamic_programming/matrix_chain_order.c
David Leal db3d6e2886
feat: add Windows CI back (#1290)
Signed-off-by: realstealthninja <realstealthninja@gmail.com>
Co-authored-by: github-actions[bot] <github-actions@users.noreply.github.com>
Co-authored-by: realstealthninja <68815218+realstealthninja@users.noreply.github.com>
2023-09-08 15:38:14 -06:00

120 lines
2.9 KiB
C

/**
* @file
* @brief [Matrix Chain
* Order](https://en.wikipedia.org/wiki/Matrix_chain_multiplication)
* @details
* From Wikipedia: Matrix chain multiplication (or the matrix chain ordering
* problem) is an optimization problem concerning the most efficient way to
* multiply a given sequence of matrices. The problem is not actually to perform
* the multiplications, but merely to decide the sequence of the matrix
* multiplications involved.
* @author [CascadingCascade](https://github.com/CascadingCascade)
*/
#include <assert.h> /// for assert
#include <limits.h> /// for INT_MAX macro
#include <stdio.h> /// for IO operations
#include <stdlib.h> /// for malloc() and free()
/**
* @brief Finds the optimal sequence using the classic O(n^3) algorithm.
* @param l length of cost array
* @param p costs of each matrix
* @param s location to store results
* @returns number of operations
*/
int matrixChainOrder(int l, const int *p, int *s)
{
// mat stores the cost for a chain that starts at i and ends on j (inclusive
// on both ends)
int **mat = malloc(l * sizeof(int *));
for (int i = 0; i < l; ++i)
{
mat[i] = malloc(l * sizeof(int));
}
for (int i = 0; i < l; ++i)
{
mat[i][i] = 0;
}
// cl denotes the difference between start / end indices, cl + 1 would be
// chain length.
for (int cl = 1; cl < l; ++cl)
{
for (int i = 0; i < l - cl; ++i)
{
int j = i + cl;
mat[i][j] = INT_MAX;
for (int div = i; div < j; ++div)
{
int q = mat[i][div] + mat[div + 1][j] + p[i] * p[div] * p[j];
if (q < mat[i][j])
{
mat[i][j] = q;
s[i * l + j] = div;
}
}
}
}
int result = mat[0][l - 1];
// Free dynamically allocated memory
for (int i = 0; i < l; ++i)
{
free(mat[i]);
}
free(mat);
return result;
}
/**
* @brief Recursively prints the solution
* @param l dimension of the solutions array
* @param s solutions
* @param i starting index
* @param j ending index
* @returns void
*/
void printSolution(int l, int *s, int i, int j)
{
if (i == j)
{
printf("A%d", i);
return;
}
putchar('(');
printSolution(l, s, i, s[i * l + j]);
printSolution(l, s, s[i * l + j] + 1, j);
putchar(')');
}
/**
* @brief Self-test implementations
* @returns void
*/
static void test()
{
int sizes[] = {35, 15, 5, 10, 20, 25};
int len = 6;
int *sol = malloc(len * len * sizeof(int));
int r = matrixChainOrder(len, sizes, sol);
assert(r == 18625);
printf("Result : %d\n", r);
printf("Optimal ordering : ");
printSolution(len, sol, 0, 5);
free(sol);
printf("\n");
}
/**
* @brief Main function
* @returns 0
*/
int main()
{
test(); // run self-test implementations
return 0;
}