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* feat: add Matrix Chain Order * updating DIRECTORY.md * Update dynamic_programming/matrix_chain_order.c Co-authored-by: David Leal <halfpacho@gmail.com> * Update dynamic_programming/matrix_chain_order.c Co-authored-by: David Leal <halfpacho@gmail.com> * Update dynamic_programming/matrix_chain_order.c Co-authored-by: David Leal <halfpacho@gmail.com> * Update matrix_chain_order.c * chore: apply suggestions from code review * chore: apply suggestions from code review * updating DIRECTORY.md * Update dynamic_programming/matrix_chain_order.c Co-authored-by: Taj <tjgurwara99@users.noreply.github.com> * updating DIRECTORY.md * updating DIRECTORY.md --------- Co-authored-by: github-actions[bot] <github-actions@users.noreply.github.com> Co-authored-by: David Leal <halfpacho@gmail.com> Co-authored-by: Taj <tjgurwara99@users.noreply.github.com>
92 lines
2.5 KiB
C
92 lines
2.5 KiB
C
/**
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* @file
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* @brief [Matrix Chain Order](https://en.wikipedia.org/wiki/Matrix_chain_multiplication)
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* @details
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* From Wikipedia: Matrix chain multiplication (or the matrix chain ordering problem)
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* is an optimization problem concerning the most efficient way to multiply a given sequence of matrices.
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* The problem is not actually to perform the multiplications,
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* but merely to decide the sequence of the matrix multiplications involved.
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* @author [CascadingCascade](https://github.com/CascadingCascade)
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*/
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#include <assert.h> /// for assert
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#include <stdio.h> /// for IO operations
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#include <limits.h> /// for INT_MAX macro
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#include <stdlib.h> /// for malloc() and free()
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/**
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* @brief Finds the optimal sequence using the classic O(n^3) algorithm.
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* @param l length of cost array
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* @param p costs of each matrix
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* @param s location to store results
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* @returns number of operations
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*/
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int matrixChainOrder(int l,const int *p, int *s) {
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// mat stores the cost for a chain that starts at i and ends on j (inclusive on both ends)
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int mat[l][l];
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for (int i = 0; i < l; ++i) {
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mat[i][i] = 0;
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}
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// cl denotes the difference between start / end indices, cl + 1 would be chain length.
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for (int cl = 1; cl < l; ++cl) {
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for (int i = 0; i < l - cl; ++i) {
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int j = i + cl;
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mat[i][j] = INT_MAX;
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for (int div = i; div < j; ++div) {
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int q = mat[i][div] + mat[div + 1][j] + p[i] * p[div] * p[j];
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if (q < mat[i][j]) {
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mat[i][j] = q;
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s[i * l + j] = div;
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}
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}
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}
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}
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return mat[0][l - 1];
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}
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/**
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* @brief Recursively prints the solution
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* @param l dimension of the solutions array
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* @param s solutions
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* @param i starting index
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* @param j ending index
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* @returns void
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*/
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void printSolution(int l,int *s,int i,int j) {
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if(i == j) {
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printf("A%d",i);
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return
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}
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putchar('(');
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printSolution(l,s,i,s[i * l + j]);
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printSolution(l,s,s[i * l + j] + 1,j);
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putchar(')');
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}
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/**
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* @brief Self-test implementations
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* @returns void
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*/
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static void test() {
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int sizes[] = {35,15,5,10,20,25};
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int len = 6;
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int *sol = malloc(len * len * sizeof(int));
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int r = matrixChainOrder(len,sizes,sol);
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assert(r == 18625);
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printf("Result : %d\n",r);
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printf("Optimal ordering : ");
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printSolution(len,sol,0,5);
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free(sol);
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printf("\n");
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}
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/**
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* @brief Main function
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* @returns 0
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*/
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int main() {
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test(); // run self-test implementations
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return 0;
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}
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