C Program for Matrix Chain Multiplication | DP-8
Given a sequence of matrices, find the most efficient way to multiply these matrices together. The problem is not actually to perform the multiplications, but merely to decide in which order to perform the multiplications.
We have many options to multiply a chain of matrices because matrix multiplication is associative. In other words, no matter how we parenthesize the product, the result will be the same. For example, if we had four matrices A, B, C, and D, we would have:
(ABC)D = (AB)(CD) = A(BCD) = ....
However, the order in which we parenthesize the product affects the number of simple arithmetic operations needed to compute the product, or the efficiency. For example, suppose A is a 10 × 30 matrix, B is a 30 × 5 matrix, and C is a 5 × 60 matrix. Then,
(AB)C = (10×30×5) + (10×5×60) = 1500 + 3000 = 4500 operations
A(BC) = (30×5×60) + (10×30×60) = 9000 + 18000 = 27000 operations.
Clearly the first parenthesization requires less number of operations.
Given an array p[] which represents the chain of matrices such that the ith matrix Ai is of dimension p[i-1] x p[i]. We need to write a function MatrixChainOrder() that should return the minimum number of multiplications needed to multiply the chain.
Input: p[] = {40, 20, 30, 10, 30}
Output: 26000
There are 4 matrices of dimensions 40x20, 20x30, 30x10 and 10x30.
Let the input 4 matrices be A, B, C and D. The minimum number of
multiplications are obtained by putting parenthesis in following way
(A(BC))D --> 20*30*10 + 40*20*10 + 40*10*30
Input: p[] = {10, 20, 30, 40, 30}
Output: 30000
There are 4 matrices of dimensions 10x20, 20x30, 30x40 and 40x30.
Let the input 4 matrices be A, B, C and D. The minimum number of
multiplications are obtained by putting parenthesis in following way
((AB)C)D --> 10*20*30 + 10*30*40 + 10*40*30
Input: p[] = {10, 20, 30}
Output: 6000
There are only two matrices of dimensions 10x20 and 20x30. So there
is only one way to multiply the matrices, cost of which is 10*20*30
Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.
Following is a recursive implementation that simply follows the above optimal substructure property.
C
/* A naive recursive implementation that simply follows the above optimal substructure property */#include <limits.h>#include <stdio.h> // Matrix Ai has dimension p[i-1] x p[i] for i = 1..nint MatrixChainOrder(int p[], int i, int j){ if (i == j) return 0; int k; int min = INT_MAX; int count; // place parenthesis at different places between first // and last matrix, recursively calculate count of // multiplications for each parenthesis placement and // return the minimum count for (k = i; k < j; k++) { count = MatrixChainOrder(p, i, k) + MatrixChainOrder(p, k + 1, j) + p[i - 1] * p[k] * p[j]; if (count < min) min = count; } // Return minimum count return min;} // Driver program to test above functionint main(){ int arr[] = { 1, 2, 3, 4, 3 }; int n = sizeof(arr) / sizeof(arr[0]); printf("Minimum number of multiplications is %d ", MatrixChainOrder(arr, 1, n - 1)); getchar(); return 0;} |
Minimum number of multiplications is 30
Dynamic Programming Solution
C
// See the Cormen book for details of the following algorithm#include <limits.h>#include <stdio.h> // Matrix Ai has dimension p[i-1] x p[i] for i = 1..nint MatrixChainOrder(int p[], int n){ /* For simplicity of the program, one extra row and one extra column are allocated in m[][]. 0th row and 0th column of m[][] are not used */ int m[n][n]; int i, j, k, L, q; /* m[i, j] = Minimum number of scalar multiplications needed to compute the matrix A[i]A[i+1]...A[j] = A[i..j] where dimension of A[i] is p[i-1] x p[i] */ // cost is zero when multiplying one matrix. for (i = 1; i < n; i++) m[i][i] = 0; // L is chain length. for (L = 2; L < n; L++) { for (i = 1; i < n - L + 1; i++) { j = i + L - 1; m[i][j] = INT_MAX; for (k = i; k <= j - 1; k++) { // q = cost/scalar multiplications q = m[i][k] + m[k + 1][j] + p[i - 1] * p[k] * p[j]; if (q < m[i][j]) m[i][j] = q; } } } return m[1][n - 1];} int main(){ int arr[] = { 1, 2, 3, 4 }; int size = sizeof(arr) / sizeof(arr[0]); printf("Minimum number of multiplications is %d ", MatrixChainOrder(arr, size)); getchar(); return 0;} |
Minimum number of multiplications is 18
Please refer complete article on Matrix Chain Multiplication | DP-8 for more details!
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