Matrix Chain Multiplication (A O(N^3) Solution) in C++

C++Server Side ProgrammingProgramming

If a chain of matrices is given, we have to find minimum number of correct sequence of matrices to multiply.

We know that the matrix multiplication is associative, so four matrices ABCD, we can multiply A(BCD), (AB)(CD), (ABC)D, A(BC)D, in these sequences. Like these sequences, our task is to find which ordering is efficient to multiply.

In the given input there is an array say arr, which contains arr[] = {1, 2, 3, 4}. It means the matrices are of the order (1 x 2), (2 x 3), (3 x 4).

Input − The orders of the input matrices. {1, 2, 3, 4}. It means the matrices are

{(1 x 2), (2 x 3), (3 x 4)}.

Output − Minimum number of operations need multiply these three matrices. Here the result is 18.

Algorithm

matOrder(array, n)
Input: List of matrices, the number of matrices in the list.
Output: Minimum number of matrix multiplication.
Begin
   define table minMul of size n x n, initially fill with all 0s
   for length := 2 to n, do
      for i:=1 to n-length, do
         j := i + length – 1
         minMul[i, j] := ∞
         for k := i to j-1, do
            q := minMul[i, k] + minMul[k+1, j] + array[i-1]*array[k]*array[j]
            if q < minMul[i, j], then
               minMul[i, j] := q
            done
         done
      done
   return minMul[1, n-1]
End

Example

#include<iostream>
using namespace std;
int matOrder(int array[], int n){
   int minMul[n][n]; //holds the number of scalar multiplication needed
   for (int i=1; i<n; i++)
      minMul[i][i] = 0; //for multiplication with 1 matrix, cost is 0
      for (int length=2; length<n; length++){ //find the chain length starting from 2
         for (int i=1; i<n-length+1; i++){
            int j = i+length-1;
            minMul[i][j] = INT_MAX; //set to infinity
            for (int k=i; k<=j-1; k++){
               //store cost per multiplications
               int q = minMul[i][k] + minMul[k+1][j] + array[i-1]*array[k]*array[j];
               if (q < minMul[i][j])
                  minMul[i][j] = q;
            }
      }
   }
   return minMul[1][n-1];
}
int main(){
   int arr[] = {1, 2, 3, 4};
   int size = 4;
   cout << "Minimum number of matrix multiplications: "<<matOrder(arr, size);
}

Output

Minimum number of matrix multiplications: 18
raja
Published on 25-Sep-2019 12:21:32
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