Strassens Matrix Multiplication

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Strassen's Matrix Multiplication is the divide and conquer approach to solve the matrix multiplication problems. The usual matrix multiplication method multiplies each row with each column to achieve the product matrix. The time complexity taken by this approach is O(n3), since it takes two loops to multiply. Strassens method was introduced to reduce the time complexity from O(n3) to O(nlog 7).

Naive Method

First, we will discuss Naive method and its complexity. Here, we are calculating Z=X Y. Using Naive method, two matrices (X and Y) can be multiplied if the order of these matrices are p q and q r and the resultant matrix will be of order p r. The following pseudocode describes the Naive multiplication −

Algorithm: Matrix-Multiplication (X, Y, Z) 
for i = 1 to p do 
   for j = 1 to r do 
      Z[i,j] := 0 
      for k = 1 to q do 
         Z[i,j] := Z[i,j] + X[i,k] × Y[k,j]

Complexity

Here, we assume that integer operations take O(1) time. There are three for loops in this algorithm and one is nested in other. Hence, the algorithm takes O(n3) time to execute.

Strassens Matrix Multiplication Algorithm

In this context, using Strassens Matrix multiplication algorithm, the time consumption can be improved a little bit.

Strassens Matrix multiplication can be performed only on square matrices where n is a power of 2. Order of both of the matrices are n × n.

Divide X, Y and Z into four (n/2)×(n/2) matrices as represented below −

$Z = \begin{bmatrix}I & J \\K & L \end{bmatrix}$ $X = \begin{bmatrix}A & B \\C & D \end{bmatrix}$ and $Y = \begin{bmatrix}E & F \\G & H \end{bmatrix}$

Using Strassens Algorithm compute the following −

$$M_{1} \: \colon= (A+C) \times (E+F)$$

$$M_{2} \: \colon= (B+D) \times (G+H)$$

$$M_{3} \: \colon= (A-D) \times (E+H)$$

$$M_{4} \: \colon= A \times (F-H)$$

$$M_{5} \: \colon= (C+D) \times (E)$$

$$M_{6} \: \colon= (A+B) \times (H)$$

$$M_{7} \: \colon= D \times (G-E)$$

Then,

$$I \: \colon= M_{2} + M_{3} - M_{6} - M_{7}$$

$$J \: \colon= M_{4} + M_{6}$$

$$K \: \colon= M_{5} + M_{7}$$

$$L \: \colon= M_{1} - M_{3} - M_{4} - M_{5}$$

Analysis

$$T(n)=\begin{cases}c & if\:n= 1\\7\:x\:T(\frac{n}{2})+d\:x\:n^2 & otherwise\end{cases} \:where\: c\: and \:d\:are\: constants$$

Using this recurrence relation, we get $T(n) = O(n^{log7})$

Hence, the complexity of Strassens matrix multiplication algorithm is $O(n^{log7})$.

Example

Let us look at the implementation of Strassen's Matrix Multiplication in various programming languages: C, C++, Java, Python.

C C++ Java Python 
#include<stdio.h>
int main(){
   int z[2][2];
   int i, j;
   int m1, m2, m3, m4 , m5, m6, m7;
   int x[2][2] = {
       {12, 34}, 
       {22, 10}
       };
   int y[2][2] = {
       {3, 4}, 
       {2, 1}
   };
   printf("The first matrix is: ");
   for(i = 0; i < 2; i++) {
      printf("\n");
      for(j = 0; j < 2; j++)
         printf("%d\t", x[i][j]);
   }
   printf("\nThe second matrix is: ");
   for(i = 0; i < 2; i++) {
      printf("\n");
      for(j = 0; j < 2; j++)
         printf("%d\t", y[i][j]);
   }
   m1= (x[0][0] + x[1][1]) * (y[0][0] + y[1][1]);
   m2= (x[1][0] + x[1][1]) * y[0][0];
   m3= x[0][0] * (y[0][1] - y[1][1]);
   m4= x[1][1] * (y[1][0] - y[0][0]);
   m5= (x[0][0] + x[0][1]) * y[1][1];
   m6= (x[1][0] - x[0][0]) * (y[0][0]+y[0][1]);
   m7= (x[0][1] - x[1][1]) * (y[1][0]+y[1][1]);
   z[0][0] = m1 + m4- m5 + m7;
   z[0][1] = m3 + m5;
   z[1][0] = m2 + m4;
   z[1][1] = m1 - m2 + m3 + m6;
   printf("\nProduct achieved using Strassen's algorithm: ");
   for(i = 0; i < 2 ; i++) {
      printf("\n");
      for(j = 0; j < 2; j++)
         printf("%d\t", z[i][j]);
   }
   return 0;
}

Output

The first matrix is: 
12	34	
22	10	
The second matrix is: 
3	4	
2	1	
Product achieved using Strassen's algorithm: 
104	82	
86	98

#include<iostream>
using namespace std;
int main() {
   int z[2][2];
   int i, j;
   int m1, m2, m3, m4 , m5, m6, m7;
      int x[2][2] = {
         {12, 34}, 
         {22, 10}
      };
   int y[2][2] = {
      {3, 4}, 
      {2, 1}
   };
   cout<<"The first matrix is: ";
   for(i = 0; i < 2; i++) {
      cout<<endl;
      for(j = 0; j < 2; j++)
         cout<<x[i][j]<<" ";
   }
   cout<<"\nThe second matrix is: ";
   for(i = 0;i < 2; i++){
      cout<<endl;
      for(j = 0;j < 2; j++)
         cout<<y[i][j]<<" ";
   }

   m1 = (x[0][0] + x[1][1]) * (y[0][0] + y[1][1]);
   m2 = (x[1][0] + x[1][1]) * y[0][0];
   m3 = x[0][0] * (y[0][1] - y[1][1]);
   m4 = x[1][1] * (y[1][0] - y[0][0]);
   m5 = (x[0][0] + x[0][1]) * y[1][1];
   m6 = (x[1][0] - x[0][0]) * (y[0][0]+y[0][1]);
   m7 = (x[0][1] - x[1][1]) * (y[1][0]+y[1][1]);

   z[0][0] = m1 + m4- m5 + m7;
   z[0][1] = m3 + m5;
   z[1][0] = m2 + m4;
   z[1][1] = m1 - m2 + m3 + m6;

   cout<<"\nProduct achieved using Strassen's algorithm: ";
   for(i = 0; i < 2 ; i++) {
      cout<<endl;
      for(j = 0; j < 2; j++)
         cout<<z[i][j]<<" ";
   }
   return 0;
}

Output

The first matrix is: 
12 34 
22 10 
The second matrix is: 
3 4 
2 1 
Product achieved using Strassen's algorithm: 
104 82 
86 98

public class Strassens {
   public static void main(String[] args) {
      int[][] x = {{12, 34}, {22, 10}};
      int[][] y = {{3, 4}, {2, 1}};
      int z[][] = new int[2][2];
      int m1, m2, m3, m4 , m5, m6, m7;
      System.out.print("The first matrix is: ");
      for(int i = 0; i<2; i++) {
         System.out.println();//new line
         for(int j = 0; j<2; j++) {
            System.out.print(x[i][j] + "\t");
         }
      }
      System.out.print("\nThe second matrix is: ");
      for(int i = 0; i<2; i++) {
         System.out.println();//new line
         for(int j = 0; j<2; j++) {
            System.out.print(y[i][j] + "\t");
         }
      }
      m1 = (x[0][0] + x[1][1]) * (y[0][0] + y[1][1]);
      m2 = (x[1][0] + x[1][1]) * y[0][0];
      m3 = x[0][0] * (y[0][1] - y[1][1]);
      m4 = x[1][1] * (y[1][0] - y[0][0]);
      m5 = (x[0][0] + x[0][1]) * y[1][1];
      m6 = (x[1][0] - x[0][0]) * (y[0][0]+y[0][1]);
      m7 = (x[0][1] - x[1][1]) * (y[1][0]+y[1][1]);
      z[0][0] = m1 + m4- m5 + m7;
      z[0][1] = m3 + m5;
      z[1][0] = m2 + m4;
      z[1][1] = m1 - m2 + m3 + m6;
      System.out.print("\nProduct achieved using Strassen's algorithm: ");
      for(int i = 0; i<2; i++) {
         System.out.println();//new line
         for(int j = 0; j<2; j++) {
            System.out.print(z[i][j] + "\t");
         }
      }
   }
}

Output

The first matrix is: 
12	34	
22	10	
The second matrix is: 
3	4	
2	1	
Product achieved using Strassen's algorithm: 
104	82	
86	98	

import numpy as np
x = np.array([[12, 34], [22, 10]])
y = np.array([[3, 4], [2, 1]])
z = np.zeros((2, 2))
m1, m2, m3, m4, m5, m6, m7 = 0, 0, 0, 0, 0, 0, 0
print("The first matrix is: ")
for i in range(2):
    print()
    for j in range(2):
        print(x[i][j], end="\t")
print("\nThe second matrix is: ")
for i in range(2):
    print()
    for j in range(2):
        print(y[i][j], end="\t")
m1 = (x[0][0] + x[1][1]) * (y[0][0] + y[1][1])
m2 = (x[1][0] + x[1][1]) * y[0][0]
m3 = x[0][0] * (y[0][1] - y[1][1])
m4 = x[1][1] * (y[1][0] - y[0][0])
m5 = (x[0][0] + x[0][1]) * y[1][1]
m6 = (x[1][0] - x[0][0]) * (y[0][0] + y[0][1])
m7 = (x[0][1] - x[1][1]) * (y[1][0] + y[1][1])

z[0][0] = m1 + m4 - m5 + m7
z[0][1] = m3 + m5
z[1][0] = m2 + m4
z[1][1] = m1 - m2 + m3 + m6

print("\nProduct achieved using Strassen's algorithm: ")
for i in range(2):
    print()
    for j in range(2):
        print(z[i][j], end="\t")

Output

The first matrix is: 

12	34	
22	10	
The second matrix is: 

3	4	
2	1	
Product achieved using Strassen's algorithm: 

104.0	82.0	
86.0	98.0
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