# Strassen’s Matrix Multiplication in algorithms

In this article, we are going to discuss about the strassen matrix multiplication, formula of matrix multiplication and algorithms for strassen matrix multiplication.
Submitted by Prerana Jain, on June 22, 2018

Introduction

Strassen in 1969 which gives an overview that how we can find the multiplication of two 2*2 dimension matrix by the brute-force algorithm. But by using divide and conquer technique the overall complexity for multiplication two matrices is reduced. This happens by decreasing the total number if multiplication performed at the expenses of a slight increase in the number of addition.

For multiplying the two 2*2 dimension matrices Strassen's used some formulas in which there are seven multiplication and eighteen addition, subtraction, and in brute force algorithm, there is eight multiplication and four addition. The utility of Strassen's formula is shown by its asymptotic superiority when order n of matrix reaches infinity. Let us consider two matrices A and B, n*n dimension, where n is a power of two. It can be observed that we can contain four n/2*n/2 submatrices from A, B and their product C. C is the resultant matrix of A and B.

### Procedure of Strassen matrix multiplication

There are some procedures:

1. Divide a matrix of order of 2*2 recursively till we get the matrix of 2*2.
2. Use the previous set of formulas to carry out 2*2 matrix multiplication.
3. In this eight multiplication and four additions, subtraction are performed.
4. Combine the result of two matrixes to find the final product or final matrix.

### Formulas for Stassen’s matrix multiplication

In Strassen’s matrix multiplication there are seven multiplication and four addition, subtraction in total.

```    1.	D1 =  (a11 + a22) (b11 + b22)
2.	D2 =  (a21 + a22).b11
3.	D3 =  (b12 – b22).a11
4.	D4 =  (b21 – b11).a22
5.	D5 =  (a11 + a12).b22
6.	D6 =  (a21 – a11) . (b11 + b12)
7.	D7 =  (a12 – a22) . (b21 + b22)

C11 = d1 + d4 – d5 + d7
C12 = d3 + d5
C21 = d2 + d4
C22 = d1 + d3 – d2 – d6
```

## Algorithm for Strassen’s matrix multiplication

Algorithm Strassen(n, a, b, d)

```begin
If n = threshold then compute
C = a * b is a conventional matrix.
Else
Partition a into four sub matrices  a11, a12, a21, a22.
Partition b into four sub matrices b11, b12, b21, b22.
Strassen ( n/2, a11 + a22, b11 + b22, d1)
Strassen ( n/2, a21 + a22, b11, d2)
Strassen ( n/2, a11, b12 – b22, d3)
Strassen ( n/2, a22, b21 – b11, d4)
Strassen ( n/2, a11 + a12, b22, d5)
Strassen (n/2, a21 – a11, b11 + b22, d6)
Strassen (n/2, a12 – a22, b21 + b22, d7)

C = d1+d4-d5+d7     d3+d5
d2+d4           d1+d3-d2-d6

end if

return (C)
end.
```

### Code for Strassen matrix multiplication

```#include <stdio.h>

int main()
{
int a,b,c,i,j;
int m1,m2,m3,m4,m5,m6,m7;

printf("Enter the 4 elements of first matrix: ");
for(i=0;i<2;i++)
for(j=0;j<2;j++)
scanf("%d",&a[i][j]);

printf("Enter the 4 elements of second matrix: ");
for(i=0;i<2;i++)
for(j=0;j<2;j++)
scanf("%d",&b[i][j]);

printf("\nThe first matrix is\n");
for(i=0;i<2;i++)
{
printf("\n");
for(j=0;j<2;j++)
printf("%d\t",a[i][j]);
}

printf("\nThe second matrix is\n");
for(i=0;i<2;i++)
{
printf("\n");
for(j=0;j<2;j++)
printf("%d\t",b[i][j]);
}

m1= (a + a)*(b+b);
m2= (a+a)*b;
m3= a*(b-b);
m4= a*(b-b);
m5= (a+a)*b;
m6= (a-a)*(b+b);
m7= (a-a)*(b+b);

c=m1+m4-m5+m7;
c=m3+m5;
c=m2+m4;
c=m1-m2+m3+m6;

printf("\nAfter multiplication using \n");
for(i=0;i<2;i++)
{
printf("\n");
for(j=0;j<2;j++)
printf("%d\t",c[i][j]);
}

return 0;
}
```

Output:

```    Enter the 4 elements of first matrix:
5 6 1 7
Enter the 4 element of second matrix:
6 2 8 7
The first matrix  is
5       6
1       7
The second matrix is
6       2
8      7
After multiplication
78         52
62         51
```

Complexity: The time complexity is O(N2.8074).