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Easy way to remember Strassen’s Matrix Equation

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  • Difficulty Level : Medium
  • Last Updated : 27 May, 2022

Strassen’s matrix is a Divide and Conquer method that helps us to multiply two matrices(of size n X n). You can refer to the link, for having the knowledge about Strassen’s Matrix first : Divide and Conquer | Set 5 (Strassen’s Matrix Multiplication) But this method needs to cram few equations, so I’ll tell you the simplest way to remember those : stressen_formula_new_new You just need to remember 4 Rules :

  • AHED (Learn it as ‘Ahead’)
  • Diagonal
  • Last CR
  • First CR

Also, consider X as (Row +) and Y as (Column -) matrix Follow the Steps :

  • Write P1 = A; P2 = H; P3 = E; P4 = D
  • For P5 we will use Diagonal Rule i.e. (Sum the Diagonal Elements Of Matrix X ) * (Sum the Diagonal Elements Of Matrix Y ), we get P5 = (A + D)* (E + H)
  • For P6 we will use Last CR Rule i.e. Last Column of X and Last Row of Y and remember that Row+ and Column- so i.e. (B – D) * (G + H), we get P6 = (B – D) * (G + H)
  • For P7 we will use First CR Rule i.e. First Column of X and First Row of Y and remember that Row+ and Column- so i.e. (A – C) * (E + F), we get P7 = (A – C) * (E + F)
  • Come Back to P1 : we have A there and it’s adjacent element in Y Matrix is E, since Y is Column Matrix so we select a column in Y such that E won’t come, we find F H Column, so multiply A with (F – H) So, finally P1 = A * (F – H)
  • Come Back to P2 : we have H there and it’s adjacent element in X Matrix is D, since X is Row Matrix so we select a Row in X such that D won’t come, we find A B Column, so multiply H with (A + B) So, finally P2 = (A + B) * H
  • Come Back to P3 : we have E there and it’s adjacent element in X Matrix is A, since X is Row Matrix so we select a Row in X such that A won’t come, we find C D Column, so multiply E with (C + D) So, finally P3 = (C + D) * E
  • Come Back to P4 : we have D there and it’s adjacent element in Y Matrix is H, since Y is Column Matrix so we select a column in Y such that H won’t come, we find G E Column, so multiply D with (G – E) So, finally P4 = D * (G – E)
  • Remember Counting : Write P1 + P2 at C2
  • Write P3 + P4 at its diagonal Position i.e. at C3
  • Write P4 + P5 + P6 at 1st position and subtract P2 i.e. C1 = P4 + P5 + P6 – P2
  • Write odd values at last Position with alternating – and + sign i.e. P1 P3 P5 P7 becomes C4 = P1 – P3 + P5 – P7

C++




#include <bits/stdc++.h>
#include <cmath>
#define vi vector<int>
#define vii vector<vi>
using namespace std;
/* finding next square of 2*/
int nextPowerOf2(int k)
{
    return pow(2, int(ceil(log2(k))));
}
// printing matrix
void display(vii C, int m, int n)
{
    for (int i = 0; i < m; i++)
    {
        cout << "|"
             << " ";
        for (int j = 0; j < n; j++)
        {
            cout << C[i][j] << " ";
        }
        cout << "|" << endl;
    }
}
//! addition and subtraction
void add(vii &A, vii &B, vii &C, int size)
{
    for (int i = 0; i < size; i++)
    {
        for (int j = 0; j < size; j++)
        {
            C[i][j] = A[i][j] + B[i][j];
        }
    }
}
void sub(vii &A, vii &B, vii &C, int size)
{
    for (int i = 0; i < size; i++)
    {
        for (int j = 0; j < size; j++)
        {
            C[i][j] = A[i][j] - B[i][j];
        }
    }
}
//!-----------------------------
void Strassen_algorithm(vii &A, vii &B, vii &C, int size)
{
    if (size == 1)
    {
        C[0][0] = A[0][0] * B[0][0];
        return;
    }
    else
    {
        int newSize = size / 2;
        vi z(newSize);
        vii a(newSize, z), b(newSize, z), c(newSize, z), d(newSize, z),
            e(newSize, z), f(newSize, z), g(newSize, z), h(newSize, z),
            c11(newSize, z), c12(newSize, z), c21(newSize, z), c22(newSize, z),
            p1(newSize, z), p2(newSize, z), p3(newSize, z), p4(newSize, z),
            p5(newSize, z), p6(newSize, z), p7(newSize, z), fResult(newSize, z),
            sResult(newSize, z);
        int i, j;
 
        //! divide the matrix in equal parts
        for (i = 0; i < newSize; i++)
        {
            for (j = 0; j < newSize; j++)
            {
                a[i][j] = A[i][j];
                b[i][j] = A[i][j + newSize];
                c[i][j] = A[i + newSize][j];
                d[i][j] = A[i + newSize][j + newSize];
 
                e[i][j] = B[i][j];
                f[i][j] = B[i][j + newSize];
                g[i][j] = B[i + newSize][j];
                h[i][j] = B[i + newSize][j + newSize];
            }
        }
        /*
             A         B           C
           [a b]   * [e f]   =  [c11 c12]
                [g h]      [c21 c22]
           p1,p2,p3,p4=AHED for this: A:Row(+) and B:Column(-)
           p5=Diagonal :both +ve
           p6=Last CR  :A:Row(-) B:Column(+)
           p7=First CR :A:Row(-) B:Column(+)
        */
        //! calculating all strassen formulas
        //*p1=a*(f-h)
        sub(f, h, sResult, newSize);
        Strassen_algorithm(a, sResult, p1, newSize);
 
        //*p2=h*(a+b)
        add(a, b, fResult, newSize);
        Strassen_algorithm(fResult, h, p2, newSize);
 
        //*p3=e*(c+d)
        add(c, d, fResult, newSize);
        Strassen_algorithm(fResult, e, p3, newSize);
 
        //*p4=d*(g-e)
        sub(g, e, sResult, newSize);
        Strassen_algorithm(d, sResult, p4, newSize);
 
        //*p5=(a+d)*(e+h)
        add(a, d, fResult, newSize);
        add(e, h, sResult, newSize);
        Strassen_algorithm(fResult, sResult, p5, newSize);
 
        //*p6=(b-d)*(g+h)
        sub(b, d, fResult, newSize);
        add(g, h, sResult, newSize);
        Strassen_algorithm(fResult, sResult, p6, newSize);
 
        //*p7=(a-c)*(e+f)
        sub(a, c, fResult, newSize);
        add(e, f, sResult, newSize);
        Strassen_algorithm(fResult, sResult, p7, newSize);
 
        /* calculating all elements of C by p1,p2,p3
        c11=p4+p5+p6-p2
        c12=p1+p2
        c21=p3+p4
        c22=p1-p3+p5-p7
        */
        add(p1, p2, c12, newSize); //!
        add(p3, p4, c21, newSize); //!
 
        add(p4, p5, fResult, newSize);
        add(fResult, p6, sResult, newSize);
        sub(sResult, p2, c11, newSize); //!
 
        sub(p1, p3, fResult, newSize);
        add(fResult, p5, sResult, newSize);
        sub(sResult, p7, c22, newSize); //!
 
        // Grouping the results obtained in a single matrix:
        for (i = 0; i < newSize; i++)
        {
            for (j = 0; j < newSize; j++)
            {
                C[i][j] = c11[i][j];
                C[i][j + newSize] = c12[i][j];
                C[i + newSize][j] = c21[i][j];
                C[i + newSize][j + newSize] = c22[i][j];
            }
        }
    }
}
/*for converting matrix to square matrix*/
void ConvertToSquareMat(vii &A, vii &B, int r1, int c1, int r2, int c2)
{
    int maxSize = max({r1, c1, r2, c2});
    int size = nextPowerOf2(maxSize);
 
    vi z(size);
    vii Aa(size, z), Bb(size, z), Cc(size, z);
 
    for (unsigned int i = 0; i < r1; i++)
    {
        for (unsigned int j = 0; j < c1; j++)
        {
            Aa[i][j] = A[i][j];
        }
    }
    for (unsigned int i = 0; i < r2; i++)
    {
        for (unsigned int j = 0; j < c2; j++)
        {
            Bb[i][j] = B[i][j];
        }
    }
    Strassen_algorithm(Aa, Bb, Cc, size);
    vi temp1(c2);
    vii C(r1, temp1);
    for (unsigned int i = 0; i < r1; i++)
    {
        for (unsigned int j = 0; j < c2; j++)
        {
            C[i][j] = Cc[i][j];
        }
    }
    display(C, r1, c1);
}
int main()
{
    vii a = {
        {1, 2, 3},
        {1, 2, 3},
        {0, 0, 2}};
    vii b = {
        {1, 0, 0},
        {0, 1, 0},
        {0, 0, 1}};
    ConvertToSquareMat(a, b, 3, 3, 3, 3); // A[][],B[][],R1,C1,R2,C2
    return 0;
}


Output

| 1 2 3 |
| 1 2 3 |
| 0 0 2 |

This article is contributed by Mohit Gupta 🙂. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.


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