✨ อัปเดตล่าสุด: web//https://www.geeksforgeeks.org/freivalds algorithm

Given three matrices A, B and C, find if C is a product of A and B.

Examples:

Input : A = 1 1
            1 1
        B = 1 1
            1 1
        C = 2  2
             2 2
Output : Yes
C = A x B

Input : A = 1 1 1
            1 1 1
            1 1 1
        B = 1 1 1
            1 1 1
            1 1 1
        C = 3 3 3
            3 1 2
            3 3 3 
Output : No

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to find product of A and B and then check if product is equal to C or not. A possible time complexity of this method is O(n^2.8874) using Stression’s matrix multiplication.

Freivalds’ algorithm is a probabilistic randomized algorithm that works in time O(n²) with high probability. In O(kn²) time the algorithm can verify a matrix product with probability of failure less than 2^-k. Since the output is not always correct, it is a Monte Carlo randomized algorithm.

Steps :

Generate an n × 1 random 0/1 vector r⃗.
Compute P⃗ = A × (Br)⃗ – Cr⃗.
Return true if P⃗ = ( 0, 0, …, 0 )^T, return false otherwise.

The idea is based on the fact that if C is actually a product, then value of A × (Br)⃗ – Cr⃗ will always be 0. If the value is non-zero, then C can not be a product. The error condition is that the value may be 0 even when C is not a product.

C++

// CPP code to implement Freivald’s Algorithm 
#include <bits/stdc++.h> 
using namespace std; 
  
#define N 2 
  
// Function to check if ABx = Cx 
int freivald(int a[][N], int b[][N], int c[][N]) 
{ 
    // Generate a random vector 
    bool r[N]; 
    for (int i = 0; i < N; i++) 
        r[i] = random() % 2; 
  
    // Now comput B*r for evaluating 
    // expression A * (B*r) - (C*r) 
    int br[N] = { 0 }; 
    for (int i = 0; i < N; i++) 
        for (int j = 0; j < N; j++) 
            br[i] = br[i] + b[i][j] * r[j]; 
  
    // Now comput C*r for evaluating 
    // expression A * (B*r) - (C*r) 
    int cr[N] = { 0 }; 
    for (int i = 0; i < N; i++) 
        for (int j = 0; j < N; j++) 
            cr[i] = cr[i] + c[i][j] * r[j]; 
  
    // Now comput A* (B*r) for evaluating 
    // expression A * (B*r) - (C*r) 
    int axbr[N] = { 0 }; 
    for (int i = 0; i < N; i++) 
        for (int j = 0; j < N; j++) 
            axbr[i] = axbr[i] + a[i][j] * br[j]; 
  
    // Finally check if value of expression 
    // A * (B*r) - (C*r) is 0 or not 
    for (int i = 0; i < N; i++) 
        if (axbr[i] - cr[i] != 0) 
            false; 
  
    return true; 
} 
  
// Runs k iterations Freivald. The value 
// of k determines accuracy. Higher value 
// means higher accuracy. 
bool isProduct(int a[][N], int b[][N], 
               int c[][N], int k) 
{ 
    for (int i=0; i<k; i++) 
        if (freivald(a, b, c) == false) 
            return false; 
    return true; 
} 
  
// Driver code 
int main() 
{ 
    int a[N][N] = { { 1, 1 }, { 1, 1 } }; 
    int b[N][N] = { { 1, 1 }, { 1, 1 } }; 
    int c[N][N] = { { 2, 2 }, { 2, 2 } }; 
    int k = 2; 
    if (isProduct(a, b, c, k)) 
        printf("Yes"); 
    else
        printf("No"); 
    return 0; 
} 

Java

// Java code to implement  
// Freivald's Algorithm 
import java.io.*; 
import java.util.*; 
import java.math.*; 
  
class GFG { 
    static int N = 2; 
  
    // Function to check if ABx = Cx 
    static boolean freivald(int a[][], int b[][], 
                                       int c[][]) 
    { 
        // Generate a random vector 
        int r[] = new int[N]; 
        for (int i = 0; i < N; i++) 
        r[i] = (int)(Math.random()) % 2; 
  
        // Now comput B*r for evaluating 
        // expression A * (B*r) - (C*r) 
        int br[] = new int[N]; 
        Arrays.fill(br, 0); 
        for (int i = 0; i < N; i++) 
            for (int j = 0; j < N; j++) 
                br[i] = br[i] + b[i][j] * r[j]; 
  
        // Now comput C*r for evaluating 
        // expression A * (B*r) - (C*r) 
        int cr[] = new int[N]; 
        Arrays.fill(cr, 0); 
        for (int i = 0; i < N; i++) 
            for (int j = 0; j < N; j++) 
                cr[i] = cr[i] + c[i][j] * r[j]; 
  
        // Now comput A* (B*r) for evaluating 
        // expression A * (B*r) - (C*r) 
        int axbr[] = new int[N]; 
        Arrays.fill(axbr, 0); 
        for (int i = 0; i < N; i++) 
            for (int j = 0; j < N; j++) 
                axbr[i] = axbr[i] + a[i][j] * br[j]; 
  
        // Finally check if value of expression 
        // A * (B*r) - (C*r) is 0 or not 
        for (int i = 0; i < N; i++) 
            if (axbr[i] - cr[i] != 0) 
                return false; 
  
        return true; 
    } 
  
    // Runs k iterations Freivald. The value 
    // of k determines accuracy. Higher value 
    // means higher accuracy. 
    static boolean isProduct(int a[][], int b[][],  
                                 int c[][], int k) 
    { 
        for (int i = 0; i < k; i++) 
            if (freivald(a, b, c) == false) 
                return false; 
        return true; 
    } 
  
    // Driver code 
    public static void main(String args[]) 
    { 
        int a[][] = { { 1, 1 }, { 1, 1 } }; 
        int b[][] = { { 1, 1 }, { 1, 1 } }; 
        int c[][] = { { 2, 2 }, { 2, 2 } }; 
        int k = 2; 
        if (isProduct(a, b, c, k)) 
            System.out.println("Yes"); 
        else
            System.out.println("No"); 
    } 
} 
  
/*This code is contributed by Nikita Tiwari.*/

Python3

# Python3 code to implement Freivald’s Algorithm 
from random import *
N = 2
  
# Function to check if ABx = Cx 
def freivald(a, b, c) : 
      
    # Generate a random vector 
    r = [0] * N 
      
    for i in range(0, N) : 
        r[i] = (int)(random()) % 2
  
    # Now comput B*r for evaluating 
    # expression A * (B*r) - (C*r) 
    br = [0] * N 
      
    for i in range(0, N) : 
        for j in range(0, N) : 
            br[i] = br[i] + b[i][j] * r[j] 
  
    # Now comput C*r for evaluating 
    # expression A * (B*r) - (C*r) 
    cr = [0] * N 
    for i in range(0, N) : 
        for j in range(0, N) : 
            cr[i] = cr[i] + c[i][j] * r[j] 
  
    # Now comput A* (B*r) for evaluating 
    # expression A * (B*r) - (C*r) 
    axbr = [0] * N 
    for i in range(0, N) : 
        for j in range(0, N) : 
            axbr[i] = axbr[i] + a[i][j] * br[j] 
  
    # Finally check if value of expression 
    # A * (B*r) - (C*r) is 0 or not 
    for i in range(0, N) : 
        if (axbr[i] - cr[i] != 0) : 
            return False
              
    return True
  
# Runs k iterations Freivald. The value 
# of k determines accuracy. Higher value 
# means higher accuracy. 
def isProduct(a, b, c, k) : 
      
    for i in range(0, k) : 
        if (freivald(a, b, c) == False) : 
            return False
    return True
  
# Driver code 
a = [ [ 1, 1 ], [ 1, 1 ] ] 
b = [ [ 1, 1 ], [ 1, 1 ] ] 
c = [ [ 2, 2 ], [ 2, 2 ] ] 
k = 2
  
if (isProduct(a, b, c, k)) : 
    print("Yes") 
else : 
    print("No") 
  
# This code is contributed by Nikita Tiwari 

C#

// C# code to implement  
// Freivald's Algorithm 
using System; 
  
class GFG  
{ 
    static int N = 2; 
  
    // Function to check 
    // if ABx = Cx 
    static bool freivald(int [,]a,  
                         int [,]b, 
                         int [,]c) 
    { 
        // Generate a  
        // random vector 
        Random rand = new Random(); 
        int []r = new int[N]; 
          
        for (int i = 0; i < N; i++) 
        r[i] = (int)(rand.Next()) % 2; 
  
        // Now compute B*r for  
        // evaluating expression 
        // A * (B*r) - (C*r) 
        int []br = new int[N]; 
          
        for (int i = 0; i < N; i++) 
            for (int j = 0; j < N; j++) 
                br[i] = br[i] +  
                        b[i, j] * r[j]; 
  
        // Now compute C*r for  
        // evaluating expression  
        // A * (B*r) - (C*r) 
        int []cr = new int[N]; 
          
        for (int i = 0; i < N; i++) 
            for (int j = 0; j < N; j++) 
                cr[i] = cr[i] +  
                        c[i, j] * r[j]; 
  
        // Now compute A* (B*r) for 
        // evaluating expression  
        // A * (B*r) - (C*r) 
        int []axbr = new int[N]; 
          
        for (int i = 0; i < N; i++) 
            for (int j = 0; j < N; j++) 
                axbr[i] = axbr[i] +  
                          a[i, j] * br[j]; 
  
        // Finally check if value  
        // of expression A * (B*r) -  
        // (C*r) is 0 or not 
        for (int i = 0; i < N; i++) 
            if (axbr[i] - cr[i] != 0) 
                return false; 
  
        return true; 
    } 
  
    // Runs k iterations Freivald. 
    // The value of k determines 
    // accuracy. Higher value 
    // means higher accuracy. 
    static bool isProduct(int [,]a, int [,]b,  
                          int [,]c, int k) 
    { 
        for (int i = 0; i < k; i++) 
            if (freivald(a, b, c) == false) 
                return false; 
        return true; 
    } 
  
    // Driver code 
    static void Main() 
    { 
        int [,]a = new int[,]{ { 1, 1 },  
                               { 1, 1 }}; 
        int [,]b = new int[,]{ { 1, 1 },  
                               { 1, 1 }}; 
        int [,]c = new int[,]{ { 2, 2 },  
                               { 2, 2 }}; 
        int k = 2; 
        if (isProduct(a, b, c, k)) 
            Console.WriteLine("Yes"); 
        else
            Console.WriteLine("No"); 
    } 
} 
// This code is contributed  
// by Manish Shaw(manishshaw1) 

PHP

<?php 
// PHP code to implement 
// Freivald’s Algorithm 
$N = 2; 
  
// Function to check  
// if ABx = Cx 
function freivald($a, $b, $c) 
{ 
    global $N; 
      
    // Generate a  
    // random vector 
    $r = array(); 
    $br = array(); 
    $cr = array(); 
    $axbr = array(); 
      
    for ($i = 0; $i < $N; $i++)  
    { 
        $r[$i] = mt_rand() % 2; 
        $br[$i] = 0;  
        $cr[$i] = 0; 
        $axbr[$i] = 0; 
    } 
  
    // Now comput B*r for  
    // evaluating expression  
    // A * (B*r) - (C*r) 
    for ($i = 0; $i < $N; $i++)  
    { 
        for ($j = 0; $j < $N; $j++) 
            $br[$i] = $br[$i] +  
                      $b[$i][$j] *  
                      $r[$j]; 
    } 
      
    // Now comput C*r for  
    // evaluating expression 
    // A * (B*r) - (C*r) 
    for ($i = 0; $i < $N; $i++)  
    { 
        for ($j = 0; $j < $N; $j++) 
            $cr[$i] = $cr[$i] +  
                      $c[$i][$j] * 
                      $r[$j]; 
    } 
  
    // Now comput A* (B*r) for  
    // evaluating expression  
    // A * (B*r) - (C*r) 
    for ($i = 0; $i < $N; $i++)  
    { 
        for ($j = 0; $j < $N; $j++) 
            $axbr[$i] = $axbr[$i] +  
                        $a[$i][$j] *  
                        $br[$j]; 
    } 
  
    // Finally check if value  
    // of expression A * (B*r) - 
    // (C*r) is 0 or not 
    for ($i = 0; $i < $N; $i++) 
        if ($axbr[$i] - $cr[$i] != 0) 
            return false; 
  
    return true; 
} 
  
// Runs k iterations Freivald.  
// The value of k determines  
// accuracy. Higher value 
// means higher accuracy. 
function isProduct($a, $b, $c, $k) 
{ 
    for ($i = 0; $i < $k; $i++) 
        if (freivald($a,  
                     $b, $c) == false) 
            return false; 
    return true; 
} 
  
// Driver code 
$a = array(array(1, 1),  
           array(1, 1)); 
$b = array(array(1, 1),  
           array(1, 1)); 
$c = array(array(2, 2),  
           array(2, 2)); 
$k = 2; 
if (isProduct($a, $b,  
              $c, $k)) 
    echo ("Yes"); 
else
    echo ("No"); 
      
// This code is contributed  
// by Manish Shaw(manishshaw1) 
?> 

Output:

Yes

My Personal Notes

Sep	OCT	Nov
	21
2018	2019	2020

Freivald’s Algorithm to check if a matrix is product of two

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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