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
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(n2.8874) 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 r?.
- Compute P? = A × (Br)? – Cr?.
- Return true if P? = ( 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)? – Cr? 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.
Below is the implementation of the above approach:
C++
// CPP code to implement Freivald’s Algorithm#include <bits/stdc++.h>using namespace std;#define N 2// Function to check if ABx = Cxint 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 compute 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 compute 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 compute 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 codeint 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 Algorithmimport 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 compute 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 compute 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 compute 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 Algorithmimport random N = 2# Function to check if ABx = Cxdef freivald(a, b, c) : # Generate a random vector r = [0] * N for i in range(0, N) : r[i] = (int)(random.randrange(509090009) % 2) # Now compute 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 compute 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 compute 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 codea = [ [ 1, 1 ], [ 1, 1 ] ]b = [ [ 1, 1 ], [ 1, 1 ] ]c = [ [ 2, 2 ], [ 2, 2 ] ]k = 2if (isProduct(a, b, c, k)) : print("Yes")else : print("No")# This code is contributed by Nikita Tiwari |
C#
// C# code to implement // Freivald's Algorithmusing 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$N = 2;// Function to check // if ABx = Cxfunction freivald($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 compute 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 compute 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 compute 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)?> |
Javascript
<script> // JavaScript code to implement Freivald’s Algorithm let N = 2; // Function to check if ABx = Cx function freivald(a,b,c) { // Generate a random vector let r = new Array(N); for (let i = 0; i < N; i++) r[i] = Math.random() % 2; // Now compute B*r for evaluating // expression A * (B*r) - (C*r) let br = new Array(N); br.fill(0); for (let i = 0; i < N; i++) for (let 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) let cr = new Array(N); cr.fill(0); for (let i = 0; i < N; i++) for (let 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) let axbr = new Array(N); axbr.fill(0); for (let i = 0; i < N; i++) for (let 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 (let 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. function isProduct(a,b,c,k) { for (let i=0; i<k; i++) if (freivald(a, b, c) == false) return false; return true; } // Driver code let a = [[ 1, 1 ], [ 1, 1 ]]; let b = [[ 1, 1 ], [ 1, 1 ]]; let c = [[ 2, 2 ], [ 2, 2 ]]; let k = 2; if (isProduct(a, b, c, k)) document.write("Yes"); else document.write("No");</script> |
Output:
Yes
Time Complexity: O(N ^ 2)
Auxiliary Space: O(N )
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
