Dadas tres arrays A, B y C, encuentre si C es un producto de A y B.
Ejemplos:
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
Una solución simple es encontrar el producto de A y B y luego verificar si el producto es igual a C o no. Una posible complejidad temporal de este método es O(n 2.8874 ) usando la multiplicación de arrays de Stression .
El algoritmo de Freivalds es un algoritmo aleatorio probabilístico que trabaja en el tiempo O(n 2 ) con alta probabilidad. En tiempo O(kn 2 ) el algoritmo puede verificar un producto matricial con probabilidad de falla menor a 2 -k . Dado que la salida no siempre es correcta, es un algoritmo aleatorio de Monte Carlo .
Pasos :
- Genere un vector aleatorio n × 1 0/1 r⃗ .
- Calcule P⃗ = A × (Br)⃗ – Cr⃗ .
- Devuelve verdadero si P⃗ = ( 0, 0, …, 0 ) T , de lo contrario devuelve falso.
La idea se basa en el hecho de que si C es realmente un producto, entonces el valor de A × (Br)⃗ – Cr⃗ siempre será 0. Si el valor es distinto de cero, entonces C no puede ser un producto. La condición de error es que el valor puede ser 0 incluso cuando C no es un producto.
A continuación se muestra la implementación del enfoque anterior:
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 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 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 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 Algorithm
import random
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.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 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 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>
Producción:
Yes
Complejidad temporal: O(N ^ 2)
Espacio auxiliar: O(N ^ 2)
Publicación traducida automáticamente
Artículo escrito por nickhilrawat y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA