Cartesian Product of Two Sets
Last Updated :
19 Oct, 2022
Let A and B be two sets, Cartesian productA × B is the set of all ordered pair of elements from A and B
A × B = {{x, y} : x ? A, y ? B}
Let A = {a, b, c} and B = {d, e, f}
The Cartesian product of two sets is
A x B = {a, d}, {a, e}, {a, f}, {b, d}, {b, e}, {b, f}, {c, d}, {c, e}, {c, f}}
A has 3 elements and B also has 3 elements. The Cartesian Product has 3 x 3 = 9 elements.
In general, if there are m elements in set A and n elements in B, the number of elements in the Cartesian Product is m x n
Given two finite non-empty sets, write a program to print Cartesian Product.
Examples :
Input : A = {1, 2}, B = {3, 4}
Output : A × B = {{1, 3}, {1, 4}, {2, 3}, {2, 4}}
Input : A = {1, 2, 3} B = {4, 5, 6}
Output : A × B = {{1, 4}, {1, 5}, {1, 6}, {2, 4},
{2, 5}, {2, 6}, {3, 4}, {3, 5}, {3, 6}}
CPP
#include <stdio.h>
void findCart( int arr1[], int arr2[], int n, int n1)
{
for ( int i = 0; i < n; i++)
for ( int j = 0; j < n1; j++)
printf ( "{%d, %d}, " , arr1[i], arr2[j]);
}
int main()
{
int arr1[] = { 1, 2, 3 };
int arr2[] = { 4, 5, 6 };
int n1 = sizeof (arr1) / sizeof (arr1[0]);
int n2 = sizeof (arr2) / sizeof (arr2[0]);
findCart(arr1, arr2, n1, n2);
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG {
static void findCart( int arr1[], int arr2[], int n,
int n1)
{
for ( int i = 0 ; i < n; i++)
for ( int j = 0 ; j < n1; j++)
System.out.print( "{" + arr1[i] + ", "
+ arr2[j] + "}, " );
}
public static void main(String[] args)
{
int arr1[] = { 1 , 2 , 3 };
int arr2[] = { 4 , 5 , 6 };
int n1 = arr1.length;
int n2 = arr2.length;
findCart(arr1, arr2, n1, n2);
}
}
|
Python3
def findCart(arr1, arr2, n, n1):
for i in range ( 0 , n):
for j in range ( 0 , n1):
print ( "{" , arr1[i], ", " , arr2[j], "}, " , sep = " ", end=" ")
arr1 = [ 1 , 2 , 3 ]
arr2 = [ 4 , 5 , 6 ]
n1 = len (arr1)
n2 = len (arr2)
findCart(arr1, arr2, n1, n2)
|
C#
using System;
class GFG {
static void findCart( int [] arr1, int [] arr2, int n,
int n1)
{
for ( int i = 0; i < n; i++)
for ( int j = 0; j < n1; j++)
Console.Write( "{" + arr1[i] + ", " + arr2[j]
+ "}, " );
}
public static void Main()
{
int [] arr1 = { 1, 2, 3 };
int [] arr2 = { 4, 5, 6 };
int n1 = arr1.Length;
int n2 = arr2.Length;
findCart(arr1, arr2, n1, n2);
}
}
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PHP
<?php
function findCart( $arr1 , $arr2 , $n , $n1 )
{
for ( $i = 0; $i < $n ; $i ++)
for ( $j = 0; $j < $n1 ; $j ++)
echo "{" , $arr1 [ $i ] , " , " ,
$arr2 [ $j ], "}" , "," ;
}
$arr1 = array ( 1, 2, 3 );
$arr2 = array ( 4, 5, 6 );
$n1 = sizeof( $arr1 ) ;
$n2 = sizeof( $arr2 );
findCart( $arr1 , $arr2 , $n1 , $n2 );
?>
|
Javascript
<script>
function findCart(arr1, arr2, n, n1)
{
for (let i = 0; i < n; i++)
for (let j = 0; j < n1; j++)
document.write( "{" + arr1[i]+ ", "
+ arr2[j]+ "}, " );
}
let arr1 = [ 1, 2, 3 ];
let arr2 = [4, 5, 6 ];
let n1 = arr1.length;
let n2 = arr2.length;
findCart(arr1, arr2, n1, n2);
</script>
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Output
{1, 4}, {1, 5}, {1, 6}, {2, 4}, {2, 5}, {2, 6}, {3, 4}, {3, 5}, {3, 6},
Time complexity: O(M*N) where M and N are size of given sets
Auxiliary space: O(1) because it is using constant space for variables
Practical Examples:
1) A set of playing cards is Cartesian product of a four element set to a set of 13 elements.
2) A two dimensional coordinate system is a Cartesian product of two sets of real numbers.
Reference:
https://en.wikipedia.org/wiki/Cartesian_product
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