Skip to content
Related Articles
Open in App
Not now

Related Articles

Cartesian Product of Two Sets

Improve Article
Save Article
  • Difficulty Level : Basic
  • Last Updated : 19 Oct, 2022
Improve Article
Save Article

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




// C++ Program to find the Cartesian Product of Two Sets
#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 }; // first set
    int arr2[] = { 4, 5, 6 }; // second set
    int n1 = sizeof(arr1) / sizeof(arr1[0]);
    int n2 = sizeof(arr2) / sizeof(arr2[0]);
    findCart(arr1, arr2, n1, n2);
    return 0;
}

Java




// Java Program to find the
// Cartesian Product of Two Sets
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] + "}, ");
    }
    // Driver code
    public static void main(String[] args)
    {
 
        // first set
        int arr1[] = { 1, 2, 3 };
 
        // second set
        int arr2[] = { 4, 5, 6 };
 
        int n1 = arr1.length;
        int n2 = arr2.length;
        findCart(arr1, arr2, n1, n2);
    }
}
 
// This code is contributed by Nikita Tiwari.

Python3




# Python3 Program to find the
# Cartesian Product of Two Sets
 
 
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="")
 
 
# Driver code
arr1 = [1, 2, 3# first set
arr2 = [4, 5, 6# second set
 
n1 = len(arr1)  # sizeof(arr1[0])
n2 = len(arr2)  # sizeof(arr2[0]);
 
findCart(arr1, arr2, n1, n2)
 
# This code is contributed
# by Smitha Dinesh Semwal

C#




// C# Program to find the
// Cartesian Product of Two Sets
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]
                              + "}, ");
    }
 
    // Driver code
    public static void Main()
    {
 
        // first set
        int[] arr1 = { 1, 2, 3 };
 
        // second set
        int[] arr2 = { 4, 5, 6 };
 
        int n1 = arr1.Length;
        int n2 = arr2.Length;
 
        findCart(arr1, arr2, n1, n2);
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP Program to find the
// Cartesian Product of Two Sets
 
function findCart($arr1, $arr2, $n, $n1)
{
    for ($i = 0; $i < $n; $i++)
        for ( $j = 0; $j < $n1; $j++)
            echo "{", $arr1[$i] ," , ",
                      $arr2[$j], "}",",";
}
 
// Driver Code
 
// first set
$arr1 = array ( 1, 2, 3 );
 
// second set
$arr2 = array ( 4, 5, 6 );
$n1 = sizeof($arr1) ;
$n2 = sizeof($arr2);
findCart($arr1, $arr2, $n1, $n2);
 
// This code is contributed by m_kit.
?>

Javascript




<script>
 
// JavaScript Program to find the
// Cartesian Product of Two Set
 
    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]+"}, ");
    }
   
// Driver Code
 
        // first set
        let arr1 = [ 1, 2, 3 ];
           
        // second set
        let arr2 = [4, 5, 6 ];
           
        let n1 = arr1.length;
        let n2 = arr2.length;
        findCart(arr1, arr2, n1, n2);
        
       // This code is contributed by chinmoy1997pal.
</script>

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
 


My Personal Notes arrow_drop_up
Related Articles

Start Your Coding Journey Now!