Cartesian Product of Two Sets (original) (raw)

Last Updated : 29 Jul, 2024

Let A and B be two sets, Cartesian product**A × 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.

JavaScript

PHP

`

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