# Number of Symmetric Relations on a Set

• Difficulty Level : Medium
• Last Updated : 19 May, 2022

Given a number n, find out the number of Symmetric Relations on a set of first n natural numbers {1, 2, ..n}.

Examples:

Input  : n = 2
Output : 8

Given set is {1, 2}. Below are all symmetric relation.
{}
{(1, 1)},
{(2, 2)},
{(1, 1), (2, 2)},
{(1, 2), (2, 1)}
{(1, 1), (1, 2), (2, 1)},
{(2, 2), (1, 2), (2, 1)},
{(1, 1), (2, 2), (2, 1), (1, 2)}

Input  : n = 3
Output : 64

A Relation ‘R’ on Set A is said be Symmetric if xRy then yRx for every x, y âˆˆ A
or if (x, y) âˆˆ R, then (y, x) âˆˆ R for every x, y?A

Total number of symmetric relations is 2n(n+1)/2.
How does this formula work?
A relation R is symmetric if the value of every cell (i, j) is same as that cell (j, i). The diagonals can have any value.

There are n diagonal values, total possible combination of diagonal values = 2n
There are n2 – n non-diagonal values. We can only choose different value for half of them, because when we choose a value for cell (i, j), cell (j, i) gets same value.
So combination of non-diagonal values = 2(n2 – n)/2
Overall combination = 2n * 2(n2 – n)/2 = 2n(n+1)/2

## C++

 // C++ program to count total symmetric relations// on a set of natural numbers.#include  // function find the square of nunsigned int countSymmetric(unsigned int n){    // Base case    if (n == 0)        return 1;    // Return 2^(n(n + 1)/2)   return 1 << ((n * (n + 1))/2);} // Driver codeint main(){    unsigned int n = 3;     printf("%u", countSymmetric(n));    return 0;}

## Java

 // Java program to count total symmetric// relations on a set of natural numbers.import java.io.*;import java.util.*; class GFG {     // function find the square of n    static int countSymmetric(int n)    {        // Base case        if (n == 0)            return 1;         // Return 2^(n(n + 1)/2)    return 1 << ((n * (n + 1)) / 2);    }     // Driver code    public static void main (String[] args)    {        int n = 3;        System.out.println(countSymmetric(n));    }}  // This code is contributed by Nikita Tiwari.

## Python3

 # Python 3 program to count# total symmetric relations# on a set of natural numbers. # function find the square of ndef countSymmetric(n) :    # Base case    if (n == 0) :        return 1      # Return 2^(n(n + 1)/2)    return (1 << ((n * (n + 1))//2))   # Driver code n = 3print(countSymmetric(n)) # This code is contributed# by Nikita Tiwari.

## C#

 // C# program to count total symmetric// relations on a set of natural numbers.using System; class GFG {     // function find the square of n    static int countSymmetric(int n)    {        // Base case        if (n == 0)            return 1;         // Return 2^(n(n + 1)/2)    return 1 << ((n * (n + 1)) / 2);    }     // Driver code    public static void Main ()    {        int n = 3;        Console.WriteLine(countSymmetric(n));    }}  // This code is contributed by vt_m.



## Javascript



Output

64

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