# Number of Symmetric Relations on a Set

Given a number n, find out 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), (1, 2), (2, 1), (1, 2)}

Input  : n = 3
Output : 64
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A Relation ‘R’ on Set A is said be Symmetric if xRy then yRx for every x, y ∈ A
or if (x, y) ∈ R, then (y, x) ∈ 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 n ` `unsigned ``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 code ` `int` `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 n ` `def` `countSymmetric(n) : ` `    ``# Base case ` `    ``if` `(n ``=``=` `0``) : ` `        ``return` `1` `  `  `    ``# Return 2^(n(n + 1)/2) ` `    ``return` `(``1` `<< ((n ``*` `(n ``+` `1``))``/``/``2``)) ` ` `  `  `  `# Driver code ` ` `  `n ``=` `3` `print``(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. `

## PHP

 ` `

Output:

```64
```

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