# Number of relations that are neither Reflexive nor Irreflexive on a Set

• Last Updated : 13 Oct, 2022

Given a positive integer N, the task is to find the number of relations that are neither reflexive nor irreflexive on a set of first N natural numbers. Since the count of relations can be very large, print it to modulo 109 + 7.

A relation R on a set A is called reflexive, if no (a, a) â‚¬ R holds for every element a â‚¬ A.
For Example: If set A = {a, b} then R = {(a, b), (b, a)} is irreflexive relation.

Examples:

Input: N = 2
Output: 8
Explanation: Considering the set {1, 2}, the total possible relations that are neither reflexive nor irreflexive are:

1. {(1, 1)}
2. {(1, 1), (1, 2)}
3. {(1, 1), (2, 1)}
4. {(1, 1), (1, 2), (2, 1)}
5. {(2, 2)}
6. {(2, 2), (2, 1)}
7. {(2, 2), (1, 2)}
8. {(2, 2), (2, 1), (1, 2)}

Therefore, the total count is 8.

Input: N = 3
Output: 384

Approach: The given problem can be solved based on the following observations:

• A relation R on a set A is a subset of the cartesian product of a set, i.e., A * A with N2 elements.
• A relation will be non-reflexive if it doesn’t contain at least one pair of (x, x) and relation will be non-irreflexive if it contains at least one pair of (x, x) where x â‚¬ R.
• It can be concluded that the relation will be non-reflexive and non-irreflexive if it contains at least one pair of (x, x) and at most (N – 1) pairs of (x, x).
• Among N pairs of (x, x), the total number of possibilities of choosing any number of pairs except 0 and N – 1 is (2N – 2). For the remaining (N2 – N) elements, each element has two choices i.e., either to include or exclude it in the subset.

From the above observations, the total number of relations that are neither reflexive nor irreflexive on a set of first N natural numbers is given by .

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `const` `int` `mod = 1000000007;` `// Function to calculate x^y``// modulo 10^9 + 7 in O(log y)``int` `power(``long` `long` `x, unsigned ``int` `y)``{``    ``// Stores the result of (x^y)``    ``int` `res = 1;` `    ``// Update x, if it exceeds mod``    ``x = x % mod;` `    ``// If x is divisible by mod``    ``if` `(x == 0)``        ``return` `0;` `    ``while` `(y > 0) {` `        ``// If y is odd, then``        ``// multiply x with res``        ``if` `(y & 1)``            ``res = (res * x) % mod;` `        ``// Divide y by 2``        ``y = y >> 1;` `        ``// Update the value of x``        ``x = (x * x) % mod;``    ``}` `    ``// Return the value of x^y``    ``return` `res;``}` `// Function to count the number``// of relations that are neither``// reflexive nor irreflexive``void` `countRelations(``int` `N)``{``    ``// Return the resultant count``    ``cout << (power(2, N) - 2)``                ``* power(2, N * N - N);``}` `// Driver Code``int` `main()``{``    ``int` `N = 2;``    ``countRelations(N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.io.*;``import` `java.lang.*;``import` `java.util.*;` `class` `GFG{` `static` `int` `mod = ``1000000007``;` `// Function to calculate x^y``// modulo 10^9 + 7 in O(log y)``static` `int` `power(``int` `x, ``int` `y)``{``    ` `    ``// Stores the result of (x^y)``    ``int` `res = ``1``;` `    ``// Update x, if it exceeds mod``    ``x = x % mod;` `    ``// If x is divisible by mod``    ``if` `(x == ``0``)``        ``return` `0``;` `    ``while` `(y > ``0``)``    ``{``        ` `        ``// If y is odd, then``        ``// multiply x with res``        ``if` `((y & ``1``) != ``0``)``            ``res = (res * x) % mod;` `        ``// Divide y by 2``        ``y = y >> ``1``;` `        ``// Update the value of x``        ``x = (x * x) % mod;``    ``}` `    ``// Return the value of x^y``    ``return` `res;``}` `// Function to count the number``// of relations that are neither``// reflexive nor irreflexive``static` `void` `countRelations(``int` `N)``{``    ` `    ``// Return the resultant count``    ``System.out.print((power(``2``, N) - ``2``) *``                      ``power(``2``, N * N - N));``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `N = ``2``;``    ` `    ``countRelations(N);``}``}` `// This code is contributed by susmitakundugoaldanga`

## Python3

 `# Python program for the above approach``mod ``=` `1000000007` `# Function to calculate x^y``# modulo 10^9 + 7 in O(log y)``def` `power(x, y):` `    ``# Stores the result of (x^y)``    ``res ``=` `1` `    ``# Update x, if it exceeds mod``    ``x ``=` `x ``%` `mod` `    ``# If x is divisible by mod``    ``if``(x ``=``=` `0``):``        ``return` `0` `    ``while``(y > ``0``):` `        ``# If y is odd, then``        ``# multiply x with res``        ``if` `(y ``%` `2` `=``=` `1``):``            ``res ``=` `(res ``*` `x) ``%` `mod` `        ``# Divide y by 2``        ``y ``=` `y >> ``1` `        ``# Update the value of x``        ``x ``=` `(x ``*` `x) ``%` `mod` `    ``# Return the value of x^y``    ``return` `res` `# Function to count the number``# of relations that are neither``# reflexive nor irreflexive``def` `countRelations(N):` `    ``# Return the resultant count``    ``print``((power(``2``, N) ``-` `2``) ``*` `power(``2``, N ``*` `N ``-` `N))` `# Driver Code``N ``=` `2``countRelations(N)` `# This code is contributed by abhinavjain194`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{` `static` `int` `mod = 1000000007;` `// Function to calculate x^y``// modulo 10^9 + 7 in O(log y)``static` `int` `power(``int` `x, ``int` `y)``{``    ` `    ``// Stores the result of (x^y)``    ``int` `res = 1;` `    ``// Update x, if it exceeds mod``    ``x = x % mod;` `    ``// If x is divisible by mod``    ``if` `(x == 0)``        ``return` `0;` `    ``while` `(y > 0)``    ``{``        ` `        ``// If y is odd, then``        ``// multiply x with res``        ``if` `((y & 1) != 0)``            ``res = (res * x) % mod;` `        ``// Divide y by 2``        ``y = y >> 1;` `        ``// Update the value of x``        ``x = (x * x) % mod;``    ``}` `    ``// Return the value of x^y``    ``return` `res;``}` `// Function to count the number``// of relations that are neither``// reflexive nor irreflexive``static` `void` `countRelations(``int` `N)``{``    ` `    ``// Return the resultant count``    ``Console.Write((power(2, N) - 2) *``                   ``power(2, N * N - N));``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `N = 2;``    ``countRelations(N);``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output:

`8`

Time Complexity: O(log N)
Auxiliary Space: O(1)

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