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

• Last Updated : 27 May, 2021

Given a positive intege 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) R holds for every element a € 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 € 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 irreflexivevoid countRelations(int N){    // Return the resultant count    cout << (power(2, N) - 2)                * power(2, N * N - N);} // Driver Codeint main(){    int N = 2;    countRelations(N);     return 0;}

Java

 // Java program for the above approachimport 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 irreflexivestatic void countRelations(int N){         // Return the resultant count    System.out.print((power(2, N) - 2) *                      power(2, N * N - N));} // Driver Codepublic static void main(String[] args){    int N = 2;         countRelations(N);}} // This code is contributed by susmitakundugoaldanga

Python3

 # Python program for the above approachmod = 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 irreflexivedef countRelations(N):     # Return the resultant count    print((power(2, N) - 2) * power(2, N * N - N)) # Driver CodeN = 2countRelations(N) # This code is contributed by abhinavjain194

C#

 // C# program for the above approachusing 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 irreflexivestatic void countRelations(int N){         // Return the resultant count    Console.Write((power(2, N) - 2) *                   power(2, N * N - N));} // Driver Codepublic 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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