Given a positive integer N, the task is to find the number of Asymmetric Relations in a set of N elements. Since the number of relations can be very large, print it modulo 109+7.
A relation R on a set A is called Asymmetric if and only if x R y exists, then y
Rx for every (x, y) € A.
For Example: If set A = {a, b}, then R = {(a, b)} is asymmetric relation.
Examples:
Input: N = 2
Output: 3
Explanation: Considering the set {1, 2}, the total number of possible asymmetric relations are {{}, {(1, 2)}, {(2, 1)}}.Input: N = 5
Output: 59049
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.
- There are total N pairs of type (x, x) that are present in the Cartesian product, where any of (x, x) should not be included in the subset.
- Now, one is left with (N2 – N) elements of the Cartesian product.
- To satisfy the property of asymmetric relation, one has three possibilities of either to include only of type (x, y) or only of type (y, x) or none from a single group into the subset.
- Hence, the total number of possible asymmetric relations is equal to 3 (N2 – N) / 2.
Therefore, the idea is to print the value of 3(N2 – N)/2 modulo 109 + 7 as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <iostream>using namespace std;const int mod = 1000000007;// Function to calculate// x^y modulo (10^9 + 7)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 result 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 final // value of x ^ y return res;}// Function to count the number of// asymmetric relations in a set// consisting of N elementsint asymmetricRelation(int N){ // Return the resultant count return power(3, (N * N - N) / 2);}// Driver Codeint main(){ int N = 2; cout << asymmetricRelation(N); return 0;} |
Java
// Java program for the above approachimport java.io.*;class GFG{ final static int mod = 1000000007;// Function to calculate// x^y modulo (10^9 + 7)public 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 result 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 final // value of x ^ y return res;} // Function to count the number of// asymmetric relations in a set// consisting of N elementspublic static int asymmetricRelation(int N){ // Return the resultant count return power(3, (N * N - N) / 2);} // Driver codepublic static void main (String[] args){ int N = 2; System.out.print(asymmetricRelation(N));}}// This code is contributed by user_qa7r |
Python3
# Python3 program for the above approachmod = 1000000007# Function to calculate# x^y modulo (10^9 + 7)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 result 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 final # value of x ^ y return res# Function to count the number of# asymmetric relations in a set# consisting of N elementsdef asymmetricRelation(N): # Return the resultant count return power(3, (N * N - N) // 2)# Driver Codeif __name__ == '__main__': N = 2 print(asymmetricRelation(N))# This code is contributed by SURENDRA_GANGWAR |
C#
// C# program for the above approachusing System;class GFG{ const int mod = 1000000007;// Function to calculate// x^y modulo (10^9 + 7)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 result 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 final // value of x ^ y return res;}// Function to count the number of// asymmetric relations in a set// consisting of N elementsstatic int asymmetricRelation(int N){ // Return the resultant count return power(3, (N * N - N) / 2);}// Driver Codepublic static void Main(string[] args){ int N = 2; Console.WriteLine(asymmetricRelation(N));}}// This code is contributed by ukasp |
Javascript
<script>// Javascript program for the above approachvar mod = 1000000007;// Function to calculate// x^y modulo (10^9 + 7)function power(x, y){ // Stores the result of x^y var 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 result 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 final // value of x ^ y return res;}// Function to count the number of// asymmetric relations in a set// consisting of N elementsfunction asymmetricRelation( N){ // Return the resultant count return power(3, (N * N - N) / 2);}// Driver Codevar N = 2;document.write( asymmetricRelation(N));</script> |
Output:
3
Time Complexity: O(N*log N)
Auxiliary Space: O(1)
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