一组对称关系的数量

给定数字n，找出前n个自然数{1，2，..n}集合上的对称关系数。
例子：

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

如果xRy则yRx对每个x，y∈A，则说说集合A上的关系’R’是对称的
或者如果(x，y)∈R，那么每x，y(y，x)∈R

Total number of symmetric relations is 2^n(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} = 2^n(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.

Javascript

 
// JavaScript program to count total symmetric
// relations on a set of natural numbers.
 
// function find the square of n
    function countSymmetric(n)
    {
        // Base case
        if (n == 0)
            return 1;
       
    // Return 2^(n(n + 1)/2)
    return 1 << ((n * (n + 1)) / 2);
    }
   
 
// Driver Code
 
        let n = 3;
        document.write(countSymmetric(n));
        

输出：

64