Number of handshakes such that a person shakes hands only once

There are N number of persons in a party, find the total number of handshake such that a person can handshake only once.

Examples:

Input : 5
Output : 10

Input : 9
Output : 36

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

We can see a recursive nature in the problem.

// n-th person has (n-1) choices and after
// n-th person chooses a person, problem
// recurs for n-1.
handshake(n) = (n-1) + handshake(n-1)

// Base case
handshake(0) = 0

Below is implementation of above recursive formula.

C++

// Recursive CPP program to count total  
// number of  handshakes when a person 
// can shake hand with only one. 
#include <stdio.h> 
  
// function to find all possible handshakes 
int handshake(int n)  
{ 
    // when n becomes 0 that means all the  
    // persons have done handshake with other 
    if (n == 0)  
        return 0; 
    else
        return (n - 1) + handshake(n - 1);  
} 
  
int main() 
{ 
    int n = 9; 
    printf("%d", handshake(n)); 
    return 0; 
} 

Java

// Recursive Java program to  
// count total number of  
// handshakes when a person 
// can shake hand with only one. 
import java.io.*; 
  
class GFG  
{ 
  
// function to find all 
// possible handshakes 
static int handshake(int n)  
{ 
    // when n becomes 0 that  
    // means all the persons  
    // have done handshake 
    // with other 
    if (n == 0)  
        return 0; 
    else
        return (n - 1) + handshake(n - 1);  
} 
  
// Driver Code 
public static void main (String[] args)  
{ 
    int n = 9; 
    System.out.print(handshake(n)); 
} 
} 
  
// This code is contributed  
// by chandan_jnu 

Python3

# Recursive Python program  
# to count total number of 
# handshakes when a person 
# can shake hand with only one. 
  
# function to find all  
# possible handshakes 
def handshake(n):  
  
    # when n becomes 0 that means  
    # all the persons have done 
    # handshake with other 
    if (n == 0): 
        return 0
    else: 
        return (n - 1) + handshake(n - 1)  
  
# Driver Code 
n = 9
print(handshake(n)) 
  
# This code is contributed  
# by Shivi_Aggarwal 

C#

// Recursive C# program to  
// count total number of  
// handshakes when a person 
// can shake hand with only one. 
using System; 
  
class GFG  
{ 
  
// function to find all 
// possible handshakes 
static int handshake(int n)  
{ 
    // when n becomes 0 that  
    // means all the persons  
    // have done handshake 
    // with other 
    if (n == 0)  
        return 0; 
    else
        return (n - 1) + handshake(n - 1);  
} 
  
// Driver Code 
public static void Main (String []args)  
{ 
    int n = 9; 
    Console.WriteLine(handshake(n)); 
} 
} 
  
// This code is contributed  
// by Arnab Kundu 

PHP

<?php 
// Recursive PHP program to   
// count total number of 
// handshakes when a person 
// can shake hand with only one. 
  
// function to find all  
// possible handshakes 
function handshake($n)  
{ 
    // when n becomes 0 that means  
    // all the persons have done 
    // handshake with other 
    if ($n == 0)  
        return 0; 
    else
        return ($n - 1) + handshake($n - 1);  
} 
  
// Driver Code 
$n = 9; 
echo(handshake($n)); 
  
// This code is contributed  
// by Shivi_Aggarwal 
?> 

Output:

We can come up with a direct formula by expanding the recursion.

handshake(n) = (n-1) + handshake(n-1)
             = (n-1) + (n-2) + handshake(n-2)
             = (n-1) + (n-2) + .... 1 + 0
             = n * (n - 1)/2

C++

// Recursive CPP program to count total  
// number of  handshakes when a person 
// can shake hand with only one. 
#include <stdio.h> 
  
// function to find all possible handshakes 
int handshake(int n)  
{ 
   return n * (n - 1)/2; 
} 
  
int main() 
{ 
    int n = 9; 
    printf("%d", handshake(n)); 
    return 0; 
} 

Java

// Recursive Java program to  
// count total number of 
// handshakes when a person 
// can shake hand with only one. 
class GFG 
{ 
  
// function to find all  
// possible handshakes 
static int handshake(int n)  
{ 
    return n * (n - 1) / 2; 
} 
  
// Driver code 
public static void main(String args[]) 
{ 
    int n = 9; 
    System.out.println(handshake(n)); 
} 
} 
  
// This code is contributed  
// by Arnab Kundu 

Python3

# Recursive Python program  
# to count total number of 
# handshakes when a person 
# can shake hand with only one. 
  
# function to find all  
# possible handshakes 
def handshake(n):  
      
    return int(n * (n - 1) / 2) 
  
# Driver Code 
n = 9
print(handshake(n)) 
      
# This code is contributed  
# by Shivi_Aggarwal 

C#

// Recursive C# program to  
// count total number of  
// handshakes when a person  
// can shake hand with only one.  
using System; 
  
class GFG 
{ 
      
// function to find all  
// possible handshakes  
static int handshake(int n)  
{  
    return n * (n - 1) / 2;  
}  
  
// Driver code  
static public void Main () 
{ 
    int n = 9;  
    Console.WriteLine(handshake(n));  
}  
}  
  
// This code is contributed by Sachin 

PHP

<?php 
// Recursive PHP program to  
// count total number of  
// handshakes when a person 
// can shake hand with only one. 
  
// function to find all  
// possible handshakes 
function handshake($n)  
{ 
    return $n * ($n - 1) / 2; 
} 
  
// Driver Code 
$n = 9; 
echo(handshake($n)); 
  
// This code is contributed 
// by Shivi_Aggarwal 
?> 

Output:

My Personal Notes

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

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

PHP

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