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In how many ways the ball will come back to the first boy after N turns

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  • Last Updated : 03 Mar, 2022

Four boys are playing a game with a ball. In each turn, the player (who has the ball currently) passes it to a different player randomly. Bob always starts the game. The task is to find in how many ways the ball will come back to Bob after N passes.
Examples: 
 

Input: N = 3 
Output:
Here are all the possible ways: 
Bob -> Boy1 -> Boy2 -> Bob 
Bob -> Boy1 -> Boy3 -> Bob 
Bob -> Boy2 -> Boy1 -> Bob 
Bob -> Boy2 -> Boy3 -> Bob 
Bob -> Boy3 -> Boy1 -> Bob 
Bob -> Boy3 -> Boy2 -> Bob
Input: N = 10 
Output: 14763 
 

 

Approach: Let the number of sequences that get back to Bob after N passes are P(N). The are two cases, either pass N – 2 is to Bob or it is not. Note that Bob can’t have the ball at (N – 1)th pass because then he won’t have the ball at the Nth pass.
 

  1. Case 1: If pass N – 2 is to Bob then the pass N – 1 can be to any of the other 3 boys. Thus, the number of such sequences is 3 * P(N – 2).
  2. Case 2: If pass N – 2 is not to Bob then pass N – 1 is to one of the 2 boys other than Bob and the one who got the ball in hand. So, substitute Bob for the receiver of pass N – 1, and get a unique N – 1 sequence. So, the number of such sequences are 2 * P(N – 1).

Hence the recurrence relation will be P(N) = 2 * P(N – 1) + 3 * P(N – 2) where P(0) = 1 and P(1) = 0.
After solving the recurrence relation, P(N) = (3N + 3 * (-1N)) / 4
Below is the implementation of the above approach: 
 

C++




// Function to return the number of
// sequences that get back to Bob
#include <bits/stdc++.h>
using namespace std;
 
int numSeq(int n)
{
    return (pow(3, n) + 3 * pow(-1, n)) / 4;
}
 
// Driver code
int main()
{
    int N = 10;
    printf("%d", numSeq(N));
    return 0;
}
 
// This code is contributed by Mohit kumar

Java




// Function to return the number of
// sequences that get back to Bob
import java.util.*;
 
class GFG
{
 
static int numSeq(int n)
{
    return (int) ((Math.pow(3, n) + 3 *
                    Math.pow(-1, n)) / 4);
}
 
// Driver code
public static void main(String[] args)
{
    int N = 10;
    System.out.printf("%d", numSeq(N));
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Function to return the number of
# sequences that get back to Bob
def numSeq(n):
    return (pow(3, n) + 3 * pow(-1, n))//4
     
# Driver code
N = 10
print(numSeq(N))

C#




// C# implementation of the above approach
using System;
 
// Function to return the number of
// sequences that get back to Bob
class GFG
{
 
    static int numSeq(int n)
    {
        return (int) ((Math.Pow(3, n) + 3 *
                       Math.Pow(-1, n)) / 4);
    }
     
    // Driver code
    public static void Main()
    {
        int N = 10;
        Console.WriteLine(numSeq(N));
    }
}
 
// This code is contributed by AnkitRai01

Javascript




<script>
 
// Function to return the number of
// sequences that get back to Bob
function numSeq(n)
{
    return  Math.floor((Math.pow(3, n) + 3 * Math.pow(-1, n)) / 4);
}
 
// Driver code
 
    let N = 10;
    document.write(numSeq(N));
 
// This code is contributed by Surbhi Tyagi.
 
</script>
Output: 
14763

 

Time Complexity: O(log n)

Auxiliary Space: O(1)


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