In how many ways the ball will come back to the first boy after N turns
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: 6
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). There 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.
- 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).
- 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++
#include <bits/stdc++.h>
using namespace std;
int numSeq( int n)
{
return ( pow (3, n) + 3 * pow (-1, n)) / 4;
}
int main()
{
int N = 10;
printf ( "%d" , numSeq(N));
return 0;
}
|
Java
import java.util.*;
class GFG
{
static int numSeq( int n)
{
return ( int ) ((Math.pow( 3 , n) + 3 *
Math.pow(- 1 , n)) / 4 );
}
public static void main(String[] args)
{
int N = 10 ;
System.out.printf( "%d" , numSeq(N));
}
}
|
Python3
def numSeq(n):
return ( pow ( 3 , n) + 3 * pow ( - 1 , n)) / / 4
N = 10
print (numSeq(N))
|
C#
using System;
class GFG
{
static int numSeq( int n)
{
return ( int ) ((Math.Pow(3, n) + 3 *
Math.Pow(-1, n)) / 4);
}
public static void Main()
{
int N = 10;
Console.WriteLine(numSeq(N));
}
}
|
Javascript
<script>
function numSeq(n)
{
return Math.floor((Math.pow(3, n) + 3 * Math.pow(-1, n)) / 4);
}
let N = 10;
document.write(numSeq(N));
</script>
|
Time Complexity: O(log n)
Auxiliary Space: O(1)
Last Updated :
20 Aug, 2022
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