Josephus problem | Set 1 (A O(n) Solution)

In computer science and mathematics, the Josephus Problem (or Josephus permutation) is a theoretical problem. Following is the problem statement:

There are n people standing in a circle waiting to be executed. The counting out begins at some point in the circle and proceeds around the circle in a fixed direction. In each step, a certain number of people are skipped and the next person is executed. The elimination proceeds around the circle (which is becoming smaller and smaller as the executed people are removed), until only the last person remains, who is given freedom. Given the total number of persons n and a number k which indicates that k-1 persons are skipped and kth person is killed in circle. The task is to choose the place in the initial circle so that you are the last one remaining and so survive.

For example, if n = 5 and k = 2, then the safe position is 3. Firstly, the person at position 2 is killed, then person at position 4 is killed, then person at position 1 is killed. Finally, the person at position 5 is killed. So the person at position 3 survives.
If n = 7 and k = 3, then the safe position is 4. The persons at positions 3, 6, 2, 7, 5, 1 are killed in order, and person at position 4 survives.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

The problem has following recursive structure.

  josephus(n, k) = (josephus(n - 1, k) + k-1) % n + 1
  josephus(1, k) = 1

After the first person (kth from beginning) is killed, n-1 persons are left. So we call josephus(n – 1, k) to get the position with n-1 persons. But the position returned by josephus(n – 1, k) will consider the position starting from k%n + 1. So, we must make adjustments to the position returned by josephus(n – 1, k).

Following is simple recursive implementation of the Josephus problem. The implementation simply follows the recursive structure mentioned above.

C++

#include <iostream> 
using namespace std; 
  
int josephus(int n, int k) 
{ 
    if (n == 1) 
        return 1; 
    else
        /* The position returned by josephus(n - 1, k) 
        is adjusted because the recursive call  
        josephus(n - 1, k) considers the  
        original position k % n + 1 as position 1 */
        return (josephus(n - 1, k) + k-1) % n + 1; 
} 
  
// Driver Program to test above function 
int main() 
{ 
    int n = 14; 
    int k = 2; 
    cout << "The chosen place is " << josephus(n, k); 
    return 0; 
} 
  
// This code is contributed by shubhamsingh10 

C

#include <stdio.h> 
  
int josephus(int n, int k) 
{ 
  if (n == 1) 
    return 1; 
  else
    /* The position returned by josephus(n - 1, k) is adjusted because the 
       recursive call josephus(n - 1, k) considers the original position  
       k%n + 1 as position 1 */
    return (josephus(n - 1, k) + k-1) % n + 1; 
} 
  
// Driver Program to test above function 
int main() 
{ 
  int n = 14; 
  int k = 2; 
  printf("The chosen place is %d", josephus(n, k)); 
  return 0; 
} 

Java

// Java code for Josephus Problem 
import java.io.*; 
  
class GFG { 
  
static int josephus(int n, int k) 
{ 
if (n == 1) 
    return 1; 
else
    /* The position returned by josephus(n - 1, k)  
    is adjusted because the recursive call  
    josephus(n - 1, k) considers the original  
    position k%n + 1 as position 1 */
    return (josephus(n - 1, k) + k-1) % n + 1; 
} 
  
// Driver Program to test above function 
public static void main(String[] args) 
{ 
int n = 14; 
int k = 2; 
System.out.println("The chosen place is " + josephus(n, k)); 
} 
} 
  
// This code is contributed by Prerna Saini 

Python3

# Python code for Josephus Problem 
  
def josephus(n, k): 
  
      if (n == 1): 
          return 1
      else: 
      
      
          # The position returned by  
          # josephus(n - 1, k) is adjusted 
          # because the recursive call 
          # josephus(n - 1, k) considers 
          # the original position  
          # k%n + 1 as position 1  
          return (josephus(n - 1, k) + k-1) % n + 1
  
# Driver Program to test above function 
  
n = 14
k = 2
  
print("The chosen place is ", josephus(n, k)) 
  
# This code is contributed by 
# Sumit Sadhakar 

C#

// C# code for Josephus Problem 
using System; 
  
class GFG { 
      
    static int josephus(int n, int k) 
    { 
        if (n == 1) 
            return 1; 
        else
            /* The position returned  
            by josephus(n - 1, k) is 
            adjusted because the 
            recursive call josephus(n 
            - 1, k) considers the  
            original position k%n + 1 
            as position 1 */
            return (josephus(n - 1, k) 
                       + k-1) % n + 1; 
    } 
      
    // Driver Program to test above 
    // function 
    public static void Main() 
    { 
        int n = 14; 
        int k = 2; 
        Console.WriteLine("The chosen "
        + "place is " + josephus(n, k)); 
    } 
} 
  
// This code is contributed by anuj_67. 

PHP

<?php 
// PHP code for  
// Josephus Problem 
  
function josephus($n, $k) 
{ 
    if ($n == 1) 
        return 1; 
    else
        /* The position returned by  
           josephus(n - 1, k) is  
           adjusted because the 
           recursive call josephus 
           (n - 1, k) considers the  
           original position k%n + 1  
           as position 1 */
        return (josephus($n - 1, $k) +  
                    $k - 1) % $n + 1; 
} 
  
    // Driver Code 
    $n = 14; 
    $k = 2; 
    echo "The chosen place is ", josephus($n, $k); 
  
// This code is contributed by ajit. 
?> 

Output:

The chosen place is 13

Time Complexity: O(n)

Josephus problem | Set 2 (A Simple Solution when k = 2)

Source:
http://en.wikipedia.org/wiki/Josephus_problem

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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My Personal Notes

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

C

Java

Python3

C#

PHP

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