Josephus Problem in Algorithm Design

Josephus Problem

What is the Josephus Problem?

The Josephus Problem is a theoretical problem related to a certain elimination game. It involves a group of people standing in a circle, where every k-th person is eliminated until only one person remains. The challenge is to determine the position of the last person standing.

• n: The number of people in the circle.
• k: The number of steps to count before eliminating a person.
• The task is to find the position of the last remaining person after repeated elimination.

Josephus Problem: Recursive Solution

The Josephus problem can be solved recursively. The key observation is that once a person is eliminated, the problem size reduces, and the solution for n-1 people can be related to the solution for n people.

Recursive Formula:

$$ Josephus(n, k) = (Josephus(n-1, k) + k) \% n $$

Where:

• Josephus(n, k) gives the position of the survivor for n people and step size k.
• The + k accounts for the step size.
• The modulo operation % n ensures the circular nature of the problem.

Josephus Problem: Implementation in Code

Here is a recursive implementation of the Josephus problem in C++:

#include <iostream>
using namespace std;

class Solution {
public:
    int josephus(int n, int k) {
        if (n == 1) return 0;  // Base case: Only one person remains, position is 0 (0-based index)
        return (josephus(n - 1, k) + k) % n;
    }
};

int main() {
    Solution sol;
    int n = 5, k = 2; // Example: 5 people, every 2nd person is eliminated
    cout << "The survivor's position is: " << sol.josephus(n, k) + 1 << endl; // Output in 1-based index
    return 0;
}

Josephus Problem: Explanation of the Code {#josephus-code-explanation}

• Base Case: When there’s only one person left (n == 1), the survivor is the first person (position 0 in 0-based index).

• Recursive Call: For n > 1, the recursive formula is applied to find the survivor’s position for n-1 people, then adjusted for the step size k using the modulo operation.

• Output: The result is printed as josephus(n, k) + 1 to convert the 0-based result to a 1-based index.

Example Walkthrough {#josephus-example}

Let’s consider an example where n = 5 and k = 2:

Step 1: Find the result for n = 1

• When n = 1, the survivor is at position 0.

Step 2: Find the result for n = 2

• Recursively, we find the survivor for n = 1, which is 0.

• The formula:

  $$ (Josephus(1,2)+2) \% 2 = (0+2) \% 2 = 0 $$

So, the survivor for n = 2 is person 1.

Step 3: Find the result for n = 3

• Recursively, we find the survivor for n = 2, which is 0.

• The formula:

  $$ (Josephus(2,2)+2) \% 3 = (0+2) \% 3 = 2 $$

So, the survivor for n = 3 is person 3.

Step 4: Find the result for n = 4

• Recursively, we find the survivor for n = 3, which is 2.

• The formula:

  $$ (Josephus(3,2)+2) \% 4 = (2+2) \% 4 = 0 $$

So, the survivor for n = 4 is person 1.

Step 5: Find the result for n = 5

• Recursively, we find the survivor for n = 4, which is 0.

• The formula:

  $$ (Josephus(4,2)+2) \% 5 = (0+2) \% 5 = 2 $$

So, the survivor for n = 5 is person 3.

Thus, the last person standing for n = 5 and k = 2 is person 3 (using 1-based index).