# How to solve Josephus problem using pointer in C program

## Introduction

We will see how to solve Josephus problem using pointer in C program. So we are going to use C program as well as pointer to solve Josephus problem.

The Josephus problem (or Josephus permutation) is a theoretical problem related to a certain counting-out game. The problem is described as below.

People are standing in a circle waiting to be executed. Counting begins at a specified point in the circle and proceeds around the circle in a specified direction. After a specified number of people are skipped, the next person is executed.

The procedure is repeated with the remaining people, starting with the next person, going in the same direction and skipping the same number of people, until only one person remains, and is freed.

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## Implementation of Josephus Problem

The implementation of Josephus problem is given below using C program and pointer.

The `n` is the number of persons or people in the circle and `k` is the start position of execution.

``````#include <stdio.h>
#include <stdlib.h>

typedef struct person {
int position;
struct person *next;
}person;

int main() {
int w = findWinner(5, 3);
printf("Winner: %d", w);
w = findWinner(10, 3);
printf("\nWinner: %d", w);
w = findWinner(5, 2);
printf("\nWinner: %d", w);
w = findWinner(7, 3);
printf("\nWinner: %d", w);

return 0;
}

int findWinner(int n, int k) {
int i, counter;
person *p, *q, *r;

p = (person*) malloc(sizeof(person));
p->position = 1;
p->next = p;
q = p;

for(i=2; i<=n; i++) {
r = (person*) malloc(sizeof(person));
r->position = i;
r->next = p;
q->next = r;
q = r;
}

while(p->next != p) {
q = p;
r = q;
for(counter = 1; counter<k; counter++) {
r = q;
q = q->next;
}
r->next = q->next;
p = r->next;
q->next = NULL;
free(q);
q = NULL;
}

return p->position;
}``````

## Testing the Program

By executing the above C program you will get the following output:

``````Winner: 4
Winner: 4
Winner: 3
Winner: 4``````

## Ananlysis of the logic

Let’s take an example of there are 5 people in a circle and execution starts clockwise at position 3, so we will find out the winner of the game.

So there are five people – 1 2 3 4 5 and execution starts clockwise at position 3. Therefore if we kill 3 then are left with the list something like – 1 2 4 5.

Now again we have to kill the person at position 3 clockwise counting from the last killed person.

Now we need to execute the person at position 1. So executing the person at 1 we are left with the list 2 4 5. After executing another person we get list 2 4.

Now we are left with 2 persons in the list. So we have to calculate the position wisely to execute the person.

So finally the winner is 4.