Question

Need different C code then what has been posted here

Problem

In this assignment, you have to simulate the Josephus problem. There are n number of prisoners 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.

Example

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.

Input: n and k

Output:

The position number who will survive.

Requirement:

You must use the concept of circular queue while inserting and it should be implemented using circular doubly linked list. You don’t need to use front and rear as you are using linked list. Root and last node automatically handle that.

Hint: After getting the value of n, generate n numbers (1, 2, 3., …, n) and insert into the doubly circular linked list. Then start deleting nodes based on the value of k until the list has only one node remaining.

Answer 1

/* this code is just for your reference and it is working on Circulary linked list concept

*/

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

struct node // create the linked list
{
int num;
struct node *next;
struct node *prev;
};

void create(struct node **); // it will create the linked list
int survivor(struct node **, int); // it will get the index of survivor

int main()
{
struct node *head = NULL;
int survive, skip;

create(&head); // go to the creation of the linked list
printf("Enter the number of persons to be skipped: "); // the number of persons(k) which we have to skip
scanf("%d", &skip);
survive = survivor(&head, skip); // will return the index of survivour
printf("The person to survive is : %d\n", survive); // print out the name
free(head); // free all memory whatever we store in malloc

return 0;
}

int survivor(struct node **head, int k)
{
struct node *p, *q;
int i;

q = p = *head;
while (p->next != p)
{
for (i = 0; i < k - 1; i++) // process till (k-1) person
{
q = p;
p = p->next;
}
q->next = p->next;   
free(p); // free or execute the k person
p = q->next; // linked list will store next address
}
*head = p;

return (p->num);
}

void create (struct node **head)
{
struct node *temp, *rear;
int n, k;
printf("Enter a number of prisoners : "); // number of prisoners as n
scanf("%d", &n);
for(int i=1;i<=n;i++){ // run for loop till n time to create the linked list of n persons
temp = (struct node *)malloc(sizeof(struct node));
temp->num = i;
temp->next = NULL;
if (*head == NULL)
{
*head = temp;
}
else
{
rear->next = temp;
}
rear = temp;
}
rear->next = *head;
}