Solving Josephus with linked lists

Question

I've been tryin for a while but i cant figure out how to make the program below take N as input and generate an M in order that the last soldier that dies is the 13th(N>13);

 int main()
 {
 int N, M;
 struct node { int player_id; struct node *next; };
 struct node *p, *q;
 int i, count;

 printf("Enter N (number of players): "); scanf("%d", &N);
 printf("Enter M (every M-th payer gets eliminated): "); scanf("%d", &M);

// Create circular linked list containing all the players:

p = q = malloc(sizeof(struct node));

p->player_id = 1;

for (i = 2; i <= N; ++i) {
    p->next = malloc(sizeof(struct node));
    p = p->next;
    p->player_id = i;
}

p->next = q;//Close the circular linkedlist by having the last node point to the 1st   

// Eliminate every M-th player as long as more than one player remains:

for (count = N; count > 1; --count) {
   for (i = 0; i < M - 1; ++i)

 p = p->next;       p->next = p->next->next;

  // Remove the eiminated player from the circular linked list.
     }    printf("Last player left standing is %d\n.", p->player_id);

   return 0;
  }

the result should be the same as this one(but I need it in linked lists cos i dont understand that one):>.

Answer 1

I didn't read all the code above, I think it can find the last item for given N and M

According to the original problem, 12<N<100. So, probably it can be solved in the given time limit simply with brute force.

• You read in N
• start a loop for finding m from 1
• in the loop:
  • run the algorithm, using the loop variable as m. If the last item is 13, return the loop variable.

EDIT: You don't have to work a lot. You just simply start a loop instead of reading M.

M=1;
while(1)
{
//your code goes here: 
//build up the linked list with N element
//eliminate the nodes, until only one remains
//instead of writing out the last number, you check it: if it equals 13 print M, if doesn't, increment `M` and continue the loop.
 if(p->player_id==13)
 {
   printf("The minimal M is: %d\n", M);
   break;
 }
 else
   M++;
}

If You want to do this for multiple N-s, You can put this loop into a function. In this case insted of printing M, the function should return it. The funny thing is: The linked list part is done by You. Maybe You should try simpler exercises first.

EDIT 2: HERE is my final answer, check the structure and the output, I hope it is understandable.

NOTE: I think if You really want to learn, You should do something like this instead of jumping in a not trivial problem:

• understand pointers
• understand structs
• understand linked lists
• implement insert/update/delete to the head/tail/certain position of a linked list
• solve Josephus problem by Yourself in 10 minutes