Dining Philosophers Problem

It is a simple representation of the need to allocate several resources among several processes in a deadlock-free and starvation-free manner.

Problem Statement

Five philosophers sit around a circular table. Between each pair is a fork...

Solution to Dining-Philosophers Problem

- One simple solution is to represent each chopstick with a semaphore.

- A philosopher tries to grab a chopstick by executing a wait() operation on that semaphore.

- She releases her chopsticks by executing the signal() operation on the appropriate semaphores.

- Thus, the shared data are semaphore chopstick[5]; where all the elements of chopstick are initialized to 1.

Interactive Simulation

Click the button below to start the simulation and observe how philosophers avoid deadlock.

Dining Philosophers Code

Select a language to view the implementation:

#include <stdio.h>
#include <pthread.h>
#include <semaphore.h>

#define N 5
sem_t forks[N];

void* philosopher(void* num) {
    int id = *(int*)num;
    while (1) {
        printf("Philosopher %d is thinking\n", id);
        sem_wait(&forks[id]);
        sem_wait(&forks[(id + 1) % N]);
        printf("Philosopher %d is eating\n", id);
        sem_post(&forks[id]);
        sem_post(&forks[(id + 1) % N]);
    }
    return NULL;
}

int main() {
    pthread_t philosophers[N];
    int ids[N];
    for (int i = 0; i < N; i++) {
        sem_init(&forks[i], 0, 1);
        ids[i] = i;
    }
    for (int i = 0; i < N; i++) {
        pthread_create(&philosophers[i], NULL, philosopher, &ids[i]);
    }
    for (int i = 0; i < N; i++) {
        pthread_join(philosophers[i], NULL);
    }
    return 0;
}