Dining Philosophers Problem

The Dining Philosophers Problem is a classic problem in computer science and synchronization theory, often used to illustrate synchronization issues and techniques for resolving them. It was formulated by Edsger Dijkstra in 1965.

Here's the scenario with five philosophers (P0 to P4):

  1. Five philosophers sit around a circular table.

  2. Each philosopher thinks and occasionally needs to eat.

  3. Between each pair of philosophers, there is a single chopstick (a total of five chopsticks).

  4. In order to eat, a philosopher must use two chopsticks - the one to their left and the one to their right.

  5. A philosopher can only pick up one chopstick at a time and cannot pick up a chopstick already in use by another philosopher.

  6. After eating, the philosopher puts down both chopsticks and starts thinking again.

The problem illustrates the challenges of resource allocation and avoiding deadlock, where no progress is possible (e.g., if each philosopher picks up the chopstick on their left, then no one can pick up their second chopstick).

Here's a simple pseudo-code to solve this problem:

semaphore chopsticks[5]  # One semaphore per chopstick, initially available

def philosopher(id):
    while True:
        think()  # Philosopher is thinking
        wait(chopsticks[id])  # Pick up left chopstick
        wait(chopsticks[(id + 1) % 5])  # Pick up right chopstick
        eat()  # Philosopher is eating
        signal(chopsticks[id])  # Put down left chopstick
        signal(chopsticks[(id + 1) % 5])  # Put down right chopstick

for i in range(5):
    start_thread(philosopher, i)
  • wait(semaphore) is a function that decrements the semaphore. If the semaphore value is zero, the philosopher must wait.

  • signal(semaphore) increments the semaphore, potentially waking up other waiting philosophers.

  • Each philosopher is represented as a thread running the philosopher function.

  • chopsticks is an array of semaphores, one for each chopstick.

Deadlock Scenario

Deadlock can occur if each philosopher grabs the chopstick on their left simultaneously. In this case, each philosopher is waiting for the chopstick to their right, which is held by another philosopher. Since no philosopher can grab both chopsticks, no one can eat, and the system is in a deadlock.

To demonstrate this:

  1. Each philosopher executes wait(chopsticks[id]) and successfully acquires their left chopstick.

  2. Now, every philosopher tries to execute wait(chopsticks[(id + 1) % 5]) to acquire their right chopstick.

  3. Since all right chopsticks are already held by the next philosopher, all philosophers are now waiting.

  4. No philosopher can proceed to eat() or signal(), leading to a deadlock.

Avoiding Deadlock

One common solution to avoid deadlock is to ensure that not all philosophers can pick up chopsticks simultaneously. For example, you can enforce that at most four philosophers can try to acquire chopsticks at any time. This can be done by introducing an additional semaphore to control the number of philosophers allowed to pick up chopsticks. Alternatively, you can modify the protocol for acquiring chopsticks, such as having one philosopher pick up the right chopstick first, while others pick up the left first.