Computing Thoughts

Bruce Eckel's Programming Blog

Dec 29, 2016 - 8 minute read

Dining Philosophers in Java 8

Because tasks can become blocked, it’s possible for one task to get stuck waiting for another task, which in turn waits for another task, and so on, until the chain leads back to a task waiting on the first one. You get a continuous loop of tasks waiting on each other, and no one can move. This is called deadlock.1

If you try running a program and it deadlocks right away, you can immediately track down the bug. The real problem is when your program seems to be working fine but has the hidden potential to deadlock. Here, you might not get any indication that deadlocking is possible, so the flaw is latent in your program until it unexpectedly happens—typically to a customer (in a way almost certainly difficult to reproduce). Thus, preventing deadlock through careful program design is a critical part of developing concurrent systems.

The Dining Philosophers problem, invented by Edsger Dijkstra, is the classic demonstration of deadlock. The basic description specifies five philosophers (the example shown here allows any number). These philosophers spend part of their time thinking and part of their time eating. While they are thinking, they don’t need any shared resources, but they eat using a limited number of utensils. In the original problem description, the utensils are forks, and two forks are required to get spaghetti from a bowl in the middle of the table. A more convincing version uses chopsticks; clearly, each philosopher requires two chopsticks to eat.

A difficulty is introduced: As philosophers, they have very little money, so they can only afford five chopsticks (more generally, the same number of chopsticks as philosophers). These are spaced around the table between them. When a philosopher wants to eat, that philosopher must pick up the chopstick to the left and the one to the right. If the philosopher on either side is using a desired chopstick, our philosopher must wait until the necessary chopsticks become available.

A StickHolder class manages a single Chopstick by keeping it in a BlockingQueue of size one. A BlockingQueue is a collection, designed to be safely used in concurrent programs, that blocks (waits) if you call take() and the queue is empty. Once a new element is placed in the queue, the block is released and that value is returned:

// concurrent/
import java.util.concurrent.*;

public class StickHolder {
  private static class Chopstick {}
  private Chopstick stick = new Chopstick();
  private BlockingQueue<Chopstick> holder =
    new ArrayBlockingQueue<>(1);
  public StickHolder() { putDown(); }
  public void pickUp() {
    try {
      holder.take(); // Blocks if unavailable
    } catch(InterruptedException e) {
      throw new RuntimeException(e);
  public void putDown() {
    try {
    } catch(InterruptedException e) {
      throw new RuntimeException(e);

For simplicity, the Chopstick is never actually produced by the StickHolder, but kept private within the class. If you call pickUp() and the stick is unavailable, pickUp() blocks until the stick is returned by another Philosopher calling putDown(). Note that all the thread safety in this class is achieved through the use of the BlockingQueue.

Each Philosopher is a task that attempts to pickUp() the chopstick to both its right and left so it can eat, then releases those chopsticks with putDown():

// concurrent/

public class Philosopher implements Runnable {
  private final int seat;
  private final StickHolder left, right;
  public Philosopher(int seat,
    StickHolder left, StickHolder right) { = seat;
    this.left = left;
    this.right = right;
  public String toString() {
    return "P" + seat;
  public void run() {
    while(true) {
      // System.out.println("Thinking");   // [1]
      System.out.println(this + " eating");

No two Philosophers can successfully take() the same chopstick at the same time. In addition, if a chopstick has already been taken by one Philosopher, the next Philosopher that tries to take that same chopstick will block, waiting for it to be released.

The result is a seemingly-innocent program that deadlocks. I’ve used arrays here instead of collections only because the resulting syntax is cleaner:

// concurrent/
// Deadlock can be hidden in a program
import java.util.*;
import java.util.concurrent.*;
import static java.util.concurrent.TimeUnit.*;

public class DiningPhilosophers {
  private StickHolder[] sticks;
  private Philosopher[] philosophers;
  public DiningPhilosophers(int n) {
    sticks = new StickHolder[n];
    Arrays.setAll(sticks, i -> new StickHolder());
    philosophers = new Philosopher[n];
    Arrays.setAll(philosophers, i ->
      new Philosopher(i,
        sticks[i], sticks[(i + 1) % n]));    // [1]
    // Fix by reversing stick order:
    // philosophers[1] =                     // [2]
    //   new Philosopher(0, sticks[0], sticks[1]);
      .forEach(CompletableFuture::runAsync); // [3]
  public static void main(String[] args) {
    // Returns right away:
    new DiningPhilosophers(5);               // [4]
    // Keeps main() from exiting:
    ScheduledExecutorService sched =
    sched.schedule( () -> {
    }, 3, SECONDS);

Test this program by hand. When you stop seeing output, that means the program is deadlocked.

In the DiningPhilosophers constructor, each Philosopher is given a reference to a left and right StickHolder. Every Philosopher except the last one is initialized by situating that Philosopher between the next pair of chopsticks. The last Philosopher is given the zeroth chopstick for its right chopstick, so the round table is completed. That’s because the last Philosopher is sitting right next to the first one, and they both share that zeroth chopstick. [1] shows the right-hand stick selected with a modulus of n, wrapping the last Philosopher around to be next to the first one.

Now all Philosophers can try to eat, each waiting on the Philosopher next to them to put down its chopstick.

To start each Philosopher running at [3], I call runAsync() which means that the DiningPhilosophers constructor returns right away at [4]. Without anything to keep main() from completing, the program simply exits and doesn’t do much. To prevent this, I create a ScheduledExecutorService which holds main() open until it is shutdown(). Then I schedule a task which shuts down the Executor after three seconds, at which point all tasks automatically terminate as the program exits.

In the configuration as given, the Philosophers spend virtually no time thinking. Thus they all compete for chopsticks while trying to eat, and deadlock tends to happen quickly. You can change this:

  1. Add more Philosophers by increasing the value at [4].

  2. Uncomment line [1] in

Either one will make deadlock much less likely, which shows the danger of writing a concurrent program and believing it’s safe because it seems to “run OK on my machine.” You can easily convince yourself the program is deadlock free, even though it isn’t. This example is interesting precisely because it demonstrates that a program can appear to run correctly while still prone to deadlock.

To repair the problem, we observe that deadlock occurs when four conditions are simultaneously met:

  1. Mutual exclusion. At least one resource used by the tasks must not be shareable. Here, a chopstick can be used by only one Philosopher at a time.

  2. At least one task must hold a resource and wait to acquire a resource currently held by another task. That is, for deadlock to occur, a Philosopher must be hold one chopstick and wait for a second one.

  3. A resource cannot be preemptively taken away from a task. Tasks only release resources as a normal event. Our Philosophers are polite and they don’t grab chopsticks from other Philosophers.

  4. A circular wait can happen, whereby a task waits on a resource held by another task, which in turn is waiting on a resource held by another task, and so on, until one of the tasks is waiting on a resource held by the first task, thus gridlocking everything. In, the circular wait happens because each Philosopher tries to get the right chopstick first, then the left.

Because all these conditions must be met to cause deadlock, you must only prevent one of them to prohibit deadlock. In this program, an easy way to prevent deadlock is to break the fourth condition. This condition happens because each Philosopher tries to pick up its chopsticks in a particular sequence: first right, then left. Because of that, it’s possible for each Philosopher to hold its right chopstick while waiting for the left, causing the circular wait condition. However, if the one of the Philosophers is initialized to try to instead get the left chopstick first, that Philosopher never prevents the Philosopher on the immediate right from picking up a chopstick, precluding the circular wait.

In, uncomment the line at [1] and the one following it. This replaces the original philosophers[1] with a Philosopher that has its chopsticks reversed. By ensuring that the second Philosopher picks up and puts down the left chopstick before the right, we remove the potential for deadlock.

This is only one solution to the problem. You can also solve it by preventing one of the other conditions (see advanced threading books for more details).

There is no language support to help prevent deadlock; it’s up to you to avoid it by careful design. These are not comforting words to the person who’s trying to debug a deadlocking program. And of course the easiest and best way to avoid concurrency problems is never share resources—unfortunately that’s not always possible.

  1. You can also have livelock when two tasks are able to change their state (they don’t block) but they never make any useful progress. [return]