Non-blocking solution to the dining philosophers

algorithm, blocking, dining-philosopher, python

Solution

Here is a definition of a non-blocking algorithm: http://en.wikipedia.org/wiki/Non-blocking_algorithm.

A pseudo code of a non-blocking solution to this problem:

# The number of forks.
FORKS_COUNT = ... 

# Indicates if the i-th fork is taken or not.
taken = new bool[FORKS_COUNT] 

# The philosopherId is a position at the table.
def haveDinner(philosopherId):
    leftFork = philosopherId
    rightFork = (philosopherId + 1) % FORKS_COUNT
    if leftFork > rightFork:
        swap(leftFork, rightFork)
    while true:
        # Tries to take the left fork.
        while not compare_and_swap(taken[leftFork], false, true):
            # Do nothing.
        # Tries to take the right fork.
        while not compare_and_swap(taken[rightFork], false, true):
            # Do nothing.
        # Eats.
        ...
        # Returns the forks to the table.
        compare_and_swap(taken[leftFork], true, false)
        compare_and_swap(taken[rigthFork], true, false)

This solution uses the compare-and-swap idiom.

Problem

I have been asked to write a simple solution to the dining philosophers problem in python. That itself seems quite straight forward but am some what confused since I am asked to write a non-blocking solution. I am unsure what is meant by this in this context. Is anyone able to give any hints or point me in the right direction?

Original source