CSI 450 - Homework 4

 In the dining philosphers, 5 philosophers are sitting around a
 table. Each has their own bowl of rice, but between each philosopher
 there is a single chopstick. To eat, a philosopher must have both
 chopsticks, to her left and right, but another philosopher may
 already have them (there is only one chopstick between them and that
 is the only one that the philosopher may use). Philsopher's may not
 communicate with one another.
 
 You must solve the dining philosopher problem without starvation or
 deadlock. Philosopher's think and eat in a loop for all time. For
 example you may think of this as the following:

while(true)
{
   think();
   eat();
}

 

 

Part 1: - due Thursday 11/19 (although I recommend you complete this by 11/17)

As a convention, you may assume that a dining philosopher knows his
own identifier "i" and the identifier of his two chopsticks "i" and
"j" (we know that j is ( (i+1) % 5 ) and thus not necessarily simply
i+1. The 4th philosopher has access to chopstick 4 and chopstick 0 for
example.

On paper, write the function for eat. You must use a mutex for each
chopstick and solve the problems listed above given the restraints
listed above. Include proper code for all pieces, although you should
write this by hand and not worry so much about if it precisely
compiles.