Any deadlock-free solution of Dinning-philosopher does not guarantee that there will be no starvation. You are required to provide a starvation-free code that allows a philosopher to pick-up chopsticks for ith time only if the other four philosophers have eaten (i-1) time(s). Example:

Philosopher#1 can be allowed to pick-up chopsticks to eat 2nd time only if other philosophers have eaten at least 1 time. Assume that all of the five philosophers are coming for eating after thinking for some finite amount of time.

Pseudocode of a function the philosopher threads would be extended from: Void philosopher (int n) // n is the philosopher number { While(1) { Pickup chopsticks Eat Putdown chopsticks Think } }

Please note this is a coding solution and should be answered in C.