Using the semaphores you have implemented, detect deadlocks in Dining Philosophers problem.
2.Test case:
The first line of the test case contains two numbers
A,B
A is the number of philosophers, and B is the number of entries coming after the first line
The number of chopsticks is the same as number of philosophers
Your program should have an array of semaphores where each semaphore (each element of the array) corresponds to each chopstick.
Each philosopher gets access to “the right” chopstick by doing
P(Sem[(philosopher ID – 1)% (number of Chopsticks)])
Then yields to the next philosopher
When the philosopher comes back the philosopher attempts to access “the left” chopstick by doing
P(Sem[philosopher ID % (number of Chopsticks) ])
Then yields to the next philosopher
When it comes back it prints
Printf(“\n Philosopher %d is eating \n”, philosopher ID);
Then it releases the right chopstick by doing
V(Sem[(philosopher ID – 1)% (number of Chopsticks)])
Then it releases the left chopstick by doing
V(Sem[(philosopher ID)% (number of Chopsticks)])
Then it deletes itself from the readyQ.
Then it yields to the next philosopher
The next B lines has philosopher numbers in order to populate the Ready Queue.
The program starts with the first element of the Ready Queue.
Test case example
4,2
1
3
Output
Philosopher 1 is eating
Philosopher 3 is eating
Test case example 2
4,2
1
2
Output
Philosopher 2 is eating
Philosopher 1 is eating
