![]() Similarly, suppose now Philosopher P1 wants to eat, it will enter in Philosopher() function, and execute Wait( take_chopstickC ) by doing this it will try to hold C1 chopstick but will not be able to do that, since the value of semaphore C1 has already been set to 0 by philosopher P0, therefore it will enter into an infinite loop because of which philosopher P1 will not be able to pick chopstick C1 whereas if Philosopher P2 wants to eat, it will enter in Philosopher() function, and execute Wait( take_chopstickC ) by doing this it holds C2 chopstick and reduces semaphore C2 to 0, after that, it executes Wait( take_chopstickC ) by doing this it holds C3 chopstick( since i =2, therefore (2 + 1) % 5 = 3) and reduces semaphore C3 to 0. Let value of i = 0( initial value ), Suppose Philosopher P0 wants to eat, it will enter in Philosopher() function, and execute Wait( take_chopstickC ) by doing this it holds C0 chopstick and reduces semaphore C0 to 0, after that it execute Wait( take_chopstickC ) by doing this it holds C1 chopstick( since i =0, therefore (0 + 1) % 5 = 1) and reduces semaphore C1 to 0 Let's understand how the above code is giving a solution to the dining philosopher problem? Finally, the philosopher starts thinking again. This shows that the philosopher i have eaten and put down both the left and right chopsticks. ![]() ![]() On completion of eating by philosopher i the, signal operation is performed on take_chopstickC and take_chopstickC. The eating function is performed after that. This shows philosopher i have picked up the chopsticks from its left and right. In the above code, first wait operation is performed on take_chopstickC and take_chopstickC. Semaphore: A semaphore is an integer variable in S, that apart from initialization is accessed by only two standard atomic operations - wait and signal, whose definitions are as follows: Wait and Signal operations will be used for the solution of the Dining Philosophers Problem, for picking a chopstick wait operation can be executed while for releasing a chopstick signal semaphore can be executed. We use a semaphore to represent a chopstick and this truly acts as a solution of the Dining Philosophers Problem. The solution of the Dining Philosophers Problem Similarly suppose now Philosopher P1 wants to eat, it will enter in Philosopher() function, and execute take_chopstick by doing this it holds C1 chopstick after that it execute take_chopstick by doing this it holds C2 chopstick( since i =1, therefore (1 + 1) % 5 = 2)īut Practically Chopstick C1 is not available as it has already been taken by philosopher P0, hence the above code generates problems and produces race condition. Suppose Philosopher P0 wants to eat, it will enter in Philosopher() function, and execute take_chopstick by doing this it holds C0 chopstick after that it execute take_chopstick by doing this it holds C1 chopstick( since i =0, therefore (0 + 1) % 5 = 1) The five Philosophers are represented as P0, P1, P2, P3, and P4 and five chopsticks by C0, C1, C2, C3, and C4. The dining philosopher demonstrates a large class of concurrency control problems hence it's a classic synchronization problem.įive Philosophers sitting around the tableĭining Philosophers Problem- Let's understand the Dining Philosophers Problem with the below code, we have used fig 1 as a reference to make you understand the problem exactly. In case if both immediate left and right chopsticks of the philosopher are not available then the philosopher puts down their (either left or right) chopstick and starts thinking again. A philosopher can only eat if both immediate left and right chopsticks of the philosopher is available. To eat a philosopher needs both their right and a left chopstick. A bowl of noodles is placed at the center of the table along with five chopsticks for each of the philosophers. The dining philosopher's problem is the classical problem of synchronization which says that Five philosophers are sitting around a circular table and their job is to think and eat alternatively. Next → ← prev THE DINING PHILOSOPHERS PROBLEM
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