Dining philosopher problem in operating system

 

The dining philosopher problem is a classic example of a synchronization problem in computer science. It was first described by Edsger Dijkstra in 1965 and is used to illustrate the challenges of designing a concurrent system. The problem involves a number of philosophers seated at a round table with a bowl of rice and chopsticks in front of each philosopher. The philosophers are supposed to eat the rice by alternating between thinking and eating. However, they need to use chopsticks to pick up the rice, and only two chopsticks are available between each philosopher. The goal of the problem is to design an algorithm that prevents deadlocks, starvation, and other synchronization problems.

The dining philosopher problem has been used to demonstrate various synchronization techniques, including semaphores, monitors, and message passing. In this problem, there are five philosophers, each with a unique number from 0 to 4. Each philosopher thinks for a while, then tries to pick up the chopsticks on their left and right to eat. If the chopsticks are not available, the philosopher puts them down and waits until they are available. After eating, the philosopher puts down the chopsticks and thinks again.

One common solution to the dining philosopher problem is the semaphore solution. A semaphore is a synchronization tool that can be used to coordinate access to a shared resource. In this case, the semaphore is used to control access to the chopsticks. Before a philosopher can pick up the chopsticks, they must wait for the semaphore. After eating, the philosopher releases the semaphore to allow another philosopher to pick up the chopsticks.

The following is an example of a semaphore solution to the dining philosopher problem in C:

#include <stdio.h>

#include <pthread.h>

#include <semaphore.h>

#define N 5

#define LEFT (i + N - 1) % N

#define RIGHT (i + 1) % N

#define THINKING 0

#define HUNGRY 1

#define EATING 2

int state[N];

sem_t mutex;

sem_t s[N];

void take_chopsticks(int i)

{

sem_wait(&mutex);

state[i] = HUNGRY;

printf("Philosopher %d is hungry\n", i);

test(i);

sem_post(&mutex);

sem_wait(&s[i]);

sleep(1);

}

void put_chopsticks(int i)

{

sem_wait(&mutex);

state[i] = THINKING;

printf("Philosopher %d is thinking\n", i);

test(LEFT);

test(RIGHT);

sem_post(&mutex);

}

void test(int i)

{

if (state[i] == HUNGRY && state[LEFT] != EATING && state[RIGHT] != EATING)

{

state[i] = EATING;

printf("Philosopher %d is eating\n", i);

sem_post(&s[i]);

}

}

void* philosopher(void* num)

{

int i = ((int)num);

for (;;)

{

sleep(1);

take_chopsticks(i);

sleep(0);

put_chopsticks(i);

}

}

int main(){...}

To solve the Dining Philosophers Problem, there are several approaches that can be used. One common solution is the Resource Allocation Graph (RAG) approach. In this approach, the chopsticks are represented as resources and each philosopher is represented as a process that requests and releases these resources. When a philosopher wants to eat, they request the chopsticks to their left and right. If both chopsticks are available, the philosopher will eat and then release the chopsticks. If one or both chopsticks are in use, the philosopher will wait until they become available.

Another solution to the Dining Philosophers Problem is the Semaphore approach. In this approach, semaphores are used to control access to the chopsticks. A semaphore is a variable that is used to control access to a shared resource. In this case, the semaphore controls access to the chopsticks. When a philosopher wants to eat, they first request access to the semaphore for the chopstick to their left. If the chopstick is available, they are granted access and can pick it up. Then they request access to the semaphore for the chopstick to their right. If both chopsticks are available, the philosopher can eat and then release the chopsticks. If one or both chopsticks are in use, the philosopher will wait until they become available.

A third solution to the Dining Philosophers Problem is the Mutex approach. In this approach, a Mutex (or Mutual Exclusion) lock is used to control access to the chopsticks. A Mutex lock ensures that only one process can access a shared resource at a time. When a philosopher wants to eat, they request access to the Mutex lock for the chopstick to their left. If the chopstick is available, they are granted access and can pick it up. Then they request access to the Mutex lock for the chopstick to their right. If both chopsticks are available, the philosopher can eat and then release the chopsticks. If one or both chopsticks are in use, the philosopher will wait until they become available.

Each of these solutions has its own advantages and disadvantages. The Resource Allocation Graph approach is simple and straightforward, but it can be difficult to implement in practice. The Semaphore approach is more flexible, as it allows multiple processes to access the chopsticks at the same time, but it requires more complex logic to ensure that deadlocks are avoided. The Mutex approach is the most widely used solution, as it is easy to implement and ensures that deadlocks are avoided, but it is also the most restrictive, as it only allows one process to access the chopsticks at a time.

In conclusion, the Dining Philosophers Problem is an important problem in computer science and operating systems that is used to illustrate the principles of process synchronization and deadlock avoidance. The Resource Allocation Graph, Semaphore, and Mutex approaches are all solutions to this problem, each with its own advantages and disadvantages. Ultimately, the choice of which approach to use will depend on the specific requirements and constraints of the problem being solved.