Dining Philosopher Problem Using Semaphores

Q.1: What is the main challenge in the dining philosophers problem?

Answers:

The main challenge in the dining philosophers’ problem is preventing deadlocks. Deadlock occurs when each philosopher picks up one chopstick and is waiting indefinitely for the other chopstick. This situation leads to a system-wide halt, and no philosopher can make progress. Resolving the deadlock issue while allowing all philosophers to eat without starvation requires careful synchronization and resource allocation.

Q.2: How can the dining philosophers problem be solved?

Answer:

Several solutions exist to solve the dining philosophers problem and prevent deadlocks. Some common approaches include:

Resource hierarchy: Assign a total order to the chopsticks and have each philosopher pick up the lower-numbered chopstick first. This prevents circular dependencies and guarantees that deadlock cannot occur.

Chandy/Misra solution: Introduce an arbitrator or central authority that manages the allocation of chopsticks. Philosophers request permission from the arbitrator before attempting to pick up the chopsticks, ensuring that only a limited number of philosophers can hold chopsticks simultaneously.

Dijkstra’s solution using semaphores: Assign a semaphore to each chopstick, representing its availability. Philosophers must acquire both chopsticks by locking their corresponding semaphores. If a philosopher cannot acquire both chopsticks, they release the acquired chopstick and wait before trying again.

Timeout-based solution: Set a timer or timeout for each philosopher. If a philosopher cannot acquire both chopsticks within a certain time limit, they release the acquired chopstick and retry later. This approach avoids deadlocks but may result in starvation if philosophers keep getting interrupted.

Q.3: Can the dining philosophers problem have multiple solutions?

Answer:

Yes, the dining philosophers problem can have multiple solutions. The choice of solution depends on the specific requirements of the system and the desired behavior. Different algorithms and approaches can be employed to ensure the philosophers can eat without deadlocks and minimize the occurrence of starvation. The suitability of a particular solution depends on factors such as the number of philosophers, the desired fairness, efficiency, and the overall system architecture.

Dining Philosopher Problem Using Semaphores

The Dining Philosopher Problem states that K philosophers are seated around a circular table with one chopstick between each pair of philosophers. There is one chopstick between each philosopher. A philosopher may eat if he can pick up the two chopsticks adjacent to him. One chopstick may be picked up by any one of its adjacent followers but not both.

Dining Philosopher

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C++ #include #include #include #include #define N 5 #define THINKING 2 #define HUNGRY 1 #define EATING 0 #define LEFT (phnum + 4) % N #define RIGHT (phnum + 1) % N int state[N]; int phil[N] = { 0, 1, 2, 3, 4 }; std::mutex mutex; std::condition_variable S[N]; void test(int phnum) { if (state[phnum] == HUNGRY && state[LEFT] != EATING && state[RIGHT] != EATING) { // state that eating state[phnum] = EATING; std::this_thread::sleep_for(std::chrono::milliseconds(2000)); std::cout << "Philosopher " << phnum + 1 << " takes fork " << LEFT + 1 << " and " << phnum + 1 << std::endl; std::cout << "Philosopher " << phnum + 1 << " is Eating" << std::endl; // notify the philosopher that he can start eating S[phnum].notify_all(); } } // take up chopsticks void take_fork(int phnum) { std::unique_lock lock(mutex); // state that hungry state[phnum] = HUNGRY; std::cout << "Philosopher " << phnum + 1 << " is Hungry" << std::endl; // eat if neighbours are not eating test(phnum); // if unable to eat wait to be signalled S[phnum].wait(lock); std::this_thread::sleep_for(std::chrono::milliseconds(1000)); } // put down chopsticks void put_fork(int phnum) { std::unique_lock lock(mutex); // state that thinking state[phnum] = THINKING; std::cout << "Philosopher " << phnum + 1 << " putting fork " << LEFT + 1 << " and " << phnum + 1 << " down" << std::endl; std::cout << "Philosopher " << phnum + 1 << " is thinking" << std::endl; test(LEFT); test(RIGHT); } void philosopher(int num) { while (true) { take_fork(num); put_fork(num); } } int main() { std::thread threads[N]; for (int i = 0; i < N; i++) { threads[i] = std::thread(philosopher, i); std::cout << "Philosopher " << i + 1 << " is thinking" << std::endl; } for (int i = 0; i < N; i++) threads[i].join(); return 0; } C #include #include #include #define N 5 #define THINKING 2 #define HUNGRY 1 #define EATING 0 #define LEFT (phnum + 4) % N #define RIGHT (phnum + 1) % N int state[N]; int phil[N] = { 0, 1, 2, 3, 4 }; sem_t mutex; sem_t S[N]; void test(int phnum) { if (state[phnum] == HUNGRY && state[LEFT] != EATING && state[RIGHT] != EATING) { // state that eating state[phnum] = EATING; sleep(2); printf("Philosopher %d takes fork %d and %d\n", phnum + 1, LEFT + 1, phnum + 1); printf("Philosopher %d is Eating\n", phnum + 1); // sem_post(&S[phnum]) has no effect // during takefork // used to wake up hungry philosophers // during putfork sem_post(&S[phnum]); } } // take up chopsticks void take_fork(int phnum) { sem_wait(&mutex); // state that hungry state[phnum] = HUNGRY; printf("Philosopher %d is Hungry\n", phnum + 1); // eat if neighbours are not eating test(phnum); sem_post(&mutex); // if unable to eat wait to be signalled sem_wait(&S[phnum]); sleep(1); } // put down chopsticks void put_fork(int phnum) { sem_wait(&mutex); // state that thinking state[phnum] = THINKING; printf("Philosopher %d putting fork %d and %d down\n", phnum + 1, LEFT + 1, phnum + 1); printf("Philosopher %d is thinking\n", phnum + 1); test(LEFT); test(RIGHT); sem_post(&mutex); } void* philosopher(void* num) { while (1) { int* i = num; sleep(1); take_fork(*i); sleep(0); put_fork(*i); } } int main() { int i; pthread_t thread_id[N]; // initialize the semaphores sem_init(&mutex, 0, 1); for (i = 0; i < N; i++) sem_init(&S[i], 0, 0); for (i = 0; i < N; i++) { // create philosopher processes pthread_create(&thread_id[i], NULL, philosopher, &phil[i]); printf("Philosopher %d is thinking\n", i + 1); } for (i = 0; i < N; i++) pthread_join(thread_id[i], NULL); }...

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Problem with Dining Philosopher