Semaphore Solution to Dining Philosopher

Each philosopher is represented by the following pseudocode:  

process P[i]
while true do
{ THINK;
PICKUP(CHOPSTICK[i], CHOPSTICK[i+1 mod 5]);
EAT;
PUTDOWN(CHOPSTICK[i], CHOPSTICK[i+1 mod 5])
}

There are three states of the philosopher: THINKING, HUNGRY, and EATING. Here there are two semaphores: Mutex and a semaphore array for the philosophers. Mutex is used such that no two philosophers may access the pickup or put it down at the same time. The array is used to control the behavior of each philosopher. But, semaphores can result in deadlock due to programming errors.

Outline of a philosopher process:

Var successful: boolean;
repeat
successful:= false;
while (not successful)

if both forks are available then
lift the forks one at a time;
successful:= true;

if successful = false
then
block(Pi);
{eat}

put down both forks;

if left neighbor is waiting for his right fork
then
activate (left neighbor);
if right neighbor is waiting for his left fork
then
activate( right neighbor);
{think}
forever

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


Similar Reads

Semaphore Solution to Dining Philosopher

Each philosopher is represented by the following pseudocode:...

Code

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); }...

Problem with Dining Philosopher

We have demonstrated that no two nearby philosophers can eat at the same time from the aforementioned solution to the dining philosopher problem. The problem with the above solution is that it might result in a deadlock situation. If every philosopher picks their left chopstick simultaneously, a deadlock results, and no philosopher can eat. This situation occurs when this happens....

FAQs on Dining Philosopher Problem Using Semaphores

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