ME doubt

Question

Here answers says removal of 3 and 4 will not cause any problem

but in galvin book there is no wait and signal of locks ,other then chopstic semaphore

so my question is both 1,2,3,4 can be removed without any proble...

Answer 1

See the name "Dining Philosopher's Problem" is used intentionally , but you have to examine the code to check whether the properties of mutual exclusion , progress and bounded waiting is satisfied or not.So now let's discuss in light of the given code :
 

Let for the time being we do not use wait(lock) statement and here lock is a binary semaphore.Now let's see what happens.In light of the classical dining philosopher's problem , a solution to this can be considered as correct if a person can have both left and right chopsticks at the same time.

Now consider a scenario where wait(lock) is not used.Let the processes be P₀ , P₁ , ...P₄..Let P₀ execute wait(fork[0]) where fork[0] is also a binary semaphore.Now we preempt P₀ and go to P₁ , execute wait(fork[1]) . We repeat the same process till P₄ executes wait(fork[4]).Now all of the processes have executed one wait() operation each of which are successful wait() operation.Now value of fork[i] = 0 for 0 <= i <= 4 since wait() has been performed on each fork[i].

Now when we try to execute wait(fork[(0+1) % 5]) for P₀ which is wait([fork[1]) which is already held by P₁.Similarly P₁ tries to execute wait(fork[2]) but that is already held by P₂.In short , no process is able to execute its 2nd wait() operation and hence no one can go into critical section and eat().In other words the processes are deadlocked.So this is not the correct solution.

Hence , the wait(lock) and signal(lock) is necessary.We consider the preemption in this case here , so wait(lock) is say performed by P₀ then lock = 0 and hence other processes are not allowed to do wait(fork[i]) unless and until signal(lock) is performed by that particular process after which other process can enter and try to do wait(fork[i]) operation.

In short wait(lock) and signal(lock) are necessary hence they cannot be removed.

Hence option B) is correct.So in such questions , we should always go with implementation which is given in the question not simply name as it may lead us to wrong answers

Also one more point to add that preemption can be taken anywhere in the code if it is a concurrent process in order to check whether the code works fine , gives proper result or not.

I hope I have addressed your doubt.

Comments

@habib khan I think that removal of 3 and 4 is necessary bcoz the solution including 3 and 4 is not correct. consider 2 enters cs and acquires 2 and 3 fork. now P2 has signalled on LOCK and eating after that P3 comes and wait(LOCK) is performed then it cannot acquire 3 fork bcoz P2 cannot release it since for release it has to wait(LOCK) which will block P2 and P3 also blocked and any other process will also be blocked so deadlock has occured. Therefore option must 3 and 4 "must be removed" for correct operation.

Answer 2

Lets say  (1) and (2) are removed . If lets say all the processes acquire their left fork and each one is waiting for acquiring the right fork , then this is deadlock.

Now, lets consider the problem with this code. Consider that P2 and P4 have acquired forks and eating yummy Gaulab jamun :P

Now, P3 wants to eat . So, it executes wait(lock). Now, its waiting for acquirin the forks.

Now, lets say P4 and P2 have eaten so much (that their stomach will burst) and want to release forks.

Now, now, since they cant execute wait(lock), they will be deadlocked with P3. P3 is waiting for acquiring resources and P2,P4 waiting to release resources.

So, issue is with statements (3) and (4). So, remove them. Now, it will work fine.