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+37 votes

The following program consists of $3$ concurrent processes and $3$ binary semaphores. The semaphores are initialized as $S0=1, S1=0$ and $S2=0.$

$$\begin{array}{|l|l|}\hline \text{Process P0} & \text{Process P1} & \text{Process P2} \\ \hline \text{while (true) \{} & \text{wait (S1);} & \text{wait (S2);} \\ \text{ wait (S0);} & \text{release (S0);} & \text{release (S0);} \\ \text{ print ‘0';} & & \\ \text{ release (S1);} & & \\ \text{ release (S2);} & \text{} & \text{} \\ \text{\}} & \text{} \\\hline \end{array}$$

How many times will process $P0$ print '$0$'?

- At least twice
- Exactly twice
- Exactly thrice
- Exactly once

0

if s is a binary semaphore and the initial value of s is 1 then can 'signal' be performed on s ? what will happen if signal(s) is performed?

0

because of the word concurrent we can execute both the process together, but my doubt is if s=1 then can we release(s) again? because it is a binary semaphore & already its value is 1. please someone explain

0

when p0 execute second time it will free s1,and s2 so agian there will chance to run p0 again n this process continue forever and infinite zero will be printed ?

0

Given a binary semaphore 's' with initial value = 1. V(s) / UP(s) / signal(s) can still be performed. The process performing signal(s) on such a semaphore will check if there's some other process in the blocked queue (which is not possible since value of s is 1) and then it will set the value of s to 1. i.e. The value of 's' in such a scenario doesn't change.

+36 votes

First P0 will enter the while loop as S0 is 1. Now, it releases both S1 and S2 and one of them must execute next. Let that be P1. Now, P0 will be waiting for P1 to finish. But in the mean time P2 can also start execution. So, there is a chance that before P0 enters the second iteration both P1 and P2 would have done release (S0) which would make S1 1 only (as it is a binary semaphore). So, P0 can do only 1 more iteration printing '0' two times.

If P2 does release (S0) only after P0 starts its second iteration, then P0 would do three iterations printing '0' three times.

If the semaphore had 3 values possible (an integer semaphore and not a binary one), exactly three '0's would have been printed.

0

Can't understand "If the semaphore had 3 values possible, exactly three '0's would have been printed" How we can comment "exactly 3 0's " ?

+2

@Gate-Keeda At least 1 is possible (because it holds for exactly 3 also). But less than 2 is not possible here.

+3

@shree Here semaphore can have value either 0 or 1 only (because it is a binary semaphore). If we allow, 0, 1 and 2 as the values then both P1 and P2 will both release S0 (thus incrementing it twice), and thus P0 will execute totally 3 times. Thus exactly 3 0's will be printed.

0

sorry to bother you much. But I'm not getting why atleast once is not possible? Because it isn't in the options?

+13

It will print 0 at least 2 times (in any circumstance). Now, even if "at least one" is there in option, at least 2 is the better answer.

+2

sorry i didn't get it.why exaxtly 3 0's ?.it will keep on executing wait and release and a infinte number of 0's can be printed.

+13

Infinite times 0 could not printed bcoz process P1 and P2 are not under while loop means they are executed only once. when process P0 execute second time it release s1 and s2 but process P1 and P2 not executed second time and program terminated.

0

because of the word concurrent we can execute both the process together, but my doubt is if s=1 then can we release(s) again? because it is a binary semaphore & already its value is 1. please someone explain.

0

Can anyone please tell why there is no preemption here? ....means I understood both the cases and both are without consideration of preemption....please someone help...

Thanks

Thanks

0

@Arjun sir plz make that bracket in bold trust me half of the doubts will be vanished.

(as it is a binary semaphore)

Rest are cleared already.

0

@Arjun sir Can someone please explain the answer is given as atleast 2, it means that infinite zeroes can also be printed, but my question is how can p1 and p2 be executed again and again because there is no while loop in those processes.(They will be executed once and complete the processes) P.S. :- I know one condition can be until all semaphores are zero they can be executed. But I want to know is this the method they are following???

+16 votes

Option A is True.

Initially `P0`

will execute because only `S0=1`

. It will print single 0.

Now when `S1`

and `S2`

are releases by `P0`

then any one of them can be executed.

Let us suppose P1 executes and releases S0(Now value of S0 is 1).

Now there are two possibilities either `P0`

or `P2`

can execute.

Let us take `P2`

executes and releases `S0`

, so at the end P0 execute and print 0 (means two 0's) but if `P0`

executes before `P2`

then total of 3 0's will print(one at the time of `P0`

and then `P2`

which releases `S0`

so `P0`

executes again).

So the perfect answer is at least two 0's.

[email protected] http://stackoverflow.com/questions/12069305/how-many-times-will-process-p0-print-0

+3 votes

Minimum no. of time 0 printed is twice when execute in this order (p0 p1 p2 p0)

Maximum no. of time 0 printed is thrice when execute in this order (p0 p1 p0 p2 p0)

Maximum no. of time 0 printed is thrice when execute in this order (p0 p1 p0 p2 p0)

0 votes

0

You are saying that the least possibility is 1 rt?

First, we need to also have a look at the options and answer accordingly.

We have to try the maximum times we can execute P0.

Here we can execute P0 atleast twice. i.e. >=2.

First, we need to also have a look at the options and answer accordingly.

We have to try the maximum times we can execute P0.

Here we can execute P0 atleast twice. i.e. >=2.

0

@MINIPanda,

I think you are asking for P_{1} and P_{2 }. If they will not execute then P_{0} will be executed only once and it will wait for release(S0).

0

Oh yes it will be P1 and P2.

And if that is the case then had "at least once" been in the options, it would have been more appropriate right?

And if that is the case then had "at least once" been in the options, it would have been more appropriate right?

0

@MINIPanda,

as you are saying P1 and P2 don't execute, then after the first iteration of P0 who will perform release(S0) for next iteration of it :)

So for your case "only one time" will be the answer.

as you are saying P1 and P2 don't execute, then after the first iteration of P0 who will perform release(S0) for next iteration of it :)

So for your case "only one time" will be the answer.

0

@MINIPanda,

don't get confused with "atleast".In the options "atleast" is there only bcoz P0 can be executed either twice or thrice so merging both choices we are getting "atleast twice"

don't get confused with "atleast".In the options "atleast" is there only bcoz P0 can be executed either twice or thrice so merging both choices we are getting "atleast twice"

0 votes

0

0

Hi, here atleast 1 cannot be the answer because, in any order we try to execute the 3 processes, say if P0 executes first then after its execution, S1=S2=1, and S0=0, so if P0 executes again since its in while I can execute again, it gets blocked and waiting for someone to releaseS0.

then whenever P1 or P2 get a chance they will not be blocked since S1=S2=1, so they will release S0 and make S0=1,

Now there r 2 cases, either P1 and P2 occur one after the other then finally P0 comes in this case S1=1 only so only once P0 can execute and it will print second 0.

Or P1 then P0 then again P2 then P0, in this way P0 gets two more times to run so total 3 zeros. So anyhow atleast 2 two times its printed.

then whenever P1 or P2 get a chance they will not be blocked since S1=S2=1, so they will release S0 and make S0=1,

Now there r 2 cases, either P1 and P2 occur one after the other then finally P0 comes in this case S1=1 only so only once P0 can execute and it will print second 0.

Or P1 then P0 then again P2 then P0, in this way P0 gets two more times to run so total 3 zeros. So anyhow atleast 2 two times its printed.

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