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Question

Q1. Consider the methods used by process P1 and P2 for accessing their critical sections whenever needed, as given below. The initial values of shared Boolean variables, S1 and S2 are random assigned.

For the program to guarantee mutual exclusion, the predicate P and Q in the while loop should beA

A.

B.  

C.

D.

ans is option B. My  doubt is what is the case in which option D will not satisfy mutual exclusion? I am confused between B and D.

Comments

yes option (D) also satisfy mutual exclusion .

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@Shaik Masthan ...how to solve such type of questions?

if the condition in the while loop evaluates to true...then it will wait or execute CS?

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while loop evaluates to true

then it will in continuously in loop ( i.e., it is waiting )

 

while loop evaluates to false

 then it will out of loop ( i.e., it is entered in CS )

 

how to solve such type of questions?

you have to take 4 cases here, initially

1) S1=0 and S2=0 

2) S1=0 and S2=1 

3) S1=1 and S2=0 

4) S1=1 and S2=1

then take each option and check these 4 cases with 

i) first P execute then Q execute

ii) First Q execute then P execute.

with in these cases, if any option giving ME in all cases... then check it with instruction level code ( as we do in LOCK variable synchronizing method. )

Answer 1

a) P:flag[j]=true and S₁!=S₂

    Q:flag[i]=true and S₁==S₂

 Initially , flag[i]=flag[j]=false.

 

P₁ will make flag[i] =false, true and Make S₁=S₂ so , if P:flag[j]=true and S₁==S₂ then P will be false hence no waiting in loop. So P₁ will enter the CS.

Now check that P2 can enter the CS or not when P1 is in CS.

P2 will make flag[j]= false ,true and make S₁=S₂+1, so ,if    Q:flag[i]=true and S₁==S₂ then Q will be false hence no waiting in loop. So, P2 will enter the CS.

So, both processes will enter the critical section.

Hence, option (a) is false.

 

b) P:flag[j]=true and S₁==S₂

    Q:flag[i]=true and S₁!=S₂

P₁ will make flag[i] =false, true and Make S₁=S₂ so , if P:flag[j]=true and S₁==S₂ then P will be false as flag[j]=false, hence no waiting in loop. So P₁ will enter the CS.

Now check that P2 can enter the CS or not when P1 is in CS.

P2 will make flag[j]= false ,true and make S₁=S₂+1, so ,if    Q:flag[i]=true and S₁!=S₂ then Q will be true as (flag[i]=true and S₁!=S₂)=TRUE, hence,P2 will wait in the loop.

So, only one  process will enter the critical section.

Hence, option (b) is true.

 

c) P:flag[i]=true and S₁!=S₂

    Q:flag[j]=true and S₁==S₂

P₁ will make flag[i] =false, true and Make S₁=S₂ so, if P:flag[i]=true and S₁!=S₂ then P will be false hence no waiting in loop. So P₁ will enter the CS.

Now check that P2 can enter the CS or not when P1 is in CS.

P2 will make flag[j]= false ,true and make S₁=S₂+1, so ,if    Q: flag[j]=true and S₁!=S₂then Q will be false hence no waiting in loop. So, P2 will enter the CS.

So, both processes will enter the critical section.

Hence, option (c) is false.

 

d) P:flag[i]=true and S₁==S₂

    Q:flag[j]=true and S₁!=S₂

 

P₁ will make flag[i] =false, true and Make S₁=S₂ so, if P:flag[i]=true and S₁==S₂ then P will be true hence  waiting in loop. So P₁ will not enter the CS.

Now check for P2 ,

P2 will make flag[j]= false ,true and make S₁=S₂+1, so ,if    Q: flag[j]=true and S₁!=S₂then Q will be TRUE hence waiting in loop. So, P2 cannot enter the CS.

But Now, if again preemption happens, then P2 has made S₁!=S₂ So,P1 can enter the CS and P2 will wait in the loop. So, mutual exclusion holds.  

Hence, option (d) is true.

Correct me if i am wrong