GATE CSE 1993 | Question: 22

Question

Write a concurrent program using $\text{parbegin-parend}$ and semaphores to represent the precedence constraints of the statements $S_1$ to $S_6$, as shown in figure below.

Comments

Hello

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nice one

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any resources to learn this topic?

Answer 1

parbegin

    begin   S1  parbegin    V(a)    V(b)    parend  end
    
    begin   P(a)    S2  parbegin    V(c)    V(e)    parend  end
    
    begin   P(b)    S3  V(d)    end
    
    begin   P(f)    P(c)    S4  end
    
    begin   P(g)    P(d)    P(e)    S5  end
    
    begin   S6  parbegin    V(f)    V(g)    parend  end
    
parend

Here, the statement between parbegin and parend can execute in any order. But the precedence graph shows the order in which the statements should be executed. This strict ordering is achieved using the semaphores.

Initially all the semaphores are $0.$

For $S_1$ there is no need of semaphore because it is the first one to execute.

Next $S_2$ can execute only when $S_1$ finishes. For this we have a semaphore $a$ which on signal executed by $S_1$, gets value $1.$ Now $S_2$ which is doing a wait on $a$ can continue execution making $a=0$;

Likewise this is followed for all other statements.

Comments

Hey bro..I think there is typo in the 3rd line ..we should write

P(a) S2 V(c) V(e)..

Like wise see further...

Make me correct if..I m wrong

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Please Clarify this

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@krishn.jh  statements within parbegin-parend have no predefined order for execution. The answer is correct because wait() operation is performed on semaphores f and g, which guarantees that s4, s5 will execute after s6, because after execution of s6 only f and g are signaled.

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@Sachin Mittal 1 sir... You haven't covered this topic in course..?

Answer 2

begin

S1;

          perbegin

           S3;

           begin

            S2;

                         perbegin

                          S4;

                          S5;

                          perend;

             end

             perend

end

begin

S7

                   perbegin

                    S4;

                    S5;

                   perend

end

Comments

you have to use wait and signal constructs to ensure this order. Your answer is correct but is less parallel and not using semaphore.

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signal and wait for cobegin and coend structure

not for parbegin and parend structure

right?

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There's typo in this answer. It should be S6 and not S7.

Answer 3

IT IS EASY

Comments

Is there any proof of number of edges should be equal to binary sem used?

 

Answer 4

Begin

Cobegin

begin S₁:V(a) ; V(b) ; end

begin P(a):S₂ ;V(c); V(d) ; end

begin P(b) : S₃ ; V(e) ; end

begin P(c) : S₄ ; end

begin P(d) : P(e) : S₅ ; end

Coend

End

Begin

Cobegin

begin S₆:V(f) ;V(g) ;end

begin P(f) : S₄ ; end

begin P(g) : S₅ ; end

Coend

End

Comments

You should combine all into one.S4 will do two down operations.similarly S5 will do 3 down operations.There is no need to have two being end blocks

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@srestha  

 WHAT DO THEY MEAN BY PAREND AND PARBEGIN?

MAYBE IT IS SILLY QUESTION . BUT I NEVER CAME ACROSS THIS CONCEPT

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Same here @Deepanshu

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parbegin = parallel begin and parend= parallel end

Answer 5

begin:

      parbegin:

            begin:

                  S1;

                  parbegin:

                        S3; V(a);

                        begin:

                                S2;

                                parbegin:

                                     P(b); S4;

                                     P(a); P(b); S5;

                                parend

                          end

                    parend

               end

               S6; V(b);

       parend

​​​​​​​end

 

                        

Answer 6

https://gateoverflow.in/?qa=blob&qa_blobid=10305005738767994331