1.8k views

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. edited | 1.8k views
0
THEY ALSO HAVE TO WRITE THAT ALL SEMAPHORS ARE INTIALIZED TO ZERO.
0

Is this concept (parbegin and parend) there in the syllabus any more ?

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.

by Active (3.6k points)
edited
+1
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
0

Please Clarify this +3

@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.

begin

S1;

perbegin

S3;

begin

S2;

perbegin

S4;

S5;

perend;

end

perend

end

begin

S7

perbegin

S4;

S5;

perend

end
by Veteran (117k points)
+4
you have to use wait and signal constructs to ensure this order. Your answer is correct but is less parallel and not using semaphore.
0
signal and wait for cobegin and coend structure

not for parbegin and parend structure

right?

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

by Veteran (117k points)
edited by
0
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
+2

WHAT DO THEY MEAN BY PAREND AND PARBEGIN?

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

0
Same here @Deepanshu

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

by Active (1.1k points)