Predict serializability of given schedule

Question

S:r1(P),r2(Q),r1(Q),r2(P),w1(Q),w2(P)

Predict serializability of  given schedule:

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a) It is view serializable NOT conflict serializble

b) It is view serializable but NOT conflict serializble only if P= Q

c) It is view serializable and conflict serializble

d) It is neither view serializable nor conflict serializble

Comments

is it option D?

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there is a cycle in the graph and no Blind Write..   option D rt??

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Precedence graph please!!

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Why is it not View serializable ? @sh!va

Answer 1

In the above given schedule ,

whe have a cycle in a precedence graph  so it is not conflict serializable .

How is Cycle formed ?

Consider T1 and T2 as nodes , you have an  edge from T1 to T2 beacuse there is a conflict operation of R1(P)-->W2(P)

you have an edge from T2 to T1 beacuse there is an conflict operation between R2(Q) --> W1(Q)

Hence a presence of cycle mean it is not Conflict serilizable

Let us check for View Serializable .

1 . Check for blind write ?

 since there is no Blind write so it not view serializable also ,

if there would have been blind write we would have follow polygraph technique and check for cycle too . However this is not required here

So the answer for above question is option D :)