made easy - advance level full test - serializability

Question

Consider the following schedule:

R2(A), R1(A), R3(B), W3(B), W2(A), R1(A)

Is the above schedule view serializable?

Comments

my thinking –

Initial reads – Since T2 is doing the initial read on A before T1, T2 will come before T1. T3 does the initial read on B and no other transaction reads B. So, our schedule could be T2 → T1 → T3 or T3 → T2 → T1.

Updated reads – T1 reads the value of A updated by T2, hence T1 will come after T2. No violation to our orders above.

Final write – T3 writes B last, T2 writes A last. But since no other transactions write A or B, our orders are still valid and hence the schedule is view serializable to both T2 → T1 → T3 and T3 → T2 → T1.

However, the answer key says that the schedule isn’t view serializable. What is wrong with my approach?

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@atulcse The given schedule is not view serializable schedule. Since there is no blind write hence not view serializable .

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Blind write is a necessary but not sufficient condition for a schedule to be view serializable .

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Hii @atulcse Initial read for every transactions should be match. In case of T2-->T1-->T3 you are reading updated value,updated by T2 transaction similarly in T3-->T2-->T1

Answer 1

It is not conflict serializable, and no blind write so not View serializable.