Consider the following transactions with data items P and Q initialized to zero:T1: read (P) ;read (Q) ;if P = 0 then Q : = Q + 1 ;write (Q) ;T2: read (Q) ;read (P) ;if Q = 0 then P : = P + 1 ;write (P) ;Any non-serial interleaving of T1 and T2 for concurrent execution leads toA serializable scheduleA schedule that is not conflict serializableA conflict serializable scheduleA schedule for which a precedence graph cannot be drawn
Question
Consider the following transactions with data items P and Q initialized to zero:T1: read (P) ;read (Q) ;if P = 0 then Q : = Q + 1 ;write (Q) ;T2: read (Q) ;read (P) ;if Q = 0 then P : = P + 1 ;write (P) ;Any non-serial interleaving of T1 and T2 for concurrent execution leads toA serializable scheduleA schedule that is not conflict serializableA conflict serializable scheduleA schedule for which a precedence graph cannot be drawn
Solution
The given transactions T1 and T2 can lead to a schedule that is not conflict serializable.
Here's why:
Conflict serializability is a form of equivalence between schedules. Two schedules are conflict equivalent if they have the same set of transactions, and the order of any two conflicting operations is the same in both schedules.
In the given transactions, T1 and T2, both are reading and writing to data items P and Q. If we interleave these transactions, there can be a situation where T1 reads the initial value of P (which is 0), then T2 reads the initial value of Q (which is also 0). After that, T1 increments Q by 1 and writes it back. Then T2 increments P by 1 and writes it back.
In this case, the final state of the system is P=1, Q=1. However, if we execute T1 and T2 serially in the order T1->T2, the final state of the system would be P=0, Q=1. And if we execute them in the order T2->T1, the final state would be P=1, Q=0.
So, the interleaved execution does not match with any of the serial executions, hence it is not conflict serializable.
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