Consider the following transactions with data items P and Q initialized to…

Answer: a schedule that is not conflict serializable

Reasoning:

• Initial values: P = 0, Q = 0.

• Transaction T1 (in order): read P; read Q; if P = 0 then Q := Q + 1; write Q.

• Transaction T2 (in order): read Q; read P; if Q = 0 then P := P + 1; write P.

• In any non-serial interleaving where the transactions overlap, T1's read of P happens before T2's eventual write of P, producing a read–write conflict and an edge from the transaction that read P to the transaction that later writes P.

• Similarly, T2's read of Q happens before T1's eventual write of Q, producing a read–write conflict and an edge from the transaction that read Q to the transaction that later writes Q.

• Concretely, the precedence (conflict) graph has an edge from T1 to T2 and an edge from T2 to T1, forming a cycle.

• A cycle in the precedence graph means the schedule cannot be transformed into any serial schedule by swapping non-conflicting operations; therefore it is not conflict serializable.

Conclusion: Any non-serial interleaving of the two transactions yields the read/write conflicts that produce a cycle in the precedence graph, so such a schedule is not conflict serializable.