View serializability check of schedule

Two schedules are view equivalent if (1) each transaction reads the initial value of an item in both or in neither, (2) each read reads the value written by the same transaction in both, and (3) the final write of each item is done by the same transaction in both. A schedule is view serializable if it is view equivalent to some serial schedule.

T1: r1(Q), w1(Q); T2: w2(Q); T3: w3(Q) S: r1(Q), w2(Q), w1(Q), w3(Q)

Step 1: conflict serializability (for comparison)

Conflicts: r1(Q) before w2(Q) gives T1 → T2; w2(Q) before w1(Q) gives T2 → T1. This is a cycle, so S is not conflict serializable. We must test view serializability directly.

Step 2: view properties of S

• Initial read: T1 reads the initial value of Q in r1(Q).
• Reads-from: there is only one read, and it reads the initial value, so there is no "reads from" relation.
• Final write: w3(Q) is the last write on Q, so T3 does the final write.

Step 3: compare with the serial schedule T1, T2, T3

Serial: r1(Q), w1(Q), w2(Q), w3(Q)

• Initial read: T1 reads the initial Q. Same.
• Reads-from: no other reads. Same.
• Final write: T3. Same.

All three conditions are met.

Conclusion: S is view serializable, view equivalent to the serial schedule <T1, T2, T3>, even though it is not conflict serializable. This is possible because T2 and T3 perform blind writes (writes without reading). The final write w3(Q) overwrites w2(Q) and w1(Q), and no transaction reads those values, so their order does not matter.