Chapter 18: Problem 20

Prove that the basic two-phase locking protocol guarantees conflict serializability of schedules. (Hint: Show that, if a serializability graph for a schedule has a cycle, then at least one of the transactions participating in the schedule does not obey the two-phase locking protocol.)

Short Answer

Expert verified

The basic Two-Phase Locking protocol guarantees conflict serializability because a cycle cannot exist in its serializability graph, meaning the schedule is conflict serializable. If a cycle exists, it indicates that at least one transaction in the schedule is not following the Two-Phase Locking protocol.

Step by step solution

Understand Two-Phase Locking Protocol

The Two-Phase Locking protocol has two phases, the Expanding phase and Shrinking phase. In the first phase (Expanding), a transaction may obtain locks, but cannot release any lock. In the second phase (Shrinking), it may release locks, but cannot obtain any new locks. This means once a transaction releases its first lock, it is in the second phase and cannot request more locks.

Analyze Serializability Graph

A serializability graph is a directed graph where nodes represent transactions and an edge from node T1 to T2 means that T1 is scheduled before T2. A schedule is considered conflict serializable if its serializability graph is acyclic.

Finding Cycle in Serializability Graph

For any two transactions T1 and T2, if T1 must wait for T2 (T1 -> T2 in the graph), it means T1 has requested for a data item that T2 has locked. If there is a cycle, it means there is a conflicting operation. Therefore, for every pair of transactions (T1, T2) that T1 -> T2, if T2 releases the lock before T1, then T1 and T2 are not following the Two-Phase Locking protocol.

Conclude the Proof

If a schedule follows the Two-phase Locking protocol, it is guaranteed to be conflict serializable, because a cycle cannot exist in its serializability graph. This is because, once a transaction releases its first lock (enters the shrinking phase), it cannot request a new lock, avoiding the possibility of a cycle. Hence, if a serializability graph for a schedule has a cycle, then at least one of the transactions participating in the schedule does not obey the two-phase locking protocol.

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Prove that the basic two-phase locking protocol guarantees conflict serializability of schedules. (Hint: Show that, if a serializability graph for a schedule has a cycle, then at least one of the transactions participating in the schedule does not obey the two-phase locking protocol.)

Short Answer

Step by step solution

Understand Two-Phase Locking Protocol

Analyze Serializability Graph

Finding Cycle in Serializability Graph

Conclude the Proof

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