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A' is set of all possible schedules

'C' is set of all possible schedules that are guaranteed to produce a correct final result

'S' is the set of all serializable schedules

'P' is the set of all schedules possible under 2-phase locking protocol

Which is FALSE?

- $P\subseteq C$
- $S\subset P$
- $S\subseteq P$
- $P\subset C$

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A transaction schedule is **serializable** if its outcome (e.g., the resulting database state) is equal to the outcome of its transactions executed serially, i.e. without overlapping in time.

Seraializable is a superset of P and is equivalent to C.

Not all serializable schedules are allowed by 2PL, but all 2PL are serialisable, so S must be a superset of P.

Example S1: w1(x) w3(x) w2(y) w1(y) The lock by T1 for y must occur after w2(y), so the unlock by T1 for x must also occur after w2(y) (according to 2PL).

Because of the schedule legality, w3(x) cannot occur where shown in S1 because T1 holds the x lock at that point. However, S1 serializable (equivalent to T2, T1, T3).

So: B and C both are false according to me.

For diag and read: https://www2.cs.sfu.ca/CourseCentral/454/bzhou/documents/s7.pdf

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In databases and transaction processing, **two-phase locking** (**2PL**) is a concurrency controlmethod that guarantees serializability.

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then what about this schedule??https://www.gateoverflow.in/235026/test-series?show=243203

Is it serializable??

but 2PL is satisfying.

rt??

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