GATE Overflow Test Series | Mock GATE | Test 5 | Question: 11

Question

Which of the following statements is/are correct? (Mark all the appropriate choices)

• A. With strict $2PL,$ transactions can be serialized in the order in which they commit.
• B. With rigorous $2PL,$ transactions can not be always serialized in the order in which they commit.
• C. With strict $2PL,$ only exclusive locks are held until the end of the transaction.
• D. With rigorous $2PL,$ both shared and exclusive locks are held until the end of the transaction.

Comments

strict 2pl can be serialized in order of commit and also in the order of lockpoint it depends on the scenario so option A is also true please update.

Answer 1

With rigorous $2PL,$ locks are released only at the end of the transaction. It can be shown that with rigorous $2PL,$ transactions can be serialized in the order in which they commit. This is not the case for Strict $2PL.$ So, options A and B are False.

Strict $2PL,$ holds only exclusive locks until the end of the transaction.

So, only options C;D are TRUE.

Comments

if schedule has only shared lock for every data item  then in case of Strict 2 PL Schedule can be serialized in any order But then too you can serialize it in order in which the transaction commit given stmt 1 would have been wrong iff you have atleast 1 case where you cannot serialize the transactions in order in whic they commit is there any such case?

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Here’s an counter example for option A:

	$T_1$	$T_2$
	read_lock(X)
	write_lock(Y)
	R(X)
Locking point of $T_1\rightarrow$	release_lock(X)
		write_lock(X)
		R(X)
		W(X)
		commit
		release_lock(X)	$\leftarrow$ locking point of $T_2$
	W(Y)
	commit
	release_lock(Y)

 

In this schedule order in which they are committing is  $T_2 \rightarrow T_1 $, but it is not the equivalent serial schedule, (This schedule is not conflict equivalent to $T_2 \rightarrow T_1$, and we know 2PL ensures conflict serial schedule)

 

equivalent serial schedule in all variant of 2PL will be the order of locking point of transactions,

It so happens that for rigorous 2PL, locking point is same as the commit point, unlocking start after committing only.

and in conservative 2PL locking point is the starting point of each transaction..

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Still, the option $A$ says With strict $2PL$, transactions can be serialized…..

Which means $A$ will be satisfied even with a single example.

Implies Option A is also correct?

@Deepak Poonia Sir please check this