Made-easy-test-series closed

Question

Consider the following schedule involving two transactions:

S1: r1(A); w1(A); w2(B); w3(C); r1(D); w2(D).

S2:r3(A); w4(B); r1(C); r3(D); w3(B); w2(D); r3(A); w1(D); r3(B); r2(C); r1(A).

A) only S1 is followed under 2PL.

B) only s2 is followed under 2PL.

C) both s1 and s2 are allowed under 2PL.

D) neither s1 nor s2 are allowed under 2PL.

Comments

Both are allowed under 2PL. For 2 i am getting T4, T3, T1, T2 as a sequence of execution.

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Utkarsh Joshi please check again

i'm getting T4 -> T3 -> T2 -> T1

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Yes that is also possible :)

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Utkarsh Joshi but answer given is D

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@Utakrsh joshi

Can u please check my doubt in that close as duplicate link

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Mk Utkarsh You are also getting both allowed under 2PL right?? if yes why to bother about the given solution :)

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@jatin khachane 1 no idea about that bro :(

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For the set of transactions to be in 2PL, we need to have 2 phases

phase 1 : locking 

phase 2 : unlocking

for now, conside s() as shared lock and x() as exclusive lock.

in s1 :  X(A);r1(A); w1(A); X(B)w2(B); X(C)w3(C); S(D)r1(D);  *here we cannot gain exclusive lock until lock is released* w2(D).

hence first one is not 2PL.

similarly the next one also.

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@Utkarsh Joshi

What is difference here between 'followed under' and 'allowed under'