Open In App
Related Articles

GATE | GATE-CS-2014-(Set-3) | Question 65

Improve Article
Save Article
Like Article

Consider the transactions T1, T2, and T3 and the schedules S1 and S2 given below.

T1: r1(X); r1(Z); w1(X); w1(Z)
T2: r2(Y); r2(Z); w2(Z)
T3: r3(Y); r3(X); w3(Y)
S1: r1(X); r3(Y); r3(X); r2(Y); r2(Z);
    w3(Y); w2(Z); r1(Z); w1(X); w1(Z)
S2: r1(X); r3(Y); r2(Y); r3(X); r1(Z);
    r2(Z); w3(Y); w1(X); w2(Z); w1(Z) 

Which one of the following statements about the schedules is TRUE?
(A) Only S1 is conflict-serializable.
(B) Only S2 is conflict-serializable.
(C) Both S1 and S2 are conflict-serializable.
(D) Neither S1 nor S2 is conflict-serializable.

Answer: (A)

Explanation: For conflict serializability of a schedule( which gives same effect as a serial schedule ) we should check for conflict operations, which are Read-Write, Write-Read and Write-Write between each pair of transactions, and based on those conflicts we make a precedence graph, if the graph contains a cycle, it’s not a conflict serializable schedule.

To make a precedence graph: if Read(X) in Ti followed by Write(X) in Tj ( hence a conflict ), then we draw an edge from Ti to Tj ( Ti -> Tj)

If we make a precedence graph for S1 and S2 , we would get directed edges for S1 as T2->T1, T2->T3, T3->T1, and for S2 as T2->T1, T2->T3, T3->T1, T1->T2. In S1 there is no cycle, but S2 has a cycle. Hence only S1 is conflict serializable.

Note : The serial order for S1 is T2 -> T3 -> T1.

Quiz of this Question

Level Up Your GATE Prep!
Embark on a transformative journey towards GATE success by choosing Data Science & AI as your second paper choice with our specialized course. If you find yourself lost in the vast landscape of the GATE syllabus, our program is the compass you need.

Last Updated : 28 Jun, 2021
Like Article
Save Article
Similar Reads