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

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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.

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Last Updated : 28 Jun, 2021
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