Consider transactions T1: r1(x); r1(z); w1(z) T2: r2(y); r2(z); w2(z) T3: r3(y); r3(x); w3(y), and schedules S1: r1(x); r3(y); r3(x); r2(y); r2(z); w3(y); w2(z); r1(z); w1(x); w1(z) and S2: r1(x); r3(y); r2(y); r3(x); r1(z); r2(z); w3(y); w1(x); w2(z); w1(z). Which statement is TRUE?

1. AOnly S₁ is conflict-serializable
2. BOnly S₂ is conflict-serializable
3. CBoth S₁ and S₂ are conflict-serializable
4. DNeither S₁ nor S₂ is conflict-serializable

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Show answer & explanation

Correct answer: A. Only S₁ is conflict-serializable

Explanation

S1: r2(y) before w3(y) gives T2 → T3; r3(x) before w1(x) gives T3 → T1; w2(z) before r1(z)/w1(z) gives T2 → T1. No cycle, so S1 is conflict-serializable (T2, T3, T1). S2: r1(z) before w2(z) gives T1 → T2, but r2(z) before w1(z) gives T2 → T1 — a cycle, so S2 is not.