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Consider the following two schedules: S1: r₃(X); r₁(X); r₁(Z); r₂(Z); r₃(Y); W₁(X); C₁; W₃(Y); C₃;…

In both S1 and S2, the conflicting operations give the precedence edges T3 -> T1 (from r3(X) before W1(X)), T1 -> T2 (from r1(Z) before W2(Z)), and T3 -> T2 (from W3(Y)…

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Consider the following two schedules:
S1: r₃(X); r₁(X); r₁(Z); r₂(Z); r₃(Y); W₁(X); C₁; W₃(Y); C₃; r₂(Y); W₂(Y); W₂(Z); C₂;
S2: r₁(Z); r₂(Z); r₃(Y); r₁(X); r₃(X); W₁(X); W₃(Y); r₂(Y); W₂(Y); W₂(Z); C₁; C₂; C₃;
Which of the following is true regarding above schedules?

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Answer and explanation

Correct answer A

In both S1 and S2, the conflicting operations give the precedence edges T3 -> T1 (from r3(X) before W1(X)), T1 -> T2 (from r1(Z) before W2(Z)), and T3 -> T2 (from W3(Y) before r2(Y) and W2(Y)), which is acyclic, so both S1 and S2 are conflict serializable in the order T3 -> T1 -> T2. In S1, T2 reads Y written by T3 only after T3 commits (C3 precedes r2(Y)), making S1 both cascadeless and recoverable. In S2, T2 reads Y written by T3 before T3 commits and commits (C2) before C3, so S2 is neither recoverable nor cascadeless.

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