Consider schedules \(S_{3}, S_{4},\) and \(S_{5}\) below. Determine whether each schedule is strict, cascadeless, recoverable, or nonrecoverable. (Determine the strictest recoverability condition that each schedule satisfies.) $$\begin{array}{l} S_{3}: r_{1}(X) ; r_{2}(Z) ; r_{1}(Z) ; r_{3}(X) ; r_{3}(Y) ; w_{1}(X) ; c_{1} ; w_{3}(Y) ; c_{3} ; r_{2}(Y) ; w_{2}(Z) ; w_{2}(Y) ; c_{2} \\ S_{4}: r_{1}(X) ; r_{2}(Z) ; r_{1}(Z) ; r_{3}(X) ; r_{3}(Y) ; w_{1}(X) ; w_{3}(Y) ; r_{2}(Y) ; w_{2}(Z) ; w_{2}(Y) ; c_{1} ; c_{2} ; c_{3} \\ S_{5}: r_{1}(X) ; r_{2}(Z) ; r_{3}(X) ; r_{1}(Z) ; r_{2}(Y) ; r_{3}(Y) ; w_{1}(X) ; c_{1} ; w_{2}(Z) ; w_{3}(Y) ; w_{2}(Y) ; c_{3} ; c_{2} \end{array}$$

Step by step solution

01

Analyze Schedule \(S_{3}\)

In \(S_{3}\), \(t_{1}\) writes item X after \(t_{3}\) reads it. Additionally, \(t_{2}\) reads item Y and then writes it after \(t_{3}\) writes it. Therefore, \(S_{3}\) is a nonrecoverable schedule as there are uncommitted changes.

02

Analyze Schedule \(S_{4}\)

In \(S_{4}\), every transaction that writes an item commits before another transaction reads that item. It means that, no transaction reads an uncommitted change from another transaction and every transaction reads only committed changes. Hence, \(S_{4}\) is a strict schedule.

03

Analyze Schedule \(S_{5}\)

In \(S_{5}\), every transaction which writes an item, commits after all other transactions have finished their reading and before they start any writing operation. There is no read operation that is reading uncommitted change from another transaction, however there are cases where a transaction reads a value that is later updated by another transaction. Therefore, \(S_{5}\) is recoverable but not cascadeless or strict.

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