Consider the three transactions \(T_{1}, T_{2},\) and \(T_{3},\) and the schedules \(S_{1}\) and \(S_{2}\) given below. Draw the serializability (precedence) graphs for \(S_{1}\) and \(S_{2},\) and state whether each schedule is serializable or not. If a schedule is serializable, write down the equivalent serial schedule(s). $$\begin{array}{l} T_{1}: r_{1}(X) ; r_{1}(Z) ; w_{1}(X) \\ T_{2}: r_{2}(Z) ; r_{2}(Y) ; w_{2}(Z) ; w_{2}(Y) \\ T_{3}: r_{3}(X) ; r_{3}(Y) ; w_{3}(Y) ; \\ S_{1}: r_{1}(X) ; r_{2}(Z) ; r_{1}(Z) ; r_{3}(X) ; r_{3}(Y) ; w_{1}(X) ; w_{3}(Y) ; r_{2}(Y) ; w_{2}(Z) ; w_{2}(Y) \\ S_{2}: r_{1}(X) ; r_{2}(Z) ; r_{3}(X) ; r_{1}(Z) ; r_{2}(Y) ; r_{3}(Y) ; w_{1}(X) ; w_{2}(Z) ; w_{3}(Y) ; w_{2}(Y) \end{array}$$

Step by step solution

01

Constructing the Serializability Graph for \(S_1\)

A serializability graph is built by considering each transaction as a node and creating directed edges between nodes based on read-write conflicts. In \(S_1\), \(T_1\) reads 'X' before \(T_3\) writes 'X', hence \(T_3\) precedes \(T_1\) (an edge from \(T_1\) to \(T_3\)). Similarly, \(T_2\) writes 'Y' before \(T_3\) reads 'Y', hence \(T_3\) also precedes \(T_2\) (an edge from \(T_2\) to \(T_3\)). Since there are no cycles, \(S_1\) is serializable.

02

Equivalent Serial Schedules for \(S_1\)

A serial schedule is equivalent to the given schedule if it maintains the order of conflicting read and write operations. Based on the constructed graph, possible equivalent serial schedules for \(S_1\) could be: \(T_1\), \(T_2\), \(T_3\) or \(T_1\), \(T_3\), \(T_2\).

03

Constructing the Serializability Graph for \(S_2\)

Following the same process as in Step 1 to analyze \(S_2\), \(T_1\) reads 'X' before \(T_3\) writes 'X', hence an edge from \(T_1\) to \(T_3\). \(T_2\) writes 'Y' before \(T_3\) reads 'Y', hence an edge from \(T_2\) to \(T_3\). But here, \(T_3\) writes 'Y' before \(T_2\) reads 'Y', creating an edge from \(T_3\) to \(T_2\), thus forming a cycle. Therefore, \(S_2\) is not serializable.

04

Equivalent Serial Schedules for \(S_2\)

Since \(S_2\) is not serializable, no equivalent serial schedules exist.

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