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Consider the following four schedules due to three transactions (indicated by the subscript) using read and write on a data item x, denoted by r(x) and w(x) respectively. Which one of them is conflict serializable?

  • a)
    r1(x); r2(x); w1(x); r3(x); w2(x)

  • b)
    r2(x); r1(x); w2(x); r3(x); w1(x)

  • c)
    r3(x); r2(x); w1(x); r2(x); w1(x)

  • d)
    r2(x); r3(x); w2(x); r1(x); w1(x)

Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Consider the following four schedules due to three transactions (indic...
r1(x); r2(x); w1(x); r3(x); w2(x)



Contains cycle; Not conflict serializable



Not contains cycle; conflict serializable.
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Most Upvoted Answer
Consider the following four schedules due to three transactions (indic...
Explanation:

  • Conflict serializability means that the schedule is equivalent to a serial schedule in which the order of any two conflicting operations is the same in both schedules.

  • Conflicting operations are those that operate on the same data item and at least one of them is a write operation.

  • We can use a precedence graph to determine if a schedule is conflict serializable.

  • If the precedence graph is acyclic, then the schedule is conflict serializable.



Precedence graphs:

  • The nodes represent transactions or operations.

  • If Ti performs a write operation on a data item that is later read by Tj, then there is a directed edge from Ti to Tj.

  • If there is a cycle in the graph, then the schedule is not conflict serializable.



Schedule analysis:

  • Option A: r1(x); r2(x);w1(x); r3(x); w2(x)

    • Precedence graph: T1 --> T3, T2 --> T3

    • Not conflict serializable due to cycle in graph.



  • Option B: r2(x); r1(x);w3(x); r2(x); w1(x)

    • Precedence graph: T1 --> T2, T1 --> T4, T3 --> T2, T3 --> T4

    • Not conflict serializable due to cycle in graph.



  • Option C: r3(x); r2(x);w1(x); r2(x); w1(x)

    • Precedence graph: T2 --> T1, T2 --> T4, T3 --> T1, T3 --> T4

    • Not conflict serializable due to cycle in graph.



  • Option D: r2(x); r2(x);w3(x); r1(x); w1(x)

    • Precedence graph: T2 --> T1, T4 --> T1, T4 --> T3

    • Conflict serializable since the graph is acyclic.





Therefore, option D is the correct answer as it is conflict serializable.
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