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Consider the following schedules involving two transactions:
S1 : r1 (X); r1 (Y); r2 (X); r2 (Y); w2 (Y); W1 (X)
S2 : r1 (X); r2 (Y); r2 (X); w2 (Y); r1 (Y); W1 (X)
S2 : r1 (X); r2 (Y); r2 (X); w2 (Y); r1 (Y); W1 (X)
Which of the following statements is TRUE?
- a)Both S1 and S2 are conflict serializable.
- b)S1 is conflict serializable and S2 is not conflict serializable.
- c)S1 is not conflict serializable and S2 is conflict serializable.
- d)Both S1 and S2 are not conflict serializable.
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
Consider the following schedules involving two transactions:S1 : r1 (X...
Consider S1 schedule:
S1
r1(X) → Read X process 1
r1(Y) → Read Y process 1
r2(X) → Read X process 2
r2(Y) → Read Y process 2
w2(Y) → But this write cause r1(Y)
w1(X) → The value is obsolete, so S1 is not conflict serializable.
S1
r1(X) → Read X process 1
r1(Y) → Read Y process 1
r2(X) → Read X process 2
r2(Y) → Read Y process 2
w2(Y) → But this write cause r1(Y)
w1(X) → The value is obsolete, so S1 is not conflict serializable.
Consider S2 schedule:
S2
r1(X) → Read X by process 1
r2(X) → Read X by process 2
r2(Y) → Read Y by process 3
w2(Y) →
Even here, there is no problem, since only transaction 2 has read Y and w1(X) is manipulating.
r1(Y) takes place after the above step, and S2 is thus conflict serializable.
S2
r1(X) → Read X by process 1
r2(X) → Read X by process 2
r2(Y) → Read Y by process 3
w2(Y) →
Even here, there is no problem, since only transaction 2 has read Y and w1(X) is manipulating.
r1(Y) takes place after the above step, and S2 is thus conflict serializable.
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Community Answer
Consider the following schedules involving two transactions:S1 : r1 (X...
Explanation:
Determine Conflict Serializable:
- To determine if a schedule is conflict serializable, we need to check if the schedule can be transformed into an equivalent serial schedule by swapping non-conflicting operations.
S1 Analysis:
- S1: r1(X); r1(Y); r2(X); r2(Y); w2(Y); w1(X)
- The conflicting pairs are r1(X) and w1(X) which creates a conflict.
- As there is a conflict in S1, it is not conflict serializable.
S2 Analysis:
- S2: r1(X); r2(Y); r2(X); w2(Y); r1(Y); w1(X)
- The conflicting pairs are r1(X) and w1(X), r2(Y) and w2(Y) which create conflicts.
- By swapping non-conflicting operations, we can transform S2 into an equivalent serial schedule: r1(X); w1(X); r2(Y); w2(Y); r1(Y); r2(X)
- Since S2 can be transformed into an equivalent serial schedule, it is conflict serializable.
Therefore, the correct statement is:
S1 is not conflict serializable and S2 is conflict serializable (Option C).
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Consider the following schedules involving two transactions:S1 : r1 (X); r1 (Y); r2 (X); r2 (Y); w2 (Y); W1 (X)S2 : r1 (X); r2 (Y); r2 (X); w2 (Y); r1 (Y); W1 (X)Which of the following statements is TRUE?a)Both S1 and S2 are conflict serializable.b)S1 is conflict serializable and S2 is not conflict serializable.c)S1 is not conflict serializable and S2 is conflict serializable.d)Both S1 and S2 are not conflict serializable.Correct answer is option 'C'. Can you explain this answer?
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