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Let ri(z) and wi(z) denote read and write operations respectively on a data item z by a transaction Ti. Consider the following two schedules.
S1 : r1(x) r1(y) r2(x) r2(y) w2(y) w1(x)
S2 : r1(x) r2(x) r2(y) w2(y) r1(y) w1(x)
Which one of the following options is correct?
  • a)
    S1 is conflict serializable, and S2 is not conflict serializable.
  • b)
    S1 is not conflict serializable, and S2 is conflict serializable.
  • c)
    Both S1 and S2 are conflict serializable.
  • d)
    Neither S1 nor S2 is conflict serializable.
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
Let ri(z) and wi(z) denote read and write operations respectively on a...
Explanation:

Both schedules S1 and S2 are not conflict serializable because they contain conflicting operations.

Schedule S1:

S1 : r1(x) r1(y) r2(x) r2(y) w2(y) w1(x)

Conflict Serializability:

A schedule is conflict serializable if its precedence graph is acyclic. The precedence graph can be constructed by considering the conflicting operations in the schedule.

Precedence Graph for S1:


r1(x) r2(x) w2(y)
/ \ /
/ \ /
r1(y) r2(y)
\ / \
\ / \
w1(x)


Cycle in Precedence Graph:

From the precedence graph, it is clear that there is a cycle present in the graph (r1(x) -> w1(x) -> r1(y) -> r2(y) -> w2(y) -> r2(x) -> r1(x)). Therefore, schedule S1 is not conflict serializable.

Schedule S2:

S2 : r1(x) r2(x) r2(y) w2(y) r1(y) w1(x)

Precedence Graph for S2:


r1(x) r2(x) w2(y)
/ \ /
/ \ /
r1(y) r2(y)
\ / \
\ / \
w1(x)


No Cycle in Precedence Graph:

From the precedence graph, it is clear that there is no cycle present in the graph. Therefore, schedule S2 is conflict serializable.

Conclusion:

Based on the analysis, the correct option is (b) S1 is not conflict serializable, and S2 is conflict serializable.
Free Test
Community Answer
Let ri(z) and wi(z) denote read and write operations respectively on a...
S1: r1(x) r1(y) r2(x) r2(y) w2(y) w1(x)
Precedence Graph:
S2:r1(x) r2(x) r2(y) w2(y) r1(y) w1(x)
Precedence Graph:
There exists a Cycle in Precedence of S1. Hence S1 is not Conflict Serializable.
Whereas S2 is Conflict Serializable.
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Let ri(z) and wi(z) denote read and write operations respectively on a data item z by a transaction Ti. Consider the following two schedules.S1 : r1(x) r1(y) r2(x) r2(y) w2(y) w1(x)S2 : r1(x) r2(x) r2(y) w2(y) r1(y) w1(x)Which one of the following options is correct?a)S1 is conflict serializable, and S2 is not conflict serializable.b)S1 is not conflict serializable, and S2 is conflict serializable.c)Both S1 and S2 are conflict serializable.d)Neither S1 nor S2 is conflict serializable.Correct answer is option 'B'. Can you explain this answer?
Question Description
Let ri(z) and wi(z) denote read and write operations respectively on a data item z by a transaction Ti. Consider the following two schedules.S1 : r1(x) r1(y) r2(x) r2(y) w2(y) w1(x)S2 : r1(x) r2(x) r2(y) w2(y) r1(y) w1(x)Which one of the following options is correct?a)S1 is conflict serializable, and S2 is not conflict serializable.b)S1 is not conflict serializable, and S2 is conflict serializable.c)Both S1 and S2 are conflict serializable.d)Neither S1 nor S2 is conflict serializable.Correct answer is option 'B'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Let ri(z) and wi(z) denote read and write operations respectively on a data item z by a transaction Ti. Consider the following two schedules.S1 : r1(x) r1(y) r2(x) r2(y) w2(y) w1(x)S2 : r1(x) r2(x) r2(y) w2(y) r1(y) w1(x)Which one of the following options is correct?a)S1 is conflict serializable, and S2 is not conflict serializable.b)S1 is not conflict serializable, and S2 is conflict serializable.c)Both S1 and S2 are conflict serializable.d)Neither S1 nor S2 is conflict serializable.Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let ri(z) and wi(z) denote read and write operations respectively on a data item z by a transaction Ti. Consider the following two schedules.S1 : r1(x) r1(y) r2(x) r2(y) w2(y) w1(x)S2 : r1(x) r2(x) r2(y) w2(y) r1(y) w1(x)Which one of the following options is correct?a)S1 is conflict serializable, and S2 is not conflict serializable.b)S1 is not conflict serializable, and S2 is conflict serializable.c)Both S1 and S2 are conflict serializable.d)Neither S1 nor S2 is conflict serializable.Correct answer is option 'B'. Can you explain this answer?.
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