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Consider the following schedules involving three transaction:
S1 : W2(x), W1(x), R3(x), W2(y), R3(y), R3(z), R2(x), R1(y)
S2 : R2(z), W2(x), W2(y), R1(x), R3(x), R2(z), R3(y), W1(x)
Which of the above schedules are conflict serializable?
  • a)
    Both S1 and S2
  • b)
    Only S1
  • c)
    Only S2
  • d)
    Neither S1 nor S2
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Consider the following schedules involving three transaction:S1 : W2(...
A schedule is serializable if the precedence graph does not contain any cycle.
It contains cycles so it is not serializable.
It contains no cycle so is serializable.
So option (C) is correct.
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Most Upvoted Answer
Consider the following schedules involving three transaction:S1 : W2(...
Conflict serializability is a property of schedules in database systems. A schedule is said to be conflict serializable if it is equivalent to some serial schedule, where the order of conflicting operations is preserved.

To determine whether a schedule is conflict serializable, we can use the precedence graph method. In this method, we construct a directed graph called the precedence graph, where each transaction is represented by a node and there is an edge from Ti to Tj if there is a conflict between an operation of Ti and an operation of Tj.

Let's analyze each schedule using the precedence graph method:

S1 : W2(x), W1(x), R3(x), W2(y), R3(y), R3(z), R2(x), R1(y)

Precedence graph for S1:

T1 -> T2 -> T3
| |
V V
T1 -> T3 T2

In the precedence graph for S1, there are cycles (T1 -> T2 -> T3 -> T1) which indicates that the schedule is not conflict serializable.

S2 : R2(z), W2(x), W2(y), R1(x), R3(x), R2(z), R3(y), W1(x)

Precedence graph for S2:

T1 -> T2 -> T3
| |
V V
T1 <- t2="">

In the precedence graph for S2, there are no cycles and the graph is acyclic, indicating that the schedule is conflict serializable.

Therefore, only schedule S2 is conflict serializable.
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Consider the following schedules involving three transaction:S1 : W2(x), W1(x), R3(x), W2(y), R3(y), R3(z), R2(x), R1(y)S2 : R2(z), W2(x), W2(y), R1(x), R3(x), R2(z), R3(y), W1(x)Which of the above schedules are conflict serializable?a)Both S1 and S2b)Only S1c)Only S2d)Neither S1 nor S2Correct answer is option 'C'. Can you explain this answer?
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