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Consider the following schedules involving two transactions. Which one of the following statements is TRUE?

• $S_1 :r_1(X); r_1(Y); r_2(X); r_2(Y); w_2(Y); w_1(X)$
• $S_2 :r_1(X); r_2(X); r_2(Y); w_2(Y); r_1(Y); w_1(X)$
1. Both $S_1$ and $S_2$ are conflict serializable.

2. $S_1$ is conflict serializable and $S_2$ is not conflict serializable.

3. $S_1$ is not conflict serializable and $S_2$ is conflict serializable.

4. Both $S_1$ and $S_2$ are not conflict serializable.

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Also S1: not a view serializable as no blind write present

For $S1$ : it is not conflict serializable

for $S2$ : it is conflict serializable

edited
You can make a Precedence graph for the two schedules.And check for the conflicts pairs making a Cycle.

S1 makes a cycle. T1-->T2 and T2-->T1. And hence is not Conflict Serializable.

S2 does not makes cycle and has transition from T2-->T1 only. And hence is Conflict Serializable.

Conflict Pairs: RW, WR, WW.
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These WR RW WW are adjacent in the transactions T1 and T2 or any where (in cross or different) in the transaction. please mention.

Ans - Option C

In S1: RW conflict from 1-2 and 2-1 so It is not C.S.

In S2: RW conflict AND WR conflict from 2-1 only so it is C.S.
As per the precedence graph there is a cycle in schedule S1 T1--->T2 and T2---->T1
so that's why it is not conflict serializable.

Schedule S1
T1            T2
---------------------
r1(X)
r1(Y)
r2(X)
r2(Y)
w2(Y)
w1(X)

Schedule S2
T1            T2
---------------------
r1(X)
r2(X)
r2(Y)
w2(Y)
r1(Y)
w1(X)
In precedence graph for S2 there is only one edge that is T2--->T1(no cycle) so S2 is conflict serializable

OPTION IS A
OPTION IS B IS APPROPIATE ONE

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