Consider the following schedules involving two transactions.

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 statements is correct with respect to above?

This question was previously asked in
UGC NET Computer Science (Paper 2) July 2018 Paper
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  1. Both S1 and S2 are conflict serializable
  2. Both S1 and S2 are not conflict serializable
  3. S1 is conflict serializable and S2 is not conflict serializable
  4. S1 is not conflict serializable and S2 is conflict serializable

Answer (Detailed Solution Below)

Option 4 : S1 is not conflict serializable and S2 is conflict serializable
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Detailed Solution

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Concept:

Draw the precedence graph for the above two schedules: 

1) If precedence graph contains cycle, then that schedule will not be conflict serializable

2) If graph does not contain any cycle, then schedule will be conflict serializable.

Explanation:

S1: r1(X); r1(Y); r2(X) ; r2(Y) ; w2(Y) ; w1(X)

There are two transaction involved.

T1

T2

r1(x)

 

r1(y)

 

 

r2(x)

 

r2(y)

 

w2(y)

w1(x)

 

 

Precedence graph:

F2 Raju.S 24-04-2020 Savita D3

S2: r1(X) ; r2(X) ; r2(Y) ; w2(Y) ; r1(Y) ; w1(X)

T1

T2

r1(x)

 

 

r2(x)

 

r2(y)

 

w2(y)

r1(y)

 

w1(x)

 

 

Precedence graph :

F2 Raju.S 24-04-2020 Savita D4

From here, it is clear that S1 is not conflict serializable and S2 is conflict serializable.
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