GATE CSE 2009 | Question: 43

Question

Consider two transactions $T_1$ and $T_2$, and four schedules $S_1, S_2, S_3, S_4$,  of  $T_1$  and  $T_2$ as given below:

$T_1: R_1[x]W_1[x]W_1[y]$

$T_2: R_2[x]R_2[y]W_2[y]$

$S_1: R_1[x]R_2[x]R_2[y] W_1[x] W_1[y] W_2[y]$

$S_2: R_1[x]R_2[x]R_2[y] W_1[x] W_2[y] W_1[y]$

$S_3: R_1[x]W_1[x]R_2[x] W_1[y] R_2[y] W_2[y]$

$S_4: R_2[x]R_2[y]R_1[x] W_1[x] W_1[y] W_2[y]$

Which of the above schedules are conflict-serializable?

• A. $S_1 \text{ and } S_2$
• B. $S_2 \text{ and } S_3$
• C. $S_3$ only
• D. $S_4$ only

Comments

Can someone please explain in S1 how T2-> T1 is causing conflict? And making a cycle

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bro write-write is also a conflict. That’s why S1 is not conflict serilizable

Answer 1

The answer is B.

• S1 has a cycle from $T1\to T2$ and $T2 \to T1.$
• S2-- It is uni-directional and has only $T2\to T1.$
• S3-- It is uni-directional and has only $T1\to T2.$
• S4-- same as S1.

A schedule is conflict serializable if there is no cycle in the directed graph made by the schedules.

In the schedules we check for RW, WR, WW conflicts between the schedules and only these conflicts contribute in the edges of the graph.

Comments

I got the answere by drawing schedules but why T1 and T2 is given...??

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to describe what belongs to T1 and wat to T2

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in S2 there is a blind write..then how it could be

Answer 2

option b

Answer 3

BEST ANSWER

 

The answer is B.