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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?

  1. $S_1 \text{ and } S_2$
  2. $S_2 \text{ and } S_3$
  3. $S_3$ only
  4. $S_4$ only
asked in Databases by Veteran (59.7k points)
edited by | 1.4k views

1 Answer

+20 votes
Best answer

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.

answered by Boss (19.9k points)
edited by
–2
Ans will be c because
0
no. it's b) you can check it by precednce graph.
0
Thanks @Sandeep for correction
0
I got the answere by drawing schedules but why T1 and T2 is given...??
0
to describe what belongs to T1 and wat to T2
Answer:

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