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Consider the transactions $T1, T2, \:\text{and} \:T3$ and the schedules $S1 \:\text{and} \:S2$ given below. 

$T1: r1(X); r1(Z); w1(X); w1(Z) $

$T2: r2(Y); r2(Z); w2(Z) $

$T3: r3(Y); r3(X); w3(Y) $

$S1: r1(X); r3(Y); r3(X); r2(Y); r2(Z); w3(Y); w2(Z); r1(Z); w1(X); w1(Z) $

$S2: r1(X); r3(Y); r2(Y); r3(X); r1(Z); r2(Z); w3(Y); w1(X); w2(Z); w1(Z) $

Which one of the following statements about the schedules is TRUE?

  1. Only $S1$ is conflict-serializable.
  2. Only $S2$ is conflict-serializable.
  3. Both $S1$ and $S2$ are conflict-serializable.
  4. Neither $S1$ nor $S2$ is conflict-serializable.
in Databases by Veteran (105k points) | 2.3k views

2 Answers

+33 votes
Best answer

     

$S_1$ has no cycle hence, Conflict-Serializable

$S_2$ has cycle hence NOT Conflict-Serializable

Answer is option A.

by Boss (30.8k points)
edited by
0
In schedule S1: r1(z), w1(x), w1(z) are not taken into consideration for conflicts.

Is it because they are present at the last and all other operations have happened before them?
+1

@Gupta731

Is it because they are present at the last and all other operations have happened before them?

yes,

0
Precedence graph is always between conflicting pairs. right?

like R(x)W(x), W(x)R(x) & W(x)W(x).
+4 votes
In s1 there is no cycle ..serial order of execution for s1 is t2 t3 and t1 in s2 there is cycle in precedence graph so not conflict serializable so ans is a
by Boss (31.4k points)
Answer:

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