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+18 votes

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.
asked in Databases by Veteran (96.1k points) | 2k views

3 Answers

+30 votes
Best answer


$S_1$ has no cycle hence, Conflict-Serializable

$S_2$ has cycle hence NOT Conflict-Serializable

Answer is option A.

answered by Boss (30.6k points)
edited by
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?


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


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
answered by Boss (30.9k points)
–6 votes

(D) Neither S1 nor S2 is conflict-serializable.

we if draw a dependency graph for both schedule it forms a cyclic graph, so we can say they are no conflict seriallzable.

answered by Active (3.3k points) 1 flag:
✌ Edit necessary (Lakshman Patel RJIT “Please edit you solution”)
S1 is CS.
You are right.
How you conclude that it wrong

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