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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.
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$S_1$ has no cycle hence, Conflict-Serializable

$S_2$ has cycle hence NOT Conflict-Serializable

by Boss (30.8k points)
edited
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

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).
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)