Consider the transactions $T_1, T_2, T_3$ and the schedules $S_1$ and $S_2$ given below.

$T_1 : r_1(x); r_1(z); w_1(z)$
$T_2 : r_2(y); r_2(z); w_2(z)$
$T_3 : r_3(y); r_3(x); w_3(y)$

$S_1$ :
$r_1(x); r_3(y); r_3(x); r_2(y); r_2(z); w_3(y); w_3(z); r_1(z); w_1(x); w_1(z)$

$S_2$ :
$r_1(x); r_3(y); r_2(y); r_3(x); r_1(z); r_2(z); w_3(y); w_1(x); w_2(z); w_1(z)$

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

To check conflict serializability we build precedence graph.

For $S_1$

Conflicts:

$T_1 \rightarrow T_3$
$T_3 \rightarrow T_2$
$T_2 \rightarrow T_1$

Cycle exists:

$T_1 \rightarrow T_3 \rightarrow T_2 \rightarrow T_1$

Therefore

$S_1$ not conflict-serializable

For $S_2$

Conflicts:

$T_3 \rightarrow T_1$
$T_2 \rightarrow T_1$
$T_3 \rightarrow T_2$

Graph has no cycle

Serializable order:

$T_3 \rightarrow T_2 \rightarrow T_1$

Thus

$S_2$ is conflict-serializable