Let $r_i(z)$ and $w_i(z)$ denote read and write operations respectively on a data item $z$ by a transaction $T_i$. Consider the following two schedules. $$S_1 : r_1(x)\ r_1(y)\ r_2(x)\ r_2(y)\ w_2(y)\ w_1(x)$$ $$S_2 : r_1(x)\ r_2(x)\ r_2(y)\ w_2(y)\ r_1(y)\ w_1(x)$$ Which one of the following options is correct?

To check for conflict serializability, we construct the precedence (serialization) graph for each schedule:

1. Schedule $S_1$:

   • Conflicting operations on item $y$: $r_1(y)$ is executed before $w_2(y)$, which creates a directed edge $T_1 \rightarrow T_2$.

   • Conflicting operations on item $x$: $r_2(x)$ is executed before $w_1(x)$, which creates a directed edge $T_2 \rightarrow T_1$.

   • The precedence graph contains a cycle: $T_1 \leftrightarrow T_2$. Therefore, $S_1$ is not conflict serializable.

2. Schedule $S_2$:

   • Conflicting operations on item $y$: $w_2(y)$ is executed before $r_1(y)$, which creates a directed edge $T_2 \rightarrow T_1$.

   • Conflicting operations on item $x$: $r_2(x)$ is executed before $w_1(x)$, which creates a directed edge $T_2 \rightarrow T_1$.

   • The precedence graph contains no cycle (only the edge $T_2 \rightarrow T_1$). Therefore, $S_2$ is conflict serializable.

Hence, $S_1$ is not conflict serializable, and $S_2$ is conflict serializable, which corresponds to option B.