Let $R_i(z)$ and $W_i(z)$ denote read and write operations on a data element $z$ by a transaction $T_i$, respectively. Consider the schedule $S$ with four transactions. $S$: $R_4(x)R_2(x)R_3(x)R_1(y)W_1(y)W_2(x)W_3(y)R_4(y)$ Which one of the following serial schedules is conflict equivalent to $S$?

To find the conflict equivalent serial schedule, we identify all conflicting operation pairs (operations on the same data item by different transactions, where at least one is a write) in schedule $S$:

1. Conflicts on $x$:

   • $R_4(x)$ occurs before $W_2(x) \implies T_4 \rightarrow T_2$

   • $R_3(x)$ occurs before $W_2(x) \implies T_3 \rightarrow T_2$

2. Conflicts on $y$:

   • $R_1(y)$ occurs before $W_3(y) \implies T_1 \rightarrow T_3$

   • $W_1(y)$ occurs before $W_3(y) \implies T_1 \rightarrow T_3$

   • $W_1(y)$ occurs before $R_4(y) \implies T_1 \rightarrow T_4$

   • $W_3(y)$ occurs before $R_4(y) \implies T_3 \rightarrow T_4$

3. Precedence Graph Edges:

   • $T_1 \rightarrow T_3$

   • $T_3 \rightarrow T_4$

   • $T_4 \rightarrow T_2$

   • (along with $T_1 \rightarrow T_4$ and $T_3 \rightarrow T_2$, which are consistent with this order)

Performing a topological sort on this acyclic graph gives the unique serial order:

Thus, option (A) is the conflict equivalent serial schedule.