Let S be the following schedule of operations of three transactions $T_1$, $T_2$ and $T_3$ in a relational database system: $$R_2(Y), R_1(X), R_3(Z), R_1(Y), W_1(X), R_2(Z), W_2(Y), R_3(X), W_3(Z)$$ Consider the statements P and Q below: P: S is conflict-serializable. Q: If $T_3$ commits before $T_1$ finishes, then S is recoverable. Which one of the following choices is correct?

To evaluate statement P (Conflict Serializability), we identify all conflicting pairs of operations belonging to different transactions:

1. On data item $X$:

   • $W_1(X)$ occurs before $R_3(X)$, giving the directed edge $T_1 \to T_3$.

2. On data item $Y$:

   • $R_1(Y)$ occurs before $W_2(Y)$, giving the directed edge $T_1 \to T_2$.

3. On data item $Z$:

   • $R_2(Z)$ occurs before $W_3(Z)$, giving the directed edge $T_2 \to T_3$.

The edges in the precedence (serialization) graph are:

Since there are no cycles in the precedence graph, the schedule $S$ is conflict-serializable (with serializability order $T_1 \to T_2 \to T_3$). Hence, P is true.

To evaluate statement Q (Recoverability):

• Transaction $T_3$ reads $X$ after $T_1$ writes $X$ ($W_1(X) \dots R_3(X)$), which means $T_3$ reads dirty (uncommitted) data written by $T_1$.

• For a schedule to be recoverable, any transaction $T_j$ that reads a value written by transaction $T_i$ must commit only after $T_i$ commits.

• If $T_3$ commits before $T_1$ commits/finishes and $T_1$ subsequently aborts, the database enters an inconsistent, unrecoverable state.

• Thus, the schedule is not recoverable under this condition. Hence, Q is false.

Therefore, P is true and Q is false (Option B).