Question 46

Checking conflicting operations (pairs of operations from different transactions on the same data item where at least one is a write):

• Data item $z$: Accessed only by $T_1$ ($r_1(z), w_1(z)$) $\rightarrow$ no conflicts.
• Data item $x$: Accessed only by $T_2$ ($r_2(x), w_2(x)$) $\rightarrow$ no conflicts.
• Data item $y$: $T_3$ performs $r_3(y); w_3(y)$ before $T_2$ performs $r_2(y); w_2(y)$. Thus the conflicts $(r_3(y), w_2(y))$, $(w_3(y), r_2(y))$, and $(w_3(y), w_2(y))$ all yield the single directed edge $T_3 \rightarrow T_2$ in the precedence graph.

Since $T_1$ has no edges and $T_3$ must precede $T_2$, the serial schedules conflict-equivalent to $S$ are all topological sorts with $T_3$ before $T_2$: $T_1 T_3 T_2$ (B), $T_3 T_2 T_1$ (C), and $T_3 T_1 T_2$ (D).