Consider the following read-write schedule $S$ over three transactions $T_1$, $T_2$, and $T_3$, where the subscripts in the schedule indicate transaction IDs: $$S: r_1(z); w_1(z); r_2(x); r_3(y); w_3(y); r_2(y); w_2(x); w_2(y);$$ Which of the following transaction schedules is/are conflict equivalent to $S$ ?

To determine the serial schedules conflict equivalent to $S$, we construct the precedence (serialization) graph by finding all pairs of conflicting operations (operations on the same data item by different transactions where at least one is a write):

1. Operations on $z$:

   • Only $T_1$ accesses $z$ ($r_1(z)$, $w_1(z)$). There are no conflicts with other transactions involving data item $z$.

2. Operations on $x$:

   • Only $T_2$ accesses $x$ ($r_2(x)$, $w_2(x)$). There are no conflicts involving $x$.

3. Operations on $y$:

   • $T_3$ has $r_3(y)$ and $w_3(y)$.

   • $T_2$ has $r_2(y)$ and $w_2(y)$.

   • In schedule $S$, $w_3(y)$ precedes $r_2(y)$ and $w_2(y)$, which creates a conflict dependency from $T_3$ to $T_2$ ($T_3 \to T_2$).

Thus, the only edge in the precedence graph is:

$T_1$ has no edges connected to it, so any topological ordering must only satisfy that $T_3$ appears before $T_2$.

The valid topological sortings (conflict-equivalent serial schedules) of $\{T_1, T_2, T_3\}$ are:

• $T_1 T_3 T_2$ (Option B)

• $T_3 T_1 T_2$ (Option D)

• $T_3 T_2 T_1$ (Option C)

Option A ($T_1 T_2 T_3$) violates the precedence constraint $T_3 \to T_2$. Hence, the correct options are B, C, and D.