Consider the following transactions with data items $P$ and $Q$ initialized to zero: $T_1$ : read (P);read (Q);if P = 0 then Q := Q + 1 ;write (Q). $T_2$ : read (Q);read (P);if Q = 0 then P := P + 1 ;write (P). Any non-serial interleaving of $T_1$ and $T_2$ for concurrent execution leads to

Let us analyze the read and write operations of transactions $T_1$ and $T_2$:

• $T_1$: $R_1(P)$, $R_1(Q)$, $W_1(Q)$

• $T_2$: $R_2(Q)$, $R_2(P)$, $W_2(P)$

Conflicting pairs between $T_1$ and $T_2$ are:

1. On item $Q$: $R_2(Q)$ and $W_1(Q)$, $W_1(Q)$ and $R_2(Q)$

2. On item $P$: $R_1(P)$ and $W_2(P)$, $W_2(P)$ and $R_1(P)$

In any non-serial interleaving:

• $T_1$ reads $P$ before $T_2$ writes $P$, creating a dependency edge $T_1 \to T_2$.

• Simultaneously, $T_2$ reads $Q$ before $T_1$ writes $Q$, creating a dependency edge $T_2 \to T_1$.

Because the precedence (serialization) graph contains a cycle ($T_1 \to T_2$ and $T_2 \to T_1$), the resulting schedule is not conflict serializable.

Therefore, option (B) is the correct choice.