GATE CS 2010 Q42 — Transactions and Concurrency Control

To find the correct serialization order of the given schedule, we construct its precedence graph by identifying conflicting operations. Two operations conflict if they belong to different transactions, access the same data item, and at least one of them is a Write operation.

Let's analyze the conflicting operations chronologically:

1. Conflicts on data item $Y$:
   • $T3: \text{Read}(Y)$ occurs before $T2: \text{Write}(Y)$. This gives the precedence edge: $$T3 \to T2$$
2. Conflicts on data item $X$:
   • $T1: \text{Read}(X)$ occurs before $T3: \text{Write}(X)$. This gives: $$T1 \to T3$$
   • $T1: \text{Write}(X)$ occurs before $T3: \text{Write}(X)$. This also gives: $$T1 \to T3$$
   • $T3: \text{Write}(X)$ occurs before $T2: \text{Read}(X)$. This gives: $$T3 \to T2$$
   • $T3: \text{Write}(X)$ occurs before $T2: \text{Write}(X)$. This also gives: $$T3 \to T2$$

Combining all the precedence constraints, we get the following directed edges in the precedence graph:

• $T1 \to T3$
• $T3 \to T2$

Since there are no cycles, the schedule is conflict-serializable. Performing a topological sort on this graph yields the unique serialization order:

Therefore, the correct option is A.