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:
\[ P:\ S \text{ is conflict-serializable.} \] \[ Q:\ \text{If } T_3 \text{ commits before } T_1 \text{ finishes, then } S \text{ is recoverable.} \] Which one of the following choices is correct?

Question

Let $R(z)$ and $W(z)$ denote read and write operations on a data element $z$ by a transaction $T_i$, respectively. Consider the schedule $S$ with four transactions.
\[ S: R_1(x) R_2(x) R_3(y) W_1(y) W_2(x) W_3(y) R_4(y) \] Which one of the following serial schedules is conflict equivalent to $S$?

Answer 1

To determine the correctness of the statements $P$ and $Q$, we need to analyze the schedule given for the operations of transactions $T_1, T_2, T_3$.

The schedule provided is:

$R_2(Y), R_1(X), R_3(Z), R_1(Y), W_1(X), R_2(Z), W_2(Y), R_3(X), W_3(Z)$

Step 1: Conflict Serializability

For a schedule to be conflict-serializable, it must be possible to transform the schedule into a serial schedule by swapping non-conflicting operations.

Operations conflict if they are from different transactions, involve the same data item, and at least one of these operations is a write.

Reading Conflicts: These do not cause conflicts unless followed by a writing operation.
Write after Read (WAR), Write after Write (WAW), or Read after Write (RAW): All these can cause conflicts.

Let's analyze the conflicts:

$R_2(Y)$ vs $W_2(Y)$ (Same transaction, ignored)
$R_1(X)$ vs $W_1(X)$ (Same transaction, ignored)
$R_2(Z)$ vs $W_3(Z)$ (Conflict between $T_2$ and $T_3$ due to write after read)
$R_3(X)$ vs $W_1(X)$ (Conflict between $T_3$ and $T_1$ due to read after write)

From	To
$T_3$	$T_1$
$T_2$	$T_3$

Since the conflict graph has no cycles, the schedule is conflict-serializable. So, statement $P$ is true.

Step 2: Recoverability

For a schedule to be recoverable, no transaction should commit before another transaction from which it has read data (and that data has been modified).

Let's analyze the condition given:

If $T_3$ commits before $T_1$ finishes:

$T_3$ reads $X$ written by $T_1$ ($R_3(X)$ after $W_1(X)$)
If $T_3$ commits before $T_1$, then it leads to a non-recoverable schedule because $T_3$ has read $X$ from $T_1$, which hasn't yet committed.

This makes the schedule non-recoverable if $T_3$ commits before $T_1$ finishes. Thus, statement $Q$ is false.

Conclusion

Therefore, the correct answer is that $P$ is true and $Q$ is false.

From	To
\(T_3\)	\(T_1\)
\(T_2\)	\(T_3\)

Show Hint

The Correct Option is B

Solution and Explanation

Step 1: Conflict Serializability

Step 2: Recoverability

Conclusion

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