Question

Consider the following four schedules due to three transactions (indicated by the subscript) using read and write on a data item x, denoted by r(x) and w(x) respectively. Which one of them is conflict serializable?

s1: r1(x); r2(x); w1(x); r3(x); w2(x)
s2: r2(x); r1(x); w2(x); r3(x); w1(x)
s3: r3(x); r2(x); r1(x); w2(x); w1(x)
s4: r2(x); w2(x); r3(x); r1(x); w1(x) (1) s1 (2) s2 (3) s3 (4) s4 answer:
o S1
o S2
o S3
o S4

Answer 1

Among the given schedules, S2, S3, and S4 are conflict serializable. However, S2 is the most straightforwardly conflict serializable, corresponding to the serial order 2 -> 1 -> 3.

Step-by-step explanation:

To determine which of the given schedules are conflict serializable, we have to look for conflicts between transactions involving the same data item. A conflict occurs when two operations are from different transactions, at least one operation is a write, and they are on the same data item.

Let's analyze the schedules:

1. S1: r1(x); r2(x); w1(x); r3(x); w2(x) - There is a conflict between w1(x) and w2(x), and the ordering r2(x) before w1(x) cannot be serialized, so S1 is not conflict serializable.
2. S2: r2(x); r1(x); w2(x); r3(x); w1(x) - This schedule has conflicts, such as between w2(x) and w1(x), which can be serialized as 2 -> 1 -> 3, making S2 conflict serializable.
3. S3: r3(x); r2(x); r1(x); w2(x); w1(x) - Similar to S2, conflicts can be managed in a serial order of 3 -> 2 -> 1, so S3 is also conflict serializable.
4. S4: r2(x); w2(x); r3(x); r1(x); w1(x) - This schedule contains writes by both transaction 1 and 2 on x but can be ordered as 2 -> 3 -> 1, making S4 conflict serializable.

Among these options, both S2, S3, and S4 are conflict serializable. However, since the question is looking for a single correct option and considering the given sequence of operations, S2 is the most straightforward conflict serializable, as it easily translates into the serial order 2 -> 1 -> 3.