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While reading this solved quiz on concurrency control, I got stuck on exercise 2, which reads as follows:

2) Consider the following schedule:

S: r1(X), w2(X), r3(X), r1(Y), w2(Z), r3(Y), w3(Z), w1(Y)

(a) (2 points) Draw the precedence graph.

enter image description here

(b) (1 point) Is S conflict serializable?

No.

After reading that S is not conflict-serializable, my question is: can we say that S is serializable? Any suggestions?

Thanks to anyone who help me.

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    $\begingroup$ Please include all necessary material in the question. People shouldn't have to follow a link to find out what you're even asking. $\endgroup$ – David Richerby Aug 14 '15 at 19:11
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A schedule is considered serializable if it is equivalent to a serial one. The conflict graph that appears on the answer to item (a) is proof that the schedule is not conflict-serializable, and it is also not view-serializable, because there is still a cycle between $T_1$ and $T_3$, when considering just the history of the data they visualize. That means the schedule cannot be considered serializable at all.

Conflict-serializability is stronger than view-serializability. It is the criterion used for the implementation of most schedulers, and that is why the quiz emphasizes this point.

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