blind writes

Question

Consider the following schedule $S1$.

$S_{1}: R_{1}(C) R_{2}(C) W_{1}(A) W_{2}(A) W_{1}(C) R_{1}(B) R_{2}(B) W_{1}(B) W_{1}(D) W_{2}(B) W_{2}(D) R_{1}(F) W_{3}(E) R_{3}(F)$

Let $'X'$ be the number of 'blind-write' operations in the given schedule and $'z'$ be the number of conflict equivalent serial schedules to $S_{1}$. The value of $X*Z$ is_________.

Answer 1

The Schedule is not conflict serializable. This can be checked by the precedence graph, it contains a cycle. So the number of conflict equivalent serial schedule is $0$. It has 5 blind writes that are the writes without any prior read of that data item.

So the value of $Z*X= 0*5=0$

Comments

Is not there 5 blind writes ?

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oops yes !