Suppose these three transactions are given: T1: r1(a); r1(c); w1(c); T2: r2(a);

Question

Suppose these three transactions are given:
T1: r1(a); r1(c); w1(c);
T2: r2(a); r2(b); w2(a);
T3: r3(b); r3(c); w3(b); w3(c);
A "blind write" is is a write operation by a transaction of a data item that it never read.

Now for each of the schedules below, tell whether or not it is conflict-serializable and whether or not it is view-serializable. Give reasons. And if any answer is "yes," then provide the serial schedule to which it is equivalent:
S1: r2(a); r3(b); r2(b); r1(a); r1(c); w2(a); w1(c); r3(c); w3(b); w3(c);
S2: r2(a); r3(b); r1(a); r2(b); r3(c); r1(c); w2(a); w3(b); w1(c); w3(c);

Explanation / Answer

S1. Not Confict Serializable.

Not a blind write so Not view Serializable.

S2.  Not Confict Serializable.

Not a blind write so Not view Serializable.

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