Serializability

Question

 

Comments

there are 2 $read(Y)$ in $T_{2}$

The first read(Y) read the initial value of Y. Isn't it?

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yes there is 2 read in T2 .

But in serial schedule only one read in T2 .

Ao it it not view equvalent????

i am canfused  here ????

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@ abhishekmehta4u it is also not view serializable because y is reading in T2 which is updated by T3 so it will make execution T3---->T2

initial read of Y is in T1 as well as T2 and write of Y is T3 ,so execution must (T1 T2)--->T3

so we can see from above both of order, no order is possible.

Answer 1

D - Not conflict serializable, Not view serializable, Cascading and Can't say about recoverable.

Not Conflict Serializable: We have cycle in precedence graph for the conflicts, T2 R(Y) ---> T3 W(Y) and T3 W(Y) ---> T2 R(Y). Thus, it is not conflict serializable.

Not View Serializable: Check for blind writes, yes we have T3 W(Y). So, we will draw polygraph for it. We add two dummy transactions Tb and Tf with W(X) W(Y) and R(X) R(Y) respectively. For WR(Y) from Tb ---> T2, so T3 W(Y) must occur after T2 is over, so we have edge from T2--->T3. For WR(Y) we already have an edge from T3--->T2. So, this forms a cycle, hence not view serializable.

Cascading: Cascadeless means data item should not be read from non-committed transactions, but in the above schedule, we can clearly see, ex: T3 W(Y) ---> T2 R(Y). So, it is cascading.

Can't say about recoverable: Recoverable means transaction should not commit until the transaction from which data item has been read has been committed. Here, we are not given any details about commit of the transactions, so we can't say anything.

Comments

Someone please confirm the answer...

 Isn't it view serialisable ??