View Serializability

Question

Consider the following schedules: 

Which of the following is correct about above schedule?

• A. Only view serializable
• B. Only conflict serializable
• C. Both conflict and view serializable
• D.  Neither view serializable nor conflict serializable.

Please tell me the approach for checking view serializability ?

Answer 1

• if a schedule is conflict serializable then it is view also.
• if schedule is non conflict then it may or may not view. 
• view serializable schedule is sufficent and necessary condition for serializability condition.i.e if a schedule is view then it must be a serialzable schedule.
• for checking view we need to check three condition.

1. final write 
2. initial read
3. updated read

• in this example final write  in transaction T1 on data item a. then this schedule must be execute in last(T2 T3----->T1). Simillarlly final write in transaction T3 on data item on b. then this transaction must be execute in last (T1 T2----> T3)
• NOW we make a serial schedule. there are 3 transaction so total 3!=6 serial schedule is possible. 
• here first schedule is T1 T2 T3 . but according to final write  T1 in the last. (T2 T3---->T1) . so this cant be equvalent to this serial schedule.
• simillarlly you are check for all serial schedule . then we get 2 schedule . 
• simillarly we check for initial read . then we get it is not equvalent to any serial schedule.and we not need to check for updated write.

so this is nigher conflict nor view

Comments

Yes your answer is correct but in final write why are u writing T1 T2->T3 ??because B is writing only in T3 (not in T,1,T2) so writing of B doesnot effect because it is alone.

And initial read of A is also in T2.

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yes brother, because of B it's not getting conflicting due to no where else B is writing.... because of initial reads (T_initial -> T₁, T_initial -> T2  and for  final writing T₂ -> T₁ here there is no loop )  then it is view serializable... correct me if i am wrong

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According to question T1 has initial read and last update on A.

Now ,

If T1 executes last then T1's initial read will be lost.

If T1 executes first then T2's initial read and T1's last update will be lost...

So, not view serializable