Concurrency Control (View Serializability)

Question

Read(Q)                                                  

                          Write(Q)                                    

Write(Q)

Is this schedule View serializable????

in <T1,T2> it violates the final Write criteria

in <T2,T1> does it violate initial Read criteria?

Answer 1

not view serilalizable.coz

fot two trannsation possible schedules are

1 in <T1,T2> it violates the final Write criteria

2.  in <T2,T1>  it violate initial Read criteria coz

         the write of t2 will come first the t1 will read it

so we r not getting any view equivalance bw two transcation.so not vs

Comments

But in the video of nptel prof. says

For each Q , if T_i READS the *initial* value of Q in S1, then T_i MUST in S2 also READ the *initial* value of Q.

Is it the fact that it won't be reading *initial* value as it is being written that is why it is not vs? If not that what does *initial* mean here??? Prof. PPC says it is vs. i am a little confused here.

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this line is not clear to me:

For each Q , if Ti READS the *initial* value of Q in S1, then Ti MUST in S2 also READ the *initial* value of Q

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That means, if there are two schedules S1 and S2, then a transaction Ti will be doing the initial read i.e. first READ of a memory variable in S2 if it is doing the initial read of that variable in S1.