Conflict and View serializable

Question

Comments

It should be I and VI. is it?

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yes, correct.

it is only view equivalent to T1->T2->T3

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@joshi,@Aswani T2 is blindwriting and then value is read by T3 what difference will it make if i schedule w(a) before T2`s write(a) do i have to assume that T2 reading input or have read input previously?

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what difference will it make if i schedule w(a) before T2`s write(a)

whose w(a) ?? 

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Transaction A write on (A).

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why cant it be T2->T1->T3 ?

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@Parshu

because if you do T2->T1->T3, then initial read R1(A) will not be preserved.

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@joshi and what about T1->T2->T3?

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@saxena

yes, T1->T2->T3 is only view equivalent to above schedule.

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@ Parshu gate

For this schedule T2->T1->T3

Yes T1 is reading the data item for the first time but it is not actually reading the value of the data item that was before the start of all the transactions as it will be overwritten by T2.

So, this is not VS

Answer 1

i and vi

No cycle in polygraph and possible order is T1->T2->T3 according to the topological order of polygraph.