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Number of schedules view equal to following  schedule :-

 r1(A), w1(B), r2(A), w2(B), r3(A), w3(B)

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there are total 30 possible view equivalent schedules, 1 is given to us hence there will be 29 view equivalent schedules to given schedule:

Note: It is "CANNOT arrange" in the attached pic.

Every transaction is writing B hence W3(B) has to be executed at the end, remaining operations can be arranged.

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6 votes
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Here blind write perform by w1(B) and w2(B) , blind write means write before read. 

now, w3(B) is fixed due to the rule "last operation on some item should be same in view equivalence schedule."

so leaving w3(B) we have 5 schedule, we can arrange them in 5! = 120 way .

r3(A) can be arranging 2! way 

r2(A) can be arranging 2! way

so total number of view equal schedule possible (120 / 2! 2! ) = 30

in all these 30 schedules  w3(B) is last transaction , it is fixed.

so number of view equal schedule is 30 .

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4 votes
4 votes

Every conflict serializable schedule is also view serializable ,  but every view serializale schedule is not conflict serializable .

That means conflict serializable is subset of view serializable. More number of vew serial schedule is possible.

Also in this question , number of view-equivalent schedules of given schedule is asked .

There is a difference between view equal and view equal to a serial schedule.

View equal schedule = (concurrent + serial )  schedule

View view-equivalent schedules to a serial schedule = only serial schedule is consider 

If a schedule S is view-equivalent to a serial schedule, we say S is view-serializable , Clearly view equal and view serializable are different from this line .

If it is given view equal only then it can include concurrent schedules also .That is the reason we can't draw polygraph here..as in polygraph we draw it for serial schedules only .

We assume either one transaction executes before or after another in serial schedule but the operations of two can be interleaved in between in concurrent schedule.

Here blind write perform by w1(B) and w2(B) , blind write means write before read. 

now, w3(B) is fixed due to the rule "last operation on some item should be same in view equivalence schedule."

so leaving w3(B) we have 5 schedule, we can arrange them in 5! = 120 way .

r3(A) can be arranging 2! way 

r2(A) can be arranging 2! way

so total number of view equal schedule possible (120 / 2! 2! ) = 30

in all these 30 schedules  w3(B) is last transaction , it is fixed.

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