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Best answer
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14 votes

Given schedule s: w1(a) r2(a) w3(a) r4(a) w5(a) r6(a)

The Definition to become view equivalent serial schedule 

1) Initial read must be same.

2)  Any write by Ti  followed by a read in Tj , the same condition should be followed

[ that means We need to preserve Write-Read dependency( W-R ) . ]

3) last operation on some item should be same in view equivalence schedule.

Now, in this question there is blind write by T1 and T3.

and  WR is maintained by T1T2 , T3T4  and  

final / last operation is  T5 T6 

according to definition T5 T6 should be same hence we can arrange T1T2 , T3T4 in 2 ways

first order -     T1T2 , T3T4

second order - T3T4 , T1T2 

so in total there are 2  view equivalent serial schedules are possible where last operation T5T6 is fixed.

Note:

http://quiz.geeksforgeeks.org/dbms-how-to-test-two-schedule-are-view-equal-or-not-2/

Similar problem:

https://gateoverflow.in/13348/solve

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4 votes
4 votes

First of all thanks to Bikram Sir for clearing my doubt and giving me source for it.

So, I solved it using Polygraph Method, which may easy to understand.

First Image: Illustration of Polygraph Method by example

Second Image: Solution of Both Questions(including https://gateoverflow.in/13348/solve)

Sir, can you please check my Polygraphs and correct me!

–1 votes
–1 votes
is the answer c?
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