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Consider following Schedule S with data item x :

S : W1(X) R2(X) W3(X) R4(X) W5(X) R6(X) W7(X) R8(X) W9(X) R10(X)

The number of schedule view equivalent to Schedule S but not conflict equivalent to Schedule S ?
in Databases 499 views

1 Answer

1 vote
Conditions for view equal:

Initial  Read : -

Updated Read: 1>2,3>4,5>6,7>8,9>10

Final write : 9

So 9>10 transaction order is at the final ,with the other 4 precedences permutations.

Total permutations of 1>2,3>4,5>6,7>8 = 4! = 24 ways ( including the one given in the question)

We can observe that since every transaction is conflicting with every adjacent transaction , no other schedule is conflict equivalent.
So total number of view equal schedules to above and not conflict equivalent = 23
0
at the end why we are doing -1 from 24
1
As there is only one schedule possible which is conflict serializable i.e. is presently u ar seeing on screen. And any change in it will lead to transaction order change or loop .
0
And what about initial read?
2
why you have not fixed initial read??
0
because there is blind write before initial read so initial read condition doesn’t hold anymore

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