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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 ?
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1 Answer

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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

4 Comments

And what about initial read?
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why you have not fixed initial read??
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2
because there is blind write before initial read so initial read condition doesn’t hold anymore
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