# View Serializability

1 vote
499 views
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 ?

1 vote
Conditions for view equal:

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
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

## Related questions

1
1.2k views
I am looking for some clarity on this topic. Here is some random schedule as an example: $r1(x) w1(x) r2(x) w2(x) r3(y) r3(x) w3(x) c3 a1 c2$ I was told, that for conflict serializablity we simply ignore the aborted transaction, ... not ignored for recoverability, strictness or avoidance of cascading rollbacks. I would be grateful if the question would be answered since this really confuses me.
1 vote