The given schedule S as follows....
For a serial schedule to be view equivalent to the given schedule , the initial read and final writes
on all the data items (here x and y) must be the same
Lets consider the serial schedule S1 (T1 --> T2 --> T3) .....
here the final write operation on y must be done by transaction T1 but in this
case it is done by T3 (the red highlighted ones are the other violations) ,
so this schedule is not view equivalent to S
Lets consider the serial schedule S2 (T1 --> T3 --> T2).....
here the final write on y is done by Transaction T2
Similarly we can find that the serial schedules T2 --> T1 --> T3 , T2 --> T3 --> T1 , T3 --> T1 --> T2
are not view equivalent to schedule S
The only schedule which is view equivalent to S as follows...
So there are 5 serial shedules which are not view equivalent to S....
