Two schedules S and S' are said to be view equivalent if the following three conditions hold:
1. The same set of transactions participates in S and S', and S and S' include the
same operations of those transactions.
2. For any operation ri(X) of Ti in S, if the value of X read by the operation has
been written by an operation wj(X) of Tj (or if it is the original value of X
before the schedule started), the same condition must hold for the value of X
read by operation ri(X) of Ti in S'.
3. If the operation wk(Y) of Tk is the last operation to write item Y in S, then
w
k(Y) of Tk must also be the last operation to write item Y in S'.
Point 1 is very obvious.
Point 2 - simply says that reads should be similar
i.e. if in a Schedule S a Trans Ti reads a Data item 'A' which is updated by Trans Tj then in S' also Trans Ti must read data item 'A' modified by Tj only.Also initial reads should be same i.e. If Trans Ti reads 'A' from initial database in Schedule S then in S' also Ti should read 'A' from initial database only.
Point 3 - Final write should be same.If Ti updates 'A' finally(last updation) in S then in S' also 'A' must be finally updated by same Ti only
Now for the given question
Final Writes
A is written by T3,T4(final write).So in any View equal schedule T4 W(A) must be after T3 W(A).For View equal serial schedule we can say T4 must be executed after T3 ( denoted by T3->T4)
B is written by T1,T4.B is finally written by T4. So T1->T4
C and D are written by only one transactions so they can't cause final updation violation.
Now checking for Reads
T3 reads C which was updated by T1 so T1->T3
T2 reads B which was updated by T1 so T1->T2
T1->T4,T1->T3,T1->T2.So we can say T1 should be the First Trans
T4->T2 is not possible as in that case read R(B) of T2 will be from updation of Write of T4( instead of T1)
so T2->T4
with T3->T4 T1->4 and T2->T4 it is clear that T4 must be last trans
to check for T2 and T3 see that T3->T2 is not possible as T2 reads A from initial database.
so T2->T3
Final Answer T1->T2->T3->T4