For any 2 schedule S1 and S2 to be View Equivalent 3 steps to be followed:
Step 1: Initial Reads: This means that the data item is been read from the database directly.
All the initial reads of all the transactions of Si schedule should be same as all the initial reads of all the transactions of Sj schedule. For example: If R1(X) and R2(Y) of S1 are initial reads then in S2 also R1(X) and R2(Y) should be initial reads.
Step 2: Updated reads: This means that the data item is been read from the updated data from the write transaction just above this read.
All the updated reads of all the transactions of Si schedule should be same as all the updated reads of all the transactions of Sj schedule. For example: If R1(X) reads after W2(X) and R2(Y) reads after W3(Y) of S1 are updated reads then in S2 also R1(X) should read after W2(X) and R2(Y) should read after W3(Y)
Step 3: Final Write: The transaction which is writing the data item at last in any schedule.
All the final write of all the transactions of Si schedule should be same as all the final write of all the transactions of Sj schedule. For example: If W1(X) writes data item X at last and W2(Y) writes data item Y at last of S1 then in S2 also W1(X) writes data item X at last and W2(Y) writes data item Y at last.
Here in the given example
Considering S1 and S2:
Initial reads: no initial reads in S1 and S2--------------------no problem
Updated Reads: In S1 - R3(X) reads after W2(X)and in S2 - R3(X) reads after W1(X)------------ not samesame
Final Write: In S1 - W2(X) and in S2 - W2(X)-------------------same
Therefore S1 and S2 are not view equal
so not view serializable also.
Considering S3 and S4:
Initial reads: in S3- R1(X) and S4- R1(X)--------------------same
Updated Reads: In S3 and S4 not present----------------no problem
Final Write: In S3 - W1(X) and in S4 - W2(X)-------------------not same
Therefore S3 and S4 are not view equal
so not view serializable also
so option b and d are correct
