Given schedule s: w1(a) r2(a) w3(a) r4(a) w5(a) r6(a)
The Definition to become view equivalent serial schedule
1) Initial read must be same.
2) Any write by Ti followed by a read in Tj , the same condition should be followed
[ that means We need to preserve Write-Read dependency( W-R ) . ]
3) last operation on some item should be same in view equivalence schedule.
Now, in this question there is blind write by T1 and T3.
and WR is maintained by T1T2 , T3T4 and
final / last operation is T5 T6
according to definition T5 T6 should be same hence we can arrange T1T2 , T3T4 in 2 ways
first order - T1T2 , T3T4
second order - T3T4 , T1T2
so in total there are 2 view equivalent serial schedules are possible where last operation T5T6 is fixed.
Note:
http://quiz.geeksforgeeks.org/dbms-how-to-test-two-schedule-are-view-equal-or-not-2/
Similar problem:
https://gateoverflow.in/13348/solve