842 views
1 1 vote

Problem pic

I want to confirm my answers..

For 19.2.1

a) VIEW equal serial order T1 T2T3 or T2T1T3 

b) VIEW equal serial order 6 possible orders

c) 2 VIEW equal serial orders

d) 2 VIEW equal serial orders

For 19.2.2

A) 2 VIEW equal serial orders and one conflict equal one

b) 1 VIEW equal serial orders and 1 conflict equal..

{ all schedules have 1 conflict equal serial order}

Plz point out my mistakes.

1 Answer

0 0 votes
For a) It is not view serializable because if you take any other serial schedule, then order changes (eg. T2T3T1) i.e. initial reads order changes .
Position:
Show:

Related questions

10 10 votes
1 answers 1 answer
5.1k
5.1k views
flow2539 asked May 28, 2018
5,128 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...
3 3 votes
1 1 answer
2.1k
2.1k views
1 1 vote
1 1 answer
2.3k
2.3k views
Na462 asked Jan 13, 2019
2,329 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 n...
3 3 votes
3 3 answers
6.3k
6.3k views
shivangi5 asked Dec 6, 2017
6,253 views
Consider the 2 transactionsT1: R(A) W(A) W(B)T2: R(A) W(A) R(B) W(B)How many view serializable schedules are possible which are not conflict serializable?(A) 0(B) 1(C) 2(...