The Gateway to Computer Science Excellence
0 votes
166 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 not conflict equivalent to Schedule S ?
in Databases by Loyal (7k points) | 166 views

1 Answer

0 votes
Conditions for view equal:

Initial  Read : -

Updated Read: 1>2,3>4,5>6,7>8,9>10

Final write : 9

So 9>10 transaction order is at the final ,with the other 4 precedences permutations.

Total permutations of 1>2,3>4,5>6,7>8 = 4! = 24 ways ( including the one given in the question)

We can observe that since every transaction is conflicting with every adjacent transaction , no other schedule is conflict equivalent.
So total number of view equal schedules to above and not conflict equivalent = 23
by (199 points)
0
at the end why we are doing -1 from 24
0
As there is only one schedule possible which is conflict serializable i.e. is presently u ar seeing on screen. And any change in it will lead to transaction order change or loop .
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
50,737 questions
57,321 answers
198,395 comments
105,145 users