search
Log In
2 votes
765 views

how many view equivalent schedules are possible for the Sch given below:

in Databases 765 views
0
only serial schedules or both serial and non serial?
0
serial
0
if only

serial then 6
0
Will you please mention those 6 serial schedules
2
1234

1324

2134

2314

3124

3214
0
Thanks it helped :)
1
T1 T2 T3 T4,

T2 T1 T3 T4,

T1 T3 T2 T4,

T2 T3 T1 T4,

T3 T1 T2 T4,

T3 T2 T1 T4
0
What if we replace W(B) in all transactions with W(A)?
0
then only 2 would be possible

1234 and 1324
0
Then 2 view equivalent schdule.

1) t1 t2 t3 t4,

2)t1 t3 t2 t4
0

I think the Schedule will not be View Serial as their will be no Blind Read isn't it

0

@Nandkishor3939 what is the actual answer given for above problem?

0
There will be no View equivalent schedule if all W(B) are replaced with W(A)
0

Its a self doubt question BTW, for the given question the answer should be 6

0
There are 2 view equivalent schedules possible for this

T1 T2 T3 T4

T1 T3 T2 T4

3 Answers

3 votes

Non-serial schedules which are view equiv. to given schedule = 630 (7! / (2! * 2! * 2! ))

 
Serial schedules view equiv. to given schedule = 6

0
Hi, it should be ( 8! / (2*2*2*2) ) isn’t it.
1

@ bro we have to fix the last W(B) because the final writer has to be same for all data. Please see view serializable conditions from here https://www.geeksforgeeks.org/view-serializability-in-dbms-transactions/

 

0 votes

polygraph for the given schedule: (View serializable)

$T_{0}$$\overset{A}{\rightarrow}$ $T_{1}$

$T_{0}$$\overset{A}{\rightarrow}$ $T_{2}$

$T_{0}$$\overset{A}{\rightarrow}$ $T_{3}$

$T_{0}$$\overset{A}{\rightarrow}$ $T_{4}$

$T_{4}$$\overset{B}{\rightarrow}$ $T_{f}$

$T_{1}$ $\rightarrow$$T_{4}$

$T_{2}$ $\rightarrow$$T_{4}$

$T_{3}$ $\rightarrow$$T_{4}$

 

 

($T_{1}T_{2}T_{3}$) $\,!$ $\rightarrow$$T_{4}$

$\therefore$ Total 6 cases

0 votes

T1 is fixed in first position due to first rule of view serializibility of firt read 

T4 is fixed in last position due to rule of last write 

there is no write – read dependency to maintain in T2 and T3 SO they can be any order 

so possible serial schedules are T1-T2-T3-T4 AND T1-T3-T2-T4

Related questions

3 votes
1 answer
1
1.1k views
No. Of serial schedules view equal to S S: R1(A), R3(D), W1(B), R2(B), W3(B), R4(B), W2(C), R5(C), W4(E), R5(E), W5(B) . If possible please provide detailed solutions. Thank you in advance
asked Jan 7, 2018 in Databases thepeeyoosh 1.1k views
9 votes
2 answers
2
4.8k views
Number of schedules view equal to following schedule :- r1(A), w1(B), r2(A), w2(B), r3(A), w3(B)
asked Jul 10, 2017 in Databases rahul sharma 5 4.8k views
4 votes
1 answer
3
591 views
Consider thw following transactions:- T1 :- r1(A) w1(A) r1(B) w1(B) T1 :- r2(A) w2(A) r2(B) w2(B) a) Number of schedules serializable as t1->t2? b) Number of schedules serializable as t2->t1?
asked Jul 12, 2017 in Databases rahul sharma 5 591 views
9 votes
2 answers
4
2k views
1) T1: R(X), T2: W(X), T1: W(X), T2: Abort, T1: Commit 2) T1: W(X), T2: R(X), T1: W(X), T2: Abort, T1: Commit 3) T1: W(X), T2: R(X), T1: W(X), T2: Commit, T1: Abort Can anyone explain whether these schedules are serializable, conflict-serializable, view serializable, recoverable, avoids-cascading-aborts, and strict? The abort operation actually bugging me.
asked Sep 9, 2016 in Databases vix28 2k views
...