No. of View equivalent schedules

Question

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

Comments

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

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There are 2 view equivalent schedules possible for this

T1 T2 T3 T4

T1 T3 T2 T4

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@Deepak Poonia can you please solve the problem ?

Answer 1

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

 
Serial schedules view equiv. to given schedule = 6

Comments

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

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@Joey 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/

 

Answer 2

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

Answer 3

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

Answer 4

Non serial schedules :-   fix the last write on B 

$\frac{7!}{2!\times 2!\times 2!} = 630$

serial schedules :-3! = 6