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

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

 
Serial schedules view equiv. to given schedule = 6

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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

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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

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Non serial schedules :-   fix the last write on B 

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

serial schedules :-3! = 6 

edited by

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