Raghuramkrishnan

Question

T1	T2	T3
R(X) W(X) commit	W(X) commit	R(X) commit

 

is this transaction conflict serializable and view serializable?

 

Comments

View Serializable??

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how?

Answer 1

It is not Conflict Serializable as there is dependency between $T_1$ and $T_2$ in both the directions for conflicting operations (cycle in precedence graph).

For View Equivalence:

1. If transaction $T_i$ reads initial value of any data item $X$ in one schedule it must do the same in all the equivalent schedules
2. If transaction $T_i$ writes the final value of any data item $X$ in one schedule, it must do the same in all the view-equivalent schedules. 
3. If transaction $T_i$ reads data item $X$ written by transaction $T_j$, it must do the same in all the view equivalent schedules

So, the given schedule is not View Serializable as it requires $T_1$ to come before $T_2$ to ensure $T_1$ reads first and $T_2$ to come before $T_1$ to ensure $T_1$ writes last, which is not possible. 

Comments

same question is found in my notes where it is also given Not VS and not CS 

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Sorry, I was using the wrong definition for VS - the condition is not to ensure the transaction reading first remaining the same but the value being read must remain the same. So, the given one is not VS - I have edited and added the definition too.

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@Gurdeep Saini

@Arjun Sir,

How you are getting T2$\rightarrow$T1 in polygraph??

This is what I am getting :

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

$T_{1}$$\rightarrow$ $T_{2}$

$T_{1}$$\rightarrow$ $T_{3}$

$T_{1}$$\rightarrow$ $T_{f}$

$T_{2}$ $\cdots$> $T_{1}$

$T_{3}$ $\cdots$> $T_{2}$

& out of them $T_{2}$ $\cdots$> $T_{1}$ , $T_{3}$ $\cdots$> $T_{2}$ if I consider only $T_{3}$ $\cdots$> $T_{2}$ then I will not get any cycle & which will lead to VSS. 

Please point out my mistake if any.