Seriaizability

Question

Answer 1

There is a cycle formation between T3 to T4 due to W(B) and T4 to T3 due to R4(A) and W3(A) . Hence the schedule is non serializable . We cannot say anything about about view serializibility here becoz for that we need atleast 2 schedules. Therefore ans should be D.

Comments

There is a full-proof method to find view serializability. I cant access my camera now. I will post it later. But just get the trick that...

For each data variable A, B, C in given schedule .. Find

- First read by which Tx

- Last write by which Tx

- Dirty Reads sequences.

Now, take some serial schedule like T1->T2->T3     and try drawing this above similar table.. If every entry matches.. Then given schedule is View Serializable to this schedule. (You can also try some other serial schedule for comparison)

PS. : It may seem a longer method than the method for conflict serializability. But, its an easy method when u get comfortable with it. I was also very confused about how to solve View-Ser. qstns because there was ni exact method and just guesses. And I had lost several marks because of it. Practice above method.

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Yes plz provide the method @Tushar ....i am waiting for an answer . Thanks :)

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Checkout the answer posted ..

Answer 2

what i think its hard to make  a polygraph for this. as we know they are not going to ask a question taking lot of your time . just give it a try and check the view equivelnce for s1,s2,s3,s4. and i think it is view equivalent to this serial schedule . so according to me b  just check if its is not .

Comments

yes,@tendua u r right there is a blind write present on B between T2 T3 T4 .T2 read B write B ,and T3 ,T4 write B without reading it. It is view serializable.

Answer 3

Plz check this !!

Comments

i don' know whether u are thanking me or insulting me . u want to be "NOKIA " sure go that way. You are 100 % right . Best of luck . and yes no need to even listen to a fool like me who told u there were 4 schedule . i was talking about transitions. i never mentioned schedules.

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Yes @Tendua i appreciated you for making me learn .... Thanks that you pointed that mistake of mine ..... And i indeed need good wishes from experts like you :) Thank you once again .

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@ Arayana , so here in this question, we can check view serializability by taking the given schedule & checking its view equality with serial schedules possible..

If it becomes view equal to some serial schedule ,then we say view serializable..

But here it's not view serializable..

Answer 4

Option (b). I have stated one trick to check view serializability in the comment. Here's the pic for above schedule... 

Step 1) Write down table like this for given schedule

Step 2) Consider some serial schedule (Here matching one is T1-->T2-->T3) and draw similar table for it.

Step 3) Compare both tables, if every entry matches, Hurrray.. It's a view-ser. schedule :)

Comments

Final write of B says :  T₄  should be below than T3 & T2 .. 

But now t4 & t3 read A , written by t2 . So, T2 should be above both..

So, this gives   t2 > t3 > t4

But, t3 can't be in between t2 & t4 becoz  t4 reads A which is modified by t2, so, t3 can't be in between them as t3 also writes A..

So, I guess D should be the correct answer here.