PROBLEM ON VIEW SERIALIZABILITY

Question

check whether it is view serializable or not?

schedule S:R1(X),W2(X),W1(X)

Comments

Not view serializable as it violates the first point which says if transaction ti reads the initial value of x then transaction tj must read the initial value of x.  But instead of that transaction t2 has started writing x without reading it  

Can it be concluded like this

Answer 1

no it is not view serializable..
reason is there may b 2 schedule possible :
1. T1--->T2
2. T2----->T1z
 in case 1 final write of x is done by T2, but in given question final write of x is done by T1. so it violates condition of view serializability.
in case 2 initial read of T1 is updated value by T2 but in original question it is directly reads database.. again condition of view serialization failed..

so not serializable schedule..

Comments

can we say like this there is no serial schedule available so it is not view searializable..

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a particular schedule may not be serial but it may be serialisable. but reverse is not always true..
see following implications :

1. serial -----> serialisable
2. view serialisable <-------> serialisable
3. conflict serialisable ------> view serialisable
4. not serialisable ---------> not serial

A schedule which is not serial but it may be View serialisable .. Serialisable too..

Answer 2

The ans can be viewed here

Answer 3

This is definitely not CS as the precedence graph bears cycle.

This schedule is VS,as when we try to draw the view serializable graph ,we find no loops.

Comments

A schedule is view serial iff :
A. schedule is serial.
B. schedule is not serial but it is view equal to any one serial schedule..
Obviously given schedule is not serial.
If it is View Serial,  Can anyone gives serial schedule which is view equal to given schedule ???

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according to me it is view serializable . i have already given explanation .. now you should ask from arjun sir..

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sonam vyas in the link provided , after drawing polygraph in 2nd example we see there is a cycle b/w T1-T4 but still it says it is view-serializable and also gives a serial schedule .

can u please explain it?

Answer 4

schedule will be view serializable if 1. it is conflict  serializable then surly its vs.

                                                    2. if is not cs than we have to check blind write . if there is blind write then it may be or may not be vs. for that we have to draw poly graph . if there is cycle in the poly graph we can say its not vs other wise it is vs.or not serializable schedule also.

so this schedule is vs bcoz there is no cycle in poly graph.

Comments

https://en.wikipedia.org/wiki/Blind_write i think you should read it...

http://infolab.stanford.edu/~ullman/dscb/vs-old.pdf 

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consider this schedule Sg: r1 (X); w2(X); W1(X); W3(X); CI; C2; C3; T1 T2 T3 r1(X) w2(X) W1(X) W3(X) C1 C2 C3 in the above schedule W2(x) and W3(X) are BLIND WRITE .!! but in the given question there is no blind write as LAST wirte{W1(X)} is done in T1 in which there is presence of R1(X). so in this question there is no BLIND WRITE therefore not VIEW SERIALIZABLE!!

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you missed out something .. i said if there is blind write then we check it is vs or not but if there is not blind write then we does not check it... and in this question blind write is present in t2 transaction .and if there is blind write ,then we have to draw poly graph ,if in poly graph cycle is present then it will not vs otherwise it will be... so we cant say surly ,if there is blind write it must be vs.. we have to check by poly graph..

in this question firstly blind write is present and when we draw poly graph for it there is no cycle thats why it is vs...

Answer 5

its not a view serilizable..because if read t1 the t2 can also read the transaction..

Answer 6

SCHEDULE IS VIEW SERIALIZABLE BUT NOT CONFLICT SERIALIZABLE SCHEDULE