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While checking whether a schedule is view serialisable or not. I read that if there is no blind write then schedule is not view serialisable.
I am not able to figure out reason behind this.
in Databases by (85 points) | 47 views
Where did you read this?
Please follow standard books at least as a reference. The sentence given there is with respect to the previous sentence given. The question you asked should be answered in most DBMS textbooks.

Yes sir, I did that only 
here was my original question –
But still I read it again and I have figured it out something. I am writing explanation in answer please tell me whether I am right or not.

Yes, that's correct. Only blind writes can make a view serializable schedule not conflict serializable.

2 Answers

+1 vote
All the conditions of view serialisablity ( intial reader, write-read sequence and final writer ) also comes under conflict serialisability.
But if a schedule is not conflict serializable then there is only one possibility where view serialisability can differ from conflict serialisable schedule i.e write-write conflict i.e here one transaction will write on a item and other transaction will just write again on that data item without reading previous (blind write).  So, we only have to preserve final write.

If the write is blind then only we can think of swapping it because we have to only keep final write same.
If there is a blind write but still there may occur some condition where we will not be able to generate serial schedule.
by (85 points)
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Writing is more powerful than reading

Will, you agree to me


Suppose T1 reads on data item A first, then many instructions perform Write on A one after other let us say T2, T3, …, Tn-1 and Tn does that write operation on it.

What T2 does is not a blind write

but what T3 to Tn transactions do is known as a blind write.

Now why do we say given a schedule, if it's not conflict serialisable, there may be chances that the schedule is view serialisable?

We say this because,

What matters to the user is 

  1. who read the value of a first
  2. who read the intermediate values
  3. who wrote the final value in the database

Now for consistency,

you tell me, 

If a schedule is having cycles in the precedence graph, but there are blind writes in it.

this means

T1 T2 T3 T4


After seeing this schedule,

what matters to the user is that T1 read the first value, so in any  non serial view equivalent schedule, T1 must read value of A first. Then only, we would be able to achieve consistency.

And no matter what T2 and T3 write to Database, their value is not going to be final, so basically they can execute in any order,

But what matters to the user, T4 has performed the final write, which will be visible in the database, so T4 must happen after every transaction.

Now there may be many view equivalent schedules

But suppose, there are no blind writes like T3, T4 and suppose later on T1 again writes on the database

Consider this example.

T1 T2

Now there is no blind write here and you can see, this schedule is not conflict serialisable, so it cannot even be view serialisable because, in the polygraph also, there will be cycles. as T1 has to execute first as it is reading on A first. but at the same time, it has to execute last also.

Having blind writes also does not guarantee, that you will get view serialisable. But if there are no blind writes, you can straightaway declare that as you do not have conflict serialisable schedule, and also that you are not having any blind write, then you can say that the schedule is not even view serialisable.


I hope this in detail discussion would have cleared your doubt


If you need any help, do let me know.




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