#DBMS Blind Write

Question

Please explain the concept of Blind Write in a simplified manner w/ a small example! 

Comments

Consider a transaction, that starts writing a particular data item without reading it's present value. 

$T_1$	$T_2$	$T_3$
	$R_2(X)$
$R_1(X)$
		$W_3(X)$
	$W_2(X)$

Here $W_3(X)$ is a blind write since there's no $R_3(X)$ before it. Whereas, $W_2(X)$ has $R_2(X)$ before it. Any view serializable schedule that is not conflict serializable must contain a blind write.

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Consider a transaction, that starts writing a particular data item without reading it's present value. 

T1	T2	T3
		W3(X)
R1(X)
	R2(X)
	W2(X)

is W3(X) a blind write now? @krish__

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You need not consider the existence of other transactions in the system. I came up with the schedule just to show you the difference between a normal one and a blind one.

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is W3(X) a blind write now? yes or no

@krish__

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Yes, it's a blind write.

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blind write is simply when a transaction writes without reading. i.e a transaction have WRITE(Q), but no READ(Q) before it. So, the transaction is writing to the database "blindly" without reading previous value.

Yes. W3(X) is a blind write.

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

In T3 is W(X) is still a blind write? as there's no read on data item X.

Also, is W(X) in T3 to R(X) in T2 is a producer - consumer scenario or W(X) is a blind write therefore not  a PC?

@Rishabh Gupta 2 @krish__

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It doesn't make sense to have R(X) in T1 in the same row as W(X) in T3. The rows indicate a point in time. 2 operation cannot happen at the same time as per our usual outlook.

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	T2	T3
		R(Y)
		W(X)
	R(X)
	W(X)

In T3 is W(X) is still a blind write? as there's no read on data item X.

Also, is W(X) in T3 to R(X) in T2 is a producer - consumer scenario or W(X) is a blind write therefore not  a PC?@krish__

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$W(X)$ by $T_3$ is still a blind write and $W(X)$ is a producer to $T_2$.

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HOW COME BLIND WRITE BECOME A PRODUCER AS IT IS SIMPLY WRITING WITHOUT READING!@krish__ 

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Is there any constraint of being a producer? that it should first read then produce?   iarnav

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T1	T2
	R(X)
R(X)
	W(X)
W(X)

This is not conflict serializable but it is view serializable and it does not contain any blind write. This contradicts what you said

 

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Ninja

It is not view serozable .

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@ninja its not view serializable

its due to final write of x is done by transaction T1 so it should come after T2 as T2 is also writing on X..so T2->T1 should be the sequence of exn

second... updated read... T2 is reading updated value from T1 transaction so this says sequence as T1->T2

so its not view serializable as no possible sequence you can give. So you need atleast one blind write if something thats not conflict serializable to be a view serializable.

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It is not view serializable as it does not maintain the intial reads done on the data item. To preserve R(x) @ T1 we need the execution sequence as T1--> T2. But to preserve R(x) @ T2 we need the execution sequence as T2--> T1. Both of which are not possible.

Answer 1

blind write is simply when a transaction writes without reading. i.e a transaction have WRITE(Q), but no READ(Q) before it. So, the transaction is writing to the database "blindly" without reading previous value.

If there is no read that happens prior to the first write then it is said to be a blind write.

• W3(X) is a blind write, as there is no read before write [R3(X) before W3(X)]

• W2(X) is not a blind write, as a read happens before write [R2(X) before W2(X)]