Lossless Join and Decomposition

Question

Let R(A, B, C, D) be a relational schema with the following functional dependencies:
 

{ A --> B, B --> C, C --> D, D --> B }
 

The decomposition of R into (A, B), (B, C) and (B, D):

(A) gives a lossless join, and is dependency preserving

(B) gives a lossless join, but is not dependency preserving

(C) does not give a lossless join, but is dependency preserving

(D) does not give a lossless join and is not dependency preserving
 

In this question, given decomposition is Lossless, But i have confusion about Dependency Preserving or Not.

Here, according to me C -> D is lost because, there is no any decomposed releation has 'C' and 'D' both.

But, i searched about it and asked others also, they are saying that C -> D is also preserved.

Can anyone from you help me??

Comments

R2(B,C) should have B-> C and C -> B because original FD has C -> D and D-> B.

R3(B,D) should have B-> D and D-> B.

 

So, C -> D preserving through C -> B and B -> D.

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Thank you!

Answer 1

A --> B, B --> C, and D --> B are preserved directly, while C -- >D is preserved indirectly using C --> B and B --> D; hence the decomposition is dependency preserving.
 

Comments

Thank you!
I initially thought that for an FD to be preserved, both its LHS and RHS must appear in a single decomposed relation. I didn’t realize that transitivity can also be used to preserve dependencies.

Answer 2

Step 1: Join R1(A, B) and R2(B, C)

Is B a key for R1 or R2?

Look at the FDs: We have B-->C . This means B is a key for R2.

 

 

Result: This first join is Lossless. We now have a combined intermediate table R12(A, B, C).

Step 2: Join the result R12(A, B, C) with R3(B, D)

Common Attribute: {B} (Since B is in R12 and also in R3).

Is B a key for R12 or R3?
Look at the FDs again: We have B-->C and C-->D By transitivity (B-->C--D), we know Because B-->D , B is a key for R3.

Result: This second join is also Lossless.

Why it works in simple lines:

• A and B are linked by B.
• B and C are linked by B.
• B and D are linked by B.

Comments

Thank You!