dependency preserving

Question

Let R (A, B, C, D) be a relational schema with the following functional dependencies : 
A → B, B → C, C → D and D → B. The decomposition of R into (A, B), (B, C), (B, D)

1. gives a lossless join, and is dependency preserving
2. gives a lossless join, but is not dependency preserving
3. does not give a lossless join, but is dependency preserving
4. does not give a lossless join and is not dependency preserving

Comments

@ARJUN SIR @BIKRAM SIR

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1) is correct,,,???

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PLS EXPLAIN THE PART OF DEPENDENCY PRESERVING ?

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explained, see my comment below after ur answer .

Answer 1

In this can we apply transitivity rule

A->B B->C THEN A->C

B->C C->D THEN B->D

C->D D->B THEN C->B

NOW FROM DECOMPOSITION IT IS LOSSLESS AND ALL DEPENDENCIES ARE PRESERVED

BUT FOR C->D WE CAN DERIVE BY JOINING C->B AND B->D

Comments

option A  gives a lossless join, and is dependency preserving is correct .

(A,B) (B,C) -> common attribute is B and due to B->C, B is a key for (B,C) and hence ABC can be losslessly decomposed into (A,B) and (B,C).  Thus the given decomposition is lossless.

 

A->B is present in (A, B),

B->C  and C->B is present in (B, C),

B->D is present in (B, D) and C->D is indirectly we get  via C->B [ C->B and B->D so  C->D]   in (B, C) and D->B in (B, D).

so original relation have  A->B , B->C , C->D , D>B  hence dependency is preseved.

Answer 2

here C---->B and B --->D

SO C--->D also  implied

so DP perserving

Comments

https://gateoverflow.in/?qa=blob&qa_blobid=6852447450858925440

see this why B->A does not hold ?

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B->A   or    A->B .....????

in first example na???

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got it now u r right