GATE2001-1.23, UGCNET-June2012-III: 18

Question

Consider a schema $R(A,B,C,D)$ and functional dependencies $A \rightarrow B$ and $C \rightarrow D$. Then the decomposition of R into $R_1 (A,B)$ and $R_2(C,D)$ is

• A. dependency preserving and lossless join
• B. lossless join but not dependency preserving
• C. dependency preserving but not lossless join
• D. not dependency preserving and not lossless join

Comments

I want to ask that the decomposed relation will be in which normal form if it has not lossless decomposition?

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@vijay kumar 2

It can be any NF of above 3NF power

i.e;

4NF,5NF, ...

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$\color{red}{\text{Detailed Video Solution:}}$

Detailed Video Solution

Answer 1

Answer is C.

Here, no common attribute in R1 and R2, therefore lossy join will be there.

and both the dependencies are preserved in composed relations so, dependency preserving.

Comments

Its 1NF .

Candidate key = {AC}

Not 2NF because both dependencies have partial dependency.

    eg : In A-->B ,  A is proper subset of Candidate key and B is a non-prime attribute.

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They are asking about decomposed relation. So it's BCNF

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@Arjun sir

The answer key is wrong in GO pdf for this question. kindly check once!

Answer 2

A decomposition {R1, R2} is a lossless-join decomposition if  R1 ∩ R2 → R1 (R1 should be key) or R1 ∩ R2 → R2 (R2 should be key)   but  (A,B)  ∩ (C,D) = ∅  so lossy join 

FD:1         A→B    

FD:2         C→D

R1(A,B) have all attributes of FD1 and R2(C,D) have all attributes of FD2 so ,dependency preserved decompostion

Reference : - question no. 8.1 Korth  http://codex.cs.yale.edu/avi/db-book/db6/practice-exer-dir/8s.pdf 

Comments

if R1 and R2 have common attributes then it need not be lossless join

for being in lossless join common attributes should be key in either of relations

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Yes u r right @gate Ranker

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@rishi

edit ur answer

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Now ok? GateRanker