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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

1. dependency preserving and lossless join
2. lossless join but not dependency preserving
3. dependency preserving but not lossless join
4. not dependency preserving and not lossless join
recategorized | 1.9k views

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.

edited
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right
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is this decomposition in 2NF?
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Its BCNF

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

edited
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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
+1
@rishi

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Now ok? GateRanker

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