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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
asked in Databases by Veteran (59.5k points)
edited by | 1.4k views

2 Answers

+25 votes
Best answer

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.

answered by Loyal (8.2k points)
edited by
0
right
+7 votes

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 

answered by Loyal (6.6k points)
edited by
+4
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
0
Yes u r right @gate Ranker
+1
@rishi

edit ur answer
0
Now ok? GateRanker


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