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1 vote
1 vote
Decompose below relation into 2NF,3NF and BCNF with lossless join and Dependency Preserving.

$R(ABCDE)$

$F=\{AB\rightarrow C,DE \rightarrow C,B \rightarrow D\}$

 

According to my analysis

The only key here is $ABE$

So, Dependencies $AB\rightarrow C,B\rightarrow D$ are partial dependencies.

I decomposed as below

$R_1(ABC) \{AB\rightarrow C\}$

$R_2(BD)\{B \rightarrow D\}$

$R_3(CDE) \{DE \rightarrow C\}$

And All three above are in BCNF as well.

Is my decomposition correct?
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4 Comments

make one more table, (A,B,E) to make it lossless
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Which relation should be joined with R3 at last to obtain the original relation ?
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(A,B,E) with (A,B,C)

then (A,B,C,E) with (B,D)

then (A,B,C,D,E) with (C,D,E)
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1 Answer

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0 votes
It is Lossy Decomposition. Dependency preserving decomposition is not possible till BCNF.

4 Comments

@YesAbhisa-Sorry I wrote it wrong. Please check now.

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@Ayush Upadhyaya Yes,it is correct as the dependencies are preserved and LHS of each FD is a key to the table(BCNF).
For making decomposition lossless, we generally create a relation taking the candidate key, which has been done in  [

R4(ABE)] . It helps in joining all the tables.

I had not decomposed it to 
R3(CDE) {DE→C} taking CDE as a result of which the dependency was not preserved.
Thanks for feedback. :)
 

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For making relation lossless we always take candidate key.kindly reply.
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