1,231 views
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?

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

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

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

@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. :)

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

1 vote