NORMALIZATION

Question

The Highest Normal Form of the Following Relation?

R(ABCDE)

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

 

EDIT :- Decompose it into BCNF if possible

Comments

2NF?

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Hemanth_13 

It is not even in 2NF itself because there exist partial dependencies here..

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Yes  :P

thanks  akash.dinkar12 :)

Answer 1

Given Relation is ABCDE ..now we have some of given FD as

FD={AB→C,DE→C,B→D} ..now we can see that on LHS only 2 attributes are there , which are  C and D , so there is no way that we can have A,B,E on LHS .

So every key must have ABE as part of it , so that we can have all attributes in the RHS

So basic minimal set is

ABE ...taking closure of it

(ABE)+=ABCDE

Now prime attributes are A,B,E but as B->D FD is there , a part of key is determining the nonkey , which is voilation of 2NF

so given relation is in 1NF only

Comments

@Shaik Masthan How did u get relation (ABE)and relation (DEC)  has no  key common with other relation.So Lossless is not satisfied.

I didn't get the decomposition :(

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why it is lossless?

i already gave the procedure to merge the relations to get loss-less

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Thanks.Understood :)