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

Shubham why have you consider multivalued attributes? there was no need to consider them.

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@subham can't we decompose it  further to get the higher normal form?

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why have you consider multivalued attributes? there was no need to consider them.

i didn't get you brother

 

@Kshitij Hansda

can't we decompose it  further to get the higher normal form?

you are not allowed to decompose the relation... if you do that, then we can get atleast 3NF for every relation

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@Shaik, shubham has written AB->->C this shows that there is multivalued dependency.

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brother it is typing mistake, it occurs due to copy and paste

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oh sorry

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yes Shaik Masthan is correct

 

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@Shaik

Could you show the decomposition ?The purpose of posting the query was actually to know the decomposition(should have changed the query :P )

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(A,B,C), (D,E,C), (B,D) and (A,B,E)

1st Step :-  (A,B,C), combined with (B,D) ===> (A,B,C,D) due to common attribute forms key in (B,D)

2nd Step :- (A,B,C,D) combined with (A,B,E) ===> (A,B,E,C,D) due to common attribute forms key in (A,B,C,D)

3rd Step :- (A,B,E,C,D) combined with (D,E,C) ===> (A,B,E,C,D) due to common attribute forms key (may be Super Key ) in (D,E,C)

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@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 :)