# Thread: Decomposing this relation to 3NF

1. Registered User
Join Date
Apr 2012
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## Unanswered: Decomposing this relation to 3NF

Hi all! I don't know how to compose this relation to BCNF form, spent along time and I decided to join this forum to help me solve the problem

My relation is:
R(ABDEF), with the minimal cover

S = {A->B, B->E, BE->AF}

This has candidate keys DA, DB, DE and is in 3NF (you agree?)

I want to decompose this to BCNF. These are the instructions I have been given to decompose into BCNF form

So we can see that A->B and B->E violate BCNF form. Agreed?
So now, let's apply the procedure given in the link so we get
R(AB) {A->B}

but we now have a FD B->E in the second remind relation, and there is no B attribute in the relation. What should I do?

Thanks
Thomas

2. Registered User
Join Date
Sep 2014
Posts
1
Hi Thomas,

I am facing a similar problem. Did you find on how to go about such functional dependencies?

Thanks,

3. Registered User
Join Date
Oct 2014
Posts
5
Originally Posted by thomas49th
Hi all! I don't know how to compose this relation to BCNF form, spent along time and I decided to join this forum to help me solve the problem

My relation is:
R(ABDEF), with the minimal cover

S = {A->B, B->E, BE->AF}

This has candidate keys DA, DB, DE and is in 3NF (you agree?)
I don't agree. I take it that S pictures all the functional dependencies in the relation? Why then are DA, DB and DE candidate keys? That does not make sense to me.

According to the functional dependencies you've given (I've given them names to more easily refer to them):

FD1: {A} -> B
FD2: {B} -> E
FD3: {B,E} -> A
FD4: {B,E} -> F

According to these FDs the relation R is not even in 2NF, let alone 3NF. Are you sure you gave all the information? On what basis did you give the candidate keys? According to the FDs above, candidate keys in relation R are {A}, {B} and {B,E}.

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