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Editorial Team
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Editorial Team
Asked: June 5, 20262026-06-05T03:56:07+00:00 2026-06-05T03:56:07+00:00

I’m practicing taking as input a set of functional dependencies and output candidate key(s).

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I’m practicing taking as input a set of functional dependencies and output candidate key(s). Is there an algorithm and how come in such case there is no web-based implementation where I can input my FD:s and as output get a list of superkeys / candidate keys?

I practice on what I find here on SO and a suitable question is
how to find the highest normal form for a given relation
where the functional dependencies mentioned are

B->G

BI->CD

EH-> AG

G-> DE

Please check if I’m doing this right when I try to find that the candidate key is BFHI:

The FD B->G can be rewritten as ABCDEFHI->ABCDEFGHI and therefore ABCDEFHI is a superkey.
The FD BI->CD can be rewritten as ABEFGHI->ABCDEFGHI and therefore ABEFGHI is a superkey.
The FD EH->AG can be rewritten as BCDEEFHI->ABCDEFGHI and therefore BCDEEFHI is a superkey.
The FD G->DE can be rewritten as ABCFGHI->ABCDEFGHI and therefore ABCFGHI is a superkey.

In our superkeys, BFHI is in every one. Therefore BFHI is the candidate key and it cannot be reduced further which can be seen from inspection(?)

Am I reasoning this the right way?

There is another question the augmenting algorithm can handle, if it works,
Database extraneous attributes and decomposition

Here, the FD:s are

A->BCD

BC->DE

B->D

D->A

Here the FB A->BCD can be written as AEF->ABCDEF and therefore AEF is a superkey. The FD BC->DE can be rewritten as ABCF->ABCDEF and therefore ABCF is a superkey. The FD B->D can be rewritten as ABCEF->ABCDEF and therefore ABCEF is a superkey. The FD D->A can be rewritten as BCDEF->ABCDEF and therefore BCDEF is a superkey. For all superkeys, F is the only member(s) that is in every superkey and therefore F is the only candidate key.

Does this work?

Thanks for any answer/comment

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  1. Editorial Team
    Editorial Team
    2026-06-05T03:56:10+00:00Added an answer on June 5, 2026 at 3:56 am
    No, but as F is not in any of the FD:s then it has to be a member of every candidate key.
    
    Also, A->BCD, BC->DE, B->D, D->A give us 
    A+ (the cover of A) = ABCDE
    B+ = ABCDE
    C+ = C
    D+ = ABCDE so the 
    E+ = E
    F+ = F.
    
    The combinations giving ABCDEF are
    AF
    BF
    DF
    and hence the candidate keys are {AF, BF, DF}
    and every enhancement of any of those three are the superkeys
    
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