Question

Given the following functional dependendies on a relation R(A B C D E F G)

AB → CF
BG → C
AEF → C
ABG → ED
CF → AE
A → CG
AD → FE
AC → B

I have worked out the candidate keys by using the method where you put the attribute in either a left, middle, right column depending if it is seen on the left hand side of a dependency, right hand side or both. Left means that the attribute is necessary, middle is unknown and right means not part of a key.

I got this:

L | M       | R
--|---------|----
- | ABCDEFG | - 

From here I worked out the closures for each individual attribute and the permutations: BC, BD, BE, BF, BG, CD, CF…

I found that only the closure of A and CF contain all attributes and therefore are candidate keys however the solution the problem also has BFG.

Can someone explain what I am doing wrong in calculating candidate keys?
Thanks

Answer 1

This algorithm tries to find shortcuts (pg 3), but in your case it doesn’t find any. To determine whether any particular combination of attributes is a key, you try to figure out whether that combination determines every other attribute. In your case, you’ve done all the work; you’re just missing something about BFG.

Logic                                    Attributes
--
BFG -> BFG, ∴ ...                       { B   FG}
BG -> C, ∴ BFG -> CF                    { BC  FG}
BFG -> CF and CF -> AE, ∴ BFG -> AE     {ABC EFG}
BFG -> AE,  ∴  BFG -> A
BFG -> A and ABG -> ED, ∴ BFG -> ED     {ABCDEFG}

So BFG is a candidate key.