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VW->XYZ, Z->W, Y->VW

it is Funtional dependency relation S{V,W,X,Y,Z}. candidate key is 'Y'?? or another?? I have a question. Why Z->W break normal form? and what type is broken normal form?

analog_by
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1 Answers1

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The relation has three candidate keys:

1. Y
2. VW
3. VZ

and you can verify this by calculating the closure of each of them.

The relation is not in Boyce-Codd Normal Form (BCNF), since in the dependency Z → W the determinant (Z) is not a superkey (while in the other two dependencies the determinants are keys).

Finally, note that the relation is in Third Normal Form (3NF), since W is a prime attribute (that is, is an attribute of a candidate key).

Renzo
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  • Thanks you for your reply. But, candidate key means minimal + unique. Because of it, I think vw, vz are not candidate key doesn,t it?? – analog_by Mar 14 '17 at 06:06
  • @analog_by, they **are** minimal, since minimal means “no other attribute can removed without losing the property of implying all the attributes of the table’ (it *does not* mean “constituted by a single attribute”!). So, `vw` and `vz` are minimal since you cannot remove any attribute from them without losing the fact of being keys (`v`, `w` and `z` alone **are not** keys). – Renzo Mar 14 '17 at 09:30