BCNF和3NF等效的条件

Question

Statement - "If a table contains only one candidate key, the 3NF and the BCNF are equivalent."
In the below image, relation is 3nf but not bcnf. But it has only one candidate key ie AB, so according to above statement it should be either both (3nf, bcnf) or nothing. Can somebody explain what i am missing here ?

Answer 1

In,simple words the given words can be written as

AB->C

AB->D

C->B

In above relations,A,B,C are prime attributes and D is non-prime attributes

Here,AB and AC are minimal Super keys

Condition for 3NF and BCNF

X->Y

For BCNF, X should be Super key whether Y is prime or non-prime attribute

For 3NF, X should be Super key only when Y is non-prime ,otherwise it is a optional one

 BCNF and 3NF are equivalent when there are no **non-prime** attributes

Hope,it helps.

Answer 2

"If a table contains only one candidate key, the 3NF and the BCNF are equivalent."

The second half, "the 3NF and the BCNF are equivalent", doesn't mean anything. What do you think it means? The normal forms are different things. BCNF requires a more restrictive condition than 3NF. So when a relation is in BCNF it has to also be in 3NF. So a relation can be in both normal forms at the same time.

Presumably the quote is trying to say that, "if a table contains only one candidate key", then if it is in 3NF then it is also in BCNF. And this is correct, because a table that is in 3NF but not in BCNF has to have overlapping CKs (candidate keys). Here the CKs, ie sets of attributes that determine all attrbutes that don't contain a smaller such set, are AB & AC. Which overlap. This table is in 3NF, but it's not in BCNF.

(Note that the FDs that hold are not just ones you are explicitly told hold but all the ones that must hold when they do per Armstrong's axioms.)