Exactly what is 2NF and 3NF?

Question

What's the main point of Normalization?

I mean if a normal form is not in 2NF, it is because of partial dependency i.e. a non key attribute is dependent on a part of a candidate key. So, let's say, for a relation R(A,B,C) with FDs:

AB->C, B->C

Clearly, AB is the candidate key and B->C is the partial dependency.

Solution: Decompose the relation such that (B,C) forms a new relation with B as the key.

Now, if a relation is not in 3NF, it is because a non key attribute is dependent on another non key attribute i.e. to say if FDs for a relation R(A,B,C) are:

A->B,B->C

Clearly, A is the key and B->C shows transitive dependency, so not in 3NF.

Solution: Decompose the relation such that (B,C) forms a new relation with B as the key.

So, what's the exact difference? I mean, why such a marked distinction? Essentially in both of the cases the action is same. Decompose the relation using the dependency where the determinant (B here) is either PART of a key or not. Why have separate terms like partial dependency or transitive dependency? Why not just see, if there exists a dependency wherein a non prime attribute is determined by a something which is NOT a candidate key( no matter whether it is a partial key or another non prime attribute )

Why can't we implement a method like this:

• 1 NF -- having all elements in the atomic form
• X NF -- if there's any dependency of the form non_key -> non_prime_attribute(s) , decompose the relation with one of the new relation having this particular "non_key" as the key with those non_prime_attributes.
• BCNF : Where for all the dependencies of the form X->Y, X is a superkey?

Can we have such NF condition format? Does it combine all the conditions?

Answer 1

So, what's the exact difference?

2NF is not 3NF & definitions of 2NF are not definitions of 3NF. There isn't any particular semantic or syntactic structural similarity that would leave some kind of "difference" other than that a 2NF relation can have the sort of problem FD (functional dependency) that violates 3NF that a 3NF relation doesn't have. You can find definitions all over the place. You almost give them correctly here yourself. But a NF (normal form) is a condition, not a process. What do you mean "actions are the same"? Being in 3NF implies being in 2NF, so naturally decomposing to 3NF also gives 2NF. But there are relations that are in 2NF but not in 3NF, and there may be decompositions for a relation to 2NF that don't get to 3NF. Those decompositions will involve in a removal of all problem partial FDs that does not result in the removal of all problem transitive FDs.

(Because 3NF is always achievable and there are no other disadvantages compared to 2NF, 2NF isn't even useful. It's just a condition that was discovered first that is not as strong as 3NF.)

(3NF is frequently defined in terms of 2NF plus no transitive dependencies of non-prime attributes on CKs, but actually no such FDs implies no partial FDs of non-prime attributes on CKs, hence 2NF, so the first condition is redundant.)

Why not just see, if there exists a dependency wherein a non prime attribute is determined by a something which is NOT a candidate key

Why should that condition be helpful? It is not a description of just getting rid of the problem FDs of 2NF & 3NF--that's what putting into 3NF does.

Getting rid of non-trivial FDs that are not determined by superkeys happens to give BCNF. It implies 2NF & 3NF. But it is different from both of them. A BCNF relation exhibits no FD-based update anomalies. It is always achievable. However 3NF is alway achievable while "preserving FDs", whereas BCNF is not. There are cases where in order for a FD that held in the original to be enforced in a view/query that gives it via constraints on its components we need an EQD (equality dependency) constraint. That says two column sets have the same set of subrow values, which is more expensive to enforce than a FD. Either you have BCNF & an EQD & fewer update anomalies or you have 3NF/EKNF & a FD & certain update anomalies.

The NF that really matters is 5NF, which implies BCNF, with no update anomalies & with other benefits. (We might then decide to denormalize for performance reasons.)

PS Normalization to a given NF does not necessarily involve normalization to lower NFs.

Answer 2

It almost sounds as though you want to know why they called these two normal forms by different names instead of inventing just one form that covers both cases. If that's not the case, please ignore this answer.

Part of the answer is that the forms weren't discovered at the same time. And part of the answer is that the problem with 1NF that gave rise to 2NF is not the same as the problem with 2NF that gave rise to 3NF, even though they both exhibit harmful redundancy.

What might satisfy you a little more is BCNF. BCNF was actually discovered later than 4NF, so that name was already in use. But BCNF has to be placed between 3NF and 4NF, because it is more restrictive than 3NF but less restrictive than 4NF. So it was discovered "out of sequence", so to speak.

In BCNF, every (non trivial) determinant is a candidate key. That seems to be what you are looking for. I conjecture that any relation that is in 1NF and where every determinant is a candidate key, could be shown to be in 2NF and 3NF. But the proof is beyond me.

Answer 3

2NF and 3NF are essentially historical concepts and your question is a reasonable one. There is no real reason to apply them in practical database design because better tools exist today.

When it comes to teaching there is possibly some justification for mentioning 2NF and 3NF. Doing so allows students to explore the concepts involved (as you have done) while also teaching them a bit about the origins and rationale of design theory. In school maths lessons I was taught long division and differentiation from first principles. No one uses those techniques in practice, they are just teaching aids.

Answer 4

Before checking for 2NF the relation should be in 1NF. In simple words 2NF have only full dependencies only, no partial dependencies in relation. Full dependency means if x gives y, then by removal of any element in x, then y is not having any relation. If by removal of x, you are having relation with y then it is partial dependency. For 3NF we have to check for the 2NF, in 3NF we should not have any transitive relations like if x gives z, then there is no relation like x gives y and y gives z.

Solution for 2NF create a table for the partial dependcies and add foreign key in new relation which is primary key on the previous relation.

Solution for 3NF create a relation for both x gives y and y gives z. Add keys to relations.