Skipping steps in Normalization?

Question

Just curious: is there some reason why one cannot do all necessary normalizations in a single step? Isnt normalization ultimately the redrawing of the Functional Dependency (FD)

Answer 1

Yes: Normalization can be characterized by rearranging (hyper)graphs. It does not have to be done by moving through normal forms in some order. (It's just a common misconception that it is.)

The normal forms on the continuum from 1NF to 6NF are those dealing with problematic FDs (functional dependencies) and JDs (join dependencies). They can be ordered so that if a relation value or variable satisfies a form then it satisfies the forms before but not necessarily after. Currently: 1NF, 2NF, 3NF, EKNF, BCNF, 4NF, ETNF, RFNF, SKNF, 5NF aka PJ/NF, Overstrong PJ/NF, 6NF. This ordering has nothing to do per se with decomposing to relation values or variables that are in higher normal forms. It is not necessary to decompose through a sequence of forms.

The normal forms are just different conditions that have been found with helpful properties. Moreover, the normal forms are just those that have been discovered; there may well be other helpful properties to be distinguished. We don't pass through them to normalize now. ETNF is 2012!

As to your graph characterization:

A FD has a set of attributes as determinant. Which determines another set. But since the one determines the other if and only if the one determines each of the sets that contain exactly one member of the other, informally but unambiguously we also talk about a set of attributes determining an attribute. A FD {...} -> a holds iff a = f(...). (There can be zero or more determinant attributes.) BCNF is the highest normal form re problematic FDs, but there are higher normal forms re problematic JDs. A JD with given components holds in a relation iff it is always their join. Ie its meaning/predicate can be expressed as the AND of the components'. So a FD {...} -> A holds iff a JD holds corresponding to a meaning/predicate with conjunct A = F(...)! A MVD (multi-valued dependency) corresponds to a certain binary JD. 5NF means that every JD that holds is "implied by the keys" (a technical term).

There are algorithms that starting with FDs decompose directly to 2NF, directly to 3NF and directly to BCNF (with various other properties like preservation of FDs). See the Alice book. One can decompose to 6NF simply by decomposing until there are no nontrivial JDs, without regard to FDs.

(See C. J. Date's Database Design and Relational Theory: Normal Forms and All That Jazz.)