Given the functional dependencies of a relation is there more then one way to normalize it

There's a little aphorism that goes:

• The key (1NF)
• The whole key (2NF)
• And nothing but the key (3NF)

. . . So help me Codd

In your example we can assume 1NF to begin with as the relational structure doesn't imply any repeating groups within the row (i.e. no D₁, D₂, D₃ etc.).

R = {a, b, c, d, e, f, g} F = {AB --> C, A --> DE, B --> F, F --> GH, D --> IJ }

2NF deals with 'the whole key' - if you have a composite key and some members of the relation are dependent on a part of that key then they should be split off into their own relations. In your case:

• R1 = AB --> C is in 2NF as 'AB' is 'the whole key'

• R2 = A --> DEIJ is 2NF but not 3NF as A --> DE, but A --> IJ is not 3NF as the real dependency of I and J is D --> IJ. D is not a key of this relation (A --> DEIJ), so A --> IJ violates the 'nothing but the key' princple of 3NF.

• R3 = B --> FGH is in 2NF but not 3NF as B --> F is a correct relation and F is dependent on the entirety of the key 'B'. However, B --> GH is not in 3NF as the correct relation is F --> GH. B --> GH violates the 'nothing but the key' principle, as G ahd H are correctly dependent on the non-key attribute F.

Your decomposition to 2NF is correct. Decomposition to 3NF requires taking the non-key attributes that have their own dependencies into separate relations.

The relation in 3NF would look like:

• R1 = AB --> C

• R2 = A --> DE (I and J are dependent on the non-key attribute D)

• R3 = B --> F (G and H are dependent on the non-key attribute F)

• R4 = D --> IJ

• R5 = F --> GH

2NF: Remove Partial Dependencies

R = {A, B, C, D, E, F, G, H, I, J} includes partial dependencies.

D and E depend only on A, F depends only on B, G, H, I and J don't depend on the key (directly) at all.

R0 = {A, B, C}

R1 = {A, D, E, I, J}

R2 = {B, F, G, H}

R0, R1, and R2 contain no partial dependencies (or repeating groups) so they are 2NF. However R1 and R2 are still an issue, because they contain transitive dependencies.

3NF: Remove Transitive Dependencies

I and J depend on D, not on the key of R1. Therefore you need to further normalize R1 as follows:

R1 = {A, D, E, I, J}

R1a = {A, D, E}

R1b = {D, I, J}

Similarly, G and H depend only on F so R2 must be decomposed as follows:

R2 = {B, F, G, H}

R2a = {B, F}

R2b = {F, G, H}

Now all of your remaining relations (R0, R1a, R1b, R2a, R2b) are devoid of repeating groups, partial dependencies and transitive dependencies. That means your relations are in 3NF.

When you are looking at an relation that hasn't been normalized and a series of dependencies, you can often normalize by inspection just by recognizing what your primary keys are going to be. Any attribute or combination of attributes that functionally determine other attributes are going to end up as primary keys. Once you've got your primary keys defined, you just need to figure out which non-key attributes go with each key. This is obvious from the statement of what your functional dependencies are.