Can we decompose a schema into BCNF when it is not even in 1NF

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.

To simplify, say we have

R(A,B,C,D)

AB -> C
A  -> D

Relation R is NOT in 2NF -- the KEY is {AB}, but {D} is dependent only on part of the key.

Now decomposition

• Step 1. R1 {ABC} R2 {AD}

• Done.

So, both R1 and R2 are in 6NF and therefore in 5,4,3,2,1. There is no in-between -- it may happen that in some specific example you may see "progress" 1, 2, 3, BCNF; but that is an exception, not the rule.

... looking at a systematic approach to move from 1 NF to 2 NF to 3 NF to BCNF ...

is a common misunderstanding.