3NF Normalization Algorithm Step: Relation with SuperKey - database-normalization

Question

I was wondering about the last step of the 3NF normalization algorithm where it states:
4) If none of the relations obtained in previous steps contains a superkey of R, then add a new relation whose schema is a key for R.
My specific question is, what happens with the semantic of that relation? Why have only one relation and not many single-attribute relations (one for each attribute of the key)?
I found that in some examples that extra relation makes sense, but in others it seems to "mix" attributes that aren't related...

Answer 1

The last step of the 3NF normalization algorithm is needed to guarantee that the decomposition generated by the algorithm is lossless.
In fact there is a theorem that states that if a decomposition preserves the dependencies, and one of the decomposed schema is a superkey of the original relation, then the decomposition is also lossless.
The algorithm, with the previous steps, guarantees that every functional dependency is present in some of the decomposed relation. The introduction of a relation that contains a key, if no key is already present in some of the other relation, guarantees that the algorithm produces a decomposition that preserves both data and dependencies.
Added
Here is a simple example that shows the need of this last step. Suppose that an instance of the relation R(A, B, C, D), with A->B, C->D, (with key A,C) is:
R
A | B | C | D
-------------
1 2 2 3
1 2 3 4
2 3 2 3
The decomposition in R1(A,B), R2(C,D) is in third normal form but is lossy (additive). In fact, projecting that instance on the decomposition produces:
R1 R2
A | B C | D
----- -----
1 2 2 3
2 3 3 4
The additive property of this decomposition is clear if we perform a natural join of the decomposed relation, which produces an instance different from the original one:
R1 ⨝ R2 =
A | B | C | D
-------------
1 2 2 3
1 2 3 4
2 3 2 3
2 3 3 4
The situation does not change if you decompose R in R1(A,B), R2(C,D), R3(A), R4(C): in fact, recomposing it with R1 ⨝ R2 ⨝ R3 ⨝ R4 produces exactly the same relation as above with 4 rows:
R1 R2 R3 R4 R1 ⨝ R2 ⨝ R3 ⨝ R4 =
A | B C | D A C A | B | C | D
----- ----- -- --- --------------
1 2 2 3 1 2 1 2 2 3
2 3 3 4 2 3 1 2 3 4
2 3 2 3
2 3 3 4
Instead, the situation changes completely with the decomposition in R1(A,B), R2(C,D), R3(A, C). When you recompose with the natural join you obtain the original instance:
R1 R2 R3 R1 ⨝ R2 ⨝ R3 =
A | B C | D A | C A | B | C | D
----- ----- ------ --------------
1 2 2 3 1 2 1 2 2 3
2 3 3 4 1 3 1 2 3 4
2 2 2 3 2 3
So, in summary, in the first two cases you have a loss of information (the original instance is not obtained), while in the third case you have a 3NF and a lossless (nonadditive) decomposition.