无损，依赖保持第三范式中关系的分解

Question

I am learning about normalization of databases and I came across the following problem:

 Given the following relation, BOSQID and its functional dependencies
 F={S->D, I->B, IS->Q, B->O},

How would I go about finding a lossless, dependency preserving decomposition of the above into 3rd normal form?

Not really sure what 3rd normal form is, I looked it up and found the following requirements:

1. The relation R (table) is in second normal form (2NF)

2. Every non-prime attribute of R is non-transitively dependent on every superkey of R.

After reading the above, I'm not too sure on where to begin. Also, not sure if it is helpful or relevant to answering the question, but I found the candidate key to be

IS

If someone could point me in the right direction, I would greatly appreciate it.

Answer 1

We have a relation, R(BOSQID) and its functional dependencies:

F={S→D, I→B, IS→Q, B→O}

As, you have concluded that only {SI} is candidate key, since {SI}+ = SIDBQO=BOSQID and no subset of {SI} is computed all attributes of relation R.

According to definition of partial dependency {S→D, I→B} are partial dependencies, since subset of candidate keys map non-prime attribute(s).

According to 2NF, partial dependencies (ie, {S→D, I→B}) not allowed. But {IS→Q, B→O} both are allowed in 2NF. Finally, relation R is not in 2NF.

To achieve 2NF, we decompose the relation R such that decomposition should satisfy lossless join and dependency preserving relations.

I decompose it in relations R1(ISQ) with F1={IS→Q}, R2(SD) with F2={S→D}, and R3(IBO) with F3={I→B, B→O}.

Now, candidate key(s) and normal form of relations R1, R2, and R3, are {IS} with BCNF, {S} with BCNF, and {I} with 2 NF respectively.

Since, decomposition R3 is not in 3NF, since FD {B→O}, neither B is Candidate of R3 nor O is prime attribute of R3. So, we need to decompose again to achieve 3NF in relation R3 such that decomposition should satisfy lossless join and dependency preserving relations.

I decompose R3 in R31(IB) with F31={I→B}, and R32(BO) with F32={B→O}. Now candidate key(s) and normal form of relation R31 and R32 are, {I} with BCNF and {B} with BCNF respectively.

Therefore, final BCNF as well as 3NF decomposition is R1(ISQ), R2(SD), R31(IB) and R32(BO).