Database Relations | Heighest normal form

Question

MSQ

 

A relation R(A,B,C,D) has only trivial functional dependencies of the form ( A→A, AB→AB,ABC→A, etc)

Then consider the following options:

• A. The relation is surely in BCNF
• B. The relation is surely in 3NF
• C. The relation is surely in 2NF
• D. None of the above

 

Answer 1

If no valid FDs holds (a part from the trivial dependencies), then the only candidate key is ABCD, and the relation is all in 1NF, 2NF, 3NF and BCNF.
Hence A,B and C are all correct.

Comments

You are correct that if no non-trivial functional dependencies hold in a relation, then the only candidate key is ABCD and the relation is in all four normal forms (1NF, 2NF, 3NF, and BCNF).

A relation is in First Normal Form (1NF) if and only if it does not contain any repeating groups. Since the only functional dependencies in the relation are trivial, there are no repeating groups, and the relation is in 1NF.

A relation is in Second Normal Form (2NF) if and only if it is in First Normal Form (1NF) and every non-key attribute is fully dependent on the primary key. In this case, the only candidate key is ABCD, and all non-key attributes are fully dependent on the primary key, so the relation is in 2NF.

A relation is in Third Normal Form (3NF) if and only if it is in Second Normal Form (2NF) and every determinant in every non-trivial functional dependency is a superkey of the relation. Since there are no non-trivial functional dependencies in the relation, the relation is in 3NF.

A relation is in Boyce-Codd Normal Form (BCNF) if and only if every determinant in every non-trivial functional dependency is a superkey of the relation. Since there are no non-trivial functional dependencies in the relation, the relation is in BCNF.

Therefore, all three options (Option 1: The relation is surely in BCNF, Option 2: The relation is surely in 3NF, and Option 3: The relation is surely in 2NF) are correct.

Answer 2

Option 1 is correct. A relation is in BCNF (Boyce-Codd Normal Form) if and only if every determinant in the relation is a candidate key. In other words, if a functional dependency X → Y holds in a relation and X is not a superkey, then the relation is not in BCNF.

In this case, the functional dependencies given in the question are all trivial, meaning that they are of the form X → X or X → XY, where X is a subset of the attributes in the relation. These functional dependencies do not violate the BCNF condition because the determinant (X) is always a superkey. Therefore, the relation is surely in BCNF.

Option 2 (The relation is surely in 3NF) is not necessarily true because a relation can be in 3NF without being in BCNF. 3NF (Third Normal Form) is a less strict normal form than BCNF and requires that all non-prime attributes in the relation are fully functionally dependent on the relation's primary key.

Option 3 (The relation is surely in 2NF) is also not necessarily true because a relation can be in 2NF without being in BCNF. 2NF (Second Normal Form) is a less strict normal form than BCNF and requires that all non-prime attributes in the relation are fully functionally dependent on the relation's primary key.

Option 4 (None of the above) is not correct because one of the options (Option 1) is correct.

Comments

@rsansiya111 thanks for the detailed answer. But, B and C are also be right, since if a relation is in BCNF, it’s surely in 3NF and 2NF

Answer 3

Trivial FD = when the RHS is a subset of the LHS.

• Example: A → A, AB → AB, ABC → A, etc. Since there are no non-trivial FDs, no single attribute or subset can determine all other attributes.

• So the only key is the whole set of attributes {A, B, C, D}.

• Hence, every proper subset is NOT a key

•  The relation is surely in BCNF
   The relation is surely in 3NF
   The relation is surely in 2N

only trivial FDs = no violation of any normal form → highest NF (BCNF). If you’re told “only trivial FDs exist”, the relation is surely in BCNF → so also in 3NF & 2NF.

Does “only trivial FDs” guarantee 1NF?

No, it doesn’t guarantee 1NF by itself.

• Functional dependencies talk about how attributes relate,

• 1NF is about how values are stored (atomic vs multi-valued).

So:

• If the relation has only trivial FDs AND attributes are atomic →  it’s in 1NF (and BCNF).

• But if attributes have multi-valued or composite values, it may not be in 1NF even if FDs are trivial.

Whenever we talk about 2NF, 3NF, BCNF, we assume the table is already in 1NF.
So if the question is about higher NFs, we implicitly assume:

 Attributes are atomic → so 1NF is satisfied.

I assume this all in this logic 

• Trivial FDs alone don’t guarantee 1NF (because 1NF is about atomicity).

• But in normalization questions, we assume 1NF.

• Once 1NF is assumed, trivial FDs ⇒ BCNF ⇒ also 2NF & 3NF.