Test: Functional Dependency - Computer Science Engineering (CSE) MCQ

By applying these rules repeatedly, we can find all of F⁺, given F.

Lossy-join decomposition is the decomposition used here.

This can be computed as the canonical cover.

The query can be satisfied by any of the two options.

The value of m must lie between the min value of r and s and 0.

Boyce-Codd Normal Form should satisfy the following two conditions:

• It should be in the Third Normal Form.
• And, for any dependency A → B, A should be a super key.

Relationship between 1NF, 2NF, 3NF and BCNF:

I. Every relation in 3NF is also in BCNF
False, Every relation in BCNF is also in 3NF. A relation schema that is in the third normal form then that table may or may not be in BCNF. Here 3NF is a subset of BCNF hence the given statement is False.
II. Every relation in BCNF is also in 3NF
True, Every relation in BCNF is also in 3NF. Here BCNF is a superset of 3NF hence the given statement is True.
Hence the correct answer is Only II is correct.

Concept:
The closure of F, denoted as F⁺, is the set of all regular FD, that can be derived from.
For trivial functional dependency,
Let A and be two sets consists of attributes of a relation
A → B
A ⊇ B 
Option 1: 
{X, Z} → {Z, W}
{X, Z}  ⊉ {Z, W}
Not a trivial functional dependency

Option 2: 
{X, V} → {V}
{X, V} ⊇ {V}
It is a trivial functional dependency

Option 3: 
{X, W, V} → {Y}
{X, W, V} ⊉  {Y}
Not a trivial functional dependency

Option 4: 
{Y, W} → {Y, X}
{Y, W} ⊉ {Y, X}
Not a trivial functional dependency
NOTE:
⊇ → superset
⊉ → not superset

Concept:
A candidate key is a column, or set of columns, in a table that can uniquely identify any database record without referring to any other data
The following functional dependencies,
U → V, VW → X, Y → W, X → U
S = (U, V, W, X, Y, Z)
Here YZ is an independent key. So, for every key, the YZ should their. Hence only option 4 is containing YZ for all keys.
∴ Hence the correct answer is UYZ, VYZ, XYZ.

1NF:

• There are only Single Valued Attributes.
• There is a unique name for every Attribute/Column.

2NF:

• A relation that is in First Normal Form 
• Every non-primary-key attribute is fully functionally dependent on the primary key 

3NF

• The third normal form is based on the concept of transitive dependency. A functional dependency X->Y in a relation schema R is a transitive dependency if there exists a set of attributed Z in R that is neither a candidate key nor a subset of any key of R.
• A relation is in 3NF if it satisfies 2NF and no prime attribute of R is transitively dependent on the primary key.
• If X - > A is a functional dependency, then A should be a prime attribute or X should be a candidate key.
• Lossless join and dependency preservation is always possible in 3NF. 3NF decomposition is always lossless join and dependency preserving.

BCNF:

• It stands for Boyce Codd normal form.
• A relation R is in BCNF if whenever a non-trivial functional dependency X -> A holds in R, then X is a superkey of R. Any relation with two attributes is always in BCNF. Because when a relation contains only two attributes then one attribute determines another and the left side of the functional dependency will always be a candidate key in that case.
• BCNF is not always dependency preserving.

Therefore the answer is Option 1.

It is possible to use Armstrong’s axioms to prove that Pseudotransitivity rule is sound.