Database Normalization Problem Analysis

The problem asks us to identify the normal form of a relation R(A, B, C, D, E) given a set of functional dependencies F.

The relation has the following attributes: R(A, B, C, D, E).

The given functional dependencies are: F = {A → BC, D → E, BC → D, A → D}.

Understanding Normal Forms (NF)

Database normal forms are rules that help eliminate data redundancy and improve data integrity. The most common normal forms are 1NF, 2NF, 3NF, and BCNF.

Step 1: Check for First Normal Form (1NF)

A relation is in 1NF if all attribute values are atomic (indivisible) and single-valued.

The problem statement explicitly mentions, "all attributes are simple and take atomic values only." This directly tells us that the relation R satisfies the conditions for 1NF.

Conclusion for 1NF: The relation R is in 1NF.

Step 2: Find the Candidate Key(s)

A candidate key is a minimal superkey. A superkey is a set of attributes that can uniquely identify any tuple (row) in the relation.

To find candidate keys, we can calculate the closure of different attribute sets using the given functional dependencies:

• A → BC
• D → E
• BC → D
• A → D

Let's calculate the closure of {A}, denoted as A$^+$:

• Start with A: {A}
• Using A → BC, we add B and C: {A, B, C}
• Using A → D, we add D: {A, B, C, D}
• Now we have D. Using D → E, we add E: {A, B, C, D, E}

So, A$^+$ = {A, B, C, D, E}, which includes all attributes of R. This means A is a superkey.

Is A minimal? Can any attribute be removed from A and still have the closure be R? No, because A is a single attribute. Thus, A is a candidate key.

Are there other candidate keys? Let's check other attribute closures:

• BC$^+$ = {B, C}. Using BC → D, we get {B, C, D}. Using D → E, we get {B, C, D, E}. This is not all attributes of R.
• D$^+$ = {D}. Using D → E, we get {D, E}. This is not all attributes of R.
• Consider combining attributes. For example, AB$^+$? Since A alone determines everything, any set including A will also determine everything. So AB, AC, AD, AE, ABC, etc., are superkeys, but not minimal.

Conclusion for Candidate Key: The only candidate key for R is {A}.

From this, we can identify prime and non-prime attributes:

• Prime attributes: Attributes that are part of *any* candidate key. In this case, only A.
• Non-prime attributes: Attributes that are not part of any candidate key. In this case, B, C, D, E.

Step 3: Check for Second Normal Form (2NF)

A relation is in 2NF if it is in 1NF and every non-prime attribute is fully functionally dependent on every candidate key.

Full functional dependency means that a non-prime attribute cannot be dependent on a *proper subset* of a candidate key.

Our only candidate key is {A}. Since A is a single attribute, it has no proper subsets other than the empty set (which doesn't determine other attributes). Therefore, any dependency where A determines a non-prime attribute (like A → B, A → C, A → D, A → E) is considered a full dependency on the key A.

Let's check the dependencies involving the key A:

• A → BC (which implies A → B and A → C): B and C are non-prime. A is the full key. These are full dependencies.
• A → D: D is non-prime. A is the full key. This is a full dependency.
• From A → D and D → E, we can deduce A → E. E is non-prime. A is the full key. This is a full dependency.

All non-prime attributes (B, C, D, E) are fully functionally dependent on the candidate key A.

Conclusion for 2NF: The relation R is in 2NF.

Step 4: Check for Third Normal Form (3NF)

A relation is in 3NF if it is in 2NF and there are no transitive dependencies of non-prime attributes on a candidate key.

A transitive dependency exists when X → Z exists because of X → Y and Y → Z, where X is a candidate key, Y is *not* a superkey, and Z is a non-prime attribute.

Our candidate key is A. Non-prime attributes are B, C, D, E.

Let's look at the dependencies again:

• A → BC
• D → E
• BC → D
• A → D

Consider the dependency A → E. We deduced this from A → D and D → E.

• X = A (Candidate Key)
• Y = D
• Z = E (Non-prime attribute)

Is Y (D) a superkey? We found that D$^+$ = {D, E}, which is not the full set of attributes {A, B, C, D, E}. So, D is not a superkey.

Since A → D and D → E, where A is the key, D is not a superkey, and E is a non-prime attribute, this is a transitive dependency A → E via D.

Because there is a transitive dependency, the relation R is not in 3NF.

Conclusion for 3NF: The relation R is not in 3NF.

Summary of Normal Forms

Based on our analysis:

• The relation R is in 1NF.
• The relation R is in 2NF.
• The relation R is NOT in 3NF.

Therefore, the highest normal form the relation R belongs to is 2NF.

Revision Table: Normal Form Properties

Additional Information on Functional Dependencies and Keys

Understanding functional dependencies is crucial for database normalization. A functional dependency X → Y means that the value of X uniquely determines the value of Y.

Closure of an Attribute Set (X$^+$): The set of all attributes that are functionally dependent on X. It's calculated by starting with X and repeatedly adding attributes based on the given FDs until no new attributes can be added.

Determinant: In a functional dependency X → Y, X is the determinant.

Superkey: An attribute or set of attributes that uniquely identifies a tuple. The closure of a superkey must contain all attributes of the relation.

Candidate Key: A minimal superkey. No proper subset of a candidate key is a superkey.

Prime Attribute: An attribute that is part of *at least one* candidate key.

Non-Prime Attribute: An attribute that is *not* part of *any* candidate key.

Partial Dependency: A non-prime attribute is dependent on a proper subset of a candidate key (applies only if candidate key has multiple attributes).

Transitive Dependency: X → Z via Y, where X is a candidate key, Y is not a superkey, and Z is a non-prime attribute.

By carefully identifying candidate keys and checking for partial and transitive dependencies based on the definitions, we can determine the normal form of a relation.