Question

Consider the following relational schema along with all the functional dependencies that hold on them. 
R1(A, B, C, D, E): { \( D \rightarrow E \), \( EA \rightarrow B \), \( EB \rightarrow C \) } 
R2(A, B, C, D): { \( A \rightarrow D \), \( A \rightarrow B \), \( C \rightarrow A \) } 
Which of the following statement(s) is/are TRUE?

• A. R1 is in {3NF}
• B. R2 is in {3NF}
• C. R1 is {NOT} in {3NF}
• D. R2 is {NOT} in {3NF}

Hint:

To check {3NF}, ensure that all non-trivial functional dependencies have a {superkey} as their determinant, or the dependent attributes are {prime attributes}.

Answer 1

The Correct Option is C, D

A relation is in Third Normal Form (3NF) if for every functional dependency \( X \rightarrow Y \), either \( X \) is a superkey, or each attribute in \( Y \) is a prime attribute (part of a candidate key).

For R1(A, B, C, D, E):

• The dependency \( D \rightarrow E \) violates 3NF because \( D \) is not a superkey, and \( E \) is not a prime attribute.
• Since the violation of 3NF exists in this relation, R1 is NOT in 3NF.

For R2(A, B, C, D):

• The dependencies \( A \rightarrow D \) and \( A \rightarrow B \) indicate that \( A \) is not necessarily a superkey.
• The dependency \( C \rightarrow A \) further confirms that transitive dependencies exist, which violates 3NF.
• Thus, R2 is NOT in 3NF.

Thus, the correct options are (C, D).