GATE CS 1997 Q6.9 — Normal Forms

To determine the normal form of the relation $R(a, b, c, d)$, let us analyze the given conditions step-by-step:

1. First Normal Form (1NF):

   A relation is in 1NF if the domain of each attribute contains only atomic (indivisible) values. Since the problem explicitly states that the domains of $a, b, c, d$ include only atomic values, the relation $R$ is in 1NF.

2. Candidate Key:

   Let us find the attribute closure to determine the candidate key(s):

   • $\{a, b\}^+ = \{a, b, c, d\}$ (using $a \rightarrow c$ and $b \rightarrow d$)

   Since no proper subset of $\{a, b\}$ can derive all attributes, the only candidate key for $R$ is $\{a, b\}$.

   This gives us:

   • Prime attributes: $a, b$ (attributes that are part of a candidate key)
   • Non-prime attributes: $c, d$ (attributes that are not part of any candidate key)
3. Second Normal Form (2NF):

   A relation is in 2NF if it is in 1NF and no non-prime attribute is partially dependent on any candidate key. In other words, a functional dependency $X \rightarrow Y$ violates 2NF if $X$ is a proper subset of a candidate key and $Y$ is a non-prime attribute.

   Let us check the given functional dependencies:

   • $a \rightarrow c$: Here, $a$ is a proper subset of the candidate key $\{a, b\}$, and $c$ is a non-prime attribute. This is a partial dependency.
   • $b \rightarrow d$: Similarly, $b$ is a proper subset of the candidate key $\{a, b\}$, and $d$ is a non-prime attribute. This is also a partial dependency.

   Since partial dependencies exist, the relation is not in 2NF.

Thus, the relation is in first normal form but not in second normal form.