GATE CS 1999 Q1.24 — Normal Forms

To determine which option is a candidate key for the relation scheme $R(A, B, C, D, E, F)$, we compute the attribute closure for each of the given options using the set of functional dependencies:

1. $C \rightarrow F$
2. $E \rightarrow A$
3. $EC \rightarrow D$
4. $A \rightarrow B$

Let's find the closure for each option:

• Option A: $(CD)^+$
  $(CD)^+ = \{C, D, F\}$ (using $C \rightarrow F$)
  Since this does not contain all attributes of $R$, $CD$ is not a key.


• Option B: $(EC)^+$
  $(EC)^+ = \{E, C\}$
  $\Rightarrow \{E, C, A\}$ (using $E \rightarrow A$)
  $\Rightarrow \{E, C, A, B\}$ (using $A \rightarrow B$)
  $\Rightarrow \{E, C, A, B, D\}$ (using $EC \rightarrow D$)
  $\Rightarrow \{E, C, A, B, D, F\}$ (using $C \rightarrow F$)
  Since $(EC)^+ = \{A, B, C, D, E, F\}$, $EC$ is a superkey.
  Furthermore, neither $E^+ = \{E, A, B\}$ nor $C^+ = \{C, F\}$ contains all attributes, meaning $EC$ is a minimal superkey (candidate key).


• Option C: $(AE)^+$
  $(AE)^+ = \{A, E, B\}$ (using $A \rightarrow B$ and $E \rightarrow A$)
  This does not contain all attributes of $R$, so $AE$ is not a key.


• Option D: $(AC)^+$
  $(AC)^+ = \{A, C, B, F\}$ (using $A \rightarrow B$ and $C \rightarrow F$)
  This does not contain all attributes of $R$, so $AC$ is not a key.

Thus, EC is a key for the relation $R$.