Question 54

The relation $R(A, B, C, D, E, F, G, H)$ has functional dependencies:

1. Identify essential attributes:

   • Attribute $D$ does not appear on the right-hand side of any dependency (in fact, $D$ does not appear in $F$ at all).
   • Therefore, $D$ must be a component of every candidate key.

2. Check closures of candidate key combinations:

   • Closure of $(AD)$:

     $$(AD)^+ = \{A, D\} \xrightarrow{A \rightarrow BC} \{A, B, C, D\} \xrightarrow{B \rightarrow CFH} \{A, B, C, D, F, H\} \xrightarrow{F \rightarrow EG} \{A, B, C, D, E, F, G, H\} = R$$

     Since no proper subset of $AD$ determines all attributes, $AD$ is a candidate key.

   • Since $E \rightarrow A$, consider $(ED)$:

     $$(ED)^+ = \{E, D\} \xrightarrow{E \rightarrow A} \{A, E, D\} = (AD)^+ = R$$

     Thus, $ED$ is a candidate key.

   • Since $F \rightarrow EG$, consider $(FD)$:

     $$(FD)^+ = \{F, D\} \xrightarrow{F \rightarrow EG} \{D, E, F, G\} \xrightarrow{E \rightarrow A} (AD)^+ = R$$

     Thus, $FD$ is a candidate key.

   • Since $B \rightarrow CFH$, consider $(BD)$:

     $$(BD)^+ = \{B, D\} \xrightarrow{B \rightarrow CFH} \{B, C, D, F, H\} \xrightarrow{F \rightarrow EG} \{B, C, D, E, F, G, H\} \xrightarrow{E \rightarrow A} R$$

     Thus, $BD$ is a candidate key.

   • Attributes $C, G, H$ only have $CH \rightarrow G$. Even with $D$, $(CHD)^+ = \{C, D, G, H\} \ne R$, so no other combination can form a candidate key.

Therefore, the relation $R$ has exactly 4 candidate keys: $\{AD, BD, ED, FD\}$. This matches option (B).