Relation $R$ has eight attributes $ABCDEFGH$. Fields of $R$ contain only atomic values. $F=\{CH\rightarrow G, A\rightarrow BC, B\rightarrow CFH, E\rightarrow A, F\rightarrow EG\}$ is a set of functional dependencies (FDs) so that $F^+$ is exactly the set of FDs that hold for $R$. How many candidate keys does the relation $R$ have?

Let the set of attributes be $U = \{A, B, C, D, E, F, G, H\}$.

1. Identify essential attributes:

   • Attribute $D$ does not appear on the right-hand side of any functional dependency in $F$. Therefore, $D$ must be part of every candidate key.

2. Determine candidate keys:

   • $(AD)^+ = \{A, D\} \xrightarrow{A \rightarrow BC} \{A, B, C, D\} \xrightarrow{B \rightarrow CFH} \{A, B, C, D, F, H\} \xrightarrow{F \rightarrow EG} \{A, B, C, D, E, F, G, H\} = U$. So, $AD$ is a candidate key.

   • Since $E \rightarrow A$, we test $(ED)^+$: $(ED)^+ = \{E, D, A, B, C, F, H, G\} = U$. So, $ED$ is a candidate key.

   • Since $F \rightarrow EG \rightarrow E$, we test $(FD)^+$: $(FD)^+ = \{F, D, E, G, A, B, C, H\} = U$. So, $FD$ is a candidate key.

   • Since $B \rightarrow CFH \rightarrow F$, we test $(BD)^+$: $(BD)^+ = \{B, D, C, F, H, E, G, A\} = U$. So, $BD$ is a candidate key.

3. Check other combinations:

   • From $CH \rightarrow G$, $CHD$ closure gives $(CHD)^+ = \{C, H, D, G\} \neq U$.

   • No other subset combined with $D$ can derive all attributes.

Thus, there are exactly 4 candidate keys: $\{AD, BD, ED, FD\}$.

Hence, the correct option is (B).