Question-17
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) such that \(F^+\) is exactly the set of FDs that hold for \(R\).
How many candidate keys does the relation \(R\) have?
Given FD set: - \(CH \rightarrow G\) - \(A \rightarrow B\) - \(A \rightarrow C\) - \(B \rightarrow C\) - \(B \rightarrow F\) - \(B \rightarrow H\) - \(E \rightarrow A\) - \(F \rightarrow E\) - \(F \rightarrow G\)
Calculating closures: - \([A]^+ = \{ABCDEFGH\}\) - \([AD]^+ = \{ABCDEFGH\}\) - \([ED]^+ = \{ABCDEFGH\}\) - \([FD]^+ = \{ABCDEFGH\}\) - \([BD]^+ = \{ABCDEFGH\}\)
The attribute \(D\) is not present in the FD set, so whenever an attribute is not present in the FDs, adding that attribute forms a candidate key. Therefore, candidate keys = \([AD, ED, FD, BD]\), leading to \(4\) candidate keys for relation \(R\). Hence, the correct answer is \(4\).