The best normal form of relation scheme R(A, B, C, D) along with the set of functional dependencies \(F = \{AB \rightarrow C, AB \rightarrow D, C \rightarrow A, D \rightarrow B\}\) is
Third Normal form
First find the candidate keys. \(AB^+ = ABCD\), so AB is a key. Using \(C \rightarrow A\) and \(D \rightarrow B\): \(CD^+ = ABCD\), \(AD^+ = ABDC\) and \(BC^+ = ABCD\).
So the candidate keys are AB, AD, BC and CD, and every attribute A, B, C, D is prime.
3NF test: a dependency \(X \rightarrow Y\) is allowed if X is a superkey or Y is prime. AB → C and AB → D have a key on the left, while C → A and D → B have prime right sides, so R is in 3NF (and hence also in 2NF and 1NF).
BCNF test: every determinant must be a superkey. \(C^+ = CA\) and \(D^+ = DB\), so C and D are not superkeys, and C → A, D → B violate BCNF.
The highest normal form satisfied is therefore 3NF.
Hence, the answer is Third Normal form.
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