Solutions for Section 3.6
Exercise 3.6.1(a)
In the solution to Exercise 3.5.1 we found that there are 14 nontrivial dependencies, including the three given ones and 11 derived dependencies. These are: C->A, C->D, D->A, AB->D, AB-> C, AC->D, BC->A, BC->D, BD->A, BD->C, CD->A, ABC->D, ABD->C, and BCD->A.
We also learned that the three keys were AB, BC, and BD. Thus, any dependency above that does not have one of these pairs on the left is a BCNF violation. These are: C->A, C->D, D->A, AC->D, and CD->A.
One choice is to decompose using C->D. That gives us ABC and CD as decomposed relations. CD is surely in BCNF, since any two-attribute relation is. ABC is not in BCNF, since AB and BC are its only keys, but C->A is a dependency that holds in ABCD and therefore holds in ABC. We must further decompose ABC into AC and BC. Thus, the three relations of the decomposition are AC, BC, and CD.
Since all attributes are in at least one key of ABCD, that relation is already in 3NF, and no decomposition is necessary.
Exercise 3.6.1(b)
(Revised 1/19/02) The only key is AB. Thus, B->C and B->D are both BCNF violations. The derived FD's BD->C and BC->D are also BCNF violations. However, any other nontrivial, derived FD will have A and B on the left, and therefore will contain a key.
One possible BCNF decomposition is AB and BCD. It is obtained starting with any of the four violations mentioned above. AB is the only key for AB, and B is the only key for BCD.
Since there is only one key for ABCD, the 3NF violations are the same, and so is the decomposition.
Solutions for Section 3.7
Exercise 3.7.1
Since A->->B, and all