Classroom Exercise: Normalization Consider the relation R(A,B,C,D) with these given FDs: AB -> C C -> D D -> A Compute all nontrivial FDs that follow from these. Compute the key(s) for R. What are all the superkeys? What are all the BCNF violations? Decompose R into BCNF. What are all the 3NF violations (before decomposing into BCNF)? Decompose R into 3NF.

Consider the relation R(A,B,C,D) with these given FDs: AB -> C C -> D D -> A Compute all nontrivial FDs that follow from these. 1. For each set of attributes X, compute X+: A+ = A; B+ = B; C+ = CDA; D+ = DA; AB+ = ABCD; AC+ = ACD; AD+ = AD; BC+ = BCDA; BD+ = BDAC; CD+ = CDA; ABC+ = ABCD; ABD+ = ABCD; ACD+ = ACD; BCD+ = BCDA 2. New FDs are of form X -> A for all A in X+; drop trivial ones A -> A; B -> B; C -> CDA; D -> DA; AB -> ABCD; AC -> ACD; AD -> AD; BC -> BCDA; BD -> BDA; CD -> CDA; ABC -> ABCD; ABD -> ABCD; ACD -> ACD; BCD -> BCDA 3. Drop redundant ones (if we have X -> A, don't need XY -> A): C -> D; C -> A; D -> A, AB -> C; AB -> D; AC -> D; BC -> D; BC -> A; BD -> A; CD -> A; ABC -> D; ABD -> C; BCD -> A

So the relation R(A,B,C,D) has these FDs: C -> D (given originally) C -> A D -> A (given originally) AB -> C (given originally) AB -> D Compute the key(s) for R. three keys: AB, BC, and BD What are all the superkeys? every superset of a key: ABC, ABD, BCD, ABCD What are all the BCNF violations? any FD whose LHS does not contain a key: C -> D; C -> A; D -> A

Let's start with C -> D. Decompose R into BCNF. Start with violating FD X -> A. Compute X+. New relations are R1 with attributes X+ and R2 with attributes R - X+ U X. Project R's FD's onto R1 and R2. Check if need to decompose some more. Let's start with C -> D. Compute C+ = CDA. New relation R1(C,D,A): FD's for R1: C -> D, C -> A, D -> A key for R1: C D -> A violates BCNF, need to decompose New relation R2(B,C): FD's for R2: none key for R2: BC in BCNF

Decompose R1(C,D,A) with FD's C -> D; C -> A; D -> A and key C Start with D -> A: Compute D+ = DA new relation R3(D,A) with FD D -> A and key D. Is in BCNF. new relation R4(C,D) with FD C -> D and key C. Is in BCNF. Final set of BCNF relation schemas: R2(B,C), R3(D,A), and R4(C,D)

What about decomposing into 3NF? Original relation is R(A,B,C,D) We already discovered the FD's: C -> D (given originally) C -> A D -> A (given originally) AB -> C (given originally) AB -> D We already discovered the keys: AB, BC, BD What are the 3NF violations (LHS does not contain a key AND RHS is not part of a key)? None! Every attribute is part of a key, so no FD violates FD. Thus no decomposition is necessary.