Unformatted text preview: r W Y, W is not a super key but Y is part of a candidate key.
(c) R is not in BCNF. Z X causes violation, because Z is not a super key.
Therefore, decompose R into XZ and YZW.
YZW is not in BCNF. W Y causes violation, because W is not a super key.
Therefore, decompose YZW into YW and ZW.
As a result, R is replaced by XZ, YW, and ZW that are all in BCNF.
Q.7 [15 pts] Given the relation schema R(A, B, C, D) with the functional dependency set
F = {AB C, B D, A B, DA C, A DC}. Find a canonical cover of F.
Answer:
Combine A B and A DC into A BDC, the set becomes {AB C, B D, A
BDC, DA C}.
B is extraneous in AB C, since A+ using F contains C.
D is extraneous in DA C, since A+ using F contains C.
The set now becomes {A C, B D, A BDC, A C} = {B D, A BDC}.
D is extraneous in A BDC, since A+ using {B D, A BC} contains D.
There is no other extraneous attributes.
Therefore, the canonical cover is {B D, A BC} Q.8 [15 pts] Consider a relation schema R(A,B,C,D,E,F) and a set of functional
dependencies {AB C, C FA, F E} that hold on R. Using only Armstrong’s axioms show that the functional dependency AB E also holds on R. (You should
not use rules other than the Armstrong axioms, such as union or decomposition.)
An...
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 Spring '11
 ÖzgürUlusoy
 Relational model, Candidate key, Database normalization, BCNF, super key

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