For w y w is not a super key but y is part of a

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online