CPSC 304 2005 MIDTERM 1 SOLUTIONS

# Thereforethefinalanswerisr1abr3defr5acr6ade

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Unformatted text preview: it is in BCNF, so we turn to R2. DEàF violates BCNF because in R DE+ = DEF. Therefore, projected onto R2, DE+ = DEF, and DE is not a key of R2. DEàF is therefore a violation of BCNF, and we must decompose again. We get R3(D,E,F), and R4(A,C,D,E). Are they in BCNF? R3 is in BCNF because DE is a key, so it is not a BCNF violation, and there are no other non­trivial functional dependencies on R3. For R4, however, we know from above that A+ = ABC. Projected onto R4, this means that A+ = AC. Which means both that the functional dependency AàC holds on R4 and that A is not a key of R4. So we decompose to yield R5(A,C), R6(A,D,E). R5 is in BCNF since it only has 2 attributes, and since we know that the closure of A yields nothing in R6 other than itself, and similarly for the closure of DE, R6 is in BCNF. Therefore, the final answer is: R1(A,B), R3(D,E,F), R5(A,C), R6(A,D,E) b. {4 marks} Is R(A, B, C, D, E, F) in 3NF? If so, why. If not, list all violations of 3NF in R. B, C, and F are part of no keys, nor, as illustrated above, are A, DE, or B part of a superkey. Therefore...
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## This document was uploaded on 03/18/2014 for the course CPSC 304 at University of British Columbia.

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