CPSC 304 2005 MIDTERM 1 SOLUTIONS

Sinceadeabcdefadeisakeyandinfactthe onlykeyofr

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Unformatted text preview: he algorithm we covered in class and in the book; circle all relations in your final decomposition. We begin by finding all of the (candidate) keys. The easiest way to do this is to calculate the closures. When in doubt, look at all closures. But we can tell that the only interesting closures are those that appear on the left hand side of some rule, so we’ll start there: A+ = ABC B+ = BC DE+ = DEF Thus we can see that A must be part of any key (since it’s the only way to derive A), and that DE and E must similarly be part of any key. Since ADE+ = ABCDEF, ADE is a key, and in fact, the only key of R. For a relation S to be in BCNF, each non­trivial functional dependency of the form Xày (where X is a set of attributes, and y is a single attribute), it must be the case that X is a superkey of S. Therefore, since ADE is the only key, all of the given functional dependencies violate BCNF, and we must decompose R. We arbitrarily choose to begin decomposition on AàB. We decompose into R1(AB), R2(A,C,D,E,F) Now we determine if we need to continue decomposing R1 and R2. Because R1 is a 2 valued relation, we know...
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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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