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hw5sol - (5 Candidate key(s A This is a lossless BCNF...

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CS174A H OMEWORK 5 SOLUTIONS 3. a) CDE_ ACD_ BCD b) R is in 3NF because B, E and A are all parts of keys c) R is not in BCNF because none of A, BC and ED contain a key 5. (1) 6. (3) Candidate keys: A, C. Since A and C are both candidate keys for R, it is already in BCNF So from a normalization standpoint it makes no sense to decompose R further.
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Unformatted text preview: (5) Candidate key(s): A. This is a lossless BCNF Decomposition. This is however not dependency preserving (B consider Æ C). So it is not free of (implied) redundancy. This is not the best decomposition(The decomposition AB, BC, CD is better)...
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