hw5sol

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 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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