(ABF, BCF, CDF, CEF, CFG) ABF rightarrow CDEG AB rightarrow C, C rightarrow A, B

Question

(ABF, BCF, CDF, CEF, CFG) ABF rightarrow CDEG AB rightarrow C, C rightarrow A, BC rightarrow D, D rightarrow EG, BE rightarrow C, CG rightarrow B, CE rightarrow G The following is a candidate key BC BCF DEF BDE ABC Given R(A, B, C, D, E) and A rightarrow BC, B rightarrow C, D rightarrow E. Which of the following is a correct NF decomposition of R? AB, CD, DE, AE ABC, BC, DE BC, ABC, BDE ABC, AD, DE Given (R, F), Which statement is false? It is in 2NF but not in BCNF It is in 1NF but not in 3NF It is in BCNF and it is not in 3NF It is in 2Nf but not in 3NF a minimal cover has 10 fds, any set of 12fds cannot be a minimal cover True False Which anomaly is NOT avoided by the "repeatable read" isolation level? WW Phantom Dirty read none

Explanation / Answer

1. answer is option (b) BCF

candidate key is BCF.

if you take closure properties of BCF, you can get all the attributes in this relation.so it is the candidate key.

BCF+ = BCFADEG

2. answer is option d)ABC,AD,DE

this relation is currently not in 2nd normal form.because its candidate key is AD. but there is a partial dependency in the following A-> BC and D->E. so take cloure of LHS of these two dependencies, that is A+ and D+

if you take closure of A, you can get A+ = ABC

similarly, D+ = DE

therefor these ABC and DE act as two new relations and the original table is AD

finally you can get 3 relations. ABC, DE, AD,

now this is in 2nd normal form

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