·QI [1.5 pts]: For each of the following relation schemas and set of FD\'s: R(A,
ID: 3593734 • Letter: #
Question
·QI [1.5 pts]: For each of the following relation schemas and set of FD's: R(A,B,C,D) with FD's AB C, C D, and D A o R(A,B,C,D) with FD's B> C and B -> D o R(A,B,C,D) with FD's AB -> C, BC - D, CD> A and AD - B R(A,B,C,D) with FD's A B. B C, C D and D A R(A,B,C,D,E) with FD's AB C, DE C, and B D R(A,B,C,D,E) with FD's AB C, C D, D B, and D E Do the following: a) Indicate all the BCNF violations. Do not forget to consider FD's that are not in the given set, but follow from them. However, it is not necessary to give violations that have more than one attribute on the right side? b) Decompose the relations, as necessary, into collections of relations that are in BCNF?Explanation / Answer
Hello there,
a) * In the first relation, AB is Candidate key, and as A can be derived by D so BD is also key and by same reason BC is also key, so in this AB -> C is in BCNF but C->D and D-> A is violating BCNF because they are in 3 NF.
* As decomposition,(A,B,C) ,(C,D), (D,A) relation will be there.
b) * In the second relation, B can derive all except A so, AB is key. B->C and B->D both have partial dependency.
* As decomposition,(A,B,C) and (A,B,D) relation will be there.
c) * In 3rd relation all AB,BC,CD and AD are keys so relation is in BCNF, so no need to decomposition.
d) * in 4th relation all A,B,C and D are keys so relation is in BCNF, so no need to decomposition.
Hope, you got the answer, feel free to ask any queries and to give feedback.
Thank you
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