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R(A,B,C) is a relation.Which of the following don't have dependency preserving BCNF decomposition?

  1. A->B, B->C
  2. A->B,B->C,C->A
  3. AB->C,C->A
  4. None of these
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option c

See AB is the candidate key and C which is a non prime attribute is able to generate part of prime attribute A.

So ans should be C
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Answer should be c)
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3 Answers

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3 votes

In option, A relation can be clearly divided into AB and BC which would be both dependency preserving as well as in BCNF.

option B is already in BCNF .

option C has no BCNF decomposition that can preserve dependencies because to satisfy AB->C  you will need all three attributes in relation which in turn will create C->A dependency which would violate BCNF normal form

example :- one possible decomposition that is in BCNF is  AC and BC  but this does not satisfy AB->C dependency.

So answer should be option C.

3 Comments

Great answer ! @yajush
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How is it even possible to come to an answer? I feel some information is missing, like who is the candidate key in the original table? You said, option B is already in BCNF.

But we don’t know whether C is a candidate key or not
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I am giving reasoning with respect to option C

C isn’t the CK in the given relation, now if you break the relation for making it BCNF, then still dependency preserving won’t be satisfied as in the end you will be getting all the 3 attributes in one relation among the relation your break into. So, option C is appropriate. @kirtipurohit

Reference : https://www.cs.rochester.edu/courses/261/spring2018/lectures/l12.pdf

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0 votes
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In BCNF, NP->P and P->P relations are not allowed. In option C, C->A is NP->P. Hence it is not in BCNF.

Hence Option C.
0 votes
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Option C , where non prime attribute C determine A Which is the  part of the candidate key AB. It is not allowed.

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