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For the relationship R(A,B,C,D), I have the following FDs..

  • AB->C
  • BC->D
  • CD->A
  • AD->B

There's no obvious BCNF violation and when I take the closures I get..

  • AB+=ABCD
  • BC+=ABCD
  • CD+=ABCD
  • AD+=ABCD

My only thought is that I'm wrong in determining my closures but I've never seen a set of FDs like this so I'm not sure. Can someone tell me if I did anything wrong or is this already in BCNF?

1 Answers1

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Informally, you have BCNF when every arrow is an arrow out of a candidate key. I think your closures are right, and this is already in BCNF.

Mike Sherrill 'Cat Recall'
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