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Given a relation R(A,B,C) and a FD set (A->B, B->C), find the FD Closure. Enumerate all the FD's and organize them accordingly to the LHS of the FD's. Mark all the trivial FD's.

I am a little confused with the second part of the question and don't understand what exactly they mean by "organize them accordingly to the LHS of the FD's" My understanding is, Apart from the given FD's, we have A->C(on account of transitivity) and the trivial FD's will be A->A , B->B, C->C, A->AB, A->AC, B->BC. Is this the answer to the given question or am I missing something here?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Raphael
    Nov 26, 2018 at 21:48

1 Answer 1

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I suppose that in the exercise, the request “organize them accordingly to the LHS”, is just a request of sorting all the functional dependencies (maybe lexicographically) with respect first to the attributes of the left hand side (the determinant of the functional dependency) (and then maybe on the rigth hand side for all the FD with the same left part).

For what concerns the list, I think it should include all the possible functional dependencies, including those with two attributes as determinant, as well as the (only) dependency with three attributes as determinant, as well as those with all the subset of attributes determined on the right part. So something like:

A -> A
A -> AB
A -> AC
A -> BC
A -> ABC
AB -> A
...
B -> B
...

If you want to be super-formal, you can include also the trivial dependencies A -> {}, etc., as well as the trivial dependency {} -> {}.

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