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A simple question that I do not seem to find an answer for online: Is the relation that has no non-trivial functional dependencies considered to be in BCNF form and if not, how do I decompose it? Thank you in advance!

Alexander Nenartovich
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  • See [ask], hits googling 'stackexchange homework' & the voting arrow mouseover texts. Quote a definition of BCNF from a textbook & ask re the first place you get stuck showing whether your case satisfies it. PS Your question "if not, how do I decompose it?" is a case of "how do I decompose a relation to BCNF" & that's unresearched, too broad & (nevertheless) a duplicate. Right now you are just asking for us to rewrite a textbook with a bespoke tutorial & you have shown no research or other effort. PS "online" is not a good source of relational model information. Read a textbook. – philipxy Jul 14 '19 at 03:28
  • I'm afraid you simply misunderstood my question. It was anything but broad and I was not asking anybody to re-write textbooks and/or create tutorials. PS Please proofread your posts before submitting :) – Alexander Nenartovich Jul 14 '19 at 04:07
  • I did not misunderstand your question. The part that I commented as too broad was how to decompose it & I explained why that is too broad. The first part of your question is also too broad. I said to find a definition & show your work applying it. Asking us to do that is too broad. Either of these questions makes it reasonable to close your question as too broad. Also you should ask one question per post. Asking 2 also makes your post too broad. – philipxy Jul 14 '19 at 04:13
  • Asking whether a relation with trivial FDs only is in BCNF is too broad? Seriously? As the answer I posted below demonstrates, this question can be clearly answered in a single relatively short sentence by those who know the answer. – Alexander Nenartovich Jul 14 '19 at 04:22
  • Your comments do not reflect an understanding of the references I gave re protocols here or others at [help]. Not all questions are on-topic. Eg ones where you use a word but you haven't bothered to find out what it means. – philipxy Jul 14 '19 at 05:05
  • Question: Is a relation with trivial functional dependencies only in BCNF? Answer: Yes, as follows from the definition of BCNF, any relation that has only trivial functional dependencies is in BCNF. Please explain how and why this is a very broad question. – Alexander Nenartovich Jul 14 '19 at 05:18
  • All your comments are oblivious to mine & are addressed by my first comment. I'm done. – philipxy Jul 14 '19 at 05:36
  • Still failed to explain how the question is too broad. Considering the fact that it could have been answered in a single reasonably short sentence (as shown above) I still contend you did not fully understand what I was asking. – Alexander Nenartovich Jul 14 '19 at 05:45

1 Answers1

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After additional research, I finally stumbled upon this definition of BCNF:

A relational schema R is considered to be in Boyce–Codd normal form (BCNF) if, for every one of its dependencies X → Y, one of the following conditions holds true:

  • X → Y is a trivial functional dependency (i.e., Y is a subset of X)
  • X is a superkey for schema R

So to answer my own question, every relation that has trivial functional dependencies only is in BCNF by definition.

Perhaps it helps somebody who has a similar question.

Alexander Nenartovich
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    Regarding decomposition. A relation satisfying only trivial dependencies would have to consist only of a single key and no non-prime attributes at all. Such a relation could not be non-loss decomposed because a key is irreducible by definition. Did you have an example in mind? – nvogel Jul 14 '19 at 05:18
  • I believe you are correct, such a relation would have to consist of a single (probably composite) key and no other attributes. As for an example, not sure how helpful it'll be for you but the question I asked relates to the following problem: Given the following relation R(A, B, C, D, E) with FD’ s AB —> C , DE —> C , and B —> D determine if R is in BCNF and if not decompose it into smaller relations. I eventually got to a relation with attributes A, B, E that had trivial dependencies only and wasn't sure if that relation was in BCNF. – Alexander Nenartovich Jul 14 '19 at 05:27
  • @nvogel The question asks whether it can be decomposed if it isn't in BCNF, but from your comment you seem to think it is asking about decomposing it if it is in BCNF. Hm... I guess you are just exploring decomposition based on the given FDs indpendent of notions of BCNF. PS A schema whose FDs are all trivial might still be losslessly decomposable per JDs that hold that aren't associated with FDs. – philipxy Jul 14 '19 at 05:40