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Q. It is known that for R(A,B,C,D,E):

  1. R has exactly 5 superkeys.
  2. ABC is a candidate key.
  3. D is a non-prime attribute.
  4. ABE and ACE are not superkeys.

Is there another candidate key? If so, what is it?

Edit: The question of the problem is to determine if there is another candidate key in R(A,B,C,D,E) besides ABC, considering the 1,2,3,4 conditions hold true.

My approach is that, according to the 2nd condition ABC is a candidate key then the super keys are: ABC,ABCD,ABCE,ABCDE.

But the 1st condition says there are exactly 5 super keys which means the 5th super key could be the other candidate key. And according to 3rd and 4th condition the only other SK/CK could be BCE since ACE and ABE can't be SK's.

But if BCE is a candidate key BCDE should by a super key and this makes a total of 6 super keys and violates the 1st condition.

I am not exactly sure where I went wrong. Please help me analysing it right.

TheNamelessFoe
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    This has unclear claims & reasoning. Why is it that "Given, ABC as a candidate key there's possibility of only 4 super keys"? What does "That's where I think there is another candidate key" mean? Why is it that "it could be BCE since D is non prime"? What exactly is your question? PS Suggest you enumerate some possibilities relevant to CKs & non-CKs. – philipxy Dec 09 '18 at 01:46
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    PS The given conditions are impossible. But don't let that stop you. – philipxy Dec 09 '18 at 04:45
  • I am sorry for being unclear. The question of the problem is to determine if there is another candidate key in R(A,B,C,D,E) besides ABC, considering the 1,2,3,4 conditions hold true. – TheNamelessFoe Dec 09 '18 at 07:21
  • My approach is that, according to the 2nd condition ABC is a candidate key then the super keys should be ABC,ABCD,ABCE,ABCDE. Which amounts to 4 SK's. But the 1st condition says there are exactly 5 super keys which means the remaining 1 super key must be a candidate key. And taking 3rd and 4th condition into account the only other candidate key could be BCE. But if BCE is a candidate key BCDE should by a super key and this makes a total of 6 super keys and violates the 1st condition. I am not exactly sure where I went wrong. Please help if I am analysing it right. – TheNamelessFoe Dec 09 '18 at 07:26
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    Your comment's presentation is much clearer. Please clarify via post edits, not comments. But explain "taking into account". Always change a post to be the best presentation possible at the time you edit. (Please don't write vague & fuzzy things like "the superkeys should be ..." or "Which amounts to 4 SK's". Just state (possibly newly derived) facts clearly: "then the following are superkeys: ..." & "That is four of the five superkeys so the remaining ...".) Your question should *not* be the assignment question, it should be about how you are stuck. PS See my last PS. – philipxy Dec 09 '18 at 10:20
  • You added more about "according to 3rd and 4th condition" but you are still not clear. What is the reasoning that starts with that & uses "ACE and ABE can't be SK's" & ends up with "the only other SK/CK could be BCE"? – philipxy Dec 10 '18 at 07:05

1 Answers1

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Let’s try to detail the steps to solve the problem.

There are 25 (= 32) possible subsets of the attributes:

{{}, A, B, C, D, E, AB, AC, AD, AE, BC, BD, BE, CD, CE, DE, ABC, ABD, ABE,
 ACD, ACE, ADE, BCD, BCE, BDE, CDE, ABCD, ABCE, ABDE, ACDE, BCDE, ABCDE}

We are looking for candidate keys different from ABC, knowing the facts that you have mentioned.

  1. ABC is a candidate key.

If ABC is a candidate key then any subset of it cannot be a candidate key. This excludes the possibility of having {}, A, B, C, AB, AC, BC as candidate keys. Analogously, each superset of it cannot be a candidate key. This exclude ABCD, ABCE, ABCDE.

So we have now the following possible candidate keys:

{D, E, AD, AE, BD, BE, CD, CE, DE, ABC, ABD, ABE,
 ACD, ACE, ADE, BCD, BCE, BDE, CDE, ABDE, ACDE, BCDE}
  1. D is a non-prime attribute.

D is not a prime attribute so it can be included in superkeys but not in candidate keys or part of them. This means that if D is present in a set of attributes, then the set cannot be a candidate key. So this excludes D, AD, BD, CD, DE, ABD, ACD, ADE, BCD, BDE, CDE, ABCD, ABDE, ACDE, BCDE, from being candidate keys.

We have now the following possible candidate keys:

{E, AE, BE, CE, ABC, ABE, ACE, BCE}
  1. ABE and ACE are not superkeys.

Since ABE and ACE are not superkeys, they are not candidate keys as well, and of course none of their subsets are candidate keys. So, this exclude also E, AE, BE, CE, ABE, ACE.

So we have now the following possible candidate keys:

{ABC, BCE}

We know already that ABC is a candidate key, so the only remaining possibility is that BCE is a candidate key.

But we know that:

  1. R has exactly 5 superkeys. 2. ABC is a candidate key.

From these two facts, follows that ABC, ABCE, ABCD, ABCDE are four superkeys. Only one other is missing.

So, if BCE is a candidate key, as you have already noted, this would imply that also BCE and BCDE are superkeys. We have six different superkeys, and this is in contradiction with the hypothesis.

On the other hand, if BCE is not a candidate key, then we have only four superkeys, and also this in contradiction with the hypothesis.

So, we could say that it is not possible to give an answer to the question. Finally note that we cannot solve it also if we assume a non-usual meaning for “superkey”, as it is sometimes used: that is, a strict superkey, a set of attributes that strictly includes a key. In this case, if BCE is a candidate key, we have again four superkeys: ABCE, ABCD, ABCDE, and BCDE. If, on the other hand, BCE is not a candidate key, we have only three superkeys: ABCE, ABCD, and ABCDE.

Renzo
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    Interestingly, if we accept 1 + 2 as premises then already 3 is false & 4 is moot. PS See the comments on the question, it turns out the asker's commented clarifications (not yet complete or incorporated into their post) show that where they clarified they reasoned correctly & their actual question was, did they make a mistake to get that the premises are contradictory.. – philipxy Dec 09 '18 at 18:47