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Database Management

For R=(,B,C,D,E,F,G,H,I,J) and functional dependencies H,D->A D->E,F,G H,D,B->C,J H->I,J A->BC What will be all the candidate keys? (I am getting DH as the candidate key)

11/18/2018 11:53:46 AM


4 Answers

New Answer


Yes, you are getting the correct answer: DH -> A,B,C,D,E,F,G,H,I,J Thus, the relation is in the Second Normal form, because NOT-key attributes are dependent on other NOT-key attributes. P.S. You have forgotten the A in the brackets ;)


But the relation should be in First Normal Form because H->I,J and for 2NF,subset of the candidate key should not be determining non-prime attributes.


You are right about the second normal form. And I'm sorry, but I only know until the 3.normal form. Can't help you with Boyce-Codd.


Panayot Zhaltov please check once whether I am correct or not. And also how to convert the relation to BCNF