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1

Not knowing your full situation, I recommend something like this. The foreign keys should create the dependencies you need. If your SQL structure is already in a tabular format, you might consider putting this information into temporary tables--like the ones below. But it looks like your tables are there according to your assertion comment. From what you ...


2

When you have a relation R with a set of functional dependencies F and a decomposition of R in R1...Rn, you must consider two different concepts: The closure of the set of functional dependency F. This closure, called F+, is the set of all the dependencies derived from F, by applying, until possible, a set of rules called “Armstrong’s axioms”. This set can ...


5

The last two lines are an abbreviated way of solving the problem without recurring to the complete algorithm to check for dependency preservation. In particular the teacher noted that combining the dependencies that you can obtain from R1 and R3 you cannot obtain C, which is essential to get A in the dependency BC → A. This dependency can never be derived ...


1

I'd better say F¹⁺=∅ F²⁺=∅ F³⁺={AD → E, B → D, AB → D,AB → E} F⁴⁺={E → G} and (F¹⁺ U F²⁺ U F³⁺ U F⁴⁺ ) != F


4

You are correct, the relation is in 2NF, 3NF, BCNF. The reason is that the relation has two keys, A and CD. So the relation is in BCNF (which is a property stronger than 3NF and 2NF) since each determinant of the minimal conver of R1 is a key. Here is one minimal cover: A → C A → D C D → B C D → A



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