Normalisation Databases : 1NF, 2NF, 3NF Answers are provided. Just explain part

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Normalisation Databases : 1NF, 2NF, 3NF

Answers are provided. Just explain part B for each question please

Exercise 19.7 Suppose you are given a relation R with four attributes ABCD. For each of the following sets of FDs, assuming those are the only dependencies that hold for R, do the following: (a) Identify the candidate key(s) for R. (b) Identify the best normal form that R satisfies (1NF, 2NF, 3NF, or BCNF). (c) If R is not in BCNF, decompose it into a set of BCNF relations that preserve the dependencies. 2. B C, D A 3. ABC D, D- A 4, A ? B, BC ? D, A ? C 5, AB ? C, AB ? D, C ? A, D ? B

Explanation / Answer

Below are the explanation of part b. Please comment of you are still having any trouble understanding it ..

1. b) There is no partial dependency and B is the candidate key, So we can derive all attributes from B. Now we have B-> C and C-> D and because B is candidate key, B-> D is also true. So there is transitive relationship between B,C And D. Hence it is not in 3NF.

2. b) Because BD is candidate key and the FDs are B-> C and D-> A. So A and C have partial dependency on B and D. so it is not in 2NF. The attributes are assumed to be atomic so there are in 1NF.

3. b) There is a FD D-> A here in relation. But D is not the candidate key. So there is attribute which is dependent on a non key attribute. Hence it is not in BCNF.

4. b) Here A-> B and A-> C so A-> BC. Now we also have BC-> D as FD. But A is the candidate key so A-> D also holds. So there is transitive dependency between A, BC and D. So it is not in 3NF.

5.b) C-> A and D-> B are the FDs having C and D on left side. But C and D are not candidate keys. So here attributes are dependent on non key attributes. So it is not in BCNF.

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