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# Normalization Levels for Relation Assignment Answers Needed

Question 1

The relation customer(id, firstname, lastname, street, city, points, credictcardnumber) is decomposed into

Lossless-join decomposition

3NF decomposition

A B C D 1 x a 9 2 y b 9 3 x b 9 4 y a 9

A → B

Question 3

Consider the attributes ABCDEF and functional dependencies on those attributes: { A→B, C→DE, B→F }

AB, CDE, ACF

Question 4

in 3NF, but not in BCNF.

in BCNF.

# Step By Step Answers with Explanation

This type of decomposition is called:

Lossless-join decomposition.

A B C D

1 x a 9

A → B: This holds because for every unique value of A, B is the same.

B → C: This does not hold because for the same value of B (x or y), we have different values for C (a and b).

Explanation:

B → C does not hold because B does not determine a unique value for C.

A BCNF decomposition ensures that for each non-trivial functional dependency X → Y in the original relation, X should be a superkey of the decomposed relations.

Let's find the superkeys:

AB (A→B)

CDE (C→DE)

AB represents the functional dependency A→B.

CDE represents the functional dependency C→DE.

We need to determine the normal form of relation R.

Let's start by checking the highest normal form that R satisfies:

A → BC (This holds, and A is a candidate key)

B → CFH (This holds, and B is a candidate key)

In this case, we can see that E → A, and A → BC forms a transitive dependency, where E → A → BC. This violates 3NF.

BCNF: A relation is in BCNF if, for every non-trivial functional dependency X → Y, X is a superkey.

E → A (E is not a superkey, so this is not in BCNF)

F → EG (F is not a superkey, so this is not in BCNF)

R does not satisfy 2NF because not all non-prime attributes are fully functionally dependent on the candidate keys.

R does not satisfy 3NF because it has a transitive dependency.

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