Unformatted text preview: Give all the functional
dependencies that violate 3NF, and justify your answer.
Answer:
The only candidate key of R is AC.
A B violates 3NF, because A is not a super key and B is not part of the candidate key
AC.
D F violates 3NF, because D is not a super key and F is not part of the candidate key
AC. Q.5 [5 pts] Given two relation schemas: R(A, B) with functional dependency A
and S(B, C) with functional dependency B C. State whether the result of (R
in BCNF. Justify your answer. B,
S) is Answer:
The result of (R
S) which will have the functional dependencies A B and B C.
The resulting relation.is not in BCNF, because for B C, B is not a super key of that
relation.
Q.6 [25 pts] Consider the relation schema R(X, Y, Z, W) with the following set of
functional dependencies: {XY Z, XY W, Z X, W Y}
(a) [5 pts] List the candidate key(s) of R.
(b) [10 pts] State whether R is in 3NF or not, and why. If it is not, decompose it into 3NF
relations.
(c) [10 pts] State whether R is in BCNF or not, and why. If it is not, decompose it into
BCNF relations.
Answer:
(a) Candidate keys: XY, YZ, ZW, XW.
(b) R is in 3NF.
For XY Z, XY is a super key.
For XY W, XY is a super key.
For Z X, Z is not a super key but X is part of a candidate key.
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 Spring '11
 ÖzgürUlusoy
 Relational model, Candidate key, Database normalization, BCNF, super key

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