CHAPTER 11: RELATIONAL DATABASE DESIGN ALGORITHMS AND FURTHER
DEPENDENCIES
Answers to Selected Exercises
11.15 Show that the relation schemas produced by Algorithm 11.2 are in 3NF.
Answer:
We give a proof by contradiction. Suppose that one of the relations R i resulting from
Algorithm 13.1 is not in 3NF. Then a FD Y > A holds R i in where: (a) Y is not a
superkey of R, and (b) A is not a prime attribute. But according to step 2 of the
algorithm, R i will contain a set of attributes X union A 1 union A 2 union ... union A n ,
where X > A i for i=1, 2, ..., n, implying that X is a key of R i and the A i are the only
nonprime attributes of R i . Hence, if an FD Y > A holds in R i where A is nonprime and
Y is not a superkey of R i , Y must be a proper subset of X (otherwise Y would contain X
and hence be a superkey). If both Y > A and X > A hold and Y is a proper subset of X,
this contradicts that X > A is a FD in a minimal set of FDs that is input to the algorithm,
since removing an attribute from X leaves a valid FD, thus violating one of the
minimality conditions. This produces a contradiction of our assumptions. Hence, R i must
be in 3NF.
11.16 Show that if the matrix S resulting from Algorithm 11.1 does not have a row
that is all "a" symbols, then projecting S on the decomposition and joining it back will always produce
at least one spurious tuple.
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 Fall '09
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 Algorithms, Relational Database, Relational model, Candidate key, Database normalization, relation schemas

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