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# CHAPTER 11 - CHAPTER 11 RELATIONAL DATABASE DESIGN...

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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 non-prime attributes of R i . Hence, if an FD Y -> A holds in R i where A is non-prime 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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CHAPTER 11 - CHAPTER 11 RELATIONAL DATABASE DESIGN...

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