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One of the more desirable normal forms that we can obtain is Boyce Codd normal form (BCNF) A relation schema R is in BCNF with respect to a set F of functional dependencies if, for all functional dependencies in F + of the form , where R and R, at least one of the following holds: is a trivial functional dependency (that is, ) is a superkey for schema R A database design is in BCNF if each member of the set of relation schemas that constitutes the design is in BCNF As an illustration, consider the following relation schemas and their respective functional dependencies: Customer-schema = (customer-name, customer-street, customer-city) customer-name customer-street customer-city Branch-schema = (branch-name, assets, branch-city) branch-name assets branch-city Loan-info-schema = (branch-name, customer-name, loan-number, amount) loan-number amount branch-name We claim that Customer-schema is in BCNF We note that a candidate key for the schema is customer-name The only nontrivial functional dependencies that hold on Customer-schema have customer-name on the left side of the arrow Since customer-name is a candidate key, functional dependencies with customer-name on the left side do not violate the de nition of BCNF Similarly, it can be shown easily that the relation schema Branch-schema is in BCNF The schema Loan-info-schema, however, is not in BCNF First, note that loan-number is not a superkey for Loan-info-schema, since we could have a pair of tuples representing a single loan made to two people for example, (Downtown, John Bell, L-44, 1000) (Downtown, Jane Bell, L-44, 1000)

Now cos 2t oscillates between 1, so all we need to do is nd an a such that lim |cos 2t e at | = 0 Clearly, this is true for any a > 0; therefore, cos 2t is of

CHAPTER 10:

The McGraw Hill Companies, 2001

7

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Because we did not list functional dependencies that rule out the preceding case, loannumber is not a candidate key However, the functional dependency loan-number amount is nontrivial Therefore, Loan-info-schema does not satisfy the de nition of BCNF We claim that Loan-info-schema is not in a desirable form, since it suffers from the problem of repetition of information that we described in Section 72 We observe that, if there are several customer names associated with a loan, in a relation on Loan-infoschema, then we are forced to repeat the branch name and the amount once for each customer We can eliminate this redundancy by redesigning our database such that all schemas are in BCNF One approach to this problem is to take the existing nonBCNF design as a starting point, and to decompose those schemas that are not in BCNF Consider the decomposition of Loan-info-schema into two schemas: Loan-schema = (loan-number, branch-name, amount) Borrower-schema = (customer-name, loan-number) This decomposition is a lossless-join decomposition To determine whether these schemas are in BCNF, we need to determine what functional dependencies apply to them In this example, it is easy to see that loan-number amount branch-name applies to the Loan-schema, and that only trivial functional dependencies apply to Borrower-schema Although loan-number is not a superkey for Loan-info-schema, it is a candidate key for Loan-schema Thus, both schemas of our decomposition are in BCNF It is now possible to avoid redundancy in the case where there are several customers associated with a loan There is exactly one tuple for each loan in the relation on Loan-schema, and one tuple for each customer of each loan in the relation on Borrower-schema Thus, we do not have to repeat the branch name and the amount once for each customer associated with a loan Often testing of a relation to see if it satis es BCNF can be simpli ed: To check if a nontrivial dependency causes a violation of BCNF, compute + (the attribute closure of ), and verify that it includes all attributes of R; that is, it is a superkey of R To check if a relation schema R is in BCNF, it suf ces to check only the dependencies in the given set F for violation of BCNF, rather than check all dependencies in F + We can show that if none of the dependencies in F causes a violation of BCNF, then none of the dependencies in F + will cause a violation of BCNF either Unfortunately, the latter procedure does not work when a relation is decomposed That is, it does not suf ce to use F when we test a relation Ri , in a decomposition of R, for violation of BCNF For example, consider relation schema R (A, B, C, D, E), with functional dependencies F containing A B and BC D Suppose this were.

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Transform every term in the equation into the s domain Do algebraic manipulations to nd a solution in the s domain Invert the solution to nd a solution to the original equation in the time domain

The McGraw Hill Companies, 2001

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