Re: Sixth normal form

From: Jan Hidders <>
Date: Fri, 10 Aug 2007 13:08:53 -0000
Message-ID: <>

On 9 aug, 04:15, "Brian Selzer" <> wrote:
> My use of the term "independent" is not the same as that defined by
> Rissanen. I consider a projection over a set of attributes A to be
> independent if and only if every dependent set of attributes determined by
> any subset of A is also a subset of A. In other words, a projection over a
> set of attributes A is independent if and only if for each functional
> dependency X --> Y in the closure of the set of functional dependencies for
> the relation schema containing A, if X is a subset of A then Y must also be
> a subset of A.

Very good, finally a proper definition. So you apprently meant it as a unary predicate over sets of attributes and not as a binary relationship between sets of attributes. Let me see, if we have a relation R1(A,B,C) with an FD A->B then you call the following sets independent: {}, {B}, {C}, {A,B}, {B,C}, {A,B,C}. Correct? And if we have R2(A,B,C) with A->B and A->C then the following are independent: {}, {B}, {C}, {B,C}, {A,B,C}.

> Consider a simple relation schema, {A, B, C} such that A --> B and B --> C.
> Rissanen would consider both of the projections {A, B} and {B, C}
> independent. I don't.

Indeed. You only call {B,C} independent.

> The closure of the set of functional dependencies
> includes A --> C, which can only be preserved by the inclusion dependency,
> {A,B}[B] IN {B,C}[B].

Not necessarily. That depends on your definition of FDs over attributes in different relations. The usual definition in normalization theory is that they hold for a schema if they hold for the natural join of all relations in the schema. In that case the FD is preserved also without the inclusion dependency.

  • Jan Hidders
Received on Fri Aug 10 2007 - 15:08:53 CEST

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