Re: Lots of Idiotic Silly Braces?

"paul c" <toledobythesea_at_oohay.ac> wrote in message news:Lf4oi.131821$xq1.73460_at_pd7urf1no...
> Brian Selzer wrote:
> ...
>> Can rva's be keys? A relation value being the extension of a predicate,
>> the set of tuples in a relation value represents a set of positive atomic
>> formulae, and under the closed world assumption, that set implies the
>> negation of each atomic formula that conforms to the schema but is not
>> represented by a tuple. How, then, can a relation valued attribute be a
>> key? Consider, the schema R{S{A, B}}, and the following relation value,
>> r:
>>
>> r = {{S={{A=3, B=4}, {A=3, B=5}}}, {S={A=3, B=4}}}
>>
>> Now, suppose that P(A, B) is the predicate of S. The first tuple of r
>> asserts that P(3, 5) is true, but the second tuple implies that P(3, 5)
>> is false. ...
>
> I think "S" here is an attribute name, not a relation. S refers to two
> different relation values which happen to appear as tuples in r. The first
> tuple of of r asserts that the ("first", relation-valued) tuple that
> combines (P(3,4) AND P(3,5)) is true in one relation value. The second
> tuple implies that (P (3,5)) is false in the "second" relation value.
>

But that's the point. How can they both be true? And if they can, then how can it be consistent since there isn't also a tuple {S={A=3, B=5}}?

>> ... It stands to reason that P(3, 5) cannot be both true and false. As a
>> consequence, I don't think that a relation valued attribute can stand on
>> its own. Lending support to this is the fact that the duality of GROUP
>> and UNGROUP does not hold for R. r UNGROUP S yields the relation value
>> (call it r'),
>>
>> r' = {{A=3, B=4}, {A=3, B=5}}
>>
>> but r' GROUP {A, B} AS S yields either
>>
>> (1) {{S={{A=3, B=4}}}, {S={{A=3, B=5}}}}
>>
>> or
>>
>> (2) {{S={{A=3, B=4}, {A=3, B=5}}}}
>> ...
>
> If r' has two tuples and one attribute and S refers to a relation with
> attributes A and B, r' GROUP {A,B} has one tuple. I find it hard to read
> this notation, but I think I see two tuples in (1) so I would reject it
> and I think (2) is right.
>

r has two tuples and one attribute, S
r' has two tuples and two attributes, A and B.

>> I can't be sure, but neither (1) nor (2) is identical to r. So clearly
>> information is lost when you ungroup a relation expression that has only
>> an rva in its heading. Furthermore, due to the definition of a
>> functional dependency, it follows that the duality of GROUP and UNGROUP
>> can only hold when the rva is a dependent attribute.
>> ...
>
> It holds when the attributes that are not grouped are dependents.
> > For these reasons, I believe that rvas cannot be keys or components
> of
>> keys--at least not for base relations. Have I overlooked something?
>
> S is the key of r above.
>
> p
Received on Sat Jul 21 2007 - 03:31:13 CEST