Re: abnormal forms
From: Bob Badour <bbadour_at_pei.sympatico.ca>
Date: Fri, 21 Apr 2006 00:04:13 GMT
Message-ID: <1kV1g.63419$VV4.1185338_at_ursa-nb00s0.nbnet.nb.ca>
>
> Could be, once one is talking about an implementation. I didn't think I
> needed to go that far for a more elementary purpose.
>
> I had in mind that there would be no such thing as an single
> valued-attribute except that all 'canonical' values would be
> single-element non-empty sets. I would have thought this could avoid
> typing errors.
>
> Yes, I didn't want to make an example like that because to me it takes
> me toward negation and I haven't yet wrestled where that could lead.
> Same reason I tried to avoid talking about an 'equality operator'.
Date: Fri, 21 Apr 2006 00:04:13 GMT
Message-ID: <1kV1g.63419$VV4.1185338_at_ursa-nb00s0.nbnet.nb.ca>
paul c wrote:
> Bob Badour wrote:
>
>> paul c wrote: >> >>> Bob Badour wrote: >>> >>>> ... >>> >>> If you have relation >> >> I assume by relation you mean relation variable.
>
> Could be, once one is talking about an implementation. I didn't think I
> needed to go that far for a more elementary purpose.
>> ...
>>
>> If you mean 'comparable' in the sense of the relational equality
>> operator that compares two relations for equality, then the answer is
>> yes. The result of the comparison is false.
>>
>>
>>> and is it relationally comparable to
>>>
>>> 3. SP{S,P} with value
>>>
>>> S P
>>> = - (where the '=' underscore means S is a 'key')
>>> 1 1
>>> 2 1
>>>
>>> ?
>>
>>
>>
>> That would depend on the definition of the equality operator. Because
>> the two relations have different types, the equality operator can
>> either return false or cause a compile-time error.
>> ...
>
> I had in mind that there would be no such thing as an single
> valued-attribute except that all 'canonical' values would be
> single-element non-empty sets. I would have thought this could avoid
> typing errors.
Why would anyone want to deprive themselves of the benefits of a compile-time type checker?
> > ...
>
>>> For sure, #3 is in a kind of canonical form as far as Codd was
>>> concerned.
>>
>> I disagree that #3 is a canonical form of either #1 or #2. Both #1 and
>> #2 allow {{},1} but #3 does not.
>> ...
>
> Yes, I didn't want to make an example like that because to me it takes
> me toward negation and I haven't yet wrestled where that could lead.
> Same reason I tried to avoid talking about an 'equality operator'.
I suggest you wrestle with equality first--it is a central concept to the set theory upon which the relational model is based. Received on Fri Apr 21 2006 - 02:04:13 CEST