Re: Scalars & atomic values & variables

From: Lauri Pietarinen <lauri.pietarinen_at_atbusiness.com>
Date: Mon, 05 Jan 2004 23:11:35 +0200
Message-ID: <btck3p$7bt$1_at_nyytiset.pp.htv.fi>


Bob Badour wrote:

>"Lauri Pietarinen" <lauri.pietarinen_at_atbusiness.com> wrote in message
>news:3FF96A65.9010007_at_atbusiness.com...
>
>
>>Bob Badour wrote:
>>
>>
>>
>>>"Lauri Pietarinen" <lauri.pietarinen_at_atbusiness.com> wrote in message
>>>news:3FF904F7.3060207_at_atbusiness.com...
>>>
>>>
>>>
>>>
>>>>Dawn M. Wolthuis wrote:
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>>Therefore, unless everyone is discussing the exact same set of objects
>>>>>(which could be defined by a set of types) and operators, the use of
>>>>>
>>>>>
>the
>
>
>>>>>term scalar is inexact and apt to be confusing as different values will
>>>>>
>>>>>
>>>>>
>>>>>
>>>be
>>>
>>>
>>>
>>>
>>>>>scalar in different models. When the whole idea of the model is to be
>>>>>extensible, then the term scalar is also not very useful as adding in
>>>>>another function/operator into the space and/or adding in new objects
>>>>>
>>>>>
>can
>
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>>>>>change what is and is not a scalar.
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>If we think of scalars in terms of what can be seen by relational
>>>>
>>>>
>>>>
>>>>
>>>operators (union, projection, cartesian product, etc...) it would seem
>>>
>>>
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>>>>to me that defining scalars as values that cannot be "cracked open" with
>>>>
>>>>
>>>>
>>>>
>>>these operators would pretty well do it for me.
>>>
>>>That strikes me as entirely arbitrary. One can crack open a string using
>>>
>>>
>an
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>>>operation that returns a relation of position/character tuples. I guess
>>>
>>>
>that
>
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>>>makes a string non-scalar.
>>>
>>>
>>>
>>Well, I don't know what the definition of a relational operation is, but
>>if we decide that such a function
>>is not a relational operator then it is not arbitrary.
>>
>>
>
>Um, am I hearing you correctly? If we make an arbitrary decision, then it is
>not arbitrary ?!?
>
What I meant was that if we decide that the only relational operators are the ones that can
be derived from the primitive ones (or a set of primitives, such as UNION, PROJECTION,RESTRICTION)
then a function that returns a relation from a string is not a relational operator. Or it is some kind of hybrid,
which of course is perfectly OK, but not the kind I had in mind. Would that be arbitrary?

Lauri Received on Mon Jan 05 2004 - 22:11:35 CET

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