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Re: Pizza Example

From: Jan Hidders <jan.hidders_at_REMOVETHIS.pandora.be>
Date: Thu, 15 Apr 2004 19:30:20 GMT
Message-ID: <gpBfc.72317$e17.4591742@phobos.telenet-ops.be>


Dan wrote:
> Jan Hidders <jan.hidders_at_REMOVETHIS.pandora.be> wrote in message news:<l5ifc.71570$MG1.4809827_at_phobos.telenet-ops.be>...
>

>>Eric Kaun wrote:
>>
>>>"Jan Hidders" <jan.hidders_at_REMOVETHIS.pandora.be> wrote in message
>>>news:KVZec.70676$MG1.4809827_at_phobos.telenet-ops.be...
>>>
>>>>So, one would expect that the NEST and UNNEST operators of the nested
>>>>relational algebra would not be allowed, wouldn't one? 
>>>
>>>Do you mean GROUP? As far as I know, those are merely shorthand, not
>>>something new.
>>
>>?? Are you saying that the GROUP / UNGROUP operators, as Date calls 
>>them, can be expressed with the operators of the flat relational algebra?

>
> I believe these types of operators wouldn't conform to the rules of
> constructing well-formed formulas as well, which of course impacts
> FOL-based queries. In predicate calculus, the construction of a wff
> using a form G(P(x),...) is not allowed. I haven't had a chance to
> think much in terms of what the implications would be exactly however.
> Perhaps you know.

You go from first-order logic to higher-order logic, but in a very restricted way because all your sets are finite and typed (and therefore also finite in depth). Queries are still computable but can now be exponential, and deciding satisfiability was already gone in the flat relational model anyway. Note that you can always simulate this in the flat model by allowing abstract identifiers that denote the sets and a binary table that encodes the element-of relationship, plus a constraint that says that two set-identifiers are the same if they have exactly the same elements associated with them.

Received on Thu Apr 15 2004 - 14:30:20 CDT

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