[xquery-talk] Function for determining one XPath as subset of another
Adam Retter
adam.retter at googlemail.com
Thu Jan 28 04:18:09 PST 2016
Okay thanks to everyone for the replies.
I think the scope of what I was looking for got thrown out in all the
discussion, or perhaps I wasn't clear enough early in my initial post.
I am not interested in XQuery, and I am only interested in a small
subset of XPath. I am concerned with comparing two path expressions,
these path expressions may have predicates, what I didn't make clear
is that those predicates may not contain functions, only the
comparison operators eq, ne, gt, ge, lt, le, and =, !=, >, >=, <, <=.
As Christian observed, [1] is not the same as [@a eq 'b']. In my
original post, I posed these two path expressions:
1. //w
2. /x/y/z/w[@a = 'v']
In this instance I can actually ignore the predicate, as I can see
that /x/y/z/w would already produce the subset of //w and the
predicate only serves to restrict that further.
I am not looking for a formal proof in any sense. I am looking for
something practical that I can use in code.
On 28 January 2016 at 09:55, Pavel Velikhov <pavel.velikhov at gmail.com> wrote:
>
> Finally, can it be proved that /w[@a=b] is a subset of /w, taking into
> account that the filter can only contain standard operators (eq, gt, lt, etc
> as defined in the op namespace)?
>
>
> Okay, now we are close! However now formal do you want the proof to be? I
> can give this informal proof:
>
> The first path expression selects some subset of the children of the node
> that it applies, where the label is 'w'. The second one selects all
> descendants with label 'w', hence it's result contains all nodes of the
> first path expression.
>
> But if you want to go formal, you need to use the semantics of XQuery path
> expressions, as well as the formal specification of the data model. There
> used to be a formal document on XQuery data model.
>
>
>
> 2016-01-27 17:46 GMT+01:00 Christian Grün <christian.gruen at gmail.com>:
>>
>> > Well, so, to continue, let's assume that there are no user-defined
>> > functions, and in fact the only thing we want to proof is select+filter,
>> > where a filter is limited to the default operators. From that is it
>> > follows
>> > that
>> >
>> > -path1:
>> > select-child-nodes-by-name(select-child-nodes-by-name($context,'x'),'w')
>> > -path2: select-descendant-nodes-by-name($context,'w')
>>
>> Just to complete this: The predicate must not be numeric (//w[1] is
>> not equivalent to /descendant::w[1]).
>>
>>
>>
>> > Op woensdag 27 januari 2016 heeft daniela florescu <dflorescu at me.com>
>> > het
>> > volgende geschreven:
>> >>
>> >> >
>> >> > It seems to be a long-standing tradition that computer scientists,
>> >> > when
>> >> > asked to prove a difficult conjecture C, respond by giving a proof
>> >> > for a
>> >> > simplified conjecture C'. While this might lead to progress in the
>> >> > long run,
>> >> > and enables them to get papers published in the academic literature,
>> >> > it is
>> >> > totally useless to practical engineeers who want to know whether they
>> >> > can
>> >> > safely rely on C.
>> >>
>> >> Michael,
>> >>
>> >>
>> >> what you say is nice and true.
>> >>
>> >> However given that:
>> >> 1. path expressions point (syntactically hence esemantically) into
>> >> XQuery’s expressions
>> >> 2. XQuery expression language is Turing complete
>> >> 3. Subsumption for a Turing complete language is undecidable.
>> >>
>> >> Well, I can hardly see a way to decide this problem other then by
>> >> introducing SOME restrictions
>> >> of some sort… but of course some restrictions that would not nullify
>> >> the
>> >> original problem all together
>> >> and make the solution useless.
>> >>
>> >> Best,
>> >> Dana
>> >
>> >
>> >
>> > --
>> >
>> > W.S. Hager
>> > Lagua Web Solutions
>> > http://lagua.nl
>> >
>> >
>> >
>> > _______________________________________________
>> > talk at x-query.com
>> > http://x-query.com/mailman/listinfo/talk
>
>
>
>
> --
>
> W.S. Hager
> Lagua Web Solutions
> http://lagua.nl
--
Adam Retter
skype: adam.retter
tweet: adamretter
http://www.adamretter.org.uk
More information about the talk
mailing list