[xquery-talk] Function for determining one XPath as subset of another

[Adam Retter]

Okay so for those that are interested, here is the solution that I
ended up with:

1) I wrote my own XPath 2 parser as I wanted a simplified AST to start
operating from. The project I am working on needs to be under GPL2 and
I could not find a decent Java library that was compatible with that
license - https://github.com/exquery/xpath2-parser/

2) I then wrote some utility functions for descending through the Path
expressions and making subset comparisons -
https://github.com/adamretter/TEI-Completer/blob/master/src/main/java/org/humanistika/oxygen/tei/completer/XPathUtil.java#L79
Yes, this has many limitations and is most likely not complete yet,
but it serves me well for the small subset of XPath path expressions
that I want to support.

3) You can see in my test-cases how I consider one path expression as
a subset of another path expression:
https://github.com/adamretter/TEI-Completer/blob/master/src/test/java/org/humanistika/oxygen/tei/completer/XPathUtilTest.java#L40

Maybe that is of interest to some of you...

Cheers

On 28 January 2016 at 12:18, Adam Retter <adam.retter at googlemail.com> wrote:
> 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



-- 
Adam Retter

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