>> For the other approach, it would make more sense to simply have an
>> attribute with an XPath expression:
>>
>> <element name="orders" x:require="count(order) > 4">
>> <oneOrMore>
>> <ref name="order"/>
>> </oneOrMore>
>> </element>
>
> I don't think the syntax of writing assertions has something to do
> with the algorithm. This way of writing assertions can work fine with
> the "RNG-first-n-Schematron-later" style, and writing <s:rule> and
> <s:assertion> can work well with the "use-schematron-to-guide-RNG"
> style, too.
The issue isn't the syntax but the fact that the Schematron assertion
specifies an error message to be given to the user. With the
use-schematron-to-guide-RNG approach, it may be ambiguous which Schematron
constraint was not satisfied, so there's no way to determine which message
to the user. If you just have attribute (x:require) with no message, then
that doesn't matter.
> function check( Pattern p ) {
> if( p is already checked ) return;
>
> for( each e in first(p) ) {
> E' := { e' | e' \in first(p) and e ~ e' }
>
> if( E' contains more than one element with assertion )
> eureka!;
>
> check( choice of all content models in E' );
> check( residual of p by E' );
> }
> }
That doesn't look quite right to me. If we have
<choice>
<element>
<anyName/>
<empty/>
</element>
<element name="a">
<s:assert test="x">...</s:assert>
<ref name="any"/>
</element>
<element name="b">
<s:assert test="y">...</s:assert>
<ref name="any"/>
</element>
</choice>
then your algorithm would appear to make this ambiguous (when e is the
<anyName/> case).
I think it should be something like
function check( Pattern p ) {
if( p is already checked ) return;
for( each x in witnesses(first(p)) ) {
E' := { e' | e' \in first(p) and x \in e' }
if( E' contains more than one element with assertion )
eureka!;
check( choice of all content models in E' );
check( residual of p by <x/> );
}
}
where witnesses(Pattern p) returns a set of names with one name from each
equivalence class of names where two names are equivalent iff for any name
class in p the two names both belong to the name class or both do no belong.
Isn't checking this going to be exponential with something like
(a? & b? & c? & ...)*
?
James
