RE: [xml-dev] Does DTD allow deriving all possible paths in an XMLdocument?

One other point: you need to know what the root element type is. Technically
this is not part of the document type definition (=DTD), rather it is part
of the document type declaration. The two are often (understandably)
confused.

Regards,

Michael Kay
http://www.saxonica.com/
http://twitter.com/michaelhkay 
 

> -----Original Message-----
> From: Liam Quin [mailto:] 
> Sent: 13 October 2009 01:04
> To: 
> Cc: 
> Subject: Re: [xml-dev] Does DTD allow deriving all possible 
> paths in an XMLdocument?
> 
> On Mon, Oct 12, 2009 at 07:10:26PM -0400,  wrote:
> >     I have one question about XML DTD. In a paper, the authors say 
> > that DTD allows deriving all possible paths from the root to the 
> > leaves appearing in related XML documents. Does this statement 
> > correct? Based on my knowledge, DTD may not contain all possible 
> > elements in an XML document. I would like to get your 
> opinion. Thanks.
> 
> If you say that the XML document must be dtd-valid, then all 
> elements in the document must be listed (and defined) in the DTD.
> 
> A ontent model like
> <!ELEMENT mayhem ANY>
> means that "mayhem" elements may contain any elements at all 
> as children (as well as text), but for the document to be 
> valid the elemets must still be declared.
> 
> However, it is not possible to precompute all paths to leaves, because
> (1) mayhem could have any elements as children
> (2) a recursive content content model does not generate a finite
>     grammar - there's an unbounded set of possible valid input
>     documents.
> 
> e.g. <!ELEMENT doll (doll?)>
> allows
>     <doll><doll><doll /></doll></doll>
> to any depth.
> 
> These two points, ANY and cursion, are also rtue for SGML.
> 
> Liam
> 
> 
> --
> Liam Quin, W3C XML Activity Lead, 
> http://www.w3.org/People/Quin/ http://www.holoweb.net/~liam/ 
> * http://www.fromoldbooks.org/
> 
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