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In a message dated 1/6/03 2:00:04 PM, jenglish@f... writes:
>I've heard this claim asserted many, many times,
>but I still don't believe it.
As you may know, Oleg Kiselyov's SXML is an S-expression
representation of XML. (See http://ssax.sourceforge.net/)
>Can anyone demonstrate the isomorphism? That is,
>come up with a pair of functions f :: S-Exp -> XML
>and g :: XML -> S-Exp such that (f . g) = id_{XML}
>and (g . f) = id_{S-Exp}?
An important attempt to demonstrate this with SXML and
Oleg's SSAX parser exists. See a message on Oleg's mailing list:
http://sourceforge.net/mailarchive/forum.php?thread_id=1310443&forum_id=599
>Note that any such pair of functions must also satisfy
>h(x) = f(g(h(x))) = h(f(g(x))) for all x :: XML and
>h :: XML -> XML, including, for example,
>h = XPath(ancestor::*/following-sibling::*).
I suspect that this could be demonstrated in conjunction with
Oleg's SXPath, but I am not aware of an attempt to do so, per se.
I think it is safe to say, however, that a casual atttempt to treat
s-exprs as isomorphic to XML can easily fail. One needs a well-
thought-out s-expr representation (which Oleg's is).
Jim Bender
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