RE: Re: topological sort

> So your construct count be replaced by:
> 
>   not(field/type/ref[not(. = $processed/name)])
> 
> I'm not sure whether this will be more efficient - it probably depends
> on the processor.  

In Saxon it almost certainly will be. The count()=count() construct will
fully enumerate both node-sets, whereas the not(x[not(y)]) construct will
stop as soon as the first relevant element is found.

Mike Kay
> 

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