hi lars, thanks a lot for the illustration. i think this is a great approach. i would like to ask a couple of detailed questions though.
> If two nodes in the graph have the same distance to the primaryType, > following rules are applied: > 1) nt:base has precedence over nt:unstructured really? my gut feel would have been the other way around... i haven't thought it through though... did you have a particular rationale that you can share? > 2) built-in node types have precedence over nt:base since nt:base is the "mother" all built-in nodes i think i can agree to the sentence pre-pending "other " ;) > 3) user-defined node types have precedence over built-in node types ack. > 4) an alphabetical precedence ordering takes place it sounds like this is so arbitrary that i would probably just opt for a quick "implementation" defined (==undefined) resolution instead of sorting. what do you think? did you mean to specify a resolution order by (1) - (4) or is the sequence of the rules arbitrary. regards, david
