Hi Dan,

>> Oleg's solution is neater anyway though, as long as it's unlikely
>> that there'll be hordes of children of a (because if there were
>> lots of children then doing count()s would involve a lot of node
>> visits which would take time).
>
> See my earlier message. I'm not sure Oleg's solution actually
> meets the specified problem.

True; it does with a minor change though:

  b and count(b|c|d) = count(*)

Cheers,

Jeni

---
Jeni Tennison
http://www.jenitennison.com/


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