Re: Quote of the Week

"Paul" <paul_at_test.com> wrote in message news:oFSoc.3147$wI4.311658_at_wards.force9.net...
> It's a very ingenious method, but the thing that worries me a bit is
> that you're basically creating a list and specifiying an arbitrary order:
>
> 1 Albert {
> 2 Bert {
> 3 }
> 4 Chuck {
> 5 Donna {
> 6 }
> 7 Eddie {
> 8 }
> 9 Fred {
> 10 }
> 11 }
> 12 }
>
> So everyone supervises the people between the numbers of their opening
> and closing brackets. (just another way of writing what you're saying
> really and easier to write than ascii art!). Now how is this different
> to this?:
>
> 1 Albert {
> 2 Chuck {
> 3 Eddie {
> 4 }
> 5 Fred {
> 6 }
> 7 Donna {
> 8 }
> 9 }
> 10 Bert {
> 11 }
> 12 }
>
> You have two different relations representing *identical* real-world
> situations. How would you compare two of these organization structure
> relations for equality?

Aggregated total? Assume

Albert=1
Chuck=2
Donna=3
...

Then, define recursively

SubtreeWeight = sum of children weights * max(node encoding) + parent node weight

Finally, tree structure are identical if they have identical weight at the root. Received on Fri May 14 2004 - 18:27:19 CEST