[EM] Is "Inverse Nanson" better than standard Nanson?
Michael Rouse
mrouse at cdsnet.net
Tue Jul 3 21:21:16 PDT 2001
>Let me first suggest another way you could use Nanson's method to form
>a complete ranking. Instead of assuming the first dropped candidate
>is ranked last, apply Nanson until you get a winner. This winner is
>ranked first. Remove this candidate from consideration, and apply
>Nanson to the remaining candidates to get the second candidate.
>Remove it, apply Nanson, get third ranked, and so on. This method
>finds a Condorcet order as well. This is how I would use Nanson's
>method to get a full ranking. I'll call it Nanson+. If you invert
>the ballots, apply Nanson+, and invert the resulting ranking, you get
>inverse Nanson.
That is an equivalent method, and it generates the same results. I didn't
see Nanson+ when I did a web search (Google only had 34 hits for Nanson and
Borda together). The only reason I started at the bottom rather than the
top (trying to find the loser to drop rather than the winner) was because I
like to throw away candidates that affect the overall rankings the least. I
consider Nanson+ and inverse Nanson generating the same rank order of
candidates a plus; I didn't mean to claim credit for a method someone else
has already come up with.
>I generally feel that it's up to a method's proponents to prove that a
>method has desirable properties, rather than up to others to prove
>that it doesn't. Especially with a method that is rather difficult to
>carry out.
I don't consider it any harder than Nanson+ (it's much easier than the
"Crosscut" method I'm playing with! :), but it *is* superfluous considering
Nanson+ generates the same ranking list. I'm not even sure I could be
called a *proponent* of either form of Nanson's method -- I think it has
interesting properties, but I knew others probably had already looked at
similar methods and knew where the flaws were. I'd rather see what problems
it has before spending a lot of time trying to prove it is a better method.
(OK, I'm lazy ;-)
Michael Rouse
mrouse at cdsnet.net
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