[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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