[EM] Is "Inverse Nanson" better than standard Nanson?
Blake Cretney
bcretney at postmark.net
Sun Jul 8 12:59:24 PDT 2001
On Tue, 03 Jul 2001 21:21:16 -0700
Michael Rouse <mrouse at cdsnet.net> wrote:
> 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 see the technique of successively eliminating the winner to get a
full ranking as a well known way to get a ranking from a single-winner
method. I don't know if anyone has suggested this before for Nanson.
Nanson+ is a name I made up for the purposes of my email.
Inverse Nanson and Nanson+ don't generate the same order. However,
they have the connection that if you invert the input and output of
one, you get the other. This makes me suspect similar properties.
For example, if I have an example from Nanson+ where increasing a
candidate in the votes decreases it in the ranking, then I must have
an example for Inverse Nanson where decreasing in the votes increases
it in the ranking. This is the result of the property of inverting
inputs and outputs.
Similarly, if I have an example where replacing C with clones C1,C2,C3
does not result in a ranking with C replaced by C1, C2, C3 in some
order, then the same will be true of the inverse method.
So, methods and their inverses have similar properties, in some ways.
---
Blake Cretney
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