[EM] Is "Inverse Nanson" better than standard Nanson?

[Blake Cretney]

Richard Moore <rmoore4 at home.com> wrote:
> As someone -- I forget who -- pointed out here some time
> ago, you can calculate Borda results from a pairwise matrix.
> Just add the figures in each column (or row, depending on
> how you organize your matrix). This assumes the matrix
> contains counts of all votes, not just winning votes. So,
> it could be deemed a pairwise method.

In fact, that is how you would carry out Nanson in practice.  Once you
have the pairwise matrix, you can easily see the Borda scores for any
subset of the candidates.

On Sun, 01 Jul 2001 23:12:52 -0700
Michael Rouse <mrouse at cdsnet.net> wrote:

> We can also invert Nanson's method like this:
> 1. Calculate the Borda count for all candidate.
> 2. Drop the highest-scoring candidate.
> 3. Re-order the remaining candidates.
> 4. As long as there are two or more candidates remaining, return to
step 1.
--snip--
> 5. Put the final "loser" candidate in the appropriate position on a
second 
> list, and remove from the original list.
> 6. Re-order the remaining candidates.
> 7. As long as there are two or more candidates remaining on the
original 
> list, return to step 1.

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. 

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.

The inversing makes the method more confusing, and I doubt it improves
on Nanson+'s properties.  In particular, I suspect the method violates
monotonicity and clone independence.  It clearly violates reversal
symmetry.  So, I think I'll stick with Ranked Pairs, at least until
some desirable properties are proven for inverse Nanson.

---
Blake Cretney