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

Michael Rouse mrouse at cdsnet.net
Sun Jul 1 23:12:52 PDT 2001


Nanson's method -- which calculates the Borda score for each candidate, 
drops the lowest, then recalculates -- can be proven to always find the 
Condorcet winner if one exist. It is one of few non-Condorcet methods with 
this property.

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.

This method can be shown to always pick the Condorcet loser in a set, if 
there is one available. We can take this Condorcet loser and drop it from 
the original list, then re-order the candidates and recalculate. In this 
way we can generate a sorted candidate list from the bottom up, one that 
will use Condorcet order if possible but will still find a single candidate 
for each available position even if there are one or more circular ties.

The revised "inverse" Nanson method would add the following steps:
...
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.

Just as a small example of where the inverse Nanson would come up with a 
Condorcet order and standard Nanson would not, consider the following:
2     3
A    B
B    C
C    A

Borda would give the following count: A=4, B=8, C=3. With Nanson, we would 
drop C, then check B(3) vs. A(2). You would end with the order BAC. In the 
inverse Nanson, we would drop B since it is the highest, then compare A vs. 
C. A would lose. Comparing C vs. B, C would lose. We would then end up with 
the order BCA, which agrees with the Condorcet order.

Anyway, I think it has some advantages over the standard Nanson method. I's 
be interested in any thoughts on it, especially examples where it breaks 
the monotonicity criteria (Nanson does and I just want to get a feel for 
how often that would happen in comparison). Thanks!

Michael Rouse
mrouse at cdsnet.net



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