[EM] Summability and resistant set election: two other options excluded

[Kristofer Munsterhjelm]

About a month ago, I wrote in my post "Resistant set incompatibility and 
passing methods", that

> I couldn't find any such "disjoint collisions" with the same first
> preference counts *and* pairwise matrices, leaving open the
> possibility that a combination of the two could be used to create a
> summable Resistant method.

Well, here's a disproof found by linear programming, showing that a 
method using only the pairwise magnitudes and first preferences can't be 
resistant. Surprisingly, using the full positional matrix doesn't help 
either.

Election 1:

         2: A > B > C > D
         1: A > D > C > B
         1: B > A > C > D
         2: B > D > C > A
         1: C > A > D > B
         1: C > B > D > A
         2: D > A > B > C
         1: D > B > A > C

A disqualifies B and C, and B disqualifies D. Hence the resistant set 
consists of A alone.

Election 2:

	3: A > B > C > D
         2: B > A > D > C
         1: B > D > C > A
         1: C > D > A > B
         1: C > D > B > A
         1: D > A > B > C
         1: D > A > C > B
         1: D > B > C > A

This election has the same first preference counts as election 1, and 
the same Condorcet matrix. It even has the same positional matrix 
(number of first preferences, second preferences, etc, per candidate).

But D disqualifies A. Hence the two elections' resistant sets are 
disjoint despite having the same pairwise magnitudes and first 
preference counts.

A method that only uses first preferences and the pairwise matrix can't 
distinguish between the two and must get at least one of them wrong. 
This also holds for any refined subset of the resistant set, since if 
the resistant sets are disjoint, subsets are too.

This is a minimal example in the total number of voters over two 
elections. It would in theory be possible to make a "partially 
resistant" method that elects from the resistant set for every 
four-candidate election with fewer than 11 voters. Perhaps such a method 
would generalize to provide partial protection at larger voter and 
candidate numbers, but it's hard to tell. For that matter, it's hard to 
tell how to construct such a method.

With all of this, summability is looking pretty unlikely. But perhaps 
there is a particular structure that works... or a relaxation of the 
resistant set that retains its strategy resistance and admits summability.

With three candidates, pairwise + first preferences suffice, as shown by 
fpA - fpC.

-km