[EM] Benham not DMTBR after all?

[Kristofer Munsterhjelm]

Consider this election:

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

A is the Condorcet winner and the DMT set is {A, B} because C is the
Condorcet loser and the two first voter groups constitute 60% of the
voters, which is more than a third. ({A, B} is also a clone set.)

So suppose that A wins because the method passes Condorcet.

But then one of the BAC voters buries the winner A under the non-DMT
candidate C:

1: A>B>C
1: B>A>C
1: B>C>A  <-- here
2: C>A>B

This is an ABCA cycle and B wins in fpA-fpC (A's score is -1, B's score
is 1, and C's score is 0). In addition both Carey and IRV elect B.

Benham eliminates A (who has the least first preference count), then B
beats C pairwise and wins.

Minmax (even Ext-Minmax) produces a three-way tie, but can be made to
favor B like this:

10: A>B>C
22: B>A>C
20: C>A>B

before, and

10: A>B>C
11: B>A>C
11: B>C>A
20: C>A>B

after.


But if Benham doesn't pass this "minimal" version of DMTBR (if the DMT
set is {A, B}, then changing BAC to BCA shouldn't change the winner from
A to B), then full unrestricted DMTBR is much harder than I thought.

(Conversely, just meeting DMTCBR is more impressive than I thought.)

My optimally strategy resistant 5-voter 3-candidate method *also* elects
B. Which suggests that full DMTBR is incompatible with Condorcet, or
that requiring it makes the method so vulnerable to compromising that
it's not worth it.

Even the weaker clone independence related criterion:
	"If A is the DMTC, and we clone A, and one of the clones still wins,
then if someone who prefers another candidate buries the winner directly
under an adjacent non-DMT candidate, then the winner shouldn't change
[to someone the voter prefers to the winner]"

seems to be hard to satisfy, since in the example above, A and B are
clones. Removing the bracketed part produces something much stronger
which it might be possible to show outright as being impossible with a
Moulin-style proof.

-km