[EM] Condorcet-composite method DMTBR disproof

Kristofer Munsterhjelm km_elmet at t-online.de
Fri Mar 25 10:58:39 PDT 2022

On 25.03.2022 01:15, Kevin Venzke wrote:
> Hi Kristofer,
> Le jeudi 24 mars 2022, 18:03:03 UTC−5, Kristofer Munsterhjelm <km_elmet at t-online.de> a écrit :
>> I may have a disproof showing that if X is DMTBR, then Condorcet//X and
>> Smith,X are not necessarily so. But this is a pretty strong claim, so
>> I'd like to check with the list first.
>> For the (somewhat inelegant) election:
>> 14: L>C>R>X>Y>Z
>> 10: R>C>L>Y>Z>X
>> 7: C>L>R>X>Z>Y
>> 15: Y>Z>X>L>C>R
>> 15: Z>C>R>L>Y>X
>> 7: Z>X>C>R>L>Y
>> 22: Y>X>R>C>L>Z
>> can anyone check if:
>>      - the innermost DMT set is {L, C, R},
>>      - the IRV winner is L,
>>      - the Condorcet winner is C,
>>      - and if the bloc of 14 voters votes L>X>C>R>Y>Z instead, then the
>> Smith set contains every candidate?
> Those all look correct to me, although I'm not sure if "innermost DMT set" is
> always a clear term. You might have more than one I think, where neither is a
> subset of the other. I am no expert on the concept though...

A DMT set is a set of candidates who are a solid coalition of more than
1/3 of the voters, and who (strictly) beat everybody outside it
pairwise. The innermost DMT set is the smallest such set.

I don't think there can be more than one because then each set would
have to beat everyone outside it, and you can't have two sets that
properly beat each other pairwise.

If the properties I listed in my original post are true, then this
election shows that while the innermost DMT set is {L, C, R} and C is
the CW, the L-faction burying C under X expands the Smith set to all the
candidates, so that Smith//IRV (and Smith,IRV and Condorcet//IRV and...)
elects L. Hence burying under someone not in the innermost DMT set paid
off, which contradicts DMTBR.

And since IRV itself passes DMTBR (or really burial immunity period),
this shows that it's not always the case that Smith-X or Condorcet-X is
DMTBR even if X is. No wonder I had such trouble proving its compatibility!

What's interesting is that this doesn't seem to hurt the strategic
performance of the method compared to those that do pass both Condorcet
and DMTBR. See e.g. JGA's strategy evaluation. Perhaps it's like his
candidate strategy evaluation of minmax: although minmax technically
isn't cloneproof, it doesn't have much candidate exit/entry incentive,
whereas IRV, though being technically cloneproof, does have considerable
exit incentive. Perhaps in some way, what matters is whether the failure
is significant or not.

But Smith-M and Condorcet-M should pass DMTCBR whenever M does, because
if W is the DMT candidate, then it's simultaneously the C-M and M
winner, and thus burying in favor of some other candidate Z can only
expand the Smith set, which cannot help Z due to the base method passing

This should, if I'm right (do correct me! I'm feeling I'm on a bit of
uncertain ground) also hold for every set is just the CW when there is
one, and where buriers can never kick someone off the set by burying,
for the same reason. I *think* Landau has this property? I don't think
Copeland has it. Banks might.

Since there's compatibility for DMTCBR, Benham should pass both
Condorcet and DMTBR (proper). Due to how IRV works, it can't eliminate
the whole innermost DMT set. As soon as everybody but one member of the
innermost DMT set has been eliminated, that candidate wins, just like
IRV. So where does IRV fail? It fails by possibly eliminating the CW
early (consider Burlington). Benham protects against this.

Benham's elimination mechanism preserves the property of IRV's that if
the voters who prefer A to W bury W under some X, then that doesn't
change the elimination order until A has been eliminated (because these
buriers will be ranking A ahead of W anyway). Thus it should pass DMTBR
because IRV does.

And that in turn suggests that the easiest way to get DMTBR and
Condorcet is to make some kind of method that does both DMTCBR and
Condorcet, and then eliminate in some clever manner. Just simply
eliminating the Plurality loser won't work as it's not monotone. But
perhaps something based on the idea of batch eliminating everybody with
low first preferences up to a sum of 1/3 fpp, and then repeating this
procedure on what's left? It's not obvious how to incorporate Condorcet
compliance even if that turns out to be monotone, though.


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