[EM] Burial examples
Kristofer Munsterhjelm
km_elmet at t-online.de
Mon Apr 2 05:57:27 PDT 2018
On 04/01/2018 02:10 PM, Chris Benham wrote:
> On 1/04/2018 7:27 AM, Kristofer Munsterhjelm wrote:
>
>> Passing dominant mutual quarter burial resistance is really hard, if
>> even possible:
>
> I am sure that for a Condorcet method it isn't possible.
>
> Meeting dominant mutual third burial resistance isn't so easy. I don't
> know of any Condorcet method
> that does that is better than the "Hare-Condorcet hybrid".
>
> The version I like: voters strictly rank from the top as many or as few
> candidates as they wish. If there is
> a CW (or single pairwise-undefeated candidate?) x then elect x.
> Otherwise eliminate the candidate that (among remaining candidates) is
> voted top-most on the fewest
> ballots and if among remaining candidates there is pairwise-beats all
> candidate (or single pairwise-undefeated
> candidate?) x then elect x. If not, repeat and so on.
>
> But it fails mono-raise and can fail to elect a positionally dominant
> uncovered candidate, unlike my other
> favourite Condorcet methods.
My investigations seem to suggest that we can have mono-raise without
affecting strategic susceptibility too much; or rather, without doing so
as long as the Smith set is of three candidates or fewer.
Because of a combinatorial explosion, it's very hard to use the approach
I am using to go beyond three-candidate methods, however, which limits
what I can do.
I would ultimately want to find a method that is as strategically robust
as a Hare-Condorcet hybrid while retaining clone independence and
without having to give up mono-raise, and beyond that pass as many
criteria as possible. (I'd also like to find out if Smith and
mono-add-top are compatible.) Finding out if reversal symmetry is
compatible with DMTBR and Condorcet would help narrow down the criteria
such a "better method" could have.
My approach so far has been to look at what three-candidate methods my
programs find, and then try to extend them. However, they are often hard
to understand. Suppose the simulator finds out that a method where A's
score in an ABCA cycle is (2 * fpA + fpB - 2 * fpC) is a good one (where
fpX is the number of ballots ranking X first). What's going on here and
how would it be generalized? There's no obvious answer. It gets even
worse with nonlinear methods, as the methods may seem stranger still.
E.g. the simulator might say that a method where A's score in an ABCA
cycle is (max(C>A, B>C)/C>A) is a good one, but why?
Thus my attempt to find a criterion or criteria to distinguish good
methods from bad ones. And thus my interest in DMTBR.
Would Uncovered,IRV be a better method? Or "eliminate and stop when one
candidate covers everybody else"?
> It comfortably handles your four burial scenario examples. In the first
> three the burying has no effect and in
> the fourth it backfires.
>
> My other favourite Chicken Dilemma criterion complying Condorcet method,
> MMLV(erw) Margins-Sort Elimination,
> also easily elects the sincere CW in the first 3 examples but the burial
> succeeds in the fourt
>
> https://wiki.electorama.com/wiki/Chicken_Dilemma_Criterion
Chicken Dilemma doesn't seem to be that criterion, at least not on its
own, because as far as I understand it, Smith,Plurality passes it. And
Smith,Plurality's performance is even worse than that of the methods
that reduce to minmax.
Of course, it's possible to argue against my simulations by saying it
counts what can be measured rather than measures what counts; more
specifically, that CD violation is a far more serious problem as it
induces a more destructive form of strategy. But that'd be more a reason
to want both CD and DMTBR than to say CD is the criterion that
distinguishes strategy-resistant methods from methods that aren't.
> Taking your first example:
>
> 10: A>B
> 11: B>C (sincere is B>A)
> 03: C>A
>
> C>A 14-10, A>B 13-11, B>C 21-3.
>
> Scores: B11 A10 C3
>
> Both BA and AC are pairwise out of order. The score margin between B
> and A is 1 (11-10) and the score margin
> between A and C is 7 (10-3). So the pair with the smallest margin swap
> places and now no adjacent pair is pairwise
> out of order and only 3 candidates are on the list (A>B>C) so A (the
> sincere CW) now being on top of the list, wins.
That's interesting. Maybe I should try to make DMTBR failure examples
for maximal sets of the voting methods implemented by LeGrand's voting
calculator. In particular, Raynaud and Baldwin seem to have different
enough logic that examples that work on the Plurality-like methods don't
work on them.
Copeland and Small don't allow for any decisive (B wins with certainty)
three-candidate examples, but arguably fail outright because B has a
nonzero chance of winning as soon as the cycle is established.
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