# [EM] Single-candidate DMTBR idea

Kristofer Munsterhjelm km_elmet at t-online.de
Tue Mar 15 05:19:42 PDT 2022

```On 13.03.2022 07:02, Forest Simmons wrote:

> Warren and I gave an example an of a set of four factions in a single
> plane that induces a cycle geometrically. This cannot happen with only
> three factions ... if there is a cycle, then the preferences are
> inconsistent with the lengths of the sides of the triangle.
>
> A three dimensional issue space can give rise to a cycle but those
> preferences are not based on metrics/distances between factions.

I think you're using a more constrained model than me. Here's a spatial
model example with three candidates and three faction centers that
should work (unless I've miscalculated):

Let A be at (0.85, 0.8), B at (0.2, 0.8), and C at (0.5, 0.4).

Then let voter faction 1 be at (1.1, 1.5). The distance to A is ~0.75.
The distance to B is ~1.14. The distance to C is ~1.25. So A>B>C.

Let voter faction 2 be at (-0.65, 0.9). The distance to B is ~0.86. The
distance to C is ~1.25. The distance to A is ~1.50. So B>C>A.

Let voter faction 3 be at (0.8, -0.1). The distance to C is ~0.58. The
distance to A is ~0.90. The distance to B is ~1.08. So C>A>B.

> Myth #2 busted: It also is not the case that, in two dimensions with
> voters who prefer candidates closer to them in L2 (or L1) distance, a
> Condorcet winner necessarily exists at all.

The page then proceeds to give an example with three voters (faction
centers in your example) for the Manhattan distance.

But my point is that even if it were impossible to create such examples
in the plane, then we couldn't as a consequence assume thet the voters
aren't voting in some more generalized space where it *is* possible.
That would be like saying that just because a cycle can't exist with
single-peaked preferences on a line, then all Condorcet cycles must be
the result of strategy.

In addition, there may be differential constraints that imply that if

33: A>B>C
33: B>C>A
33: C>A>B

is a perfect tie, then increasing the support for A and decreasing the
support for C should not make C win. It may be (though I haven't proven
so) that making the method elect C will induce some kind of
monotonicity failure as a consequence.

(From a spatial model, there's a similar argument: suppose we have 1/3
of the voters at each point, then we move some voters from the point
closest to C to the point closest to A. Changing the initial tie into a
C win as a result seems somewhat strange.)

So a choice to elect C would have to be about strategy resistance rather
than honest performance.

>     If the cycle is false, though, then any faction could have done the
>     burial. You say the A faction is the only group that has anything to
>     win
>     by conducting the burial, so they must have done it. However, there's a
>     bit of battle-of-wits logic here. Suppose that the method did elect
>     C by
>     this logic. Then it's possible that the C voters, knowing this,
>     engineered the cycle (honest is C>B>A) in order to push their winner
>     from B (their second choice) to C (their first).
>
>
> The burial of C by A isn't the only possibility ... just the most likely
> to succeed ... so the one most in need of checking.
>
> If the check confirms B>C as sincere, then B wins, so the burial of B
> did not pay off for C.
>
>     Okay, so C can't win
>     because of second-order reasoning.
>
>
> C can and will win if it was buried by A, because it is a finalist in
> the B vs C honesty check.

I was making a point about unrestricted domain. If the method passes
unrestricted domain, then there's no such check, and you can't determine
who the burying coalition is. Because it's a simple Condorcet cycle, no
matter who the winner ends up being, there could have been a burying
coalition who benefitted.

What threw me, I think, is that you were arguing that C should be
elected the winner of the single-winner ranked voting method, based on
the given ballots.

But on rereading, it seems that you're proposing that there's an
(essentially automated) top-two runoff after the ordinary ranked method
is finished, between the winner and the runner-up; and that the ranked
method should elect C so that C is one of the two finalists in the
second round, thus deterring first-round burial.

In that case, the first-round method is not trying to figure out who the
single winner of the election should be, but rather which set of two
candidates will be the finalists for the second round - for either a
manual runoff, or an automated one.

And I would suspect that a single-winner (ranked, UD) method has a
different objective than one that's intended to pick finalists for a
runoff. A runoff finalist method can prioritize covering all the bases
while a single-winner method is constrained to that the winner must be a
reasonable compromise, and that the runner-up has these properties in
case the winner is ineligible.

A note about the automatic runoff mechanism: it's interesting in that it
has different properties than Approval/Range, as far as breaking UD
goes. It shows that not all UD-failing methods have to be ambiguous (the
way ratings are) or lead to inherent instrumental voting even by honest
voters. And that, in turn, means that not all extensions beyond UD are
susceptible to the same problems as rated voting.[1]

I would think manual runoff is better, though, for two reasons. First,
it's a very subtle point that makes the honesty for the runoff ballot
incentive compatible, and ordinary voters might not get why it's safe to
be honest.

Second, having separate rounds makes it possible to have finalist
debates to examine their differences in more detail. The voters can then
change their minds between rounds based on what they see - e.g. a weak
CW could be exposed as a weak CW between the rounds if the runner-up's
platform withstands greater scrutiny. This might be part of the reason
why plurality + top two doesn't degrade into two-party rule the way
ordinary plurality, contingent vote, and IRV does. (Then again, the
necessary honest voting in the final round might be what makes TTR
better than IRV.)

Since the three-candidate method with a runoff is comparable to a "loser
election" (where the loser is disqualified), it should be possible to
rewrite my minimal manipulability finder to construct a minimally
manipulable method in this context. It would take some work, however.

-km

[1] Maybe the relevant distinction is whether the proposed
extension/breaking of UD would allow the method to pass IIA. If it does,
*then* the blurring of context and method may follow, as it does in
Range (e.g. IIA no longer implying that the winner doesn't change if
candidates enter or exit). But this is just an idea.
```