Hi Chris,

Although all three methods satisfy Mono-add-top, the combined method doesn't:

330: D>A | C>B
302: C>A | D>B
163: B>C | A>D
126: A>B>D | C
77: B>D | C>A

A is the approval winner, but MinMax elects C (no CW present) and IRV elects D
(eliminating A right away). D has more approval than C and so D wins.

But if we add 10 more D>A|C>B ballots:

340: D>A | C>B
302: C>A | D>B
163: B>C | A>D
126: A>B>D | C
77: B>D | C>A

By Mono-add-top, this should not be able to take the win away from D, since we
added D-top ballots. But with the new ballots, now MinMax is willing to elect A
instead of C. A has more approval than IRV's D winner (unchanged) and so A wins.

Kevin
votingmethods.net

On 2025-08-10 07:44, Chris Benham via Election-Methods wrote:

This is a generally interesting idea: suppose properties X and Y are
incompatible. How close to Y can we get while retaining X?

But it's also, generally, a very hard question to answer. The closest I
can think of something that tries to do that is River's ISDA (as an
approximation to independence of covered alternatives without losing
monotonicity), and I don't think River was deliberately designed to pass it.

-km

Hello,

Kristofer Munsterhjelm via Election-Methods election-methods@lists.electorama.com a écrit :

I don't think it's so uncommon to try to do this, for example:

ICA: satisfy weak FBC and preserve as much Condorcet as possible
ACP: satisfy Later-no-harm (and -help!) and preserve as much Condorcet as possible
CDTT,___: satisfy minimal defense and preserve as much Later-no-harm as possible

In a DNA context I can search for methods that only have X failures of some type,
or match another method with only X number of deviations, but the resulting
methods can hardly ever be explained in plain English. Optimistically they could
be used to rule out certain outcomes being possible in certain scenarios, or
suggest broadly what the method would have to be like.

Kevin
votingmethods.net

On 2025-08-13 15:32, Kevin Venzke wrote:

I've done similar things with mixed integer programming to find
minimally manipulable methods with few voters and candidates. It's not
at all scalable and it's very hard to understand just what's going on,
because all it outputs is an assignment like "for this particular
election, elect A; for that particular election, elect B".

In principle similar methods could be used to find compatible criteria
(again with few voters and candidates) - a SAT solver should work. But
generalizing is indeed hard.

-km

Kevin,

Thanks for that demonstration.

A much more simple method (using the same type of ballots) definitely
does meet Mono-add-Top:

Elect whichever of the Hare winner and the most approved candidate
pairwise beats the other.

James Green-Armytage mentioned a while ago that he thought that would be
a good method. At the time I had different priorities and dismissed it
as something clunky that fails Condorcet and Mono-raise, but now I
agree. As a practical proposition it is probably doubtful that the extra
complication versus plain Hare gives enough bang for buck, and I suppose
as well as failing Condorcet it fails Double Defeat.  But nonetheless it
must be quite a bit more Condorcet efficient than Hare, while hanging on
to Mono-add-Top compliance.

Chris

On 10/08/2025 8:10 pm, Kevin Venzke wrote:

On 8/21/25 11:37, Chris Benham via Election-Methods wrote:

Here I'll put in a plug for refining IRV by eliminating pairwise losing
candidates when they occur.  It's a simple compromise between IRV and
Condorcet methods that isn't "clunky" and yields lots of "bang for the
buck."

A pairwise losing candidate is a candidate who loses every one-on-one
contest against every other remaining candidate.

Only when a counting round lacks a pairwise losing candidate does the
combined method fall back on eliminating the candidate with the fewest
transferred votes.

Richard Fobes

On 8/21/25 11:37, Chris Benham via Election-Methods wrote:

Hi Chris,

Unfortunately this doesn't seem to me to work:

358: B>C | A
305: C | A>B
187: A | B>C
148: A>C | B

A is the IRV winner, C is the approval winner, and C beats A, so C wins.

Add 35 more of the C-top ballots:

358: B>C | A
340: C | A>B
187: A | B>C
148: A>C | B

Now B is the IRV winner, C is still approval winner, but B beats C, so B wins.

Kevin
votingmethods.net

Kevin,

Interesting. I used to be in some ways quite intelligent.

What keeps getting more and more confirmed to me is that Hare (aka IRV)
is a very bad mixer and it is very difficult to compete with it on its
strengths.  The "Benham" method hangs on to Unburiable Mutual Dominant
Third but beyond that pays a significant price (at least in terms of
criterion compliances) for meeting Condorcet.

It occurred to me that could have some pretty horrible method that in
some way just combined the Approval and MMM (minimum additional votes to
become the CW) scores that would meet Mono-add-Top and probably
mono-nearly everything else.

Chris

On 22/08/2025 8:45 am, Kevin Venzke wrote:

Richard,

As I just commented in my reply to Kevin,  I think Hare makes a bad a
mixer and so is difficult to fruitfully "refine".

Can you give an example where your suggested method performs better than
plan Hare (aka IRV)?

I think your suggested method would be quite a bit more difficult to
hand-count than say Benham. With Benham when considering the candidate
that Hare would next eliminate we only have to establish that it has a
single pairwise defeat (before eliminating it), not that it loses all
its pairwise contests.

Can you give an example where your method gives a better (or just
different) result than Benham?

Chris

On 22/08/2025 6:47 am, Richard via Election-Methods wrote: