[EM] coherent, house monotonic, droop proportional

[Gustav Thorzen]

On Tue, 14 Apr 2026 08:14:43 -0500
Ross Hyman via Election-Methods <election-methods at lists.electorama.com> wrote:

> Dear all,
> Back in June I posted a preprint on arxiv describing an STV-like
> method that satisfied Droop proportionality like conventional STV,
> and, unlike conventional STV, was also coherent and house monotonic.
> I never submitted it to publication because my gut was telling me that
> there was an error. Well, it took a while, but I found the error and I
> have now revised the paper: https://arxiv.org/abs/2506.12318
> Unfortunately, the new method is cumbersome, but I believe the new
> procedure corresponds to the proof. I would greatly appreciate it if
> anyone can look it over to see if it is correct.
> Best,
> Ross
> ----

So I took a look at this (arxiv version 3 to be exact) and tried to
work my way through it.
I think I finally have a good enough grasp to check some of the
claims and proofs in it.

But first some definitions I want to make sure I understood correctly:
Coherence - Same as https://en.wikipedia.org/wiki/Coherence_(fairness)?
(also using the Oklahoma example.)
Droop proportionality - Generalization of Mutual Majority criterion for more
then 1 winner systems?
Quota-Based Phragmen - Refereing to the systems in the
(Salmi, Olli 2002) and (Woodall, Douglas R. 2003) references?

and since you never provide any sort of (explicit) name to any of the methods,
the one you propose is the one describes halfway through section 5 Top-down Phragmen?

Assuming all the above are answered yes, my results of checking your claims are as follows:
(1) Trivially true from your definition, though it feels like the definition could be simplified.
(2) I found no errors here, but I am not confident I got all possible edge cases covered.
I leave this to someone else to check.
(3) No errors in the proof, but I will go into detail about what is proven later on.
(4) Trivially false unless you use "exceeds" as a shorthand for
"greater then or equal to" rather then "strictly greater then",
which would be a first for me.
An example would an exactly equal number of rankorders ranking Alice > Bob
as thoose ranking Bob > Alice.
We get a perfect tie where the winner is determined by random tiebreaking and whoever
of the two gets lucky will be a winner with votes exactly equal to Q_1, not exceeding it.
You have not stated this is supposed to break the norm and be a fully deterministic system
without any form of randomness no matter what, so random tiebreaking is implicitly assumed.
(5) False for pretty much the same reason as (4) is false.
(6) I found no errors here, but I am not confident I got all possible edge cases covered.
I leave this to someone else to check.
(7) False for pretty much the same reason as (4) is false.
(8) False for pretty much the same reason as (4) is false.
(9) I found no errors here, but I am not confident I got all possible edge cases covered.
I leave this to someone else to check.

When it comes to claim (3) about LN-Help+Harm,
we sort of run into some problems about have ballots are counted.
You mention in passing the proofs for the claims all assumes complete rankorder,
implicitly also that they are to be strict,
as well as reducing to regular IRV for the single winner case,
which also typically assumes strict rankorders.
But voter symmetry + candidate symmetry and complete strict rankorders
forces the inference rule for incomplete ballots to cause spoilage
whenever two or more candidates are left out on the provided ballot.
This causes LN-Help+Harm failures in edge cases not covered in your proof,
for example (signle winner 4 candidates here for simplicity):
10 A>B>C>D
09 B>A>C>D
01 B>A (spoiled forcefully by voter+candidate symmetry+complete rankorder, or otherwise)
results in A winning, but if the incomplete ballot is extended to add later preferences
of C and D (B>A>C>D or B>A>D>C, does not matter), the ballot is not spoiled
leaving a tiebreaker between A and B each with 50% chance by voter+candidate symmetry.
This is a LN-Help failure for B, and a LN-Harm failure for A,
even if we use inference for only 1 missing candidate cases (which is still possible).

What I normally see is for IRV to simply allow incomplete preference orders,
which it looks like you could do to solve the above problem,
but then things get extremely sensitive to what part of the ballots are keept.
For example, how would A>{everyone else}>B be counted?
Doing it in any way other then simply turning it into A top and nothing else
leads to LN-Help/Harm failures, but it has other possibly unwanted
strategic side effects.

There are basiclly two versions of LN-Help/Harm each,
one about what happens when we add our preferences to incomplete ballots,
and one about when we reorder strictly lower preferences.
Your proof looks like it covers the later version for each,
but depending on how ballots are counted,
the first version is failed for complete preference orders
(I will not assume you are willing ditch voter of candidate symmetry
without explicitly stating such.)

Normally I deal with this by proving both cases,
or refere to whatever version others use,
or avoid any claims that depends on the difference.


Hopefully this is still worth something,
even if a bit late.
Gustav