[EM] Simplest Condorcet method to hand count?

[Kristofer Munsterhjelm]

On 2025-05-24 07:22, Etjon Basha via Election-Methods wrote:
> Thank you all, gentlemen
> 
> I greatly enjoyed the discussion and FWIW come out of it with a greater 
> appreciation of Condorcet.
> 
> A summary for my own benefit: it seems like a hand-count makes one chose 
> between two counts for races with n candidates.
> In all, there seems to be no faster and more practical way than a 
> pairwise count to compute a Condorcet winner manually.

The only way I see to break this barrier is to use randomness. You could 
randomly sample ballots if the complexity is prohibitive. E.g. you could 
count a random pairwise contest from each ballot. But then you need a 
trusted way to generate randomness.

Theoretically, you could even do this on the front-end: ask each voter 
his preference for a randomly chosen pairwise contest X>Y. This is a 
very simple ballot and would leave the voter to only have to consider 
two candidates.[1] However, it would be unfamiliar and probably wouldn't 
be received well: even in the best case, you've replaced voters having 
to trust a machine count with voters having to trust your source of 
randomness, and if the race has a few very clear frontrunners, the 
majority of the voters who don't get asked about any frontrunner contest 
might feel cheated out of their opinion.

But deterministically for Condorcet, I think v voters times n candidates 
is the best you can do (at least without seriously thinking outside the 
box). Because suppose otherwise, that you only check (n-1) candidates. 
Then it's possible that the candidate that you didn't check is the 
actual CW. There may be heuristics - e.g. if someone has a majority of 
the first preferences, that candidate is the CW - but they won't always 
hold.


On a related note, I have been thinking about how to formally prove that 
a given method has a minimum degree of summability, or that for, say, 
thre candidates, it must know the value of (e.g.) five variables. Such 
an approach could show that, yes, you need n pairwise counts for every 
known method. (It could also disprove it, but that would be more 
surprising.)

For concrete methods that are always decisive, this becomes a linear 
algebra problem.[2] I haven't found a way to solve this problem, 
however. There's an obvious upper bound using separating hyperplanes, 
but to my knowledge it's just that - an upper bound.

The problem for methods that pass some criteria is even harder. For 
instance, I suspect that determining the innermost mutual majority set - 
the set of candidates that methods passing mutual majority must elect 
from - is not summable. But there are lots of methods that do pass 
mutual majority and *are* summable. I would be surprised if my intuitive 
hunch for Condorcet were to be disproven, though.

-km

[1] A fun theoretical system for people who consider rational ignorance 
a problem would be to assign every voter a defined pairwise contest a 
sufficiently long time before the election. Then the voter would only 
have to evaluate these two candidates, and could do so in more detail. 
But in practice, I don't think it would be accepted, and would require 
significant technology in its own right.

[2] Each election can be characterized by a "vote vector" which lists, 
for each preference orde, how many voters voted that way. Then each 
candidate has a win region which is a union of convex polytopes in 
n!-dimensional space, and every point in A's win region has coordinates 
equal to a voting vector where the method declares A a winner. Then you 
have to find a minimal dimension projection so that no point in 
different candidates' win regions overlap after projecting all the win 
regions down to that dimension.