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@kodos I'll have a look to see if I can find a way to change that or if it's "hardwired" in.

@jack-waugh I see to an extent, but I would argue that your collapse of rankings is incompatible with the distinction by score. You prefer one to the other, even if they are both horrible, right?

@jack-waugh I’m not sure, I think coupling encourages consistency, which is a prerequisite to honesty. The coupled structure is also simpler and more efficient. The score winner is not necessarily the Condorcet winner for instance, and need not be the winner of a Condorcet compliant method when the Condorcet winner doesn’t exist (e.g. B2R).

The same validation logic also works in a homologous sense for Smith compliance. I think it’s less flexible if for instance two divergent Smith compliant methods were pitted against each other in the absence of a Condorcet winner. In that instance though, the Condorcet winner could not be tested with the final head-to-head. I’m not sure about extending to particular special subsets of the Smith set in the same way, depending on the subset.

@toby-pereira I would say they are declared the winner if the methods coincide. Otherwise there may be potential clone issues. It’s also simpler that way. Although if we did have an unambiguous runner up, that would be preferred to validate any potential Condorcet winner status. For example, the second highest scorer.

But if we wanted the decision process to be consistent, that has the potential to cause a recursion of successive head to heads, say, if the two methods repeatedly coincide. Unlikely but still. That would probably be the “right” way to do it in this context.

If you're pitting the winners of two methods against each other, what do you do if it's the same candidate? Are they just the winner, or does there need to be a final head-to-head between two candidates?

@jack-waugh exactly, that's the kind of thing I'm considering. Now I'm kind of puzzled though about whether a head-to-head between two winners of clone-independent methods can fail independence of clones. I actually don't think it's possible that the composite system fails independence of clones. FYI my preference for a Condorcet method would be Tideman's Bottom 2 Runoff (B2R), which I'm now certain is equivalent to BNR. For example, we could even pit the Approval winner against the B2R winner head-to-head. That would be a form of Approval-seeded Llull with a final independent head-to-head.

@jack-waugh I agree it isn’t too complex to understand, it could also be made more convenient by setting defaults. I’m actually more concerned with failing independence of clones. But I’m not sure—is it even true that this kind of runoff actually would fail independence of clones? Only the Condorcet and score winners could be pitted against each other, and those are both (roughly) clone independent methods.

I think a real second runoff would only matter in cases where voters were employing tactics. Otherwise, probably we should elect the Condorcet winner (IMO).

@jack-waugh I think I may have proved that the recursive bottom N runoff and ordinary bottom 2 runoff are equivalent: https://www.votingtheory.org/forum/topic/564/bottom-n-and-bottom-2-runoffs-are-equivalent

So actually it may make no difference, which would nice to know.

But I think maybe the concept of a second independent runoff is too complicated… I just got a bit hung up thinking about how the Condorcet winner as computed might not actually be the “real” Condorcet winner.

Bottom 2 Runoff (B2R) is a Smith-compliant cardinal-Condorcet method. We can describe it generally by considering two independent "score" and "head-to-head" functions over a candidate set.

To be precise, if Z+ is the set of natural numbers, let s:Z+ to R ("score") and hth:Z+xZ+ to R (“head-to-head”) be fixed real-valued functions, where s is injective, and where hth(y,x)=-hth(x,y) and hth(x,y)=0 if and only if x=y (“no ties”). In our case we consider a set C of “candidates” to be a finite subset of Z+.

Let B2R(C) denote the “Bottom 2 Runoff” winner in C, determined by repeatedly pitting the bottom two scoring candidates against each other in a head to head, and eliminating the loser until one candidate (the winner) remains.

Also, let RCLE(C) denote the “Recursive Condorcet Loser Elimination” winner in C, determined by repeatedly looking at the set of N lowest-scoring candidates, starting with the full set C, and eliminating any Condorcet loser among them (if none, update N↦N−1 and retry, resetting to the full remaining set with each elimination), until one candidate (the winner) remains.

Both B2R and RCLE are Condorcet compliant and Condorcet loser compliant methods. It also turns out that for fixed score and head-to-head functions with no ties, they are equivalent.

PROOF:

We can proceed by induction on the size of C. With 1 or 2 candidates, the equivalence is trivial. So we consider |C|>=3.

Without loss of generality, we can assume that there is no Condorcet winner in C—with a Condorcet winner, B2R automatically coincides with RCLE, since both methods are Condorcet compliant.

We can also assume that there is no Condorcet loser in C, because that Condorcet loser will immediately be eliminated in RCLE, and will have no material effect on B2R. More formally, if L is the Condorcet loser in C, then B2R(C)= B2R(C omitting L), and RCLE(C) = RCLE(C omitting L).

Now, consider the first candidate eliminated by each method, and call them B2R1 and RCLE1, respectively. If B2R1=RCLE1, then we will be done by the inductive hypothesis. Thus, we have to examine the case where B2R1 differs from RCLE1. In that case, the only option is that RCLE1 is the Condorcet loser in the set of bottom-N scorers for some N >= 3. That is, RCLE1 loses head-to-head against every other candidate in that bottom-N scoring set.

In particular, RCLE1 loses head-to-head against every candidate whose score is lower than its own score. This means that in B2R, no matter which of the candidates whose scores are lower than RCLE1's emerges to face RCLE1, that candidate is guaranteed to beat out RCLE1 head-to-head. Therefore, the presence of RCLE1 in the candidate pool is immaterial to the outcome of B2R.

Formally, it follows that B2R(C)=B2R(C omitting RCLE1). But we also know that by definition, RCLE(C)=RCLE(C omitting RCLE1) as well. By the inductive hypothesis, B2R(C omitting RCLE1) = RCLE(C omitting RCLE1), and this implies that B2R(C)=RCLE(C).

QED

This is true no matter which scoring or head-to-head functions are used, as long as they are static functions and no ties are involved.

This means that, while RCLE superficially appears to be more conservative than B2R, this is actually not the case. B2R is always just as conservative as RCLE, since they are equivalent. B2R is actually just a more efficient implementation of RCLE.

@jack-waugh hm I quite like this in theory. I’m also recollecting that multi round systems tend to fail independence of clones. In principle, approval-seeded Llull could be made more robust by considering larger sets of the lowest-approved candidates and iteratively removing Condorcet losers.