[EM] STAR cloneproof variant based on Score Chain Climbing

Forest Simmons forest.simmons21 at gmail.com
Fri Mar 11 21:03:12 PST 2022


Your brainstorming, if may call it that, is infectious.

As you know Score Chain Climbing produces a maximal totally ordered subset
of candidates ordered by pairwise defeat ... every candidate in the chain
pairwise defeats all of the other chain members below it, and the chain
cannot be extended upward while retaining this total order.

In fact, the chain consists of the successive values of the candidate
variable X in the following formulation of Score Chain Climbing:


Initialize a variable X as the (name of) the lowest score candidate. Then

While more than one candidate remains, eliminate all of the candidates
pairwise defeated by X, before storing a new name into X, the name of the
lowest score remaining candidate.

The last value of X (the SCC winner Xf) is one of the finalists.

The other finalist is the second to the last value of X, which we designate

But doesn't the last X defeat all of the previous X's?

Yes, according to the ballots. But there is a good chance that the only
reason Xf defeats Xf' on the ballots is that Xf' was insincerely buried
under Xf.

So how do we vindicate (or expose as fraudulent) the finalist Xf?

We could take another trip to the polls for a runoff between between Xf and

Otherwise, we can require voters to submit two ballots ... one to determine
the two finalists, and the other to choose between them.

Sincere voters simply duplicate their first ballot to produce their second
one. The strategy burdened voters adjust their insincerities to produce
their second ballot.

It is crucial that the second ballot be used exclusively for choosing the
winner between the two finalists.

However, once the final winner has been certified , these ballots can be
used for forensics.


El mar., 15 de feb. de 2022 9:48 a. m., Ted Stern <dodecatheon at gmail.com>

> Here's a proposal for a STAR variant that handles clones:
> Top two score winners (A and B), plus the score winner when you exclude
> the top score winner ballots (by score weight), that's the clone-proof
> part. Call that the exclusive winner, X. In other words, if a ballot gives
> a score of 5 out of 10 to the top score winner, remove half that ballot's
> weight.
> Eliminate any candidates defeated by X.  If more than one remains, the
> winner is the one who defeats the other.
> This follows the logic of Forest Simmons' Score Chain Climbing to resolve
> cycles.
> The clone problem is that A and B could be clones. Removing A's ballot
> contributors finds the non-clone while avoiding pushover incentive. If X
> defeats both A and B, it's likely the CW. Otherwise, whichever of A or B is
> defeated by X is "weaker" (low probability, but possible in cycles). Using
> a lower-scoring candidate as an eliminator reduces burial incentive.
> Why do I propose finding X that way instead of by a Hare or Droop quota?
> Well, for one thing, it's a summable process. Next, if A has >50% approval,
> A and B are probably the two candidates to choose from anyway. If A has
> <50% approval, X is being found with something like a Hare quota, moving
> more toward Droop as A's approval decreases.
> Your thoughts?
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