[EM] Score from Rankings

[Kevin Venzke]

Hi Forest, you could start with a 1v1 and toss in noise candidates who can beat only one of them:
26 D>A
23 A
27 C>B
24 B

Kevin

(end)


Le mercredi 4 mai 2022, 11:33:38 UTC−5, Forest Simmons <forest.simmons21 at gmail.com> a écrit : 
I just realized that whenever the Smith set has three members, the member ranked between the other two is (by this new method) a cutoff between the two extremes. So in this case, the middle category "Mid" does not get used at all.

In particular, all seven of Kevin's test cases are reduced to ordinary Approval by this DSV method.

Are there any three slot examples with four member Smith sets for us to test?

-Forest

El mar., 3 de may. de 2022 2:47 p. m., Forest Simmons <forest.simmons21 at gmail.com> escribió:
> How to automatically convert ranked ballot sets into score ballot sets.
> 
> This automatic conversion has been attempted in the past under the heading of Designated Strategy Voting (DSV) ... with mixed results ... not anything to write home about.
> 
> A very crude attempt was Borda's rank scoring system, which is sometimes used in scoring team sports tournaments, for example.
> 
> A new approach is based on two basic facts that are becoming more and more obvious: 
> 
> 1. It is easy to convert sets of ranked choice ballots to three-slot ballots in a monotonic, clone-free, decisive way.
> 
> 2. There are many good monotonic, clone free, decisive ways of scoring three-slot ballots.
> 
> In this message we will concentrate on the first fact, since it is the most recent major advancement in this topic. [Everybody and their dog has their favorite solution to the second of these two steps.
> 
> I repeat the first step from the forwarded message that generically denominates the three respective slots Top, Middle, and Bottom, for want of better nomenclature: 
> 
> First generate a table of pairwise defeats to consult  while making a second (final) pass through the input ballots.
> 
> Then for each candidate k and each ballot B, decided whether B increments the Top, Bottom, or Mid level count of candidate k as follows:
> 
> If on ballot B candidate k outranks some candidate j that defeats every candidate that outranks k, then increment the Top level count of candidate k.
> 
> Else-If some candidate j ranked above k defeats every candidate that is outranked by k, then increment the Bottom level count of candidate k.
> 
> Else increment the Mid level count of candidate k.
> 
> End-If 
> 
> How has this simple procedure gone so long without discovery?
> 
> The idea came to me when Rob Lanphier introduced me to "fear anchor" terminology. The candidate that defeats everybody you like better than k is an example of a fear anchor.
> 
> On the other hand, the candidate that outranks k and defeats everybody outranked by k, could be called a "hope anchor."
> 
> If k  outranks some fear anchor, then k belongs in the Top slot.
> 
> If k is out ranked by some hope anchor, then relegate k to the bottom slot.
> 
> Is it possible for both?: if a hope anchor h outranks k, and a fear anchor f is outranked by k, then h defeats every candidate (such as f) that k outranks, and f defeats every candidate (such as h) that outranks k.
> 
> So h defeats f, and f defeats h. 
> 
> Which means that the answer is no ... it is not possible for a ballot B to increment both Top and Bottom level counts for the same candidate k.
> 
> Note that when there is a ballot CW,  it is a fear anchor for every candidate ranked above it, and a hope anchor for every candidate that is outranks ... which means that candidates outrsnking the CW always get Top status, while candidates outranked by the CW get Bottom status.
> 
> It easy to show that Top and Bottom candidates on the (pre-converted) ranked ballots preserve their status as such under the conversion to three-slot ballots.
> 
> It is also easy to show that candidates awarded Top or Bottom status respectively by ballot B, form, respectively, Top and Bottom anchored solid coalitions on ballot B.
> 
> What's not to like?
> 
> -Forest