# [EM] What are some simple methods that accomplish the following conditions?

C.Benham cbenham at adam.com.au
Mon Jun 3 23:28:28 PDT 2019

```Kevin,

As something fairly simple I think I like this version of your "Idea 1":

1. If there is a CW using all rankings, elect the CW.

2. Otherwise flatten/discard all disapproved rankings.

3. If there is a "CW" based on the remaining rankings (i.e. the rankings
among approved plus approved over not approved)
then elect that candidate.

4. Otherwise elect the most approved candidate.

That strikes me as something not too hard to explain or sell.

Chris Benham

> *Kevin Venzke* stepjak at yahoo.fr
> Sat Jun 1 12:48:27 PDT 2019
>
>
> Hi Forest,
>
> I had two ideas.
>
> Idea 1:
> 1. If there is a CW using all rankings, elect the CW.
> 2. Otherwise flatten/discard all disapproved rankings.
> 3. Use any method that would elect C in scenario 2. (Approval,
> Bucklin, MinMax(WV).)
>
> So scenario 1 has no CW. The disapproved C>A rankings are dropped. A
> wins any method.
> In scenario 2 there is no CW but nothing is dropped, so use a method
> that picks C.
> In both versions of scenario 3 there is a CW, B.
>
> If step 3 is Approval then of course step 2 is unnecessary.
>
> In place of step 1 you could find and apply the majority-strength
> solid coalitions (using all rankings)
> to disqualify A, instead of acting based on B being a CW. I'm not sure
> if there's another elegant way
> to identify the majority coalition.
>
> Idea 2:
> 1. Using all rankings, find the strength of everyone's worst WV
> defeat. (A CW scores 0.)
> 2. Say that candidate X has a "double beatpath" to Y if X has a
> standard beatpath to Y regardless
> of whether the disapproved rankings are counted. (I don't know if it
> needs to be the *same* beatpath,
> but it shouldn't come into play with these scenarios.)
> 3. Disqualify from winning any candidate who is not in the Schwartz
> set calculated using double
> beatpaths. In other words, for every candidate Y where there exists a
> candidate X such that X has a
> double beatpath to Y and Y does not have a double beatpath to X, then
> Y is disqualified.
> 4. Elect the remaining candidate with the mildest WV defeat calculated
> earlier.
>
> So in scenario 1, A always has a beatpath to the other candidates, no
> matter whether disapproved
> rankings are counted. The other candidates only have a beatpath to A
> when the C>A win exists. So
> A has a double beatpath to B and C, and they have no path back. This
> leaves A as the only candidate
> not disqualified.
>
> In scenario 2, the defeat scores from weakest to strongest are B>C,
> A>B, C>A. Every candidate has
> a beatpath to every other candidate no matter whether the
> (nonexistent) disapproved rankings are
> counted. So no candidate is disqualified. C has the best defeat score
> and wins.
>
> In scenario 3, the first version: B has no losses. C's loss to B is
> weaker than both of A's losses. B
> beats C pairwise no matter what, so B has a double beatpath to C.
> However C has no such beatpath
> to A, nor has A one to B, nor has B one to A. The resulting Schwartz
> set disqualifies only C. (C needs
> to return B's double beatpath but can't, and neither A nor B has a
> double beatpath to the other.)
> Between A and B, B's score (as CW) is 0, so he wins.
>
> Scenario 3, second version: B again has no losses, and also has double
> beatpaths to both of A and
> C, neither of whom have double beatpaths back. So A and C are
> disqualified and B wins.
>
> I must note that this is actually a Condorcet method, since a CW could
> never get disqualified and
> would always have the best worst defeat. That observation would
> simplify the explanation of
> scenario 3.
>
> I needed the defeat strength rule because I had no way to give the win
> to B over A in scenario 3
> version 1. But I guess if it's a Condorcet rule in any case, we can
> just add that as a rule, and greatly
> simplify it to the point where it's going to look very much like idea
> 1. I guess all my ideas lead me to
> the same place with this question.
>
> Oh well, I think the ideas are interesting enough to post.
>
> Kevin
>
> >Le jeudi 30 mai 2019 à 17:32:42 UTC−5, Forest Simmons <fsimmons at
> pcc.edu> a écrit :
> >
> >In the example profiles below 100 = P+Q+R, and 50>P>Q>R>0.  One
> consequence of these constraints is that in all three profiles below
> the cycle >A>B>C>A will obtain.
> >
> >I am interested in simple methods that always ...
> >
> >(1) elect candidate A given the following profile:
> >
> >P: A
> >Q: B>>C
> >R: C,
> >and
> >(2) elect candidate C given
> >P: A
> >Q: B>C>>
> >R: C,
> >and
> >(3) elect candidate B given
>
> >
> >P: A
> >Q: B>>C  (or B>C)
> >R: C>>B. (or C>B)
> >
> >I have two such methods in mind, and I'll tell you one of them below,
> but I don't want to prejudice your creative efforts with too many ideas.
> >
> >Here's the rationale for the requirements:
> >
> >Condition (1) is needed so that when the sincere preferences are
>
> >
> >P: A
> >Q: B>C
> >R: C>B,
> >the B faction (by merely disapproving C without truncation) can
> defend itself against a "chicken" attack (truncation of B) from the C
> faction.
> >
> >Condition (3) is needed so that when the C faction realizes that the
> game of Chicken is not going to work for them, the sincere CW is elected.
> >
> >Condition (2) is needed so that when  sincere preferences are
>
> >
> >P: A>C
> >Q: B>C
> >R: C>A,
> >then the C faction (by proactively truncating A) can defend the CW
> against the A faction's potential truncation attack.
> >
> >Like I said, I have a couple of fairly simple methods in mind. The
> most obvious one is Smith\\Approval where the voters have
> >control over their own approval cutoffs (as opposed to implicit
> approval) with default approval as top rank only. The other
> >method I have in mind is not quite as
> >simple, but it has the added advantage of satisfying the FBC, while
> almost always electing from Smith.
>
>
>
>
>
>
/
/

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