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

Thu Jun 6 19:03:09 PDT 2019

```Kevin,

> Specifically should "positional dominance" have the same meaning
> whether or not the method has approval in it?

If the voters all choose to approve all the candidates they rank, then
yes.  (For a while I was wrongly assuming that Forest's suggested
default approval was for all ranked-above-bottom candidates, but then I
noticed that he specified that it was only for top voted candidates).

One of my tired examples:

25: A>B
26: B>C
23: C>A
26: C

Assuming all the ranked candidates are approved, C is by far the most
approved and the most top-voted candidate.
Normal Winning Votes (and your idea 2 in this example) elect B.

> I will go easy on these methods over failing MD, because it happens
> when some of the majority don't approve their common candidate.

For me this this type of ballot avoids the Minimal Defense versus
Chicken Dilemma dilemma, rendering those criteria inapplicable.

48: A
27: B>C
25: C

The problem has been that we don't know whether the B>C voters are
thinking "I am ranking C because above all I don't want that evil A
to win" or  "My C>A preference isn't all that strong, and I think that
my favourite could well be the sincere CW, and if  C's supporters rank
B above A then B has a good chance to win. But if they if they create a
cycle by truncating I'm not having them steal it".

With the voters able to express explicit approval we no longer have to
guess which it is.

> I don't think it's ideal if burying X under Y (both disapproved) can
> only backfire when Y is made the CW.
>
Why is that?  The post-election complaint (by any of the voters) would
be .. what?

If you don't allow voters to rank among their unapproved candidates then
arguably you are not even trying to elect the sincere CW.
Instead you are just modifying Approval to make it a lot more
Condorcet-ish.

A lot of voters like relatively expressive ballots. I think that is one
of the reasons why Approval seems to be a lot less popular than IRV.

Chris Benham

On 6/06/2019 5:34 pm, Kevin Venzke wrote:
> Hi Chris,
>
> I've been short on time so I don't actually have much thought on any
> of the methods, even my own.
>
> I suppose Idea 2 is the same as Schwartz-limited MinMax(WV) if nobody
> submits disapproved rankings. I'm not sure if it makes sense to reject
> the method over that. Specifically should "positional dominance" have
> the same meaning whether or not the method has approval in it? As a
> comparison, I will go easy on these methods over failing MD, because
> it happens when some of the majority don't approve their common candidate.
>
> I would have liked to simplify Idea 2, but actually Forest's eventual
> proposal wasn't all that simple either. As I wrote, if you add "elect
> a CW if there is one" it can become much simpler, so that it isn't
> really distinct from Idea 1. I actually tried pretty hard to present
> three "Ideas" in that post, but kept having that problem.
>
> I posted those ideas because I thought Forest posed an interesting
> challenge, and I thought I perceived that he was trying to fix a
> problem with CD. That said, I am not a fan of
> Smith//Approval(explicit). If all these methods are basically the same
> then I probably won't end up liking any of them. I don't think it's
> ideal if burying X under Y (both disapproved) can only backfire when Y
>
> Kevin
>
>
> Le mercredi 5 juin 2019 à 21:26:23 UTC−5, C.Benham
> <cbenham at adam.com.au> a écrit :
>
> Kevin,
>
> I didn't comment earlier on your "idea 2".
>
> If there no "disapproved rankings" (i.e. if the voters all approve the
> candidates they rank above bottom),
> then your suggested method is simply normal  Winning Votes, which I
> don't like because the winner can
> be uncovered and positionally dominant or pairwise-beaten and
> positionally dominated by a single other
> candidate.
>
> On top of that I don't think it really fills the bill as "simple".
> Approval Margins (using Sort or Smith//MinMax
> or equivalent or almost equivalent algorithm) would be no more complex
> and in my opinion would be better.
>
> I would also prefer the still more simple Smith//Approval.
>
> What did you think of my suggestion for a way to implement your idea 1?
>
>
> Chris
>
>
>
>> *Kevin Venzke* stepjak at yahoo.fr
>> Sat Jun 1 12:48:27 PDT 2019
>>
>>
>> Hi Forest,
>>
>>
>> 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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