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

Forest Simmons fsimmons at pcc.edu
Sat Jun 1 14:41:49 PDT 2019

```Kevin, Chris, and Kristofer:

Great ideas!.

DMC (explicit approval version) also comes to mind as fairly simple to
describe:

*Remove candidates from the bottom of the approval list until there is a
ballot pairwise beats-all candidate among the remaining candidates (to
elect). *

And analogous to Chris's equivalence between Approval Sorted Margins and
Condorcet(approval margins), we have the equivalence between DMC and
Condorcet(winning approval).

Of course we are talking explicit approval every time we say approval in
this context.

We know that none of these Condorcet efficient methods satisfies the FBC,
but how about MDDA with explicit approval instead of its implicit approval
default?

This has the problem that Chris has pointed out in MJ and other Bucklin
versions where addition of irrelevant ballots can change a majority of
ballots into a minority.

So here's my idea: do MDDA with explicit approval AND symmetric completion
of all of the ballots except for the Top Rank.  By excusing the top rank
from symmetric completion, we preserve the FBC, but we lose the Condorcet
Criterion.

[This shows how close we can get to both the CC and the FBC, if we do a
full symmetric completion including the top rank, then this version of MDDA
satisfies the CC but not the FBC. We can toggle back and forth.
Suppose that we balance on the edge (of symmetric completion in the top
rank or not) with asset voting so the proxies could finesse this
difference. The unsophisticated voters could just vote their favorites,
etc. Would this come even closer to satisfying both the CC and the FBC? In
other words mightn't this expedient externalize the strategizing enough to
satisfy both FBC and CC for all practical purposes?]

I think that the FBC is more important in this context, so suggest that we
exempt at least the top rank from symmetric completion if not all of the
approved ranks.

Now what about the irrelevant ballots problem?  I think that symmetric
completion, if only for the truncated candidates, would suffice.  If X and
Y are both truncated on N new ballots then N/2 of them count for X>Y and
N/2 of them count for Y>X so a majority defeat between X and Y is preserved.

Note that symmetric completion among only the truncated candidates is the
same as "half power truncation."

This brings me to another question.  Suppose that voters knew that
candidate B became a plurality loser because of insincere truncation.
Would they feel so bad it B won anyway?

This leads to a weak form of Plurality:  If a candidate fails Plurality,
even after the ballots have been symmetrically completed, then that
candidate must not be elected.

Would this weak form be adequate?

Thanks for your great ideas and interest in these questions!

Forest

On Sat, Jun 1, 2019 at 12:49 PM Kevin Venzke <stepjak at yahoo.fr> wrote:

> 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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