[EM] Exploring Friendly Voting

[Forest Simmons]

Continuing in the same context for a little while we introduce Methods 4 &
5, respectively as maximizing and minimizing the mined and subtrahend,
respectively, of the difference arising in method 3 ...

Method 4: Maximize the minuend, which is the sum of the Bottom totals of
the hostile candidates and the Top totals of the friendly candidates.

Method 5: Minimize the subtrahend, which is the sum of the Top totals of
the hostile (vexing) candidates and the Bottom totals of the Friendly
candidates.

Note the numerical and psychological advantages of Methods 4&5 due to all
terms being positive: ... numerical stability and linguistic congruence in
the sense tally=sum, as opposed to difference, in the public mind.

That said, let's specialize to our Example 1 context again for Method 4:
Here there is only one candidate (C) hostile to candidate A, to whom the
other two candidates A and B are friendly ... so the tally for candidate A
is ...
Top(A)+Top(B)+Bottom (C), which is just fpA+fpB+fpA in terms of faction
sizes.

For comparison purposes only, let's subtract zero in the form of n-n from
this tally, where n is the total number of ballots fpA+fpB+fox. After the
dust settles we see that A's tally is numerically equal to ...
fpA-fpC+n, the same score for A  as in methods 1 and 2, up to a constant n.

Similarly, Method 5 yields (for A) a tally of fpC-fpA+n  to be minimized.

So in the special context of Example 1, all five methods agree on the
winner.

But once we go beyond Example 1, it seems to me that Method 5 is better
strategically than Method 4 ... because Method 4, with two terms of foA in
its tally, gives too much control to the A faction ... the likely culprit
of the burial that created the ABCA beat cycle.

So I  propose tentatively Method 5:

Elect the candidate that minimizes the total tally TT given by: friendly
bottom totals plus hostile top totals.

Just one item of business to finish the five method specifications: when
rankings are not complete Top and Bottom counts must be done fractionally
to preserve clone independence.

That's it ... so now let's do an example of Method 5 with incomplete
rankings:

48 C
28 A>B
24 B

For the different TT tallies we need the following Too and Bottom
(fractional) counts:
Top(A)=28, Top(B)=24, Top(C)=48
Bot(A)=(48+24)/2=36,
Bot(B)=48/2=24,
Bot(C)=(24/2)+28=40

TT(A)=Bot(A)+Bot(B)+Top(C)=36+24+48
=108
TT(B)= Bot(B)+Bot(C)+Top(A)=24+40+28
=92
TT(C)=Bot(C)+Bot(A)+Top(B)=40+36+24=100

It appears that argmin TT(Y) is B.

Let's compare this with the total tallies TT'(Y) for method 4:
TT'(A)=Top(A)+Top(B)+Bot(C)
=28+24+40=92
TT'(B)=Top(B)+Top(C)+Bot(A)
=24+48+36=108
TT'(C)=Top(C)+Too(A)+Bot(B)
=48+28+24=100
Argmax TT'(Y)=B, the same as the method 5 winner.

The method 3 totals are the respective differences TT'-TT of the methods 4
and 5 ... 92-108=-16, 108-92=16, and 100-100=0. So B wins by method 3.

Somebody should check my work on these methods and also calculate the
Method 1 and 2 winners.

Methods 3, 4, and 5 do not appear to be Chicken Strategy resistant.

I hope that these sample calculations provide enough clarity for those who
want to explore further.

-Forest

On Sun, Feb 12, 2023, 9:42 PM Forest Simmons <forest.simmons21 at gmail.com>
wrote:

> For simplicity we'll stick with complete rankings for now ... with three
> methods that are equivalent in that context, though not in general ... like
> MinMaxPaiwiseOpposition and MaxMinPairwiseSupport are equivalent in this
> context.
>
> We say candidate X is Friendly to Y if X doesn't defeat Y pairwise ...
> otherwise X Vexes  Y.
>
> Also X has Top or Bottom Status on ballot B according to whether no
> candidate outranks it or it outranks no candidate on ballot  B.
>
> Method 1:
>
> Elect the candidate argmax S(Y) defined as Sum{Bot(X)-Top(X)| X vexes Y}
> where Bot(X) and Top(X), respectively, are the number of ballots on which X
> has Top or Bottom status.
>
> Example 1.
>
> fpA: A>B>C
> foB: B>C>A
> fpC: C>A>B
>
> C is the only candidate that vexes A, so S(A) is merely Bot(C)-Too(C),
> which equals fpA-fpC.
>
> Method 2:
>
> Elect argmax S'(Y) defined as
> Sum{Top(X)-Bot(X)| X is friendly to Y }.
>
> Method 2 applied to the same ballot profile from example 1 goes as follows:
>
> S'(A)=(Top(A)-Bot(A))+(Too(B)-Bot(B)), which equals fpA-fpB+fpB-fpC, which
> simplifies to fpA-fpC, the same exact expression given by method 1.
>
> Method 3. Elect argmax (S(X)+S'(X)).
>
> Note that S(A)+S'(A) can (by rearrangement of terms) be written as
> Bot(C)+Top(A)+Top(B) minus
> [Top(C) +Bot(A)+Bot(B)].
>
> The three positive terms cinsist in the Top counts of friendly candidates
> and the Botoom count of the unfriendly candidate.
>
> The subtrahend consists of the bracketed terms which are the Bottom counts
> of the friendly candidates and the Top count of the unfriendly candidate,
> as it should be: we want our friends to have high top counts and low bottom
> counts ... and vice-versa for those who vex us.
>
> Does this heuristic make sense as a foundation for a Friendly Voting
> Method?
>
> -Forest
>
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