[EM] Three Category Symmetric MJ

Forest Simmons forest.simmons21 at gmail.com
Fri Apr 29 20:43:55 PDT 2022

```El jue., 28 de abr. de 2022 11:34 p. m., robert bristow-johnson <
rbj at audioimagination.com> escribió:

>
>
> > On 04/28/2022 11:10 PM Forest Simmons <forest.simmons21 at gmail.com>
> wrote:
> >
> >
> > Cool how everything is related to Fourier Transforms!
> >
> >
>
> that was just one app.  I mainly used it for autocorrelation (never
> transforming into the frequency domain) and once or twice used it with an
> FFT to find a peak's location.
>
> But it's about peaks, doesn't matter what the graph is about.
>
> If y[m] > y[m-1] and y[m] > y[m+1]   (m integer)
>
> Then you have a discrete peak at m but an implied peak a little to the
> left or right of m.  I think that the more "precise" location of the peak is
>
>   n = m + (1/2)(y[m+1]-y[m-1])/(2y[m]-y[m+1]-y[m-1])
>

This function differs from the one used in our "median interpolation"
expression only by the factor of 2 in the first term of the denominator. If
you insert that factor in the contour plot expression, it changes the
rotation pivot from the point (.5, .5) to the point (1, 1) which is the
most natural choice for a different interpolation that doesn't insist that
(x, y)=(.49, 51) is a more favorable judgment than (0, .49), just because
the first point represents a candidate with Top on a (bare) 51% majority of
the ballots, even though it also is judged Bottom by a near majority of
49%. Under median rules the other point representing only 49% first place
votes is considered an inferior position, even though it was not voted last
place on any ballot. That gives a lot of weight to first preferences.

Three level Range voting moves the pivot point to (infinity, infinity), so
all of the iso-clines are parallel with the diagonal y=x, .... the mean
(rather than median) judgment interpolation that is attributed to David
Gale by Andy in his (Andy's) dissertation. That interpolation is not
continuous at the boundary of the middle region with the outer two regions,
but serves perfectly well as a second order tie breaker ... when the first
order tie breaker (the correct median interpolator) needs further
resolution.

So we have three prominent possibilities for the pivot ...
...(infinty,infinity), (1, 1), and (1, 1)/2.

In the median judgment context we have no choice ... we must use an
interpolation based on the latter pivot if we want continuity. In other
contexts the (1, 1) pivot is an excellent compromise  between median and
mean judgment.

That peak position estimation is a kind of quadratic interpolation, while
our median interpolation formula was based on uniform rotation of a line
through a pivot point. Deeper insight into the connection between these two
kinds of interpolation would be satisfying.

I remember dealing with something similar back in my digital filter days:
it turns out that if x(n)=exp(jwn), then the ratio

r=(x(n+1)-x(n-2))/(x(n)-x(n-1))

simply reduces r=2cos(w)+1, as I remember correctly. It gives you a Quick
and Dirty way of estimating the dominant frequency of a sinusoidal signal
added to a constant (dc) signal. [Any DC voltage drops iut as does any
amplitude or phase delay.]

In a noisy environment do a least squares fit of r to the over determined
set of equations ...

{x(n+1)-x(n-2)=r(x(n)-x(n-1))| n=3, 4, 5...k}

Then w is given approximately by

w=ArcCos((r-1)/2).

-Forest

> but it assumes a quadratic fit.  (But that's usually good enough.)
>
>
> > El jue., 28 de abr. de 2022 2:02 p. m., robert bristow-johnson <
> rbj at audioimagination.com> escribió:
> > >
> > >  that looks a lot like the quadratic interpolation of a peak's
> locations when you have three discrete points located at
> > >
> > >  (-1, x), (0, 1/2), and (+1, y)
> > >
> > >  where both x <= 1/2 and y <= 1/2 (there's a zero-over-zero problem if
> there is this equality x = y = 1/2.
> > >
> > >  I use this formula often with auto-correlation (used for musical
> pitch detection) and sometimes with the magnitude of spectrum that comes
> from the FFT.
> > >
> > >  interesting to see something familiar.
> > >
> > >  r b-j
> > >
> > >  > On 04/28/2022 1:55 PM Forest Simmons <forest.simmons21 at gmail.com>
> wrote:
> > >  >
> > >  >
> > >  > contourplot (.5(y-x)/(1-x-y) for x=0... .5, y=0... .5)
> > >  >
> > >  >
> > >  > Paste the above command into Wolfram Alpha to get a contour plot of
> the Middle region.
> > >  >
> > >  >
> > >  > El mié., 27 de abr. de 2022 6:10 p. m., Forest Simmons <
> forest.simmons21 at gmail.com> escribió:
> > >  > > A brief summary: The three judgment categories are Good, Middle,
> and Bad, meaning Satisfactory to Excellent, Mediocre, and Unsuitable (for
> whatever reasons, including inability to elicit an opinion from the voter,
> > >  > >
> > >  > > Candidates marked Good on more than half of the ballots are
> judged to be in the Good category. Candidates marked Bad on more than half
> of the ballots are judged to be in the Bad category. All other candidates,
> including those marked Mediocre on more than half of the ballots, are
> judged to be in the Middle category.
> > >  > >
> > >  > > It is desirable for election purposes to establish a finish order
> that respects these judgment categories ... with all candidates judged
> Good, ahead of the the other candidates, and all candidates judged Bad
> behind the other candidates.
> > >  > >
> > >  > > Within the Good category, candidate k finishes ahead of candidate
> j if the number of ballots g(k) on which k is marked Good exceeds the
> corresponding number g(j) for candidate j. If this rule needs further
> resolution because j and k are marked good on the same number of ballots,
> then k is ahead of j in the finish orde if k is marked Bad on fewer ballots
> than j is so marked.
> > >  > >
> > >  > > Similarly, within the Bad category, k finishes ahead of j if k is
> marked Bad on more ballots than k is, and when j and k are marked Bad on
> the same number of ballots k is ahead of j if k is marked Good on more
> ballots than j is.
> > >  > >
> > >  > > So far this is all clear and logical. But how to establish a
> logical, consistent finish order within the Middle category is not so
> obvious ... until we have a good graphical representation of the three
> categories and how they fit together.
> > >  > >
> > >  > > How they fit together is important because (among other reasons)
> a small change in the number of ballots can easily bump a borderline
> candidate from one category into another. If k goes from (barely) Good to
> Middle, it should enter the Middle category at the upper end of the finish
> order within the Middle category. Otherwise, the method has no hope of
> passing any reasonable form of the Participation criterion.
> > >  > >
> > >  > > The graphical representation will make this continuity
> requirement clear. One definition of "topology" is the science of
> continuity. Without respect for the relevant topology, it isn't possible to
> have the continuity needed for the Participation compliance alluded to
> above.
> > >  > >
> > >  > > So here's the picture: Let N be the total number of ballots
> submitted. For each candidate k, let g(k), b(k), and m(k) be the number of
> ballots on which k is marked Good, Bad, or Mediocre. For graphical purposes
> let the three dimensional Cartesian coordinates x, y, and z be defined as
> x=b(k)/N, y=g(k)/N, and z=m(k)/N, so that (no matter the candidate k ...
> hence suppression of k in the notation) we have the constraint equation
> x+y+z=1.
> > >  > >
> > >  > > The graph of this equation is the convex hull of the three points
> (1,0,0), (0,1,0), and (0, 0, 1), which are the vertices of an equilateral
> triangle.
> > >  > >
> > >  > > It is convenient to project this graph vertically onto the x,y
> plane, the planar region R given by the planar inequality x+y<=1. To
> understand this picture rewrite the constraint equation as x+y=1-z, which
> reveals z in the role of "slack variable" for the planar inequality.
> > >  > >
> > >  > > The intersection of R with the inequality y>50%, is a right
> triangle representing the set of candidates in the Good category:
> > >  > > {k | g(k)>50% of N}. The set of candidates tied for y=constant,
> is a horizontal line segment in the Good triangle. In other words, the
> "level curves" or "contour lines" of the equations y=c, for .5<c<1-x fill
> up the Good candidate region with horizontal level curves.
> > >  > >
> > >  > > Similarly the vertical segments given by x=c for .5<x<1-y fill up
> > >  > >
> > >  > > To fill up the Middle region we need a family of line segments
> that smoothly transition from the vertical segment V forming the leftmost
> boundary of the Bad region to the horizontal segment H at the lower
> boundary of the Good region.
> > >  > >
> > >  > > The only way to accomplish this is to rotate the segment V
> clockwise 90 degrees about the point (.5, .5).
> > >  > >
> > >  > > At time t let V(t) make a counter clockwise angle with the
> diagonal y=x of t degrees, t goes from -45 to +45 degrees. Let (x, y) be a
> point on the segment V(t). Then the angle t between the diagonal y=x and
> the segment connecting V(t) is 45 degrees minus the angle theta of V(t)
> with the negative X axis. We have tan(theta) = slope of V(t), which equals
> (.5 -y)/(.5 -x) since both (x,y) and (.5, .5) are on V(t).
> > >  > >
> > >  > > So t = 45 deg - arctan((.5 -y)/(.5 -x))
> > >  > >
> > >  > > Taking the tangent of both sides of this equation, and making use
> of the formula for the tangent of a difference, we get
> > >  > >
> > >  > > tan(t) as (tan 45deg - tan theta) divided by (1 +tan 45deg * tan
> theta).
> > >  > >
> > >  > > Since tan 45 deg equals 1, and tan theta equals the slope (.5
> -y)/(.5-x), we get
> > >  > >
> > >  > > tan(t)= (1 - slope)/(1+slope)
> > >  > >
> > >  > > Substituting slope =(.5 -y)/(.5-x), and simplifying
> algebraically, we get...
> > >  > >
> > >  > > tan(t)=(y-x)/[2( 1-x-y)].
> > >  > >
> > >  > > Recognizing (1 - x -y) as the slack variable z, we see that
> tan(t) is just
> > >  > > (y-x)/(2z). In terms of g, b, and m,
> > >  > > tan(t)= .5 (g(k)-b(k)/m(k) ....
> > >  > > which shows where the mysterious formula for the finish order
> within the Middle category came from in my previous message.
> > >  > >
> > >  > > In particular, tan 45deg = 1, which is the same as (.5
> -x)/(2(1-x-.5)) given by the formula, and tan(-45 deg) is -1, which is the
> same as(y-x)/(2z), for any point on V=V(0), where x=.5, and z=1-.5 -y, for
> any y between 0 and. 5.
> > >  > >
> > >  > > It all checks out and fits together at the boundaries of the
> three regions.
> > >  > >
> > >  > > That's it for now!
> > >  > >
> > >  > > -Forest
> > >  > >
> > >  > >
> > >  > >
> > >  > >
> > >  > >
> > >  > >
> > >  > >
> > >  > >
> > >  > >
> > >  > > El mar., 26 de abr. de 2022 7:20 p. m., Forest Simmons <
> forest.simmons21 at gmail.com> escribió:
> > >  > > > Good Work!
> > >  > > >
> > >  > > > I have come to the same conclusion about MJ being much closer
> to IIA than Range.
> > >  > > >
> > >  > > > So I've been trying to improve MJ to make it more symmetrical,
> satisfy some kind of participation, improve tie breaking all around, etc.
> > >  > > >
> > >  > > > I have a good 3-slot version that is as decisive as possible
> for a reverse symmetry method.
> > >  > > >
> > >  > > > I am reminded of our flurry of 3-slot methods from twenty years
> ago. Back then the EM list was very sure that the best proposals were
> Condorcet and Approval, but not at all sure which would be most viable. We
> thought a majoritarian 3-slot method would be plenty simple, easy to count,
> and have room to distinguish Roosevelt, Stalin, and Hitler. Three slot
> Bucklin was a popular suggestion with various tie breaking rules, but no
> version was symmetrical, or any better than (the still unheard of) MJ with
> three judgment categories.
> > >  > > >
> > >  > > > As I have recently come to understand, our lack of design
> success (i.e. inability to get reverse symmetry into three slot Bucklin)
> was due to ignorance of the underlying topology.
> > >  > > >
> > >  > > > To make a long story short, I will finish this message by
> cutting straight to the chase, saving the full explanations for tomorrow:
> > >  > > >
> > >  > > > Suppose the three categories are Good, Middle, and Bad. Good
> means a mixture of desirable qualities from competent to excellent. Bad
> means incompetent or otherwise unsuitable. Middle includes some good
> qualities but not unalloyed with baser metals ... which is different from
> "no opinion" the same way that "average" is different from "no basis for a
> grade." To avoid "dark horse" problems, we will count blank as Bad. (I'm
> sure some philosopher or lawyer could explain why a candidate able to
> generate neither appreciable support nor opposition, would be unsuitable.)
> > >  > > >
> > >  > > > For each candidate k, let g(k), m(k), & b(k) be the number of
> ballots on which candidate k is categorized as Good, Middle, or Bad,
> respectively.
> > >  > > >
> > >  > > > If some candidate is judged to be Good on more than half of the
> ballots, then among the candidates tied for the greatest g(k) value, elect
> the one categorized as Bad on the fewest ballots.
> > >  > > >
> > >  > > > In other words (and symbols) ....
> > >  > > >
> > >  > > > elect argmin{b(j)| j is in argmax[g(k)]}.
> > >  > > >
> > >  > > > If (in the other extreme) every candidate is judged to be Bad
> on more than half of the ballots, then from among the candidates tied for
> the least b(k) value, elect the one categorized as Good on the most
> ballots. In symbols ...
> > >  > > >
> > >  > > > elect argmax{g(j)| j is in argmin[b(k)]}.
> > >  > > >
> > >  > > > If neither of the two above cases obtains, then elect from
> among the candidates in the set argmax[(g(k)-b(k))/m(k)] the candidate j
> categorized as Bad on the fewest ballots or the one categorized as Good on
> the most ballots, depending on whether or not g(j) is greater than b(j). In
> symbols ... elect
> > >  > > >
> > >  > > > argmax{T(j)| candidate j is among the tied candidates, i.e. j
> is a member of argmax[(g(k)-b(k))/m(k)]},
> > >  > > >
> > >  > > > where the tie breaker function T to be maximized is given by ...
> > >  > > >
> > >  > > > T(j) = min(b(j),g(j))sign(b(j)-g(j))
> > >  > > >
> > >  > > > [Here we have made use of the fact that b(j) is minimized when
> its opposite -b(j) is maximized.]
> > >  > > >
> > >  > > > In my next message I will unfold all of these mysteries into
> plain view!
> > >  > > >
> > >  > > > At least now you have the complete 3-slot reverse symmetry
> compliant MJ recipe safely in the EM cloud.
> > >  > > >
> > >  > > > -Forest
> > >  > > >
> > >  > > >
> > >  > > >
> > >  > > >
> > >  > > >
> > >  > > >
> > >  > > > El mar., 26 de abr. de 2022 2:13 a. m., Kristofer Munsterhjelm <
> km_elmet at t-online.de> escribió:
> > >  > > > > On 26.04.2022 00:51, Forest Simmons wrote:
> > >  > > > > >
> > >  > > > > >
> > >  > > > > > El lun., 25 de abr. de 2022 4:25 a. m., Kristofer
> Munsterhjelm
> > >  > > > > > <km_elmet at t-online.de <mailto:km_elmet at t-online.de>>
> escribió:
> > >  > > > > >>
> > >  > > > > >> So the M-W strategy is: let
> > >  > > > > >> v_i be the strategic rating we want to find
> > >  > > > > >> u_i be the public utility of candidate i
> > >  > > > > >> p_ij be the voter's perceived probability that i and j will
> > >  > > > > >> be tied.
> > >  > > > > >>
> > >  > > > > >
> > >  > > > > > I could be wrong but I think it should be "tied for
> winning."
> > >  > > > >
> > >  > > > > You're right. I was looking at the paper just now, and it
> says:
> > >  > > > >
> > >  > > > > "For each pair of candidates i and j, the /pivot probability/
> p is the
> > >  > > > > probability (perceived by a voter) of the event that
> candidates i and j
> > >  > > > > will be tied for first place in the election."
> > >  > > > >
> > >  > > > > I imagine you could refine it a little by letting the p.p. be
> > >  > > > > parameterized by the vote to submit. E.g. if it's Range
> voting and i's
> > >  > > > > score minus j's score is 1, then you could flip the win from
> i to j by
> > >  > > > > voting j 10 and i 0. But this would complicate the strategy a
> lot at
> > >  > > > > (probably) only very slight benefit.
> > >  > > > >
> > >  > > > > > It is interesting that this strategy can actually result in
> non-solid
> > >  > > > > > approval coalitions on ballots ... sometimes it requires
> you to approve
> > >  > > > > > X while leaving unapproved some candidate Y rated above X
> on the same
> > >  > > > > > ballot ... i.e. insincere strategy.
> > >  > > > > >
> > >  > > > > > Furthermore, if estimates of both the utilities u_i and
> u_j, as well as
> > >  > > > > > of the probabilities p_ij in question were known with a
> high degree of
> > >  > > > > > precision, you might get away with those insincere gaps in
> the approval
> > >  > > > > > order.
> > >  > > > > >
> > >  > > > > > These facts reflect the fragility (anti-robustness) of the
> winning tie
> > >  > > > > > probability based strategy.
> > >  > > > >
> > >  > > > > Yes, I think Warren observed something similar: under
> imperfect
> > >  > > > > information, the optimal Range/Approval strategy might have
> you
> > >  > > > > approving of X and not Y even though you rank Y ahead of X.
> Under
> > >  > > > > perfect information, there's always some kind of cutoff where
> you
> > >  > > > > approve everybody above it and don't everybody below it.
> > >  > > > >
> > >  > > > > > Nevertheless, your result is highly relevant because it
> shows that on a
> > >  > > > > > fundamental level there is a meaningful, experimental way
> of defining
> > >  > > > > > individual utilities that are just as good as the
> theoretical utilities
> > >  > > > > > invoked as a basis for Approval strategy.
> > >  > > > >
> > >  > > > > I keep harping on the problem of Range and Approval to fail
> "de facto
> > >  > > > > IIA" despite passing it de jure, and I suspect it's related
> to this. If
> > >  > > > > we can't standardize a and b, then if the method behaves
> differently
> > >  > > > > when given u_i and up_i values, then you can get strange
> behavior. So
> > >  > > > > the guidelines about how to vote (mean utility, etc) are just
> > >  > > > > preprocessing steps that make your ballot expression no
> longer depend on
> > >  > > > > what a and b are. Then it's much more honest to attach these
> guidelines
> > >  > > > > to the method itself so it does so for the voter, so that
> voters don't
> > >  > > > > have to care about what society's a and b values are supposed
> to be, and
> > >  > > > > so that the method doesn't get away with sweeping de-facto
> failures
> > >  > > > > under the rug.
> > >  > > > >
> > >  > > > > At least MJ recognizes this and says "the only way we're
> going to get
> > >  > > > > IIA is if we have a and b values that are close enough to
> commensurable
> > >  > > > > that the problem doesn't occur". And then the point of using
> > >  > > > > instead of scores, and using order statistics, is to make the
> whole
> > >  > > > > process relatively insensitive to what a and b are, so that
> (hopefully)
> > >  > > > > a common grade standard can be established.
> > >  > > > >
> > >  > > > > > It is equally true for the not as sensitive strategy of
> approving the
> > >  > > > > > candidates k with above expectation utilities:
> > >  > > > > > u_k >sum P_i u_i,
> > >  > > > > > based on estimates of (non tie based) winning probabilities
> P_i, which
> > >  > > > > > are still sketchy because of rampant misinformation, not to
> mention
> > >  > > > > > intentional disinformation.
> > >  > > > >
> > >  > > > > Those are zero-info strategies, and indeed, they're also
> insensitive to
> > >  > > > > a and b.
> > >  > > > >
> > >  > > > > SARVO tries to get around the fragility/chaos problem by
> averaging over
> > >  > > > > a lot of vote orders. But it's somewhat of a hack; it's not
> particularly
> > >  > > > > elegant, and it fails scale invariance. Perhaps better is
> finding a
> > >  > > > > voting equilibrium where the mixed strategy is so that the
> distribution
> > >  > > > > of the M-W votes are stable, and then electing the candidate
> with the
> > >  > > > > highest expected score. I haven't read the M-W paper in
> detail, though,
> > >  > > > > so I don't know if finding this equilibrium is easy.
> > >  > > > >
> > >  > > > > (Another possibility, inspired by counterfactual regret
> minimization, is
> > >  > > > > to do M-W strategy by every voter, and then once everybdoy
> has submitted
> > >  > > > > a vote, pulling one of the voters from the list and having
> > >  > > > > his strategic ballot. Keep doing so over a long enough
> timeline and the
> > >  > > > > average of scores should converge to an equilibrium.)
> > >  > > > >
> > >  > > > > For the zero-info strategies, I tried to figure out what the
> optimum
> > >  > > > > zero info strategy is for Lp cumulative voting. I didn't get
> all the way
> > >  > > > > there, but this is what I figured:
> > >  > > > >
> > >  > > > > Under zero information, p_ij is equal for all pairs, and is
> (I think)
> > >  > > > > 1/n^2. So the objective for a zero-info voter is to maximize
> > >  > > > > SUM i=1..n v_i R_i
> > >  > > > > with R_i = SUM i != j: 1/(n^2) (u_i - u_j).
> > >  > > > >
> > >  > > > > We also have the constraint that SUM i=1..n |v_i|^p = 1 (due
> to Lp
> > >  > > > > normalization).
> > >  > > > >
> > >  > > > > So to use a Lagrangian:
> > >  > > > > max SUM i=1..n R_i v_i + lambda (1 - SUM i=1..n |v_i|^p)
> > >  > > > > i.e.
> > >  > > > > max SUM i=1..n (R_i v_i - lambda |v_i|^p) + lambda
> > >  > > > >
> > >  > > > > Now do a hack and use v_i^p instead because it's easier to
> differentiate
> > >  > > > > (might not be sound?), and let's consider one particular v,
> say v_1.
> > >  > > > >
> > >  > > > > The derivative wrt v_1 is
> > >  > > > > v_1 = ( -R_1/(lambda*p) )^(1/(p-1))
> > >  > > > > and wrt lambda
> > >  > > > > sum i=1..n: v_i^p = 1.
> > >  > > > >
> > >  > > > > So what that means is that the optimum is at
> > >  > > > > v_i = (R_i/k)^(1/(p-1))
> > >  > > > > where k is a constant set so that the pth powers of the
> voting variables
> > >  > > > > sum to one. (I.e. lambda is set so that -lambda p = k,
> because the
> > >  > > > > derivative wrt lambda itself places no constraint on lambda.)
> > >  > > > >
> > >  > > > > In particular, for max norm (Range), the calculation involves
> an 1/infty
> > >  > > > > norm, i.e. 0 norm, so that the scores only depend on the sign
> values of
> > >  > > > > the R variables. I don't *quite* get the right result here
> (it seems to
> > >  > > > > indicate the optimum vote would be +1 or -1 for every
> candidate), which
> > >  > > > > I think is because I turned |v_i| into v_i above.
> > >  > > > >
> > >  > > > > For ordinary cumulative voting (l1-cumulative), all R_i are
> raised to
> > >  > > > > some power that's approaching infinity. So as this power
> approaches
> > >  > > > > infinity, the k term grows to satisfy the constraint that the
> pth power
> > >  > > > > sums must be 1. This means that everything except the v_i
> corresponding
> > >  > > > > to the greatest R_i will approach zero, whereas the remaining
> one
> > >  > > > > approaches one. So the best zero-info strategy is to give max
> score to
> > >  > > > > your favorite and nobody else.
> > >  > > > >
> > >  > > > > For the quadratic norm, v_i = R_i/k, so only here is the zero
> info vote
> > >  > > > > directly proportional to R_i.
> > >  > > > >
> > >  > > > > And R_i - R_j is proportional to u_i - u_j with the same
> constant of
> > >  > > > > proportionality throughout, because:
> > >  > > > > R_i - R_j = 1/(n^2) (SUM i!=k (u_i - u_k) - SUM j!=k (u_j -
> u_k))
> > >  > > > > = 1/(n^2) ( (n-1) u_i - SUM k: (u_k) + u_i - (n-1) u_j + SUM
> k:
> > >  > > > > (u_k) - u_j)
> > >  > > > > = 1/(n^2) (n (u_i - u_j))
> > >  > > > > = 1/n (u_i - u_j)
> > >  > > > >
> > >  > > > > Hence for quadratic voting, so are the optimal zero info
> scores v_i.
> > >  > > > > Looking at R_i - R_j removes the b factor, which is probably
> why I can't
> > >  > > > > show that R_i is proportional to u_i directly.
> > >  > > > >
> > >  > > > > Again, it's not entirely sound but it indicates the general
> direction.
> > >  > > > > Do improve my calculations if you can, as they're very rough.
> > >  > > > >
> > >  > > > > (The problem with quadratic voting is that it isn't
> cloneproof. I
> > >  > > > > suspect that only Range itself is, because for every other
> p-norm >= 1,
> > >  > > > > you can imagine a two-candidate election where A gets
> 1+epsilon points,
> > >  > > > > B gets 1, then clone A to make A lose, if you just make
> epsilon small
> > >  > > > > enough.)
> > >  > > > >
> > >  > > > > -km
> > >  > > > >
> > >  > ----
> > >  > Election-Methods mailing list - see https://electorama.com/em for
> list info
> > >
> > >  --
> > >
> > >  r b-j . _ . _ . _ . _ rbj at audioimagination.com
> > >
> > >  "Imagination is more important than knowledge."
> > >
> > >  .
> > >  .
> > >  .
> > >  ----
> > >  Election-Methods mailing list - see https://electorama.com/em for
> list info
> > >
>
> --
>
> r b-j . _ . _ . _ . _ rbj at audioimagination.com
>
> "Imagination is more important than knowledge."
>
> .
> .
> .
> ----
> Election-Methods mailing list - see https://electorama.com/em for list
> info
>
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