[EM] Three Category Symmetric MJ

Forest Simmons forest.simmons21 at gmail.com
Thu Apr 28 20:10:16 PDT 2022


Cool how everything is related to Fourier Transforms!

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,
> so blank=Bad).
> > >
> > > 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 the
> bad region.
> > >
> > > 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
> grades
> > > > >  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 him
> readjust
> > > > >  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
>
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