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
>>>
>>
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