FS
Forest Simmons
Fri, Apr 29, 2022 1:03 AM
Andy,
Great dissertation!
I look forward to spending more time with it.
I started thinking about the three category version because there is an
important possible application in the Universal Domain context of ranked
choice ballots where equal ranks and truncations are allowed: instead of
Approval Sorted Margins, do MJ Sorted Margins ...using three category
reverse symmetry MJ (with our helical ramp continuity function), where the
three categories Good, Bad, and Middle, respectively, are simply Too,
Bottom, and Middle ballot listings. Of course, Top ranked on a ballot means
not outranked on the ballot. Bottom on the ballot means "does not outrank
any candidate" on the ballot, and Middle means "out ranks at least one
candidate and is outranked by at least one, as well."
This reverse symmetry basis for MJ Sorted Margins allows that Sorted
Margins method to satisfy Reverse Symmetry in a way that Implicit Approval
Sorted Margins does not.
And the (extended) tie breaking rule is so decisive it does not suffer from
the potential indecisiveness of Implicit Approval.
The one case where the tie breaking rule is not decisive... where two
candidates j and k have Top(k)=Bot(k) and Too(j)=Bot(j) ... is extremely
rare. In that case I think the tie should be broken in favor of the
candidate whose common Too, Bot value puts it closest to the midpoint (25%,
25%) of the segment connecting points (0,0) and (50%, 50%) representing
luke warm and controversial candidates, respectively.
If the j and k judgments are the same distance from the midpoint, then that
gives us an opportunity to use the pairwise defeat information inherent in
the ranked choice ballots.
If that is the only use of the pairwise preferences (stopping short of the
Sorted Margins application), then how much incentive could there be for
burial or compromise?
Best Wishes,
Forest
El jue., 28 de abr. de 2022 12:51 p. m., Andy Jennings <
elections@jenningsstory.com> escribió:
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@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@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@t-online.de> escribió:
On 26.04.2022 00:51, Forest Simmons wrote:
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
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
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
of the probabilities p_ij in question were known with a high degree
precision, you might get away with those insincere gaps in the
order.
These facts reflect the fragility (anti-robustness) of the winning
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
fundamental level there is a meaningful, experimental way of defining
individual utilities that are just as good as the theoretical
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,
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
Andy,
Great dissertation!
I look forward to spending more time with it.
I started thinking about the three category version because there is an
important possible application in the Universal Domain context of ranked
choice ballots where equal ranks and truncations are allowed: instead of
Approval Sorted Margins, do MJ Sorted Margins ...using three category
reverse symmetry MJ (with our helical ramp continuity function), where the
three categories Good, Bad, and Middle, respectively, are simply Too,
Bottom, and Middle ballot listings. Of course, Top ranked on a ballot means
not outranked on the ballot. Bottom on the ballot means "does not outrank
any candidate" on the ballot, and Middle means "out ranks at least one
candidate and is outranked by at least one, as well."
This reverse symmetry basis for MJ Sorted Margins allows that Sorted
Margins method to satisfy Reverse Symmetry in a way that Implicit Approval
Sorted Margins does not.
And the (extended) tie breaking rule is so decisive it does not suffer from
the potential indecisiveness of Implicit Approval.
The one case where the tie breaking rule is not decisive... where two
candidates j and k have Top(k)=Bot(k) and Too(j)=Bot(j) ... is extremely
rare. In that case I think the tie should be broken in favor of the
candidate whose common Too, Bot value puts it closest to the midpoint (25%,
25%) of the segment connecting points (0,0) and (50%, 50%) representing
luke warm and controversial candidates, respectively.
If the j and k judgments are the same distance from the midpoint, then that
gives us an opportunity to use the pairwise defeat information inherent in
the ranked choice ballots.
If that is the only use of the pairwise preferences (stopping short of the
Sorted Margins application), then how much incentive could there be for
burial or compromise?
Best Wishes,
Forest
El jue., 28 de abr. de 2022 12:51 p. m., Andy Jennings <
elections@jenningsstory.com> escribió:
> Forest,
>
> I like this rule. My thesis has a section about MJ tie-breaking rules and
> I came to the same conclusion:
> http://ajennings.net/dissertation.pdf
> pages 24-31
>
> ~ Andy
>
> On Thu, Apr 28, 2022 at 10:56 AM Forest Simmons <
> forest.simmons21@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@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@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@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@t-online.de <mailto:km_elmet@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
>>
>
FS
Forest Simmons
Fri, Apr 29, 2022 3:07 AM
Andy,
Thanks for pointing me to that!
Now with Andy Jennings and A. Fabre arriving at the same conclusion
independently, I have even more confidence that this method must be one in
"The Book," as Paul Erdos used to say.
The only disadvantage of "Usual Judgment", according to the Wikipedia
article (and I whole heartedly agree) is that for the general case of
several judgment categories the description of the complete tie breaking
system is too comlex for the attention span of the lay voter.
But this obstacle is nonexistent for the (arguably most important) three
category symmetric case, which is even easier to understand than standard
MJ, at any level whether heuristical or operational.
See my message to Andy Jenning for more on that tooic.
On Reverse Symmetry ...
Up until a few months ago when I took an interest in decloning the
Kendall-tau metric, I thought that the Reverse Symmetry Criterion was an
academic fetish that had no practical value.
And it wouldn't in the context of unlimited cardinal ratings, for example.
The intuitive geometry of Range Voting is flat Euclidean space of however
many dimensions. Think of the intuitively valuable Yee Ka-Ping, Brian
Olson diagrams ... flat 2-D space ... valuable in the same way that flat
maps can still be useful for local navegation.
But the topology of permutations of n symbols representing n candidates is
fundamentally spherical of dimension (n-2).
Think of Saari's arrangement of permutations of three candidates on a clock
face. In that case n-2 is 3-2=1. A one dimensional sphere is a circle. A
zero dimensional sphere is a pair of points ... corresponding to 2-2=0 .
With four candidates you get a 2-dim sphere, the surface of a ball like the
surface of the earth. Put one candidate at the North Pole, and the other
three uniformly spaced 30 degrees south of the equator.
In that context, reversal of a ballot corresponds to an antipodal point on
the sphere of possible ordinal ballots. In the context of Universal Domain
methods, there is no getting around it!
Kendall-tau is topologically correct but metrically distorted by clones,
the same way massive bodies distort the local curvature of space-time.
Now according to Universal Domain principals, the strength of first and
last preferences are not supposed to be treated as stronger than any other
preferences. But in practice, voters tend to have stronger feelings about
those first and last preferences, and given half a chance (i.e. allowed
equal rankings and truncations) they will avail themselves of the
opportunity to express those strong preferences.
So here we have it .. an important opportunity for a three candidate
reverse symmetry median judgment method ... it allows us to color slightly
outside the Universal Domain lines without breaking the official UD rules
or offending common sense. [As you have correctly guessed, in this
application of median judgment the three judgment categories are Top,
Truncation (or lowest rank if there no truncations), and the in-between
ranks.]
Thanks,
Forest
El jue., 28 de abr. de 2022 2:01 p. m., Andy Dienes andydienes@gmail.com
escribió:
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@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@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@t-online.de> escribió:
On 26.04.2022 00:51, Forest Simmons wrote:
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
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
approval coalitions on ballots ... sometimes it requires you to
X while leaving unapproved some candidate Y rated above X on the
ballot ... i.e. insincere strategy.
Furthermore, if estimates of both the utilities u_i and u_j, as
of the probabilities p_ij in question were known with a high degree
precision, you might get away with those insincere gaps in the
order.
These facts reflect the fragility (anti-robustness) of the winning
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
fundamental level there is a meaningful, experimental way of
individual utilities that are just as good as the theoretical
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
candidates k with above expectation utilities:
u_k >sum P_i u_i,
based on estimates of (non tie based) winning probabilities P_i,
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
Andy,
Thanks for pointing me to that!
Now with Andy Jennings and A. Fabre arriving at the same conclusion
independently, I have even more confidence that this method must be one in
"The Book," as Paul Erdos used to say.
The only disadvantage of "Usual Judgment", according to the Wikipedia
article (and I whole heartedly agree) is that for the general case of
several judgment categories the description of the complete tie breaking
system is too comlex for the attention span of the lay voter.
But this obstacle is nonexistent for the (arguably most important) three
category symmetric case, which is even easier to understand than standard
MJ, at any level whether heuristical or operational.
See my message to Andy Jenning for more on that tooic.
On Reverse Symmetry ...
Up until a few months ago when I took an interest in decloning the
Kendall-tau metric, I thought that the Reverse Symmetry Criterion was an
academic fetish that had no practical value.
And it wouldn't in the context of unlimited cardinal ratings, for example.
The intuitive geometry of Range Voting is flat Euclidean space of however
many dimensions. Think of the intuitively valuable Yee Ka-Ping, Brian
Olson diagrams ... flat 2-D space ... valuable in the same way that flat
maps can still be useful for local navegation.
But the topology of permutations of n symbols representing n candidates is
fundamentally spherical of dimension (n-2).
Think of Saari's arrangement of permutations of three candidates on a clock
face. In that case n-2 is 3-2=1. A one dimensional sphere is a circle. A
zero dimensional sphere is a pair of points ... corresponding to 2-2=0 .
With four candidates you get a 2-dim sphere, the surface of a ball like the
surface of the earth. Put one candidate at the North Pole, and the other
three uniformly spaced 30 degrees south of the equator.
In that context, reversal of a ballot corresponds to an antipodal point on
the sphere of possible ordinal ballots. In the context of Universal Domain
methods, there is no getting around it!
Kendall-tau is topologically correct but metrically distorted by clones,
the same way massive bodies distort the local curvature of space-time.
Now according to Universal Domain principals, the strength of first and
last preferences are not supposed to be treated as stronger than any other
preferences. But in practice, voters tend to have stronger feelings about
those first and last preferences, and given half a chance (i.e. allowed
equal rankings and truncations) they will avail themselves of the
opportunity to express those strong preferences.
So here we have it .. an important opportunity for a three candidate
reverse symmetry median judgment method ... it allows us to color slightly
outside the Universal Domain lines without breaking the official UD rules
or offending common sense. [As you have correctly guessed, in this
application of median judgment the three judgment categories are Top,
Truncation (or lowest rank if there no truncations), and the in-between
ranks.]
Thanks,
Forest
El jue., 28 de abr. de 2022 2:01 p. m., Andy Dienes <andydienes@gmail.com>
escribió:
> I believe this is also the "Usual Judgement" Usual judgment - Wikipedia
> <https://en.wikipedia.org/wiki/Usual_judgment>
>
> On Thu, Apr 28, 2022 at 3:51 PM Andy Jennings <elections@jenningsstory.com>
> wrote:
>
>> Forest,
>>
>> I like this rule. My thesis has a section about MJ tie-breaking rules and
>> I came to the same conclusion:
>> http://ajennings.net/dissertation.pdf
>> pages 24-31
>>
>> ~ Andy
>>
>> On Thu, Apr 28, 2022 at 10:56 AM Forest Simmons <
>> forest.simmons21@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@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@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@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@t-online.de <mailto:km_elmet@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
>>>
>> ----
>> Election-Methods mailing list - see https://electorama.com/em for list
>> info
>>
>>
FS
Forest Simmons
Fri, Apr 29, 2022 3:10 AM
Cool how everything is related to Fourier Transforms!
El jue., 28 de abr. de 2022 2:02 p. m., robert bristow-johnson <
rbj@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@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
El mié., 27 de abr. de 2022 6:10 p. m., Forest Simmons <
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
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
Since tan 45 deg equals 1, and tan theta equals the slope (.5
tan(t)= (1 - slope)/(1+slope)
Substituting slope =(.5 -y)/(.5-x), and simplifying algebraically, we
tan(t)=(y-x)/[2( 1-x-y)].
Recognizing (1 - x -y) as the slack variable z, we see that tan(t) is
(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
That's it for now!
-Forest
El mar., 26 de abr. de 2022 7:20 p. m., Forest Simmons <
Good Work!
I have come to the same conclusion about MJ being much closer to IIA
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
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
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 <
On 26.04.2022 00:51, Forest Simmons wrote:
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
probability (perceived by a voter) of the event that candidates i
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
score minus j's score is 1, then you could flip the win from i to
voting j 10 and i 0. But this would complicate the strategy a lot
(probably) only very slight benefit.
It is interesting that this strategy can actually result in
approval coalitions on ballots ... sometimes it requires you to
X while leaving unapproved some candidate Y rated above X on
ballot ... i.e. insincere strategy.
Furthermore, if estimates of both the utilities u_i and u_j, as
of the probabilities p_ij in question were known with a high
precision, you might get away with those insincere gaps in the
order.
These facts reflect the fragility (anti-robustness) of the
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
fundamental level there is a meaningful, experimental way of
individual utilities that are just as good as the theoretical
invoked as a basis for Approval strategy.
I keep harping on the problem of Range and Approval to fail "de
IIA" despite passing it de jure, and I suspect it's related to
we can't standardize a and b, then if the method behaves
when given u_i and up_i values, then you can get strange
the guidelines about how to vote (mean utility, etc) are just
preprocessing steps that make your ballot expression no longer
what a and b are. Then it's much more honest to attach these
to the method itself so it does so for the voter, so that voters
have to care about what society's a and b values are supposed to
so that the method doesn't get away with sweeping de-facto
under the rug.
At least MJ recognizes this and says "the only way we're going to
IIA is if we have a and b values that are close enough to
that the problem doesn't occur". And then the point of using
instead of scores, and using order statistics, is to make the
process relatively insensitive to what a and b are, so that
a common grade standard can be established.
It is equally true for the not as sensitive strategy of
candidates k with above expectation utilities:
u_k >sum P_i u_i,
based on estimates of (non tie based) winning probabilities
are still sketchy because of rampant misinformation, not to
intentional disinformation.
Those are zero-info strategies, and indeed, they're also
a and b.
SARVO tries to get around the fragility/chaos problem by
a lot of vote orders. But it's somewhat of a hack; it's not
elegant, and it fails scale invariance. Perhaps better is finding
voting equilibrium where the mixed strategy is so that the
of the M-W votes are stable, and then electing the candidate with
highest expected score. I haven't read the M-W paper in detail,
so I don't know if finding this equilibrium is easy.
(Another possibility, inspired by counterfactual regret
to do M-W strategy by every voter, and then once everybdoy has
a vote, pulling one of the voters from the list and having him
his strategic ballot. Keep doing so over a long enough timeline
average of scores should converge to an equilibrium.)
For the zero-info strategies, I tried to figure out what the
zero info strategy is for Lp cumulative voting. I didn't get all
there, but this is what I figured:
Under zero information, p_ij is equal for all pairs, and is (I
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
(might not be sound?), and let's consider one particular v, say
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
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
norm, i.e. 0 norm, so that the scores only depend on the sign
the R variables. I don't quite get the right result here (it
indicate the optimum vote would be +1 or -1 for every candidate),
I think is because I turned |v_i| into v_i above.
For ordinary cumulative voting (l1-cumulative), all R_i are
some power that's approaching infinity. So as this power
infinity, the k term grows to satisfy the constraint that the pth
sums must be 1. This means that everything except the v_i
to the greatest R_i will approach zero, whereas the remaining one
approaches one. So the best zero-info strategy is to give max
your favorite and nobody else.
For the quadratic norm, v_i = R_i/k, so only here is the zero
directly proportional to R_i.
And R_i - R_j is proportional to u_i - u_j with the same constant
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
Looking at R_i - R_j removes the b factor, which is probably why
show that R_i is proportional to u_i directly.
Again, it's not entirely sound but it indicates the general
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
you can imagine a two-candidate election where A gets 1+epsilon
B gets 1, then clone A to make A lose, if you just make epsilon
Cool how everything is related to Fourier Transforms!
El jue., 28 de abr. de 2022 2:02 p. m., robert bristow-johnson <
rbj@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@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@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@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@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@t-online.de <mailto:km_elmet@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@audioimagination.com
>
> "Imagination is more important than knowledge."
>
> .
> .
> .
> ----
> Election-Methods mailing list - see https://electorama.com/em for list
> info
>
RB
robert bristow-johnson
Fri, Apr 29, 2022 6:34 AM
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])
but it assumes a quadratic fit. (But that's usually good enough.)
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@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@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@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@t-online.de escribió:
On 26.04.2022 00:51, Forest Simmons wrote:
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
> On 04/28/2022 11:10 PM Forest Simmons <forest.simmons21@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])
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@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@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@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@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@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@t-online.de <mailto:km_elmet@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@audioimagination.com
> >
> > "Imagination is more important than knowledge."
> >
> > .
> > .
> > .
> > ----
> > Election-Methods mailing list - see https://electorama.com/em for list info
> >
--
r b-j . _ . _ . _ . _ rbj@audioimagination.com
"Imagination is more important than knowledge."
.
.
.
FS
Forest Simmons
Sat, Apr 30, 2022 3:43 AM
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 <
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
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
El mié., 27 de abr. de 2022 6:10 p. m., Forest Simmons <
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
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
Since tan 45 deg equals 1, and tan theta equals the slope (.5
tan(t)= (1 - slope)/(1+slope)
Substituting slope =(.5 -y)/(.5-x), and simplifying
tan(t)=(y-x)/[2( 1-x-y)].
Recognizing (1 - x -y) as the slack variable z, we see that
(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
That's it for now!
-Forest
El mar., 26 de abr. de 2022 7:20 p. m., Forest Simmons <
Good Work!
I have come to the same conclusion about MJ being much closer
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
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 <
On 26.04.2022 00:51, Forest Simmons wrote:
El lun., 25 de abr. de 2022 4:25 a. m., Kristofer
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
You're right. I was looking at the paper just now, and it
"For each pair of candidates i and j, the /pivot probability/
probability (perceived by a voter) of the event that
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
score minus j's score is 1, then you could flip the win from
voting j 10 and i 0. But this would complicate the strategy a
(probably) only very slight benefit.
It is interesting that this strategy can actually result in
approval coalitions on ballots ... sometimes it requires
X while leaving unapproved some candidate Y rated above X
ballot ... i.e. insincere strategy.
Furthermore, if estimates of both the utilities u_i and
of the probabilities p_ij in question were known with a
precision, you might get away with those insincere gaps in
order.
These facts reflect the fragility (anti-robustness) of the
probability based strategy.
Yes, I think Warren observed something similar: under
information, the optimal Range/Approval strategy might have
approving of X and not Y even though you rank Y ahead of X.
perfect information, there's always some kind of cutoff where
approve everybody above it and don't everybody below it.
Nevertheless, your result is highly relevant because it
fundamental level there is a meaningful, experimental way
individual utilities that are just as good as the
invoked as a basis for Approval strategy.
I keep harping on the problem of Range and Approval to fail
IIA" despite passing it de jure, and I suspect it's related
we can't standardize a and b, then if the method behaves
when given u_i and up_i values, then you can get strange
the guidelines about how to vote (mean utility, etc) are just
preprocessing steps that make your ballot expression no
what a and b are. Then it's much more honest to attach these
to the method itself so it does so for the voter, so that
have to care about what society's a and b values are supposed
so that the method doesn't get away with sweeping de-facto
under the rug.
At least MJ recognizes this and says "the only way we're
IIA is if we have a and b values that are close enough to
that the problem doesn't occur". And then the point of using
instead of scores, and using order statistics, is to make the
process relatively insensitive to what a and b are, so that
a common grade standard can be established.
It is equally true for the not as sensitive strategy of
candidates k with above expectation utilities:
u_k >sum P_i u_i,
based on estimates of (non tie based) winning probabilities
are still sketchy because of rampant misinformation, not to
intentional disinformation.
Those are zero-info strategies, and indeed, they're also
a and b.
SARVO tries to get around the fragility/chaos problem by
a lot of vote orders. But it's somewhat of a hack; it's not
elegant, and it fails scale invariance. Perhaps better is
voting equilibrium where the mixed strategy is so that the
of the M-W votes are stable, and then electing the candidate
highest expected score. I haven't read the M-W paper in
so I don't know if finding this equilibrium is easy.
(Another possibility, inspired by counterfactual regret
to do M-W strategy by every voter, and then once everybdoy
a vote, pulling one of the voters from the list and having
his strategic ballot. Keep doing so over a long enough
average of scores should converge to an equilibrium.)
For the zero-info strategies, I tried to figure out what the
zero info strategy is for Lp cumulative voting. I didn't get
there, but this is what I figured:
Under zero information, p_ij is equal for all pairs, and is
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
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
(might not be sound?), and let's consider one particular v,
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
sum to one. (I.e. lambda is set so that -lambda p = k,
derivative wrt lambda itself places no constraint on lambda.)
In particular, for max norm (Range), the calculation involves
norm, i.e. 0 norm, so that the scores only depend on the sign
the R variables. I don't quite get the right result here
indicate the optimum vote would be +1 or -1 for every
I think is because I turned |v_i| into v_i above.
For ordinary cumulative voting (l1-cumulative), all R_i are
some power that's approaching infinity. So as this power
infinity, the k term grows to satisfy the constraint that the
sums must be 1. This means that everything except the v_i
to the greatest R_i will approach zero, whereas the remaining
approaches one. So the best zero-info strategy is to give max
your favorite and nobody else.
For the quadratic norm, v_i = R_i/k, so only here is the zero
directly proportional to R_i.
And R_i - R_j is proportional to u_i - u_j with the same
proportionality throughout, because:
R_i - R_j = 1/(n^2) (SUM i!=k (u_i - u_k) - SUM j!=k (u_j -
= 1/(n^2) ( (n-1) u_i - SUM k: (u_k) + u_i - (n-1) u_j + SUM
(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
Looking at R_i - R_j removes the b factor, which is probably
show that R_i is proportional to u_i directly.
Again, it's not entirely sound but it indicates the general
Do improve my calculations if you can, as they're very rough.
(The problem with quadratic voting is that it isn't
suspect that only Range itself is, because for every other
you can imagine a two-candidate election where A gets
B gets 1, then clone A to make A lose, if you just make
El jue., 28 de abr. de 2022 11:34 p. m., robert bristow-johnson <
rbj@audioimagination.com> escribió:
>
>
> > On 04/28/2022 11:10 PM Forest Simmons <forest.simmons21@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@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@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@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@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@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@t-online.de <mailto:km_elmet@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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> list info
> > >
> > > --
> > >
> > > r b-j . _ . _ . _ . _ rbj@audioimagination.com
> > >
> > > "Imagination is more important than knowledge."
> > >
> > > .
> > > .
> > > .
> > > ----
> > > Election-Methods mailing list - see https://electorama.com/em for
> list info
> > >
>
> --
>
> r b-j . _ . _ . _ . _ rbj@audioimagination.com
>
> "Imagination is more important than knowledge."
>
> .
> .
> .
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> Election-Methods mailing list - see https://electorama.com/em for list
> info
>
