- mathfun@mailman.xmission.com - mailman.xmission.com

Re: [math-fun] How are the five justices voting against any remedy for partisan gerrymandering not "the enemy of the people"
by John Golden 28 Jun '19

28 Jun '19

The majority decision has a few claims to test with math: " > > The fact that the Court can adjudicate one-person, one-vote claims does not mean that partisan gerrymandering claims are justiciable. This Court’s one-person, one-vote cases recognize that each person is entitled to an equal say in the election of representatives. It hardly follows from that principle that a person is entitled to have his political party achieve representation commensurate to its share of statewide support. Vote dilution in the one-person, one-vote cases refers to the idea that each vote must carry equal weight. That requirement does not extend to political parties; it does not mean that each party must be influential in proportion to the number of its supporters. The racial gerrymandering cases are also inapposite: They call for the elimination of a racial classification, but a partisan gerrymandering claim cannot ask for the elimination of partisanship. Pp. 15–21. and > (c) None of the proposed “tests” for evaluating partisan gerrymandering claims meets the need for a limited and precise standard that is judicially discernible and manageable. Pp. 22–30. In particular (c) reminds me of the 538 evaluation on how the SCOTUS is not willing to use math: https://fivethirtyeight.com/features/the-supreme-court-is-allergic-to-math/ Justice Kagan leads her dissent with "For the first time ever, this Court refuses to remedy a constitutional violation because it thinks the task beyond judicial capabilities.” This to me points to their unwillingness to understand the mathematics involved. The violations she cites are the 1st and 14th amendment. Later: " > > For the first time in this Nation’s history, the majority declares that it can do nothing about an acknowledged constitutional violation because it has searched high and low and cannot find a workable legal standard to apply.The majority gives two reasons for thinking that the adjudication of partisan gerrymandering claims is beyond judicial capabilities. First and foremost, the majority says, it cannot find a neutral baseline—one not based on contestable notions of political fairness—from which to measure injury. > ...And second, the majority argues that even after establishing a baseline, a court would have no way to answer “the determinative question: ‘How much is too much?’” So if we’re going to discuss on math-fun, I’d love to hear your thoughts on how to measure, what is neutral and how much is too much. Cheers, John ps Decision at https://www.supremecourt.gov/opinions/18pdf/18-422_9ol1.pdf > On Jun 28, 2019, at 10:23 AM, math-fun-request(a)mailman.xmission.com wrote: > > Message: 1 > Date: Thu, 27 Jun 2019 16:59:34 -0700 (GMT-07:00) > From: Dan Asimov <dasimov(a)earthlink.net> > To: math-fun(a)mailman.xmission.com > Subject: [math-fun] How are the five justices voting against any > remedy for partisan gerrymandering not "the enemy of the people" > Message-ID: > <188820833.15200.1561679974743(a)wamui-fatboy.atl.sa.earthlink.net> > Content-Type: text/plain; charset=UTF-8 > > I'm curious how voting against any remedy for partisan gerrymandering does not > make the five justices who voted in the majority "the enemy of the people". > > (Reasoning: Democracy means government responsive to the will of the people, > as determined by their vote. If expressing that will is thwarted or > misrepresented, then Democracy is thwarted.) > > > Anticipating some possible responses to this question, I will say this: > > 1) This question is not political, but rather scholarly. > > 2) If I am censured or banned from math-fun for asking this question, > > ?OR? is this topic is banned from further discussion, > > then I will have nothing further to do with math-fun, as of immediately. > > Cf. https://www.chronicle.com/article/Meet-the-Math-Professor/239260 > > ?Dan

1 0

Re: [math-fun] optimal advance tournament scheduling
by Dan Asimov 28 Jun '19

28 Jun '19

Just in case anyone else on math-fun actually knows what maximum likelihood means ... ... it makes no sense whatoever to talk about maximum likelihood unless (as has no been done below) this is in the context of a PARAMETRIZED FAMILY OF DISTRIBUTIONS. Then you say under which one, as the probability model, makes what happened more likely than the others. So saying what maximum likelihood will or will not predict WITHOUT reference to a model is utterly meaningless. —Dan ---- Perhaps this is interesting: if we have three players who have played three > games and created a cycle (A beat B, B beat C, C beat A), a three-way tie > in strength is NOT the maximum-likelihood solution; -----

2 1

[math-fun] political discussion
by rcs＠xmission.com 27 Jun '19

27 Jun '19

A reminder: Political discussion is off-topic for the Math-Fun list. I decide what's political. Check with me if you're uncertain. Rich

1 0

[math-fun] How are the five justices voting against any remedy for partisan gerrymandering not "the enemy of the people"
by Dan Asimov 27 Jun '19

27 Jun '19

I'm curious how voting against any remedy for partisan gerrymandering does not make the five justices who voted in the majority "the enemy of the people". (Reasoning: Democracy means government responsive to the will of the people, as determined by their vote. If expressing that will is thwarted or misrepresented, then Democracy is thwarted.) Anticipating some possible responses to this question, I will say this: 1) This question is not political, but rather scholarly. 2) If I am censured or banned from math-fun for asking this question, —OR— is this topic is banned from further discussion, then I will have nothing further to do with math-fun, as of immediately. Cf. https://www.chronicle.com/article/Meet-the-Math-Professor/239260 —Dan

1 0

Re: [math-fun] optimal advance tournament scheduling
by Steve Witham 27 Jun '19

27 Jun '19

At 6/23/19, 5:25 PM, "Keith F. Lynch" <kfl(a)KeithLynch.net> wrote, > If all 8 players play each other, and any result is possible, > including ones that violate consistency and transitivity, that conveys > 28 bits. With consistency and transitivity, most patterns of 28 bits > are impossible. A complete ordering conveys log2(8!) bits, about 15.3 > bits. Either way, with just 16 games we can't get more than 16 bits. It seems to me prior probabilities are necessary to talk about information. An idea like that matchups *tend* to be *somewhat* consistent and transitive is the beginning of a prior. If I think that the total probability of one or more violation of transitivity or consistency is one in a binary million (2**20), then that probability gets divided between all the violating outcomes, so there would be more than 20 bits of information in any one of them. On the other hand, the more unlevel the playing field, so to speak, the less *expected* surprise. From my impression of statistics, what you do is start with a general model of how things work, which has hidden parameters, and then given an actual outcome, try to derive the hidden parameters most likely to give that outcome. (I actually don't understand why the peak of the curve is given such importance.) So if each team has a "skill level", and the odds of A wins/B wins/tie are just a function of the two skills, then once the tournament is over you can back-figure the most likely vector of 8 skills, and from that pick the most skillful team! Easy peasy. But, even having no prejudices about the teams to start with (not to mention ideas about how they interact), the amount of information the outcome represents depends on both the model (of behind-the-scenes Truth, not the tournament plan) and the actual outcome. Or, you can calculate the expected number of bits and turn off the TV satisfied. I understand the Google pagerank algorithm is equivalent to applying one (linear?) model to this problem. Of course, the information everyone wants to know is, who's the best team? Which for 8 teams is 3 bits without any prejudices. Then the question is, what are the odds a given method of matching teams gives the right answer (given some model of "best teams")? Because the less likely you're right, the fewer bits you actually delivered. On the other hand, if the best you can deliver is three bits, there are diminishing... returns to even... talking about... this. --Steve

5 4

[math-fun] Fwd: Three piece Golden Rectangle to square dissection
by Bill Gosper 26 Jun '19

26 Jun '19

[This was an afterthought to the kids, but you may like the Cartesian "chord blood" and the line intersector.] [Just kids]: (Aside: This dissection is of a very restricted type in that the pieces permute without rotating.) This should work for rectangles other than Golden. Which ones? This is what I meant when I sent the old construction and asked for the applicable range of aspect ratios. (Nobody answered.) Instead of Golden, can the rectangle be tall and skinny? Extremely squat? Same answer for both constructions? Numbers, please. Younger guys: Note that, if the rectangle is a⨉b, the radius of the 3rd arc is √a√b, the geometric mean. (Old name: mean proportional.) Fairly amazing geometry thm: When two chords intersect, the product of the "halves" of one equals the product of the "halves" of the other. Cartesian proof with four arbitrary points on the unit circle, In[599]:= Assuming[{a, b, c, d} \[Element] Reals, FullSimplify@ComplexExpand[Abs[I^a - #] Abs[# - I^b] == Abs[I^c - #] Abs[# - I^d] &@ Echo@Interseg[I^a, I^b, I^c, I^d]]] ((I^c-I^d) Im[I^b Conjugate[I^a]]+(I^a-I^b) Im[I^c Conjugate[I^d]])/Im[(I^c-I^d) (Conjugate[I^a]-Conjugate[I^b])] Out[599]= True (This takes "forever" w/o the ComplexExpand.) Note the "Echo"ed previous line is a fully general line intersector, if you replace I^a by z1, ..., I^d by z4, and z1,z2 determine the 1st line and z3, z4 determine the 2nd. Mma must offer this functionality somewhere. ? Note also: The "chords" can intersect outside the circle, in which case they are secants! —Bill On Sun, Jun 9, 2019 at 4:45 AM Bill Gosper <billgosper(a)gmail.com> wrote: > On Tue, May 21, 2019 at 6:17 PM Bill Gosper <billgosper(a)gmail.com> wrote: > >> Golden rectangle↔︎square <http://gosper.org/gold2square.gif> >> I forget where I stole this, but can you find the superfluous step I just >> noticed? >> (Not the retrace, which was to tell you a dimension.) >> —rwg >> > SPOILER: The arc centered on the top edge of the rectangle is > superfluous. > >> (Rebuild in progress) >> > Nicer golden rectangle to square <http://gosper.org/gold2square2.gif> > I can re-export this to loop if people want. > I think it loops if you download and view it with Finder. > —rwg > >> >> >

1 1

Re: [math-fun] optimal advance tournament scheduling
by Dan Asimov 26 Jun '19

26 Jun '19

That's only one of many things you can do in statistics, and it's called the "maximum likelihood" estimator. By competent statisticians it's given only the importance that it's due: If you want to know which parameter makes the known outcome as probable as it can be, according to your model, that's it. —Dan Steve Witham wrote: ----- ... From my impression of statistics, what you do is start with a general model of how things work, which has hidden parameters, and then given an actual outcome, try to derive the hidden parameters most likely to give that outcome. (I actually don't understand why the peak of the curve is given such importance.) ... -----

1 0

Re: [math-fun] ellipsoid from girths: multivariate series reversion
by Brad Klee 25 Jun '19

25 Jun '19

"The total 1-boundary is topologically a triangle, and can be flattened into the plane." Orient an ellipsoid in R^3 with symmetry planes XY, YZ, ZX, and let A, B, C be the semi-axis lengths along X,Y, and Z. Take the positive octant, and cut triangle ABC through the ellipsoid. Then define P(ABC) the curvilinear projection of ABC onto the surface (X/A)^2 + (Y/B)^2 + (Z/C)^2 = 1. Clearly the Euclidean triangle ABC satisfies the triangle inequality, and clearly the girths also satisfy the triangle inequality if P(ABC) can indeed be flattened into the plane. Now let L1, L2, L3 be the euclidean lengths along sides of ABC and L1' > L1, L2' > L2, L3' > L3 are the corresponding curvilinear lengths along P(ABC). Without loss of generality, require that L1' > L2' >= L3'. Clearly L1 < L2 + L3, but we need to prove sufficiently that L1' < L2 + L3. Now write L1' = k*L1, with at most k=(sqrt(2)/4)*pi=1.11072... Lengths L2 and L3 are minimized as the ellipsoid flattens toward a plane where L2^2 + L3^2 = L1^2 . Then write: ( k^2 - 1 )*( L2^2 + L3^2 ) < 2*L2*L3. Is this inequality always satisfied? If L2 approximately equals L3 then *obviously yes* because "k" is so close to one. Apparently in this thread, we have decided to haggle about the limit where L3 becomes small relative to L2, where also k approaches 1, but how rapidly? Can we still flatten the ellipsoidal triangles? Without loss of generality set L3<L2=1 and just graph the RHS and LHS of the inequality, ( k^2 - 1 )*(1+L3^2) < 2*L3 over 0 < L3 < 1: https://0x0.st/z2xm.png The linear red line from the RHS is always the superior, but if anyone still doubts me, they are welcome to check the series expansion of the LHS. Okay, now we have wasted a couple of hours. Why do I really think this is "obvious"? Because when you draw the triangles on the ellipse they have angles Pi/2 at every vertex. Obviously the flattening procedure does not preserve these angles, but Pi/2 is a lot of leeway, either plus or minus, and we just need to avoid going too far obtuse on the largest angel. On most days this extra 0.11072... is nothing for me to worry about. Tonight I am feeling generous. Cheers, Brad On Tue, Jun 25, 2019 at 7:59 PM Warren D Smith <warren.wds(a)gmail.com> wrote: > > > Once I got the search working, I also found that on Tue Oct 11 19:07:21 > > MDT 2011 Warren Smith conjectured that the triangle inequality is obeyed > > for girths. Isn't this also obvious? > > --obvious? Consider a triangle made of 3 quarter-girths drawn on > surface of ellipsoid. > Conjecture would then follow if surface metric is a metric (yes, it > is) and those > triangle-edge quarter-girths are geodesics and shortest paths (no, that latter > aint necessarily so). > > So, is it obvious to you, and why? > > > > -- > Warren D. Smith > http://RangeVoting.org <-- add your endorsement (by clicking > "endorse" as 1st step)

1 0

Re: [math-fun] Grooks
by Dan Asimov 25 Jun '19

25 Jun '19

I just got my copy and it looks to have the best coverage of the history of Hex of anything I've read, going into great detail on Hex history and related issues. I'm a little disappointed that there is so little about the subject of the math behind Hex, or the many Hex-like games that have been developed (often by mathematicians) —Dan Stan Isaacs wrote: ----- ... I just today received a copy of a new book I had ordered, called “Hex, The Full Story”, by Ryan B. Hayward and Bjorne Toft, which contains a lot about Piet Hein, the inventor of hex, and also Soma Cubes and Grooks. It was here I first learned that Grooks was Gruks in Danish. And the same day, I get this email about pronouncing “gruks”. The book seems very good at first skimming, with a lot about Hein as well as the games. (Toft also contributed to a book on Piet Hein published in 2015, in Danish. Much of his research for both that book and “Hex” was done at Stanford, where I met him.)

1 0

[math-fun] Short essay on Alan Frank's Muffin Problem
by James Propp 25 Jun '19

25 Jun '19

I've posted a draft of a 1000-word piece I've written for this year's Big Internet Math-Off, which I plan to expand (a little) and post as a Mathematical Enchantments essay later this year. http://mathenchant.org/muffin-draft2.pdf Comments are welcome. Jim Propp

2 1