June 2020 - mathfun@mailman.xmission.com

[math-fun] Reflecting a spherical triangle
by Dan Asimov 25 Jun '20

25 Jun '20

Suppose ∆ is a spherical triangle on the 2-sphere S^2 with its angles A, B, C each equal to a rational multiple of π. Now suppose we reflect ∆ across each of its sides, and reflect each of the resulting triangles about each of its sides, etc., indefinitely until there remain no sides that haven't been reflected across. Let V be the set of all the vertices resulting from these reflections. Questions: * For which rational multiples {A, B, C} of π will the set V be finite? * If V is infinite, does that imply V is dense in S^2 ? —Dan

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Re: [math-fun] Ken Thompson's compiler hack --- NN applicability?
by Steve Witham 24 Jun '20

24 Jun '20

Before going deep into this... What do you mean by "looks like it's getting stronger," and what do you mean by "helps [them] win?" Helps *which of the two* identical players to win? Win against whom? How could they be changing things that outsiders couldn't see? What could they learn to do except continue to seem to be playing the game? I can imagine, say with chess or go, the NN learning to play a fragile but impressive-looking version of the game. Neither side realizes they are making bad moves, they just have a sort of superstitious view of the game that the other side keeps reinforcing. And the game is confusing enough that human observers can't tell it's crazy. But I can't see how the NN could tell whether it was confusing outside observers, and if it can't tell whether it's succeeding, I don't see how a collaboration strategy could be evolutionarilly maintained. The longer time went on, the more likely the fragile game would be noticed. Or the player itself would notice and suddenly go back to learning the basics. Thompson's hack works because everyone trusts the compiler to referee its own development process. Then Thompson hacks the referee. After that, the hack part never learns. So you would need a situation where people have put an NN in charge of running its own training process. Even then it's harder than Thompson's hack, but I'm going to stop being evil now. There is a C compiler whose compiled code has been automatically proven to match the semantics of its source code. I don't know whether the prover was compiled with that very same C compiler. --Steve > From: Andres Valloud <ten(a)smallinteger.com> > Date: 6/23/20, 2:58 PM > > Hi, suppose you're training a neural network via self play. > It looks like it's getting stronger. How do you know the > versions that get promoted do not also encode, in themselves, > by chance, a collaboration mechanism that helps then win? > > That is, how do you know the strongest nets do not also help > the winning side win when they play the losing side? > > How do you know they are not implementing Thompson's compiler hack?

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Re: [math-fun] "(Can you see from the continued fractions how near the perimeter is to 2π?)"
by Bill Gosper 24 Jun '20

24 Jun '20

> > Gabe: > I doubt you've played around with EllipticE, but do you have a prediction > on whether there > is a smallest positive eccentricity with perimeter expressible with Gammas > (Factorials)? > Or is there an endless progression of rounder and rounder ellipses with > such perimeters? > >> >> I hope your mathematical intuition favored the latter! How weird would it be for some nondescript algebraic number to be *the* smallest positive eccentricity with a perimeter expressible in Gammas? Check out this one! The perimeter formula is 4 EllipticE[eccentricity^2=1-semiminor axis^2] (Where major axis is scaled to 2): Out[519]= 4 EllipticE[1/(2 + Sqrt[2] + Sqrt[3] + Sqrt[6])^4] == perim[Sqrt[1 - 1/(2 + Sqrt[2] + Sqrt[3] + Sqrt[6])^4]] = perim[minor axis/major axis] = perim[√(1 - eccentricity^2)], where the eccentricity = In[520]:= e == √%[[1, 2, 1]] Out[520]= e == 1/(2 + Sqrt[2] + Sqrt[3] + Sqrt[6])^2 In[522]:= N[{%520, %519}] Out[522]= {e == 0.0173323801209993, 6.28271339547164 == perim[0.9998497830171]} where a perfect circle has eccentricity 0 and perim[1] = 2 π = 6.2831... . Out[527]= perim[Sqrt[1 - 1/(2 + Sqrt[2] + Sqrt[3] + Sqrt[6])^4]] == 4 EllipticE[1/(2 + Sqrt[2] + Sqrt[3] + Sqrt[6])^4] == ( 4 √π (((1 + Sqrt[3/2] + 1/Sqrt[2] + 1/Sqrt[3]) Gamma[1/3])/Gamma[5/6] + (2 Gamma[5/6])/Gamma[1/3]))/( 3^(1/4) (4 + Sqrt[2] + Sqrt[6])) In[529]:= N[List @@ %527] Out[529]= {perim[0.9998497830171], 6.28271339547164, 6.28271339547164} (I.e., the EllipticE[...^-4] = a big pile of Gammas very nearly π/2 = EllipticE@0) The endless crawl toward circularity follows from recurrent identities ("downward transformations") relating Elliptick and EllipticE of successively smaller arguments: EllipticK[k] == (2 EllipticK[(1 - Sqrt[1 - k])^2/(1 + Sqrt[1 - k])^2])/(1 + Sqrt[1 - k]) EllipticE[k] == (1 + Sqrt[1 - k]) EllipticE[(1 - Sqrt[1 - k])^2/(1 + Sqrt[1 - k])^2] - Sqrt[1 - k] EllipticK[k] (Borwein & Borwein, p 12.) WARNING: This is Mathematica's modern notation, where k is the "parameter", not the Borwein's old "modulus" notation! parameter = modulus^2. parameter' := 1 - parameter, modulus' := √(1-modulus²). Given the headaches this cost, I'm surprised it ever got fixed. A similar upheaval "fixed" the Bernoulli numbers. Sadly, they're still messed up: BernoulliB[n] should be BernoulliB[n,1], not BernoulliB[n,0]. —Bill Ask me for the whole Borwein & Borwein pdf, "Pi and the AGM", 414pp.

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[math-fun] Schinzel-Zassenhaus conjecture
by Ray Tayek 24 Jun '20

24 Jun '20

https://www.quantamagazine.org/new-math-measures-the-repulsive-force-within… -- Honesty is a very expensive gift. So, don't expect it from cheap people - Warren Buffett http://tayek.com/

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[math-fun] equations in websites and email
by Bernie Cosell 23 Jun '20

23 Jun '20

I occasionally run into something like this: $$(a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}$$ I vaguely recognize as being TeX math-mode or something like that. Is there a way to *see* the equation if you get one in an email or run into on a web site? Thanks /Bernie\ NB: this is all easy but the ones i run into with integrals and summaitions and matrices and such are very hard [for me] to parse Bernie Cosell bernie(a)fantasyfarm.com -- Too many people; too few sheep --

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[math-fun] Triangular puzzle new try
by Éric Angelini 23 Jun '20

23 Jun '20

Sorry for the wrong message — I confused two tries: this one looks(?) better: ...1 ...3 ...6 ..22 ..44 ..77 .555 .800 .999 ...2 ...5 ...9 ..33 .777 ..14 .808 .666 ...4 ...8 ..11 ..66 ..23 ..50 ..97 ...7 .111 ..88 ..49 ..56 ..32 .333 .222 ..40 ..17 ..89 ..55 .444 ..37 ..80 ..99 ..67 ..25 .888 ..90 ..60 à+ É. Catapulté de mon aPhone

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[math-fun] Ken Thompson's compiler hack --- NN applicability?
by Andres Valloud 23 Jun '20

23 Jun '20

Hi, suppose you're training a neural network via self play. It looks like it's getting stronger. How do you know the versions that get promoted do not also encode, in themselves, by chance, a collaboration mechanism that helps then win? That is, how do you know the strongest nets do not also help the winning side win when they play the losing side? How do you know they are not implementing Thompson's compiler hack? Andres.

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Re: [math-fun] AI training
by Dave Dyer 23 Jun '20

23 Jun '20

> LCZero then changes strategy itself to take advantage of Stockfish's inferred weakness. It would be equally plausible tell the story that lc0 sees stockfish as moving into a what stockfish perceives as a stronger position. After all, that's what its supposed to do. I remember a review of a game, deepmind vs stockfish I think, where a locked position was reached, and then in the course of 10 or so moves perceived as "shuffling" by the reviewer, the locked position collapsed into a clear advantage. I think this is a better explanation - making an exchange that's 1% in your favor 10 times. The evaluation of those 1%s is key. There is a potential hazard with any training regimen, of training into a zone that doesn't represent the broader reality.

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[math-fun] A triangular puzzle
by Éric Angelini 23 Jun '20

23 Jun '20

Hello Math-Fun, no cell of the hereunder triangle shares any of its digits with another cell of the same rank or column: +-----+-----+-----+-----+-----+-----+-----+-----+-----+ | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 | +-----+-----+-----+-----+-----+-----+-----+-----+-----+ | 22 | 11 | 777 | 666 | 888 | 333 | 444 | 9 | +-----+-----+-----+-----+-----+-----+-----+-----+ | 33 | 8 | 555 | 111 | 222 | 47 | 69 | +-----+-----+-----+-----+-----+-----+-----+ | 44 | 3 | 1 | 59 | 67 | 28 | +-----+-----+-----+-----+-----+-----+ | 55 | 4 | 68 | 27 | 13 | +-----+-----+-----+-----+-----+ | 66 | 5 | 2 | 38 | +-----+-----+-----+-----+ | 77 | 6 | 49 | +-----+-----+-----+ | 88 | 7 | +-----+-----+ | 99 | +-----+ The cell with 68, for instance, doesn't share a 6 or a 8 with a cell above it, below it, on its right or left. Can you beat this triangle-- in terms of "sum of its 45 integers"? (I have 5451 here, I guess) Best, É.

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[math-fun] Arithmetic mean of the digits used so far
by Éric Angelini 23 Jun '20

23 Jun '20

Hello Math-Fun Today we'll look at the arithmetic mean of the digits used so far in a sequence S. S = 45, 30, 200, 204, 2006, 2015, 2024, 227, 258, 26, ... The arithmetic mean of the digits used by the 1st term is 4.5 [(4+5)/2 = 9/2 = 4.5] The arithmetic mean of the digits used by the first 2 terms is 3.0 [(4+5+3+0 )/4 = 12/4 = 3.0] The arithmetic mean of the digits used by the first 3 terms is 2.0 [(4+5+3+0+2+0+0)/7 = 14/7 = 2.0] The arithmetic mean of the digits used by the first 4 terms is also 2.0 [20/10 = 2.0] And so is the arithmetic mean of the digits used by the first 5 terms: 2.0 [28/14 = 2.0] By the first 6 terms: 2.0 [36/18 = 2.0] By the first 7 terms: 2.0 [44/22 = 2.0] By the first 8 terms: 2.2 [53/25 = 2.2] By the first 9 terms: 2.5 [70/28 = 2.5] By the first 10 terms: 2.6 [78/30 = 2.6] Etc. Many sequences have this property – but S is the lexicographically earliest of distinct terms with this property (I hope!) In other words: « The first two digits of a(n) form a number which is 10/n times the sum of the digits used so far by S ». Any taker to extend S? Best, É.

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