April 2020 - mathfun@mailman.xmission.com

[math-fun] Rolling polyhedra
by James Propp 26 Dec '20

26 Dec '20

I want to think about rolling polyhedra, and the polyhedra I want to think about rolling are polyhedra in which the sum of the angles at each vertex is a submultiple of 360 degrees. For instance, given any triangle T, we can create a tetrahedron whose four faces are all congruent to T, with angle-sum 180 degrees at each vertex (there's a name for such tetrahedra but I forget what it is). Are there other polyhedra in which the angles at each vertex sum to a submultiple of 360 degrees (not necessarily the same one at each vertex)? Jim Propp

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[math-fun] Chess variant
by David Wilson 13 Nov '20

13 Nov '20

In Paul Morphy's "Opera House" chess game, Morphy never moves a piece backward. I propose Opera House Chess, with initial position and moves as in chess, except pieces cannot move backward. Is it playable? With victory conditions modified?

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[math-fun] One parameter quintic solution
by Bill Gosper 01 Nov '20

01 Nov '20

This is probably ho-hum, but while trying to simplify a Chowla-surdberg, I found that 0 = a + 80 x + 20 x^3 + x^5 has solutions {((-a - Sqrt[4096 + a^2])^(1/5) E^((2 I \[Pi])/5))/2^( 1/5) + ((-a + Sqrt[4096 + a^2])^(1/5) E^((2 I \[Pi])/5))/2^( 1/5), ((-a + Sqrt[4096 + a^2])^(1/5) E^(-((2 I \[Pi])/5)))/2^( 1/5) + ((-a - Sqrt[4096 + a^2])^(1/5) E^(-((4 I \[Pi])/5)))/2^( 1/5), ((-a - Sqrt[4096 + a^2])^(1/5) E^(-((2 I \[Pi])/5)))/2^( 1/5) + ((-a + Sqrt[4096 + a^2])^(1/5) E^(-((4 I \[Pi])/5)))/2^( 1/5), (-a + Sqrt[4096 + a^2])^(1/5)/2^( 1/5) + ((-a - Sqrt[4096 + a^2])^(1/5) E^((4 I \[Pi])/5))/2^( 1/5), (-a - Sqrt[4096 + a^2])^(1/5)/2^( 1/5) + ((-a + Sqrt[4096 + a^2])^(1/5) E^((4 I \[Pi])/5))/2^(1/5)} Is it from a two-parameter family? C.f. https://en.wikipedia.org/wiki/Quintic_function#Other_solvable_quintics --rwg

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Re: [math-fun] FactorInteger
by Bill Gosper 20 Aug '20

20 Aug '20

Slight improvement: Mathematica's front end by default displays a centerdot for implicit multiplication. Strangely, this doesn't work in old the FI function below, which displays blanks. Blanks also (by default) separate digits in blocks of 3. So the display is sometimes ambiguous. But Wolfram CE Harry Calkins showed me this workaround: FI⨉[n : _Rational | _Integer] := Defer @@ {FactorInteger@n} /. {{a_Integer, 1} -> a, {a_Integer, b_} -> a^b} /. {a_} -> a /. List -> Times E.g. In[144]:= FI⨉@BernoulliB@50]@// tim During evaluation of In[144]:= 0.006522,1 (seconds) Out[144]= (5^2 417202699 47464429777438199)/(2 3 11) (The first Bernoulli number with a square factor!) Dirty trick. I used a tiny ⨉ symbol instead of x, so you need Front End completion to input FI⨉. You may want to edit the definition. Instead of little centerdots between the factors, there's a Preferences variable Appearances|Numbers|Multiplication|Automatically display multiplication symbol between multiplied numbers Use symbol: which you can bash to be a mini ⨉. Sorry, I don't know how to bash it under program control, as in an init file. Unfactor is way nicer than with FactorInteger: UnFI⨉@Defer[x_] := x In[153]:= UnFI⨉@%144 Out[153]= 495057205241079648212477525/66 —rwg On Wed, Jan 4, 2017 at 12:44 AM Bill Gosper <billgosper(a)gmail.com> wrote: > There are > > In[14]:= 52! > Out[14]= > 80658175170943878571660636856403766975289505440883277824000000000000 > > ways to shuffle a deck. > That's just a "little" over a googol ... > > In[366]:= Log[10.^100, %] > Out[366]= 0.679066483922048 > > to the 2/3 power. The square of the cube root. And vice versa. So if > anybody asks how many > ways to shuffle a deck, "Oh, about the cube root of a googol squared." > But this is actually low > by a fairly substantial factor of > In[372]:= (10.^100)^(.679 - 2/3) > > Out[372]= 17.1132830416181 > > The actual number of shufflings is enormously composite: > > In[15]:= FactorInteger@%14 > Out[15]= > {{2,49},{3,23},{5,12},{7,8},{11,4},{13,4},{17,3},{19,2},{23,2},{29,1},{31,1},{37,1},{41,1},{43,1},{47,1}} > > But this is ugly and hard to read. The doc suggests > > In[16]:= CenterDot@@(Superscript@@@%) > Out[16]= > 2^49\[CenterDot]3^23\[CenterDot]5^12\[CenterDot]7^8\[CenterDot]11^4\[CenterDot]13^4\[CenterDot]17^3\[CenterDot]19^2\[CenterDot]23^2\[CenterDot]29^1\[CenterDot]31^1\[CenterDot]37^1\[CenterDot]41^1\[CenterDot]43^1\[CenterDot]47^1 > > This actually looks neat. Except for the ^1s. Try it. But it's unclear > (actually, a nice little Mathematica exercise) how to get your original > number back. > So I wrote something called FI > > In[364]:= FI[52!] > Out[364]= 2^49 3^23 5^12 7^8 11^4 13^4 17^3 19^2 23^2 29 31 37 41 43 47 > > In[365]:= ReleaseHold@% > Out[365]= > 80658175170943878571660636856403766975289505440883277824000000000000 > > Here is the definition: > FI[n : _Rational | _Integer] := > Times @@ ((HoldForm @@ {#1[[1]]})^#1[[2]] &) /@ FactorInteger[n] > > And here's a spoiler for how to invert FactorInteger: > > In[367]:= Times @@ (Power @@@ FactorInteger[73 137]) > Out[367]= 10001 > > which also works on that CenterDot expression: > > In[368]:= Times @@ (Power @@@ %16) > > Out[368]= > 80658175170943878571660636856403766975289505440883277824000000000000 > > Finally, you might enjoy finding a fairly nice formula for how many 0s are > on the end of n!, and, > more generally, what are exponents of the various prime factors. > --Bill >

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[math-fun] How does a dull spoon manage to cut a stringy piece of meat
by Bill Gosper 14 Jul '20

14 Jul '20

while sliding back an forth against the inside of a slippery bowl? --rwg

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Re: [math-fun] Covid19 R_e considered misguided
by Henry Baker 22 May '20

22 May '20

Perhaps the first order of business is to strangle the old differential equation model(s) in the cradle, and start using *Monte Carlo* simulations of network-type models. Given the ubiquity of computing power these days, there's no excuse for restricting the search for the "car keys" only near the differential equation "lamp-post". At 11:18 AM 4/11/2020, Fred Lunnon wrote: > No doubt you're right about all this --- but any sharpening requires >further assumptions about the distribution of R_e among the population, >for which any credible model is presumably currently unavailable. WFL > >On 4/11/20, Henry Baker <hbaker1(a)pipeline.com> wrote: >> I've been reading about Covid19 and listening to a number >> of professional podcasts about Covid9 -- e.g., "This Week >> in Virology", which amazingly enough, is entertaining >> enough not to be a soporific. >> >> Reading between the lines, I'm coming to the conclusion >> that the concept of "R_e" ("effective" R), may be fatally >> flawed, as it tries to capture some sort of "mean" or >> "average" R in an inherently exponential setting. >> >> Thus, if one person has an R_e of 0.9 and another person >> has an R_e of 40.0, there isn't a good way to average the >> two R_e's to compute a composite R_e. What if a vanishingly >> small fraction of infected people induce the vast majority >> of cases? We know that spreading is a *network* phenomenon >> which means that it is highly likely to have a *fat tail* >> distribution. >> >> If I'm correct about this, then the standard epidemic >> model is also fatally flawed, and hence unreliable for >> making trillion-dollar decisions. What won a Nobel >> prize a century ago is today an undergraduate homework >> exercise. We desperately need a better class of models >> for today's pandemics. >> >> For example, in Massachusetts in the early going, almost >> 100% of the confirmed cases stemmed from a single meeting >> of a single company in a single downtown Boston hotel. >> >> As another example, in Chicago, most of the early cases >> stemmed from a single funeral and a single birthday >> party. >> >> So if it is true that there are "super spreaders", both >> in terms of individual people and/or individual events, >> there must be a better way to quickly identify and >> isolate these people and events other than putting the >> entire world on lockdown. >> >> I would imagine that some of the work on fractals should >> be useful to look at this superspreader phenomenon. >> >> Perhaps fractures in glass or metal may have relevance, >> as it may only take a single flaw in a crystal structure >> to destroy the entire structure.

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[math-fun] Extension of Game of Life
by Allan Wechsler 15 May '20

15 May '20

While musing about searching for Life patterns with particular properties, I had an idea for a Lifelike cellular automaton rule that surely must have been invented before. So I'm hoping somebody can give me references. The rule has three states. Two are Life's existing "dead" and "alive", and any situation involving only these states evolves exactly as in Life. The third state is like a quantum superposition of "dead" and "alive". Call it "spooky". The principle is this: a pattern P with spooky cells corresponds to a set L(P) of ordinary Life patterns. A Life pattern belongs to the set if it agrees with P on all non-spooky cells. A cell is alive in P', the successor of P, if for all patterns Q in L(P), it is alive in Q'. Similarly, it is dead in P' if it is always dead in Q' for all Q in L(P). A cell is spooky in P' in all other cases, that is, if it is sometimes alive and sometimes dead in Q' depending on the choice of Q from L(P). Spookiness is ignorance. Sometimes it disappears completely -- an isolated spooky cell in empty space dies is one step. This is a well-defined automaton, and its rules can be completely stated; an example is "a live cell with two live neighbors and one spooky neighbor will stay alive", an "a dead cell with three or more spooky neighbors and no live neighbors becomes spooky". Has anybody implemented this rule for Golly? (I am a Golly tyro and don't know the rule-specification language yet.) Does this rule have any non-trivial interesting patterns? Like -- does it have any oscillators involving spooky cells? Any spaceships? I know it has at least one still-life: the spooky block is stable.

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[math-fun] In case you hadn't heard, Wolfram has figured out physics.
by Bill Gosper 11 May '20

11 May '20

It turns out General Relativity was a mistake. https://writings.stephenwolfram.com/2020/04/finally-we-may-have-a-path-to-t… —Bill

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[math-fun] Cross-My-Primes
by Éric Angelini 06 May '20

06 May '20

Hello Math-Fun, have a look here for a clearer idea developed in the hereunder lines: https://bit.ly/2KMrmHM Best, É. --------------- Hello Math-Fun, this game (hope not old hat) is played on a Scrabble board where all squares are white, and no square has any text on it. The starting "star" square, in the center of the grid, has a 7 on it (the number, not the word, as this game is played with digits only that will form numbers – instead of letters forming words). The single player must now form prime numbers at every turn on the board – at the first turn placing the integer 1, at the second turn placing 2, then 3, then 4, then 5, etc. – this is the natural order of the positive integers as they appear (and because 7 starts the game, the player will jump from 6 to 8. No other jump will be allowed during the game). Those integers, as in the traditional game of Scrabble, must be attached to the existing structure at least by a digit (one digit per square). All visible numbers, before and after any turn, must be prime: they are read horizontally from left to right or vertically from top to bottom (again, like words in a traditional Scrabble game). Can one split the digits of a term bigger than 9 – say 10 or 11 ? Yes, as long as the split digits stay on the same line, forming one or more prime numbers within the existing structure. Example here: We start the game with 7 on the star: ....................7.................... then comes 1 and we form the prime 71; ....................71.................... then comes 2 and we form the prime 271; ...................271.................... then comes 3 and we form the prime 3271; ..................3271.................... then comes 4 and we form the prime 43271; .................43271.................... then comes 5 and we form the prime 53 (5 on top of 3 on the grid); ..................5....................... .................43271.................... (etc.) Best, É.

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[math-fun] Math humor on Twitter
by Keith F. Lynch 01 May '20

01 May '20

Math humor on Twitter: How to solve the Millennium math problems and win millions of dollars: https://twitter.com/MathematicsUCL/status/1255875396257615875

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