May 2015 - mathfun@mailman.xmission.com

[math-fun] 96 fractions
by Stuart Anderson 17 May '15

17 May '15

A teacher writes a list of 96 fractions on a board: , , ... , , He then asks a student to play a game: for every turn, he picks any two numbers and on the board, removes them, and adds to the list on the board. The question is: what will the last number be?

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[math-fun] replicoids
by Warren D Smith 17 May '15

17 May '15

In the real world, as opposed to ConwayLife, I do not think anybody has ever come close to constructing a self-replicating "Von Neumann machine." (Right?) I think we all agree it should be possible, and, e.g, such a machine could live on the moon... but trying to create it would be a huge undertaking -- how many years off?

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[math-fun] By popular demand, NeilB once again affronts your senses
by Bill Gosper 17 May '15

17 May '15

https://www.shadertoy.com/view/lljGRd

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Re: [math-fun] NeilB's latest eye pickle
by Dan Asimov 15 May '15

15 May '15

First time I recall the canceling-out Fred speaks of is noticing that a new refrigerator motor in my family home was quite audible at first, but over time became imperceptible. Until the thermostat would turn it off — at which moment someone might say, sincerely, "What was that?!" When I was away at college my parents moved to another town nearby. My first time home it was quite noticeable that a plane would fly overhead every 6 minutes. Pretty soon, whenever I was there, the planes became imperceptible — unless I thought about them, in which case they were suddenly there. ((( A different perceptual phenomenon is demonstrated by the short film De Duva (The Dove), at < https://www.youtube.com/watch?v=8X2QmLWWxq4 >, which is based on Ingmar Bergman films. I recommend not reading anything at that URL but just starting the video right away at full screen size. ))) Okay, so this frequency-canceling works for visual and auditory stimuli. What about taste, smell, and touch? ——Dan Fred wrote: ----- A quick search for "moving bar illusion" failed to find anything obviously relevant to RWG's impressively effective post. So I'll point out that it demonstrates very neatly the way our perceptive mechanisms filter out constant background frequencies, allowing through only changes in the signal spectrum --- visual, auditory, tactile, olfactory(?!). In this case, the subsequent apparent stretching of the text is patently opposite to that induced by the video, as our internal DPU continues to cancel out the assumed previous motion. I am reminded of an incident late one night when standing beside the Victorian clock in the hall of my home: completely unable to hear its normal loud tick, I inspected the pendulum to check whether it had stopped. As I watched in some bafflement, the tick gradually became perceptible, accompanied by the realisation that it had actually been present throughout. -----

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[math-fun] NeilB's latest eye pickle
by Bill Gosper 14 May '15

14 May '15

Some animations I demoed in 1986 illustrated that you can't tell if an enlarging hyperbola is scaling anisotropically. One of them was actually stretching horizontally and shrinking vertically at a lesser rate, which only became clear as the background faded up some crazy chewing gum wallpaper. Thus an apparently fixed hyperbola might be stretching one way and shrinking the other at complementary rates. This is shown by Neil's vastly subtler chewing gum: gosper.org/hyperbolicV2.gif . Staring at this for a few periods and then suddenly freezing it produces a strong and amusing variation on the moving bar illusion. The problem: how to freeze an mgif without moving your eyes? https://addons.mozilla.org/en-US/firefox/addon/toggle-animated-gifs/ enables Firefox to alternately play and pause mgifs with the keystroke Control-m. --Bill Gosper

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Re: [math-fun] Similar permutations problem
by Dave Dyer 13 May '15

13 May '15

Oops, my braino - your number is correct, and not such a big number by today's standards as you pointed out. The original question remains - is there a better that brute force way to approach it? In actuality, 20,000 is just the number I have on hand, the actual number is much larger.

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Re: [math-fun] Mensa Correctional Facility
by Colin Wright 12 May '15

12 May '15

Hi all, Here's the conversation so far - my reply is at the bottom of the email so you can read through to get the context, and then my reply: On Wed, Apr 15, 2015 at 12:55 PM, Veit Elser <ve10(a)cornell.edu> wrote: > There's a growing class of puzzles I call > Mensa Correctional Facility puzzles. In > these puzzles there is always a warden who > challenges prisoners with math/logic puzzles > and grants them freedom when they succeed in > solving them. Does anyone know the origins > of the following one: > > In this puzzle the warden challenges a pair > of Mensa inmates. The first inmate is shown > to the warden's room and the warden proceeds > to place identical coins, each one either > heads up or tails up, onto the 64 squares of > a checkerboard. He then points to one of the > coins and declares it to be the "freedom coin". > The inmate watching this is then invited to > flip one of the coins to help his partner > identify the freedom coin. Then, without > allowing the inmates to communicate the second > inmate is lead into the room, shown the coins, > and after a little head scratching points to > the freedom coin -- he is, after all, a Mensa > inmate -- and the prisoners are set free. > > How did they do it? On Wed, Apr 15, 2015 at 1:04 PM, Mike Stay <metaweta(a)gmail.com> wrote: > I first heard it last year from a coworker who > likes such puzzles, but the setup was slightly > different: the warden allows the two to agree > on a strategy in his presence first; then one > of them leaves; then the warden is free to place > the coins in whatever way he pleases and choose > any coin as the freedom coin. Then the other > prisoner leaves, the first one comes back, and > has to make his choice. Thane Plambeck <tplambeck(a)gmail.com> replied: > This puzzle, and related ones, seem to be coming > up a lot. It was performed (with 16 replacing 64) > at a recreational math conference I recently > attended in Portugal earlier this year. I was also > told about it at the last Gathering 4 Gardner in > Atlanta (spring 2014). The person who showed it to > me first was Colin Wright I think (cc'ed), who > might be able to say where he heard about it (if > he didn't make it up himself). I certainly didn't make it up. I was presented this as a puzzle based on labelling the vertices of a 64 dimensional cube, and then re-interpreted it back to the original form of deducing a number from a coin configuration. Not sure when I heard it first, but it was probably mid 2013 or so. I can perform it with 64 squares, it it takes my accomplices too long to learn the decoding, unless they've done this sort of thing before. I wrote up my "performance" from Lisbon here: http://www.solipsys.co.uk/FlippingPuzzle/AnotherFlippingPuzzle.html That includes all the relevant references I could find. The earliest reference I could find was "Two applications of a Hamming code" by Andy Liu in Mathematics Magazine, January 2009. You can find that here: http://www.ingentaconnect.com/content/maa/cmj/2009/00000040/00000001/art000… Hope that helps! Best regards, Colin -- If you never go off at a tangent, you will forever run in circles.

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[math-fun] Sloane sequence A000522
by Mike Speciner 10 May '15

10 May '15

This sequence is listed as Total number of arrangements of a set with n elements: a(n) = Sum_{k=0..n} n!/k!. But isn't the total number of arrangements of a set with n elements n! ? Isn't A000522 really the total number of arrangements of subsets of a set with n elements?

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[math-fun] polynomial puzzle
by Erich Friedman 09 May '15

09 May '15

consider polynomials with the property that the line passing through two inflection points on the graph is tangent to the graph at a third point. there are polynomials of degree 5 or more with this property. i can find one of degree 6 with integer coefficients. not hard at all. but i cannot seem to find one with degree 5 with integer coefficients. can you? erich friedman

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[math-fun] Hilbert[1/√2]
by Bill Gosper 08 May '15

08 May '15

Is it, its real or imaginary part, or magnitude algebraic? Or at least, are these quantities algebraically connected? My wildly recursive exact Hilbert function stacks out after a few convergents, raising the entertaining problem of rewriting it nonrecursively. "$RecursionLimit=Infinity removes any limit on the number of recursion levels. ... On some computer systems, your whole Wolfram Language session may crash if you allow it to use more stack space than the computer system allows. " But Mathematica is apparently unable to warn you where that precipice lurks! Meanwhile, a mild surprise: $RecursionLimit = 9999; In[22]:= Convergents[1/Sqrt[2],16] Out[22]= {0,1,2/3,5/7,12/17,29/41,70/99,169/239,408/577,985/1393,2378/3363,5741/8119,13860/19601,33461/47321,80782/114243,195025/275807} Let's Hilbert these for a succession of rational approximations to Hilbert[1/√2]: In[25]:=Hilbert/@%22 Out[25]={0, 1, 1 + I, 50/63 + (41 I)/63, 14/17 + (12 I)/17, 25/31 + (8 I)/11, 26425/32769 + (2703 I)/3641, 352121381971026539243677635070833214831945620557921683722821936356496290/441711766194596082395824375185729628956870974218904739530401550323154943 + 330867072223470133127042461326599267580738957131033875887852738752365031 I/441711766194596082395824375185729628956870974218904739530401550323154943, 3786557088089533296370/4722366482869645213697 + 3539308315540678808218 I/4722366482869645213697, <big numbers>...) Indeed, OS X Mathematica 10 quietly loses its back end on the very next (17th) convergent. Hilbert just the last four: In[23]:= tim[Hilbert[#]]&/@Take[%22,-4] During evaluation of In[23]:= 0.073786 seconds During evaluation of In[23]:= 14.515494 seconds During evaluation of In[23]:= 0.744485 seconds During evaluation of In[23]:= 50.913754 seconds <multihundred digit complex rationals suppressed> Wait, why did the penultimate one take only 3/4 second? In[24]:= Short/@% Out[24]= {(178756875476641101<<8>>535160091329216180)/(223007451985306231<<8>>272648361505980417)+<<1>>, (12783524938776490<<505>>462752734609543929)/(15948292093020135<<505>>211095044536439369)+<<1>>, 3525337771939/4398046511105+(3296808843473 I)/4398046511105, (66696048809823983<<523>>014863141630402424)/(83206401558341682<<523>>067958436720377525)+<<1>>} The numerator and denominator are comparatively tiny! In fact, the denominator is In[36]:= 2^42+1 Out[36]= 4398046511105 Douglas Adams, are you listening? Yet nothing's numerically strange about the approach to Hilbert[1/√2]: In[33]:= N[%23] Out[33]= {0.801573552 +0.749630877 I,0.801560748 +0.749590554 I,0.801569006 +0.749607544 I,0.801573528 +0.749603288 I} In octal, that 15th convergent and its Hilbert image are In[32]:= BaseForm[{%22[[-2]],%23[[-2]]},8] Out[32]//BaseForm= {Subscript[235616, 8]/Subscript[337103, 8], Subscript[63231640267643, 8]/Subscript[100000000000001, 8]+(Subscript[57763107534321, 8] I)/Subscript[100000000000001, 8]} Many other denominators (of the real parts) alternately exceed and inceed powers of two: Table[Log[2, Denominator[Re[%25[[k]]]] + (-1)^k], {k, 4, 16}] {6, 4, 5, 15, 238, 72, Fail,Fail, 44, 144, Fail, 42,Fail]} The failing denominators approximate powers of 2 divided by small integers, suggesting accidental cancellations. E.g. In[39]:= BaseForm[Denominator[Re[%25[[10]]]],8] Out[39]//BaseForm= Subscript[30303030303030303030303030303030303030303030303030303030303030303, 8] In[41]:= BaseForm[Denominator[Re[%25[[11]]]]+1,8] Out[41]//BaseForm= Subscript[1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000, 8] In[43]:= BaseForm[Denominator[Re[%25[[14]]]],8] Out[43]//BaseForm= Subscript[1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111, 8] In[44]:= BaseForm[Denominator[Re[%25[[16]]]],8] Out[44]//BaseForm= Subscript[512772650051277265005127726500512772650051277265005127726500512772650051277265005127726500512772650051277265005127726500512772650051277265005127726500512772650051277265005127726500512772650051277265005127726500512772650051277265005127726500512772650051277265005127726500512772650051277265005127726500512772650051277265005127726500512772650051277265005127726500512772650051277265005127726500512772650051277265005127726500512772650051277265005127726500512772650051277265005127726500512772650051277265005127726500512772650051277265005127726500512772650051277265005127726500512772650051277265005127726500512772650051277265, 8] Note that these 2ish denominators are not artifacts of bigfloat truncation, nor continued fraction truncation, nor are they properties of sqrt 2. It's just that Hilbert likes to make 2ish denominators. The irregular sizes correspond to the irregular periods of decimal (or whatever) expansions of rationals. Interestingly, inverse Hilbert also likes 2ish denominators. Using Julian's Clear[unbert]; unbert[z_] := piecewiserecursivefractal[z, Identity, If[0 <= Re[#] <= 1 && 0 <= Im[#] <= 1, Range[4], {}] &, {1 - 2*#/I &, 2*# - I &, 2*# - I - 1 &, (# - 1)*2/I &}, {(1 - #)/4 &, (# + 1)/4 &, (# + 2)/4 &, 1 - #/4 &}] in turn using his miraculous piecewiserecursivefractal[x_, f_, which_, iters_, fns_] := piecewiserecursivefractal[x, g_, which, iters, fns] = ((piecewiserecursivefractal[x, h_, which, iters, fns] := Block[{y}, y /. Solve[f[y] == h[y], y]]); Union @@ ((fns[[#]] /@piecewiserecursivefractal[iters[[#]][x], Composition[f, fns[[#]]], which, iters, fns]) & /@which[x])) In[51]:= $RecursionLimit=6666;unbert/@{1/89,1/97,1/89+I/97,I/89+1/97,89/97} Out[51]= {{959/4194303},{396685311379875009944667/1760625833650318613189865563}, <big>,<big>, {280122039752001/281474976710657}} Testing: In[52]:= Hilbert[#[[1]]]&/@%//tim 26.544898 Seconds,5 terms Out[52]= {1/89,1/97,1/89+I/97,1/97+I/89,89/97} The octal denominators: In[53]:= BaseForm[Denominator[%51],8] Out[53]//BaseForm= {{Subscript[17777777, 8]},{Subscript[1330133013301330133013301330133,8]}, {Subscript[470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047,8]}, {Subscript[470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047004700470047,8]}, {Subscript[10000000000000001, 8]}} --rwg (Sorry for the discourteous length. Indeed, GEICO has just sent me a Courtesy Renewal Notice.)

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