June 2014 - mathfun@mailman.xmission.com

Re: [math-fun] Spirography
by Bill Gosper 20 Jun '14

20 Jun '14

No. The gears shown in the picture, gosper.org/Moregears.png while as they are both can always be pulled apart orthogonally and don't work physically due to very slight intersection, can probably be modified so that they work physically and in some phases can't be pulled apart orthogonally. Combining this with the multilayer gear idea (so that you're always in such a phase), it would be possible to make a pair of gears which can't ever be separated orthogonally. They could probably still be separated without using the third dimension, though. Assuming that those modifications work. I might give it a try. Julian On Wed, Jun 18, 2014 at 4:54 PM, rwg <rwg(a)sdf.org> wrote: -------- Original Message -------- Subject: Re: [math-fun] Spirography Date: 2014-06-18 13:38 From: Allan Wechsler <acwacw(a)gmail.com> To: math-fun <math-fun(a)mailman.xmission.com> Reply-To: math-fun <math-fun(a)mailman.xmission.com> Gosper, is Julian all pessimistic because the classic spirograph has the rotor rolling around the inside of the stator? Does the problem persist if the rotor rolls around the outside? On Wed, Jun 18, 2014 at 4:23 PM, Steve Witham <sw(a)tiac.net> wrote: Date: Mon, 16 Jun 2014 11:54:40 -0400 [in issue 29] > From: James Propp <jamespropp(a)gmail.com> > > > Has anyone designed a mechanism that permits the spirographer to focus on > circumferential force? > > The knockoff product I have, has holes in both parts, suitable for pushing pins through, so in theory you only have to push on the pen tip. Rather than a wax tray the first thing I'd think of would be corrugated cardboard & the second, cork. I'm imagining something like a latchable/unlatchable zipper. > > That's a fascinating idea in itself. It could be used in funicular railways, for instance. Although the zipper tooth design is asymmetrical, I have a coat that unzips from both ends. --Steve

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[math-fun] Mathematical coloring book
by James Propp 19 Jun '14

19 Jun '14

Check out http://jdh.hamkins.org/math-for-seven-year-olds-graph-coloring-chromatic-nu… Does anyone know of any other good mathematical coloring books for kids? (Personally I think almost any interesting schematic mathematical image, such as a hypercube or a picture created with a spirograph, could be fun for a kid to color. The choice of colors wouldn't necessarily have any mathematical significance, but coloring a number of such sketches over time could gradually shift a child's esthetic in mathematical directions.) Jim

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[math-fun] Fwd: [LifeCA] Life still-life syntheses complete up to 17 bits
by rcs＠xmission.com 19 Jun '14

19 Jun '14

This may be of interest to some of the ex-Lifers on Math-Fun. --Rich ----- Forwarded message from LifeCA(a)yahoogroups.com ----- Date: Fri, 13 Jun 2014 19:37:28 -0700 From: "Mark Niemiec mniemiec(a)gmail.com [LifeCA]" Subject: [LifeCA] Life still-life syntheses complete up to 17 bits To: LifeCA mailing list <LifeCA(a)yahoogroups.com> Back in the 1980s and 1990s, Dave Buckingham completed the monumental task of synthesizing all stable and repeating Life objects up to 14 bits, including over a thousand still-lifes. When I put up my Life synthesis web site in the late 1990s (initially based largely on his work), I had planned to extend this to the 15-bit still-lifes. Most of these appeared to have fairly obvious syntheses, but there were around half a dozen that for which syntheses did not seem at all obvious. By around 2001, I had reduced this down to three objects without syntheses. Around 2002, I wrote an expert system that attempts to find syntheses for given lists of objects, by using a brute-force search through a library of synthesis recipes (currently numbering over 800 recipes). By filtering out 9)%+ of trivial objects, this allows one to concentrate on the few difficult and interesting remaining objects. This was able to find explicit syntheses for all 15-bit still-lifes except twelve, for which I already had syntheses for all but the "hard three", which formally confirmed what I had already deduced informally. This list remained unchanged for almost a decade; I was finally able to solve one in 2011. Last year, I discovered the conwaylife.com forums, and mentioned the unsolved still-lifes. With a great deal of help from Martin Grant (aka extrementhusiast), we were able to finish the remaining two 15-bit still-lifes in September and October. This success spurred interest into the unsolved 16-bit still-lifes, of which there were slightly under a hundred. Martin, Matthias Merzenich and myself (which some additional help of a few others) were able to solve most of these by the end of the year, with the last few being solved in the first week of January. This momentum continued to the 396 unsolved 17-bit still-lifes, which which were finally completed on May 17 of this year. <clip --Rich> ------------------------------------ Posted by: Mark Niemiec <mniemiec(a)gmail.com> ------------------------------------ <*> To visit your group on the web, go to: http://groups.yahoo.com/group/LifeCA/ ----- End forwarded message -----

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Re: [math-fun] Spirography
by Dave Dyer 19 Jun '14

19 Jun '14

"Poly-cyclo-epicycloidal & Other Geometric Curves" By Edwin W. Alabone M.D.; D.Sc; F.R.M.S. &c. Second edition. 1912 This book is full of beautiful reproductions of spirograph patterns and other fascinating history of 19'th century spirography. One use was to make non-reproducable engravings which served as marks of authenticity on stock certificates. The non-reproducability being a side effect of the fact the no one could reverse engineer to figure out what settings were used to produce a particular pattern.

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[math-fun] Gears
by Warren D Smith 19 Jun '14

19 Jun '14

It seems unlikely to me that any functioning 2-dimensional "spur gears" exist such that it is impossible to separate two interlocked gears to arbitrarily far distances.

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[math-fun] Galperin "ball collision" method to calculate pi
by Warren D Smith 19 Jun '14

19 Jun '14

The method is: Stationary brick #1 has mass=1. It is hit by sliding brick #2 of mass=100^N, which pushes it to a wall. The total number of bounces off both the wall and off brick #2, experienced by brick #1, is an integer agreeing in base 10 with the first N digits of pi. (Assume Newton laws of motion, rigid balls, no gravity, no friction, instantaneous perfect elastic collisions. He also depends on a conjecture that pi does not ever have a huge number of 9s in a row; if that is untrue then this could occasionally get the last digit off by 1.) Actually, what is quite cool about this is that all the collisions (except perhaps the final one) happen in total time span of order 1 (assuming initial unit speed for brick #2, initial travel distance 1 for brick #1 to wall), no matter how massive brick #2 is. (Hint: think about the final 2 collisions.) Can anybody devise a "physical" way to compute e, depending perhaps on the fact e = lim (1+1/n)^n ? On 6/18/14, math-fun-request(a)mailman.xmission.com <math-fun-request(a)mailman.xmission.com> wrote: > Send math-fun mailing list submissions to > math-fun(a)mailman.xmission.com > > To subscribe or unsubscribe via the World Wide Web, visit > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > or, via email, send a message with subject or body 'help' to > math-fun-request(a)mailman.xmission.com > > You can reach the person managing the list at > math-fun-owner(a)mailman.xmission.com > > When replying, please edit your Subject line so it is more specific > than "Re: Contents of math-fun digest..." > > > Today's Topics: > > 1. Rising chain mystery (Dan Asimov) > 2. Re: secretary problem (Warren D Smith) > 3. Re: secretary problem (Eugene Salamin) > 4. Re: secretary problem (Dan Asimov) > 5. Re: Rising chain mystery (Whitfield Diffie) > 6. Re: keyword frequencies in programming languages > (Whitfield Diffie) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Wed, 18 Jun 2014 16:33:23 -0700 > From: Dan Asimov <dasimov(a)earthlink.net> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: [math-fun] Rising chain mystery > Message-ID: <3C7B52FF-81FE-43A0-9F10-C71E0F404D6D(a)earthlink.net> > Content-Type: text/plain; charset=us-ascii > > Don't recall seeing this mentioned on math-fun: > > In Jan., there was an article in the NY Times about how when a chain stuffed > into a container open at the top, and one end is tossed over the side with > lots of room to fall (e.g., letting it drop over a balcony), the chain falls > into a pattern with the chain initially rising surprisingly high before it > falls -- looking like a fountain. > > If you don't want to hear the explanation right away so you can think about > it, stop this video soon after 1:00. (The explanation begins soon after > 1:15.) > > http://www.nytimes.com/video/science/100000002747620/fountains-of-chain.htm… > (1 minute 47 seconds).* > > Even with the explanation, I still find this amazing. > > --Dan > __________________________________________ > * Unfortunately, to watch the video you apparently must watch the ad before > it, but I didn't find this too painful. > > ------------------------------ > > Message: 2 > Date: Wed, 18 Jun 2014 19:44:40 -0400 > From: Warren D Smith <warren.wds(a)gmail.com> > To: math-fun(a)mailman.xmission.com > Subject: Re: [math-fun] secretary problem > Message-ID: > <CAAJP7Y2x1gLG8rdBLQW7s-jwPYuOgoHhNqCG-x3_SuM2NUsapg(a)mail.gmail.com> > Content-Type: text/plain; charset=ISO-8859-1 > > Brent Meeker: > The version of this problem that I've heard is that there are N pieces > of paper, each with > a positive integer written on its face. They are lying face down on > the table. You chose > and discard them one at a time until you decide to keep the most > recently one chosen. > Your objective is to end up with the biggest number possible. Note > that the range of > numbers is not specified. > > WDS: > Asimov tried to formulate it where the numbers are chosen from some > arbitrary distribution > over the reals he called "F" and goal was to maximize expected value. > Meeker goal as hereby interpreted by me, is to minimize expected rank > (counting > from the greatest, with rank=1, down, with rank=2,3,...,N). > > The Asimov formulation sucks in the sense that for arbitrary F there > may not exist any expectation at all. E.g. Cauchy distribution has no > expectation value. > So for now let me just consider the rank formulation. > > If you've seen n numbers so far, 0<=n<N, the only question facing you > is "should I stop now > (getting as my reward, the rank of the last number seen, among the > full set of N numbers)?" > The answer is: > Stop if the expected rank of the present number, is less than the > expected rank > you will get by continuing play. > > Let E(m) = expected rank if stop after >=m numbers revealed, using > optimal play from > m onward. > > Let E(m,k) = expected rank if stop after revealing exactly m numbers, > and the last one is ranked k among those m, 1<=k<=m. > > E(N) = (1+N)/2. > > E(m,k) = Sum[ r * Binomial[r-1, k-1] * Binomial[N-r, m-k] / > Binomial[N, m-1] , {r, k, N+k-m} ] > in the notation of mathics.net . > Except mathics.net crashed when I asked it to evaluate this sum. > I presume E(m,k) = 1 + N*k/(m-1) at least approximately. > > Presumably best play is to stop when m=N (since you have to) or stop > if, earlier, > E(m,k)<E(m+1). > > Presumably when 0<=m<N > E(m) obeys the recurrence > > E(m) = (1/m) * Sum[ min( E(m+1), E(m,k) ), {k, 1, m} ] > > > > ------------------------------ > > Message: 3 > Date: Wed, 18 Jun 2014 17:20:25 -0700 > From: Eugene Salamin <gene_salamin(a)yahoo.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] secretary problem > Message-ID: > <1403137225.51249.YahooMailNeo(a)web162102.mail.bf1.yahoo.com> > Content-Type: text/plain; charset=iso-8859-1 > > A proper formulation of this problem must include the cost of picking up > another piece of paper.? Otherwise, since there are only finitely many > pieces of paper, if there's no cost, just pick up all of them. > > ? --? Gene > > > >>________________________________ >> From: Warren D Smith <warren.wds(a)gmail.com> >>To: math-fun(a)mailman.xmission.com >>Sent: Wednesday, June 18, 2014 4:44 PM >>Subject: Re: [math-fun] secretary problem >> >> >>Brent Meeker: >>The version of this problem that I've heard is that there are N pieces >>of paper, each with >>a positive integer written on its face.? They are lying face down on >>the table.? You chose >>and discard them one at a time until you decide to keep the most >>recently one chosen. >>Your objective is to end up with the biggest number possible.? Note >>that the range of >>numbers is not specified. >> >>WDS: >>Asimov tried to formulate it where the numbers are chosen from some >>arbitrary distribution >>over the reals he called "F" and goal was to maximize expected value. >>Meeker goal as hereby interpreted by me, is to minimize expected rank >> (counting >>from the greatest, with rank=1, down, with rank=2,3,...,N). >> >>The Asimov formulation sucks in the sense that for arbitrary F there >>may not exist any expectation at all.? E.g. Cauchy distribution has no >>expectation value. >>So for now let me just consider the rank formulation. >> >>If you've seen n numbers so far, 0<=n<N, the only question facing you >>is "should I stop now >>(getting as my reward, the rank of the last number seen, among the >>full set of N numbers)?" >>The answer is: >>? Stop if the expected rank of the present number, is less than the >>expected rank >>you will get by continuing play. >> >>Let E(m) = expected rank if stop after >=m numbers revealed, using >>optimal play from >>m onward. >> >>Let E(m,k) = expected rank if stop after revealing exactly m numbers, >>and the last one is ranked k among those m, 1<=k<=m. >> >>E(N) = (1+N)/2. >> >>E(m,k) = Sum[ r * Binomial[r-1, k-1] *? Binomial[N-r, m-k] / >>Binomial[N, m-1] ,? {r, k, N+k-m} ] >>in the notation of? mathics.net . >>Except mathics.net crashed when I asked it to evaluate this sum. >>I presume? E(m,k) = 1 + N*k/(m-1)? at least approximately. >> >>Presumably best play is to stop when m=N (since you have to) or stop >>if, earlier, >>E(m,k)<E(m+1). >> >>Presumably when 0<=m<N >>E(m) obeys the recurrence >> >>? ? E(m) = (1/m) *? Sum[ min( E(m+1), E(m,k) ),? ? {k, 1, m} ] >> >> > > ------------------------------ > > Message: 4 > Date: Wed, 18 Jun 2014 18:21:50 -0700 > From: Dan Asimov <dasimov(a)earthlink.net> > To: Eugene Salamin <gene_salamin(a)yahoo.com>, math-fun > <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] secretary problem > Message-ID: <EC5930C6-337C-4710-A6F7-0894645F88F9(a)earthlink.net> > Content-Type: text/plain; charset=iso-8859-1 > > Just to be clear, I didn't formulate any problem, just stated some facts > about distributions. > > --Dan > > > > > ------------------------------ > > Message: 5 > Date: Wed, 18 Jun 2014 18:24:54 -0700 > From: Whitfield Diffie <whitfield.diffie(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] Rising chain mystery > Message-ID: > <CAF+O-CXn6Lm2ijQ8jzcMOWwGPEV+JcTBe3cKP9_+Tqbj8AhTjQ(a)mail.gmail.com> > Content-Type: text/plain; charset=ISO-8859-1 > >> Even with the explanation, I still find this amazing. > > This is indeed wonderful. I wonder if it accounts for cases of > chains hopping off of pulleys. It seems that if the explanation is > correct, the phenomenon shold depend on the compressibility of the > chain; suppose it had long thing links? Or rubber links? > > > Whit > > > > ------------------------------ > > Message: 6 > Date: Wed, 18 Jun 2014 18:44:15 -0700 > From: Whitfield Diffie <whitfield.diffie(a)gmail.com> > To: Warren D Smith <warren.wds(a)gmail.com> > Cc: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] keyword frequencies in programming languages > Message-ID: > <CAF+O-CUPRcy0wERBi-jM6arfRXvHiSEUWiswOV8MHf0JssA+4A(a)mail.gmail.com> > Content-Type: text/plain; charset=ISO-8859-1 > >> ... all the counts by me and by Diffie are somewhat flawed due to our >> laziness about not really counting things "right". > > It wasn't laziness on my part (although I have plenty); I didn't > know what you wanted. It wouldn't be difficult to produce counts of > two sorts of primitives. I could do this study on my code, counting > only the lisp commands I did not write myself or I could do it on the > system code and count only the ones defined in C. That is easy to get > from the documentation system and I believe I can send a list without > much trouble. > > I believe that Allan Wechsler is right in thinking that lots of > the Elisp primitives are about the data structures. Weeding those out > would take more judgement. Presumably plus and times are about the > fact that a language does arithmetic. As I said earlier, I think a > count of words in Common Lisp would be rather different. > > > Whit > > > > ------------------------------ > > Subject: Digest Footer > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > ------------------------------ > > End of math-fun Digest, Vol 136, Issue 37 > ***************************************** > -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] keyword frequencies in programming languages
by Warren D Smith 19 Jun '14

19 Jun '14

> Allan Wechsler <acwacw(a)gmail.com> > Such a table fails to distinguish between programmer-defined items, and > ones that are really part of the language core. I get the impression you > are interested in the latter, although I confess that I'm not sure what the > agenda is. Are we supposed to learn something about the languages in > question? What? > All Lisp dialects have an intentionally vague boundary between terms that > are defined by the language, and those defined by the programmer. Another > way to put this is that all Lisp programs are Lisp language extensions. I > would argue, for example, that "point", which features prominently in the > list above, reveals something about the application (Emacs) rather than the > language (Emacs Lisp). "Save-excursion" -- even more so. > In Lisp, there is no syntactic clue that infallibly singles out language > primitives. What are we trying to find out? Perhaps there is a better way > than just counting lemmas; or perhaps, for Lisp at least, we are asking the > wrong question. --I agree with AW (and knew that already). And all the counts by me and by Diffie are somewhat flawed due to our laziness about not really counting things "right". What am I trying to figure out? Well, there are probably many uses for keyword frequency data from real programs. But just one is: if you are trying to design a new language, then this data will be useful to you. I am on a likely fool's errand of trying to, at least sketchily, design a new language, and if you want to correspond with me about that, go ahead...

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[math-fun] secretary problem
by Bernie Cosell 18 Jun '14

18 Jun '14

I was just reading a PDF about the secretary problem and all of the analyses I've ever seen seems to start with the assumption that the optimal strategy will necessarily be of the form "examine the first K candidates and then take the first candidate better than any you've seen so far", and if you haven't picked any candidate before that you MUST pick the N'th. And the math goes from there to derive the usual N/e value for K. What got me curious [and this has actually drifted in and out of my pondering for some time now] is whether there's a proof that includes proving that the *assumption* is correct, also. I could see a strategy, for example, where you examine K candidates and then take the *second* best [that is, wait until you get one better than K, and then take the one better than that second one]. There might be other strategies more subtle or something beyond my ken. I don't see how to prove the assumption: that that's the best *possible* selection strategy. ??? /Bernie\ -- Bernie Cosell Fantasy Farm Fibers mailto:bernie@fantasyfarm.com Pearisburg, VA --> Too many people, too few sheep <--

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[math-fun] why can a line of slope 10^n determine n decimal digits of π?
by Gary Antonick 18 Jun '14

18 Jun '14

Just noticed this yesterday and can't figure out why it works. Draw a circle with radius 1 around 0,0. Draw a line with slope -10 that hits the circle at 1,0. This line will also hit the circle again and make a little arc. The length of the arc will be about 1/31 the perimeter of the circle. Now try again with a line of slope -100. The line will create an arc 1/314 of the circle. Starting to look a bit like pi. Am wondering why this is. In case I made a mistake here's what I'm trying to describe.. if the circle were a clock the first line would cut an arc from 15 minutes (3:00) to almost the 13 minute mark.

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Re: [math-fun] secretary problem
by Warren D Smith 18 Jun '14

18 Jun '14

Brent Meeker: The version of this problem that I've heard is that there are N pieces of paper, each with a positive integer written on its face. They are lying face down on the table. You chose and discard them one at a time until you decide to keep the most recently one chosen. Your objective is to end up with the biggest number possible. Note that the range of numbers is not specified. WDS: Asimov tried to formulate it where the numbers are chosen from some arbitrary distribution over the reals he called "F" and goal was to maximize expected value. Meeker goal as hereby interpreted by me, is to minimize expected rank (counting from the greatest, with rank=1, down, with rank=2,3,...,N). The Asimov formulation sucks in the sense that for arbitrary F there may not exist any expectation at all. E.g. Cauchy distribution has no expectation value. So for now let me just consider the rank formulation. If you've seen n numbers so far, 0<=n<N, the only question facing you is "should I stop now (getting as my reward, the rank of the last number seen, among the full set of N numbers)?" The answer is: Stop if the expected rank of the present number, is less than the expected rank you will get by continuing play. Let E(m) = expected rank if stop after >=m numbers revealed, using optimal play from m onward. Let E(m,k) = expected rank if stop after revealing exactly m numbers, and the last one is ranked k among those m, 1<=k<=m. E(N) = (1+N)/2. E(m,k) = Sum[ r * Binomial[r-1, k-1] * Binomial[N-r, m-k] / Binomial[N, m-1] , {r, k, N+k-m} ] in the notation of mathics.net . Except mathics.net crashed when I asked it to evaluate this sum. I presume E(m,k) = 1 + N*k/(m-1) at least approximately. Presumably best play is to stop when m=N (since you have to) or stop if, earlier, E(m,k)<E(m+1). Presumably when 0<=m<N E(m) obeys the recurrence E(m) = (1/m) * Sum[ min( E(m+1), E(m,k) ), {k, 1, m} ]

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