December 2011 - mathfun@mailman.xmission.com

Re: [math-fun] math-fun Digest, Vol 106, Issue 11
by Christian G. Bower 07 Dec '11

07 Dec '11

1, 1, 2, 3, 4, 5, 8, 10, 16, 23, 36, 53, 88, 135, 225, 366, 616, 1025, 1759, 2987, 5174, 8929, 15578, 27158, 47703, ... and yes not in OEIS. Perhaps it would be more correct to call them "bracelet" partitions rather than necklace partitions since you are allowing these necklaces to be "turned over." The necklace version is A093305. As far as a program: consider the one I wrote in PARI for the OEIS https://oeis.org/transforms_pari.txt ? V=vector(25) %1 = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ? V[2]=V[3]=V[4]=1 %2 = 1 ? trv_polygon(V) %3 = [1, 1, 2, 3, 4, 5, 8, 10, 16, 23, 36, 53, 88, 135, 225, 366, 616, 1025, 1759, 2987, 5174, 8929, 15578, 27158, 47703] ------ Original Message ------ Received: Wed, 07 Dec 2011 01:13:39 PM PST From: Allan Wechsler <acwacw(a)gmail.com> To: math-fun <math-fun(a)mailman.xmission.com> Subject: Re: [math-fun] math-fun Digest, Vol 106, Issue 11 > Excellent, did you use a program or just debug my attempt by eye? > > On Wed, Dec 7, 2011 at 3:26 PM, Veit Elser <ve10(a)cornell.edu> wrote: > > > For n=8 you left out 1223. I got a(8)=16 and a(9)=23. This sequence is not > > in OEIS. > > > > Veit > > > > On Dec 7, 2011, at 2:28 PM, Allan Wechsler wrote: > > > > > Define a(n) as the number of "necklace partitions" of n with elements not > > > exceeding 3. > > > > > > If my calculations are right, then OEIS does not contain this sequence. > > I > > > have: > > > > > > [1]: a(1) = 1 > > > > > > [11, 2]: a(2) = 2 > > > > > > [111, 12, 3]: a(3) = 3 > > > > > > [1111, 112, 13, 22]: a(4) = 4 > > > > > > [11111, 1112, 113, 122, 23]: a(5) = 5 > > > > > > [111111, 11112, 1113, 1122, 1212, 123, 222, 33]: a(6) = 8 > > > > > > [1111111, 111112, 11113, 11122, 11212, 1123, 1213, 1222, 133, 223]: a(7) > > = > > > 10 > > > > > > [11111111, 1111112, 111113, 111122, 111212, 11123, 112112, 11213, 11222, > > > 1133, 12122, 1232, 1313, 2222, 233]: a(8) = 15 > > > > > > [111111111, 11111112, 1111113, 1111122, 1111212, 111123, 1112112, 111213, > > > 111222, 11133, 112113, 112122, 11232, 11313, 121212, 12132, 12213, 12222, > > > 1233, 1323, 2223, 333]: a(9) = 22 > > > > > > And OEIS returns no hits for 1,2,3,4,5,8,10,15,22. Perhaps I've > > > miscounted, and I would appreciate confirmation before we submit a new > > > sequence. > > > > > > On Wed, Dec 7, 2011 at 12:31 PM, Dan Asimov <dasimov(a)earthlink.net> > > wrote: > > > > > >> Another math-fun member, using a brief Mathematica program, > > >> also found 88. It looks as though that's the correct answer. > > >> > > >> (Though, I haven't yet compared the two sets of output, the other > > >> of which is in terms of 1's, 2's, and 3's.) > > >> > > >> --Dan > > >> > > >> > > >> Tom Duff wrote: > > >> > > >> << > > >> On Tue, 6 Dec 2011, math-fun-request(a)mailman.xmission.com wrote: > > >> > > >>> Date: Tue, 6 Dec 2011 16:17:33 -0800 (GMT-08:00) > > >>> From: Dan Asimov <dasimov(a)earthlink.net> > > >>> To: math-fun <math-fun(a)mailman.xmission.com> > > >>> Subject: Re: [math-fun] Counting problem > > >>> Message-ID: > > >>> < > > >> > > 10736092.1323217054054.JavaMail.root(a)elwamui-norfolk.atl.sa.earthlink.net> > > >>> > > >>> Content-Type: text/plain; charset=UTF-8 > > >>> > > >>> Ha -- and in my spare moments today I counted 87 (by hand). > > >>> > > >>> So, can we split the difference and agree on 86 ? > > >>> > > >>> --Dan > > >>> > > >>> > > >>> Allan wrote: > > >>> > > >>> << > > >>> As luck would have it, this afternoon I had a boring staff meeting in > > >> which > > >>> I enumerated all of them ... I think. And the answer is 85 ... I > > think. > > >> > > >> I found 88. Here they are. Where is my mistake? > > >> (Each string has 3 for a 30 degree angle, 6 for 60, 9 for 90. > > >> These are the lexicographically smallest representatives > > >> of their equivalence classes.) > > >>>> > > >> > > >> > > >> > > ________________________________________________________________________________________ > > >> It goes without saying that > > >> . > > >> > > >> _______________________________________________ > > >> math-fun mailing list > > >> math-fun(a)mailman.xmission.com > > >> http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > >> > > > _______________________________________________ > > > math-fun mailing list > > > math-fun(a)mailman.xmission.com > > > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > > > > _______________________________________________ > > math-fun mailing list > > math-fun(a)mailman.xmission.com > > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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Re: [math-fun] Life's Adam & Eve
by Dan Asimov 07 Dec '11

07 Dec '11

I wrote: << By the way, so far we've been speaking only about finite configurations. But there's another problem of at least theoretical interest: Does there exist a planar pattern (i.e., a map Z^2 ->{0,1}) that has every reasonable planar pattern as a descendant? Perhaps reasonable can be defined as those planar patterns each having arbitrarily long lines of ancestors. Or, what is a priori a different condition, maybe reasonable should mean those planar patterns each having at least one infinite line of ancestors. >> Both of these options are impossible. First tile the plane by NxN squares for big enough N. By using either a small fixed finite periodic configuration, or a cemetery, in each NxN square, this shows the existence of uncountably many maps Z^2 -> {0,1} each having at least one infinite line of ancestors, which covers both questions. But any one map Z^2 -> {0,1} can only have countably many descendants. QED --Dan ________________________________________________________________________________________ It goes without saying that .

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Re: [math-fun] [EXTERNAL] Re: spherical pendulums?
by Henry Baker 07 Dec '11

07 Dec '11

Several years ago, I was considering small perturbations of geostationary orbits. Assuming that the satellites are small enough not to have any attraction for one another, the orbits are isochronous only when their semimajor axes are the same length. So if one is in a circular orbit, and the other is in a very slightly elliptical orbit, then one will act as an approximately day-long yoyo, going down & then coming back up, but never going past on the up portion. Of course, the perturbations from variations in the Earth's gravity are more significant; apparently, Mt Everest pulls enough harder to cause all the geostationary objects to eventually park themselves above the Indian Ocean. Q: suppose 2 approx geostationary objects are in the same plane with the same (extremely small) eccentricity, but their semimajor axes are at right angles to one another. Furthermore, the objects are never very far apart -- e.g., max 1 mile. In the frame of reference of one of the objects, what path does the other object take? At 07:03 AM 12/7/2011, Cordwell, William R wrote: >For the cone, if the particle is in a stable circle, I would expect that a small impulse (down or up) would give it an oscillatory behavior, similar to being in an orbit with constant angular momentum, with the resulting ellipse being viewed as an oscillation about the circular orbit. Harmonic is not clear. > >-----Original Message----- >From: math-fun-bounces(a)mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Henry Baker >Sent: Monday, December 05, 2011 8:28 AM >To: Bill Gosper >Cc: math-fun(a)mailman.xmission.com >Subject: [EXTERNAL] Re: [math-fun] spherical pendulums? > >Here's a related problem: > >A point particle of mass M (i.e., zero moment of inertia) is sliding frictionlessly around inside a vertical cylinder; gravity is downwards of strength G. What are the paths? > >Ditto with a vertical cone instead of a cylinder. > >At 12:44 AM 12/5/2011, Bill Gosper wrote: >>Are they chaotic or just hairy? The Wolfram demonstration seems quasiperiodic.? >>Gene once disabused me of the folly of trying to resolve the motion into x and y. >>But suppose we hung the string from the inside of an upward cusp of a >>cycloid of revolution. Would the pendulum simply describe an ellipse? >>--rwg

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Re: [math-fun] spherical pendulums?
by Henry Baker 07 Dec '11

07 Dec '11

Here's a related problem: A point particle of mass M (i.e., zero moment of inertia) is sliding frictionlessly around inside a vertical cylinder; gravity is downwards of strength G. What are the paths? Ditto with a vertical cone instead of a cylinder. At 12:44 AM 12/5/2011, Bill Gosper wrote: >Are they chaotic or just hairy? The Wolfram demonstration seems quasiperiodic.? >Gene once disabused me of the folly of trying to resolve the motion into x and y. >But suppose we hung the string from the inside of an upward cusp of a cycloid of >revolution. Would the pendulum simply describe an ellipse? >--rwg

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[math-fun] Life universality -- the nature of the proof
by Warren Smith 07 Dec '11

07 Dec '11

The construction by Conway of a universal computer inside LIFE (it's in his book with Berlekamp & Guy on games) had a (1) a processor and (2) a memory, and the memory was extendible (via software command) arbitrarily much. But it also could fire gliders into (3) a third region at, I think, arbitrary delays (controllable in software) and at arbitrary (achievable) angles, this region again being extendible via software command to make it grow as large as desired. Thus it was both a universal computer with infinite memory (despite initial state being finitely bounded), and also a glider "experimental particle physicist." Of course, region (3) could not be "isolated" from regions (1) and (2) because no known isolation method has ever been invented. Conway's construction was extremely inefficient, but faster versions with only polynomial slowdown not ultra-exponential, were cooked up later by others, who also worked out full details (Conway did not). Incidentally this computer can self-reproduce like actual "life" if it wants. My "immortality" question seems basically equivalent to whether it is possible to make the Conway computer "error correcting" and "fault tolerant" enough to probably last exponentially long even in the presence of "noise." This question, as far as I know, is wide open.

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[math-fun] L'Hospital's Rule
by James Propp 07 Dec '11

07 Dec '11

A student in my honors calculus class just showed me his solutions to two homework problems I assigned; in each case he got the right answer for the wrong reason. I thought I could improvise an example to show him why his method (aside from including steps with no logical justification) would give the wrong answer if he tried applying it to other problems, but I failed, and I'm wondering if someone can provide such an example, or else explain why no such example exists. The first homework problem was to evaluate the limit of (f(x+h)-f(x-h))/2h as h approaches 0. The student differentiated on top and bottom with respect to x (rather than h), obtaining lim (f'(x+h)(1+h')-f'(x-h)(1-h'))/2h' where h'=dh/dx. Substituting h=0 inside the limit gives (f'(x)(1+h')-f'(x)(1-h'))/2h' which simplifies to f'(x). The second homework problem was to evaluate the limit of (f(x+h)-2f(x)+f(x-h))/h^2 as h approaches 0. The student again differentiated on top and bottom with respect to x (rather than h), obtaining lim (f'(x+h)(1+h') - 2f'(x) + f'(x-h)(1-h'))/2hh' The student then "applied L'Hospital" again to rewrite the limit as lim (f"(x+h)(1+h')(1+h') + f'(x+h)(h") - 2f"(x) + f"(x-h)(1-h')(1-h') + f'(x-h)(-h"))/(2hh"+2h'h'). Substituting h=0 inside the limit gives (f"(x)(1+h')(1+h') + f'(x)(h") - 2f"(x) + f"(x)(1-h')(1-h') + f'(x)(-h"))/(2h'h') which simplifies to f"(x). The student half-understood my explanation of why both derivations are wrong, but I don't think he'll fully accept that his method is wrong until I show him an example where it leads to a wrong answer. I just sent him an email, asking him to use his method to compute the limit of (x+2h)/(x+h) as h approaches 0. I suspect that he'll either get 1 or 1/2 as his answer, as opposed to the correct answer, which is "1 if x is non-zero, 2 if x is zero." But then he'll still feel that his approach has merit, because it gives the right answer for at least some values of x. Is there a better example that shows that taking derivatives with respect to the wrong variable can give really, really wrong answers? Jim Propp

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Re: [math-fun] Life's Adam & Eve
by Dan Asimov 07 Dec '11

07 Dec '11

Scott wrote: << Isn't the existence of Garden of Eden configurations an easy proof of non-existence of Adam & Eve configurations? >> Yes. Duhhhhhh. << FWIW, it feels like a direct Adam/Eve nonexistence proof might be simpler than a direct Garden of Eden existence proof. >> Good point. Now let me make a feeble attempt to rescue the Adam & Eve question. Consider those patterns each having an infinite sequence of ancestors. Is there a starting pattern for which every one of these eventually appears somewhere among its descendants? (Alternatively, consider each pattern that, for every N in Z+, has a sequence of N ancestors. Same question.) --Dan ________________________________________________________________________________________ It goes without saying that .

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Re: [math-fun] Life's Adam & Eve
by Dan Asimov 07 Dec '11

07 Dec '11

By the way, so far we've been speaking only about finite configurations. But there's another problem of at least theoretical interest: Does there exist a planar pattern (i.e., a map Z^2 ->{0,1}) that has every reasonable planar pattern as a descendant? Perhaps reasonable can be defined as those planar patterns each having arbitrarily long lines of ancestors. Or, what is a priori a different condition, maybe reasonable should mean those planar patterns each having at least one infinite line of ancestors. --Dan ________________________________________________________________________________________ It goes without saying that .

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[math-fun] Life experiments
by rcs＠xmission.com 06 Dec '11

06 Dec '11

I did a few experiments in Conway Life a long time ago. I tried adding a low-probability of error to the rule -- after the regular rule executed, I modified each cell with a small probability. I've forgotten most of the details, but generally .... An initially 50% random bit field decays to about 2.8% of the cells ON. The timescale is very roughly ~10K generations. The final percentage of ON cells has a weak dependence on the starting %. (If P is near 0 or 1, the field is blank after a few steps.) [Several Lifers have done extensive experiments, measuring the probability of various patterns in the 'final' configurations.] If ~.01% of the cells are randomly forced to OFF after each time step, or, " " " forced to ON, or, " " " complemented, the eventual outcome is a blank pattern. I did one experiment about the speed and distance of 'change propagation': I made two random 8Kx8K fields, with the left halves matching and the right halves random. I ran a number of steps, and looked at how well the left halves matched. The 'change front' propagated about 500 cells into the left half, gradually slowing, and seemed to have roughly stabilized after O(10K) steps. Rich

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[math-fun] Life adam & eve
by Warren Smith 06 Dec '11

06 Dec '11

mbgreen(a)cis.upenn.edu: I have seen the claim in several posts that because Life is Turing complete, you can program a "Life Computer" to enumerate all possible configurations. Is this true? --yes. You can program it to do any computation. At all. Andy Latto: When I say that a Life computer could "enumerate all finite life configurations", I mean that it could enumerate them in symbolic form, which would need to be decoded. Translating this symbolic form into an actual realization of the configuration is what I was referring to as producing a "printer" for the computer. --right. Andy Latto has now in several posts, tried to re-say what I already had said in my post. Glad to see great minds think alike. Latto: Thinking about it a bit more, To produce all produceable configurations (those with ancestors arbitrarily far back in time), I don't think the right approach would be "for each configuration, try to find a configuration of gliders that will produce it"; this sounds like a uncomputable task. Instead, we could have the Life Turing machine enumerate all the configurations of gliders, and try them all. --exactly. As I'd already said. Latto: Except that it would have to exclude those configurations that would destroy the computer, and that may be an insoluble problem. --no it is not, as I'd already said; computer can enumerate all configs both synthesizable by, and killable by, gliders. Then as it notices them, it synthesizes and then kills them. Latto: If we only cared about keeping the computer alive for a finite amount if time, the Life-Turing-machine could simulate the glider configuration, and only actually execute it if simulation proved that it didn't destroy the computer in that amount of time. But since we want to produce all of the infinitely many configurations, we want to keep the computer alive forever, and I suspect that the question "Will this life configuration over here ever reach this spot over here?" is uncomputable. --it does not matter that it is uncomputable. We simply enumerate them all. It is like this: we can enumerate and generate all Turing machines that halt, and only those that halt -- even though we cannot decide, given any one, whether it halts. Can you see that? Latto: So I'm not sure an Adam & Eve configuration exists, or at any rate, this angle of attack will succeed in proving that it does. --It's annoying to see this stream of posts from Andy regurgitating my own post rather delayed and imperfectly, but nice to see he's think on the same line, I guess. As I said, a single "genesis seed" configuration exists that will generate every configuration of every finite size that is both synthesizable by and killable by, gliders.

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