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[math-fun] "Pushing" and "popping" for deductions about inter-referential sentences
by James Propp 16 Nov '16

16 Nov '16

I just realized (painfully close to my deadline) that my essay on self-referential sentences could really benefit from a terminological distinction that I may have to coin. It's the distinction between Deduction style #1: If P says "Q is true", and P is true, then Q is true. and Deduction style #2: If P says "Q is true", and Q is true, then P is true. I'm tempted to call them "pushing" and "popping", respectively. And now that I think about it, I'm wondering whether Hofstadter used exactly this nomenclature in "Godel, Escher, Bach". I can't decide whether it's merely the sort of thing Hofstadter would do, or something he actually did! And I can't seem to locate my copy of GEB. Can someone with a handy copy look up "push" and "pop" in the index? Thanks, Jim Propp

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[math-fun] Not particularly math, possibly fun, particularly particles
by Bill Gosper 15 Nov '16

15 Nov '16

Maybe we can at least get Minsky-style reversibility by updating the particles in reverse order if time runs negative. --rwg I somehow missed the Load from file button! My file is small enough to fit in a url. Is there a way for it to override the default initial configuration? Date: 2016-11-12 13:44 From: Mike Speciner <ms(a)alum.mit.edu> To: math-fun <math-fun(a)mailman.xmission.com> Reply-To: math-fun <math-fun(a)mailman.xmission.com> If you SHOW AS TEXT, you can cut and paste to save in a file of your choosing (and edit it as you choose). You can load such a file with the "Load from file" function. (The LOAD and SAVE functions use cookies, with a nonstandard encoding to reduce their size.) Good question about perfect computation with rationals. There are an awful lot of square roots in the calculations; I'm not sure the conditions under which they don't produce irrationals. I suspect one dimensional scenarios are more likely to be able to produce exact results. (It's been a long time since I looked at this code. I do remember deliberately not adding in gravity because that made reversibility harder. The PDP-6 version had gravity. And I vaguely remember having some code to make sure energy was [nearly] conserved, but I'm not sure it's in this new version.) --ms On 12-Nov-16 01:44, Bill Gosper wrote: >> Date: 2016-11-05 09:51 >> From: Mike Speciner <ms(a)alum.mit.edu> >> https://ms0.github.io/gas.html > This is a great toy! I delete everything except six identical balls (and > maybe one wall), set one ball > to bounce around for a while before disrupting the others, run it half a > minute into the disruption, save > it out, and show it time-reversed to some unsuspecting person. Or if I'm > afraid they'll notice the > negative time passage, I manually negate all the velocities before saving. > (Is there a way to load > one of those text files?) > > Unfortunately, neither of these reversal stunts is perfect, due to > accumulated numerical errors, > which I assume you've worked hard to minimize. > > So let's get particularly math-y and ask: Given rational positions and > velocities, can we simulate > one of these six-ball scenarios exactly and indefinitely, without runaway > denominator buildup, > nor irrational collision results? --rwg

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Re: [math-fun] Landau's function of 2^16,
by Tomas Rokicki 14 Nov '16

14 Nov '16

Sorry for the brevity I'm on a phone. This code computes g(n,p) which is the best g(n) where the factors are prime powers with the prime le p. It builds on this prime by prime. Then g(n,n) is g(n). On Nov 13, 2016 2:19 PM, "Tomas Rokicki" <rokicki(a)gmail.com> wrote: Read the code carefully. On Nov 13, 2016 2:17 PM, "Allan Wechsler" <acwacw(a)gmail.com> wrote: > Ooh, even worse: suppose N = p^k + M, and g(M) happens to be divisible by > p^k? Then the product p^k*g(M) would look like a really good candidate for > g(N), but would in fact have a spurious factor of p^k. > > On Sun, Nov 13, 2016 at 5:15 PM, Allan Wechsler <acwacw(a)gmail.com> wrote: > > > It looks like this code depends on a lemma whose truth is not obvious to > > me. > > > > "For any N, there exists a prime power p^k such that g(N) = g(N-p^k) * > > p^k." > > > > It is conceivable to me that for any component p^k of g(N), g(N-p^k) > might > > not be relatively prime to p^k. That is, it's conceivable that none of > the > > cofactors of prime powers in g(N) are themselves optimal in the Landau > > sense. I admit that it doesn't seem _likely_, because you would have to > > have bad luck for every component p^k in g(N). But still conceivable. Am > I > > missing an easy demonstration of this lemma? > > > > On Sun, Nov 13, 2016 at 12:34 PM, Tomas Rokicki <rokicki(a)gmail.com> > wrote: > > > >> Perhaps this code will be simple enough for most to follow. > >> > >> Note that I use doubles and don't report the actual partition. > >> This is trivial to do (just add an array storing the last prime > >> power that improved a result, and then walk backwards from > >> there). > >> > >> This code easily handles 65536 and gives 2.41074e+388. > >> Running it in Python (with its superior number stack) should > >> give you an exact result pretty quickly; I can provide that if > >> anyone is interested. > >> > >> for (int i=0; i<N; i++) > >> maxv[i] = 1 ; > >> for (int i=2; i<=N; i++) { > >> if (!isprime(i)) > >> continue ; > >> for (int j=N-1; j>=i; j--) { > >> long double hi = maxv[j] ; > >> for (long long pp=i; pp<=j; pp=pp*i) > >> hi = max(hi, maxv[j-pp]*pp) ; > >> maxv[j] = hi ; > >> } > >> } > >> for (int i=0; i<N; i++) > >> cout << i << " " << maxv[i] << endl ; > >> > >> > >> On Sun, Nov 13, 2016 at 9:09 AM, Tomas Rokicki <rokicki(a)gmail.com> > wrote: > >> > >> > This is a pretty common programming contest problem. > >> > You can make an additional restriction as follows: > >> > > >> > Every factor (except perhaps the last) should be a power of a > >> prime. > >> > (If a factor is composite, you can factor it into prime powers > and > >> > get equivalent impact with fewer base elements.) > >> > > >> > Then, you can solve this problem using dynamic programming. > >> > First you solve it for n=1..65536 using only powers of 2. Then based > >> > on that you solve it for n=1..65536 using powers of 2 and powers of > >> > 3. Then you extend that with powers of 5, and so on and so forth; > >> > pretty straightforward. > >> > > >> > You do need to use a programming language or library that supports > >> > bigints, or else you can represent the product by its logarithm. > >> > > >> > This is not considered a difficult problem for modern college > >> > programming contests. > >> > > >> > On Sun, Nov 13, 2016 at 4:02 AM, Fred Lunnon <fred.lunnon(a)gmail.com> > >> > wrote: > >> > > >> >> I'm suffering some disconnection here: compare JA below > >> >> > >> >> >> The upper bound used for the downward search is truly enormous for > >> n = > >> >> 2^16. > >> >> > >> >> with https://oeis.org/draft/A000793 comments > >> >> > >> >> >> Grantham mentions that he computed a(n) for n <= 500000. > >> >> > >> >> --- citing a 1995 reference given later. > >> >> > >> >> Yet 2^16 = 65536 < 500000 , and JG's hardware 20 years older! > >> >> > >> >> WFL > >> >> > >> >> > >> >> > >> >> On 11/13/16, Joerg Arndt <arndt(a)jjj.de> wrote: > >> >> > * Allan Wechsler <acwacw(a)gmail.com> [Nov 13. 2016 08:33]: > >> >> >> [...] > >> >> >> > >> >> >> (Of all the code snippets in the relevant page at A000793, all of > >> them > >> >> >> except one are clearly just enumerating partitions and comparing > >> LCM's. > >> >> >> But > >> >> >> Charles Greathouse's PARI code is not doing anything like that, > and > >> I > >> >> >> don't > >> >> >> have a clue what it *is* doing.) > >> >> > > >> >> > Charles' code seems to use what I just entered as a formula in > >> A000793: > >> >> > For n>=2, A000793(n) = max_{k} A008475(k) <= n. > >> >> > Cf. https://oeis.org/draft/A000793 > >> >> > > >> >> > Is this correct? > >> >> > > >> >> > The upper bound used for the downward search is truly enormous for > n > >> = > >> >> > 2^16. > >> >> > > >> >> > > >> >> >> _______________________________________________ > >> >> >> math-fun mailing list > >> >> >> math-fun(a)mailman.xmission.com > >> >> >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > >> >> > > >> >> > _______________________________________________ > >> >> > math-fun mailing list > >> >> > math-fun(a)mailman.xmission.com > >> >> > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > >> >> > > >> >> > >> >> _______________________________________________ > >> >> math-fun mailing list > >> >> math-fun(a)mailman.xmission.com > >> >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > >> >> > >> > > >> > > >> > > >> > -- > >> > -- http://cube20.org/ -- [ <http://golly.sf.net/>Golly link > >> > suppressed; ask me why] -- > >> > > >> > >> > >> > >> -- > >> -- http://cube20.org/ -- [ <http://golly.sf.net/>Golly link > >> suppressed; > >> ask me why] -- > >> _______________________________________________ > >> math-fun mailing list > >> math-fun(a)mailman.xmission.com > >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > >> > > > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >

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[math-fun] Draft of my November 2016 blog post
by James Propp 12 Nov '16

12 Nov '16

I started writing a new draft titled "Breaking Logic with Self-Referential Sentences" and would love to get your feedback. I plan on publishing it on the 17th and will continue to tinker with it over the coming week. Suggestions for references, and comments of all kinds, are welcome. I appreciate candid criticism from people who are sympathetic to my aims but think I've fallen short of them. My prose style is designed to make the material accessible, but my breeziness should not be interpreted as indicating indifference to scholarly correctness. I'm especially in need of a good illustration or two for this post. I like each of my essays to contain at least one visual. And I *don't* want my readers to have to see Escher's "Drawing Hands" for the gazillionth time. Surely there's something less familiar that I could include. Please leave your feedback at https://mathenchant.wordpress.com?p=1338&shareadraft=581cfacf126c7 keeping in mind that the comments of EVERYONE on math-fun will get funneled through the same email conduit; if you want me to know who you are, please sign your comments! If you sign your comments I'll assume (unless you indicate otherwise) that you don't mind my acknowledging your contribution. Title: Breaking logic with self-referential sentences Beginning: I don't especially like people who talk a lot about themselves, but I have a soft spot for sentences that do. Case in point: the self-referential sentence "This sentence is false." I really like that one. You may think you've seen this sentence before, but in fact you've seen other sentences, each m... Read more: https://mathenchant.wordpress.com?p=1338&shareadraft= 581cfacf126c7 Thanks, Jim Propp

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Re: [math-fun] Landau's function of 2^16
by Dan Asimov 12 Nov '16

12 Nov '16

An interesting question! Did not know of Landau's function. Any partition with repeated sizes has the same LCM as if all the extra duplicates are replaced by ones. So at least partitions with repetition — except for ones — can be ignored. Yikes, this seems a Herculean task. Can anyone come up with reasonable upper or lower bounds, even? Or some kind of asymptotic bounds? Here is at least some relevant info: https://oeis.org/A000793. —Dan ----- . . . a reasonable way to evaluate Landau's function (A000793) at 2^16 = 65536. This is the maximum least common multiple of any partition of 65536; equivalently it is the maximum order of any element of Sym(65536). I am sure there are lots of shortcuts, that save one from enumerating all the partitions of 65536. Can anyone suggest one? . . . -----

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[math-fun] Not particularly math, possibly fun, particularly particles
by Bill Gosper 12 Nov '16

12 Nov '16

>Date: 2016-11-05 09:51 >From: Mike Speciner <ms(a)alum.mit.edu> >https://ms0.github.io/gas.html This is a great toy! I delete everything except six identical balls (and maybe one wall), set one ball to bounce around for a while before disrupting the others, run it half a minute into the disruption, save it out, and show it time-reversed to some unsuspecting person. Or if I'm afraid they'll notice the negative time passage, I manually negate all the velocities before saving. (Is there a way to load one of those text files?) Unfortunately, neither of these reversal stunts is perfect, due to accumulated numerical errors, which I assume you've worked hard to minimize. So let's get particularly math-y and ask: Given rational positions and velocities, can we simulate one of these six-ball scenarios exactly and indefinitely, without runaway denominator buildup, nor irrational collision results? --rwg

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[math-fun] The base 2-i, digits 0,i^0..3 spacefill
by Bill Gosper 11 Nov '16

11 Nov '16

(whose?) underlying the pretty gosper.org/dana.PNG has only moderately interesting Fourier gosper.org/rt5-324.png and polygonal gosper.org/rt5.png approximations. --rwg

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[math-fun] Differential Privacy and Poll Sampling
by Henry Baker 11 Nov '16

11 Nov '16

The pollsters in the recent presidential election didn't do a very good job of predicting the result. Apparently, one of the better polls -- the LATimes poll -- was heavily criticized for having one sample that appeared to be an outlier, but who affected the result quite significantly in a way that produced (in this particular case) a more *correct* prediction. It occurred to me that it might be a good idea to use some of the ideas from *differential privacy* to reduce the possibility of having a single sample point significantly affect the result. After all, the whole point of differential privacy is to make sure that a single sample point can't be identified using a small number of questions. The polls have also been criticized for "oversampling" -- i.e., applying the wrong weighting factors to different sample points. So, my modest proposal is to *oversample* rare combinations, but then *underweight* these oversamplings in such a way as to produce the overall correct weighting. The amount of oversampling should be calculable from the basic equations of differential privacy, which try to minimize any change due to the inclusion or elimination of any single sample point. https://en.wikipedia.org/wiki/Differential_privacy "The definition [of differential privacy] gives a strong guarantee that presence or absence of an individual will not affect the final output of the algorithm significantly."

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[math-fun] The bombshell from yesterday
by Henry Baker 09 Nov '16

09 Nov '16

No, not THAT bombshell, but this one, regarding *dark matter* and the Holographic Principle: https://science.slashdot.org/story/16/11/09/0031212/new-theory-of-gravity-m… https://arxiv.org/pdf/1611.02269.pdf New Theory of Gravity Might Explain Dark Matter (phys.org) 111 Posted by BeauHD on Wednesday November 09, 2016 @02:00AM from the spacetime dept. vikingpower writes: Dutch prodigy and Amsterdam University Professor Erik Verlinde published a paper on arXiv yesterday, November 7, titled "Emergent Gravity and the Dark Universe." In the paper, Verlinde derives gravity from the so-called Holographic Principle, which -- simply put -- states that gravity emerges from the interplay between and entropy re-arrangement of sub-atomic "strings" that live in a negatively curved spacetime. At that level [...] spacetime and gravity are emergent from an underlying microscopic description in which they have no a priori meaning." Most importantly, Verlinde's paper has as a consequence that dark matter, nemesis of many an astronomer, is nothing more than an illusion.

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[math-fun] A visual riddle (hard)
by Joerg Arndt 09 Nov '16

09 Nov '16

The arrangement in the square grid shown below has four-fold rotational symmetry and none of its squares (apart from the central one) has more than two neighbors. Of course there are very many such arrangements. This one, however, has a very special property, unique within all such(*) arrangements of 289 squares. (*) edge-to-edge connected, four fold symmetry, and fewest possible number of neighbors. Which? > ................................................... > ................................................... > ................................................... > ................................................... > ..........................#####.................... > ..........................#...#.................... > ..........................#...#.................... > ..........................#...#.................... > ..........................#...#.................... > ..........................#...#.................... > ..........................#...#.................... > ..................#########...#.................... > ..................#...........########............. > ............###...#..................#............. > ............#.#...#..........#########............. > ............#.#...#..........#..................... > ............#.#...#..........#..................... > ............#.#...########...#..................... > ............#.#..........#...#...#######........... > ............#.#..........#...#...#.....#........... > ....#########.#..........#...#...#.....#........... > ....#.........########...#...#...#.....#........... > ....#....................#.......#.....#........... > ....#....................#.......#.....#........... > ....########.............#.......#.....#........... > ...........#.....#################.....#........... > ...........#.....#.......#.............########.... > ...........#.....#.......#....................#.... > ...........#.....#.......#....................#.... > ...........#.....#...#...#...########.........#.... > ...........#.....#...#...#..........#.#########.... > ...........#.....#...#...#..........#.#............ > ...........#######...#...#..........#.#............ > .....................#...########...#.#............ > .....................#..........#...#.#............ > .....................#..........#...#.#............ > .............#########..........#...#.#............ > .............#..................#...###............ > .............########...........#.................. > ....................#...#########.................. > ....................#...#.......................... > ....................#...#.......................... > ....................#...#.......................... > ....................#...#.......................... > ....................#...#.......................... > ....................#...#.......................... > ....................#####.......................... > ................................................... > ................................................... > ................................................... > ................................................... Best regards, jj Hints below, after some several blank lines. Hints: A) 289 = 15^2 + 8^2 B) Consider two copies of the arrangement C) Consider 289 copies of the arrangement P.S.: almost-spoiler at http://jjj.de/visual-riddle/visual-riddle.png

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