April 2018 - mathfun@mailman.xmission.com

[math-fun] YouTube: Making a sphere on a lathe
by Henry Baker 21 Apr '18

21 Apr '18

FYI -- several months ago, we discussed how to make a spherical object with a lathe, as might have been done circa 1500AD. Here's an 11-minute YouTube video showing exactly this process: https://www.youtube.com/watch?v=hZT0ZTvRFxI #40 Hybrid Sphere Burl & Resin "Red Dawn"

1 0

[math-fun] Maximal density of finite patters in Z/nZ
by Dan Asimov 21 Apr '18

21 Apr '18

Recent posts reminded me of this question first suggested to me by my late brother Simon (1955-2005). Given an integer n > 0, what are the finite patterns that can partition the discrete circle C_n = Z/nZ ??? A finite pattern is simply a finite subset [X] of C_n = Z/nZ *up to isometry* I.e., X and x + K are considered the same subset of X, via the corresponding equivalence relation on certain subsets of the power set P(C_n) of C_n. In this case, *translations* of A finite pattern [X] in C_n "can partition C_n" if: *there exists for some k >= 1 a finite sets of isometric copies X_j \sub C_n, j-1..., k of X in C_n, such that a) the X_j are disjoint, 1 <= j <= n and b) the union of the X_j is all of C_n. Question: --------- For each n, classify the finite patterns that can partition C_n. —Dan

1 0

Re: [math-fun] Optimally sorting a cyclic group
by Henry Baker 20 Apr '18

20 Apr '18

Yes, a radix-style sort will work; I had forgotten about that. I'm working on a model where every access is expensive, and RAM access is not even remotely constant time. In this particular problem, I can *preprocess for free* to determine the complete permutation, but then I have to pay full retail price for actually applying this permutation to rearrange the data records. I was hoping to minimize rearrangements in the cases where the data was already "almost sorted". I was thinking along the lines of Quicksort, which is pretty good when the data is already almost sorted. At 10:05 AM 4/20/2018, Tomas Rokicki wrote: >A linear walk can notice and preserve the cyclic walk else abort. For what >disorder distribution do you want to do better? > >If you have such a field as you describe radix sort is linear time, as >discussed in TAOCP. > >On Fri, Apr 20, 2018 at 12:56 PM Henry Baker <hbaker1(a)pipeline.com> wrote: >> Suppose you have N records in which one of >> the fields is a number 0..(N-1), and there >> is exactly one record for each number >> 0..(N-1). >> >> It is trivial to sort such a set using >> traditional Knuth-TAOCP algorithms. >> >> However, if I don't care that the first >> record is labelled "0", but only care >> that the final ordering preserves the >> *cyclic* ordering of the group, can I >> do any better? >> >> Obviously, I can't do a lot better, >> since cyclicly rotating the ordering >> costs O(N), while sorting costs O(NlogN), >> but is there a clever way to sort w/o >> putting "0" at the front and "(N-1)" >> at the end? In particular, is there >> a way to sort that automatically >> notices when the records are already >> in cyclic order, and doesn't bother >> making any further changes. >-- http://cube20.org/ -- http://golly.sf.net/ --

4 6

Re: [math-fun] Optimally sorting a cyclic group
by Dan Asimov 20 Apr '18

20 Apr '18

Let C_n denote the cyclic group Z/nZ, and let s : C_n —> C_n be any permutation of C_n. Question: Is there in some sense always a "nearest" rotation of C_n (i.e., a translation by a group element) to the permutation s, in a consistent manner. I.e., such that for any rotation R of C_n we have nearest(s) = R*nearest(R*s). or in group notation nearest(s) + g = g + nearest(L_g+s) That is, the nearest rotation to [a permutation s followed by a rotation R] is the result of applying the rotation R to [the nearest rotation to the permutation s]. —Dan Henry Baker wrote: ----- Suppose you have N records in which one of the fields is a number 0..(N-1), and there is exactly one record for each number 0..(N-1). ... However, if I don't care that the first record is labelled "0", but only care that the final ordering preserves the *cyclic* ordering of the group, can I do any better? -----

1 0

[math-fun] Optimally sorting a cyclic group
by Henry Baker 20 Apr '18

20 Apr '18

Suppose you have N records in which one of the fields is a number 0..(N-1), and there is exactly one record for each number 0..(N-1). It is trivial to sort such a set using traditional Knuth-TAOCP algorithms. However, if I don't care that the first record is labelled "0", but only care that the final ordering preserves the *cyclic* ordering of the group, can I do any better? Obviously, I can't do a lot better, since cyclicly rotating the ordering costs O(N), while sorting costs O(NlogN), but is there a clever way to sort w/o putting "0" at the front and "(N-1)" at the end? In particular, is there a way to sort that automatically notices when the records are already in cyclic order, and doesn't bother making any further changes.

3 2

[math-fun] approximation
by françois mendzina essomba2 19 Apr '18

19 Apr '18

Hello.... I noted that : abs(sqrt(1-x^2)/acos(x))=((x+1)/2)*(1/2+1/sqrt(2+2*x))*(1/2+1/sqrt(2+sqrt(2+2*x)))*(1/2+1/sqrt(2+sqrt(2+sqrt(2+2*x)))) *(1/2+1/sqrt(2+sqrt(2+sqrt(2+sqrt(2+2*x)))))*(1/2+1/sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+2*x)))))) * cdots ; abs(x/asin(x))=((sqrt(1-x^2)+1)/2)*(1/2+1/sqrt(2+2*sqrt(1-x^2)))*(1/2+1/sqrt(2+sqrt(2+2*sqrt(1-x^2))))*(1/2+1/sqrt(2+sqrt(2+sqrt(2+2*sqrt(1-x^2))))) *(1/2+1/sqrt(2+sqrt(2+sqrt(2+sqrt(2+2*sqrt(1-x^2)))))) *cdots ; abs(x/asinh(x))=((sqrt(1+x^2)+1)/2)*(1/2+1/sqrt(2+2*sqrt(1+x^2)))*(1/2+1/sqrt(2+sqrt(2+2*sqrt(1+x^2))))*(1/2+1/sqrt(2+sqrt(2+sqrt(2+2*sqrt(1+x^2))))) *(1/2+1/sqrt(2+sqrt(2+sqrt(2+sqrt(2+2*sqrt(1+x^2)))))) *cdots ; abs((x/sqrt(1+x^2))*(1/atan(x)))=((1/sqrt(1+x^2)+1)/2)*(1/2+1/sqrt(2+2*(1/sqrt(1+x^2))))*(1/2+1/sqrt(2+sqrt(2+2*(1/sqrt(1+x^2)))))*(1/2+1/sqrt(2+sqrt(2+sqrt(2+2*(1/sqrt(1+x^2))))))*(1/2+1/sqrt(2+sqrt(2+sqrt(2+sqrt(2+2*(1/sqrt(1+x^2))))))) * cdots ; abs((x/sqrt(1-x^2))*(1/atanh(x)))=((1/sqrt(1-x^2)+1)/2)*(1/2+1/sqrt(2+2*(1/sqrt(1-x^2))))*(1/2+1/sqrt(2+sqrt(2+2*(1/sqrt(1-x^2)))))*(1/2+1/sqrt(2+sqrt(2+sqrt(2+2*(1/sqrt(1-x^2))))))*(1/2+1/sqrt(2+sqrt(2+sqrt(2+sqrt(2+2*(1/sqrt(1-x^2)))))))*cdots; abs((1/sqrt(x^2-1))*(1/acoth(x)))=((x/sqrt(x^2-1)+1)/2)*(1/2+1/sqrt(2+2*x*(1/sqrt(x^2-1))))*(1/2+1/sqrt(2+sqrt(2+2*x*(1/sqrt(x^2-1)))))*(1/2+1/sqrt(2+sqrt(2+sqrt(2+2*x*(1/sqrt(x^2-1))))))*(1/2+1/sqrt(2+sqrt(2+sqrt(2+sqrt(2+2*x*(1/sqrt(x^2-1)))))))*cdots; have a nice day FME...

1 0

Re: [math-fun] Radial disk-dissection
by Bill Gosper 19 Apr '18

19 Apr '18

-------- Original Message ------- > Subject: [math-fun] Radial disk-dissection > Date: 2018-04-13 07:49 > From: James Propp <jamespropp(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Reply-To: math-fun <math-fun(a)mailman.xmission.com> > > If you dissect a unit disk radially into a large number of equal wedges, > it’s well known that you can reassemble them to form a shape that in the > limit converges to a 1-by-pi rectangle. > gosper.org/picfzoom.gif gosper.org/semizoom.gif --rwg I don't see how to get anything other than allowing unequal wedges. > > What other limiting shapes can we form in the limit from patterns that join > the wedges edge-to-edge? > > I’m guessing that you get a class of curvy pseudoquadrilaterals ABCD where > AB is a unit segment, CD is a unit segment, and the curves BC and AD > satisfy some sort of curvature-compatibility condition. > > Jim Propp > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >

2 1

[math-fun] Aubrey de Grey
by Adam P. Goucher 18 Apr '18

18 Apr '18

Did you hear the exciting news that Aubrey de Grey has increased the lower bound on the chromatic number of the plane from 4 to 5? https://cp4space.wordpress.com/2018/04/15/assorted-news-3/ The best upper bound is 7 (from colouring the cells of a hexagonal lattice of appropriate size). The previous lower bound of 4 is from the Moser's spindle. Best wishes, Adam P. Goucher

4 5

Re: [math-fun] Aubrey de Grey
by Dan Asimov 18 Apr '18

18 Apr '18

In fact, I just heard that yesterday, without any details. Cool! The chromatic number of the unit-distance graph of R^n is a quantity that is both utterly natural and deeply inscrutable. (I bet it turns out to be 7 for R^2, and 15 for R^3.) —Dan Adam Goucher wrote: ------------------- Did you hear the exciting news that Aubrey de Grey has increased the lower bound on the chromatic number of the plane from 4 to 5? https://cp4space.wordpress.com/2018/04/15/assorted-news-3/ The best upper bound is 7 (from colouring the cells of a hexagonal lattice of appropriate size). The previous lower bound of 4 is from the Moser's spindle. ---------

1 0

[math-fun] LRU stack sequences
by Henry Baker 17 Apr '18

17 Apr '18

Has anyone studied the following types of sequences as *purely mathematical* objects? Start with an infinite sequence of non-negative integers in their natural order: 0,1,2,3,4,5,6,... Let me call this sequence the "identity sequence". Given a non-negative integer d, *permute* this infinite sequence by moving the d'th element to the front. Thus, given 5, we get 4,0,1,2,3,5,6,... Notice that when d=0, nothing changes, so it is the identity. We can define -d to be the *inverse* permutation to +d. We can now define a *sequence* of integers operating on our initial sequence; we call these sequences of "d" values, "d-sequences". Thus, the d-sequence 5,-5 produces the original sequence, as does -5,5. Any 0,0,0 subsequence can be eliminated from a d-sequence, as it doesn't do anything. Here's the result of the first 10 digits of pi operating as a d-sequence on the identity sequence: 4,2,1,8,3,0,5,6,7,9,10,11,... We note that when d operates on a sequence, the elements indexed from d and greater don't change; i.e., only the elements indexed by <d are permuted. Here are some obvious properties of d-sequences: The d-sequence 0,1,2,3,...,n *reverses* the first n elements of the original sequence. The d-sequence k^n cyclically permutes the first k elements by n; thus k^k is the identity. Here are some obvious questions: * Are these d-sequences related to some other types of sequences? Perhaps this definition of d-sequences isn't the most elegant? * Clearly -reverse(seq) cancels seq; what about things like palindromes? * Consider d-sequences *sums*. Any interesting properties?

2 2