August 2017 - mathfun@mailman.xmission.com

Re: [math-fun] ISO a perfect pangram
by Keith F. Lynch 11 Aug '17

11 Aug '17

Veit Elser <ve10(a)cornell.edu> wrote: > 1. Extract sequences of letters, including spaces (as word > separators), from actual text. I tried that decades ago. It picks up far too much junk: Proper names, abbreviations, acronyms, jargon, computer codes, ham radio codes, misspelled words, foreign words, misspelled foreign words, etc. If I were to try it again today, now that lots of people send HTML email, it would probably tell me that "msonormal" was the most common English word. I've tried searching for word lists online that list each word by its frequency. In a perverse equivalent of Godel's theorem, it appears that every such list is either incomplete or contains trash. For instance http://norvig.com/ngrams/count_1w.txt starts promisingly: the 23135851162 of 13151942776 and 12997637966 to 12136980858 a 9081174698 in 8469404971 for 5933321709 is 4705743816 but if, for instance, I search it for the anagrams of "post" I get: post 392956436 stop 77749471 spot 26750929 tops 11771127 pots 3854743 opts 662207 tsop 205591 tpos 43379 ostp 41988 ptos 38390 otps 23858 ptso 21839 Any idea where I can find a clean and complete list? Thanks.

3 2

Re: [math-fun] ISO a perfect pangram
by Dan Asimov 11 Aug '17

11 Aug '17

There are some terrific just plain word lists online, e.g., at puzzlers.org go to "Solving tools of the Enigma" and then "grep dictionary search". >From there you can download the list (or lists; there are a lot of choices) and toss out all the items in the otherwise-promising data that are not found in the list. Kind of a pain but doable. —Dan >From: "Keith F. Lynch" <kfl(a)KeithLynch.net> >Veit Elser <ve10(a)cornell.edu> wrote: >> 1. Extract sequences of letters, including spaces (as word >> separators), from actual text. > >I tried that decades ago. It picks up far too much junk: Proper >names, abbreviations, acronyms, jargon, computer codes, ham radio >codes, misspelled words, foreign words, misspelled foreign words, etc. >If I were to try it again today, now that lots of people send HTML >email, it would probably tell me that "msonormal" was the most common >English word. > >I've tried searching for word lists online that list each word by its >frequency. In a perverse equivalent of Godel's theorem, it appears >that every such list is either incomplete or contains trash. > >For instance http://norvig.com/ngrams/count_1w.txt starts promisingly: > >the 23135851162 >of 13151942776 >and 12997637966 >to 12136980858 >a 9081174698 >in 8469404971 >for 5933321709 >is 4705743816 > >but if, for instance, I search it for the anagrams of "post" I get: > >post 392956436 >stop 77749471 >spot 26750929 >tops 11771127 >pots 3854743 >opts 662207 >tsop 205591 >tpos 43379 >ostp 41988 >ptos 38390 >otps 23858 >ptso 21839 > >Any idea where I can find a clean and complete list? Thanks. > >_______________________________________________ >math-fun mailing list >math-fun(a)mailman.xmission.com >https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

1 0

Re: [math-fun] Generalized Fibonacci sequences and arithmetic progressions
by Dan Asimov 11 Aug '17

11 Aug '17

Would you be so kind as to define what a "two-sided generalized Fibonacci sequence means"? Sorry, I'm too dumb to be able to figure out what the general term is from "n-m,m,n,m+n". —Dan ----- Given a two-sided generalized Fibonacci sequence ...,n-m,m,n,m+n,... (with m,n in Z and not both zero), must there exist a two-sided arithmetic progression ...,a-d,a,a+d,... (with a,d in Z and d nonzero) that is disjoint from it? -----

4 5

[math-fun] Para-uniform polytopes
by Allan Wechsler 10 Aug '17

10 Aug '17

There is something about the classic definition of a uniform polytope that has always bugged me. There are three clauses to the definition: 1. The polytope must be vertex-transitive; informally, there is only one "kind" of vertex. 2. The edges must all be of the same length. 3. All the facets (that is, (n-1)-dimensional cells) of the polytope must themselves be uniform. Clause 3 establishes a recursive requirement that _all_ the cells be uniform, but these cells aren't required to be _identical_ until you get down to dimension 1 and clause 2 takes over. It feels to me like clause 2 is a weird addendum. I define a para-uniform polytope as follows: 1. The polytope must be vertex-transitive. 2. All its facets must themselves be para-uniform. This makes the edges behave like all the other cells: there are allowed to be different kinds. The para-uniform polygons, for instance, are all the equi-angular polygons that alternate two edge lengths. For odd numbers of sides, the two edge lengths are forced to be equal by global structure. Here's my question: do we get any novel structures? Or are all para-uniform polytopes just uniform polytopes with variously-stretched edges?

3 3

[math-fun] Why do professors make matrix algebra so inscrutable?
by Henry Baker 10 Aug '17

10 Aug '17

Re rational canonical form: Most of you probably know this already, but I looked at perhaps 25 presentations, and *no one* made it this easy. M is an arbitrary matrix (over the rationals, for example). C is the companion matrix for the characteristic polynomial of M. We here assume that the characteristic polynomial is *irreducible*. We need to find an invertible T such that (T^^-1).M.T = C. Choose T such that the first column T1 is random, the second column is T2=M.T1, and the third column is T3=(M^^2).T1=M.M.T1=M.T2, etc. If this matrix T1:T2:T3 is invertible, it will work. Unless you're exceedingly unlucky, it will work with the first random choice of T1. So, once you've got the characteristic polynomial and know that it doesn't factor, the rest is trivial. BTW, Cayley, Sylvester, etc. would have understood this argument; they would have found all that stuff about "modules" to be completely inscrutable and extremely unhelpful -- i.e., "abstract nonsense". (%i2) M; [ a b c ] [ ] (%o2) [ d e f ] [ ] [ g h i ] (%i3) C:polytocompanion(charpoly(M,x),x),expand; [ 0 0 a e i - b d i - a f h + c d h + b f g - c e g ] [ ] (%o3) [ 1 0 - e i - a i + f h + c g - a e + b d ] [ ] [ 0 1 i + e + a ] (%i4) T; [ x ] [ c z + b y + a x ] [ ] [ ] (%o4) Col 1 = [ y ] Col 2 = [ f z + e y + d x ] [ ] [ ] [ z ] [ i z + h y + g x ] [ c (i z + h y + g x) + b (f z + e y + d x) + a (c z + b y + a x) ] [ ] Col 3 = [ f (i z + h y + g x) + e (f z + e y + d x) + d (c z + b y + a x) ] [ ] [ i (i z + h y + g x) + h (f z + e y + d x) + g (c z + b y + a x) ] (%i5) M.T-T.C,expand; [ 0 0 0 ] [ ] (%o5) [ 0 0 0 ] [ ] [ 0 0 0 ]

1 0

[math-fun] Fwd: There are no equilateral triangles in the Integer Grid.
by Bill Gosper 09 Aug '17

09 Aug '17

gosper.org/rectarith12.pdf (penultimate page). --rwg On 2017-08-08 17:17, Eugene Salamin via math-fun wrote: > Suppose, to the contrary, that such a triangle exists. We may > translate one vertex to the origin, the remaining vertices being (x1, > y1) and (x2, y2). If all 4 of these numbers are even, there exists a > half-size triangle. So we may assume that one of these numbers is > odd, and by rotation and reflection, we can take x1 to be odd. We > have > > x1^2 + y1^2 = x2^2 + y2^2 = (x1 - x2)^2 + (y1 - y2)^2 = s^2, > > and thus that > > 2 (x1 x2 + y1 y2) = s^2. > > Since s^2 is even, x1 and y1 have the same parity, and likewise for x2 > and y2. So y1 is odd. But then s^2 = x1^2 + y1^2 = 2 mod 4. Hence > x2 and y2 must also be odd. With all 4 numbers odd, x1 x2 + y1 y2 = 2 > mod 4. But s^2 is twice this, so s^2 = 0 mod 4, and we have a > contradiction. > > -- Gene > > > > On Monday, August 7, 2017, 4:04:20 AM PDT, Bill Gosper > <billgosper(a)gmail.com> wrote: > > More generally, there is no threefold rotational symmetry. > Julian's function can produce only gridpoints: > ListPlot[{3, 1} # & /@ > NestWhileList[{#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &@ > {#[[1]], #[[2]] + Floor[381 #[[1]]/128]} &@ > {#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &, > {2, 1}, # != -{1, 4} &], > AspectRatio -> 1, Axes -> False, Frame -> True] > gosper.org/trinsky3spiral.png > And all the arguments to the Floors are merely rational. --rwg

3 2

Re: [math-fun] Fwd: There are no equilateral triangles in the Integer Grid.
by Dan Asimov 09 Aug '17

09 Aug '17

On the other hand, it is the case that the group PSL(2,Z) thought of as fractional linear transformations of S^2 = C u {oo} via z |—> (az + b) / (cz + d), (a, b, c, d in Z) does preserve the rational points Q + Qi. Q + Qi is of course the union of (1/n)Z + (1/n)Zi over n = 1,2,3,..., which are all *square* lattices. So it's at least a little surprising that PSL(2,Z) has elements of order 3. —Dan I wrote: ----- In no sense, apparently! ----- ----- On Aug 9, 2017 04:32, "Dan Asimov" <dasimov(a)earthlink.net> wrote: And paradoxically, the conformal bijection z |—> -1/(1+z) of the Riemann sphere S^2 = C u {oo} to itself that preserves the integer points Z^2 ... Andy Latto wrote: ----- In what sense does this preserve the integer points? F(17) = -1/18 ----- (i.e., a fractional linear transformation of form z |—> (az + b)/(cz + d), a,b,c,d in Z, ad - bc = 1, i.e., a member of the group PSL(2,Z) = SL(2,Z) / {+-I}) ... is of order three: f(f(f(z))) == z -----

1 0

Re: [math-fun] Fwd: There are no equilateral triangles in the Integer Grid.
by Dan Asimov 09 Aug '17

09 Aug '17

In no sense, apparently! –Dan ----- On Aug 9, 2017 04:32, "Dan Asimov" <dasimov(a)earthlink.net> wrote: And paradoxically, the conformal bijection z |—> -1/(1+z) of the Riemann sphere S^2 = C u {oo} to itself that preserves the integer points Z^2 ... In what sense does this preserve the integer points? F(17) = -1/18 Andy (i.e., a fractional linear transformation of form z |—> (az + b)/(cz + d), a,b,c,d in Z, ad - bc = 1, i.e., a member of the group PSL(2,Z) = SL(2,Z) / {+-I}) ... is of order three: f(f(f(z))) == z -----

1 0

[math-fun] Stirling triangles mod 2..7
by Bill Gosper 09 Aug '17

09 Aug '17

gosper.org/stirlingcycle2-7.png gosper.org/stirlingsubset2-7.png --rwg

2 1

Re: [math-fun] Fwd: There are no equilateral triangles in the Integer Grid.
by Andy Latto 09 Aug '17

09 Aug '17

On Aug 9, 2017 04:32, "Dan Asimov" <dasimov(a)earthlink.net> wrote: And paradoxically, the conformal bijection z |—> -1/(1+z) of the Riemann sphere S^2 = C u {oo} to itself that preserves the integer points Z^2 ... In what sense does this preserve the integer points? F(17) = -1/18 Andy (i.e., a fractional linear transformation of form z |—> (az + b)/(cz + d), a,b,c,d in Z, ad - bc = 1, i.e., a member of the group PSL(2,Z) = SL(2,Z) / {+-I}) ... is of order three: f(f(f(z))) == z (for example). —Dan >From: Eugene Salamin via math-fun <math-fun(a)mailman.xmission.com> >Sent: Aug 8, 2017 5:17 PM Suppose, to the contrary, that such a triangle exists. We may translate one vertex to the origin, the remaining vertices being (x1, y1) and (x2, y2). If all 4 of these numbers are even, there exists a half-size triangle. So we may assume that one of these numbers is odd, and by rotation and reflection, we can take x1 to be odd. We have x1^2 + y1^2 = x2^2 + y2^2 = (x1 - x2)^2 + (y1 - y2)^2 = s^2, and thus that 2 (x1 x2 + y1 y2) = s^2. Since s^2 is even, x1 and y1 have the same parity, and likewise for x2 and y2. So y1 is odd. But then s^2 = x1^2 + y1^2 = 2 mod 4. Hence x2 and y2 must also be odd. With all 4 numbers odd, x1 x2 + y1 y2 = 2 mod 4. But s^2 is twice this, so s^2 = 0 mod 4, and we have a contradiction. On Monday, August 7, 2017, 4:04:20 AM PDT, Bill Gosper <billgosper(a)gmail.com> wrote: More generally, there is no threefold rotational symmetry. Julian's function can produce only gridpoints: ListPlot[{3, 1} # & /@ NestWhileList[{#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &@ {#[[1]], #[[2]] + Floor[381 #[[1]]/128]} &@ {#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &, {2, 1}, # != -{1, 4} &], AspectRatio -> 1, Axes -> False, Frame -> True] gosper.org/trinsky3spiral.png And all the arguments to the Floors are merely rational. --rwg _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

1 0