August 2016 - mathfun@mailman.xmission.com

[math-fun] A Mathematical History of Taffy Pullers
by Thomas Colthurst 23 Aug '16

23 Aug '16

http://arxiv.org/pdf/1608.00152v1.pdf by Jean-Luc Thiffeault "We describe a number of devices for pulling candy, called taffy pullers, that are related to pseudo-Anosov mapping classes. Though the mathematical connection has long been known for the two most common taffy puller designs, we unearth a rich variety of early designs from the patent literature." The article is interesting, even if like me you knew nothing about pseudo-Anosov mapping classes at the start. Also, a former U.S. president makes a surprise appearance and is commended by the author for "a keen grasp of topological dynamics". -Thomas C

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Re: [math-fun] Julian has constructed a spacefill dense with sextuple points
by Bill Gosper 23 Aug '16

23 Aug '16

No. This is why I hate calling these functions "curves". They map closed intervals onto closed areas of C. This is why they are dense with double and triple points. --rwg On 2016-08-22 16:20, James Propp wrote: > I misunderstood what game you all were playing, so my question was a bit > off base. > > Joerg and Bill et al. are looking at ways to fill all of a discrete plane; > I was thinking more about Peano curves and the like, which fill a finite > portion of a continuous plane. > > Jim > > On Monday, August 22, 2016, James Propp <jamespropp(a)gmail.com> wrote: > >> Mulling over what it might mean to say that a spacefilling curve lives on >> a grid, I came up with a nonstandard topology to put on R^2. >> >> The basic open sets are open line segments. This is a finer topology than >> the standard one. What mappings from [0,1] into R^2 are continuous in this >> topology? Piecewise-polygonal ones; what else? Does this topology have any >> uses? >> >> But this doesn't address the question that started me down this line of >> thought, namely: what might it mean to say that a spacefilling curve lives >> on a grid, or prefers certain directions? The approximating polygonal >> curves may use only certain directions, but how does this property manifest >> itself with regard to the limit-object? >> >> Jim Propp >> >> On Mon, Aug 22, 2016 at 3:26 AM, Joerg Arndt <arndt(a)jjj.de >> <javascript:_e(%7B%7D,'cvml','arndt(a)jjj.de');>> wrote: >> >>> For similar curves (that exist for orders that are odd squares) >>> on the (3.6.3.6)-grid see >>> http://jjj.de/3frac/p6/all-r25-b-curve-decompositions.pdf (last two >>> pages) >>> and >>> http://jjj.de/3frac/p6/all-r25-b-tiles.pdf (last four pages). >>> >>> Sextuple and triple points are easy to see in these. >>> >>> I cannot decode the Mathca, does you curve live >>> on the (3.6.3.6)-grid as well (morally it should)? >>> >>> Best regards, jj >>> >>> >>> * Bill Gosper <billgosper(a)gmail.com >>> <javascript:_e(%7B%7D,'cvml','billgosper(a)gmail.com');>> [Aug 22. 2016 >>> 08:17]: >>> > using a 25-way conditional. >>> > ClearAll[piecewiserecursivefractal]; >>> > piecewiserecursivefractal[x_, f_, which_, iters_, fns_] := >>> > piecewiserecursivefractal[x, g_, which, iters, >>> > fns] = ((piecewiserecursivefractal[x, h_, which, iters, fns] := >>> > Block[{y}, y /. Solve[f[y] == h[y], y]]); >>> > Union @@ ((fns[[#]] /@piecewiserecursivefractal[iters[[#]][x], >>> > Composition[f, fns[[#]]], which, iters, fns]) & /@which[x])); >>> > >>> > (I cannot praise this function enough.) >>> > >>> > In[363]:= trifil25[t_] := piecewiserecursivefractal[t, Identity, >>> > Piecewise[{{{1}, #1 == 0}, {{Ceiling[25*#1]}, >>> > Inequality[0, Less, #1, LessEqual, 1]}}, {}] & , >>> > {25*#1 + 0 & , -25*#1 + 2 & , 25*#1 - 2 & , 25*#1 - 3 & , >>> > 25*#1 - 4 & , 25*#1 - 5 & , 25*#1 - 6 & , 25*#1 - 7 & , >>> > 25*#1 - 8 & , -25*#1 + 10 & , -25*#1 + 11 & , >>> > 25*#1 - 11 & , -25*#1 + 13 & , -25*#1 + 14 & , >>> > 25*#1 - 14 & , -25*#1 + 16 & , 25*#1 - 16 & , >>> > -25*#1 + 18 & , 25*#1 - 18 & , 25*#1 - 19 & , >>> > -25*#1 + 21 & , -25*#1 + 22 & , 25*#1 - 22 & , >>> > -25*#1 + 24 & , 25*#1 - 24 & }, >>> > {#1/5 & , (1/10)*(3 + I*Sqrt[3] + (-1 - I*Sqrt[3])* >>> > #1) & , (1/10)*(3 + I*Sqrt[3] + #1 - >>> > I*Sqrt[3]*#1) & , (2 + #1)/5 & , >>> > (1/10)*(6 + #1 + I*Sqrt[3]*#1) & , >>> > (1/10)*(7 + I*Sqrt[3] + I*(I + Sqrt[3])*#1) & , >>> > (1/5)*(3 + I*Sqrt[3] - #1) & , >>> > (1/10)*(4 + 2*I*Sqrt[3] + (-1 - I*Sqrt[3])*#1) & , >>> > (1/10)*(3 + I*Sqrt[3] - 2*#1) & , >>> > (-(1/10))*I*(-I + Sqrt[3])*(-2 + #1) & , >>> > (1/5)*(2 + I*Sqrt[3] - #1) & , >>> > (1/5)*(2 + I*Sqrt[3] + #1) & , >>> > (1/10)*(5 + 3*I*Sqrt[3] + #1 - I*Sqrt[3]*#1) & , >>> > (1/10)*(4 + 2*I*Sqrt[3] + #1 + I*Sqrt[3]*#1) & , >>> > (1/10)*(4 + 2*I*Sqrt[3] + I*(I + Sqrt[3])*#1) & , >>> > (-(1/10))*I*(-I + Sqrt[3])*(-4 + #1) & , >>> > (1/5)*(2 + 2*I*Sqrt[3] + #1) & , >>> > (1/10)*(5 + 3*I*Sqrt[3] + #1 + I*Sqrt[3]*#1) & , >>> > (1/10)*(5 + 3*I*Sqrt[3] + 2*#1) & , >>> > (1/10)*(7 + 3*I*Sqrt[3] + (-1 - I*Sqrt[3])*#1) & , >>> > (1/5)*(4 + I*Sqrt[3] - #1) & , >>> > (1/10)*(9 + I*Sqrt[3] + I*(I + Sqrt[3])*#1) & , >>> > (1/10)*(9 + I*Sqrt[3] - 2*#1) & , >>> > (1/10)*(8 + I*(I + Sqrt[3])*#1) & , (4 + #1)/5 & }][[1]] >>> > E.g., >>> > In[361]:= trifil25[2/25 + 2/75] // ComplexExpand >>> > >>> > Out[361]= {1/2 + (I Sqrt[3])/10} >>> > >>> > In[367]:= Graphics[{Hue@#, >>> > Line[ReIm@{trifil25@#, trifil25[# + 5^-5]}]} & /@ Range[0, 1 - >>> 5^-5, >>> > 5^-5]] >>> > gosper.org/trifil25.png >>> > (Those little white hexagons are sextuple points.) >>> > >>> > The inverse fractal, constructed more or less automatically: >>> > In[380]:= untrifil25[t_] := >>> > piecewiserecursivefractal[t, Identity, >>> > If[0 <= Re[#1] <= 1 && 0 <= Im[#1] <= Sqrt[3]/2, {1, 2, 3, 4, 5, 6, >>> > 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, >>> > 24, 25}, {}] &, {ComplexExpand[5 #1] &, >>> > ComplexExpand[1/4 (6 - 2 I Sqrt[3] - 10 #1 + 10 I Sqrt[3] #1)] &, >>> > ComplexExpand[1/4 (-4 I Sqrt[3] + 10 #1 + 10 I Sqrt[3] #1)] &, >>> > ComplexExpand[-2 + 5 #1] &, >>> > ComplexExpand[1/4 (-6 + 6 I Sqrt[3] + 10 #1 - 10 I Sqrt[3] #1)] &, >>> > ComplexExpand[1/4 (4 + 8 I Sqrt[3] - 10 #1 - 10 I Sqrt[3] #1)] &, >>> > ComplexExpand[3 + I Sqrt[3] - 5 #1] &, >>> > ComplexExpand[1/4 (10 - 2 I Sqrt[3] - 10 #1 + 10 I Sqrt[3] #1)] &, >>> > ComplexExpand[1/2 (3 + I Sqrt[3] - 10 #1)] &, >>> > ComplexExpand[1/4 (8 - 10 #1 + 10 I Sqrt[3] #1)] &, >>> > ComplexExpand[2 + I Sqrt[3] - 5 #1] &, >>> > ComplexExpand[-2 - I Sqrt[3] + 5 #1] &, >>> > ComplexExpand[1/4 (4 - 8 I Sqrt[3] + 10 #1 + 10 I Sqrt[3] #1)] &, >>> > ComplexExpand[1/4 (-10 + 2 I Sqrt[3] + 10 #1 - 10 I Sqrt[3] #1)] &, >>> > ComplexExpand[1/4 (-2 + 6 I Sqrt[3] - 10 #1 - 10 I Sqrt[3] #1)] &, >>> > ComplexExpand[1/4 (16 - 10 #1 + 10 I Sqrt[3] #1)] &, >>> > ComplexExpand[-2 - 2 I Sqrt[3] + 5 #1] &, >>> > ComplexExpand[1/4 (-14 + 2 I Sqrt[3] + 10 #1 - 10 I Sqrt[3] #1)] &, >>> > ComplexExpand[1/2 (-3 I Sqrt[3] + 5 (-1 + 2 #1))] &, >>> > ComplexExpand[1/4 (16 - 4 I Sqrt[3] - 10 #1 + 10 I Sqrt[3] #1)] &, >>> > ComplexExpand[4 + I Sqrt[3] - 5 #1] &, >>> > ComplexExpand[1/4 (6 + 10 I Sqrt[3] - 10 #1 - 10 I Sqrt[3] #1)] &, >>> > ComplexExpand[1/2 (9 + I Sqrt[3] - 10 #1)] &, >>> > ComplexExpand[1/4 (8 + 8 I Sqrt[3] - 10 #1 - 10 I Sqrt[3] #1)] &, >>> > ComplexExpand[-4 + 5 #1] &}, {#1/25 &, (2 - #1)/25 &, (2 + #1)/ >>> > 25 &, (3 + #1)/25 &, (4 + #1)/25 &, (5 + #1)/25 &, (6 + #1)/ >>> > 25 &, (7 + #1)/25 &, (8 + #1)/25 &, (10 - #1)/25 &, (11 - #1)/ >>> > 25 &, (11 + #1)/25 &, (13 - #1)/25 &, (14 - #1)/25 &, (14 + #1)/ >>> > 25 &, (16 - #1)/25 &, (16 + #1)/25 &, (18 - #1)/25 &, (18 + #1)/ >>> > 25 &, (19 + #1)/25 &, (21 - #1)/25 &, (22 - #1)/25 &, (22 + #1)/ >>> > 25 &, (24 - #1)/25 &, (24 + #1)/25 &}] >>> > >>> > Here it finds the preimages of the simplest sextuple point: >>> > >>> > In[381]:= untrifil25[1/2 + I Sqrt[3]/10] >>> > >>> > Out[381]= {8/75, 11/75, 14/75, 17/75, 4/15, 23/75} >>> > --rwg

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[math-fun] Anisotropies of space-filling curves
by James Propp 22 Aug '16

22 Aug '16

What sorts of anisotropy does the Peano curve exhibit, from which its origins as a limit of paths on ever-finer square grids could be deduced? I mention the Peano curve for definiteness, but I intend my question to be taken in a more general context. Jim Propp

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[math-fun] Triangle question
by David Wilson 22 Aug '16

22 Aug '16

Let triangle T have integer sides and area. Is it true that T is right or can be dissected into two right triangles?

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Re: [math-fun] Chris Patuzzo and Sentient
by Lee Sallows 22 Aug '16

22 Aug '16

Mathfunners may well enjoy <http://whyarecomputers.com/4>http://whyarecomputers.com/4, an entertaining interview with Chris Patuzzo, who answered a daunting challenge of mine by coming up with: This sentence is dedicated to Lee Sallows and to within one decimal place four point five percent of the letters in this sentence are a's, zero point one percent are b's, four point three percent are c's, zero point nine percent are d's, twenty point one percent are e's, one point five percent are f's, zero point four percent are g's, one point five percent are h's, six point eight percent are i's, zero point one percent are j's, zero point one percent are k's, one point one percent are l's, zero point three percent are m's, twelve point one percent are n's, eight point one percent are o's, seven point three percent are p's, zero point one percent are q's, nine point nine percent are r's, five point six percent are s's, nine point nine percent are t's, zero point seven percent are u's, one point four percent are v's, zero point seven percent are w's, zero point five percent are x's, zero point three percent are y's and one point six percent are z's. However, as emerges during the interview, this feat was really no more than a stepping stone along Patuzzo's way to creating Sentient, his new declarative computer language, which you can read about here: http://sentient-lang.org/. Lee

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Re: [math-fun] Julian has constructed a spacefill dense with sextuple points
by Bill Gosper 22 Aug '16

22 Aug '16

So, e.g., the last figure of the first paper describes a "twicefilling" function, hitting every interior point at least twice by retracing most of itself. (Back in the 70s, Robert Maas called this "sausaging a tree".) But then we seem to have 12th order points!? It would be amazing to see piecewiserecursivefractal cough up a twelvetuple. On 2016-08-22 00:26, Joerg Arndt wrote: > For similar curves (that exist for orders that are odd squares) > on the (3.6.3.6)-grid see > http://jjj.de/3frac/p6/all-r25-b-curve-decompositions.pdf (last two pages) > and > http://jjj.de/3frac/p6/all-r25-b-tiles.pdf (last four pages). > > Sextuple and triple points are easy to see in these. > > I cannot decode the Mathca, does you curve live > on the (3.6.3.6)-grid as well (morally it should)? > > Best regards, jj I don't understand your grid notation, but it should be clear from the png. If you missed the url, you should be impolitely questioning my "little white hexagons". Do my posts usually sound this delusional? --rwg > > > * Bill Gosper <billgosper(a)gmail.com> [Aug 22. 2016 08:17]: >> using a 25-way conditional. >> ClearAll[piecewiserecursivefractal]; >> piecewiserecursivefractal[x_, f_, which_, iters_, fns_] := >> piecewiserecursivefractal[x, g_, which, iters, >> fns] = ((piecewiserecursivefractal[x, h_, which, iters, fns] := >> Block[{y}, y /. Solve[f[y] == h[y], y]]); >> Union @@ ((fns[[#]] /@piecewiserecursivefractal[iters[[#]][x], >> Composition[f, fns[[#]]], which, iters, fns]) & /@which[x])); >> >> (I cannot praise this function enough.) >> >> In[363]:= trifil25[t_] := piecewiserecursivefractal[t, Identity, >> Piecewise[{{{1}, #1 == 0}, {{Ceiling[25*#1]}, >> Inequality[0, Less, #1, LessEqual, 1]}}, {}] & , >> {25*#1 + 0 & , -25*#1 + 2 & , 25*#1 - 2 & , 25*#1 - 3 & , >> 25*#1 - 4 & , 25*#1 - 5 & , 25*#1 - 6 & , 25*#1 - 7 & , >> 25*#1 - 8 & , -25*#1 + 10 & , -25*#1 + 11 & , >> 25*#1 - 11 & , -25*#1 + 13 & , -25*#1 + 14 & , >> 25*#1 - 14 & , -25*#1 + 16 & , 25*#1 - 16 & , >> -25*#1 + 18 & , 25*#1 - 18 & , 25*#1 - 19 & , >> -25*#1 + 21 & , -25*#1 + 22 & , 25*#1 - 22 & , >> -25*#1 + 24 & , 25*#1 - 24 & }, >> {#1/5 & , (1/10)*(3 + I*Sqrt[3] + (-1 - I*Sqrt[3])* >> #1) & , (1/10)*(3 + I*Sqrt[3] + #1 - >> I*Sqrt[3]*#1) & , (2 + #1)/5 & , >> (1/10)*(6 + #1 + I*Sqrt[3]*#1) & , >> (1/10)*(7 + I*Sqrt[3] + I*(I + Sqrt[3])*#1) & , >> (1/5)*(3 + I*Sqrt[3] - #1) & , >> (1/10)*(4 + 2*I*Sqrt[3] + (-1 - I*Sqrt[3])*#1) & , >> (1/10)*(3 + I*Sqrt[3] - 2*#1) & , >> (-(1/10))*I*(-I + Sqrt[3])*(-2 + #1) & , >> (1/5)*(2 + I*Sqrt[3] - #1) & , >> (1/5)*(2 + I*Sqrt[3] + #1) & , >> (1/10)*(5 + 3*I*Sqrt[3] + #1 - I*Sqrt[3]*#1) & , >> (1/10)*(4 + 2*I*Sqrt[3] + #1 + I*Sqrt[3]*#1) & , >> (1/10)*(4 + 2*I*Sqrt[3] + I*(I + Sqrt[3])*#1) & , >> (-(1/10))*I*(-I + Sqrt[3])*(-4 + #1) & , >> (1/5)*(2 + 2*I*Sqrt[3] + #1) & , >> (1/10)*(5 + 3*I*Sqrt[3] + #1 + I*Sqrt[3]*#1) & , >> (1/10)*(5 + 3*I*Sqrt[3] + 2*#1) & , >> (1/10)*(7 + 3*I*Sqrt[3] + (-1 - I*Sqrt[3])*#1) & , >> (1/5)*(4 + I*Sqrt[3] - #1) & , >> (1/10)*(9 + I*Sqrt[3] + I*(I + Sqrt[3])*#1) & , >> (1/10)*(9 + I*Sqrt[3] - 2*#1) & , >> (1/10)*(8 + I*(I + Sqrt[3])*#1) & , (4 + #1)/5 & }][[1]] >> E.g., >> In[361]:= trifil25[2/25 + 2/75] // ComplexExpand >> >> Out[361]= {1/2 + (I Sqrt[3])/10} >> >> In[367]:= Graphics[{Hue@#, >> Line[ReIm@{trifil25@#, trifil25[# + 5^-5]}]} & /@ Range[0, 1 - 5^-5, >> 5^-5]] >> gosper.org/trifil25.png >> (Those little white hexagons are sextuple points.) >> >> The inverse fractal, constructed more or less automatically: >> In[380]:= untrifil25[t_] := >> piecewiserecursivefractal[t, Identity, >> If[0 <= Re[#1] <= 1 && 0 <= Im[#1] <= Sqrt[3]/2, {1, 2, 3, 4, 5, 6, >> 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, >> 24, 25}, {}] &, {ComplexExpand[5 #1] &, >> ComplexExpand[1/4 (6 - 2 I Sqrt[3] - 10 #1 + 10 I Sqrt[3] #1)] &, >> ComplexExpand[1/4 (-4 I Sqrt[3] + 10 #1 + 10 I Sqrt[3] #1)] &, >> ComplexExpand[-2 + 5 #1] &, >> ComplexExpand[1/4 (-6 + 6 I Sqrt[3] + 10 #1 - 10 I Sqrt[3] #1)] &, >> ComplexExpand[1/4 (4 + 8 I Sqrt[3] - 10 #1 - 10 I Sqrt[3] #1)] &, >> ComplexExpand[3 + I Sqrt[3] - 5 #1] &, >> ComplexExpand[1/4 (10 - 2 I Sqrt[3] - 10 #1 + 10 I Sqrt[3] #1)] &, >> ComplexExpand[1/2 (3 + I Sqrt[3] - 10 #1)] &, >> ComplexExpand[1/4 (8 - 10 #1 + 10 I Sqrt[3] #1)] &, >> ComplexExpand[2 + I Sqrt[3] - 5 #1] &, >> ComplexExpand[-2 - I Sqrt[3] + 5 #1] &, >> ComplexExpand[1/4 (4 - 8 I Sqrt[3] + 10 #1 + 10 I Sqrt[3] #1)] &, >> ComplexExpand[1/4 (-10 + 2 I Sqrt[3] + 10 #1 - 10 I Sqrt[3] #1)] &, >> ComplexExpand[1/4 (-2 + 6 I Sqrt[3] - 10 #1 - 10 I Sqrt[3] #1)] &, >> ComplexExpand[1/4 (16 - 10 #1 + 10 I Sqrt[3] #1)] &, >> ComplexExpand[-2 - 2 I Sqrt[3] + 5 #1] &, >> ComplexExpand[1/4 (-14 + 2 I Sqrt[3] + 10 #1 - 10 I Sqrt[3] #1)] &, >> ComplexExpand[1/2 (-3 I Sqrt[3] + 5 (-1 + 2 #1))] &, >> ComplexExpand[1/4 (16 - 4 I Sqrt[3] - 10 #1 + 10 I Sqrt[3] #1)] &, >> ComplexExpand[4 + I Sqrt[3] - 5 #1] &, >> ComplexExpand[1/4 (6 + 10 I Sqrt[3] - 10 #1 - 10 I Sqrt[3] #1)] &, >> ComplexExpand[1/2 (9 + I Sqrt[3] - 10 #1)] &, >> ComplexExpand[1/4 (8 + 8 I Sqrt[3] - 10 #1 - 10 I Sqrt[3] #1)] &, >> ComplexExpand[-4 + 5 #1] &}, {#1/25 &, (2 - #1)/25 &, (2 + #1)/ >> 25 &, (3 + #1)/25 &, (4 + #1)/25 &, (5 + #1)/25 &, (6 + #1)/ >> 25 &, (7 + #1)/25 &, (8 + #1)/25 &, (10 - #1)/25 &, (11 - #1)/ >> 25 &, (11 + #1)/25 &, (13 - #1)/25 &, (14 - #1)/25 &, (14 + #1)/ >> 25 &, (16 - #1)/25 &, (16 + #1)/25 &, (18 - #1)/25 &, (18 + #1)/ >> 25 &, (19 + #1)/25 &, (21 - #1)/25 &, (22 - #1)/25 &, (22 + #1)/ >> 25 &, (24 - #1)/25 &, (24 + #1)/25 &}] >> >> Here it finds the preimages of the simplest sextuple point: >> >> In[381]:= untrifil25[1/2 + I Sqrt[3]/10] >> >> Out[381]= {8/75, 11/75, 14/75, 17/75, 4/15, 23/75} >> --rwg

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Re: [math-fun] Magic squares of primes(?)
by Walter Trump 21 Aug '16

21 Aug '16

In 2013 Al Zimmermann and Natalia Makarova started a contest about pandiagonal magic squares of primes: http://azspcs.com/Contest/PandiagonalMagicSquares Jarek Wroblewski found the best solution for each problem: http://azspcs.com/Contest/PandiagonalMagicSquares/FinalReport Walter

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[math-fun] Greedy Queens question
by Neil Sloane 21 Aug '16

21 Aug '16

Dear Math Fun, Take an infinite chessboard, and starting at the top left corner, scan it by upwards anti-diagonals, placing a Queen whenever you reach a square that is not attacked by an existing Queen, The top left portion of the board is Qxxxxx... xxxQxx... xQxxxx... xxxxQx... xxQxxx... xxxxxx... xxxxxx... xxxxxQ... ...... The n-th term of sequence https://oeis.org/A065188 tells which row contains the Queen in column n. (It is easy to show that every column eventually contains a Queen.) If you click the "graph" button in A065188 you will see that the points appear to lie on two roughly straight lines, of slopes phi and 1/phi, phi being the golden ratio! The question is, why? How do these queens know about the golden ratio? Presumably this is related to budding sequences and sunflower seeds. Notes. Let c(n) = A065188(n). A199134 lists n such that c(n)<n (the points on the line of slope 1/phi), and A275884 lists n such that c(n)>=n (the line of slope phi). A275885 gives lengths of runs of consecutive terms in A193134. Initially I thought these lengths were always 1 or 2, which might have been significant, but that is false. A275888 gives first differences of A275884. Initially I thought these terms were always 1, 2, or 3, but that is also false. I can't correct these errors until Monday, because the OEIS is in read-only mode for a server upgrade. I can send a larger picture of the chess-board to anyone who is interested.

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[math-fun] Magic squares of primes(?)
by Zak Seidov 20 Aug '16

20 Aug '16

Fill NxN square with decimal digits such that all rows/columns and 2 main diagonals read primes. Simple(st?) 3x3 case: 227 727 733 with 227,727,733,277,223,773,223 and 727 all primes. Cases with all primes distinct? Larger (largest?) cases? Old hat? zak

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[math-fun] wifi planning and log conics ?
by Henry Baker 20 Aug '16

20 Aug '16

I was reading about wireless network cell planning for my home wifi (with $10-20 wifi routers, we're *all* wireless network engineers now!), and "distances" are sometimes "measured" in decibels: http://www.cisco.com/c/dam/en/us/td/i/400001-500000/400001-410000/400001-40… http://www.cisco.com/c/en/us/td/docs/wireless/controller/technotes/8-0/iPho… Decibels are a logarithmic function; if x dB can reach distance D, then x+6 DB can reach distance 2*D. Assuming that wireless antennae emit radiation uniformly in all directions (very bad assumption!), then on a flat surface, one simple "cell" plan would place antennae in the centers of a uniform hexagonal grid. Of course, things are never this simple -- especially indoors. --- So the question is: are there interesting/elegant/surprising geometrical questions when the distances are measured with logarithms instead of normally? For example, some wireless questions involve having to "see" two cell towers simultaneously (e.g., wireless repeaters, roaming handoffs). In some cases, a roaming handoff may involve 2 signals which are exactly 6 dB apart. With normal distances, this locus would be a hyperbola, but since we're now talking about log distances, what is the locus then? --- Another problem involves distance/power/rate tradeoffs; this level of sophistication is beyond current home wifi, because typical home wifi systems can't use precise (and different) power levels & modulation schemes for each packet. https://en.wikipedia.org/wiki/Eb/N0 So as the distance increases, the data rate has to fall. So other loci involve these ratios; perhaps other interesting curves follow constant data rates, etc. --- Note that my interest here is *not* in the electrical engineering aspects of the system, but in the shapes of curves that might be generated by such calculations.

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