April 2016 - mathfun@mailman.xmission.com

Re: [math-fun] scissor congruence
by Bill Gosper 15 Apr '16

15 Apr '16

On 2016-04-15 10:11, rcs(a)xmission.com wrote: > My note about a circular dissection is discussing constructions with > a minimal number of pieces. If you allow extra pieces, then every > piece can be a polygon. > > I've seen the theorem about polygon interdissectability (any two polygons > with the same area are interdissectable) attributed to Hilbert, likely in > either a Gardner column or in the intro to a Dissections book. You can adjust a side of a triangle to any desired length via gosper.org/A=f(bh).gif . If this is the sidelength of another triangle with the same area, then they will have the same height, and repeating the trick finishes the proof for triangles. gosper.org/shardway.PNG --rwg > I was amazed that Hilbert would be concerned with something so light. > Later I learned about his posing the same problem for 3D, and Dehn's > negative answer. > And much later I realized that this is relevant to the existence of > alternative measures for area & volume. I don't know the details, but > perhaps some dissections are constraints on alternate measures. > > I have a query on the intended meaning of "scissors dissection": > Does this mean simply that the pieces must have straight edges? > Or that the dissection can be created using straight cuts entirely > across the pieces (and more cuts across the resulting pieces: the > 4 piece triangle->square would qualify). Other possible (dis)allowances > are cuts that terminate without completely crossing the piece, but > meet up with other cuts to separate pieces; cuts which turn at vertices > (hard to actually cut), and cuts that make holes, which must be initiated > by piercing the piece (such as my example with a triangular hole). > And maybe cuts that aren't entirely used, that extend into a piece > without separating it. > > Rich > > ----- > Quoting Warren D Smith <warren.wds(a)gmail.com>: >> Any polygon in Euclidean plane is scissor-congruent to any other of >> the same area, and >> this is accomplishable entirely with straight cuts (no curves ever needed). >> I think that was first shown by Bolyai. >> >> So if you want an example where curved cuts are needed, then you need >> the two shapes to include at least one with a curved boundary. >> And then it is trivial. >> >> Make shape #1 be a square. Shape #2 is got by cutting square in two >> via an S-curve, >> rotate pieces, glue along common straight line. Obviously a curved cut is >> needed to get from 1 to 2. >> >> >> -- >> Warren D. Smith

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[math-fun] Infinite Sequential Ammann Chairs
by Ed Pegg Jr 15 Apr '16

15 Apr '16

Infinite sequentially-sized Ammann chairs can be packed perfectly into a rectangle, with a scaling factor of sqrt(phi). http://i.imgur.com/rLnbRiU.gif Are there nice infinite sequential packings with a lower scaling factor? There seems to be many rectangles that infinite Ammann chairs will fit into. What rectangle would be closest to a square? Ed Pegg Jr

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[math-fun] scissor congruence
by Warren D Smith 15 Apr '16

15 Apr '16

Any polygon in Euclidean plane is scissor-congruent to any other of the same area, and this is accomplishable entirely with straight cuts (no curves ever needed). I think that was first shown by Bolyai. So if you want an example where curved cuts are needed, then you need the two shapes to include at least one with a curved boundary. And then it is trivial. Make shape #1 be a square. Shape #2 is got by cutting square in two via an S-curve, rotate pieces, glue along common straight line. Obviously a curved cut is needed to get from 1 to 2. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Rot13 and other substitution ciphers (was Re: Quickies)
by Keith F. Lynch 15 Apr '16

15 Apr '16

Andy Latto <andy.latto(a)pobox.com> wrote: > So there being no discussion on my fallacious proof other than > discussion of rot13, .... As has been mentioned, rot13 is a simple substitution cipher which maps the first half of the alphabet onto the second half and vice versa. It's intended, not for security, but for spoiler warnings and for other stuff that people may not want to see, e.g. offensive jokes. The fact that not everyone here was familiar with it is evidence that the Internet has grown so quickly that many are not familiar with Internet culture and conventions. (This is most evident in this mailing list by the prevalence of quoting in reverse order. You'd think it would be obvious that this is backwards, given that English is read top to bottom, not vice versa, but apparently not.) The longest words that remain words after being rot13'd are Chechen <-> Purpura. Rot13 has a period of two, meaning that encryption and decryption are the same operation. A period of N means that if you encrypt (or decrypt) N times, you'll get the original text back, but never any sooner (unless you avoid certain letters). What periods can simple substitution ciphers have on an alphabet of 26 letters? How many are there of each? In particular, what's the shortest period that doesn't exist, and what's the longest period that does? And how would you compute those for alphabets of other sizes? I'm especially interested in period-two, as I'm interested in bitwise palindromes, i.e. text that remains the same after having its bit pattern time-reversed. Of course the results depend on the choice of character code. Morse code works well, except that some letters don't have reverses. "Footstool" and "taint" are the two longest Morse code single-word palindromes. The old five-bit code, ITA2, often wrongly called Baudot, also works well. ASCII and EBCDIC, not so much. The five bit code seems tailor-made for palindromes. Space maps into itself. Every letter maps into a letter. Common letters map into common letters at a rate that can't possibly be coincidence. The most common letters in English language text, in order, are ETAOIN SHRDLU. The ITA2 palindrome mapping is E<->T, A<->O, I<->N, S<->H, R<->R, D<->L. As if this weren't remarkable enough, <CR> maps into <LF>, so the standard teletype linebreak sequence, <CR><LF>, maps into itself. A little online research explains this in two ways. One, wear on mechanical equipment was proportional to the number of 1 bits, so they were minimized by giving the most common characters the fewest 1 bits. Two, punched paper tapes were sometimes accidentally loaded reversed in one or both dimensions, and they wanted this to be at least somewhat recoverable. (Of course this means the letters are in a weird order if you arrange the codes in the usual binary sequence (E A SIU DRJNFCKTZLWHYPQOBG MXV). You can't have everything.) But suppose I wanted to design a character code for bitwise palindromes. Can I do even better? I don't just want to map common letters into common letters, I want to map common digrams, trigrams, etc. into common digrams, trigrams, etc. What would be a good algorithm? Would it be practical to simply have an automated search over all period-two substitution ciphers, searching each for palindromes? (Actually assigning bit patterns is unnecessary for this purpose. If, say, A maps into B, that means that the bit pattern of A is the time-reverse of the bit pattern of B, but says nothing about what either bit pattern actually is, except that that pattern itself can't be a palindrome.)

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Re: [math-fun] Puzzle related to Yet another irritating elementary geometry puzzle
by Dan Asimov 15 Apr '16

15 Apr '16

Yes — thanks for the correction, Tom: "...will divide S^n into 2^(n+1) compartments" is most definitely what I should have written. —Dan > On Apr 14, 2016, at 1:16 PM, Tom Karzes <karzes(a)sonic.net> wrote: > > Hi Dan, > > In the following: > >> Clearly, doing the same with the right spherical n-simplex (i.e., >> having all dihedral angles equal to 90º) in the n-sphere S^n will >> divide S^n into 2^n compartments. > > Didn't you mean "...will divide S^n into 2^(n+1) compartments"? > > Tom > > > Dan Asimov writes: >> Given a regular (or any) tetrahedron in R^3 with its face planes >> extended indefinitely will divide R^3 into a certain number of >> compartments. >> >> A related situation is that of the 2-sphere S^2, where the right >> spherical triangle (i.e., having all angles equal to 90º) has its >> 1-faces extended indefinitely to become great circles. Clearly >> that will divide S^2 into 2^3 = 8 compartments. >> >> Clearly, doing the same with the right spherical n-simplex (i.e., >> having all dihedral angles equal to 90º) in the n-sphere S^n will >> divide S^n into 2^n compartments. NO! This should be 2^(n+1) compartments, as Tom points out. >> >> Puzzle: >> ------- >> Given an n-simplex in R^n with its face-hyperplanes extended >> indefinitely, how many compartments will they divide R^n into ? >> >> —Dan >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>

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Re: [math-fun] Quickies
by Keith F. Lynch 15 Apr '16

15 Apr '16

Andy Latto <andy.latto(a)pobox.com> wrote: > So there being no discussion on my fallacious proof other than > discussion of rot13, I figured it was time to post what's wrong > with the proof and the correct solution. As you saw, I did post my response about an hour before you posted that. > The problem is that even if the set of N points is in general > position, the set of N(N-1)/2 perpendicular bisectors is not! > In particular, given any 3 of the points, A, B, and C, the three > perpendicular bisectors these generate all meet in a point. Right. > To look at it another way, the perpendicular bisector of AB divides > the plane into the "A before B" and "B before A" half-planes, but > if you look at the 7 regions generated by 3 lines in the plane that > don't intersect in a point, the bounded triangular region is the > set of points that are closer to A than B, closer to B than C, and > closer to C than A, and that doesn't exist. I had realized that the bisectors meet at a point, but hadn't thought of what it would mean if they didn't. Your non-transitive ordering is a very intuitive reductio-ad-absurdum proof that perpendicular bisectors of a triangle meet at a point, one that makes it obvious that this rule generalizes to higher dimensions and to non-Euclidian spaces.

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[math-fun] Minsky "starfish"
by Bill Gosper 15 Apr '16

15 Apr '16

For x₀ = 34, y₀ = 55, δ > 0, and ε = 4 (sin π/5]^2/δ, without the Floor operation, the Minsky iteration simply repeats five points on an ellipse whose shape depends on δ. If we include the Floors, we get a variety of fivish orbits. δ = 1 produces those intriguing fractal pentagrams gosper.org/rastar.pdf discovered by Corey and Julian, and perhaps earlier by Steve Witham. A portion of the x-y rug plot is Figure 31 (p17) in http://www.blurb.com/books/2172660-minskys-trinskys-3rd-edition . A few months ago, Adam Goucher made the startling observation that this plot defines a legal kites-and-darts Penrose tiling. (See the last three images in gosper.org/g12a.pdf (102MB!), the stills from my (and Rohan Ridenour's) recent G4G12 talk. (Rohan did the animations on a separate screen.)) Varying δ gives a variety of orbits. δ = 33/100 simply loops over ten points in five close pairs. But δ = 17/50 makes five fuzzy patches of 100 points each. And δ = 17/50 + 1/googol makes a pretty doily with 30071 points! gosper.org/mnskplt.png Similarly, δ = 14/25 gives a 5794 point starfish, while δ = 14/25 + 1/googol gives a 72721 point obese starfish. gosper.org/mnskplt1.png . Various other starfish are in gosper.org/mnskplt2.png . Warning: The third one is actually five sea-cucumbers having a little fun at your expense. The last two (δ =.62 and .63) are just printouts of merely period 5 loops illustrating the experimental fact that every perturbation of δ by as little as .01 seems to produce a different orbit. The first image (δ = 43/100) is the first 10 million points of a 31 billion point orbit! Here is every 5001st point of it: gosper.org/starfish43.png . Note the radius grew from 55 to > 200000. But something is (star)fishy. Finite Minsky orbits are unicursal. How can there be a graph vertex of degree 5 in the center?? --rwg

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[math-fun] 1000x1000 pictures from OEIS
by Warren D Smith 14 Apr '16

14 Apr '16

> From: Allan Wechsler <acwacw(a)gmail.com> > I just had a weird thought. I would like to see two 1000x1000 bit rasters, > where the bit (x,y) is marked in the first if Y immediately follows X in > any sequence in OEIS, and in the second if Y *eventually* follows X in any > sequence. > --Unfortunately I think both ACW's rasters will be entirely black -- or if they are not, then they will be as soon as OEIS adds just one more sequence! I mean, a sequence such that it includes every pair x,y as adjacent elements. (Such as the 2D square spiral...) More interesting 1000x1000 pictures can be imagined. Example, if integer x is followed by integer y in N sequences in OEIS, then color (x,y) with greyscale=MonotoneFunction(N). This picture then will tell us how "related" any two integers x and y are. If you want to be even fancier get all three colors involved, e.g. using RedScale as I just said, BlueScale based on 2-apart-adjacency, GreenScale based on 3-apart.

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[math-fun] Sequential triangle packing.
by Ed Pegg Jr 14 Apr '16

14 Apr '16

I've looked at 45-45-90 isosceles right triangle packing lately. The sequential square packing sequence is https://oeis.org/A081287 -- tightest packings in a rectangle for squares of size 1 to n. Triangles seem to pack better. For example, the order 3 set has 6 triangles with small side 1 to 3, and small side 1 to 3 times sqrt(2). These pack perfectly in a 3x7 rectangle (a nice little puzzle). The order 11 set of 22 triangles packs very nicely. --Ed Pegg Jr

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[math-fun] Number affinity in OEIS
by Allan Wechsler 14 Apr '16

14 Apr '16

I just had a weird thought. I would like to see two 1000x1000 bit rasters, where the bit (x,y) is marked in the first if Y immediately follows X in any sequence in OEIS, and in the second if Y *eventually* follows X in any sequence. I don't know how one would make one of these, though.

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