- mathfun@mailman.xmission.com - mailman.xmission.com

[math-fun] Fwd: Brownie problem
by James Propp 10 Nov '20

10 Nov '20

Here we go again. :-) ---------- Forwarded message --------- From: Alan Frank <alan(a)8wheels.org> Date: Sun, Nov 8, 2020 at 10:19 AM Subject: Brownie problem To: Jim Propp <JimPropp(a)gmail.com>, Jim Propp <jpropp(a)cs.uml.edu>, William Gasarch <gasarch(a)cs.umd.edu>, <laurenrose2(a)gmail.com> This e-mail originated from outside the UMass Lowell network. ---------------------------------------------------------------------- Math folks, On the one hand, I'd probably be in better shape if I didn't buy sweet pastries. On the other hand, they seem to lead to interesting problems. I was walking home yesterday, eating a brownie. They had given me an edge piece; I prefer inside pieces. I was wondering which there were more of; of course it depends on how the tray is cut. So here is a multi-part problem: 1. If exactly half the pieces are interior, what are the possible dimensions? 2. Same question if you extend the problem to three dimensions. 3. Continuing to higher dimensions, is there a closed-form expression for the number of possible solutions as a function of the dimension? I put the numbers that I knew into OEIS and did not see likely candidates. 4. Are there any interesting results if you vary the ratio? In two dimensions, there are six additional solutions for other integral ratios. --Alan

6 9

[math-fun] Cyclic Difference Sets
by ed pegg 10 Nov '20

10 Nov '20

A perfect cyclic difference set with n values has length L = (n-1)^2 + n and works mod L. All differences up to L-1 exist uniquely. 3 values is structurally unique, 1 solution. 4 values has 2 solutions. 5 values has 1 solution. 6 values has 5 solutions. 7 values has 0 solutions. 8 values has 1 solution. Do any subsequent values have more than 1 solution? {{{7,3},{0,1,3}}, {{13,4},{0,1,3,9}},{{13,4},{0,1,4,6}}, {{21,5},{0,1,4,14,16}}, {{31,6},{0,1,3,8,12,18}},{{31,6},{0,1,3,10,14,26}},{{31,6},{0,1,4,6,13,21}},{{31,6},{0,1,4,10,12,17}},{{31,6},{0,1,8,11,13,17}}, {{57,8},{0,1,3,13,32,36,43,52}}, {{73,9},{0,1,12,20,26,30,33,35,57}},{{91,10},{0,1,5,7,27,35,44,67,77,80}},{{133,12},{0,1,3,17,21,58,65,73,100,105,111,124}},{{183,14},{0,1,3,24,41,52,57,66,70,96,102,149,164,176}},{{273,17},{0,1,22,33,83,122,135,141,145,159,175,200,226,229,231,238,246}},{{307,18},{0,1,3,30,37,50,55,76,98,117,129,133,157,189,199,222,293,299}},{{381,20},{0,1,13,16,21,45,75,82,86,92,128,142,180,206,229,231,264,286,354,363}}}

2 1

Re: [math-fun] Fwd: Brownie problem
by Dan Asimov 08 Nov '20

08 Nov '20

Suppose the entire n-brownie (meaning: n-dimensional brownie) is the unit cube [0,1]^n, and the volume *not* containing any end pieces is the inner cube [x, 1-x]^n. Then the crossover point is where (1-2x)^n = 1 - (1-2x)^n, i.e., where 1 - 2x = 1/2^(1/n), or x = (1 - 1/2^(1/n))/2. I think a pretty case is where x = 1/3. For x = 1/3 we have 3^n = 2 so n = log_3(2). I think that would be my favorite brownie. —Dan

1 0

[math-fun] Two topology puzzles
by Dan Asimov 08 Nov '20

08 Nov '20

1. A familiar way to define a torus is to start with a square, say [0,1]x[0,1] in the plane, and identify corresponding points on the right and left edges, and also corresponding points on the top and bottom edges. (I.e., identify (t,0) with (t,1), and identify (0,t) with (1,t), for all 0 ≤ t ≤ 1.) Suppose that *in addition to those identifications*, we also identify (x,y) with (y,x) for all points (x,y) of the square (not just the edges). What familiar shape is the result? Note that we're only concerned with the topology, not the geometry. 2. Start again with a fresh, unidentified square [0,1]x[0,1]. Now: * Identify (t,0) with (t+1/2 (mod 1), 1) for all 0 ≤ t ≤ 1, and * Identify (0,t) with (1, t+1/2 (mod 1)) for all 0 ≤ t ≤ 1. What familiar shape is the result, topologically? —Dan

2 1

[math-fun] Sylvester's ax+xb=c Equation in Quaternions
by Henry Baker 06 Nov '20

06 Nov '20

If he were alive today, Sylvester might have tweeted: The quaternion equation ax+xb=c is the simplest nontrivial linear equation in this noncommutative division ring. It provides a "canary in the coal mine" moment for newbies to noncommutative algebra. Yes, we can translate this equation into a 4D matrix problem, and utilize all of the tools of matrix algebra to solve it, but it doesn't seem very satisfying to not solve the equation *within* the noncommutative algebra of quaternions. The problem with Sylvester's equation is that since we can't get both the a and b coefficients on the same side of the unknown x, we can't utilize the distributive law to put the unknown x into a single term. But by using an extremely clever feline trick, we can group the terms in such a way that we can commute *scalar* numbers such as N(a)=aa', N(b)=bb', S(a)=(a+a')/2, S(b)=(b+b')/2 with the quaternion variables to achieve the same thing. Here we go. We first multiply our equation on the left by a: a(ax+xb) = ac aax + axb = ac (1) We then multiply our equation on the right by b': (ax+xb)b' = cb' axb' + xbb' = cb' (2) Adding (1)+(2): aax + axb + axb' + xbb' = ac + cb' aax + ax(b+b') + bb'x = ac + cb' aax + (b+b')ax + bb'x = ac + cb' (aa + (b+b')a + bb') x = ac + cb' x = (aa + (b+b')a + bb')^-1 (ac + cb') which makes sense, so long as the coefficient of x isn't zero. But the equation is symmetrical, so we can work instead with a' & b: a'(ax + xb) = a'c a'ax + a'xb = a'c (3) (ax + xb)b = cb axb + xbb = cb (4) Adding (3)+(4): a'ax + a'xb + axb + xbb = a'c + cb a'ax + (a'+a)xb + xbb = a'c + cb xaa' + x(a+a')b + xbb = a'c + cb x (aa' + (a+a')b + bb) = a'c + cb x = (a'c + cb) / (bb + (a+a')b + aa') We thus have solutions for our Sylvester equation. But wait, there's more! I claim that the case where N(a)=N(b) is a most important case. Suppose that we have a 'particular' solution xp of ax+xb=c, i.e., a xp + xp b = c. If there are any other solutions of ax+xb=c, then they would involve 'homogeneous' solutions xh of the 'homogeneous' equation ax+xb=0, i.e., a xh + xh b = c. We can now add these two equations: a xp + xp b + a xh + xh b = c a(xp+xh) + (xp+xh)b = c which shows that xp+xh is also a solution of our original equation ax+xb=c. Let's better understand homogeneous solutions xh: a xh + xh b = 0 <=> a xh = - xh b => N(a xh) = N(- xh b) <=> N(a)N(xh) = N(-1)N(xh)N(b) <=> N(a)N(xh) = N(xh)(b) Cancelling N(xh) from both sides, we find that we require N(a)=N(b) for nontrivial solutions xh. But if N(a)=aa'=N(b)=bb'=N, then our particular solutions from above look like: x = (aa + (b+b')a + bb')^-1 (ac + cb') = (aa + (b+b')a + a'a)^-1 (ac + cb') = ((a + (b+b') + a')a)^-1 (ac + cb') = ((a+a'+b+b')a)^-1 (ac + cb') = ((2 S(a) + 2 S(b))a)^-1 (ac + cb') = (1/a) (ac + cb') / (2S(a)+2S(b)) = ((1/a)ac + (1/a)cb') / (2S(a)+2S(b)) = (c + a'cb'/N(a)) / (S(a)+S(b))/2 = (c + a'cb'/N) / (2S(a)+2S(b)) And x = (a'c + cb) / (bb + (a+a')b + aa') = (a'c + cb) / (bb + (a+a')b + b'b) = (a'c + cb) / ((b + (a+a') + b')b) = (a'c + cb) / ((a+a'+b+b')b) = (a'c + cb) / ((2S(a)+2S(b))b) = (a'c + cb) / (b(2S(a)+2S(b)) = (a'c/b + cb/b) / (2S(a)+2S(b)) = (a'cb'/bb' + c) / (2S(a)+2S(b)) = (a'cb'/N + c) / (2S(a)+2S(b)) = (c + a'cb'/N) / (2S(a)+2S(b)) In other words, *both particular solutions are identical* ! If we wish, we can write our equation in a slightly more symmetrical fashion for understanding eigenvectors: (S(a)+S(b)) x = (c + a'cb'/N)/2 Here, the coefficient of the unknown quaternion x is simply the scalar number (S(a)+S(b)); the 4D quaternion (c+a'cb'/N)/2 gives the appropriate *direction* in 4D. This form also makes obvious the problem that occurs when S(a)=-S(b); we can't divide by zero, so our 'solution' formula doesn't work in this case. We utilize (c+a'cb'/N)/2 instead of c+a'cb'/N because we will soon find that the map c -> (c+a'cb'/N)/2 is a useful *projection* operation, which preserves quaternions c which are oriented in certain directions.

1 0

[math-fun] Spoonerisms
by James Propp 06 Nov '20

06 Nov '20

Do there exist dictionaries of spoonerisms analogous to rhyming dictionaries? One could imagine a program that would go through a phonetic dictionary of English words, looking at each pair of words and trying to spoonerize them; usually this would fail, but if n is the number of words in the dictionary, there would be n-choose-2 pairs to try, so I'm guessing there'd be lots of hits. Jim Propp

5 4

Re: [math-fun] More than you wanted to know about rotations of 4-dimensional space
by Dan Asimov 04 Nov '20

04 Nov '20

I have that book and I partially recommend it. It's full of fascinating things. But there are parts of it where I found the writing particularly opaque. For instance, where they tried to explain triality for the octonions. —Dan Adam Goucher wrote: ----- On this topic of 4-dimensional rotations, I thoroughly recommend: 'On Quaternions and Octonions' by J. H. Conway and D. A. Smith. Amongst other things, it includes a classification of all finite subgroups of SO(4). -----

1 0

[math-fun] More than you wanted to know about rotations of 4-dimensional space
by Dan Asimov 04 Nov '20

04 Nov '20

I've been reading the Wikipedia article about rotations in four dimensional space, and it has some interesting differences from any other dimension. The group of all rotations of R^n is the "special orthogonal group" denoted as SO(n). It's a Lie group of dimension equal to (n-1) + (n-1) + ... + 1 = n((n-1)/2. SO(1) and SO(2) are respectively the group of order 2 and the circle group. For n ≥ 3, n ≠ 4: ----------------- * If n is odd, n = 3, 5, 7, ...: SO(n) is a simple group*. * If n is even, n = 6, 8, 10, ...: SO(n)/{-I,I} is simple. Here {-I, I} is the group generated by the central inversion -I, and this group of order 2 is the center of SO(n) (the normal subgroup of the elements that commute with the entire group). ----------------- But for SO(4) the situation is unique. Let S^3 denote the unit sphere in 4-space, i.e., the unit quaternions. Then any rotation of R^4 (a map h : R^4 —> R^4 that preserves all distances and orientation) can be described as x —> p x q for some p, q in S^3, where x is any element of R^4 thought of as a quaternion. Given h, the ordered pair (p, q) is almost unique: the only other pair also describing h is (-p, -q). (Because quaternions are associative, no parentheses are needed in x —> p x q.) An "isoclinic" rotation of R^4 is one where all nonzero points rotate by the same angle. It turns out that all isoclinic rotations can be described by either x —> p x (a left isoclinic rotation), or x —> x q (a right isoclinic rotation). The left isoclinic rotations form a group, isomorphic with S^3, as do also the right isoclinic rotations. And each of these groups is a *normal subgroup* of SO(4). Just like SO(n) with n even, n ≥ 6, the center of SO(4) is {-I, I}. But S(4)/{-I, I} is *not* a simple group. It is isomorphic to the cartesian product SO(3)×SO(3), and this is not simple because SO(3)×{I} and {I}×SO(3) are each a nontrivial closed normal subgroup. More at <https://en.wikipedia.org/wiki/Rotations_in_4-dimensional_Euclidean_space>. —Dan ————— * SO(n) is a group and a locally Euclidean topological space such that multiplication and inversion are continuous: i.e., it is a Lie group. Saying that a Lie group is "simple" means it has no nontrivial closed normal subgroup.

2 1

Re: [math-fun] Quaternion eigenvectors, Part Deux
by Dan Asimov 04 Nov '20

04 Nov '20

Henry Baker wrote ----- It is also well known that every 4D rotation can be expressed as the *independent* rotations of 2 planes orthogonal to one another, each with their own separate rotation. ----- In fact any rotation of R^n (orientation-preserving isometry taking the origin to itself) is in fact the result of floor(n/2) rotations on mutually orthogonal 2-dimensional planes. If the angles are all distinct and not 0 or π, then this decomposition is unique. Coxeter wrote a great paper on quaternions and rotations and reflections of 4-dimensional space: "Quaternions and Reflections", the Monthly, Vol. 53, No. 3 (Mar., 1946), pp. 136-146. —Dan

6 9

[math-fun] Morse and bitwise palindromes and reversals (was Re: 73)
by Keith F. Lynch 03 Nov '20

03 Nov '20

M F Hasler <mhasler(a)dsi972.fr> wrote: > Keith F. Lynch wrote: >> Dan Asimov wrote: >>> Or consider the number 73. Not particularly conducive to >>> symmetry, you might think. >> 73 is a palindrome in Morse Code: --... ...-- > This Morse code symmetry comes from the base-10 complement > palindromic symmetry, ... Right. It's also interesting to look into Morse Code text palindromes and words that are reversals of other words. The longest single-word Morse palindrome is FOOTSTOOL. Since the commonest letters have the shortest codes, the reverses of common letters tend to be common letters, so there are lots of words that have reversals. Spaces between words are simply pauses, so the reverses of spaces are spaces. It's fun to find sentences whose reverses are sentences. Such searches are hard to automate, since computers aren't good at distinguishing between valid sentences and meaningless word salad. (Or at least they weren't last time I checked.) There are 30 possible Morse characters of one, two, three, or four "elements" (dots and/or dashes), all of which are either assigned to letters or unassigned, so the reverses of letters are usually letters. Unfortunately, the letters C, J, and Z can't be used, since their reverses are unassigned, at least in English. (I think their reverses are assigned to accented letters in various foreign variants of Morse.) Also rich in meaningful reversals and palindromes is the five-bit ITA2 teletype code, often wrongly called Baudot. To minimize wear on mechanical teletypes, it was designed so that the commonest letters have the fewest 1-bits, so once again the reverses of common letters are common letters. The reverses of letters are always letters, and the reverse of a space is a space. The ASCII code isn't very good at this. Most letters don't reverse into letters. It might be fun to design a character code for the sole purpose of making palindromes and reversals as common as possible. There's no need to actually assign bit patterns, but just to choose which letters will be the reverse of which letters. The space will of course remain the space. With 26 letters, there are 25!! (i.e. 25 * 23 * 21 * ... * 5 * 3 * 1) = 7905853580625 such codes, so it's not completely intractable to simply try them all. In addition to making common letters reverse into common letters, you'd probably want to make common digrams (e.g. TH, HE, IN, ER, etc.) reverse into common digrams. For instance TH would be its own reverse if T and H were the reverse of each other. Also see my article on this subject in the November 2005 WSFA Journal, http://www.wsfa.org/journal/j05/b/index.htm#bp

1 0