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[math-fun] Re: Tetraflexagon question
by mcintosh＠servidor.unam.mx 10 Dec '02

10 Dec '02

Just under a year ago we had an exchange of messages in which JHC outlined his ideas and constructions for making tetraflexagons, with the proviso that they were all applicable to any kind of polygon. It is instructive to follow out his recipe for four-square flexagons, being about as short as they come. The first requirement is a frieze, generated from an initial polygon by reflecting the last polygon in one of its edges; JHC said N, S, E, W, but I prefer +, ++, - or numbers if there are lots of edges. The best way to follow this out is to orient the polygon, and advance so many edges from the last on to make the new reflection. The next requirement is a permutation, which says where a given polygon on the frieze ends up on the stack which is gotten by folding the frieze about its connecting edges. Since the polygon gets turned over with each new fold, it is convenient to enumerate the faces in a 2xn array, distinguishing top and bottom. The image, that is the corresponding permutation, of the ith polygon is written in the ith cell, alternating top and bottom row. That number plus 1 is written in the paired cell. JHC didn't say this; it is the Tukey triangle [n-gon] strip, and the matrix is mentioned in some of the flexagon articles. Actually it has three rows, because the turn in passing from one polygon to the next is noted in the extra row; that is where +'s, -'s, or whatever, go. A useful picture is to draw a polygon with enough edges for the length of the frieze and to draw labelled diagonals to show the permutation. This isn't the Tuckerman Tree, but it is related. One difference is that the labels are the turns. As JHC remarked, a basic requirement is that the sum of the turns be 360 degrees so that the final edge is parallel to the starting edge. That ``adding 1'' rule needs justifying, but presumably comes into play later on. When the assembled flexagon is opened up there are two consecutive members of the stack being separated, and they had better have the same colors to give a uniform appearance to the laid out flexagon. Anyway, if we accept this preparation, there are essentially two four-cycle permutations for a four square frieze: (1 2 3 4) and (1 2 4 3), the reasoning being that we can start where we want and go off in the direction we want. Turning sequences which add up to zero (ie. full circle) are (+ + + +). (+ + - -), (+ ++ -- -), (+ ++ -- -). There are others, such as {++ ++ ++ ++), but as JHC says, ``there are some additional restrictions.'' Some eliminate silly combinations, others eliminate impossible combinations such as hard knots. These combinations give sign sequence ---------------------------------------------------------- permutation | + + + + | + + - - | + ++ -- - | + ++ - -- | ---------------------------------------------------------- (1 2 3 4) | Natural | flaps |flap-scroll| Good f'gon| ---------------------------------------------------------- (1 2 4 3) |tubulate |tubulating| scrolling | crossed | ---------------------------------------------------------- The point here is that there are lots of things which can either be called flexagons or near flexagons. What I would call natural flexagons are only a part of the panorama; even so there is a second quite reasonable full flexagon, and four more objects with reasonable behavior. The scroll, in particular, admits of generalization and was more or less known before flexagons came around. So I guess we can see that many things can be promoted in the flexagon or origami books as ``flexagons.'' In a way, they are. - hvm ------------------------------------------------- Obtén tu correo en www.correo.unam.mx UNAMonos Comunicándonos

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Re: [math-fun] Thought up while drawing polygons ...
by Dan Hoey 09 Dec '02

09 Dec '02

"Steve Gray" <stevebg(a)adelphia.net> asks: > In the unit square pick 4 points A,B,C,D at random (x,y each > uniformly distributed). Draw lines AB, BC, CD, DA. What is the > probability that two of these lines will cross (that is, we get a > reflexive quadrilateral)? It's easy to see that the answer is two thirds of the probability that the convex hull of {A,B,C,D} has four sides. That holds for any i.i.d. selection of the points (consider the points drawn from a circle). > With 5 points there are several possible topologies: no > crossings, one, two, three, or five. Is 4 crossings possible? No, but I forget the good proof, and hesitate to give you a tedious one. For N points, just characterizing the topologies gets interesting. Do we even have a closed form for the maximum number of self-intersections of the tour {p1, p2, ..., pn, p1}? Dan Hoey(a)AIC.NRL.Navy.Mil

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Re: [math-fun] unimodality question
by asimovd＠aol.com 09 Dec '02

09 Dec '02

Jim Propp writes: << Here's a fun question from Hunter Snevily (snevily(a)uidaho.edu; not a member of math-fun): Let x1,....xn be n real numbers that sum to 0 (i.e. x1+x2+...+xn=0) For each of the 2^n possible subsets of the xi's we associate the sum of the elements in that set. By def we assign the empty set the value negative infinity. Let ri be the number of subsets of size i that have sum >=0. (note that r0=0 and that rn=1) Claim 1. r0,r1,r2,...,rn is a unimodal sequence. Claim 2. For any 1<=j<=n, Bigsum i=0 to i=j ri >= Bigsum i=1 to i=j binom(n-1,i-1). >> Let's call the real numbers a1,...,an so we can use x1,...,xn as the coordinates of n-space. Let's assume that none of the sums of the numbers a1,...,an are exactly 0, except for a1 + ... + an, as given. Here's a geometrical way to look at this question. Let the corners of the unit n-cube be denoted as C_n = {0,1}^n. Let H be an arbitrary open halfspace in R^n whose boundary plane intersects C_n in precisely { (0,0,...,0), (1,1,...,1) }. Then the subsets of the indices {1,2,...,n} of size k correspond to those points S(k,n) of C_n lying on the plane x1+...+xn = k. The possible counts rk, k = 0,...n referred to in the question are the cardinalities #( S(k,n) int H ), k = 0,...,n over possible halfspaces H as above. Maybe someone else will take it from here. --Dan

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[math-fun] Clarification and correction
by Dan Hoey 09 Dec '02

09 Dec '02

Dan Asimov noted that I was somewhat unclear when I said > << > It's easy to see that the answer is two thirds of the probability that > the convex hull of {A,B,C,D} has four sides. That holds for any > i.i.d. selection of the points (consider the points drawn from a > circle). > >> I meant something like "drawn independently from any nonsingular probability distribution on a circle." And I chose a circle (not a disc) just to make the probability of the hull having four sides be 1, so that the probability of forming a reflexive quadrilateral would be 2/3. It's a symmetry argument--the three configurations A---B A---D A C | | \ / |\ /| | | X | X | | | / \ |/ \| D---C C---B D B are equally likely. Also, in my other recent message, I forgot and misquoted the Bayeux Tapestry. The correct phrase is "Isti mirant stella" ("They wondered at a star.") Dan

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[math-fun] unimodality question
by James Propp 09 Dec '02

09 Dec '02

Here's a fun question from Hunter Snevily (snevily(a)uidaho.edu; not a member of math-fun): Let x1,....xn be n real numbers that sum to 0 (i.e. x1+x2+...+xn=0) For each of the 2^n possible subsets of the xi's we associate the sum of the elements in that set. By def we assign the empty set the value negative infinity. Let ri be the number of subsets of size i that have sum >=0. (note that r0=0 and that rn=1) Claim 1. r0,r1,r2,...,rn is a unimodal sequence. Claim 2. For any 1<=j<=n, Bigsum i=0 to i=j ri >= Bigsum i=1 to i=j binom(n-1,i-1). Both results would be sharp. Example 0 (sum of all xi's) -1 -1 -1 -1 -1 4 (4-subset sums) 3 3 3 3 -2 -2 -2 -2 -2 -2 (3-subset sums) 2 2 2 2 2 2 -3 -3 -3 -3 (2-subset sums) 1 1 1 1 -4 (this row is our xi's) -infinity (empty set) Compare with 3,3,3,-4,-5 as second set of xi's 0 -3 -3 -3 4 5 -6 -6 -6 1 1 1 2 2 2 9 6 6 6 -1 -1 -1 -2 -2 -2 -9 3 3 3 -4 -5 -infinity Do you have any suggestions? Is this known? Is this trivial? Is this false? Thanks for your help. Hunter

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[math-fun] Berloquin 5 heads/5 tails puzzle
by James Propp 09 Dec '02

09 Dec '02

Tim Chow (tchow(a)alum.mit.edu) asks the following question, which he has already posted to rec.puzzles and sci.math and has asked me to post here as well (please send replies directly to him as well as to the math-fun list, since he is not a current subscriber): Subject: Berloquin 5 heads/5 tails puzzle In one of Pierre Berloquin's books (_100_Perceptual_Puzzles_ I think), he describes the following puzzle. We have ten coins in a row, five heads and five tails, tightly packed with no gaps between adjacent coins. H H H H H T T T T T The object is to rearrange the coins into an alternating heads/tails pattern, i.e., one of the following two patterns: H T H T H T H T H T T H T H T H T H T H Again there are no gaps between adjacent coins. The rearrangement must be accomplished in five "moves," where a move consists of taking any two touching coins and moving them into a two-coin gap, if such a gap exists, or to one end of the row, if no such gap exists. For example, initially there is no two-coin gap, so a possible move would be to take the "H T" coins in the center and move them to the right end, yielding H H H H T T T T H T Now there is a two-coin gap, so on the next move we would have to take two adjacent coins and move them into the gap, e.g., we could try H H H H T H T T T T As these examples illustrate, the coins are not allowed to switch places during the move, i.e., of the two moved coins, the one on the left before the move must remain on the left after the move. Also, moving coins to the end of the row is illegal if there is a two-coin interior gap. The two coins that are moved must be touching; so for example the isolated "T" at the right end in the last picture above cannot be moved on the next move. Berloquin gives a solution, but I will not spoil your fun by giving it here. I got quite interested in this puzzle many years ago when I first learned it (from a friend actually, not from Berloquin's book), and I believe I managed to show that the obvious generalization to 2n coins can be solved in n moves for all n >= 3. In my solutions I never needed to move a coin that was immediately adjacent to the gap, although I don't think this constraint was imposed by Berloquin. Unfortunately, I did not save my solution. So here are my questions. 1. Is Berloquin the inventor of this puzzle? 2. Has a general solution for n >= 3 been published somewhere? 3. How many solutions are there as a function of n? (Possibly intractable.) 4. Is the puzzle still solvable if we impose some extra "elegance" constraints? For example, we might want to require that the moves alternate "left, right, left, right, ..." (where by a "left" move I mean a move that takes the moved coins to a location that lies to the left of their location before the move, and similarly for "right"). (Jim Propp)

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[math-fun] triangoops, solid angle from three wedges
by R. William Gosper 09 Dec '02

09 Dec '02

Nick Baxter points out that in answering >what is the smallest constant c so that the graph of the function >f(x) = x^3 - c x contains the vertices of an equilateral triangle? I somehow edited sin((%pi/9)) into sin(pi)/9. Please imagine c = sqrt(8 - 8 sqrt(3) sin(pi/9)) ~ 1.80578 . Claim: The midpoint of one triangle side is the origin. Otherwise, centering some side and rotating its endpoints back onto the curve will intersect the other two sides with the curve, indicating that c is non-minimal. So let y(x) := x^3 - c x, and the vertices be z := x + i y(x), -z, and z sqrt(-3) = w + i y(w). Eliminate w from the last eqn, leaving a polynomial p(x,c) = 0 biquartic in x and cubic in c. For minimality, dc/dx = 0. This is equivalent to resultant(p,dp/dx,x) = 0, which gives a bicubic in c. Experimentally, c=1 gives no real roots x. c=2 gives two pairs. The critical c should be where these merge into double roots, i.e., when p and dp/dx have a common factor. Which is what the resultant finds. ------- A useful formula I can't find in MathWorld is for the solid angle Omega of a trihedral vertex with wedge (or sphere arc) angles a, b, c: 2 (cos(c) + cos(b) + cos(a) + 1) Omega = acos(-------------------------------------- - 1) (cos(a) + 1) (cos(b) + 1) (cos(c) + 1) With this you can readily solve for the edge or inradius of regular and nearly regular n-sided solids by setting Omega = 4 pi/n. Sucker bet: Which is greater, the edge of a regular icosahedron or its circumradius? Answer: e/r = sqrt(2-2/sqrt(5)) . I had to stare at a picture to believe this. --rwg

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[math-fun] Simultaneous rational approximation of two irrationals
by Eugene Salamin 09 Dec '02

09 Dec '02

If x is irrational, and I wish to best approximate the ratio x:1 by integers a:b, then I would use continued fractions. Suppose x and y are irrational, and I wish to best approximate the ratio x:y:1 by integers a:b:c. Is it known how to do this? __________________________________________________ Do you Yahoo!? Yahoo! Mail Plus - Powerful. Affordable. Sign up now. http://mailplus.yahoo.com

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Re: [math-fun] Simultaneous rational approximation of two irrationals
by Mike Stay 08 Dec '02

08 Dec '02

>Gene asks: > ><< >If x is irrational, and I wish to best approximate the ratio x:1 by integers >a:b, then I would use continued fractions. Suppose x and y are irrational, >and I wish to best approximate the ratio x:y:1 by integers a:b:c. Is it >known how to do this? >>> Depends what you mean by 'best'. For two irrationals, you might want to look into cubic continued fractions. Here are two papers on bifurcating continued fractions that give good approximations; the determinant of three consecutive approximations is 1 (whereas normal CF's give +/-1), so in that sense they're optimal. (arxiv.org/pdf/math.GM/0002227 and /0008060) -- Mike Stay staym(a)clear.net.nz -- Mike Stay staym(a)clear.net.nz

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Re: [math-fun] Simultaneous rational approximation of two irrationals
by asimovd＠aol.com 08 Dec '02

08 Dec '02

Gene asks: << If x is irrational, and I wish to best approximate the ratio x:1 by integers a:b, then I would use continued fractions. Suppose x and y are irrational, and I wish to best approximate the ratio x:y:1 by integers a:b:c. Is it known how to do this? >> This is a very interesting question, and I don't have much useful to contribute, except that 1) The family of ratios x:y:1 is perhaps best naturally encoded as the projective space P^2 = R^3 - {0} / (x,y,z) ~ (ax,ay,az) where a runs over all nonzero reals, aka the set of lines through the origin in R^3. 2) The answer probably depends on how the goodness of is measured; the most natural way seems to me to be the smallness of the angle between two lines, or equivalently the nearness of cos(angle) to 1. 3) The points in question will then be of the form +-(K,L,1)/sqrt(K^2+L^2+1) for integers K,L. The asymmetry of this expression suggests the last refuge of the scoundrel -- changing the question slightly. Perhaps we should allow the approximation to be via (K,L,M) for any integers K,L,M, leading to all points of the form +-(K,L,M) / ||(K,L,M)||, with sign(s) chosen so that a dot product is always positive. 4) An even more radical change-of-problem, but perhaps equally interesting, might be to look at the (dense set of) rational points +-(p,q,r) on P^2 (which especially John Conway discussed vis-a-vis spheres a few months ago) as possible approximators of an arbitrary ratio (x:y:1). Clearing denominators shows this is a subset of the points +-(K,L,M) / ||(K,L,M)|| of 3). --Dan

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