September 2020 - mathfun@mailman.xmission.com

Re: [math-fun] Scrambled necklaces
by Dan Asimov 25 Sep '20

25 Sep '20

A few possibly dumb questions: 1. Does the circular character string have marked "ends", or merely a circular ordering? 2. Is it assigned a direction (orientation)? 3. It seems that the reversals are performed consecutively. Does it matter what order they're performed in? If so, what order should they be performed in? —Dan Keith Lynch wrote: ----- What, if anything, is known about the following process? Take a circular character string, i.e. the end is attached to the beginning (a "necklace"). For each instance of each character in it, in lexical order, reverse all characters between successive instances of that character. For instance: reverse [e] revesre [r] rsevere [s] esrerev [v] erersev Here's an example using *this* sentence: isexample'sinte anceng ere:Hen s* us *th Questions include: * For each size of necklace, what's the most such processes it could take to return to the original state? And how large an alphabet would that require? For instance "reverse" takes 14 steps: reverse erersev rsereve severer everser rerseve serever everers versere ersever erevers vererse erserev rsevere reverse For my sentence example above, 16320 steps are required to return to the original sentence. * Can you ever get a reverse of the original string? * Can you ever get a rotation of the original string? * What invariants are there, other than the obvious one that the quantities of each character are unchanged? * If someone were to alternate several times between this process and exclusive-oring the text with as many repeats of a secret string as it takes to cover the text, is there any way to decrypt the result that's faster than trying all possible secret strings up to the length of the actual secret string? -----

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[math-fun] Scrambled necklaces
by Keith F. Lynch 25 Sep '20

25 Sep '20

What, if anything, is known about the following process? Take a circular character string, i.e. the end is attached to the beginning (a "necklace"). For each instance of each character in it, in lexical order, reverse all characters between successive instances of that character. For instance: reverse [e] revesre [r] rsevere [s] esrerev [v] erersev Here's an example using *this* sentence: isexample'sinte anceng ere:Hen s* us *th Questions include: * For each size of necklace, what's the most such processes it could take to return to the original state? And how large an alphabet would that require? For instance "reverse" takes 14 steps: reverse erersev rsereve severer everser rerseve serever everers versere ersever erevers vererse erserev rsevere reverse For my sentence example above, 16320 steps are required to return to the original sentence. * Can you ever get a reverse of the original string? * Can you ever get a rotation of the original string? * What invariants are there, other than the obvious one that the quantities of each character are unchanged? * If someone were to alternate several times between this process and exclusive-oring the text with as many repeats of a secret string as it takes to cover the text, is there any way to decrypt the result that's faster than trying all possible secret strings up to the length of the actual secret string?

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[math-fun] Celebration of Richard Guy next week
by Neil Sloane 25 Sep '20

25 Sep '20

Dear Math Fun: Next Thurs/Friday there is a meeting to celebrate Richard Guy at the Univ. Calgary: https://science.ucalgary.ca/mathematics-statistics/about/celebrating-richar… (I'm an invited speaker for the Friday "Problems" section, but there are some legal problems that are yet to be straightened out.) Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane(a)gmail.com

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[math-fun] release forms for Zoom talks question
by Neil Sloane 23 Sep '20

23 Sep '20

Was asked to give a Zoom talk on Oct 2 at conference honoring an old friend. Nothing was said about recording it. So (quoting Dorothy Sayers) I pushed a tentative pawn forward, and asked if there were plans to record the talks and publish them (and were they going to ask my permission first?) , because the answer would determine what I would include in the talk. A couple of days later they send me an optional consent form to sign, perhaps written in a hurry, asking me to give them permission to record the talk, to publish it, and to give them ownership of everything I talked about. Including ownership of copies of all my slides. (An arm and a leg, in short.) I said No. So my question is, has anyone seen a reasonable consent form that one might be willing to sign when one gives a zoom talk somewhere, that does not not give up all rights? It might say, OK, you can record my talk and put it on your web site, but I retain all rights to the material, the slides, the images, the music, the content, the ideas, the algorithms, the computer code, etc. Zoom has been around long enough now that someone must have solved this problem. Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane(a)gmail.com

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Re: [math-fun] Newbie questions about Kleinian groups
by Dan Asimov 21 Sep '20

21 Sep '20

All I know is that a Kleinian group is defined as a discrete subgroup of the group PSL(2,C). Here C = the field of a complex numbers. PSL(2,C) is defined as the group of linear fractional transformations PSL(2,C) = {z —> (az+b)/(cz+d) | a,b,c,z in C with ad-bc ≠ 0} PSL(2,C) can be identified with the group of 2x2 complex matrices a b { } c d such that ad-bc = 1 under matrix multiplication, IF matrices M and -M are identified with each other. I.e., PSL(2,C) = SL(2,C) / {±I}. plane PSL(2,C) is of particular interest because it is exactly the group of conformal maps of the sphere S^2 to itself. Conformal means angle- and sense-preserving. The rotations of the sphere, which form the group SO(3), are certainly all conformal. So the rotation groups of the regular polyhedra are discrete subgroups of PSL(2,C), i.e., Kleinian groups. Note_1: The group of conformal transformations of the open disk, which is also the group of isometries of the hyperbolic plane, can be identified with the subgroup PSL(2,R) of PSL(2,C), defined as PSL(2,R) = {z —> (az+b)/(cz+d) | a,b,c,d in R with ad-bc ≠ 0}. Here the hyperbolic plane H^2 is identified with the upper half plane, i.e., the set H+ = {z in C | Im(z) > 0}. (S^2 is the unique simply connected surface with constant curvature = 1, and H^2 is the unique simply connected surface with constant curvature = -1.) (When a discrete subgroup of PSL(2,C) lies entirely within PSL(2,R), it is usually called a "Fuchsian" group, and the term Kleinian group is usually reserved for discrete subgroups of PSL(2,C) that are not Fuchsian.) Note_2: PSL(2,C) can also be identified with the isometry group of 3-dimensional hyperbolic space H^3, which makes it important in the study of 3-manifolds. This can also be identified with the conformal transformations of the open unit ball in R^3. Finite groups of rotations of S^2 can be the three isometry groups of regular solids, or a finite cyclic group, or else a dihedral group (the rotation group in 3D of a regular polygon). But a finite Kleinian group need not consist of only rotations of S^2. For instance, the element z —> 1 - 1/z is of order 3 in PSL(2,C) but it's not a rotation. But: Any finite subgroup of PSL(2,C) is known to be *conjugate* to a cyclic or dihedral group or the tetrahedral or octahedral or icosahedral group. (Theorem 6.3.1 in http://www.matcuer.unam.mx/~jseade/Libro-Klein.pdf .) —Dan Allan Wechsler wrote: ----- Because I'm just an amateur, I've never really engaged with the concept of Kleinian groups before, so I'm only just starting with elementary concepts. So I have a couple of really basic questions. I think I read that all the finite spherical groups (including the symmetry groups of all polyhedra) occur as Kleinian groups. Is that correct? If so, is it because PSL(2,C) has a subgroup that is equivalent to the rotation group of a sphere? Are any _other_ finite groups Kleinian? That is, is there a finite Kleinian group that is _not_ the symmetry group of a polyhedron? (When I say "is", I hope it's clear that I mean "is isomorphic to". When I say "finite", I mean a finite number of elements, not the concept of "geometrically finite" that I haven't yet wrapped my head around in the context of Kleinian groups.) -----

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[math-fun] Newbie questions about Kleinian groups
by Allan Wechsler 21 Sep '20

21 Sep '20

Because I'm just an amateur, I've never really engaged with the concept of Kleinian groups before, so I'm only just starting with elementary concepts. So I have a couple of really basic questions. I think I read that all the finite spherical groups (including the symmetry groups of all polyhedra) occur as Kleinian groups. Is that correct? If so, is it because PSL(2,C) has a subgroup that is equivalent to the rotation group of a sphere? Are any _other_ finite groups Kleinian? That is, is there a finite Kleinian group that is _not_ the symmetry group of a polyhedron? (When I say "is", I hope it's clear that I mean "is isomorphic to". When I say "finite", I mean a finite number of elements, not the concept of "geometrically finite" that I haven't yet wrapped my head around in the context of Kleinian groups.)

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Re: [math-fun] A symmetrical rotation of Euclidean space
by Dan Asimov 21 Sep '20

21 Sep '20

----- Okay, that makes things easier than T an arbitrary rotation around the same axis (it is possible to generalize though). ----- What does the word "axis" mean here? —Dan

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[math-fun] theorem about normal matrices
by Jean Gallier 21 Sep '20

21 Sep '20

Hello everybody, There is a spectral decomposition theorem about real normal matrices (AA^T = A^T A) which says that there is an orthogonal matrix P and a block diagonal matrix D such that A = P^ D P, where the blocks of D are either a scalar (real), or a 2 x 2 matrix of the form a -b b a with b > 0. So indeed this amounts to finding an orthogonal basis and in some pairwise orthogonal planes, A behaves like a rotation composed with a positive scaling. In the special case of an orthogonal matrix, the scalars are +1 or -1 and the 2 x 2 matrices are 2D rotation matrices cos theta -sin theta sin theta cos theta 0 < theta < pi. The corresponding vectors in P give you a plane. The angle theta correspond to the two eigenvalues cos theta + i sin theta and cos theta - i sin theta. If det(A) = 1, then the number of -1 is even, and you can group them together in 2 x 2 matrices that correspond to 2D rotations of angle pi. I don’t know how old the theorem is. It know that it is proven in Gantmaker (Theory of matrices), Berger (Geometry I), and I have a version in my recent book https://www.amazon.com/Algebra-Optimization-Applications-Machine-Learning/d… <https://www.amazon.com/Algebra-Optimization-Applications-Machine-Learning/d…> Please email me if you want a pdf. Best, — Jean

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Re: [math-fun] A symmetrical rotation of Euclidean space
by Dan Asimov 21 Sep '20

21 Sep '20

Fred, thanks for your note. (How are you doing, by the way?) Yes, it would have been better to write 0 ≤ j ≤ n initially, but isn't 1 ≤ j ≤ n+1 redeemed by "(indices are modulo n+1)"? As for the later reference to R^n, isn't that actually R^n because it's the orthogonal complement in R^(N+1) of the line in the (1,1,...,1) direction? —Dan ----- To Dan --- Multiple notational confusion alert! << For all j, 1 ≤ j ≤ n+1, T(e_j) = e_(j+1) (indices are modulo n+1). >> should read 0 <= j <= n ; also elsewhere, eg. O-P only for for n even! Then half-way thru', |R(n+1) mysteriously transmogrifies into |R(n) ... That theorem is not "Chasles' theorem"; however it might well be due to Chasles, or (more likely) Euler. Fred On 9/21/20, Dan Asimov <dasimov(a)earthlink.net> wrote: > I worked this out last fall and found it surprising: > > Let {e_0,...,e_n} be the standard basis of Euclidean n-space R^(n+1) > (e_j has a 1 as its jth coordinate and 0's elsewhere). > > Define the linear mapping T : R^(n+1) —> R^(n+1) as follows > by its effect on basis vectors: > > For all j, 1 ≤ j ≤ n+1, T(e_j) = e_(j+1) (indices are modulo n+1). > > It's easy to check that T is orientation-preserving only for n even, > and that T is always an isometry of R^(n+1). > > It's also clear that the vector (1,1,...,1) of all 1's is a fixed point > of T. This means that its orthogonal complement (1,1,...,1)^⊥, namely > > (1,1,...,1)^⊥ = {(x_0,...,x_n) in R^(n+1) | Sum x_j = 0} > > (which is a copy of R^n), is carried to itself by T. > > A nice theorem states that any rotation T of R^n for n even has n/2 > mutually orthogonal 2-planes that are each taken into themselves > by a rotation by some angle. (I don't know the name of this theorem; > can anyone tell me its name or who first discovered it?) > > So for n even, say n = 2k, our T as defined above, when restricted > to (1,1,...,1)^⊥, is actually a collection of rotations by angles > that we'll call theta_j, 1 ≤ j ≤ n/2. > > Question: What are these angles theta_j ? > > Suggestion: Try to guess before doing a calculation. > > —Dan -----

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[math-fun] A symmetrical rotation of Euclidean space
by Dan Asimov 21 Sep '20

21 Sep '20

I worked this out last fall and found it surprising: Let {e_0,...,e_n} be the standard basis of Euclidean n-space R^(n+1) (e_j has a 1 as its jth coordinate and 0's elsewhere). Define the linear mapping T : R^(n+1) —> R^(n+1) as follows by its effect on basis vectors: For all j, 1 ≤ j ≤ n+1, T(e_j) = e_(j+1) (indices are modulo n+1). It's easy to check that T is orientation-preserving only for n even, and that T is always an isometry of R^(n+1). It's also clear that the vector (1,1,...,1) of all 1's is a fixed point of T. This means that its orthogonal complement (1,1,...,1)^⊥, namely (1,1,...,1)^⊥ = {(x_0,...,x_n) in R^(n+1) | Sum x_j = 0} (which is a copy of R^n), is carried to itself by T. A nice theorem states that any rotation T of R^n for n even has n/2 mutually orthogonal 2-planes that are each taken into themselves by a rotation by some angle. (I don't know the name of this theorem; can anyone tell me its name or who first discovered it?) So for n even, say n = 2k, our T as defined above, when restricted to (1,1,...,1)^⊥, is actually a collection of rotations by angles that we'll call theta_j, 1 ≤ j ≤ n/2. Question: What are these angles theta_j ? Suggestion: Try to guess before doing a calculation. —Dan

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