April 2016 - mathfun@mailman.xmission.com

Re: [math-fun] Simple finite group problem
by Keith F. Lynch 24 Apr '16

24 Apr '16

This is a reply to four people, so don't don't skip this unless you're skipping the whole topic. Please don't quote this whole message if you reply to it. Quoting is intended only to establish context. Would you soak a textbook in highlighter ink? I miss the days when mail programs would reject any message containing more quoted text than new text. A good rule of thumb is to only quote what you would manually retype if your mail program didn't support automatic quoting. Andy Latto <andy.latto(a)pobox.com> wrote: > Keith F. Lynch <kfl(a)keithlynch.net> wrote: >> Then, for each identity element I, multiply it by all other numbers >> 2 through N. If I*J is not J, it's not the right identity element >> for J. Otherwise, keep multiplying J by itself until I either get >> I back (win) or I get 0 (fail). If I didn't fail, add the group to >> the list. At the end, eliminate duplicates and list them all. > aren't you only finding the cyclic subgroups with this algorithm? Oops, yes. One reason it took me a few days to reply was because I've been trying to figure out how to fix this. (Another mistake in that paragraph is that if it fails, it wouldn't necessarily ever hit 0. But that's easy to fix. Most simply, if it finds more than N elements, fail, as there must be duplicates.) One way would be to add to my list every union of every subset of the cyclic multiplicative groups mod N with a common identity element, while eliminating any duplicates. The problem with this is that it would soon become very slow, like my first program, especially since I'd have to check every new group to make sure it was a group, and if not, add elements until it was. (The union of two groups isn't necessarily a group.) I ran my first program through N=36. That meant checking 2^36 subsets. It took more than a day to run. N=50 would have taken decades, and N=70 would have taken geological ages. The new program would be better, but the practical limit would be N=194. (For N=195, I=1, there are 40 cyclic groups, hence 2^40 subsets to check. That's more than a trillion.) I'm hoping to get to at least N=1000. My second program gets to N=1000 in about 2 minutes. But, as you point out, it only finds cyclic groups. For the record, the largest set of cyclic groups it's found (for N<1000) is 168 for N=819, I=1. That's probably because phi(819) is 2^4 * 3^3, i.e. it has lots of factors. Another approach would be to find just the generating set, i.e. the cyclic groups containing elements not in any other cyclic groups for the same N and same I. I've just written a program to find all of them. Again, it takes about a minute to get to N=1000. Again, N=819, I=1 holds the record, with a generating set of 130. The practical limit is probably N=272, as N=273, I=1 has a generating set of 40, so 2^40 subsets to check. (The values of N for which there's an I for which there's a generating set with more than one element, i.e. for which there's a group that's not cyclic, appears to be given by A033949. Also, it appears that if it's true for any I, it's true for I=1.) Okay, I think I've got it now. For every *pair* of cyclic groups containing an element not in any other cyclic group, i.e. single- generator groups, I'll generate a new group by taking their union and completing it. That will get me every group with 2 generators. Then I'll take the pairwise union of every single-generator group with every 2-generator group, and that will give me every 3-generator group, etc. I'd continue like that until a step found no more. Let me try it by hand, for N=30, I=16: Here are all the single-generator groups: A: 2 4 8 16 B: 14 16 C: 4 16 22 28 D: 16 26 Now let me take the pairwise unions of them, then complete them: A U B: 2 4 8 14 16 compl: 2 4 8 14 16 22 26 28 A U C: 2 4 8 16 22 28 compl: 2 4 8 14 16 22 26 28 -- duplicate A U D: 2 4 8 16 26 compl: 2 4 8 14 16 22 26 28 -- duplicate B U C: 4 14 16 22 28 compl: 2 4 8 14 16 22 26 28 -- duplicate B U D: 14 16 26 compl: 4 14 16 26 C U D: 4 16 22 26 28 compl: 2 4 8 14 16 22 26 28 -- duplicate That didn't quite work, as it missed two groups: {16} and {4,16}. Maybe if I also took the pairwise intersections? Sigh. > {1, 7, 13, 19} is not a group; 7 * 13 = 11. As Tom Rokicki pointed out, it is. (10-3)(10+3) = 100-9 = 91 = 1 mod 30. That came from my first program, which I designed for extreme reliability at the expense of speed. I would be astonished if it made any mistakes. I use its results to check the results from my faster programs. > This is not a big deal if you're generating the groups by hand, but > if you're doing it programatically, it means there's something wrong > with your program. I was generating them by a program. I'm not going to check 2^30 subsets by hand! That's more than a billion. Thane Plambeck <tplambeck(a)gmail.com> wrote: > Ie, there are 8 idempotents. So there will be 8 maximal subgroups. > Here they are. > In[1173]:= Map[allMutuals, idempotents] > Out[1173]= {{0}, {1, 7, 11, 13, 17, 19, 23, 29}, {6, 12, 18, 24}, {10, > 20}, {15}, {2, 4, 8, 14, 16, 22, 26, 28}, {3, 9, 21, 27}, {5, 25}} Interesting. > Listing all the subgroups is more work, but they're easily found via this > divide and conquer technique. How would you divide and conquer? Testing each of the 2^(M-1) identity-element-containing subsets of the length-M maximal subgroups would be practical for N=30, where M is at most 8, but completely impractical for much larger N. For instance for N=997 you'd have 2^995 subsets. That would take incomparably more eons than there are nanoseconds in a trillion trillion eons. You could pair the identity element with each other element in the maximal subgroup to make a new group, but that would only get you back to my list of all cyclic groups, and you'd have to build from there. Dan Asimov <dasimov(a)earthlink.net> wrote: > I like this a lot, I'm glad to hear it. > and am going to rewrite the maximal subgroups in my preferred form > ... for symmetry's sake: Don't be so negative. :-) > P.S. Now I'd like to know the same thing for the Gaussian integers > Z[i] and the Eisenstein integers Z[w] where w = exp(2pi*i/3). > These apparently have some complicated factor rings. See > e.g. http://home.wlu.edu/~dresdeng/papers/factorrings.pdf. Maybe I'll work on that after I finish this problem. Thane Plambeck <tplambeck(a)gmail.com> wrote: > possible theorem (?): the union of elements in the maximal subgroups > exhaust all moduli (as they do for thirty) iff N (ie 30, here) is squarefree I made the same conjecture here on the 20th: Mod N, the number of numbers that *don't* appear in any multiplicative group appears to be given by A055654. It appears to be always 0 for square-free N, and never 0 if N is not square free. Those groupless numbers are always divisors of N. Tom Karzes <karzes(a)sonic.net> wrote: > An additional property that holds when N is even but not a multiple > of 4: Each maximal subgroup consists solely of either even numbers > or odd numbers, never both. You can take it a lot farther than that. That property holds for every group (not just maximal ones) for every even N (not just ones not divisible by 4). [Update: I see that in a later message you noticed the latter.]

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[math-fun] The Dime and Penny Switcheroo
by James Propp 24 Apr '16

24 Apr '16

Now that my daughter and I have both solved "The Dime and Penny Switcheroo" (see page 21 of Martin Gardner's "Perplexing Puzzles and Tantalizing Teasers, or google "dime and penny switcheroo"), I'm wondering what the general story is. What if we start with n pennies and n dimes on a strip of length 2n+1? How many moves does it take to switch the pennies and the dimes? More generally, given two arbitrary configurations on the strip of length 2n+1, is there an easy way to determine the minimum number of moves required to turn one into the other? Surely this has been studied. (Is it explained in Winning Ways? I don't have my copy handy.) Jim Propp

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[math-fun] Polytopes centrally linearly Rupertable iff dual polytope is; same expansion factor
by Warren D Smith 23 Apr '16

23 Apr '16

(sorry, may have accidentally sent you a junk email, here comes the real one) LOVELY DUALITY (Rupert/Trepur?) THEOREM: A convex polytope in D dimensions, is centrally linearly Rupertable, if and only if its dual polytope is. Furthermore, the max-allowed expansion factors are equal, and the rotation matrices identical. DEFINITIONS: "convex polytope" = a compact set of nonzero D-volume that is an intersection of halfspaces. (Usual use assumes a finite number of halfspaces, but these theorems will not require that.) A convex polytope is "centrally linearly rupertable" = a rotated copied version of it can be passed thru a hole drilled thru it, where the motion is translation along a line and the two centers of mass coincide at some moment during the motion. The rotated copied version can still be passed thru even if scaled by factor E>=1, then we can consider the supremum of all allowed E, that's the "max expansion factor." If a convex polytope is the intersection of halfspaces x.v<=1 for various v, then the "dual" polytope is the convex hull of those vectors v, (the v are its vertices). Note this "duality" notion has also been called "polar" or "geometric reciprocation." The dual of a convex object is always another. The dual dual of a convex polytope is the original polytope back again (a fairly famous theorem whose proof is a short exercise in linear algebra) provided that (0,0,0,...,0) lies inside the original polytope. PROOF OF OUR THEOREM: Notation: let TruncIdent(m,n) mean the nXn diagonal matrix with first m diagonal entries all 1, remaining entries all 0. Let Pr be a matrix whose columns are the vertices of the primal polytope with center of mass (0,0,0,...,0). Let Du be a matrix whose columns are the vertices of the dual polytope. We are going to ASSUME that the origin (0,0,0,...0) is the crux point for Rupertability -- that is, if the moving rotated copy of Pr clearly lies inside a straight tunnel drilled thru P AT THE MOMENT both centerpoints coincide at (0,0,0,...,0) then it is Rupertable. This assumption seems valid if we are considering centrosymmetric Pr (and hence Du). But for non-centrosymmetric polytopes, most notably the regular simplex, the best Rupertization probably never features the two centers of mass coinciding. However, the argument we shall give actually is valid if the origin is ANY interior point of Pr, NOT necessarily the center of mass, provided we are considering only motions which cause the origin for Pr, to coincide with the origin for Du, at some one moment, and this moment is crux. So: We claim Pr is linearly rupertable if and only if there exists two DxD orthogonal matrices (a real square matrix is "orthogonal" if its transpose equals its inverse) Q1 and Q2 such that all entries of M = Du^T Q2^T TruncIdent(D-1,D) Q1 Pr (where Z^T denotes the transpose of matrix Z) are less than 1. Indeed, the max allowed expansion factor equals the reciprocal of M's maximum entry. This formalizes the idea that the orthogonal projection of Pr (after rotation by Q1) into the first D-1 dimensions, fits inside a copy of Pr rotated by Q2. Once you see that, then note that M^T = Pr^T Q1^T TruncIdent(D-1,D) Q2 Du which is the SAME THING except Pr and Du (and Q1 and Q2) interchanged! Since the entries of M and M^T are the same, Pr is linearly centrally rupertable iff Du is -- and the expansion factors, and rotation matrices Q1,Q2, are the same. QED. So, this seems to be a beautiful theorem. There are, however, some assumptions lurking around behind the scenes in the present argument and theorem statement, which it might be nice to clean up a bit. Hopefully my computer will soon start using this theorem to seek Rupertings for all the regular polytopes... (PS. I now also have a proof the regular simplex is Rupertable in all dimensions.) -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] Simple finite group problem (Thane Plambeck)
by Warren D Smith 23 Apr '16

23 Apr '16

> From: Thane Plambeck <tplambeck(a)gmail.com> > In every finite commutative monoid M, you get its maximal subgroups by.... > > (1) Computing first its set of idempotents X (ie those elements x such > that x^2 =x). > (2) Then, for each x in X, compute the subset S(x) of y's such that x > divides y and y divides x. > > Each set S(x) is a maximal subgroup of M with identity the idempotent x. > > For the particular case of the integers modulo 30, I got these idempotents > and maximal subgroups > > In[1164]:= idempotents = Select[elements, mult[#, #] == # &] > Out[1164]= {0, 1, 6, 10, 15, 16, 21, 25} > > Ie, there are 8 idempotents. So there will be 8 maximal subgroups. Here > they are. > > In[1173]:= Map[allMutuals, idempotents] > > Out[1173]= {{0}, {1, 7, 11, 13, 17, 19, 23, 29}, {6, 12, 18, 24}, {10, > 20}, {15}, {2, 4, 8, 14, 16, 22, 26, 28}, {3, 9, 21, 27}, {5, 25}} > > Listing all the subgroups is more work, but they're easily found via this > divide and conquer technique. --Mod p^e, with p=prime, the only idempotents are 0 and 1. (Proof: consider the final digit in the radix-p representation of the idempotent...) Therefore mod N, where N has known factorization into K prime powers, there will be 2^K idempotents, and hence 2^K maximal subgroups. Also, the largest subgroup will arise from the idempotent (1,1,1,...,1)=1 as identity, and it will have cardinality=EulerTotient(N) and consist of all residues relatively prime to N. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] Simple finite group problem
by Keith F. Lynch 22 Apr '16

22 Apr '16

This is a reply to three people, so don't don't skip this unless you're skipping the whole topic. Dan Asimov <asimov(a)msri.org> wrote: > Argh. Maybe this has fewer mistakes. > Tricky, that 5. ---------------------------- {0}, {1}, {5}, {6} {-1, 1}, {-4, 4} {-3, -1, 1, 3}, {-4, -2, 2, 4} {-4, -3, -2, -1, 1, 2, 3, 4} ---------------------------- The last one is not a group. The way I found them was, for every subset of {0,1,2,3,4,5,6,7,8,9}, to create a multiplication table and check every row to make sure each number in the maybe-group appeared once and only once. For a small number of small groups, it's practical to do this by hand. I also made sure that in at least one row the column headers weren't permuted, i.e. that there's an identity element. I didn't need to make sure that it's associative, since of course multiplication is associative. I didn't need to make sure each element had an inverse, since the Jth row contains the identity element for all J; the column that's in says what its inverse is. I plan to rewrite it to be much faster. For each number I, 2 through N-1, check to see if it equals its square mod N. If so, add it to the identity element table. (0 and 1 are put in the table without checking.) Then, for each identity element I, multiply it by all other numbers 2 through N. If I*J is not J, it's not the right identity element for J. Otherwise, keep multiplying J by itself until I either get I back (win) or I get 0 (fail). If I didn't fail, add the group to the list. At the end, eliminate duplicates and list them all. Of course, as with the first program, I will precompute the whole mod N table. There's no point in doing the same calculations multiple times. Tom Karzes <karzes(a)sonic.net> wrote: > Are you counting groups as distinct that are identical to each > other, but happen to use different numbers for the elements? For > example, {1, 3, 7, 9} is really the same group as {6, 2, 8, 4} (the > elements correspond in the order listed). Does it makes sense to > count it twice? I'm not sure what you're saying. Are you pointing out that those groups are isomorphic? Yes. In that sense there are only two groups of order 4, and only one group of order N whenever N is prime. In that same sense, the group of addition of reals is the same as the group of multiplication of positive reals; we've just relabeled all the elements. But nobody would suggest that means that multiplication isn't worthy of study because it's just addition with funny names and we already understand addition. > If so, then we're really identifying subsets of integers that > generate these groups, rather than the groups themselves. Well, yes. There's very little to be said about small groups themselves. They're useful tools for studying other things. Joerg Arndt <arndt(a)jjj.de> wrote: > "Keith F. Lynch" <kfl(a)KeithLynch.net> wrote: >> Mod N, the identity elements are those numbers I such that I^2 >> = I mod N. For N=10, those are 0, 1, 5, and 6. The numbers of >> identity elements mod N appear to be given by A034444. Mod 30, >> there are 8 of them, and 28 groups. > A group G has one neutral ("identity") element e by definition: > a * e = a = e * a for all a \in G *A* group, yes. But as I said, mod 30 there are 28 groups: Identity element 0: {0} Identity element 1: {1} {1,11} {1,19} {1,7,13,19} {1,17,19,23} {1,29} {1,11,19,29} {1,7,11,13,17,19,23,29} Identity element 6: {6} {6,24} {6,12,18,24} Identity element 10: {10} {10,20} Identity element 15: {15} Identity element 16: {16} {4,16} {2,4,8,16} {14,16} {16,26} {4,14,16,26} {4,16,22,28} {2,4,8,14,16,22,26,28} Identity element 21: {21} {9,21} {3,9,21,27} Identity element 25: {25} {5,25} > Usually the multiplicative group mod n consists of all u, 1<=u<n, > such that gcd(u,n)=1 (the "units"). ... > You seem to have something different in mind. I'd like to warn > about using existing terminology with non-standard meanings. What would you call what I'm doing? >> I think the order of any of these groups always divides phi(N). > The order of any subgroup of G is a divisor of the order of G. > As the order of the multiplicative group mod N is phi(N), so > your statement follows. But that applies only to the groups with an identity element of 1. The groups with different identity elements are obviously not subgroups of a group with an identity element of 1. So it's not obvious to me that they all must have orders that divide phi(N), but they do all appear to. > First step: define everything. I was hoping this already has a name. "Multiplicative groups mod N" is the obvious name, but if that's already taken for a subset of what I'm talking about, I'll have to use something else. Something short, so it will fit in a title of an OEIS entry.

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[math-fun] Possible relation between QR matrix factorization, ruperting simplexes, and GI problem
by Warren D Smith 22 Apr '16

22 Apr '16

Let the N+1 vertices of an N-simplex be the columns of an Nx(N+1) matrix. We can demand this matrix be upper triangular by translating the simplex so first column is (0,0,0,...,0) and then using the R (upper triangular) factor in the QR factorization of the remaining NxN matrix (Q is orthogonal). If we then deleted the first row, then we would get an (N-1)-dimensional simplex, which generically (i.e. if that first row had all entries unequal) would have every edge length shorter than it used to be before the deletion. If this (N-1)-simplex fits inside the original N-dimensional simplex, then that proves the original N-simplex was Rupertable. Unfortunately for that argument's importance, shortening edge lengths does not necessarily imply "fits inside." For example, a triangle with sides 999, 999, 9 cannot fit inside an equilateral triangle with sides 1000. Nevertheless, for many simplices I suspect some variant of that approach will succeed in Ruperting it. I would like to establish that all simplices are Rupertable, or find a counterexample. And if we have two N-simplices and want to know whether one fits inside another, that might be a hard problem because there are (N+1)! different ways to try. This "can it fit inside" problem seems at least as hard as the graph isomorphism problem. Specifically if I had a "fit inside" oracle for simplices then I could use it to solve all GI problems trivially as follows: The graphs must have the same number of vertices, edges and same valency-tuples in order to be isomorphic (of course) so say "no" if not. Let two simplex vertices A,B have distance 1+1000^(-N) if graph has edge AB, otherwise distance is 1. Ask oracle whether simplex #1 fits inside simplex #2 scaled by 1+1000000^(-N). "Yes" <==> graphs isomorphic. Laszlo Babai has recently claimed in an 84-page "preliminary" paper, to have a proof that graph isomorphism is soluble in "quasipolynomial" time, meaning exp(k*(logN)^c) time for some positive constants c and k: http://arxiv.org/abs/1512.03547v1 If correct, Babai's result is a major advance in computer science, one of the greatest in years. Before Babai the best runtime guarantee known was N^O(sqrt(N*logN)). -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Prince Rupert
by Warren D Smith 22 Apr '16

22 Apr '16

Rupert, Count Palatine of the Rhine, Duke of Bavaria, Duke of Cumberland, Earl of Holderness (German: Ruprecht Pfalzgraf bei Rhein, Herzog von Bayern), commonly called Prince Rupert of the Rhine, KG, PC, FRS (17 December 1619 - 29 November 1682), was a noted soldier, admiral, scientist, sportsman, colonial governor and amateur artist during the 17th century. He spent most of his young adult years being a top military commander on the side of (his uncle) King Charles on England, fighting the English civil war against Cromwell. Both cavalry and later navy. Various victories, defeats, captures, escapes. By the age of three he could speak some English, Czech and French, and mastered German while still young. Educated at Oxford becoming "master of the arts" 1636. Some claim Rupert was the inventor, or a major player in inventing, "mezzotint" printing/engraving process about 1662 (others say no); and he definitely became a noted artist using that technique. He also was allegedly one of the top 5 tennis players in England. During the 1670s on, he lived in Windsor castle and become involved in science, and was made a member of (not surprisingly) the "royal society." He also was the first governor of the Hudson's Bay Company; the area round Hudson Bay Canada still is called "Prince Rupert's land." Some credit him with inventing "Prince Rupert's drops," an early discovery of tempered glass. Others say no. Demonstrated new device for lifting water at the Royal Society, and received attention for his process for "painting colours on marble, which, when polished, became permanent." He also invented Prince Rupert's problem about cubes passing thru cubes, and was noted for his talents as a cryptographer. He invented various military inventions such as kinds of guns, torpedo, new process for making gunpowder, new alloys. A port town in British Columbia is named after him. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Generic Rupertability, and some non-Rupertable objects
by Warren D Smith 22 Apr '16

22 Apr '16

Of course, a sphere is not Rupertable (in any dimension). But in 3D, any generic ellipsoid (axes A,B,C with 0<A<B<C) is Rupertable. Indeed ellipsoids in any dimension are generically Rupertable. The expansion factor is (at least) MIN(j) Axis[j+1]/Axis[j] if the axis lengths are assumed pre-sorted into increasing order. For the same reason hyperrectangles are generically Rupertable. Any 2D convex object with diameter>width is Rupertable. Therefore every 2D convex polygon is Rupertable. But the "Reuleaux triangle" in 2D, made of 3 circular arcs, https://en.wikipedia.org/wiki/Reuleaux_triangle is NOT Rupertable since like the circle it has diameter=width. There exist an infinite number of 2D "constant width convex objects" which all are non-Rupertable. An upper bound on the Rupert expansion factor of any convex object in any dimension, is diameter/width. This bound is tight in 2 dimensions. It is usually weak, but always valid, in higher dimensions. Meanwhile a lower bound is, of course, 1. (Definitions: diameter is furthest distance between two points. Width is closest distance between two parallel hyperplanes, which object lies between.) A usually-better upper bound on the Rupert expansion factor of any convex object Q in any dimension, is as follows. Define OP(Q,d) to be the orthogonal projection of Q in direction d. The upper bound is diam(Q) / MIN[over all directions d](diam(OP(Q,d))). (Definition: a "direction" is a unit-length vector.) But this upper bound also often seems very weak. Regard an n-simplex as determined by its (n+1)*n/2 edge lengths. In 2D any triangle is Rupertable. Does there exist any non-Rupertable simplex? In any dimension? Hell: does there exist any non-Rupertable convex polytope? Not knowing that, is pathetic. Here is an attempt to make one. Make a convex polyhedron in 3D with lots of vertices (say at least V=10^12) all lying on the sphere x^2+y^2+z^2=1, fairly uniformly randomly distributed. It is their convex hull. Any orthogonal projection, is going to be a polygon with many (say, at least about 10^6) vertices, that closely approximates unit circle but generically lies inside it. The number of combinatorially-distinct types of orthogonal projections is going to be at least about 10^24. No matter which one you choose, if seems unlikely that there will exist another which can be moved to lie strictly inside it. Because there are only going to be about 10^12 "different" ways to move it. Point is, it feels like you do not have enough control to repair about 10^6 problems with only 3 degrees of freedom worth of control. Too many constraints, too few variables. So I suspect polyhedra like that in 3D, are usually going to be non-Rupertable when V is large enough. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] 3-coloring the rationals
by Keith F. Lynch 22 Apr '16

22 Apr '16

James Propp <jamespropp(a)gmail.com> wrote: > (a) the blue points form a multiplicative group (the 2-adic units in > the rationals, aka rational numbers of the form a/b where a and b > are odd integers) Right. Also just the positive blue numbers, as there's no way to get a negative by multiplying or dividing positives. > and (b) the blue and red points together form an additive group > (rational numbers of the form a/b where b is odd). Right. Also just the red numbers, as there's no way to get a blue number (odd/odd) by adding or subtracting red numbers (even/odd).

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Re: [math-fun] Ruperting dual polytopes
by Warren D Smith 21 Apr '16

21 Apr '16

Huber sent a paper Richard P. Jerrard and John E. Wetzel: "Universal Stoppers Are Rupert" College Mathematics Journal 39,2 (March 2008) 90-94 containing the pathblazing adjective "Rupert”, meaning Rupertable, plus pointing out that the regular tetrahedron, icosahedron, dodecahedron (i.e. all 3D regular polyhedra, since cube & octahedron are too) allegedly are Rupertable. Tetrahedron: Christoph J. Scriba, Das Problem des Prinzen Ruprecht von der Pfalz, Praxis der Math. 10 (1968) 241-246. Icos & Dodec: Richard P. Jerrard and John E. Wetzel, Platonic passages, 07 T-51-2, Abstracts Amer. Math. Soc. 28 (2007) 351. Prince Rupert generally: D. J. E. Schreck: Prince Rupert’s problem and its extension by Pieter Nieuwland, Scripta Math. 16 (1950) 73-80, 261-267. Also of course the regular N-gon in 2D is Rupertable with expansion factor sec(pi/N) if N=even 2*cos(pi/(2*N)) / (cos(pi/N)+1) if N=odd. So, only the sporadic 4D regular polytopes (and perhaps high-dimensional regular simplices) remain of unclear Rupertability... unless somebody already did them too... Warning: I have not seen these papers. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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