October 2018 - mathfun@mailman.xmission.com

[math-fun] Mathematics of SETI protocols ?
by Henry Baker 01 Nov '18

01 Nov '18

I've been thinking about the following problem for a while, and hope that others have also. The problem is: how do you establish communication with someone you don't know, and someone who doesn't speak the same language ? Somehow, when we are babies, we manage to figure out how to communicate with our parents. Somehow, when we are travelling, we manage to communicate -- to some extent -- with people who don't speak the same language. Let's put things into a bit more formal context. Suppose we have some number of Turing Machines that want to establish communications with one or more other TM's. Suppose that they all have some sort of shared linear channel -- e.g., audio -- that they can both listen to and send on. This is a *public* channel, so every TM can "listen"/"read" this channel, and with some error probability associated with local conditions and clock differences, get a significant percentage of the bits from the channel. There is plenty of noise on this channel (the "ether"), so not everything that is being received is necessarily from another TM. There may also be a certain amount of passive echoing, which is caused by the channel itself, and not another TM. Let's assume that the TM's are close enough that latency isn't a big issue. Let's assume that the TM's are initially in "similar" states -- i.e., each TM is aware of its own existence, and that there *may* be 1 or more other TM's trying to establish communications with it. We can't assume that any of the TM's have clocks synchronized with any other TM's. Each TM can't assume that there is only 1 other TM that wants to establish communication -- i.e., TM#1 may get confused if TM#2 and TM#3 both try to establish communication with TM#1 at the "same" time. Each TM has its own "identity" in the form of an *independent* (infinite) random stream of bits, so with enough tests, TM#1 can eventually distinguish between TM#2 and TM#3 -- assuming that it can ever get far enough into a communications protocol to perform various tests. Presumably, TM#1 sends out some sort of message with several parts: a part saying "HI", a part saying "This is a part of my identity XXXXX", and a part indicating "This is my idea of what time it is". TM#2 somehow has to *parse* this message out of the ether, and figure out that another TM is trying to establish communication, and which part of the message is "HI", which part is the other TM's identity, and which part is the other TM's clock time. Given the noisy channel, it may take a large number of attempts before any progress is made. Indeed, one aspect of this mental exercise is to determine noise bounds which might preclude ANY communications. It is my intuition that TM#1, TM#2, etc., can *eventually* -- after some number of "rounds" -- develop good hypotheses about whichever other TM's there are, and develop some number of *shared state* with those other TM's, where "shared state" can be measured in *bits*. I would imagine that some form of a "monkey see, monkey do" protocol works here. If TM#1 sends out a semi-long message -- say 256 bits -- and it later sees a significant number of those bits again with roughly the same clock frequency -- then either it has simply seen some sort of non-sentient *echo*, or some other TM is repeating those bits as part of *its* reply message. If the number of copied bits is large enough, then the chance that this reply message is random noise is negligible. So the first order of business for TM#1 is to send out a large number of test messages, with distinguishing subsequences of random bits, and at different clock rates, but *remember them all*, so that when it "sees" some of those same bits coming back, there is some chance that it could be another TM attempting communications. --- Does any of this discussion ring a bell? Surely, someone must have studied this sort of problem before?

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[math-fun] Matrix roots of polynomials
by Dan Asimov 01 Nov '18

01 Nov '18

Maybe I once learned something about this, but if so, I can't recall now: Consider a collection M of matrices over a ring, like M = M(n, Z) or M = M(n, Q) or M = M(n, R) or M = M(n, C), meaning the ring of n x n square matrices over the integers, the rationals, the reals, or the complexes. For any one of these, call it M and consider polynomials of form P(X) = X^k + A_(k-1) X^(k-1) + ... + A_1 X + A_0 where the A_j belong to the matrix ring M. Then what is known a) about the existence of solutions X in M to the equation P(X) = 0 (where 0 denotes the 0 matrix in M) ??? b) about closed formulas for the roots of P(X) in M ??? (like the quadratic, cubic, and quartic formulae). Of course, there can't be any general quintic formula* for matrices, since that would contradict Abel's theorem on its unsolvability over C. —Dan ————————————————————————————————————————————————————————————————————— * Allowing the four arithmetic operations as well as rational powers.

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Re: [math-fun] Fwd: Johnson solids trivia challenge
by Dan Asimov 30 Oct '18

30 Oct '18

Ah, that's a shame. The world also lost Branko Grünbaum, another great polytope researcher, last month: (https://en.wikipedia.org/wiki/Branko_Grünbaum) (I was confusing his 1966 work on polyhedra with regular faces — an outgrowth of his dissertation — and his unpublished work on uniform polyhedra. Wikipedia also says it was Victor Zalgaller who proved Johnson's list of 92 "Johnson solids" was complete.) But it's good to learn that the unpublished work on uniform is almost complete. I wonder who might have a copy. —Dan ----- Norman Johnson died in 2017; his Wikipedia article mentions the status of his late work. He completed _Geometries and Transformations_, which was published this year; he left an unfinished work on uniform polytopes, which will also be published in some form. On Tue, Oct 30, 2018 at 10:09 AM Dan Asimov <dasimov(a)earthlink.net> wrote: > Btw, when I e-mailed Johnson a few years ago, he was still working on > completing > his magnum opus, which I think is this same classification (or am I > confused?). > > Anyone know if it's now complete? -----

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Re: [math-fun] Fwd: Johnson solids trivia challenge
by Dan Asimov 30 Oct '18

30 Oct '18

Btw, when I e-mailed Johnson a few years ago, he was still working on completing his magnum opus, which I think is this same classification (or am I confused?). Anyone know if it's now complete? —Dan Allan Wechsler wrote: ----- Johnson chose to enumerate Regular-faced Aplanar Convex (RAC) polyhedra pretty much exactly because it was a finite class that wasn't too hard to cover methodically. ... -----

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[math-fun] Fwd: Johnson solids trivia challenge
by Bill Gosper 30 Oct '18

30 Oct '18

1. Chain of cubes. 2. Chain of truncated octahedra. (Make toroids!) Final SPOILER: "msirPlanogaxeHdetnemguA___" Search included Archimedeans and Platonics. (But not Continental Drift. For which see Road & Track, featuring additional spoilers.) On 2018-10-28 21:19, Allan Wechsler wrote: > Johnson chose to enumerate Regular-faced Aplanar Convex (RAC) polyhedra > pretty much exactly because it was a finite class that wasn't too hard to > cover methodically. Allowing flat or concave dihedrals creates infinite > classes about which it's hard to say much. Are you asking what the largest > dihedral would be in these classes? Obviously pi for flat dihedrals; my > vague intuition is "no upper bound" for concave dihedrals, but I have only > about 60% confidence in that answer. > > On Sun, Oct 28, 2018 at 11:50 PM Bill Gosper <billgosper(a)gmail.com> wrote: > >> Mathematica knows all 92. (Regular polygonal faces, dihedrals < π, not >> ordinary prisms.) Nor antiprisms! —rwg >> Name any of the four with largest dihedral. According to ries, >> that angle satisfies sinpi(sin(x)) = cospi(1/sqrt(6)).! —rwg >> Spoiler: (Kids: Don't peek! This is a nice exercise.) >> StringReverse@"]&]]\"selgnAlardehiD\",#[ataDnordehyloP[xaM@N >> ,][ataDnordehyloP[yBlamixaM@/&]#,#@ataDnordehyloP[delebaL" >> —rwg >> Gotcha: Why does this fail without the N@? >> Easy questions: Why disallow dihedrals > π (concavities)? >> Invisible edges with dihedral = π?

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[math-fun] Whisks
by James Propp 29 Oct '18

29 Oct '18

My 8-wire balloon whisk has wires that, read clockwise, follow the pattern 1 5 2 6 3 7 4 8, where 1 stands for the shortest wire, 2 stands for the second-shortest wire, etc. But Drew Lewis has an 8-wire whisk that goes 1 5 8 4 6 2 7 3. (He also has a 1 6 2 7 3 8 4 9 5 10 and a 1 4 2 5 3 and a 1 3 2, and I have a 1 2 3 4.) Is there any rhyme or reason (or math!) behind this diversity? (I personally would like to have a 1 5 3 7 2 6 4 8 whisk, since I’m a fan of the van der Corput sequence.) Jim Propp

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[math-fun] Johnson solids trivia challenge
by Bill Gosper 29 Oct '18

29 Oct '18

Mathematica knows all 92. (Regular polygonal faces, dihedrals < π, not ordinary prisms.) Name any of the four with largest dihedral. According to ries, that angle satisfies sinpi(sin(x)) = cospi(1/sqrt(6)).! —rwg Spoiler: (Kids: Don't peek! This is a nice exercise.) StringReverse@"]&]]\"selgnAlardehiD\",#[ataDnordehyloP[xaM@N ,][ataDnordehyloP[yBlamixaM@/&]#,#@ataDnordehyloP[delebaL" —rwg Gotcha: Why does this fail without the N@? Easy questions: Why disallow dihedrals > π (concavities)? Invisible edges with dihedral = π?

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Re: [math-fun] __agramma Mirificum
by Dan Asimov 28 Oct '18

28 Oct '18

I don't think so. Traversing each 2π side will have no net effect. —Dan Don Reble wrote: ----- > Also of interest would be simple closed curve with the same property > that bounds the least area (are they the same?). Does a 2pi x epsilon rectangle have the property? -----

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[math-fun] __agramma Mirificum
by Don Reble 28 Oct '18

28 Oct '18

> Also of interest would be simple closed curve with the same property > that bounds the least area (are they the same?). Does a 2pi x epsilon rectangle have the property? -- Don Reble djr(a)nk.ca

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[math-fun] Pentagramma Mirificum
by Bill Gosper 28 Oct '18

28 Oct '18

On 2018-10-27 14:17, James Buddenhagen wrote: > To make the pentagram with all right angles and edges equal pieces > of great circles on the surface of a sphere the central angle of each > edge is t, where cos(t) = (1-sqrt(5))/2 = -0.618... , about 128.17 degrees. > Thanks. That's a lot less mirificum than I thought. But more deceptive. Maybe it's just my declining comprehension, but I think a lot of web pages have it wrong. Including a paper by Arthur Cayley! —rwg > On Sat, Oct 27, 2018 at 8:05 AM Bill Gosper <billgosper(a)gmail.com> wrote: > >> (__agramma is neuter?) >> I just learned of this. >> Choose a point on a 2-sphere and traverse 90º of a great circle. >> Hang a 90º left turn. Traverse another 90º arc. Turn. Iterate. >> When, if ever, do you return to your starting point? —rwg >>

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