June 2017 - mathfun@mailman.xmission.com

[math-fun] Big Splat theory
by James Propp 04 Sep '17

04 Sep '17

Has anyone ever proposed in the Annals of Improbable Research or a similar venue that, in addition to undergoing periodic reversals of its magnetic field, Earth is also subject to occasional short-lasting reversals of its gravitational field? Such a phenomenon would handily explain selective extinction of megafauna: a mouse could survive an upward fall for five seconds followed by a downward fall for five seconds followed by an extreme deceleration, but a mastodon would not fare as well. :-) Jim Propp

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[math-fun] I was dropped from the math-fun list from about September 2016 until yesterday
by Neil Sloane 04 Jul '17

04 Jul '17

I was wondering why I hadn't seen any math-fun messages recently, and it turns out that because of a problem with Gmail, or something, I had been dropped from the list nine months ago. I will try to look at the math-fun archives some time, but if there was anything being discussed that had to do with sequences or the OEIS in the past nine months, and you happen to have a copy, please email it to me directly! 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] Unfolding questions
by David Wilson 01 Jul '17

01 Jul '17

1. For a given polyhedron, what is the minimal number edges that need to be cut to unfold it into a connected planar surface? For example, 3 edges are necessary for a tetrahedron, I think 7 for a cube. 2. Is this minimal number a function of the number of faces, edges and vertices of the polyhedron? 3. Is this minimal number unique? Does any unfolding of the same polyhedron have the same number of edge cuts? 4. If (3) is false, is there a polyhedron where some unfolding has more edge cuts, but shorter total edge cut length, than some other unfolding? 5. Is there a convex polyhedron for which some unfolding exhibits overlapping faces in the plane? If so, what is the smallest number of faces on such a polyhedron?

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[math-fun] Philip Yasskin wants to know
by James Propp 27 Jun '17

27 Jun '17

Philip Yasskin of Texas A&M taught me a fun game yesterday. (He learned it from his son, who learned it at a Renaissance Faire.) You roll a die, and the number it shows becomes the current score. The first player then rotates the die by 90 degrees, so that one of the four faces adjacent to the old top face becomes the new top face. The score increases by this amount. The second player does the same. Play alternates in this fashion until either the score becomes exactly 31 (in which case the player who just played wins) or the score exceeds 31 (in which case the player who just played loses). In some ways it's more natural to look at the "score deficit" (31 minus the score) as opposed to the score, especially if one wants to consider other versions of the game in which some number other than 31 is the "target". Philip noticed that when you look at the behavior of the game as a function of the score-deficit, you observe periodicity with period 9 (maybe he said "eventual periodicity" --- I forget), and he wondered why the 9 shows up. Here's a proposed variant that might shed light on Philip's question (or might further enlarge the darkness). For each k, we can consider a combinatorial game whose states are specified by ordered pairs (n,i), where n is a non-negative integer and i is between 1 and 2k inclusive. (For concreteness you might want to focus on k=3, and to think of i as the number shown on the top face of a 6-sided die.) A legal move goes from (n,i) to (n-j,j) as long as n-j is non-negative and j is neither i nor 2n+1-i. (For our concrete case k=3, j can be any number that is adjacent to i on the die.) Does this game exhibit eventual periodicity with period 3k for all values of k? That is, can we say that (n,i) is a P-position iff (n-3k,i) is, for all sufficiently large n? Jim Propp

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[math-fun] discrete log in reducible polynomials over GF(2) ??
by Henry Baker 25 Jun '17

25 Jun '17

This question is related to CRC polynomials, so the degree of the polynomials I'm (currently) concerned with are on the order of 32 -- i.e., far too small for crypto applications. The coefficients are 0's and 1's; denote the coefficient field F2 with elements {0,1}. Consider the polynomials in "x" over F2, i.e., F2[x]. Suppose we have a fixed polynomial p(x) in F2[x] of degree n, where n is in the approximate range of 32. We will work in the ring F2[x]/p(x), whose representatives will be the remainders of polynomials f(x) in F2[x] after division by p(x). Clearly, if r(x) is such a remainder, degree(r(x))<degree(p(x))=n. Now consider the multiplicative group generated by x itself in the ring F2[x]/p(x). We will **NOT** assume that p(x) is irreducible, so F2[x]/p(x) is not necessarily a finite field, and x may not generate the complete multiplicative group of F2[x]/p(x). I'd like to compute logarithms base x in this ring -- i.e., given the representative for the ring element x^i, I'd like to efficiently compute i. Is there any non-table-lookup method or any non-linear-search method for computing this discrete log? What about when p(x) IS irreducible -- i.e., F2[x]/p(x) really is a finite *field* ? BTW, since 32 is reasonably small, I'm willing to do a lot of precomputation; I just don't want to have to store a large table, and I'd love the final algorithm to be extremely fast.

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[math-fun] Find a function ...
by Guy Haworth 24 Jun '17

24 Jun '17

My college magazine includes this challenge: Find a function f(x) such that for every real x not equal to 0 or 1, we have f(1/x) + f(1-x) = x It's from Oxford BNC's Konstantin Ardakov. I guess there's a clue here that 1/x and 1/(1-x) feature in the solution ... but how does one go about this? Guy

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Re: [math-fun] two problems from a Facebook thread (SPOILER)
by James Propp 23 Jun '17

23 Jun '17

(I remember the days of putting in ^L ^L ^L at the top of an email before posting a spoiler. Is there something comparable that works for modern email readers?) The prisoners can achieve a probability of success exceeding 90% if each person sees which color predominates among the 99 hats that he sees and then guesses that his hat is that same color. This only fails when there's a 50-50 split, which happens about 8% of the time. I believe that this is best possible, but I don't have even the beginnings of a proof. Jim Propp On Sun, Jun 18, 2017 at 1:42 PM, Thane Plambeck <tplambeck(a)gmail.com> wrote: > (1) (via Peter Winkler) Here's a (seemingly) new hat puzzle for you: > > 100 prisoners will be fitted with red or blue hats according to fair > coinflips. Then the lights are turned on; each prisoner sees every other > hat color and must write down a guess for his own. If a majority (at least > 51) get their own hat color right, all the prisoners will be freed. > > As usual the prisoners can't communicate once the hats are placed, but have > time to strategize beforehand. Can they achieve a 90% probability of being > freed? How about 95%? > > (2) (Johan Wästlund) > I'll just throw in here a question of (what turns out to be) similar > flavor: > Suppose you play a type of one-person tic-tac-toe where in every "move" > you select a set of unoccupied squares, and then flip a coin to decide > whether you get all those squares, or your "opponent" gets them. What > winning probability can you achieve? > > -- > Thane Plambeck > tplambeck(a)gmail.com > http://counterwave.com/ > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >

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[math-fun] Lagrange point quintics
by Bill Gosper 23 Jun '17

23 Jun '17

https://en.wikipedia.org/wiki/Quintic_function#Application_to_celestial_mec… (recently added?) intriguingly claims that the L1 and L2 points for an elliptically orbiting planet are quintic roots of its orbital parameters. How can an eccentric planet have Lagrange points, unless the eccentricity is very low? --rwg

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[math-fun] Guardian problem corner
by Fred Lunnon 19 Jun '17

19 Jun '17

Three very easy but pleasant problems from Alex Bellos in today's Guardian, at https://www.theguardian.com/science/2017/jun/19/can-you-solve-it-pythagoras… WFL

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Re: [math-fun] two problems from a Facebook thread (SPOILER)
by Dan Asimov 19 Jun '17

19 Jun '17

I expect that when the count is close enough to 50-50, each individual maximizes their chance of being right if they guess the *opposite* of the color that constitutes the majority (if any) of what they see. —Dan -----Original Message----- >From: James Propp <jamespropp(a)gmail.com> >(I remember the days of putting in ^L ^L ^L at the top of an email before >posting a spoiler. Is there something comparable that works for modern >email readers?) >The prisoners can achieve a probability of success exceeding 90% if each >person sees which color predominates among the 99 hats that he sees and >then guesses that his hat is that same color. This only fails when there's >a 50-50 split, which happens about 8% of the time. >I believe that this is best possible, but I don't have even the beginnings >of a proof. >On Sun, Jun 18, 2017 at 1:42 PM, Thane Plambeck <tplambeck(a)gmail.com> wrote: > >> (1) (via Peter Winkler) Here's a (seemingly) new hat puzzle for you: >> >> 100 prisoners will be fitted with red or blue hats according to fair >> coinflips. Then the lights are turned on; each prisoner sees every other >> hat color and must write down a guess for his own. If a majority (at least >> 51) get their own hat color right, all the prisoners will be freed. >> >> As usual the prisoners can't communicate once the hats are placed, but have >> time to strategize beforehand. Can they achieve a 90% probability of being >> freed? How about 95%? >> >> (2) (Johan Wästlund) >> I'll just throw in here a question of (what turns out to be) similar >> flavor: >> Suppose you play a type of one-person tic-tac-toe where in every "move" >> you select a set of unoccupied squares, and then flip a coin to decide >> whether you get all those squares, or your "opponent" gets them. What >> winning probability can you achieve?

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