February 2017 - mathfun@mailman.xmission.com

[math-fun] Draft of my February 2017 blog post
by Bill Gosper 14 Feb '17

14 Feb '17

Almost(?) as bad: The prestigious Nueva School is using "factor" to mean "divisor" (to avoid a three syllable word?) so that "1 is a factor of 6". I actually saw a TV quiz show screw a contestant out of $1M by switching whether 1 could be a factor. --rwg Date: 2017-02-13 18:38 From: Eugene Salamin via math-fun <math-fun(a)mailman.xmission.com> To: math-fun <math-fun(a)mailman.xmission.com> Reply-To: Eugene Salamin <gene_salamin(a)yahoo.com>, math-fun < math-fun(a)mailman.xmission.com> It is unfortunate for someone knowledgeable in mathematics to lower himself to the level of the uninformed, instead of making an attempt to raise them up to a higher level. This reminds me of something I was told by a friend who volunteers in the Santa Cruz CA Public School. Apparently it's school doctrine that 1 is a prime number. The teacher told the class that 1 is prime, but added the qualification that mathematicians consider 1 to not be prime. Anyhow, James, it's your First Amendment right to say that π varies with the situation, and it's my First Amendment right to disagree with you. -- Gene

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[math-fun] Magic squares -- and surfaces
by Eric Angelini 13 Feb '17

13 Feb '17

This paper is interesting, I guess, for some of us: https://carresmagiques.blogspot.be/2017/01/area-magic-squares-and-tori-of-or der-3.html Best, É.

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[math-fun] Draft of my February 2017 blog post
by James Propp 12 Feb '17

12 Feb '17

Hi, I started writing a new draft titled "Three-point-one cheers for pi !" and would love to get your feedback. I plan on publishing it on the 17th. I can always use more links to content (images or videos) that are likely to interest readers of my essay. I am particularly interested in a link that would present Archimedes' method of calculating the volume of a ball in an accessible way. For that matter, if anyone has risen to Wigner's tacit challenge of explaining to a lay audience why pi is relevant to statistics, I'd love to know about it. Keep in mind that all math-fun feedback goes into one mail-feed, so I won't know whose feedback is whose unless you sign your comment. Also, all substantive suggestions that I use will be acknowledged (unless you specifically ask me not to do this). Please leave your feedback here: https://mathenchant.wordpress.com?p=1471&shareadraft=589f987f95cf7 Title: Three-point-one cheers for pi ! Beginning: Pi, that most celebrated of mathematical constants, leads a curiously double life. On the one hand, we define it as the ratio of the circumference of a circle to that circle's diameter, and, to the extent that we can imagine a world draped over the armature of a different geometry, we can conceive ... Read more: https://mathenchant.wordpress.com?p=1471&shareadraft=589f987f95cf7 Thanks, Jim Propp

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Re: [math-fun] math-fun Digest, Vol 168, Issue 9 re: Lifted from Littlewood, hacked from Hacking (Fred Lunnon)
by Stuart Anderson 09 Feb '17

09 Feb '17

> > > > On Tuesday, February 7, 2017, Fred Lunnon <fred.lunnon(a)gmail.com> wrote: > > > I had seen this before, long ago, but then forgotten it. > > > > A square can be dissected into finitely many unequal squares, > > but a cube cannot be dissected into finitely many unequal cubes. > > > > Why not? > > > > > this is spot on. *From:* dmoews(a)xraysgi.ims.uconn.edu (David Moews)*Date:* 11 Mar 1998 19:29:42 -0800*Newsgroups:* rec.puzzles,sci.math*Subject:* Re: No CUBE of cubes. In article <6e7eb1$2am$2(a)cantuc.canterbury.ac.nz>, Bill Taylor <mathwft(a)math.canterbury.ac.nz> wrote: |There is a standard puzzle - show it is impossible to have a cube of integer |side partitioned into a finite number of other cubes, also of integer sides, |and all of DIFFERENT sizes. ... |Why doesn't the same proof extend to "no square of unequal smaller squares", |which we know IS possible. Why can't we just look at the top edge |and the smallest square there, go across, then carry on as before? ... |What's specially different between 2D and 3D here? In this posting, by an Xd X I will mean an X partitioned into a finite number of Xs of unequal sizes. The proof is as follows: Look at the bottom of the cubed cube; you will see a squared square. *The smallest square, S, in this squared square cannot be on the boundary.* Therefore, the cube with S for a face is surrounded by four larger cubes, so its opposite face abuts another squared square. We can now look at the smallest square in this squared square. It is then the face of some cube surrounded by four larger cubes, which we can look at the opposite face of, and so on: since this process can be continued indefinitely, there can be no cubed cube. The difference between 2D and 3D is in the starred statement. The smallest line in a segmented segment can be on the boundary, but the smallest square in a squared square cannot be. -- David Moews dmoews(a)xraysgi.ims.uconn.edu

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[math-fun] Lifted from Littlewood, hacked from Hacking
by Fred Lunnon 08 Feb '17

08 Feb '17

I had seen this before, long ago, but then forgotten it. A square can be dissected into finitely many unequal squares, but a cube cannot be dissected into finitely many unequal cubes. Why not?

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Re: [math-fun] A sandpile game (SPOILER)
by Don Reble 07 Feb '17

07 Feb '17

> If ... you must play in the [only] zero pile, winning the game, does > that eliminate all loops, yielding a game that has Grundy values? Many plays reduce the zero population. The exceptions are plays into an Edge Sequence: a non-empty sequence of ones at an edge of the playing board. Such a play moves the bounding zero closer to the edge, and shrinks the Edge Sequence. If there is only one zero, that move expands or creates the other Edge Sequence: the game can loop. If there are more zeroes, the move doesn't affect the other Edge Sequence, and reduces the sum of those two sizes. The number of such moves is bounded, and then one must reduce the zero population. To do a slowest game, always move to the leftmost box, (N^2+N)/2 times. ... Ah, I didn't solve Jim's game before the deadline. But I did solve the vastly simpler game played on a circle instead of a line. >:-> -- Don Reble djr(a)nk.ca

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[math-fun] Proof of the Riemann Hypothesis
by Eric Angelini 07 Feb '17

07 Feb '17

... utilizing the Conway AF theory: http://bit.ly/2kigEKj Best, É.

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Re: [math-fun] A sandpile game (SPOILER)
by James Propp 07 Feb '17

07 Feb '17

The right move is to add a token to the already-occupied seventh box (causing it to fire to its two neighbors), turning 0 1 0 0 0 0 1 0 1 into 0 1 0 0 0 1 0 1 1; any other move loses against an ideal adversary. The general strategy is guided by the numerical sum of the “place-values” of the tokens, where a token in the 1st (leftmost) box has place value 1, a token in the 2nd box has place value 2, etc. You can check that when a firing occurs, the total value of the configuration doesn’t change, except when the 9th box fires, in which case the total value goes down by 10. So the winning strategy is, compute the value v of the current position, and then place a token in the unique position i such that v+i is congruent to 45 mod 10. (Why 45? Because it’s the value of 1+2+3+…+8+9, which is the value of the goal-position.) In combinatorial game-theory terms, the “P-positions” are those whose last decimal digit is a 5. It’s easy to show that if you follow this strategy (assuming that you aren’t faced with a position whose value already ends in a 5 when it’s your first turn to move) your opponent can’t win. But is there an easy way to show that the game must eventually end? One could cite lemmas from sandpile-theory, but that’d be cheating for my purposes. One needs to show that eventually there must be at least n-1 tokens on the board, no matter what the players do. From there, it’s easy to win in 1 move (unless someone has already won). I know a self-contained proof, but it’s not as simple as I’d like. Anyway, once you show that the game eventually ends, and you show that your opponent can’t win if you follow the place-value strategy (starting from a position whose value doesn't end in a 5), it follows that the strategy lets you win! I’m curious how hard it is to figure all this out. I never had a chance to figure it out for myself, because I reverse-engineered the game from my knowledge of the theory of sandpiles. My guess is that it would be fun and satisfying to work this out (by piecing together small patterns into larger ones), but I could be wrong. On Mon, Feb 6, 2017 at 7:39 AM, James Propp <jamespropp(a)gmail.com> wrote: > Here's a hint: Find a numerical function of the state of the board that > doesn't change when you fire, and changes in a predictable way when you add > a token (depending only on the location of the added token). > > Jim > > > On Friday, February 3, 2017, James Propp <jamespropp(a)gmail.com> wrote: > >> Since some people (myself included!) relish a specific challenge rather >> than a diffuse one, let me ask who can figure out the unique right move to >> make when the state of the board is >> >> 0 1 0 0 0 0 1 0 1 >> >> and it's your turn to move. >> >> I'll post an answer (and an explanation) on Monday if nobody beats me to >> it. >> >> (One feature of this game is that when there's a winning move, it's >> always unique. Is there a general theory of games having that property?) >> >> Jim >> >> On Fri, Feb 3, 2017 at 12:22 PM, James Propp <jamespropp(a)gmail.com> >> wrote: >> >>> I’ve devised a combinatorial game based on the theory of the abelian >>> sandpile model and I’m trying to come up with a good name for it. >>> >>> The game is played using identical tokens on a board consisting of nine >>> consecutive boxes, each of which is capable of holding one or two tokens. >>> Players take turns. When it’s your turn to move, you add a token to any of >>> the nine boxes; if the box already contains a token, then the box “fires” >>> (by sending one token to the left and one token to the right), and if that >>> leads one or more boxes to contain more than one token, then those boxes >>> fire (by sending one token to the left and one token to the right), and so >>> on; when the leftmost or rightmost box fires, the token that gets fired off >>> the board disappears from the game. Eventually the board stabilizes so that >>> each box contains at most one token. >>> >>> Example: If the state of the board is >>> >>> 1 1 1 0 1 0 0 1 0 >>> >>> and a player adds a token to the second square from the left, the state >>> of the board evolves as follows: >>> >>> 1 2 1 0 1 0 0 1 0 >>> 2 0 2 0 1 0 0 1 0 >>> 0 2 0 1 1 0 0 1 0 >>> 1 0 1 1 1 0 0 1 0 >>> >>> The goal of the game is to be the player who achieves the configuration >>> >>> 1 1 1 1 1 1 1 1 1 >>> >>> (I’ll post a video sometime soon that will make the nature of game-play >>> much clearer than the above, but I’d like to have a good name for the game >>> first.) >>> >>> There’s a very nice (but non-obvious) numerical way to look at the game >>> that makes it quite easy to recognize, from any given starting position, >>> whether it’s a win for the first player or the second player, and the >>> winning strategy is also quite simple. >>> >>> I’m pretty sure this is new. Can anyone think of a good name for it? >>> >>> Please give me feedback on this game. (It might be more apt to call it a >>> puzzle in which the goal is to figure out how to play the game well.) >>> >>> Jim Propp >>> >> >>

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[math-fun] A sandpile game
by James Propp 06 Feb '17

06 Feb '17

I’ve devised a combinatorial game based on the theory of the abelian sandpile model and I’m trying to come up with a good name for it. The game is played using identical tokens on a board consisting of nine consecutive boxes, each of which is capable of holding one or two tokens. Players take turns. When it’s your turn to move, you add a token to any of the nine boxes; if the box already contains a token, then the box “fires” (by sending one token to the left and one token to the right), and if that leads one or more boxes to contain more than one token, then those boxes fire (by sending one token to the left and one token to the right), and so on; when the leftmost or rightmost box fires, the token that gets fired off the board disappears from the game. Eventually the board stabilizes so that each box contains at most one token. Example: If the state of the board is 1 1 1 0 1 0 0 1 0 and a player adds a token to the second square from the left, the state of the board evolves as follows: 1 2 1 0 1 0 0 1 0 2 0 2 0 1 0 0 1 0 0 2 0 1 1 0 0 1 0 1 0 1 1 1 0 0 1 0 The goal of the game is to be the player who achieves the configuration 1 1 1 1 1 1 1 1 1 (I’ll post a video sometime soon that will make the nature of game-play much clearer than the above, but I’d like to have a good name for the game first.) There’s a very nice (but non-obvious) numerical way to look at the game that makes it quite easy to recognize, from any given starting position, whether it’s a win for the first player or the second player, and the winning strategy is also quite simple. I’m pretty sure this is new. Can anyone think of a good name for it? Please give me feedback on this game. (It might be more apt to call it a puzzle in which the goal is to figure out how to play the game well.) Jim Propp

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[math-fun] Question on cubefree sequences
by David Wilson 05 Feb '17

05 Feb '17

This follows on a recent discussion of the lexically first cubefree sequence. I have computed what I believe to be a(0..10000) of the lexically first infinite cubefree word over alphabet {0,1} (treating 0 < 1). I started to submit this sequence, however, when I got to the comments section, I began to write Since an infinite cubefree word over {0,1} exists, a lexically first such word must exist, so this sequence is well-defined. Then I caught myself, because I realized the reasoning was unsound. For example, there is an infinite word over {0, 1} that includes the digit 1, but no lexically first such word. So, do we actually know that there is a lexically first cubefree word over {0, 1}?

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