May 2011 - mathfun@mailman.xmission.com

[math-fun] Atomtrap
by Adam P. Goucher 16 May '11

16 May '11

Here's a PSPACE-complete puzzle game I created, if anyone is interested: https://docs.google.com/leafid=0B7zc3Wqisi4jNTE5NTU0NGQtZjdiOS00OWIxLWFlMjc… You can design your own levels (and upload!), which can be as complex as any linear-bounded automaton (Turing machine with bounded memory). One of my default levels contains a 15-puzzle built into it. There are 40 default levels, and you can upload an arbitrary number to the Internet from within the program. Also, if anyone here has Golly, the levels are in Golly's Extended RLE format, so you can additionally create and edit levels from within Golly. Please enter a *unique* username when prompted, so that you don't overwrite someone else's highscores, etc. That's happened before with two people called 'Dave'... Have fun! Sincerely, Adam P. Goucher

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[math-fun] Minskys and Trinskys
by Bill Gosper 15 May '11

15 May '11

Through the heroic efforts of Jack Holloway, the book mentioned in Neil's Minsky Circle Algorithm blogpost (http://nbickford.wordpress.com/) is now commercially available: http://www.blurb.com/books/2172660 , $59.95 (hardcover, 70pp). Also, Neil has tweaked some pictures and captions, with one yet pending: http://gosper.org/x=1_y=1o2_d=9_e=1o3.png , clarifying that x=1,y=1/2,δ=9 and ε=1/3 makes an infinite triangular spiral, despite 0 < δ ε < 4. --rwg

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[math-fun] Algorithmic traders gone wild: the $24 million dollar book on Amazon
by Henry Baker 15 May '11

15 May '11

FYI -- http://www.michaeleisen.org/blog/?p=358 AmazonÂs $23,698,655.93 book about flies By Michael Eisen | April 22, 2011 A few weeks ago a postdoc in my lab logged on to Amazon to buy the lab an extra copy of Peter LawrenceÂs The Making of a Fly Â a classic work in developmental biology that we Â and most other Drosophila developmental biologists Â consult regularly. The book, published in 1992, is out of print. But Amazon listed 17 copies for sale: 15 used from $35.54, and 2 new from $1,730,045.91 (+$3.99 shipping). I sent a screen capture to the author - who was appropriate amused and intrigued. But I doubt even he would argue the book is worth THAT much. At first I thought it was a joke Â a graduate student with too much time on their hands. But there were TWO new copies for sale, each be offered for well over a million dollars. And the two sellers seemed not only legit, but fairly big time (over 8,000 and 125,000 ratings in the last year respectively). The prices looked random Â suggesting they were set by a computer. But how did they get so out of whack? Amazingly, when I reloaded the page the next day, both priced had gone UP! Each was now nearly $2.8 million. And whereas previously the prices were $400,000 apart, they were now within $5,000 of each other. Now I was intrigued, and I started to follow the page incessantly. By the end of the day the higher priced copy had gone up again. This time to $3,536,675.57. And now a pattern was emerging. On the day we discovered the million dollar prices, the copy offered by bordeebook was 1.270589 times the price of the copy offered by profnath. And now the bordeebook copy was 1.270589 times profnath again. So clearly at least one of the sellers was setting their price algorithmically in response to changes in the otherÂs price. I continued to watch carefully and the full pattern emerged. Once a day profnath set their price to be 0.9983 times bordeebookÂs price. The prices would remain close for several hours, until bordeebook ÂnoticedÂ profnathÂs change and elevated their price to 1.270589 times profnathÂs higher price. The pattern continued perfectly for the next week. But two questions remained. Why were they doing this, and how long would it go on before they noticed? As I amusedly watched the price rise every day, I learned that Amazon retailers are increasingly using algorithmic pricing (something Amazon itself does on a large scale), with a number of companies offering pricing algorithms/services to retailers. Both profnath and bordeebook were clearly using automatic pricing Â employing algorithms that didnÂt have a built-in sanity check on the prices they produced. But the two retailers were clearly employing different strategies. The behavior of profnath is easy to deconstruct. They presumably have a new copy of the book, and want to make sure theirs is the lowest priced Â but only by a tiny bit ($9.98 compared to $10.00). Why though would bordeebook want to make sure theirs is always more expensive? Since the prices of all the sellers are posted, this would seem to guarantee they would get no sales. But maybe this isnÂt right Â they have a huge volume of positive feedback Â far more than most others. And some buyers might choose to pay a few extra dollars for the level of confidence in the transaction this might impart. Nonetheless this seems like a fairly risky thing to rely on Â most people probably donÂt behave that way Â and meanwhile youÂve got a book sitting on the shelf collecting dust. Unless, of course, you donÂt actually have the bookÂ. My preferred explanation for bordeebookÂs pricing is that they do not actually possess the book. Rather, they noticed that someone else listed a copy for sale, and so they put it up as well Â relying on their better feedback record to attract buyers. But, of course, if someone actually orders the book, they have to get it Â so they have to set their price significantly higher Â say 1.27059 times higher Â than the price theyÂd have to pay to get the book elsewhere. WhatÂs fascinating about all this is both the seemingly endless possibilities for both chaos and mischief. It seems impossible that we stumbled onto the only example of this kind of upward pricing spiral Â all it took were two sellers adjusting their prices in response to each other by factors whose products were greater than 1. And while it might have been more difficult to deconstruct, one can easily see how even more bizarre things could happen when more than two sellers are in the game. And as soon as it was clear what was going on here, I and the people I talked to about this couldnÂt help but start thinking about ways to exploit our ability to predict how others would price their books down to the 5th significant digit Â especially when they were clearly not paying careful attention to what their algorithms were doing. But, alas, somebody ultimately noticed. The price peaked on April 18th, but on April 19th profnathÂs price dropped to $106.23, and bordeebook soon followed suit to the predictable $106.23 * 1.27059 = $134.97. But Peter Lawrence can now comfortably boast that one of the biggest and most respected companies on Earth valued his great book at $23,698,655.93 (plus $3.99 shipping).

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Re: [math-fun] Mathematica help sought
by Dan Asimov 14 May '11

14 May '11

Thanks everyone for all the Mma help. What I just don't get is how this product can be so mature (ca. 20 years old) and so expensive, yet need so much manual help before it can get anywhere close to solving a relatively simple problem (which in fact is an applied problem translated into equations). THE ORIGINAL PROBLEM, in case anyone is interested, probably has an elegant solution with minimal computation: >From a recent NY Times article in the science section: PUZZLE: --------------------------------------------------------------- Let there be a lake on the plane filling up the unit circle centered at the origin. The rest of the plane is level ground. Find the quickest path from (-2,0) to (3,1), possibly passing through the lake, for an amphibious vehicle whose land speed is 1 unit/hour, and whose water speed is 1.25 units/hour. --------------------------------------------------------------- --Dan Rich writes: << You can usually lose all the trig functions with cos u -> (1 - u^2)/(1 + u^2), sin u -> 2 u / (1 + u^2), and similar for v. The new u has range [-1,1], assuming the original u was real. This is the ancient tan(u/2) -> u substitution. In your case, you also want sin((u-v)/2) to be a rational function, so u & v should be halved: cos(u/2) -> (1-u^2)/(1+u^2) etc. cos(u) -> (1 - 6 u^2 + u^4)/(1+u^2)^2 sin(u) -> (4 u - 4 u^3)/(1+u^2)^2 sin((u-v)/2) = sin(u/2 - v/2) = sin(u/2) cos(v/2) - sin(v/2) cos(u/2) -> [ 2u(1-v^2) - 2v(1-u^2) ] / [(1+u^2)(1+v^2)] The sqrts will still mess up the derivatives. If you are looking for critical points, you can multiply through by (1+u^2)(1+v^2) before differentiating to simplify things a little. >> Sometimes the brain has a mind of its own.

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[math-fun] Mathematica help sought
by Dan Asimov 14 May '11

14 May '11

I've used Mma on a very elementary basis for years, but often when I try to do the least thing beyond what I'm familiar with, it hangs or gets confused, etc. Case in point: Let f(u,v) = sqrt(5+4cos(u)) + 2.5*sin((u-v)/2) + sqrt(11-6cos(v)-2sin(v)). I wanted it to find the critical points, so I tried: Solve[{D[f[u,v],u] == 0, D[f[u,v],v] == 0}, {u,v}] and it seemed to have no idea what I was talking about. So I figured the partials by hand and confirmed them with Mma, and just used them directly instead of Df[u,v],u], etc. This time Mma canceled the calculation after 10-20 minutes, saying its memory had been used up. (Now I'm trying to get it to at least NSolve for the critical points.) But is there any way to get the original symbolic equation to come up with something useful, or does this just indicate the limits of symbolic computation? Thanks, Dan Sometimes the brain has a mind of its own.

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[math-fun] Onward and upward with the arts
by Hilarie Orman 14 May '11

14 May '11

If you like woodwork and cellular automata, and who doesn't, drop by the Wolfrule Art Opening - March 14 - Los Angeles. http://www.wolfrule.com/ (website requires Javascript; click on the XXX to avoid the annoying 5-second refresh). Hilarie

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[math-fun] Correction Re: Mathematica help sought
by Dan Asimov 14 May '11

14 May '11

No, no, no, no, no. The land speed and water speed should be the opposites of what I wrote far below: CORRECTION: The water speed is 1 unit/hour, and the land speed is 1.25 units/hour. --Dan <From a recent NY Times article in the science section: PUZZLE: --------------------------------------------------------------- Let there be a lake on the plane filling up the unit circle centered at the origin. The rest of the plane is level ground. Find the quickest path from (-2,0) to (3,1), possibly passing through the lake, for an amphibious vehicle whose land speed is 1 unit/hour, and whose water speed is 1.25 units/hour. --------------------------------------------------------------- ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ OOPS OOPS OOPS OOPS OOPS OOPS OOPS OOPS OOPS OOPS OOPS OOPS Sometimes the brain has a mind of its own.

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Re: [math-fun] Mathematica help sought
by Dan Asimov 13 May '11

13 May '11

Why Mathematica drives me up the wall: All I wanted was for it to square the expression for f[u,v] here. Finally (after the following) I tried Expand first, then Simplify, and at last it actually squared the thing. Sheesh. --Dan ****************************************************************** In[1]:= f[u_,v_] := Sqrt[5+4Cos[u]] + 2.5*Sin[(u-v)/2] + Sqrt[11-6Cos[v]-2Sin[v]] In[2]:= g[u_,v_] := f[u,v]^2 In[3]:= g[u,v] u - v Out[3]= Power[Sqrt[5 + 4 Cos[u]] + 2.5 Sin[-----] + 2 > Sqrt[11 - 6 Cos[v] - 2 Sin[v]], 2] In[4]:= FullSimplify[g[u,v]] u - v Out[4]= Power[Sqrt[5 + 4 Cos[u]] + 2.5 Sin[-----] + 2 > Sqrt[11 - 6 Cos[v] - 2 Sin[v]], 2] In[5]:= Clear[g] In[6]:= g[u_,v_] := f[u,v]*f[u,v] In[7]:= g[u,v] u - v Out[7]= Power[Sqrt[5 + 4 Cos[u]] + 2.5 Sin[-----] + 2 > Sqrt[11 - 6 Cos[v] - 2 Sin[v]], 2] In[8]:= Simplify[%] u - v Out[8]= Power[Sqrt[5 + 4 Cos[u]] + 2.5 Sin[-----] + 2 > Sqrt[11 - 6 Cos[v] - 2 Sin[v]], 2] In[9]:= In[9]:= (Sqrt[5+4Cos[u]] + 2.5*Sin[(u-v)/2] + Sqrt[11-6Cos[v]-2Sin[v]])*(Sqrt[5+4Cos[u]] + 2.5*Sin[(u-v)/2] + Sqrt[11-6Cos[v]-2Sin[v]]) u - v Out[9]= Power[Sqrt[5 + 4 Cos[u]] + 2.5 Sin[-----] + 2 > Sqrt[11 - 6 Cos[v] - 2 Sin[v]], 2] In[10]:= Simplify[%] u - v Out[10]= (Sqrt[5 + 4 Cos[u]] + 2.5 Sin[-----] + 2 2 > Sqrt[11 - 6 Cos[v] - 2 Sin[v]]) ****************************************************************** Sometimes the brain has a mind of its own.

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Re: [math-fun] true self-referential sentences with false contrapositives
by Dan Asimov 12 May '11

12 May '11

I understand why the second sentence is called the literal contrapositive, but the "this" in that sentence refers of course to something other than the "this" in the first sentence. Which is at least one reason it's only a "literal" but not actual contrapositive. --Dan Jim wrote: << The sentence "If two plus two equals four, then the premise of this sentence contains numbers" is true, while its literal contrapositive "If the premise of this sentence does not contain numbers, then two plus two does not equal four" is false. Can anyone find a more elegant example of a true self-referential sentence of the form "If ..., then ..." whose literal contrapositive is false? >> Sometimes the brain has a mind of its own.

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Re: [math-fun] Farey-ish fractions with weighted mediants
by James Propp 12 May '11

12 May '11

Back in December I wrote: >It's well-known that if you start with the list of fractions 0/1, 1/1 and >successively interpose the mediant (a+c)/(b+d) between successive elements >a/b and c/d of the list, every rational between 0 and 1 will eventually >appear. > >I just tried a variant of this today in which I interpose the weighted mediants >(2a+c)/(2b+d) and (a+2c)/(b+2d) instead, like this: > >0/1, 1/1 >0/1, 1/3, 2/3, 1/1 >0/1, 1/5, 2/7, 1/3, 4/9, 5/9, 2/3, 5/7, 4/5, 1/1 >... > >Does every rational between 0 and 1 with odd denominator eventually appear? > >I've checked this up through denominator 9. > >... At Heidi Burgiel's suggestion, I tried starting with 0/1, 1/0 instead: Stage 0: 0/1, 1/0 Stage 1: 0/1, 1/2, 2/1, 1/0 Stage 2: 0/1, 1/4, 2/5, 1/2, 4/5, 5/4, 2/1, 5/2, 4/1, 1/0 ... With this rule, it appears that we get every fraction in which the numerator and denominator have opposite parity. Moreover, it appears that for all k > 2, every fraction with numerator and denominator summing to 2k+1 occurs by stage k. I've checked this up through k = 15. The exceptional case is k = 2: the fractions 1/4 and 4/1 occur at stage 2 but the fractions 2/3 and 3/2 don't occur until stage 3. Can anyone prove this conjecture? Jim Propp

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