March 2012 - mathfun@mailman.xmission.com

Re: [math-fun] Fwd: Re simple perfect squared tori
by Dan Asimov 05 Mar '12

05 Mar '12

Summary: From Hans and Stuart, I've seen ian Gambini's thesis, in which he describes doing an exhaustive computer search for simple perfect squared square tori, of side lengths between 6 and 450. Exactly one such was found! It uses 24 squares, to make a squared torus of side length 181. He also found 107 simple perfect squared cylinders, which if made into a torus would have a "circular fault line". Finally, I made yet another mistake in the original query: Although the simple perfect squared square using the smallest number of tiles is indeed 112x112 with 21 tiles, 112 is not the smallest possible side length. There are 3 distinct simple perfect squared squared of side length 110 (2 using 22 tiles and the other using 23). 110 has been shown to be the smallest side length possible. --Dan ________________________________________________________________________________________ It goes without saying that .

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[math-fun] Oracle algorithms -- any useful ones?
by Warren Smith 04 Mar '12

04 Mar '12

The concept of an "oracle algorithm" (an algorithm that gets to ask questions to an "oracle", thus hopefully increasing its computational power) has been useful for abstract computational complexity arguments. My QUESTION: has it ever been actually *directly* useful as an algorithm? Reason I'm asking is, my alleged new integer factoring algorithm SMODA is an oracle algorithm, and does claim to be directly useful. Its oracle is a quite small -- and nice to have, but it's apparently rather hard to produce it -- database. So I wasn't sure what to make of that, and then I thought a bit and wasn't able to think of anybody else who'd ever made an algorithm with these properties. Although quite likely that is merely because I am ignorant.

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[math-fun] A nifty theorem in plane geometry
by Warren Smith 04 Mar '12

04 Mar '12

I recently discovered this theorem in geometry (which also solves a number theory problem about "Prouhet Tarry Escott" set pairs). Take an equilateral triangle with side=z. Cut it with a line segment (of length=t) from a vertex to opposite side. This divides that side into line segments of lengths x and y. Of course x+y=z. Theorem: 2*t^2 = x^2 + y^2 + z^2 and 2*t^4 = x^4 + y^4 + z^4. Picture: http://dl.dropbox.com/u/3507527/SmithPTETriangleThm.png

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[math-fun] Diophantine equation a^4+b^4=d*c^2 and the specialness of 17, 68, 82...
by Warren Smith 04 Mar '12

04 Mar '12

Fermat in about 1640 famously proved a^4+-b^4=c^2 has no positive integer solutions. Also a^4-b^4=2*c^2 has no positive integer solutions. These results are shown in T.Nagell: Intro to number theory, Wiley 1951, pp.227-230. However, a^4+b^4=2*c^2 trivially has a ton of solutions a=b, c=a^2. If we exclude the trivial by demanding b>a>0 though, I don't know of any solutions. Let us generalize. Ask about a^4 + b^4 = d*c^2 with a,b,c,d integers and b>a>0. My computer looked for solutions. Search I: a^4+b^4=d*c^2 with 0<a<b<65536 and 0<d<65. Result: a ton of solutions for d=17, but zero solutions for any other d. a=1, b=2, d=17, c=1 a=2, b=4, d=17, c=4 a=2, b=13, d=17, c=41 a=3, b=6, d=17, c=9 a=4, b=8, d=17, c=16 ... Well, that's interesting: 17 is special. Search II: a^4+b^4=d*c^2 with 0<a<b<=1024 and 0<d<=1024. Result: The following d are "special": 17, 68, 82, 97, 113, 153, 193, 257, 272, 274, 328, 337, 388, 425, 433, 452, 514, 577, 593, 612, 626, 641, 673, 706, 738, 772, 833, 873, 881, 914, 1017. [This sequence is not in OEIS.] What is going on here? I do not really know, but an initial speculation is this. When d=17, there is the easy solution a=1, b=2, d=17, c=1. But here is the thing: if there is one solution, then it can usually be used to generate an infinite set of solutions. I got that from Tito Piezas's online book http://sites.google.com/site/tpiezas/018 where he attributes it to A.Desboves (probably about 1880?). So there are "special" d with an infinity of solutions, and there are "usual" d with zero solutions (and conceivably there are a few "crazy" d with a finite nonzero number of solutions because the below Desboves method for generating more solutions happens merely to regenerate old ones, which would be an amazing miracle). How to generate more solutions from one: Piezas says Desboves said (as edited by me): if p^4+q^4 = d*r^2, then x^4+y^4 = d*z^2 where x = p*(4*m^2-3*(m+n)^2), y = q*(4*n^2-3*(m+n)^2), z = r*(4*(m+n)^4-3*(m-n)^4) where m=p^4 and n=q^4. So for example starting from p=1, q=2, d=17, r=1 we get x=863, y=314, z=182209 and sure enough 863^4 + 314^4 = 17*182209^2 is another solution. Questions: Q1. Can you find a parameterized family of d=17 solutions? Q2. Can you understand the set of special d? Q3. Can you prove there are no "crazy" d (or find them all)?

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Re: [math-fun] math-fun Digest, Vol 109, Issue 3
by Warren Smith 04 Mar '12

04 Mar '12

> From: Dan Asimov <dasimov(a)earthlink.net> > I saw this problem in some stuff I was reading: > Consider the standard binary tree with infinitely many levels. > Suppose each edge is colored green with probability = p. > What is the probability f(p) that there exists an infinite green path > starting at the root? --Let q=1-p and g(p)=1-f(p). Then g(p) obeys g(p) = q^2 + p*q*2*g(p) + p^2 * g(p)^2 you may now solve the quadratic equation to deduce g(p) and hence f(p). Don't choose the wrong root.

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[math-fun] Re simple perfect squared tori
by Stuart Errol Anderson 04 Mar '12

04 Mar '12

Ian Gambini wrote a thesis on squared squares, in 1999 (I think), he solved this problem from memory the smallest simple perfect tori has side 180 and order 20 (but I will need to check that) there is a graphic somewhere on Ed Peggs marhs puzzle site

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[math-fun] length + width + height ??
by Henry Baker 04 Mar '12

04 Mar '12

Some airlines & postal systems charge by the curious measure "length + width + height". 1. Is there a standard name for this measure? 2. What is the scientific/mathematical rationale for using this measure for charging? -- If there were such a mathematical rationale, I would think that it would somehow be based upon the statistics of packing many dissimilar boxes into a standard box -- e.g., a standard shipping container or a UPS truck. What do we know about the statistics of packing various numbers of differently shaped boxes into a large cubical box? I guess we could start with all items being the exact same shape (same L, same W, same H) and packing them into a large cube. Then consider 2 types of items, with different L's, W's and H's, and then vary the percentage of the 2 types of items & see what happens to the packing density. Etc. -- A related backwards problem might be that of taking a large standard cube & cutting it into smaller pieces of different L's, W's and H's. If we made "random" cuts in all three dimensions, what are the statistics of the different shapes that fall out? These shapes obviously pack perfectly.

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[math-fun] Cute probability problem
by Dan Asimov 04 Mar '12

04 Mar '12

I saw this problem in some stuff I was reading: Consider the standard binary tree with infinitely many levels. Suppose each edge is colored green with probability = p. What is the probability f(p) that there exists an infinite green path starting at the root? --Dan ________________________________________________________________________________________ It goes without saying that .

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[math-fun] Simple perfect squared square tori?
by Dan Asimov 03 Mar '12

03 Mar '12

A "squared square" is called perfect if no two tiles are the same size, and simple if there is no tiled subrectangle. Same terminology for a squared square torus. (But we require all tiles to be aligned with the sides of the square from which we can get the torus by identifying opposite sides. Otherwise, e.g., a square torus of side 5 is tiled by two square tiles of sides 3 and 4.) The smallest-sided perfect squared square (side = 112) is simple, and uses 21 square tiles of distinct sides. This is known to be the unique example with both the smallest side length and the fewest tiles. I thought I once read that there is a square torus that can be tiled by even fewer square tiles (aligned with the torus), but now I'm unable to locate any reference to such a thing. Does anyone know about perfect squared square tori (that don't come from identifying opposite sides of squared squares)? Of course it would be nicest if such a thing were also simple. Thanks, Dan ________________________________________________________________________________________ It goes without saying that .

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[math-fun] Lamberts to the slaughter
by Bill Gosper 02 Mar '12

02 Mar '12

rwg>[...] Prettier still (except for taking x -> +or-1) > (fourth:) > > (-(1/2))*x*Sum[(-q)^n/(1 + q^n*x), {n, 0, Infinity}] == > (1/2)*Sum[(-x)^j/(1 + q^j), {j, 1, Infinity}] > > These Lambert transformations are all the same easy trick, > which I'm not stating because I've already asked Funster > Neil (who owes us his sliding block report) to figure it > out for tomorrow. > > The trick is simply to expand the summands as infinite geometric series (which Mathematica steadfastly prohibits), and then interchange the orders of summation. As you should always do the same thing to both sides of an equation, this leaves you with what you started with, except with the sides switched. Thus you fail to prove anything. This is similar to when you try to put your left shoe on your right foot. How do you know which is wrong--the shoe or the foot? The only safe procedure is to switch both, and place your right shoe on your left foot. When this too fails, you conclude as above that the problem has no solution. This is why Julian goes barefoot. --rwg PS: As Neil describes in his blog, the later "hardest" 4x4 puzzles are no harder, and sometimes easier, than in Nick's interactive collection<http://www.puzzleworld.org/SlidingBlockPuzzles/bickford.htm>, which includes puzzles where the goal piece is nonconvex, and technically fails to make a full diagonal traverse, due to misfitting a corner. This is a shame, because that four-piece 26-mover is a gem.

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