March 2012 - mathfun@mailman.xmission.com

[math-fun] Hitting 5
by Eric Angelini 11 Mar '12

11 Mar '12

Start sequence S with any integer. If a(n) is the sum of the last two digits of S, S will always hit 5 at some point. Best, É.

9 20

[math-fun] Rank of an elliptic curve
by ed pegg 07 Mar '12

07 Mar '12

A while back, I made a list of elliptic curves that hit a lot of simple integer coordinates. I made a demo of it at http://demonstrations.wolfram.com/EllipticCurvesOnASmallLattice/ A good info page - http://en.wikipedia.org/wiki/Elliptic_curve . Here's one of the curves, it goes through 21 points in that space. Outside of (-9,-9) to (9,9),it also goes through the points {-42, 7}, {-20, 1}, and {121, -38}. eq = x^3 + 3 x^2 y + 16 x^2 - 2 x y^2 - 13 x y - 17 x + y^3 + 6 y^2 - 37 y - 210; pts = Select[Tuples[Range[-10, 10], {2}], Evaluate[eq /. {x -> #[[1]], y -> #[[2]]} ] == 0 &] ; Graphics[{Point[pts], First[ContourPlot[eq == 0, {x, -10, 10}, {y, -10, 10}]]}] Is there a convenient way to find the rank of this curve? A a glance, it looks like it might be rank 7-9 or so. Not a record setter by any means, but I'm curious. The way I found these curves with lots of lattice points was to just pick 9 points at random and fed them into the http://demonstrations.wolfram.com/NinePointCubic/ equation, and kept the resulting curve if it went through a lot of other lattice points. With more careful selection of the random points, it seems like it would be easy to generate a number of rank 9 elliptic curves. Or am I just being naive? Ed Pegg Jr

2 2

Re: [math-fun] Fwd: Re simple perfect squared tori (David Wilson)
by Stuart Anderson 06 Mar '12

06 Mar '12

> Message: 1 > Date: Mon, 05 Mar 2012 20:08:32 -0500 > From: David Wilson <davidwwilson(a)comcast.net> > To: Dan Asimov <dasimov(a)earthlink.net>, math-fun > <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] Fwd: Re simple perfect squared tori > Message-ID: <4F556390.5060208(a)comcast.net> > Content-Type: text/plain; charset=ISO-8859-1; format=flowed > > The 70x70 square cannot be tiled with squares 1x1 through 24x24, but > maybe the torus? > > On 3/4/2012 9:35 PM, Dan Asimov wrote: >> You're saying this is the smallest solution you know of? >> >> << >> That was 24 squares and a side of 181, that's memory for you > Possibly, thanks to Eric Friedman and Guenter Stertenbrink we have these http://www.squaring.net/140x70-2x24squares.png http://www.squaring.net/140x140-4x24squares.jpg so perhaps a torus with 2,3 or 4 squares of each size is achievable I'm looking into using the electrical network method with Thom Sulankes surftri program http://hep.physics.indiana.edu/~tsulanke/graphs/surftri/ to see what else is possible with other surfaces. Squared cylinders and tori were suggested by Brooks et al. (1940, p.339). Michael Goldberg said "The squaring of a torus, however, is accomplished only by a distortion of the area since a torus is not developable." (M. Goldberg, The Squaring of Developable Surfaces, Scripta Mathematica (1952) vol.18 pp. 17-24). David Gale's book 'Tracking the Automatic Ant' has a chapter Tiling a Torus: Cutting a Cake has an intersting discussion of this topic Tiling of Surfaces by Unequal Squares "The question is, or rather was, which rectangles can be tiled by squares no two of which have the same size .... (discussion of squared rectangles and examples of 33x32 and 69x61) ... Now such a "squaring" of a rectangle can be converted in a trivial way to a squaring of the cylinder, torus, Mobius strip, or Klein bottle by the usual identification of opposite sides. But there are also nontrvial squarings of these other surfaces, even simple squarings, meaning squarings in which there is no subset of tiles whose union is a rectangle. Until recently, however, it was not known whether there might be squarings of these surfaces requiring fewer than 9 squares. Then in 1991, Bracewell found a squaring of the Mobius strip using only 8 squares. Very recently the question has been completely settled by S.J. Chapman. Perhaps the simplest but most surprising result is that a 1 x 5 Mobius strip can be tiled by 2 squares as becomes obvious from Figure 9.3. To accommodate this example, one must extend the notion of tiling to allow the mapping of the squares into the surface to self-intersect on their boundaries. Chapman shows that there are no 3- or 4-squarings of the strip but that there is a unique 5-squaring (Fig 9.4) see http://mathworld.wolfram.com/MoebiusStripDissection.html For cylinders, the situation is interesting. Again it turns out that 9 squares are necessary. There are exactly two nontrivial 9-squarings of the cylinder, and these use exactly the squares of Figures 9.1 and 9.2 but in a different arrangement. The tiling corresponding to Figure 9.1 is shown in Figure 9.5. Note for example that, that the squares of size 10 and 15 are disjoint in the rectangle, but they are contiguous on the cylinder. The cases considered so far involve surfaces with a boundary, which forces one to orient the squares with one side parallel to the boundary. This is no longer the case with the torus and Klein bottle. If one allows arbitrary orientation, then, in fact, any two squares can tile some torus. Namely, let the squares have sides a and b, and consider the torus obtained from identifying opposite sides of a square side c, where c^2 = a^2 + b^2. Figure 9.6 shows how to do it and at the same time provides a new (?) proof of the Pythagorean Theorem. A more symmetrical representation is given in Figure 9.7. or here http://mathworld.wolfram.com/TorusDissection.html If one allows only tiles that are parallel to the sides of the square, then it turns out there are no nontrivial 9-squarings of the the torus. Any squaring of the Mobius strip gives a squaring of the Klein bottle. For 6 or fewer tiles these are the only ones, but in the case of 7 or 8 tiles, this is not known. Also it is not known whether there are tilings of the Klein bottle in which the tiles need not be parallel to the sides of the big square. Chapman's techniques for these problems are quite different from and simpler than those of Tutte et al, and depend on a clever encoding of tilings by matrices of 0s, 1s, and -1s." scans of the figures in the chapter; http://www.squaring.net/IMG_0211.JPG http://www.squaring.net/IMG_0212.JPG Geoffrey Morley wrote to me recently; Join opposite longer sides of rectangle 12x26 (11,1x7)(1x8)(7x8,4)(5)(6x7,1)(6), after half-twist, to create a simple perfect squared Möbius band (in imagination, not 3D space!) of order 7 with boundary square of side 6. As with SPSRs, side 5 boundary square may be possible but example hard to find. In my alternative notation each square appears twice: (9,7)(5,2)(3,5,3)(2,9)(7). It is easy to see that a square of side 8 can be added to give a 6 square solution (from an 8x29 rectangle): (9,7)(5,2)(3,8)(8)(5,3)(2,9)(7). These squares can be rearranged to give another solution: (9,7)(5,2)(8,3)(8)(3,5)(2,9)(7). Both have 7 on the boundary. The 7-square solution with 6 on the boundary is (11,7,6)(1,5)(8)(4,8,4)(5,11)(7,1)(6). A square of side 12 can be added to give an 8-square solution (from a 12x38 rectangle): (11,7,6)(1,5)(8)(4,12)(12)(8,4)(5,11)(7,1)(6), also with 6 on the boundary. Stuart

1 0

Re: [math-fun] length + width + height ??
by Henry Baker 06 Mar '12

06 Mar '12

Let's assume that these services are motivated by profits, and that there is enough competition to ensure that the prices track carefully to the costs. Let's further assume (for the moment) that weight doesn't matter; that the real problem is merely fitting the packages into the container for shipping. Let f(L,W,H) be a function that approximates the cost of shipping, and we will price based on some "cost plus" model. What would be a decent function to use that would maximize the value that one could carry? Clearly, if the standard container is 20'x8'x8', we can completely fill it with 4'x4'x4' cubes. But if we can charge more (per unit volume) for non-cubical shapes, we could still maximize the shipping value of the container, even though the container still might have some empty spaces. So what would an "optimal" f(L,W,H) look like? I would guess that f() is symmetric in all of its arguments, but perhaps that isn't even obvious. f() would somehow charge more per unit volume, the "odder" the shape is, because I assume that the statistics of the types of objects would make it less and less likely to be able to pack the container completely full. Note that mere simplicity of computation isn't all that important if the shipping costs are high enough; one would quickly implement a more sophisticated pricing function if there was significantly more profit in it. So, given these constraints, is f(L,W,H)=C*(L+W+H) even close to optimal? At 11:48 AM 3/4/2012, Robert Munafo wrote: >I think it's a lot simpler than that. > >They're trying to give a "volume discount" for customers with lage >packages. To some extent, if the customer had to pay 8 times as much for >something that has 8 times the volume (2x in each direction), they'd >probably consider it overpriced. > >There is also an intentional surcharge for oddly-shaped packages, for the >packing reason you mentioned and for other reasons. > >"length + width + height" fairly neatly encapsulates both of these >objectives. > >On Sun, Mar 4, 2012 at 11:09, Henry Baker <hbaker1(a)pipeline.com> wrote: > >> 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? >> >> [...] > >-- > Robert Munafo -- mrob.com > Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - >mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com

4 4

[math-fun] Identity (4)
by Bill Gosper 06 Mar '12

06 Mar '12

in http://gosper.org/fst.pdf, namely ((-1 + QPochhammer[a, p])*QPochhammer[q, q])/a == Sum[(q^((1/2)*n*(1 + n))* Product[1 - a*p^k + c*p^k*q^k - q^(k - n), {k, 0, Infinity}])/ ((-1)^n*((a - c*q^n)*QPochhammer[q, q, n])), {n, 0, Infinity}] eluded our attempts to rederive it, despite the hints near the end of the paper. Those led me instead to ((-1 + QPochhammer[a, p])*QPochhammer[q, q])/a == Sum[-(((-1)^n*q^((1/2)*n*(3 + n))* Product[1 + (-1 + a*p^k)*q^(k - n) + (a*c*p^k)/q^n, {k, 0, Infinity}])/(a*(c + q^n)* QPochhammer[q, q, n])), {n, 0, Infinity}] Series sez they both work! The summands are utterly unlike, even for n = 0. (Idiot Mma takes ~minutes to plug n=0 into the ratio of the summands! No sums, just products.) Changing c->-a/c in the former, even though unmentioned in the paper, ((-1 + QPochhammer[a, p])*QPochhammer[q, q])/a == Sum[((-1)^n*c*q^((1/2)*n*(3 + n))* Product[1 + (-1 + a*p^k)*q^(k - n) - (a^2*p^k)/(q^n*c), {k, 0, Infinity}])/(a*(a - c*q^n)* QPochhammer[q, q, n]), {n, 0, Infinity}] did zilch to reconcile the summands. So, I don't know where the bleep (4) came from, but I've regained some facility with nonlocal derangement, and now we have (4a) and (4b). --rwg

1 1

[math-fun] Smith's SMODA factorer self-destructs
by Warren Smith 06 Mar '12

06 Mar '12

My factoring algorithm SMODA has self-destructed to the extent it now seems less attractive than the quadratic sieve as a factorer. ----More detailed explanation: The problem was embarrassingly stupid and would have been noticed as soon as an implementation was built but Andy JENNINGS saved time by spotting the problem before that. To review, SMODA depended on the first idea that, if there were a magic oracle whom you could ask "O Oracle, here is B and N, now please generate for me a set of B-smooth numbers near to N or near to multiples of N... such that these numbers are (i) otherwise pretty random, (ii) come approximately as close (in terms of additive error) to being a multiple of N as possible existencewise, (iii) come at a small cost, e.g. polynomial in logN cost, per number found", then factoring would be possible at speeds rivalling or exceeding the Number Field Sieve. The second idea was that instead of depending on magic, it was actually possible to build the oracle by combining a certain fairly small N-independent database with some simple greedy construction algorithms. The database was hard to find, hence still somewhat oracle-like, but it seemed possible to find it fast enough that challenging world records might be feasible. Now various potential problems were pointed out, which actually did not seem to be problems, such as 1) database too expensive to find? but actually just today my computer run is ending, and has constructed preliminary database up to 2^400 size. Jim White & I were building datasets of the "world's most extensive Stormer pair collection" and it seems clear it with feasible number of computers our current codes could reach 2^500 -- and likely a good deal further with as-yet-unimplemented algorithms. 2) Why not use LLL lattice reduction methods to construct database? As far as I can tell they are an uncompetitively poor idea. The real problem is this. The greedoid which uses the database to find smooth numbers (call them S) near to N or near to multiples of N, works in a manner somewhat analogous to the way binary numbers can be constructed lying arbitrarily near to any real. One keeps on adding smaller and smaller "correction vectors" in a certain vector space, and proves that a certain natural vector logarithmic distance measure keeps getting smaller, indeed at least a factor 2 smaller each whack. That works. BUT, the problem is, when you make that natural logarithmic distance smaller, that does NOT imply that the actual difference |S - multipleofN| keeps getting smaller back in the ordinary world of integer numbers, rather than the vector world. Instead, this difference gets smaller for a while, but once |S-N| is of order about squareroot(N), it almost always would start to grow again. Simple example: if I had, say, 10111 =approx= 10000 with error=111 and I used multiplication by the rational 101/100 to try to improve the approximation, I'd get 1011100 =approx= 1010000 with error=1100 which is the opposite of an improvement in the additive error (although it does improve the *relative* error). Improving the relative error is easy and was essentially what I was doing. Although in principle it basically is always possible to improve the additive error, (e.g. if we use 271/274 not 100/101 then get 2740081 which is 81 mod 10000, or if we use 991/1002 we get 10020001 which is 1 mod 10000) I don't see how SMODA can gaurantee that without an unfeasibly enormous database far larger than I had in mind. So the net effect is that SMODA should stop working approximately at the point where you try to exceed the performance of the quadratic sieve. And it probably will be worse than QS since it is more complicated. This is a pity because SMODA and associated theory and data was quite big and aesthetic looking, but now is seen to be a "castle in the air." Is there any hope to salvage SMODA? I'm not sure. You'd need to invent a way to use the database that would be far more powerful than the simple greedoids I had in mind. (But it would be permissible to consume a good deal more compute-time than the greedoid consumed.) One way to try do that would be to find a rational number (such as 271/274 in example above) that yields improvement (this is easy by Euclid alg.), but doing so in such a way its numerator was smooth (not so easy). For example based on the two good rationals above 271/264 and 991/1002, and 91/92 which is also good, we could look for smooth numbers 271a+991b+91c in various ways such as sieving or lattice techniques. Another salvage attempt would be to have stronger guarantees about the database than those I'd had in mind. For example, we believe based on both theoretical models and computer evidence that very small databases based on very small prime sets [of order only about (logN)^2 primes] should *exist* but I had given up on trying to construct databases that small because they asymptotically seemed much too hard to find. If, however, we nevertheless assumed SMODA had access to such a super-small super-good database, then SMODA might be able to become a lot more powerful due to "cancellations" arising from common prime factors re-occuring enough of the time. So, salvaging SMODA may or may not be possible. It is worth some thought. If anybody wants to have my SMODA parts I/II paper draft/shipwrecks and/or our the Smith+White datasets of Stormer pairs, let me know. Warren D. Smith http://rangevoting.org

1 0

[math-fun] Dan Hoey
by Thane Plambeck 05 Mar '12

05 Mar '12

I just learned of the death of Dan Hoey last fall. http://www.washingtonpost.com/local/obituaries/daniel-j-hoey-computer-speci… -- Thane Plambeck tplambeck(a)gmail.com http://counterwave.com/

8 15

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

05 Mar '12

You're saying this is the smallest solution you know of? << That was 24 squares and a side of 181, that's memory for you >> ________________________________________________________________________________________ It goes without saying that .

2 1

[math-fun] Identity (5)
by Bill Gosper 05 Mar '12

05 Mar '12

in http://gosper.org/fst.pdf, namely Sum[((-1)^n* Cos[Sqrt[(n + 1/2)^2 + a^2]*Pi])/((n + 1/2)*(n - a + 1/2)*(n + a + 1/2)), {n, 0, Infinity}] == (Pi*(Sin[a*Pi]^2 - 2*Sin[(a*Pi)/Sqrt[2]]^2))/(2*a^2*Cos[a*Pi]) gives oo=oo for a = 1/2+ (integer k), but gives (6*(-1)^k*Cos[((1 + 2*k)*Pi)/Sqrt[2]])/(1 + 2*k)^3 + ((-1)^k*Sqrt[2]*Pi*Sin[((1 + 2*k)*Pi)/Sqrt[2]])/(1 + 2*k)^2 if you skip the n=k term. --rwg With help from Corey and Julian, and buffoonery from Mathematica. >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 ----------------------------------------------------------------------- f(p) = (2p - 1)/p^2 for p > 1/2, f(p) = 0 for p <= 1/2. -- Gene Yow, slope 8 at p=1/2+ ! Maybe building a tree is a good way to test coins for fairness. --rwg

2 2

[math-fun] simple perfect squared tori
by Warren Smith 05 Mar '12

05 Mar '12

can we see a picture please? What about squared klein bottles and squared moebius strips?

1 0