September 2003 - mathfun@mailman.xmission.com

Re: [math-fun] A use for Jud's method
by Allan C. Wechsler 29 Sep '03

29 Sep '03

At 05:59 PM 9/28/03 -0400, John Conway wrote: > After some more reflection, I've come to the conclusion that >Allan's "Tightness Conjecture" probably WILL fail in 3 dimensions >as well as in 4, and that Jud can probably redeem himself (so to >speak) by establishing this. Ah. I misread you (JHC) in your last message, or you thought better of it. > Namely, among the many packing configurations that Sloane et al >have produced with the "Gosset" program, there are probably several >records that aren't (currently) achieved by any "tight" configuration. >Now if Jud were to run his program for spherical caps of the appropriate >radius (minus, for safety, a very small epsilon), it should prove this. ... The expectation being that JM's program would fail to find any (tight) configurations, even though non-tight solutions for those cases are known to exist. Yes, that would be very cool. > It would also be of interest to run it for the case of 4-dimensional >unit spheres, to see just how far short of 24 it falls. Let me just >conjecture here that it won't even find 23. > > I'm sure the program would also find a few unforeseen things that >would amply justify the trouble of writing it. How about it, Jud? Well, it's certainly a fairly tough programming challenge. I'm not volunteering to write it! Can JHC say a few words more about the "Gosset program"? I have other ideas for computer experiments in this area, and would like to know if they've been tried already. -A

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Re: [math-fun] More on sphere packing
by Allan C. Wechsler 28 Sep '03

28 Sep '03

At 04:05 PM 9/28/03 -0400, John Conway wrote: >On Sun, 28 Sep 2003, Allan C. Wechsler wrote: > > > JM proposed, if I read him right, to > > prune this space by considering only configurations [that can] be built > > by adding each new sphere in the 'socket' formed by three previous spheres. > > > > JHC expresses doubt that this program could work, because of the > > possibility of losing the baby with the bathwater. It might be possible > > for 25 non-overlapping 4-spheres to kiss a central one, but only in > > configurations that are _not_ in the restricted class JM proposes. > > Not just doubt, but moral certainty [of the negative]. The problem is > not that there >might be 25-sphere configurations, but no tight ones, but that there IS >a 24-sphere configuration that is presumably the only one, and it ISN'T >tight. THIS is the crucial piece of information I was missing when I wrote my previous message. Glad I didn't bet. I didn't realize that the known 24-sphere configuration was (a) believed to be unique, and (b) was not tight. >[...] If Musin's proof is correct (as I expect it will be), and proves >the 24-sphere configuration unique (as I expect it to), then that does >disprove the TC. Indeed. > This reminds me that I had intended to ask, but forgot, a question >that I can now rephrase as "is there any radius r for which there is >a packing of N circular caps of radius r on the unit sphere, but >no tight one?" I don't expect there will be, but unlike the two >expectations above (and like the TC), this expectation has no real basis. In other words, your intuition is that the TC is true for the case of 3 dimensions; is that right? -A

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Re: [math-fun] non-square products of squares?
by Dan Hoey 28 Sep '03

28 Sep '03

I wrote: > Up to order 255, the only group that exceeds 1/6 is > SmallGroup(48,50), or > <a,b,c : a^3 = b^2 = c^2 = (a b)^3 = (a c)^3 = (b c)^2 > = (a b a^-1 c)^2 = 1> . > It has a probability of 5/24 that a,b,ab are square,square,nonsquare. I looked at your classification of groups of order 16 and verified your translation of the smaller anonymous groups I found. In hopes of identifying this group, I found its derived subgroup 2^4 . Of course G/G' is C3. But the direct product C3 x 2^4 is not G. That seems similar to the case for A4: A4/C3 = 2^2 but 2^2x3 is not A4. Somehow I thought that was the way to decompose and reconstitute groups, but I guess I'm mistaken. As for identifying the group, I have checked that its order spectrum [1,15,32,0,...,0] is unique. Its center is 1. Dan

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[math-fun] More on sphere packing
by Allan C. Wechsler 28 Sep '03

28 Sep '03

I've been having some email problems, but despite them I've been following the discussion of kissing numbers and how hard it is to establish them. In particular, I'm wondering whether we may not be missing some possible low-hanging fruit that might result from the sort of greedy packing procedure that Jud McCranie sketched, and about which John Conway expressed skepticism. The central problem is the size of the configuration space that must be explored: 24 or 25 spheres jostling around, each contributing 3 degrees of freedom to the configuration space. JM proposed, if I read him right, to prune this space by considering only configurations of a certain special kind, in which the configuration may be built by adding each new sphere in the 'socket' formed by three previously-placed spheres. (This is hard to visualize, but in four-space, you need three other spheres to form a well-defined socket with no degrees of freedom.) Enumerating such configurations would be much easier than searching an enormous 75-dimensional configuration space. JHC expresses doubt that this program could work, because of the possibility of losing the baby with the bathwater. It might be possible for 25 non-overlapping 4-spheres to kiss a central one, but only in configurations that are _not_ in the restricted class JM proposes. Let us call these JM-type configurations _tight_ configurations; forgive me if I do not give a rigorous definition -- I'll be willing to ante up later, but for the moment I think we all know what we're talking about. Obviously, the JM program depends on an unstated conjecture, one that JHC believes to be false. If forced to place bets at this point, I'd wager that JHC is right ... unless JHC offered me attractive odds. Because I'm not yet morally certain he's right. The conjecture, which I will call the Tightening Conjecture, is: if there exists ANY configuration of n non-overlapping radius-R k-spheres kissing a given unit k-sphere, then there exists a TIGHT such configuration. JHC puts forward what I believe is intended to be a counterexample to this conjecture, but I claim it isn't really a counterexample. In the case he presents, k=3, n=8, and R=(sqrt3+1)/2. He points out that it works to position each of the 8 peripheral spheres at the corners of a cube of side 2R, whose center is the center of the unit sphere. This configuration is _not_ tight, in the sense given above, and JHC implies by his use of the phrase "the correct configuration" that it is the only one. If this were true, it would certainly follow that a search confined to tight configurations would fail to find one, and if we believed the Tightening Conjecture we would incorrectly conclude that the 8 peripheral spheres simply would not fit. However, the configuration described by JHC is _not_ the only one that works. Begin with the cubical configuration described by JHC. Run a plane through the center of the central sphere, so that all eight peripheral spheres are tangent to that plane. Now it is clear that four peripheral spheres on one side of the plane (call it the north side) can be rotated independently of the other four, and after they have been rotated 45 degrees, they can move south a little bit, across the equator, and settle into the four northward-facing sockets offered between the four southern spheres. The resulting crownlike configuration still has one degree of freedom, but I am reasonably sure that this extra flex can be expunged by puckering the square formed by the four southern spheres. Eventually something will bump, at which point the configuration will be tight. The Tightening Conjecture may still be false, but JHC's cubical example does not kill it outright. My guess is that JHC has a stronger counterexample in his arsenal, that _will_ disprove the TC; my only real point in this message is that the offered example doesn't do the job. -A

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[math-fun] lifted partitions
by Jon Perry 28 Sep '03

28 Sep '03

The number of unordered partitions of n into k parts is equal to the sum of the unordered partitions of n-k into i parts, 1<=i<=k. e.g. P(9,4) is based on P(5,1-4), i.e. 5 41 32 311 221 2111 Adding a [1,1,1,1] mask gives P(9,5); 6111 5211 4311 4221 3321 3222 Jon Perry perry(a)globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com

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[math-fun] partitions to pascal
by Jon Perry 28 Sep '03

28 Sep '03

The number of ordered partitions of n into exactly k parts corresponds to Pascal's triangle. This can be seen although not immediately proved by the PT property that each entry is the partial sum of the NE-diagonal beginning at it's NW predecessor, and with a simple transform. Jon Perry perry(a)globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com

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Re: [math-fun] Re: Kissing number.
by asimovd＠aol.com 26 Sep '03

26 Sep '03

Steve Gray writes: << In "Kepler's Conjecture" by George Szpiro, it says that the naive angle-occupying calculation in 3-D gives an absolute upper bound of 14.9 spheres. This is just a solid-angle calculation. This should be easy to recompute by hand. >> Right. One unit sphere tangent to another subtends a zone on the surface of the second consisting of all points within pi/6 of a single point. Since zone area is given by 2 pi R h = 2 pi h = 2 pi (1 - sqrt(3)/2), the upper bound based only on subtended spherical area is given by 4 pi divided by 2 pi (1 - sqrt(3)/2), or 4/(2-sqrt(3) = 14.928203... --Dan

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[math-fun] Re: [math-fun] non-square products of squares?
by Dan Hoey 26 Sep '03

26 Sep '03

John Conway <conway(a)math.princeton.edu> writes: > Since the two best examples are A(4) = 2^2 : 3 and 2^4 : 3 > it would seem that 2^6 : 3 deserves examination. Let me try > to do that now (despite the fact that I have to lecture in a few > minutes). > ... I presume 7/32 beats 5/24, but don't have time to check. Well, of course it does. It seems that 2^6:3 is SmallGroup(192,1541), and yes, it has a s.s=n density of 5/32. I haven't got a way of forming the semidirect product os something that large yet. And pretty much by luck, scanning the 1090235 groups of order 768 from the highest numbered down, the first one that had a derived subgroup of 2^8 was SmallGroup(768,1085321), and it's got a s.s=n density of 85/384 . Assuming that group is 2^8:3, we've got density of 2(1-2^-k)/9 in 2^k:3 for 0(?),2,4,6,8 . I didn't find anything over 1/6 in groups of size 80 or 448. I couldn't check 320. Dan

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[math-fun] another online textbook
by Richard Schroeppel 26 Sep '03

26 Sep '03

We may get to the point of having a decent set of math texts online via a back-door route. Rich ----- Date: Fri, 26 Sep 2003 12:55:36 -0400 Reply-To: "Prof. J. E. Cremona" <John.Cremona(a)nottingham.ac.uk> Sender: Number Theory List <NMBRTHRY(a)LISTSERV.NODAK.EDU> Subject: Algorithms for modular elliptic curves - online Algorithms for Modular Elliptic Curves J. E. Cremona Online Edition Cambridge University Press, who published the First Edition (1992) and Second Edition (1997) do not plan to reprint my book, and I have no plans to write a third edition. Instead, CUP have allowed me to post the text of the second edition (with corrections) on the web. You can find it at http://www.maths.nott.ac.uk/personal/jec/book/fulltext/index.html There are pdf files of the whole book, in sections. It is identical to the second published edition, except that various errata have been corrected, and a couple of references updated. John Cremona -- Prof. J. E. Cremona | University of Nottingham | Tel.: +44-115-9514920 School of Mathematical Sciences | Fax: +44-115-9514951 University Park | Email: John.Cremona(a)nottingham.ac.uk Nottingham NG7 2RD, UK |

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Re: [math-fun] Kissing number (again).
by Jud McCranie 26 Sep '03

26 Sep '03

At 01:04 PM 9/26/2003, you wrote: > To see why, let's use the usual reduction of the problem to >packing spherical caps into a sphere, and let's change the size >of these caps to that at which there's room for just 8 mutually >tangent ones, centered at the vertices of a cube. Then after >starting with two touching ones of these, your method would >place a third so as to touch both of them, which would prevent >it from finding the correct configuration. Thank you, that's a very clear counter-argument. My assumption was that you need to pack them as tightly as you can as you go, but that's not true. Thanks to you and the others for the insight and discussions.

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