August 2014 - mathfun@mailman.xmission.com

Re: [math-fun] http://en.wikipedia.org/wiki/Steiner_chain n=4
by Bill Gosper 11 Aug '14

11 Aug '14

Fred Lunnon> Panic on again --- my contrived "explanation" for the Jacobian rank being 5 instead of 4 is utter tripe. The polynomial rank equals the dimension of the tangent space at a generic point of the variety, and is the same whether we are in projective space or Euclidean: the rank for any Steiner constraint basis has to equal n-1 no matter what. So I contrived a (non-canonical) numerical example. The first relation set k1^2 + k3*k2 + k3*k5 + k2*k4 = k3^2 + k1*k2 + k1*k4 + k2*k5 etc. have rank 2, and checks out to 10 decimals; the second set FAILS !! Phew ... The counterexample used above had [k1,...,k5,k6,k0] = [-19.62529364, -5.201811345, -27.16907840, -55.16907840, -50.50676305, -37.32621846, 4.034461316]; full details of numerical test cases can be provided on request. A minor omission in my earlier theorem on opposite pairs for n even: proof should read << Modulo similarity, any n-ring D equals C transformed by some Moebius inversion X , and conversely. <WFL I'm confused. How is In[5]:= k1^2 + k2 k3 + k2 k4 + k3 k5 -> k1 k2 + k3^2 + k1 k4 + k2 k5 /. Thread[{k1, k2, k3, k4, k5, k6, k0} -> {-19.62529364, -5.201811345, -27.16907840, -55.16907840, -50.50676305, -37.32621846, 4.034461316}] Out[5]=2185.6819136605263 -> 2185.681912691753 a failure? --rwg [Massive chop]

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Re: [math-fun] http://en.wikipedia.org/wiki/Steiner_chain n=4
by Bill Gosper 11 Aug '14

11 Aug '14

WFL>Red-faces all round alert: the claims made in my previous post concerning dimension and basis of the constraint set are unconditionally withdrawn. The only useful fact to emerge from my Gröbner explorations is that RWG's longer relation is equivalent to the more elegant and succinct (4*k5^2 - 6*k4*k1 + 2*k2*k3 - k0*k5 + k6*k5) = (4*k3^2 - 6*k2*k4 + 2*k5*k1 - k0*k3 + k6*k3) . However, even this minuscule improvement was called into question when it occurred to me to compute the dimension of the relation set via the rank of its Jacobian matrix [[d f_i / d k_j]] , where f_i and k_j range over the 10 relations and 7 variables resp. Disaster! The Jacobian rank is 5 ; but Steiner rings have dimension 3 modulo isometry, and there are only n+2 = 7 variables. Since 3+5 > 7 , I conclude that the relations cannot possibly hold for all Steiner 5-rings. Fred Lunnon My face remains a healthy shade of gray. gosper.org/steiner5.pdf suggests that your pentanomial is invariant under cyclic permutation of k1..k5. (Note k0,k6 are the annulus.) I confirm your claim of five, not ten, relations. --rwg On 8/10/14, Fred Lunnon <fred.lunnon(a)gmail.com> wrote: For n = 5 --- assuming their validity, which currently remains unproved --- RWG's experimental quadratic relations constitute a polynomial basis for constraints on Steiner ring curvatures, with dimension 4. Also there are no linear constraints. [Established using Gröbner bases.] The constraints comprise cyclic shifts (fixing k0,k6 ) of k1^2 + k3*k2 + k3*k5 + k2*k4 = k3^2 + k1*k2 + k1*k4 + k2*k5 ; k1*k0 + k2*k0 - k3*k0 - k4*k0 + 2*k1*k3 - k2^2 + k1*k5 + k3*k4 + k3*k5 = k1*k6 + k2*k6 - k3*k6 - k4*k6 + 2*k2*k4 - k3^2 + k2*k1 + k4*k5 + k5*k2 ; there are only 5 (not 10) in each orbit, since symmetry causes reflections to be equivalent to shifts. One relation in each orbit is redundant. WFL On 8/10/14, Fred Lunnon <fred.lunnon(a)gmail.com> wrote: RWG: Wait, you can do those weirdly unsymmetrical n=5's? In principle, but it's a more complicated on account of involving cyclotomy / trigonometry (as well as quadratic polynomials). Yeah, but in the States you need a prescription for that. In the end I resorted to pentacyclic (Lie-sphere) rather than tetracyclic (Moebius) coordinates, mostly because the geometric algebra machinery was to hand. Guess you need security clearance for that ... WFL On 8/10/14, Fred Lunnon <fred.lunnon(a)gmail.com> wrote: RWG: << Apparently for composite cycle length n = j h, break the list of n curvatures into j bursts of length h. Weight them by a(1),a(2),...,a(h-1),a(h), with sum_i a(i) = 0. Then sum(a(i) k_i)=0 But there are likely other relations involving as well the annulus a(0),a(n+1). >> All curvatures are determined by any three consecutive; hence there are n-1 independent constraints. Does this polynomial ideal have a linear basis? For n = 4 , 2 out of 3 are given by RWG's linear relations (k_0 + k_5) - (k_1 + k_3) = (k_0 + k_5) - (k_2 + k_4) = 0 ; what is some third independent relation? WFL On 8/10/14, Bill Gosper <billgosper(a)gmail.com> wrote: On Fri, Aug 8, 2014 at 8:39 PM, Bill Gosper <billgosper(a)gmail.com> wrote: On Fri, Aug 8, 2014 at 3:35 PM, Bill Gosper <billgosper(a)gmail.com> wrote: On Fri, Aug 8, 2014 at 7:05 AM, Bill Gosper <billgosper(a)gmail.com> wrote: If the four curvatures are k1, k2, k3, k4, and the bounding circles are k0 and k5, then empirically, k1 + k3 = k2 + k4 = k0 + k5 for some assignment of signs to k0 and k5. --rwg Found nothing for n=5, k1^2 + k2 k3 + k2 k4 + k3 k5 == k1 k2 + k3^2 + k1 k4 + k2 k5 plus nine more, closing over D_5. Huh? And apparently ten more like 0 == k0 k1 + k0 k2 - k1 k2 - k2^2 - k0 k3 + 2 k1 k3 + k3^2 - k0 k4 - 2 k2 k4 + k3 k4 + k1 k5 - k2 k5 + k3 k5 - k4 k5 - k1 k6 - k2 k6 + k3 k6 + k4 k6, where k0,k6 are the annulus. And there seems to be a very weird one involving k0 and k6, which may well fail under scrutiny. --rwg Failed. I've been finding these numerically (69D) using assorted transcendental homographic coefficients, figuring if I get a false positive, I'll be famous for finding a relation among π,e^-2, and EulerGamma, e.g. But I'm getting dozens of false positives. Boy am I confused. but for n=6, {k1,...,k6} bounded by k0 and k7, k1+k4=k2+k5=k3+k6 (opposite pairs)=3(k0-k7), *and* k1+k3+k5=k2+k4+k6. (Four relations.) --rwg Apparently for composite cycle length n = j h, break the list of n curvatures into j bursts of length h. Weight them by a(1),a(2),...,a(h-1),a(h), with sum_i a(i) = 0. Then sum(a(i) k_i)=0 But there are likely other relations involving as well the annulus a(0),a(n+1). WFL> I can prove this via plane tetracyclic coordinates. The later conjectures can undoubtedly be decided in the same fashion, for small fixed n . Fred Lunnon Yeah, but in the States you need a prescription for that. Wait, you can do those weirdly unsymmetrical n=5's? How about n=7? I'm getting dozens of unlikely results like k3^2 + 2 k1 k4 + 2 k4 k5 == 2 k3 k4 + k4^2 + 2 k1 k5 (no k0,k8,k2,k6,k7!) _______________________________________________ math-fun mailing listmath-fun@mailman.xmission.comhttps://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] Apple bizarreness--write-once hard drive
by Bill Gosper 11 Aug '14

11 Aug '14

Not exactly math-fun, but have mercy. I just had a fruitless phonechat with an Apple nonGenius. MacBook Pro freezes and requires powerbutton reboot when Virtual Box hangs well into loading XP. OK, Apple won't touch Virtual Box, but I hoped|conjectured it was just an ungraceful consequence of only 118G of unused hard drive, so I deleted many G and emptied the trash. Now there is only 117G free. While on the phone, I sacrificed an additional 2.85G Mathematica 8.0. Free shrank a little more. Anybody?? --rwg

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Re: [math-fun] Apple bizarreness--write-once hard drive
by Henry Baker 11 Aug '14

11 Aug '14

If you're running Windows under VirtualBox, you should be aware of the Windows utility "sdelete": http://technet.microsoft.com/en-us/sysinternals/bb897443.aspx There are at least 2 different ways for VirtualBox to emulate a disk drive: using a sector-by-sector copy or by using a virtual drive that doesn't bother storing sectors with all zeros. Most people use a virtual drive, but this means that "garbage" unused sectors whose contents aren't zero still take up space. If you run the Windows defragmentation utility within the virtual machine, and then run "sdelete" within the virtual machine, these "garbage" sectors will be reset to zeros, so that the virtual disk image will be much smaller. http://dotnetworkaholic.com/2013/04/06/compacting-virtual-machines-virtualb… At 10:33 PM 8/10/2014, Bill Gosper wrote: >(Apologies for the subject matter but I really need some hand-holding here.) > >On 2014-08-10 19:44, Eugene Salamin via math-fun wrote: >> Can you isolate the problem of whether the hard drive has a hardware problem? > >The customer-oriented disk diagnostics are more about damaged file structure than >drive performance, but that might be appropriate here. They found and fixed two >damaged files and a bleepload of permissions. But they recovered negligible space, >which continues to disappear. > > > Can you disable Virtual Box, > >Believe me, it's disabled far beyond my wishes. > >> and just run normal Apple software? >> >> -- Gene > >I'm trying to. I just wrote and deleted a big file on a flash drive (heating it up!). >The freed space reappeared. So if it's software, it's low-level disk driver stuff. >But what could have altered it? > >>>________________________________ >>> From: Henry Baker <hbaker1(a)pipeline.com> >>>To: math-fun <math-fun(a)mailman.xmission.com> >>>Sent: Sunday, August 10, 2014 6:39 PM >>>Subject: Re: [math-fun] Apple bizarreness--write-once hard drive >>> >>>TByte drives are really, really cheap at Costco. >>> >>>Buy one & use it for Virtual Box, etc., at least for >>>experimentation. >>> >>>In this day & age, lack of HD space shouldn't ever be >>>an issue. >>> >>>Also, you can now get 64GByte flash drives that aren't >>>quite as fast as HD's, but are fast enough for many >>>porpoises. > >Neil Bickford once repeated that "algorithms are for people who don't >know how to buy RAM". So I guess file deletion is for those whose >Costco membership has lapsed. >--rwg >>> >>>At 06:22 PM 8/10/2014, Bill Gosper wrote: >>>>Not exactly math-fun, but have mercy. >>>>I just had a fruitless phonechat with an Apple nonGenius. >>>>MacBook Pro freezes and requires powerbutton reboot when Virtual >>>>Box hangs well into loading XP. OK, Apple won't touch Virtual Box, >>>>but I hoped|conjectured it was just an ungraceful consequence of >>>>only 118G of unused hard drive, so I deleted many G and emptied >>>>the trash. Now there is only 117G free. While on the phone, I >>>>sacrificed an additional 2.85G Mathematica 8.0. Free shrank >>>>a little more. >>>> >>>>Anybody?? --rwg

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Re: [math-fun] Apple bizarreness--write-once hard drive
by Bill Gosper 10 Aug '14

10 Aug '14

(Apologies for the subject matter but I really need some hand-holding here.) On 2014-08-10 19:44, Eugene Salamin via math-fun wrote: > Can you isolate the problem of whether the hard drive has a hardware > problem? The customer-oriented disk diagnostics are more about damaged file structure than drive performance, but that might be appropriate here. They found and fixed two damaged files and a bleepload of permissions. But they recovered negligible space, which continues to disappear. > Can you disable Virtual Box, Believe me, it's disabled far beyond my wishes. > and just run normal Apple software? > > -- Gene I'm trying to. I just wrote and deleted a big file on a flash drive (heating it up!). The freed space reappeared. So if it's software, it's low-level disk driver stuff. But what could have altered it? > > > >>________________________________ >> From: Henry Baker <hbaker1(a)pipeline.com> >>To: math-fun <math-fun(a)mailman.xmission.com> >>Sent: Sunday, August 10, 2014 6:39 PM >>Subject: Re: [math-fun] Apple bizarreness--write-once hard drive >> >> >>TByte drives are really, really cheap at Costco. >> >>Buy one & use it for Virtual Box, etc., at least for >>experimentation. >> >>In this day & age, lack of HD space shouldn't ever be >>an issue. >> >>Also, you can now get 64GByte flash drives that aren't >>quite as fast as HD's, but are fast enough for many >>porpoises. Neil Bickford once repeated that "algorithms are for people who don't know how to buy RAM". So I guess file deletion is for those whose Costco membership has lapsed. --rwg >> >>At 06:22 PM 8/10/2014, Bill Gosper wrote: >>>Not exactly math-fun, but have mercy. >>>I just had a fruitless phonechat with an Apple nonGenius. >>>MacBook Pro freezes and requires powerbutton reboot when Virtual >>>Box hangs well into loading XP. OK, Apple won't touch Virtual Box, >>>but I hoped|conjectured it was just an ungraceful consequence of >>>only 118G of unused hard drive, so I deleted many G and emptied >>>the trash. Now there is only 117G free. While on the phone, I >>>sacrificed an additional 2.85G Mathematica 8.0. Free shrank >>>a little more. >>> >>>Anybody?? --rwg >>

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Re: [math-fun] Apple bizarreness--write-once hard drive
by Henry Baker 10 Aug '14

10 Aug '14

TByte drives are really, really cheap at Costco. Buy one & use it for Virtual Box, etc., at least for experimentation. In this day & age, lack of HD space shouldn't ever be an issue. Also, you can now get 64GByte flash drives that aren't quite as fast as HD's, but are fast enough for many porpoises. At 06:22 PM 8/10/2014, Bill Gosper wrote: >Not exactly math-fun, but have mercy. >I just had a fruitless phonechat with an Apple nonGenius. >MacBook Pro freezes and requires powerbutton reboot when Virtual >Box hangs well into loading XP. OK, Apple won't touch Virtual Box, >but I hoped|conjectured it was just an ungraceful consequence of >only 118G of unused hard drive, so I deleted many G and emptied >the trash. Now there is only 117G free. While on the phone, I >sacrificed an additional 2.85G Mathematica 8.0. Free shrank >a little more. > >Anybody?? --rwg

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[math-fun] http://en.wikipedia.org/wiki/Steiner_chain n=4
by Bill Gosper 10 Aug '14

10 Aug '14

If the four curvatures are k1, k2, k3, k4, and the bounding circles are k0 and k5, then empirically, k1 + k3 = k2 + k4 = k0 + k5 for some assignment of signs to k0 and k5. --rwg

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Re: [math-fun] math-fun Digest, Vol 138, Issue 2
by Steve Witham 09 Aug '14

09 Aug '14

> Date: Tue, 5 Aug 2014 10:40:35 -0400 > From: James Propp <jamespropp(a)gmail.com> > Subject: [math-fun] Colored edge maze > > ... > > The maze I have in mind is one I saw years ago. It's a square grid with > colored edges and one is required to traverse the edges in the order red, > white, blue, red, white, blue, ... (or whatever). > > Can anyone give me a link to the design of such a maze? No, but here's a short discussion of the top maze-generation algorithms inside a cool animated page about visualizing algorithms: http://bost.ocks.org/mike/algorithms/#maze-generation --Steve

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[math-fun] Fields medal history
by James Buddenhagen 09 Aug '14

09 Aug '14

This article contained some stuff I did not know and may be of interest to some on this list. Here is a link: http://www.nytimes.com/2014/08/10/opinion/sunday/how-math-got-its-nobel-.ht… P.S. I hope you can access it without being a Times subscriber, but I'm not sure. James Buddenhagen

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Re: [math-fun] "Essentially serial" computations for PoW ?
by Henry Baker 06 Aug '14

06 Aug '14

If you believe Ernst Haeckel's phrase "ontogeny recapitulates phylogeny", then perhaps growing an embryo is indeed a form of Turing Machine diagonalization, and hence almost essentially serial ! https://en.wikipedia.org/wiki/Recapitulation_theory http://home.cyber.ee/ahtbu/Complexity_3_slides.pdf At 07:16 PM 8/4/2014, W. Edwin Clark wrote: >The most interest example was: > > The number of women will not reduce the length of pregnancy.

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