August 2014 - mathfun@mailman.xmission.com

Re: [math-fun] Steiner chain constraints
by Bill Gosper 13 Aug '14

13 Aug '14

Meanwhile, Rich points out, privately for some reason, that the curvatures in the n=6 case form a "magic rectangle": gosper.org/stein6.png --rwg WFL> Oh dear --- the outer circle had been getting the wrong sign, not the inner --- give me two options, and I unfailingly pick the wrong one. This came to light when I slotted in non-random n = 4 test case k1,k2,k3,k4,k5,k0 = 1,2,1,0,2,0 . So (third time lucky?) Steiner 4-ring constraints seem to be (k0 + k5) - (k1 + k3) , (k0 + k5) - (k2 + k4) , (k0 - k5)^2 - 4*(k3 + k4)*(k0 + k5) + 4*(k3^2 + k4^2) , (k0 - k5)^2 - 4*(k2 + k3)*(k0 + k5) + 4*(k2^2 + k3^2) , (k0 - k5)^2 - 4*(k1 + k2)*(k0 + k5) + 4*(k1^2 + k2^2) , (k0 - k5)^2 - 4*(k4 + k1)*(k0 + k5) + 4*(k4^2 + k1^2) . Their Groebner basis very neatly comprises the two linear relations, together with a single quadratic: hence the dimension of the constraints is (at most) 3 = n-1, as desired (easily it must exceed 2). [The Jacobian rank still comes out as 5 --- I must be misconstruing something elementary there, but it doesn't affect things for the present.] Note that earlier constraints and test-cases proposed for n = 5 should have signs of k0,k6 reversed. Fred Lunnon

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[math-fun] Steiner chain constraints
by Fred Lunnon 13 Aug '14

13 Aug '14

Ahah! RWG's second set of Steiner 5-ring constraints failed my numerical tests because of a wrong sign for k0 --- I assume curvatures have sign built-in, allowing for circles to evert under transformation. I didn't initially suspect this problem, since by Murphy's law the dodgy sign passed when tested on the canonical concentric ring. So the constraints for n = 5 should have read: 5 cyclic shifts (fixing k0,k6 ) of k1^2 + k3*k5 + k2*k4 + k2*k3 = k3^2 + k5*k2 + k4*k1 + k1*k2 ; and 5 shifts of 4*k5^2 - 6*k4*k1 + 2*k2*k3 - k0*k5 + k6*k5 = constant . [My updated versions --- I can't be bothered to correct all the signs in his original! Could the first set also be pruned to some equivalent LHS = const form perhaps?] Unfortunately, this makes no difference to the rank, which remains 5 rather than 4; I remain baffled by this discrepancy. Fred Lunnon

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[math-fun] Re Fields Medals
by Stuart Anderson 13 Aug '14

13 Aug '14

http://www.theguardian.com/science/alexs-adventures-in-numberland/2014/aug/… Explanations of the work of the medallists > On 14 Aug 2014, at 12:32 am, math-fun-request(a)mailman.xmission.com wrote: > > Send math-fun mailing list submissions to > math-fun(a)mailman.xmission.com > > To subscribe or unsubscribe via the World Wide Web, visit > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > or, via email, send a message with subject or body 'help' to > math-fun-request(a)mailman.xmission.com > > You can reach the person managing the list at > math-fun-owner(a)mailman.xmission.com > > When replying, please edit your Subject line so it is more specific > than "Re: Contents of math-fun digest..." > > > Today's Topics: > > 1. Re: Alternatoids? (Dan Asimov) > 2. Re: Dumping water (Warren D Smith) > 3. Re: Steiner chain constraints (Fred Lunnon) > 4. junk DNA - the case of the Antarctic Midge (Warren D Smith) > 5. Maryam Mirzakhani first female Fields Medal winner > (Tehran-Iran born prof. at Stanford) (Warren D Smith) > 6. Re: Dumping water (Eugene Salamin) > 7. Re: Steiner chain constraints (Fred Lunnon) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Tue, 12 Aug 2014 17:15:35 -0700 > From: Dan Asimov <dasimov(a)earthlink.net> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] Alternatoids? > Message-ID: <F9A6EADB-5E46-4604-9EA2-227D5EE3E7B5(a)earthlink.net> > Content-Type: text/plain; charset=us-ascii > > Actually, a special case of Moufang loops are groups. > > The statement below refers to only the Moufang loops of size <= 31 that are *not* groups -- i.e., that are not associative. > > --Dan > > >> On Aug 12, 2014, at 5:05 PM, Dan Asimov <dasimov(a)earthlink.net> wrote: >> >> A paper states: "There are 13 [Moufang loops of size <= 31]: one of order 12, five of order 16, one of order 20, five of order 24, and one of order 28. > > > > > ------------------------------ > > Message: 2 > Date: Tue, 12 Aug 2014 20:16:35 -0400 > From: Warren D Smith <warren.wds(a)gmail.com> > To: math-fun(a)mailman.xmission.com > Subject: Re: [math-fun] Dumping water > Message-ID: > <CAAJP7Y14zigtfWo9uSrSD_odnk-Sqp=N2xPuQMqe-6+pzf_2YQ(a)mail.gmail.com> > Content-Type: text/plain; charset=ISO-8859-1 > > I had ignored pressure, i.e. was regarding the water in the bottle as > both shallow and > far wider than the hole. > > Yes, Eugene is correct, if the bottle were very tall there would be high > pressure and this eventually should overcome any surface tension > attempt to hold it in. > (You could easily modify my maths to incorporate pressure...) > > > > ------------------------------ > > Message: 3 > Date: Wed, 13 Aug 2014 01:51:13 +0100 > From: Fred Lunnon <fred.lunnon(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] Steiner chain constraints > Message-ID: > <CAN57YqtsNfdj5KPnuMPrndjy1JEsP2_Q1YbNgVrwvz1Sr6w85g(a)mail.gmail.com> > Content-Type: text/plain; charset=UTF-8 > > Aaarghhh! Lost the mods --- should have read << > > So the constraints for n = 5 should have read: > 5 cyclic shifts (fixing k0,k6 ) of > k1^2 + k3*k5 + k2*k4 + k2*k3 = k3^2 + k5*k2 + k4*k1 + k1*k2 ; > and 5 shifts of > 4*k5^2 - 6*k4*k1 + 2*k2*k3 + k0*k5 + k6*k5 = constant . > >>> WFL > > > >> On 8/12/14, Fred Lunnon <fred.lunnon(a)gmail.com> wrote: >> Ahah! RWG's second set of Steiner 5-ring constraints failed my >> numerical >> tests because of a wrong sign for k0 --- I assume curvatures have sign >> built-in, allowing for circles to evert under transformation. I didn't >> initially suspect this problem, since by Murphy's law the dodgy sign passed >> when tested on the canonical concentric ring. >> >> So the constraints for n = 5 should have read: >> 5 cyclic shifts (fixing k0,k6 ) of >> k1^2 + k3*k5 + k2*k4 + k2*k3 = k3^2 + k5*k2 + k4*k1 + k1*k2 ; >> and 5 shifts of >> 4*k5^2 - 6*k4*k1 + 2*k2*k3 - k0*k5 + k6*k5 = constant . >> >> [My updated versions --- I can't be bothered to correct all the signs >> in his original! Could the first set also be pruned to some equivalent >> LHS = const form perhaps?] >> >> Unfortunately, this makes no difference to the rank, which remains 5 >> rather than 4; I remain baffled by this discrepancy. >> >> Fred Lunnon > > > > ------------------------------ > > Message: 4 > Date: Tue, 12 Aug 2014 20:53:27 -0400 > From: Warren D Smith <warren.wds(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: [math-fun] junk DNA - the case of the Antarctic Midge > Message-ID: > <CAAJP7Y2OmVDFeE=4bicmK6yQznnTjRhb2foa+e5q9YPX1O8-iA(a)mail.gmail.com> > Content-Type: text/plain; charset=ISO-8859-1 > > The Antarctic Midge, Belgica antarctica, a small insect native to antarctica, > just had its genome sequenced, result 99 million base pairs and about > 13500 genes > (7333 BPs per gene). > > The Anopheles mosquito has 278 million base pairs and approximately > 13683 genes. > (20317 BPs/gene.) > > The Mountain Grasshopper Podisma pedestris > has 16600 million base pairs and probably about the same number of > genes as mosquito. > (1213184 BP/gene.) > > Humans: 3200 million BP, about 23000 genes. > (139130 BP/gene.) > > Some of the largest animal genomes at present are the small arctic > crustacean, Ampelisca macrocephala with 63200 million base pairs, > and the marbled lungfish, Protopterus aethiopicus, with 133000 million. > > If crustacean does not have as many genes as human(?) then at least > 2747826 BPs/gene. > > So anyhow, some animals have 400 times as much DNA per gene, as other animals. > > -- > Warren D. Smith > http://RangeVoting.org <-- add your endorsement (by clicking > "endorse" as 1st step) > > > > ------------------------------ > > Message: 5 > Date: Tue, 12 Aug 2014 21:36:46 -0400 > From: Warren D Smith <warren.wds(a)gmail.com> > To: math-fun(a)mailman.xmission.com > Subject: [math-fun] Maryam Mirzakhani first female Fields Medal winner > (Tehran-Iran born prof. at Stanford) > Message-ID: > <CAAJP7Y0wBha-g-_c2b3QtzB7=tidE80e-eAk6kcr=vQK4kkt1A(a)mail.gmail.com> > Content-Type: text/plain; charset=ISO-8859-1 > > and what did she actually do? > > http://en.wikipedia.org/wiki/Maryam_Mirzakhani > says some stuff, but basically I did not understand a word of this. > > http://www.mathunion.org/fileadmin/IMU/Prizes/2014/news_release_mirzakhani.… > contains a few actually comprehensible claims. > > Despite Iran culture and education system allegedly being anti-female, > unlike the enlightened West, you'll notice the first female Fields > medalist came from Iran. > > > > ------------------------------ > > Message: 6 > Date: Tue, 12 Aug 2014 20:06:02 -0700 > From: Eugene Salamin <gene_salamin(a)yahoo.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] Dumping water > Message-ID: > <1407899162.58290.YahooMailNeo(a)web162105.mail.bf1.yahoo.com> > Content-Type: text/plain; charset=utf-8 > > The pressure difference P across a surface of radius of curvature r with surface tension ?is > > P = 2?/r. > > The hydrostatic pressure of a fluid of density ? with column height h in gravity g is > > P = ?gh. > > Suppose there is a circular hole of radius R in the bottom of the jar.? A bead of fluid extends beneath the hole, and if R << h, we may neglect the variation of hydrostatic pressure within the bead, so that the bead is a spherical cap of radius of curvature r. > > Setting the pressures equal, > > rh = 2?/?g. > > For the bead to be stable, r >= R, so there is a maximum column height H with > > > RH = 2?/?g. > > On Earth g = 9.8 m/s^2, and for water, ? = 1000 kg/m^3, ? = 0.073 N/m. > > Thus RH = 1.5e-5 m^2 = 15 mm^2. > > For example, a 1 mm radius hole should support a 15 mm column height.? Note that one must use a jar, not a tube, in order to avoid the attractive capillary force. > > > ? --? Gene > > >> ________________________________ >> From: James Propp <jamespropp(a)gmail.com> >> To: math-fun <math-fun(a)mailman.xmission.com> >> Sent: Monday, August 11, 2014 9:14 PM >> Subject: [math-fun] Dumping water >> >> >> Say I have a jar full of water covered by a thin plate. Then I turn the >> jar upside down, holding the plate firmly against the lip of the jar to >> prevent water from spilling. Then I whisk away the plate so that instead of >> the plate pushing against the water, only the air beneath the jar >> is pushing upward. >> >> Of course, the water will leave the jar. But what will the geometry of the >> process be? The water can't leave as a cylindrical slug; intuitively, it >> seems that the process has instability, so that spontaneous fingering in >> the air-water interface will break the initial cylindrical symmetry. >> >> Maybe Doc Edgerton made high-speed pictures of this? >> >> Jim Propp > > ------------------------------ > > Message: 7 > Date: Wed, 13 Aug 2014 15:31:42 +0100 > From: Fred Lunnon <fred.lunnon(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] Steiner chain constraints > Message-ID: > <CAN57YqsAsqYBJo5DpC6gx_hxugKaEOSdUESTaCT0a-H8h1G5kA(a)mail.gmail.com> > Content-Type: text/plain; charset=UTF-8 > > And so to hunt missing relations when n = 4 . > Via brute-force equation-solving I discovered that > > (k0 + k5) - (k1 + k3) , (k0 + k5) - (k2 + k4) , [RWG] > > (k0 - k5)^2 + 4*(k3 + k4)*(k0 + k5) + 4*(k3^2 + k4^2) , > (k0 - k5)^2 + 4*(k2 + k3)*(k0 + k5) + 4*(k2^2 + k3^2) , > (k0 - k5)^2 + 4*(k1 + k2)*(k0 + k5) + 4*(k1^2 + k2^2) , > (k0 - k5)^2 + 4*(k4 + k1)*(k0 + k5) + 4*(k4^2 + k1^2) , > > all vanish (verified numerically). > > But the Jacobian (again!) has rank 5, instead of 3. Flubber, flubber ... > > Fred Lunnon > > >> On 8/13/14, Fred Lunnon <fred.lunnon(a)gmail.com> wrote: >> Aaarghhh! Lost the mods --- should have read << >> >> So the constraints for n = 5 should have read: >> 5 cyclic shifts (fixing k0,k6 ) of >> k1^2 + k3*k5 + k2*k4 + k2*k3 = k3^2 + k5*k2 + k4*k1 + k1*k2 ; >> and 5 shifts of >> 4*k5^2 - 6*k4*k1 + 2*k2*k3 + k0*k5 + k6*k5 = constant . >> >>>> WFL >> >> >> >>> On 8/12/14, Fred Lunnon <fred.lunnon(a)gmail.com> wrote: >>> Ahah! RWG's second set of Steiner 5-ring constraints failed my >>> numerical >>> tests because of a wrong sign for k0 --- I assume curvatures have sign >>> built-in, allowing for circles to evert under transformation. I didn't >>> initially suspect this problem, since by Murphy's law the dodgy sign >>> passed >>> when tested on the canonical concentric ring. >>> >>> So the constraints for n = 5 should have read: >>> 5 cyclic shifts (fixing k0,k6 ) of >>> k1^2 + k3*k5 + k2*k4 + k2*k3 = k3^2 + k5*k2 + k4*k1 + k1*k2 ; >>> and 5 shifts of >>> 4*k5^2 - 6*k4*k1 + 2*k2*k3 - k0*k5 + k6*k5 = constant . >>> >>> [My updated versions --- I can't be bothered to correct all the signs >>> in his original! Could the first set also be pruned to some equivalent >>> LHS = const form perhaps?] >>> >>> Unfortunately, this makes no difference to the rank, which remains 5 >>> rather than 4; I remain baffled by this discrepancy. >>> >>> Fred Lunnon > > > > ------------------------------ > > Subject: Digest Footer > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > ------------------------------ > > End of math-fun Digest, Vol 138, Issue 14 > *****************************************

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[math-fun] Dumping water
by James Propp 12 Aug '14

12 Aug '14

Say I have a jar full of water covered by a thin plate. Then I turn the jar upside down, holding the plate firmly against the lip of the jar to prevent water from spilling. Then I whisk away the plate so that instead of the plate pushing against the water, only the air beneath the jar is pushing upward. Of course, the water will leave the jar. But what will the geometry of the process be? The water can't leave as a cylindrical slug; intuitively, it seems that the process has instability, so that spontaneous fingering in the air-water interface will break the initial cylindrical symmetry. Maybe Doc Edgerton made high-speed pictures of this? Jim Propp PS: The author of this post disclaims responsibility for all property damage caused by experiments prompted by this question, along with any matrimonial rifts resulting therefrom.

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[math-fun] junk DNA - the case of the Antarctic Midge
by Warren D Smith 12 Aug '14

12 Aug '14

The Antarctic Midge, Belgica antarctica, a small insect native to antarctica, just had its genome sequenced, result 99 million base pairs and about 13500 genes (7333 BPs per gene). The Anopheles mosquito has 278 million base pairs and approximately 13683 genes. (20317 BPs/gene.) The Mountain Grasshopper Podisma pedestris has 16600 million base pairs and probably about the same number of genes as mosquito. (1213184 BP/gene.) Humans: 3200 million BP, about 23000 genes. (139130 BP/gene.) Some of the largest animal genomes at present are the small arctic crustacean, Ampelisca macrocephala with 63200 million base pairs, and the marbled lungfish, Protopterus aethiopicus, with 133000 million. If crustacean does not have as many genes as human(?) then at least 2747826 BPs/gene. So anyhow, some animals have 400 times as much DNA per gene, as other animals. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] Dumping water
by Warren D Smith 12 Aug '14

12 Aug '14

I had ignored pressure, i.e. was regarding the water in the bottle as both shallow and far wider than the hole. Yes, Eugene is correct, if the bottle were very tall there would be high pressure and this eventually should overcome any surface tension attempt to hold it in. (You could easily modify my maths to incorporate pressure...)

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[math-fun] Dumping water
by Warren D Smith 12 Aug '14

12 Aug '14

If the opening in the jar is small enough... will water still leak out? The instability (whether you call it "Taylor" or "fingering" or whatever) exists in the absence of surface tension. However, in order for this instability to happen, the surface needs to bend away from flat, say with part pushing in and another part pushing out. This increases the surface area. This suggests that this instability becomes "stability" if the surface tension is high enough. Now let us attempt to turn this into a semiquantitative law. Let the hole be circular of radius r. Regard both the "water" and "air" as incompressible inviscid fluids of different densities. Assume the water will be perturbed vertically by epsilon along some smooth surface which differs from flat by a volume-amount of order epsilon*r^2. Its surface will thus increase from pi*r^2 to pi*r^2+C*r*epsilon^2 at leading order for some dimensionless positive constant C. (By the isoperimetric theorem there is a lower bound on C.) This costs energy C*T*epsilon^2 where T is surface tension. The water will finger downward and air upward, and if the water finger is narrower and longer than the air finger, then this gains gravitational energy proportional to g*epsilon^2*densitydiff*r^2. If the water finger is geometrically identical to the air finger, though, then no gravitational energy gain or loss. So anyhow, there clearly exists a finger shape which causes instability in the limit epsilon-->0+, and this is true no matter how large the air-water surface tension. But if T is large enough compared to g*densitydiff*r^2, then the surface energy term can be larger than the gravitational term when, say, epsilon=0.1*r. Therefore, I would predict that for small enough r, equivalently large enough T, the instability will be shut down at finite size (although it will remain for infinitesimal epsilon). This predicts (using epsilon=0.1*r for this purpose) that the critical r where the leakage will be stopped should be of order rcrit = 0.1*T/(g*densitydiff) which putting in numbers g = 9.8 meter/sec^2 densitydiff = 1000 kg/meter^3 T = 0.072 Joule/meter^2 yields the crude estimate rcrit = 0.86 millimeters. So I suggest the following experiment: 1. for all large enough holes, the water will come out. 2. but for a small enough hole (perhaps a good choice is diameter=0.5 mm), we predict the water will not leak out. 3. But if soap is in the water (which ought to cut T by about a factor of 10) then it will too leak out.

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Re: [math-fun] Dumping water
by Henry Baker 12 Aug '14

12 Aug '14

bahfest.com, perhaps? Sounds like a case of keeping up with the IgNobels... My nephew won an IgNobel, BTW. https://ig-nobel.silk.co/page/Geoffrey%20Miller At 02:38 PM 8/12/2014, James Propp wrote: >Perhaps I should confess here the background of my question: I'm thinking >that my facetious way of explaining megafauna mass extinction via >"spontaneous reversal of Earth's gravitational field" would be a good >candidate for a BAHFest talk at some point. > >A BAHFest talk is supposed to include some real science along with the >bogus kind, so, in addition to citing dimensional scaling laws to explain >why falling up and then falling down would be much more injurious to big >animals than to small ones, I'll want to discuss R-T instability and its >relevance to the question "What happened to the oceans during this >postulated event?" > >See bahfest. <http://bahfest.org>com if you don't know what BAHFest is. > >Jim Propp > >On Tuesday, August 12, 2014, Tom Rokicki <rokicki(a)gmail.com <javascript:_e(%7B%7D,'cvml','rokicki(a)gmail.com');>> wrote: > >> Interesting things happen as the diameter of the opening decreases. >> The plate might be thought of as artificially increasing surface tension. >> >> -tom >> >> On Tue, Aug 12, 2014 at 9:20 AM, meekerdb <meekerdb(a)verizon.net> wrote: >> > On 8/12/2014 1:39 AM, Gareth McCaughan wrote: >> >> >> >> On 12/08/2014 05:14 <x-apple-data-detectors://5>, James Propp wrote: >> >> >> >>> Say I have a jar full of water covered by a thin plate. Then I turn the >> >>> jar upside down, holding the plate firmly against the lip of the jar to >> >>> prevent water from spilling. Then I whisk away the plate so that instead of >> >>> the plate pushing against the water, only the air beneath the jar >> >>> is pushing upward. >> >>> >> >>> Of course, the water will leave the jar. But what will the geometry of the >> >>> process be? The water can't leave as a cylindrical slug; intuitively, it >> >>> seems that the process has instability, so that spontaneous fingering in >> >>> the air-water interface will break the initial cylindrical symmetry. >> >> >> >> http://en.wikipedia.org/wiki/Rayleigh%E2%80%93Taylor_instability >> >> (I think). >> > >> > Things are made of atoms. Atoms obey quantum mechanics. Quantum mechanics >> > includes randomness. So symmetries get broken. >> > >> > Brent Meeker

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[math-fun] Unable to reach Torsten Sillke
by Neil Sloane 12 Aug '14

12 Aug '14

Does anyone know how to reach Torsten Sillke? The email address Torsten.Sillke(a)uni-bielefeld.de and he does not answer his phone in Frankfurt a M Thanks Neil -- Dear Friends, I have now retired from AT&T. New coordinates: Neil J. A. Sloane, President, OEIS Foundation 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane(a)gmail.com

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[math-fun] The "critical distance" for binary, q-ary, and spherical codes
by Warren D Smith 11 Aug '14

11 Aug '14

A "binary code" is an M-element subset of the 2^n binary n-bit words, with minimum hamming distance d. A "q-ary code" is an M-element subset of the q^n words of n letters in alphabet of size q, words, with minimum hamming distance d. A "spherical code" is a set of M points on the unit sphere in n dimensions, wiht minimum angular separation d. If we let d depend on n in some monotone manner, then if d is too large, then necessarily M will stay small, e.g. growing at most polynomially as a function of n. On the other hand, if d is too small, then codes exist with enormously large M growing faster with n than any polynomial, and indeed if d is small enough exponentially. It thus is an interesting question what is the "critical" d(n) delineating the border between "too small" and "too large." I have not seen this question & answer stated before, although one might have thought it'd be well known. CLAIM: Let k>0 be a constant and consider the following choice of d as a function of n: for binary codes: d = n/2 - k*n^(1/2) for q-ary codes (q>1 is a fixed prime integer): d = (1-1/q)*n - k*n^(1/2) for spherical codes: sec(d) = k*n^(1/2) I claim with this choice of d, codes whose cardinalities M grow like a power of n exist, where the power may be made arbitrarily large by making k small enough. I conjecture that was best possible, i.e. there exists some k>0 which forces M to be upper bounded by a polynomial in n in each of the three cases. One might conjecture still more strongly that for any fixed k>0, M is upper bounded by a polynomial in n. PROOF (but not of the conjectural part): You can simply use the duals of BCH codes (where for spherical codes, use binary codes with alphabet {-1,+1} then rescale all n-vectors to have unit length) in view of (A) this paper: Daniel Augot, Francoise Levy-dit-Vehel: Bounds on the minimum distance of the duals of BCH codes, http://hal.archives-ouvertes.fr/docs/00/50/94/78/PS/i3e2newmat.ps (B) the fact that (extended narrow sense) BCH linear q-ary codes with blocklength n and "designed distance h+1" exist, for any fixed integer h>0, which when n-->infinity have duals having order n^floor(h/2) codewords. (It is possible that actual min distance exceeds the designed distance, but this possibility does not hurt us.) Finally, my conjecture that this claim is best possible would follow for binary linear codes from the conjecture that no binary linear codes exist with the same parameters as the "Kerdock codes." This conjecture is research problem 15.4 in MacWilliams & Sloane and as far as I know still remains open (although it has been verified by computer exhaustive search for small n). -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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