December 2012 - mathfun@mailman.xmission.com

[math-fun] Has anyone here tried Cornell's "Eureqa" program?
by Henry Baker 10 Dec '12

10 Dec '12

Eureqa is a program that takes data & attempts to find a function composed from a set of given functions by all manner of functional compositions; i.e., it isn't restricted to linear combinations or splines or anything that simple. This is _not_ Lagrange interpolation. You can download & run a free version from here: http://www.nutonian.com/ Ignore all of the instructions, and just go through the built-in example. First go to "Enter Data", where you see a bunch of data already entered. Then go to "Prepare Data", where you see the data already prepared. Then go to "Set Target", where you see the target already set. Then go to "Start Search" and hit the "Run" button to start the program searching. On Windows, you don't even have to "install" the program if you don't want to; you can download the .zip version onto a USB stick & execute the program from there. I'm just trying the program out. Eventually, I'd like to see if it can guess some anti-derivatives from evaluating the expressions at a sufficient number of points.

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Re: [math-fun] How tall can a Lego tower get before the bottom brick is crushed?
by Tom Duff 10 Dec '12

10 Dec '12

On Sat, 8 Dec 2012 Henry Baker <hbaker1(a)pipeline.com> suggested: > It would probably make sense to first build a number of identical robots > that could put Lego pieces together and then climb up on the growing > tower. More realistic (or at least more fun) might be an army of volunteer 6-to-10-year-olds building modules and a pile of Autonomous quadcopters for assembly. -- Tom Duff. Purple lines run though some of the scenes during the lake floor putt-putt ride.

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Re: [math-fun] Antikythera device -- dismissive reply from expert(Prof. Mike Edmunds)
by Henry Baker 09 Dec '12

09 Dec '12

Sea-faring navigation in the ancient world was pretty well developed, but rarely involved more than a few days at a time out of sight of land. Supposedly, the ancient Romans explored part of the west coast of Africa, and did trade (using ships) with parts of India. There was an active sea trade from the Middle East to China & Japan from the 8th century onwards. Within the last 10 years a shipwreck from approx 750AD made from African wood was found just south of China; the ship had very nice & expensive goods from China. The ancient world had "water clocks", which -- if calibrated properly -- were useful for things like astrology appointments. :-) Of course, they told only relative time, not absolute time. Knowledge of tides, sunrise & sunset would have been extremely helpful, since tides would have seriously affected navigable rivers that had tides. However, when I was in Turkey a few years ago, I noticed that Mediterranean tides appeared to be significantly less than Atlantic & Pacific tides. The max/min height of their boat docks did not have anything like the dynamic range of boat docks I've seen on the U.S. coasts. Re navigating in rowboats: I think of the famous Mutiny on the HMS Bounty: http://en.wikipedia.org/wiki/Mutiny_on_the_Bounty "The Mutiny on the Bounty was a mutiny aboard the British Royal Navy ship HMS Bounty on 28 April 1789. The mutiny was led by Fletcher Christian against commanding officer Lieutenant William Bligh. ..." "Eighteen mutineers set Lieutenant Bligh afloat in a small boat with eighteen of the twenty-two crew loyal to him. The mutineers then variously settled on Pitcairn Island or in Tahiti and burned the Bounty off Pitcairn Island, to avoid detection and to prevent desertion." "Bligh navigated the 23-foot (7 m) open launch on a 47-day voyage to Timor in the Dutch East Indies, equipped with a quadrant and pocket watch and without charts or compass. He recorded the distance as 3,618 nautical miles (6,710 km). He then returned to Britain and reported the mutiny to the Admiralty on 15 March 1790, 2 years and 11 weeks after his original departure." At 10:09 AM 12/9/2012, Warren Smith wrote: >--Prof. Mike Edmunds: > Dear Dr Smith, > Thank you for your interesting message. Although the remnants of the > Antikythera Mechanism were found on a ship, we are certain it was not a > navigational instrument. All of the functions we have so far discovered > were for astronomical or for calendar purposes. I have some sailing > experience and (modest) navigation qualifications - and I can confirm that > none of the function would be of use in day-to-day navigation of a boat. > >--WDS comment: >I'm dubious. Does this guy really navigate using chronometer, sextant, and >spherical trigonometry? I doubt anybody has done that for decades. >And one's initial reaction on seeing a chronometer (which is just a >wind-up clock with high-quality innards) might well have been "no use >for day to day navigation of boat." This judgment would >be totally wrong. The only way to realize its use is to investigate mathematical >theory, not by just looking at it and saying "I've sailed on boats." > And by the way, the whole phrase "day to day" employed by Edmunds >is misleading. >More accurate would be "week to week." >To get an idea of reality, it helps to consider the famous sea-voyage >by Shackleton with 5 others in 1914 in open rowboat 800 miles. Did >they do "day to day" navigation? No they did not, because it was >impossible. They set off 24 April. >The log of this voyage reveals that they were able to >obtain successful sun sights on 26 April, 29 April, 4 May. On 8 May >they spotted land (their destination). Landed 10 May. (They then did >some further local sailing next few days.) Thus the entire voyage >lasted 17 days during which they only accomplished 3 sun-sights. >Similarly, the plan that was being investigated by the British for >navigation based on the moon (this was before the success of the >chronometer) would have >required accurate measurement of angles from moon to horizon, stars, >sun. Again, >such measurements would probably have been impossible to accomplish every day. >For example in bad seas or cloudy. I repeat, this whole scheme was >under development >by the British admiralty as a possible navigation method until it got >superseded by the unexpected success of the chronometer which allowed >simpler method. > >--Edmunds: >Its possible use for determining longitude at eclipse times has been > suggested before, but this is totally impractical for navigation because > eclipses occur so rarely. > >--WDS: I agree eclipses rare, and I agree eclipses basically no use, >but you do not need eclipses to determine longitude. And if you were >relying on eclipses, you would be being an idiot. My whole point was >not to do that and not to be an idiot. >For example, did the British admiralty intend to rely on eclipses? Of >course not. >Edmunds appears not to have read my letter and to be highly unfamiliar with the >whole idea behind this kind of navigation. > >--Edmunds: >It has been argued that such devices might > assist in setting longitude at, say, opposite ends of the Mediterranean > sea in a campaign lasting several years, as may have been attempted by the > Chinese in other waters in the 15th century - but again this is not > day-to-day navigation. Anyway, all they would have needed is a written > compilation of eclipse months! > >--WDS: Did the Chinese even know that the Earth was round? >The Greeks were well aware of it by 5th century BC. >Wikipedia http://en.wikipedia.org/wiki/Spherical_Earth has >"In the 17th century, the idea of a spherical earth, now considerably >advanced by Western Astronomy, ultimately spread to Ming China, when >Jesuit missionaries, who held high positions as astronomers at the >imperial court, successfully challenged the Chinese belief that the >earth was flat and square." citing >http://www.uni-tuebingen.de/uni/ans/eastm/back/cs11/cs11-4-martzloff.pdf >Jean-Claude Martzloff: "Space and Time in Chinese Texts of Astronomy >and of Mathematical Astronomy in the Seventeenth and Eighteenth >Centuries", Chinese Science 11 (1993-94): 66-92 (69). > So, I'm not terribly impressed by this Edmunds quote, because (a) >eclipses are >irrelevant hence everything Edmunds says about eclipses is irrelevant, >and (b) also his >whole side-remark about the Chinese (which also is pretty irrelevant) >seems somewhere between dubious and refuted anyhow. > >--Edmunds: > We know from a study of ancient literature sources - see e.g. > http://pos.sissa.it/archive/conferences/170/019/Antikythera%20&%20SKA_019.p… > - that the device was certainly not unique, indeed we have a fair number > of references to such astronomical devices over the period approx 250 BC > to 500 AD, and no mention whatever of a similar navigational device. The > subtle mechanical details of the Antikythera device would be impossible to > design without considerable previous experience, and it must come from a > tradition of the making of such geared mechanisms. > By far the most likely explanation for its presence on the (trading) boat > is that it was being transported, possibly as part of a large consignment > of goods, or simply being delivered on its own. > >--WDS: I'm happy to agree it was on that boat merely for transport. >In fact, I had speculated the device was experimental and stolen by the Romans. >So we are in agreement. So therefore this disclosure by Edmunds has >no logical bearing at all on my entire hypothesis. Which he would >have realized had he actually read my letter. > And the URL Edmunds gave does not seem convincing to me of anything much. >First of all, before Galileo invented the pendulum, it was almost >utterly pointless to build a mechanical clock in the sense accuracy >would have been totally pathetic. >A mechanical clock making an error of say, 30 minutes per day, which >probably would have been about best possible before the year 1200, >would have been for navigational >purposes off by 500 miles per day even if portable (which they weren't). Any >clock anyone ever made pre-Galileo and especially pre-1200 AD >would have been merely as a joke, toy, or mystical >device intended to impress somebody with mysticism, not actually be useful. >I do not see the point of it. > The first pendulum clock was built by Chr.Huygens in 1656. >Second, the "sphaerae" referred to in this manuscript may not have >been a mechanical device with moving parts at all, at least for all >the evidence I see given in the manuscript! >(They mention a Cyclopaedia 1810 article, but what the heck did they >know in 1810?) > >--Edmunds: > With Best Wishes, > Prof Mike Edmunds > >--WDS: >to make a genuine assessment of whether a device like the Antikythera >could have been intended for navigation (or could have been intended >as experimental with possible future development into a workable nav. >device) the main question that needs to be answered is >an accuracy question. There is angle-measurement accuracy for >observations of moon etc. And there is predictive accuracy via the >device. All previous assessments of >predictive accuracy for the device that I have seen, were wrong-headed >and hence largely irrelevant, since they concerned multiyear >predictions. In fact, if it were >used for navigation, the device would be reset to truth at start of >voyage, then >the voyage lasts (say) 3 weeks. So what we need to assess is the accuracy >during this 3 week span assuming perfect accuracy at start. >As the observers of the Antikythera device have already noted, the >device was built >to make it easy to take apart and reset. The question is 2-fold (a) >what accuracy could >this device itself have achieved, and (b) what about a similar >device but with optimized parameters?

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Re: [math-fun] colouring of R^3
by Adam P. Goucher 09 Dec '12

09 Dec '12

> Warren Smith wrote: > > I'll take you up on offer to send description of coloring & impossibility proof. > This sounds incredible. The 'if' and 'only if' parts of the proof are completely different, so I'll present them here as separate proofs. Don't publish these proofs anywhere, since the question is one of Imre Leader's problems for undergraduates at Trinity. NB: I've liberally assumed the Axiom of Choice everywhere, but that's okay because the statement of the continuum hypothesis doesn't even make sense without it (without AC, we can't compare the sizes of infinite sets). Introduction: 2-colouring of Q^2: --------------------------------- This isn't part of the proof of the main result, but it's a similar construction to the CH-dependent construction for R^3. Let's suppose we want to colour the points of Q^2 red or blue such that each row contains finitely many red points and each column contains finitely many blue points (Q = set of rationals). To do this, define an injection f: Q --> N (set of naturals), and colour a point red if f(x) <= f(y) and blue otherwise. Each row has finitely many red points, as there are only finitely many x for a given y such that f(x) <= f(y). Every column has finitely many blue points by symmetry. Theorem 1: continuum hypothesis ==> 3-colouring of R^3: ------------------------------------------------------- Let (x,y,z) be the coordinates of an arbitrary point in R^3. Let {a,b,c} be a permutation of (x,y,z), the details of which are given below. We define an injective (it may as well be bijective, but we only require injectivity) function f from the reals to the set of all countable ordinals (ordinals indexed by the first uncountable ordinal omega_1). Now, we assume without loss of generality that f(a) >= f(b) and f(a) >= f(c). We can do this since we haven't yet fixed the permutation. For fixed a, let g_a be an injection from the ordinals less than or equal to f(a) to the natural numbers (this can be done since f(a) is a countable ordinal). Now, we assume without loss of generality that g_a(f(b)) >= g_a(f(c)). Again, since we've only fixed a in the permutation, we're allowed to do this. We then colour the point (x,y,z) according to the following rules: * (x,y,z) is red if x = c, otherwise; * (x,y,z) is green if y = c, otherwise; * (x,y,z) is blue (and z = c, by process of elimination). I'll show that every line parallel to the x-axis contains finitely many red points, and the other cases follow by symmetry. Let's fix some y and z, and assume wlog that f(z) >= f(y). Now, a point is coloured red only if g_z(f(x)) <= g_z(f(y)). Note that g_z and f are injections, and g_z(f(y)) is a positive integer. There are only finitely many positive integers below g_z(f(y)) (trivially), so only finitely many values of x correspond to the point (x,y,z) being coloured red. Q.E.D. Theorem 2: ¬ CH ==> ¬ 3-colouring of R^3: ----------------------------------------- Assume there exists a subset V of R, which has cardinality strictly intermediate between |N| and |R|. Also assume there exists a 3-colouring of R^3, with the intention of deriving a contradiction. A 3-colouring of R^3 gives a 3-colouring of N x V x R, as N and V are subsets of R. Note that we can express V as a disjoint union of countable sets (proof: well-order V, and define an equivalence relation ~ such that x ~ y if one of (x,y) can be obtained from the other by applying the 'successor' operator finitely many times). A basic corollary of this is that |N x V| = |N x N x V/N| = |N x V/N| = |V| < |R|. Since every line parallel to the z-axis contains finitely many blue points, then there must be at most |N x N x V| = |V| blue points in total. As there are |R| > |V| planes parallel to the xy-plane, at least one of them (indeed, uncountably many!) must have no blue points whatsoever. So, we get a plane of the form N x V such that each row has only finitely many red points and each column has only finitely many green points. Since every line parallel to the y-axis contains finitely many green points, there must be at most |N x N| = |N| green points in total. However, there are |V| > |N| lines parallel to the x-axis, so at least one of them must contain no green points whatsoever, and thus be entirely red. Whence, we obtain a contradiction. By contraposition, we get that a 3-colouring of R^3 implies the continuum hypothesis. Sincerely, Adam P. Goucher http://cp4space.wordpress.com

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Re: [math-fun] Antoine's necklace
by Bill Gosper 08 Dec '12

08 Dec '12

DanA>I'm not sure what "at the limit the areas disappear and everything is boundary" means exactly. rwg>We have a continuous map of a circle onto the closed region depicted in the link: http://www.tweedledum.com/rwg/rad2Image4.gif . The areas in http://www.tweedledum.com/rwg/rad2fill.htm are outlined by connect-the-dots images of (the vertices of) a regular polygon inscribed in the circular domain. Doubling the number of vertices (sample points) corresponds to one iteration of the construction suggested by the upper figure. As with most such closed-loop spacefills (i.e., a sausage encasing a tree), the largest inscribable circle shrinks to a point as the sampling frequency → ∞. Thus the space is filled twice. I.e., every point has at least two inverse images, uncountably many have four, and ℵ₀ of them have six. The polygonal boundaries of the regions are admittedly artifacts of the sampling, but they illustrate the underlying spacefilling function as they crinkle up due to increasing sampling frequency. As they crinkle, the interior area remains constant, then suddenly vanishes at the limit, while the area of the closure is that same constant until it suddenly doubles. Actually, the "constant" may increase slightly to account for the shrinkage of the hexagon in the center of the lower figure. DanA>But if each of the six regions is in the limit space-filling the very same space, Terminology. Regions already fill space. At the limit, their *boundaries* fill the area near the center four times over: twice via the closed loop property, and twice because any extended patch of the lower diagram has two colors. DanA> then isn't this a lot like the sets A_n and B_n in the reals?: Where A_n = union of [k/2^n,(k+1)/2^n] for odd integers k, B_n = union of [k/2^n,(k+1)/2^n] for even integers k. both sets having the common boundary {k/2^n | k in Z}, but for any real x as n -> oo, dist(x,A_n) -> 0 and dist(x,B_n) -> 0. ----- DanA>I agree the illustrations with the Wikipedia article are unhelpful. Here's a picture of the second stage: 4 cyclically linked solid tori going around one, and in each of those 4 there are four smaller cyclically linked tori going around *it*: < http://onionesquereality.files.wordpress.com/2011/07/antoines-necklace.jpg >. (There are a few extraneous lines, which should be ignored.) Yes, three should work. In fact any sequence of numbers of tori at various stages would work -- as long as the diameters of the tori in stage n approach 0, as n -> oo. (I wonder what happens in the case where there is only one torus in each stage, pulled longitudinally around the previous torus and made to link with itself.) --Dan Ick! I think the nth one winds and unwinds 2^n times around inside the outermost one. But we lose the shrinking to a point effect. --rwg RWG wrote: << Robert M. wrote: << . . . . . . But the necklace one bothers me. That transition from linked set of tori to unlinked set of points seems impossible. A similar thing happens with increasingly fine polygonal approximations to spacefilling functions. Inhttp://www.tweedledum.com/rwg/rad2fill.htm the upper figure defines the iteration. For any finite order, six of these equipartition the area near the center of the lower figure. But at the limit, the areas disappear and everything is boundary. I don't get the many-torus illustrations inhttp://en.wikipedia.org/wiki/Antoine%27s_necklace . Doesn't the text say four tori? Shouldn't three work? (Bogus Borromean rings.)

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Re: [math-fun] How tall can a Lego tower get before the bottom brick is crushed?
by Henry Baker 08 Dec '12

08 Dec '12

You are correct about trying to copy the Eiffel Tower too slavishly. However, it's overall shape is good for resisting the wind, which is what this shape was originally engineered to do. A Lego Tower might have to have a wider aspect ratio to have the same wind resistance, but if it were fractal in nature, it need not be excessively heavy. The following web site shows how to build fractal structures, which would be appropriate for this tower. If the tower is designed correctly, it shouldn't need superglue, although it might be more wind-resistant and/or earthquake resistant if it did glue the blocks together. http://www.london-institute.org/people/farr/fractals.shtml Using robots would be much more efficient for a Lego structure, because all of the bricks are identical, and therefore it would be much easier to build such a robot. At 08:46 AM 12/8/2012, Andy Latto wrote: >On Sat, Dec 8, 2012 at 10:53 AM, Henry Baker <hbaker1(a)pipeline.com> wrote: >> FYI -- "A Lego tower with this many pieces would be approximately 3.5km high" -- i.e., over 10,000' ! >> >> This sounds like a challenge to me. I wonder if Lego would provide enough bricks to actually build a 3.5km high Lego tower. The weight per brick is ~2.5 grams. >> >> What would the appropriate shape of such a tower be? Perhaps an Eiffel Tower lookalike, only a lot bigger. > >If you want to crush a block, all the weight of 10,000 blocks has to >be on that single block. So an Eiffel Tower lookalike wouldn't do it, >since the weight would be on at least 4 blocks. And I think it would >collapse, with the center of the tower crashing down between the four >supports, long before it got that big, unless you had something >stronger than the normal lego connection holding them together. I >suspect even supergluing them together wouldn't be enough. > >> According to the following exchange, Lego bricks cost $0.12, so just a single stack 3.5km high would cost $45,000. In order to build a self-supporting tower, the bricks alone would cost millions of dollars, and the cost of erecting such a structure would also be in the millions of dollars. >> >> It would probably make sense to first build a number of identical robots that could put Lego pieces together and then climb up on the growing tower. It might take several years to build it this way, but it would significantly reduce the cost of construction. > >People don't currently find that the cheapest way to build a >skyscraper is to build skyscraper-building robots and have them build >the skyscraper. I don't see why the economic equation should change >because the skyscraper is built out of legos rather than steel >girders. > >Andy

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Re: [math-fun] How tall can a Lego tower get before the bottom brick is crushed?
by Bill Gosper 08 Dec '12

08 Dec '12

http://news.nationalpost.com/2012/11/22/above-the-world-in-90-days-china-bu… The original question seems to me very temperature-dependent. --rwg On Sat, Dec 8, 2012 at 10:53 AM, Henry Baker <hbaker1(a)pipeline.com> wrote: FYI -- "A Lego tower with this many pieces would be approximately 3.5km high" -- i.e., over 10,000' ! This sounds like a challenge to me. I wonder if Lego would provide enough bricks to actually build a 3.5km high Lego tower. The weight per brick is ~2.5 grams. What would the appropriate shape of such a tower be? Perhaps an Eiffel Tower lookalike, only a lot bigger. AnyLatto>If you want to crush a block, all the weight of 10,000 blocks has to be on that single block. So an Eiffel Tower lookalike wouldn't do it, since the weight would be on at least 4 blocks. And I think it would collapse, with the center of the tower crashing down between the four supports, long before it got that big, unless you had something stronger than the normal lego connection holding them together. I suspect even supergluing them together wouldn't be enough. According to the following exchange, Lego bricks cost $0.12, so just a single stack 3.5km high would cost $45,000. In order to build a self-supporting tower, the bricks alone would cost millions of dollars, and the cost of erecting such a structure would also be in the millions of dollars. It would probably make sense to first build a number of identical robots that could put Lego pieces together and then climb up on the growing tower. It might take several years to build it this way, but it would significantly reduce the cost of construction. People don't currently find that the cheapest way to build a skyscraper is to build skyscraper-building robots and have them build the skyscraper. I don't see why the economic equation should change because the skyscraper is built out of legos rather than steel girders. Andy _

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Re: [math-fun] Gamow syndrome (Aleph_0 and omega)
by Adam P. Goucher 08 Dec '12

08 Dec '12

**** Partial spoiler alert **** >> "Can we colour the points in R^3 with the colours red, green and >> blue such that every line parallel to the x-axis contains only >> finitely many red points, every line parallel to the y-axis >> contains only finitely many green points, and every line parallel >> to the z-axis contains only finitely many blue points?" > > Well, I'll have to buy this one since nobody else has volunteered --- > just how does the answer to this intriguingly elementary question > depend on the continuum hypothesis? There exists such a colouring if and only if there are no sets of intermediate cardinality between |N| and |R|. In particular, you can derive a contradiction if you assume there's such a colouring of R x V x N (where V is a set of intermediate cardinality), thus there are no colourings of R x R x R. For the converse, it is not too difficult to construct a colouring of O x O x O, where O is the set of all countable ordinals (i.e. ordinals below omega_1). The continuum hypothesis asserts a bijection between O and R, so a solution for O x O x O (which definitely exists) implies an equivalent solution for R x R x R. The problem generalises to k colours in R^k. By the same argument, it's possible to show that there is such a colouring if and only if R is at most the (k-1)th smallest infinite set. I'll give the details of the construction and the reductio ad absurdum if you're interested. > [I've a feeling that I once knew Imre Leader slightly, a long time ago > when we were both postgraduate students; but can no longer recall > where or how we met.] It's a small world. Sincerely, Adam P. Goucher http://cp4space.wordpress.com

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[math-fun] How tall can a Lego tower get before the bottom brick is crushed?
by Henry Baker 08 Dec '12

08 Dec '12

FYI -- "A Lego tower with this many pieces would be approximately 3.5km high" -- i.e., over 10,000' ! This sounds like a challenge to me. I wonder if Lego would provide enough bricks to actually build a 3.5km high Lego tower. The weight per brick is ~2.5 grams. What would the appropriate shape of such a tower be? Perhaps an Eiffel Tower lookalike, only a lot bigger. According to the following exchange, Lego bricks cost $0.12, so just a single stack 3.5km high would cost $45,000. In order to build a self-supporting tower, the bricks alone would cost millions of dollars, and the cost of erecting such a structure would also be in the millions of dollars. It would probably make sense to first build a number of identical robots that could put Lego pieces together and then climb up on the growing tower. It might take several years to build it this way, but it would significantly reduce the cost of construction. http://bricks.stackexchange.com/questions/1196/what-is-the-average-price-pe… http://wiki.answers.com/Q/What_is_the_weight_of_a_Lego_brick http://www.cbc.ca/news/yourcommunity/2012/12/how-tall-can-a-lego-tower-get-… How tall can a Lego tower get before the bottom brick is crushed? by Lauren O'Neil Posted: December 4, 2012 6:34 PM Last Updated: December 4, 2012 7:04 PM Categories: Community, Science & Technology Of all the bits of information available online, the exact number of Lego pieces that could be stacked, one on top of the other, before the bottom piece was crushed under the weight of the tower itself has remained elusive -- Until now. A team of researchers from The Open University in England has finally come up with an answer to the popular query, which has been a subject of debate for years on websites like Quora and Reddit. But it wasn't as simple as building a Lego tower. According to BBC Radio statistics program More or Less, engineers used a hydraulic testing machine to calculate the strength of one standard square Lego brick. They were able to exert more than 4,000 newtons (N) of force before the square gave out into a melted shape described by the BBC as "a small square of warm camembert." The team determined that the average maximum force Lego bricks can stand is 4,240N - the equivalent of 432 kg. So how many Lego bricks does that equal? The grand total of bricks a single piece of Lego could support is 375,000. A Lego tower with this many pieces would be approximately 3.5km high, significantly higher than Mount Olympus according to the scientists.

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Re: [math-fun] Antoine's necklace
by Bill Gosper 08 Dec '12

08 Dec '12

On 12/8/12, Bill Gosper <billgosper(a)gmail.com> wrote: rwg>We have a continuous map of a circle onto the closed region depicted in the link: http://www.tweedledum.com/rwg/rad2Image4.gif . Two anomalous short segments appear in the latter graphic: one green, one cyan. WFL I think you mean http://www.tweedledum.com/rwg/rad2fill.htm . The segments aren't anomalous--just their colors. Actually, those are correct. But the corresponding red segment at 1 o'clock should be magenta! --rwg

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