January 2004 - mathfun@mailman.xmission.com

Re: [math-fun] When marketroids do science...
by Mike Stay 15 Jan '04

15 Jan '04

Yes, they make great insulin pumps; I'm sure it's just that some marketer took it upon himself to "explain" how it works. ----- Original Message Follows ----- > I'm friends with some of the technical folks at Minimed > (here in the L.A. area). One of the founders of Minimed > is a ham radio buff. I'm sure that he knows the > difference! > > At 06:48 PM 1/15/2004, Mike Stay wrote: > >http://www.minimed.com/patientfam/pf_ipt_paradigm_insulin > _pump.shtml > > >RF Technology Provides More Flexibility > >RF technology uses sound waves to "beam" information > instead >of light waves like infrared (IR) technology. To > understand >why RF is more adaptable, compare your TV > remote control to >a cell phone. Your remote control uses > infrared (IR) >technology, so you must point it directly > at your TV to beam >the light waves. Cell phones, however, > use sound waves that >travel more freely, even penetrating > solid objects. As a >result, you eliminate the > inconvenience. > > >-- > >Mike Stay > >staym(a)clear.net.nz > >http://www.cs.auckland.ac.nz/~msta039 > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun -- Mike Stay staym(a)clear.net.nz http://www.cs.auckland.ac.nz/~msta039

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Re: [math-fun] Two puns
by Mike Stay 14 Jan '04

14 Jan '04

I wrote > Iraqui which is the right way to spell it if one is writing in Spanish, and thus perfectly justified in context ;D -- Mike Stay staym(a)clear.net.nz http://www.cs.auckland.ac.nz/~msta039

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[math-fun] Two puns
by Mike Stay 14 Jan '04

14 Jan '04

Babelabelian: when you mix up your words, but everyone still understands what you mean Probabelistic: adj. Characterising supporters of a sovereign Iraqui state; see Zionist. -- Mike Stay staym(a)clear.net.nz http://www.cs.auckland.ac.nz/~msta039

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[math-fun] sl(2,z)-unique convext lattice polygons
by Chris Landauer 14 Jan '04

14 Jan '04

hihi, all - it occurs to me that in the few cases that had more than one polygon left, after applying rigid motions and SL(2,Z), i should provide the polygons for n=7, area 6.5 longest edge 3, perimeter 12 (0,0) (1,0) (3,1) (3,2) (2,3) (1,3) (0,1) longest edge 3, perimeter 12 (0,0) (1,0) (2,1) (2,2) (1,3) (-1,2) (-1,1) for n=11, area 21.5 longest edge 4, perimeter 22 (0,0) (1,0) (4,1) (5,2) (5,3) (4,4) (2,5) (1,5) (0,4) (-1,2) (-1,1) longest edge 3, perimeter 22 (0,0) (1,0) (3,1) (4,2) (4,3) (3,5) (2,6) (1,6) (0,5) (-1,3) (-1,2) for n=15, area 51.5 longest edge = 5, perimeter = 36 (0,0) (1,0) (4,1) (6,2) (7,3) (8,5) (8,6) (7,8) (6,9) (5,9) (3,8) (2,7) (0,4) (-1,2) (-1,1) longest edge = 4, perimeter = 34 (0,0) (1,0) (4,1) (6,2) (7,3) (7,4) (6,6) (5,7) (3,8) (2,8) (0,7) (-1,6) (-2,4) (-2,3) (-1,1) for n=20, area 121.0 longest edge = 5, perimeter = 54 (0,0) (1,0) (4,1) (6,2) (9,4) (10,5) (11,7) (11,8) (10,10) (9,11) (7,12) (6,12) (3,11) (1,10) (-2,8) (-3,7) (-4,5) (-4,4) (-3,2) (-2,1) longest edge = 4, perimeter = 52 (0,0) (1,0) (4,1) (6,2) (7,3) (8,5) (8,6) (7,8) (6,9) (4,10) (1,11) (0,11) (-3,10) (-5,9) (-6,8) (-7,6) (-7,5) (-6,3) (-5,2) (-3,1) for n=22, area 164.0 longest edge = 5, perimeter = 62 (0,0) (1,0) (4,1) (6,2) (9,4) (10,5) (11,7) (12,10) (12,11) (11,13) (10,14) (8,15) (7,15) (4,14) (2,13) (-1,11) (-2,10) (-3,8) (-4,5) (-4,4) (-3,2) (-2,1) longest edge = 5, perimeter = 62 (0,0) (1,0) (4,1) (6,2) (9,4) (10,5) (11,7) (11,8) (10,10) (9,11) (7,12) (4,13) (3,13) (0,12) (-2,11) (-5,9) (-6,8) (-7,6) (-7,5) (-6,3) (-5,2) (-3,1) all of these polygons and the programs will eventually be on the web site, but they need explaining and it needs reorganizing (next week probly) more soon, cal Chris Landauer Aerospace Integration Science Center The Aerospace Corporation cal(a)aero.org

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[math-fun] Re: small area convex lattice polygons
by Chris Landauer 14 Jan '04

14 Jan '04

Dan Hoey wrote: > > Chris Landauer wrote: > > > well, the systematic search program i wrote to find minimum area > > convex lattice polygons took 20 hours for n=20 and 80 for n=21, with > > results as follows.... > > My understanding is that this still does not establish a lower bound > on area even for the 11-gon, unless we can somehow prove that there is > a minimum-area 11-gon with no edge more than 10 units long. Given the > use of SL(2,Z) to force a unit edge, we would have to prove the > existence of an 11-gon with a unit edge _and_ no edge longer than 10. > ... This is of course true. I am merely encouraged by the fact that for small n, the minimum area did not decrease when the edge bound was increased from some very small value up to 50. In fact, in the higher cases (with the random search program), the minimum area found decreased substantially as the edge length bound decreased, until a threshold was passed and no polygons could be completed at all. Similary, I noticed that among all of the minimum area polygons that my program found for even n, all of them were centrally symmetric (that is, the second half edge directions are the negatives of the first half ones). I think that there is a simple proof of this (perhaps less cryptic and subscripty than the one in Barany and Tokushiga), and I am presently writing the code to use this observation in a different systematic search program. Finally, I haven't figured out anything that can usefully bound the perimeter or the maximum edge length of these polygons, but it seems that most values of n<=21 have many long skinny minimum area polygons and a few more blobby ones (which are the ones that should have smaller diameters). Oh, and by the way, the maximum euclidean distance and the maximum manhattan distance between two vertices of any of the minimum area polygons my systematic search program found occur between the same pairs of vertices (I just checked that). more later, cal Chris Landauer Aerospace Integration Science Center The Aerospace Corporation cal(a)aero.org

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[math-fun] more on small area convex lattice polygons
by Chris Landauer 14 Jan '04

14 Jan '04

hihi, all - minimum area convex lattice polygons well, this problem has really caught my attention and held it (i've been dreaming about graph paper and polygons) - every time i decide to stop working on it i think of something else that will speed up the programs, give better results, or contribute towards proving that this method works the extra normalization i mentioned before did not work usefully: polygons normalized to have vertices (0,1), (0,0), and (1,0) often have longer edge lengths, and the larger number of fractions needed to find longer edges greatly overshadows the savings on selection i used this unsurprising (but unanticipated 8-() observation to find yet another faster algorithm for even n: the program makes all of the n/2 selections from the set of 2*nf fractions representing all possible edges in one half of a centrally symmetric polygon (using the original set of possible edges for the first corner and their +90 degree rotations for the second corner), and it does not assume that (0,0) -> (1,0) is the first edge (only that (0,0) is the first vertex and all y-values are non-negative) i also realized that any polygon edge can be transformed into (0,0) -> (1,0) by a rigid motion and some SL(2,Z) element, so i changed the reduction program and reduced almost all minimum area polygons that my various programs found down to a unique polygon for each size (i'm still not convinced that i am finished with this reduction) - only n=7, 11, 15, 20, and 22 stil have two distinct minimum area polygons under these transformations for n<=50 summary of results so far min number of min min longest sl(2,z) unique n area perim edge polygons (if not 1) 3 0.5 4 2 4 1.0 4 1 5 2.5 8 2 6 3.0 8 2 7 6.5 12 3 2 8 7.0 12 2 9 10.5 16 3 10 14.0 18 3 11 21.5 22 3 2 12 24.0 24 3 13 32.5 28 4 14 40.0 30 3 15 51.5 34 4 2 16 59.0 36 3 17 75.5 42 5 18 87.0 44 4 19 106.5 50 5 20 121 52 4 2 21 144.5 60 5 22 164 62 5 2 23 24 210 72 5 25 26 274 80 5 27 28 345 90 5 29 30 430 98 5 31 32 523 108 5 33 34 632 118 5 35 36 749 128 5 37 38 890 138 5 39 40 1039 148 5 41 42 1222 162 7 43 44 1412 176 7 45 46 1620 192 7 47 48 1838 208 8 49 50 2088 218 8 i still can't figure out what B&T mean by ``the minimizing polygon'', so i have no useful comment about that paper yet more soon, cal

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[math-fun] From Littlewood's Miscellany
by Thane Plambeck 13 Jan '04

13 Jan '04

I wish there were more like books like Littlewood's Miscellany (edited by Bollobas); URL at the end of this message. Here are some of Littlewood's entries from the chapter entitled "Cross-Purposes, Unconscious Assumptions, Howlers, Misprints, Etc:" > > I once objected to an apparently obscure use of the phrase, "Let us assume for simplicity." It should mean that the writer could > do the unsimplified thing, but wishes to let the reader off; it turned out that my pupil meant that he had to simplify before he > could do it. > > A minute I wrote (about 1917) for the Ballistic Office ended with the sentence "Thus sigma should be made as small as possible." > This did not appear in the printed minute. But P. J. Gregg said, "What is that?" A speck in a blank space at the end proved to be the > tiniest sigma I have ever seen (the printers must have scoured London for it). > > The spoken word has its dangers. A famous lecture was unintelligible to most of its audience because "HÃ¡rnoo", clearly an > important character in the drama, failed to be identified in time as h \nu. > > A recent (published) paper had near the beginning the passage "The object of this paper is to prove (something very important)." > It transpired with great difficulty, and not till near the end, that the "object" was an unachieved one. > > From an excellent book on Astronomy: "Many of the spirals (galaxies), but very few of the ellipsoidals, show bright lines, due, no > doubt, to the presence or absence of gaseous nebulae." (This rich complex of horrors repays analysis. Roughly it is an illegitimate > combination of the correct 'spirals show bright lines due to the presence...' and the incorrect 'ellipsoidals don't show bright > lines due to the absence...'). > > "We all know that people can sometimes do better things than they have done, but X has done a better thing than he can do." (An > actual case, with agreement on the point among experts). > And from the Chapter entitled "People", a note on a meeting between Bertrand Russell and G E Moore: > > Moore and Russell (c.1896?) were having a philosophical discussion in Hall. Russell suddenly said: "You don't like me, Moore, do > you?" Moore replied, "No." This point disposed of, the discussion proceeded as before. > Littlewood's Miscellany at Amazon: http://www.amazon.com/exec/obidos/tg/detail/-/052133702X/qid=1074011666/ Thane Plambeck 650 321 4884 office 650 323 4928 fax http://www.plambeck.org

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[math-fun] Secret Santa problem
by Marc LeBrun 12 Jan '04

12 Jan '04

Secret Santa [*] is a common holiday procedure where each person in a set of N > 3 is secretly assigned a random different member to whom they will give a gift. Everyone is to receive one gift, and no one is to know anyone else's assignment. That is, a random derangement of the entire set is selected but each participant is only allowed to know their own assignment. Is there a good manual procedure we can use to make these secret random assignments? Drawing names from a hat is natural, but it can result in self-assignments. We could make repeated drawings until none result, but this can be tedious; the portion of derangements is 1/e. We could avoid self-assignments by ordering everyone and somehow drawing random positive "deltas", but this will often violate the one gift requirement. We could have a computer (or other outside party) generate the derangement and then have each person obtain their assignment in secret, but that's clunky and requires an omniscient non-participant. Can you think of a way to generate secret derangements that's about as natural as drawing names from a hat? [*] alternatively, Chaotic Chanukah, Random Ramadan, Kooky Kwanzaa,...

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[math-fun] Subset Problem
by Jon Perry 12 Jan '04

12 Jan '04

If we let S be a set of n integers, then how can we determine the lowest number x_k such that x_k is a sum of 1, 2, 3, ..., k distinct elements of S? 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] Secret Santa -> random derangements
by Allan C. Wechsler 10 Jan '04

10 Jan '04

As seems to be my style recently, I'm not going to actually develop an algorithm for random derangements, but instead will suggest the outline of one in the hopes that some energetic funster will do all the hard work. We want to produce a random derangement on N items. Start by figuring out how many such derangements there are; say there are M of them. Select a random number between 0 and M-1. This will be the "index" of the derangement we will generate; call it K. Initialize a variable to K: this is the "residual index", and we will be decrementing it as we leaf through the derangements in a certain order. (Eudora's spell-checker doesn't like "decrementing" (or "Eudora's"!).) Element 0 can go to 1, 2, ... M-1. Calculate the number of derangements that take 0 to 1. If this number is less than or equal to the residual index, decrement the residual index by this amount. Now calculate the number of derangements that will take 0 to 2. Repeat this process until the number of derangements that takes 0 someplace is less than the residual index, and nail that sucker: 0 will go to that place, which we may call 0'. Now we must decide where to put 1. Calculate the number of derangements that take 0 to 0' and 1 to 0. If this number is less than the residual index, decrement the residual index and go on to the next legal place that 1 can go. These legal places will be everywhere except 1 and 0'. Eventually we will decide on a 1', because the number of derangements that take 0 to 0' and 1 to 1' will be greater than the residual index. Similarly, we decide on 2', 3', 4', and so on; eventually we will reach a point where the residual index is 0 and everything else is forced. (The code probably won't have to notice this specially, this being one of those examples of something that looks like a special case to a human, but is just business as usual to a program.) Essentially we have invented a numbering scheme for derangements, and it is blatantly clear that all derangements have an equal likelihood of being picked. Missing from this description is the algorithm for calculating the number of derangements with some values given; I do not know how hairy such a procedure is. In is conceivable that there is another, simpler, more elegant technique, but the recent observation about factorizations of derangement censuses gives me an intuition that this will be hard to find. -ACW

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