November 2003 - mathfun@mailman.xmission.com

Re: [math-fun] 15-puzzle
by MCKAY＠vax2.concordia.ca 09 Nov '03

09 Nov '03

Your number of configurations i misleading since there ios plenty of symmetry. I expect someone from Donald Michie's Dept. of machine Intelligence working in the 60's will have done the computation in Edinburgh. John McKay From: IN%"math-fun(a)mailman.xmission.com" "math-fun" 9-NOV-2003 21:37:37.48 To: IN%"math-fun(a)mailman.xmission.com" "math-fun" CC: Subj: RE: [math-fun] 15-puzzle Return-path: <math-fun-bounces(a)mailman.xmission.com> Received: from clyde.concordia.ca (clyde.Concordia.CA [132.205.1.1]) by vax2.concordia.ca (PMDF V6.2 #30759) with ESMTP id <01L2UCJ5B1GA000YT5(a)vax2.concordia.ca> for mckay(a)vax2.concordia.ca (ORCPT mckay(a)vax2.concordia.ca) Sun, 09 Nov 2003 21:37:36 -0400 (EDT) Received: from mailman.xmission.com (mailman.xmission.com [198.60.22.29]) by clyde.concordia.ca (8.12.10/8.12.10) with ESMTP id hAA2bQk9172924 for <mckay(a)vax2.concordia.ca>; Sun, 09 Nov 2003 21:37:26 -0500 (EST) Received: from localhost ([127.0.0.1] helo=mailman.xmission.com) by mailman.xmission.com with esmtp (Exim 3.35 #1 (Debian)) id 1AJ1vY-0001V5-03; Sun, 09 Nov 2003 19:37:16 -0700 Received: from place.org ([65.163.18.18] helo=zachs.place.org ident=postfix) by mailman.xmission.com with esmtp (Exim 3.35 #1 (Debian)) id 1AJ1vW-0001Uy-00 for <math-fun(a)mailman.xmission.com>; Sun, 09 Nov 2003 19:37:14 -0700 Received: from ACW.alum.mit.edu (pool-129-44-175-237.bos.east.verizon.net [129.44.175.237]) by zachs.place.org (Postfix) with ESMTP id E61A8180EC for <math-fun(a)mailman.xmission.com>; Sun, 09 Nov 2003 20:36:57 -0600 (CST) Date: Sun, 09 Nov 2003 21:28:51 -0500 From: "Allan C. Wechsler" <acw(a)alum.mit.edu> Subject: Re: [math-fun] 15-puzzle In-reply-to: <200311100120.UAA84181(a)fry.research.att.com> Sender: math-fun-bounces(a)mailman.xmission.com X-Sender: acw(a)place.org To: math-fun <math-fun(a)mailman.xmission.com> Errors-to: math-fun-bounces(a)mailman.xmission.com Reply-to: math-fun <math-fun(a)mailman.xmission.com> Message-id: <5.1.0.14.0.20031109212244.023f4d90(a)place.org> MIME-version: 1.0 X-Mailer: QUALCOMM Windows Eudora Version 5.1 Content-type: text/plain; format=flowed; charset=us-ascii Precedence: list X-BeenThere: math-fun(a)mailman.xmission.com X-Mailman-Version: 2.1.1 X-Scanned-By: MIMEDefang 2.35 List-Post: <mailto:math-fun@mailman.xmission.com> List-Subscribe: <http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun>, <mailto:math-fun-request@mailman.xmission.com?subject=subscribe> List-Unsubscribe: <http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun>, <mailto:math-fun-request@mailman.xmission.com?subject=unsubscribe> List-Archive: <http://mailman.xmission.com/cgi-bin/mailman/private/math-fun> List-Help: <mailto:math-fun-request@mailman.xmission.com?subject=help> List-Id: math-fun <math-fun.mailman.xmission.com> Original-recipient: rfc822;mckay(a)vax2.concordia.ca >I'm surprised that the 4x4 case hasn't been solved - can anyone help? >NJAS Why is this surprising? There are more than 2e13 configurations. If we could search a thousand configurations per second, considering each of them would take almost a millennium. So, someone will have to do some non-trivial theory in order to get a better handle on the diameter of the 15-puzzle. I believe the diameter of Rubik's Cube is also not known, in any of the popular metrics. -A _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] 15-puzzle
by N. J. A. Sloane 09 Nov '03

09 Nov '03

Jud McCranie sent in this sequence to the OEIS: %I A087725 %S A087725 0,6,31 %N A087725 Maximum number of moves required for the n X n generalization of the Fifteen Puzzle. %C A087725 a(4) >= 66, a(5) >= 114. a(3) is from Reinefeld, using the method of Korf. The lower bounds on a(4) and a(5) are from Korf and Korf & Taylor, and are unlikely to be tight. %D A087725 Richard E. Korf, Depth-First Iterative-Deeping: An Optimal Admissible Tree Search, Artificial Intelligence, 27(1), 97-110, 1985. %D A087725 Richard E. Korf and Larry A Taylor, Finding Optimal Solutions to the Twenty-Four Puzzle, Proceedings of the 11th National Conference on Artificial Intelligence, 756-761, 1993. %H A087725 Richard E. Korf, <a href="http://citeseer.nj.nec.com/cache/papers/cs/4003/http:zSzzSzwww.cs.ucla.eduz…">Home Page</a> %H A087725 Alexander Reinefeld, Complete Solution of the Eight-Puzzle and the Benefit of Node Ordering in IDA*, International Joint Conference on Artificial Intelligence, pp. 248-253, 1993. <a href="http://www.ams.org/notices/200004/fea-hales.pdf">Reinefe H A087725 Alexander Reinefeld, <a href="http://www.ams.org/notices/200004/fea-hales.pdf">Home Page</a> %H A087725 Eric Weisstein, <a href="http://mathworld.wolfram.com/15Puzzle.html">15 Puzzle in MathWorld</a> %e A087725 All solvable configurations of the Eight Puzzle on a 3x3 matrix can be solved in 31 moves or fewer, and some configurations require 31 moves, so a(3)=31. %Y A087725 Cf. A046164. %Y A087725 Adjacent sequences: A087722 A087723 A087724 this_sequence A087726 A087727 A087728 %K A087725 hard,nonn,bref,nice %O A087725 0,2 %A A087725 Jud McCranie (j.mccranie(AT)adelphia.net) Sep 30 2003 I'm surprised that the 4x4 case hasn't been solved - can anyone help? NJAS

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[math-fun] Prentice Hall strikes again
by R. William Gosper 08 Nov '03

08 Nov '03

My tutoree is now in 8th grade, and, miraculously, didn't carry home another Prentice Hall "science" book. But he just got a smearox of three pages which smell familiar. Remember "The weak force is the key to the power of the Sun."? "Because of the air within it, a huge ocean liner can float on the surface of the ocean."? The ridiculous eclipse drawing that showed both limbs of Sun, moon, and Earth lying on just two straight lines? Now they're "doing" biology. Not, as Einstein prescribed, as simply as possible. Way simpler than that. Summary: the brighter and more curious the student, the more discouraged and confused this crap will leave him. Anyone getting an A on this material is a parrot who has been misled to believe s/he has learned something. I'll indent their text. P 88. 4-2 Respiration: Using the Energy in Food Green plants (autotrophs) I'm glad they cleared that up. I thought they meant photoautotrophs. make their own food. Animals (heterotrophs) eat green plants or other animals that eat green plants. Let's see, so fish eat mosquito larvae and mosquitos eat green plants. Carnivores have an agreement not to eat each other. You have learned that energy is stored in the bonds that link together atoms in foods, such as glucose. That comma makes the sentence wrong. You can think of the energy stored in food as a savings account. You eat food to add to your energy savings account. Huh?? You eat food to add to the energy stored in food? P89 During aerobic respiration, food enters the mitochondria. Burgers, fries, Ding-Dongs with taco sauce, oysters with Cool Whip, ... The food is broken down when it combines with oxygen. During this process, water and carbon dioxide are produced as waste products. Yeah, that water stuff is pretty useless. A worse greenhouse gas than CO2. But respiration produces BOTH! We must save the Earth by banning respiration. And club soda. Maybe if we could consume enough anti-oxidants ... The energy released from food during respiration is not always used right away by the cell. Often, it is stored in a chemical compound called ATP. But sometimes the cell just needs a hit of warm club soda? Basically, ATP is a chemical substance that can be used to store energy. How does a chemical store energy without becoming some other chemical? When a cell needs energy, it breaks down the ATP and uses the energy released. Is this like breaking down food? Although respiration occurs in a series of complex steps, the overall process can be written down as a word equation and as a chemical equation, as follows: Glucose + Oxygen --> Carbon Dioxide + Water + ATP (energy) or C6H12O6 + 6O2 --> 6CO2 + 6H2O + ATP Gee, I thought the P in ATP had something to due with phosphorus. But there's no phosphorous on the left side. Wow, cells can make phosphorus out of glucose. Nuclear transmutation! Notice anything familiar about the equation for respiration? You are quite right if you said it is the opposite of the equation for photosynthesis. Yeah except there's no chlorophyll in respiration and you didn't mention ATP in photosynthesis. Maybe that's what you mean by "opposite". p90 So how do anaerobic organisms get their energy? The answer is that anaerobic organisms undergo a different type of respiration, appropriately called anaerobic respiration. I'm sure glad you told me it was appropriately. Scientists use the term fermentation when they speak of anaerobic respiration. Do they mean the same thing, or does one occur during the other, or is one an example of the other, or are they just wishing they had a beer? Carbon dioxide is also the source of bubbles in alcoholic beverages such as beer and wine. But not soda? Hey, my dad's Chianti doesn't bubble. Must be some new kind of emissions control. 4-2 Section Review 2. Why are the mitochondria appropriately called the powerhouses of the cell? Beats the hell out of me, but I'm sure glad you told me it was appropriately.

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[math-fun] moRe: powers of PseudoAntisymmetric (-1,0,1)- Matrices
by wouter meeussen 08 Nov '03

08 Nov '03

rehash of thread 15/08/2003, with some news. OEIS?Anum=A072148 Sequence: 2,14,92,796,7672 Definitions: pseudoAntisymmetric : T(i,j)= -T(j,i) for j<i , so T = diagonal+Antisymmetric. (my definition, forgive..) powerlength: minimal p>0 so that T^p = Identity Consider the (-1,0,1)-matrices T with properties : Det[T] not zero (invertible), all powers T^k are also invertible (-1,0,1) matrices. Properties: powerlength of T divides 12, Det[t] is 1 or -1, T is pseudoAntisymmetric, the powers T^k need not be all pseudoAntisymmetric: for 4x4 matrices, all those with 8 non-zero elements have powerlength 4, and their powers 2 and 3 are not pseudoAntisymmetric; for the 5x5 matrices, all those with 9 non-zero elements have powerlength 4, and their powers 2 and 3 are not pseudoAntisymmetric; all those with 10 non-zero elements have powerlength 12, and their powers 2,3,6,7,10 and 11 are not pseudoAntisymmetric; There is a system in this madness, but this margin is too small... W. (I owe Marc LeBrun <mlb(a)fxpt.com> for help, partial insight & lots inspiration, thanx Marc) I put the 796 4-by4 and the 7672 5-by-5 on http://users.pandora.be/Wouter.Meeussen/pseudoAntisymmMatrixPowers_4.txt http://users.pandora.be/Wouter.Meeussen/pseudoAntisymmMatrixPowers_5.txt

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[math-fun] Does anyone speak these languages?
by Steve Gray 04 Nov '03

04 Nov '03

The World Council of Endangered Languages has an urgent need for speakers of the following languages, all in danger of extinction: Fubuwudschey, Bunegronk, Old Vittolubwek, Spefuffian, Minugharink, Plikurvisk, Jugbugozug, Quebzitical, Slubuferch, Lower Adefugunotie, Utyolikkikach, and Qyppuxifif. They most urgently need someone who knows Veftunichiklops, which is spoken only by only one woman who is 102 years old and deaf. If you speak at least one of these, please call 800-788-7000 immediately. There is a reward.

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[math-fun] Re: [EMHL] A question about Morley
by Richard Guy 04 Nov '03

04 Nov '03

I owe the group(s) a screed about the Lighthouse Theorem. Here's about 1% of what is known. Lighthouse theorem for k = 3: Two lighthouses each have a doubly infinite beam, rotating once per minute, but not necessarily in phase. Photos, taken at 20 second (or 10 second!) intervals, when superimposed, show 3 equally inclined beams through each lighthouse. These illuminate (intersect in) 9 points which are the vertices of 3 equilateral triangles. It's tempting to say that these are homothetic, but it's more perspicuous to say that they are successively rotated through pi/3. Conway has called these three triangles the Guy Faux triangles. The Morley Miracle is that we now have the complete Morley triangle theorem without even having a triangle to start with! A reasonably direct, and quite elementary synthetic proof of the whole kaboodle follows from `angles in the same segment' (often known, I'm told, as the `butterfly theorem'). The nine edges of the Guy Faux triangles form 3 sets of 3 parallel lines, equally inclined, forming 27 equilateral triangles, 3 triangles sharing each of 27 points. 18 of the triangles are Morley triangles for a triangle which it's left to the reader to construct. It will be found that the 27 points lie on 18 lines other than the edges of the Morley triangles. These are trisectors of angles of a triangle ABC where B and C are the original lighthouses, with two more pairs of lighthouses at A and B and at C and A. The 9 non-Morley triangles form 3 sets of 3 Guy Faux triangles, one set for each pair of lighthouses. Lighthouse Theorem for general k. k equally inclined (multiples of pi/k) beams from each of two lighthouses intersect in k^2 points which are the vertices of k regular k-gons, inclined at multiples of pi/k with each other. So, if k is odd, they appear to be homothetic, but if k is even, there are two different orientations. E.g., for k = 4, you get two pairs of parallel squares, inclined at pi/4 with each other. It's worth looking at the lighthouse theorem for k = 2, yielding 4 orthocentric points, and giving a hint to the fact that the 72 Morley triangles of the triangles ABC, BCH, CAH, ABH (H being the orthocentre of ABC) are all `homothetic'. Note: these 4 triangles have the same 9-point circle. A remarkable theorem of Bremner, Goggins, Guy & Guy is that, apart from a couple of exceptions (equilateral triangle, and a family of Pythagorean triangles), if a triangle with rational edges has a Morley triangle with a rational edge-length, then all 72 triangles have rational edge-length. [{\it Acta Arith.}, {\bf93}(2000) 177--187; {\it MR} {\bf2001f}:51024] I haven't managed a complete description of the general situation, but an outline is as follows: The k k-gons each have (k choose 2) edges (join any pair of vertices). So there are k sets of (k choose 2) edges, inclined at a multiple of pi/k with each other. We can choose 2 sets of parallels in (k choose 2) ways, so the number of points of intersection is (k choose 2)^3. These lie on other lines, which (at present) are too numerous to mention [when k = 3, k = (k choose 3), and this obscures the generality] [perhaps an OEIS editor will compose a remark to go with the other comments to sequence A059827 ?]. But with k = 6 a good deal can be seen, [k = 2 is included, so beams are perpendicular to other beams, bringing in the orthocentre], since we now not only have 72 Morley triangles, but also (I believe) 72 more for each of the 12 triangles IBC, where I is any of the 4 incentres and B, C are any pair of vertices of some triangle. I'm not sure (yet) whether one needs to take a variety of positions for A --- perhaps Io, Ia, Ib, Ic (apologies that I don't know the official notation) or H may (also) qualify. One may also be able to say something about Malfatti. R. On Tue, 4 Nov 2003, jpehrmfr wrote: > Dear Hyacinthists > with the 108 common points of the trisectors of ABC, we can form 54 > equilateral triangles. > 27 are homothetic with the standard Morley triangle > 18 other ones have their sidelines parallel to three trisectors of > the same vertex of ABC > But what about the direction of the sidelines of the 9 other ones? > Friendly. Jean-Pierre > > > ------------------------ Yahoo! Groups Sponsor ---------------------~--> > Buy Ink Cartridges or Refill Kits for your HP, Epson, Canon or Lexmark > Printer at MyInks.com. Free s/h on orders $50 or more to the US & Canada. > http://www.c1tracking.com/l.asp?cid=5511 > http://us.click.yahoo.com/mOAaAA/3exGAA/qnsNAA/6HYolB/TM > ---------------------------------------------------------------------~-> > > > > Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/ > >

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[math-fun] mu & the zeta function; sums of squares
by Richard Schroeppel 03 Nov '03

03 Nov '03

Re: mu(Large) Would taking several zeta zeros improve things? Maybe with weighted voting? Why doesn't this give the same estimates for mu(Large) and mu(Large+1) and ... ? One might also try finding primes P that don't divide Large and guestimating mu(P*Large); this would reflect on the likely value of mu(Large). ---- sums of squares: for Dan's problem: Two squares, N is x2+y2 iff all the poison divisors of N (those P=4K+3) occur to an even power. 9 is ok as an exact power divisor (quotient not a multiple of 3), but 3 and 27 are out. As N gets large, 0% are the sum of two squares; if a number is the sum of two squares, it's probably the sum many ways. 1105 = 5.13.17 = four ways, or 32 ways with +- signs and swapping. Of course the total number of sums of squares < N (counting multiplicity) is roughly pi*N (or pi/4 or pi/8, depending on sign & ordering convention). So the average number of representations of a number is pi, but the representations concentrate in a few values. Three squares, N is x2+y2+z2, if, after removing all possible divisors of 4, the leftover is not 8K+7. The typical number of representations is roughly sqrtN, but I think it can wander up & down by a slowly growing factor, logN or loglogN. When 4|N, x,y,z must be even, so 4^K has only one "unique" rep, and 4^K N and N have the same number of reps. Four squares, all N are representable, the number of reps is a simple multiple of the sum of the divisors of N, with the multiplier depending roughly on N mod 8. Divsum(N)/N >=1, and is less than a slowly growing function of N. Overall, these functions are fairly irregular, so I think Dan's program will work better in in higher dimensions. Hardy & Wright sez formulas exist for 5,6, and 8 squares. Conway & Sloane will know more, if they are telling. Rich rcs(a)cs.arizona.edu

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[math-fun] less than or equal to
by MCKAY＠vax2.concordia.ca 03 Nov '03

03 Nov '03

What is wrong with " up to" ? John

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Re: [math-fun] Simple EF (unnatural numberklaature)
by asimovd＠aol.com 03 Nov '03

03 Nov '03

Mike Speciner writes: << ... If you're going to invent new words (and I'm pretty sure I'd rather stick with "nonnegative") they really ought to be clearly better than the alternative. I understand the awkwardness of phrases like "greater than or equal to", but it's only a problem when speaking. Otherwise, I can write ">=", or, in my favorite programming language, "ge". >> But "only a problem when speaking" can be significant if one likes to talk about math, such as when collaborating or teaching. I hope that simple concepts like nonnegative can be expressed with fewer syllables in the future. Maybe using just "nonneg" and "nonpos" will catch on. --Dan

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Re: [math-fun] units
by reid＠math.arizona.edu 02 Nov '03

02 Nov '03

about a year ago, i read a news article about police radar and laser systems, which mentioned that modern systems measure not only speed, but also acceleration, and perhaps some other data. the article was of the preachy "look what they've got to catch you with" type, so had no technical details. but i was always curious as to what units they use for acceleration; miles per hour^2 seems like it would be awkward, so perhaps they use (miles per hour) per second . or they could be clever and just call it miles per square minute . does anyone know what they use? come to think of it, car ads often use mixed units (though not quite explicitly) when they boast "0 to 60 [mph] in N seconds". mike

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