July 2006 - mathfun@mailman.xmission.com

[math-fun] Buried Cable problem
by Michael Kleber 24 Jul '06

24 Jul '06

A week or two ago I heard for the first time the following enjoyable brain teaser: --------- A company has just buried a 10-mile-long cable containing 120 individual wires. Unfortunately, the project was overseen by a summer intern, and the sad result is that the wires are entirely unlabelled. Your job is to fix this mess: label each wire with the same name on both ends. The tools you have available are (1) a battery, (2) a lightbulb, and (3) an unlimited amount of patch cord. And (4) your feet, and no other form of transportation. You start at one end of the cable, and can do as much as you want there, but then you need to walk down to the other end and repeat. In how few trips down the length of the cable can you label the wires? --------- I have a second problem for follow-up, but I can only present it to people who have solved the first problem (for general n, of course, not just n=120). So tune in tomorrow for my tale of the ups and downs in solving this one, and perhaps help me out of my end state (down). --Michael Kleber p.s.: An interesting observation is that with n=2 the problem is unsolvable; there is no way to tell the two wires apart. -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.

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[math-fun] Blender help
by Bernie Cosell 22 Jul '06

22 Jul '06

My wife isn't exactly sure where to turn. She's been struggling mightily with Blender and losing at every turn. We bought the [800 page] book, but it really doesn't address the problems she's running into and she's going nuts constantly fighting with the program. As I mentioned a few months back, what she wants to do is "geometric" stuff [for example, a currrent hack: two intersecting tetrahedra and then be able to highlight and play with the octahedron that shows up in their intersection]. I'd have thought that that kind of thing would be fairly easy [extrapolating frm the similar 2-D stuff I've done with both Illustrator and Visio], but apparently not. The types of things that are making her go bonkers: 1) the edges. Since Blender deals with 3D objects, you need to have some kind of 'strut' to use for the edges. She decided on using triangular-cross-section struts. But then to make that work out properly, she apparently had to hand-compute the intersections between the structs [at the odd angles that they show up in in polyhedra] and by-hand "mitre" the struts so that they fit "just right" with no points sticking out. Does that sound right? [she and I ended up struggling to remember high-school trig and were throwing around lots of sqrt(3)s and such, but I can't imagine that an "artist type" who would want some kind of polyhedron or the like would even know HOW to do the 8-decimal- place calculations, much less want to bother to be hand-trimming each strut. She/I/We *must* be missing something. 2) Faces. Apparently, in order to do faces she needed to compute-by- hand the coordinates of the vertices and then fit a face to it, but apparently once she does that the vertices are "gone" and if she needs to remove or redo the face, she has to start over with the vertex calculations... [not to mention just having to *compute* the coordinates of the vertices of a polyhedron feels very wrong to me]. My guess is that she's missing some key intution on how to use blender for this type of application [partly extrapolating from my experience with "2D" tools, where I don't do any of that stuff: I use skews and rotates and copying/pasting and such and let the program do all the heavy-trig-lifting; I use layers to allow me to "scribble" and test fitting pieces in over stuff that was already done, but that allows me to easily wipe it away if I screw it up]. The problem is that we don't know where to turn to try to figure out what's going on -- virtually all of the Blender info is aimed at *animation* [which is not what my wife is interested in - what she wanted/needed was a 3D modeling/drawing program] and *artists* [rather than geometers or engineers or the like]. Do any of you know of any Blender resources that might help shed some light on this kind of use of the program? THANKS!!! /Bernie\ -- Bernie Cosell Fantasy Farm Fibers mailto:bernie@fantasyfarm.com Pearisburg, VA --> Too many people, too few sheep <--

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[math-fun] Coffin for sale
by Michael Kleber 22 Jul '06

22 Jul '06

If there's anyone out there who, like I did, wants to own a hard copy of Stewart Coffin's book _The Puzzling World of Polyhedral Dissections_ (available free on line), permit me to point out that abebooks.com has just listed a copy for the unusually low price of $27.50. http://www.abebooks.com/servlet/SearchResults?tn=The+Puzzling+World+of+Poly… --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.

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[math-fun] Question-sequence probability
by Guy Haworth 21 Jul '06

21 Jul '06

There are 9! sequences for the 9 questions. So there are (9!-1) 'other' sequences. There are 9*8/2 = 36 pairs to flip if one flip is required. So there are 1+36 sequences 0 or 1 'flip' away from the 1-9 sequence. Therefore the probability = 37/(9!-1) = 37/362,879, just over 1 in 10,000 g

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Re: [math-fun] How to Compute This Probability?
by Daniel Asimov 21 Jul '06

21 Jul '06

Robert Baillie asks: << . . . the same 9 topics, in the same order, except with items 6 and 9 reversed. Let's assume that there are only these 9 topics to choose from. What is the probability that . . . ? >> I'll answer the question I think was intended: Question: Given a random permutation of {1,...,9}, what is the probability that it moves at most 2 symbols? The # of perms. moving 0 symbols is 1. No perm. can move only 1 symbol. The number of pairs of symbols is 9*8/2 = 36. For each pair, the number of perms. interchanging the pair but moving no other symbols is 1. Thus the answer is 37/9! = 37/362880 = 1.01962+ x 10^(-4) or about 1 out of 10,000. ***************************************************** Statistics tries to address questions of the sort that can't really be answered, like "What is the probability that two permutations of {1,...,9} that differ in only 2 places were actually chosen independently?" The basic method is to define the "null hypothesis" as the assumption that the two perms. were in fact chosen independently. Then (before examining the data) an "N% confidence interval" is chosen to include events whose probability -- *assuming the null hypothesis* -- is at least N%. N is typically selected to be 5 or 1. The decision procedure is then to accept or reject the null hypothesis "at the N% confidence level" according as the event in question falls inside the N% confidence interval or not. This method was originally designed to deal with events having a normal distribution. In a different situation like this one, defining the "confidence interval" can be a bit of an art. Let d(k) be the number of derangements (perms without any fixed points) of k elements. According to A000166 of Neil Sloane's OEIS, d(n) for n = 0,...,9 is given by n | 0 1 2 3 4 5 6 7 8 9 d(n)| 1, 0, 1, 2, 9, 44, 265, 1854, 14833, 133496* The number of permutations of {1,...,9} moving exactly k elements, 0 <= k <= 9 is then #(k) = C(9,k)*d(k). My calculator tells me this sequence is k | 0 1 2 3 4 5 6 7 8 9 #(k) | 1 0 36 168 1134 5544 22260 66744 133497 133496 9-k | 9 8 7 6 5 4 3 2 1 0 (Here 9-k is the number of elements FIXED by #(k) different perms. of {1,...,9}.) Now, how to define an N% confidence interval? I'd start with the average number of fixed elements -- call it Favg -- and move outward to get an interval # fixed elements F lies in the interval I(t) = [Favg-t, Favg+t] for some real number t. The corresponding probabilities are of course #(9-F) for each F, and the P(t) := Prob(F lies in I(t)) = sum{F in I(t)} of #(9-F). Thus choose your confidence level of N% = c, say, and find the largest t such that P(t) <= 1-c. Call it t(c). Then I(t(c)) is your confidence interval at the N% level of confidence. --Dan __________________ * Had I checked, I'd have learned that this is part of A098825 of the OEIS *before* doing the calculation!

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[math-fun] How to Compute This Probability?
by Robert Baillie 20 Jul '06

20 Jul '06

Hypothetical situation: On July 7, a reporter (let's call her Eileen Byrne) asks Donald Rumsfeld questions on 9 topics, in this order: 1. North Korea 2. William Perry's op-ed on striking NK 3. diplomatic efforts on NK 4. stability of Iraq 5. political pressure in U.S. for withdrawal 6. U.S. troops' morale 7. American journalists giving away state secrets 8. Iraq War is being fought at home, in the media 9. Hamdan decision On July 8, a different reporter (let's call her Monica Crowley) asks Donald Rumsfeld questions on the same 9 topics, in the same order, except with items 6 and 9 reversed. Let's assume that there are only these 9 topics to choose from. What is the probability that these "reporters" chose the order of their questions independently? (That is, either the same order, or at most one switch of two questions?) Bob Baillie

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[math-fun] GF2 polynomial multiplication
by Schroeppel, Richard 19 Jul '06

19 Jul '06

(continuing a thread from June 29) I mentioned a half dozen ways of doing GF2 polynomial multiplication. One of them is to represent the polynomial by its value at several places. [Do the multiplication of the values at each place, then recover the polynomial product (or its values) by interpolation.] Gene points out that GF2 has a limited repertoire of places. FTAW offers the fix of removing the problem to GF[2^k]. I'm mentioning here that you can actually remove the problem to Z, and just use ordinary integers to do the problem, then clip the coefficients back down to 0-1 values. Another option is to do the problem with coefficients moved into Mod P or Mod N, or even a collection of Mods. You can even use mod 2+i, or phi, or 1+w (w=cbrt(1)); the last two shade over into GF[2^k]. Rich ---------- To make this work for GF(2), you have to reinterpret them as polynomials over GF(2^k) for some sufficiently large k. Franklin T. Adams-Watters -----Original Message----- From: Eugene Salamin <gene_salamin(a)yahoo.com> --- "Schroeppel, Richard" <rschroe(a)sandia.gov> wrote: ... > Another way to represent polynomials is by giving the values at enough > points. ... But for polynomials over GF2, the only possible values are 0 and 1. Gene

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[math-fun] Lagrangian for electrodynamics
by Eugene Salamin 19 Jul '06

19 Jul '06

Can one of the advanced theoretical physics people here answer this question. Is there a Lagrangian, either classical or quantum, that describes the dynamics of the electromagnetic field together with charged particles, electrons if we need to be specific? From what I see, there is a Lagrangian in which the field variables are varied, with fixed charges and currents, to give Maxwell's equations. A second Lagrangian with particle coordinates varied, for fixed fields, gives the Lorentz force law. And finally a third Lagrangian (see Landau & Lifschitz, "Classical Theory of Fields") tries to meld the two together, but is only valid to order (v/c)^2. I am looking for One Lagrangian that rules them all, one that determines the exact theory of electrodynamics. Gene __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com

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[math-fun] Philately
by Richard Guy 19 Jul '06

19 Jul '06

Apologies if this is too far off topic, but I'm going this afternoon to the launching of a commemorative Canadian postage stamp celebrating the 100th anniversary of the founding of the Alpine Club of Canada. If the response is not too overwhelming, and can be made before 15:00 Mountain Time today, Wednesday, 2006-07-19, I will endeavor to send first day cards or envelopes, or whatever is available, to any philatelically inclined munster or seqfan who sends a snailmail address. R.

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Re: [math-fun] Another physics question
by dasimov＠earthlink.net 18 Jul '06

18 Jul '06

<< The idea was to give a time-independent example of energy, but explaining the difference between mass and weight may not be clear to beginners, especially why acceleration has to be involved. >> One pedagogic loophole the keeps acceleration only implicit, would be to define work (energy E) as determined by the time t it takes to bring a (frictionless) body of mass m up to speed s (along a straight line): E = ms/t. This has a visceral aspect to it, always a good thing in teaching. --dan

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