July 2006 - mathfun@mailman.xmission.com

RE: Re: [math-fun] Buried Cable problem
by Torgerson, Mark D 18 Jul '06

18 Jul '06

For a more practical solution On one end link all the wires in pairs labeling them P1, P2, etc. (If there are an odd number of wires label the leftover uniquely) On the other end identify the pairs, them label the wires in the pairs P1a P1b, P2a P2b, etc. (and identify the odd guy if it exists.) Tie all the 'a' wires together, keeping the 'b' wires separate. March back over to the other side and separate and label a's and b's -----Original Message----- From: math-fun-bounces+mdtorge=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+mdtorge=sandia.gov@mailman.xmission.com] On Behalf Of Michael Kleber Sent: Monday, July 17, 2006 8:14 PM To: math-fun Subject: Re: Re: [math-fun] Buried Cable problem Much thanks to Franklin, who has been good enough to come up with *both* of the solutions to the cable problem. Now I can get on with my extended discussion. The executive summary of this longwinded message: The triangle number solution is really cool, but sadly, the much simpler daisy-chaining solution is better. I would dearly love to see a modification of the puzzle, for which the triangle number solution is the only one. Please please please. Okay, now for the details. When we first heard the problem, both (funster-lurker) Thomas Colthurst and then I independently arrived at the triangle-number solution, as did some colleagues at the Broad when I started to spread the problem around. The quick summary of that solution: say the number of wires is triangular, eg n=10: draw the triangle * * * * * * * * * * First patch together all wires in the same row; from the other end of the cable, you can tell the size of the equivalence class each wire is in. Now patch together all wires in the same column, and trek back to the beginning. The sizes of a wire's two equivalence classes act as unique coordinates. This can be generalized beyond triangle numbers, too. In the 10-wire version above, the coordinates given to the wires end up being 44; 43,34; 42,33,24; 41,32,23,14. All of these pairs sum to one of 8,7,6,5. So if we had, say, three more wires: * * * then we could simultaneously apply this technique to them, getting coordinates 22;12,21; there's no problem confusing the two triangles with each other. In general, then, we can use this method for n wires any time n can be written as a sum of triangular numbers such that if we use T(r) in the sum, we can't use any T(s) for r<s<=2r. Chewing on this last week, I was delighted to prove that this condition holds for all but finitely many n: there are only 39 numbers which cannot be so written; the largest is 720. (Yes, NJAS, I'll get the list to you eventually, but let this discussion play out first please!) So I was feeling pleased with my clever technique and my proof, and all was well until I shared the problem with Broad colleague Jade Vinson (another mathematician, of holyhedron fame). Jade promptly came up with a daisy-chaining method, as Franklin has already described; one construction works for all odd n, and a tiny variation works for all even n>2. This leaves the beautiful triangle number solution looking too clever by half: it's both harder and less universal. Thus my appeal to math-fun: please find a puzzle, of similarly elegant statement, to which the triangle solution is the only one. There certainly should be hope, because the daisy-chaining method queries the state of the system in a fundamentally different way from the triangle technique: for triangles, you need only figure out *how many* other wires each one is associated with, while you can only daisy-chain if you can learn the *identity* of an associated wire. Can anyone rescue "All but 39 values of n" from the dustbin of too-clever puzzle solution history? --Michael Kleber On 7/17/06, franktaw(a)netscape.net <franktaw(a)netscape.net> wrote: > Here's an entirely different approach, which works for all n > 2. > > The case where n is odd is simpler. Link all the wires in pairs, with > 1 left over. At the other end, identify all the pairs. Now link the > singleton to one of pair, link the other member of that pair to one of > a different pair, etc., until the last pair will have one wire > unlinked. Now identify these pairs at the other end. The originally > isolated wire is already identified; the wire it is now linked to can > be identified, which identifies the other member of that pair, etc. > > For n even, link all but 2 wires in pairs. At the other end, link one > of the previously unconnected wires to one of a pair, and chain > through the remaining pairs as before. The other previously > unconnected wire remains unconnected. First identify the wire which > was never connected; the remaining wires then reduce to the previous case. > > Note that this solves the problem never connecting more than 2 wires > together, so we could solve the problem with switchboards instead of > the patch cord. > > Franklin T. Adams-Watters > -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen. _______________________________________________ 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] Another physics question
by Jason Holt 18 Jul '06

18 Jul '06

I also had a physics question recently. I was trying to explain e=mc^2 to my family, and was going to try to give an intuitive example of just how much energy there is in 300,000,000^2 kg*m^2/s^2. I realized that impulse (or momentum), measured in kg*m/s, is much easier for me to conceptualize and explain: a baseball travelling at 30mph has a certain amount of kinetic energy that anyone can experience personally just by catching it. It was much more awkward to try to explain pushing a 1kg weight exactly hard enough to accelerate it 1m/s^2, but only until it had travelled 1m. So if both units can be used to express energy, why do we speak of impulse as so different from energy, and is there an intuitive way to explain the conversion from the 30mph baseball's energy into kg*m^2/s^2 without calculus? I still remember the test question in 10th grade physical science in which I was asked to calculate the work required to carry a 10kg weight across a 10m room. I argued that, at least in the ideal sense, it required no work (if you're infinitely patient), since there was no change in gravitaional potential energy, but the teacher wouldn't budge. I suppose I've resented work ever since. Oh yeah, and maybe Mike Stay could also briefly summarize quantum gravity for us in a way that makes it clear just how that principle works. I was wondering about that, too. -J

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[math-fun] primes in pi; buried cable
by Schroeppel, Richard 18 Jul '06

18 Jul '06

This is from Eric Weisstein ... Date: Mon, 17 Jul 2006 08:42:54 -0500 (CDT) From: "Eric W. Weisstein" <eric(a)weisstein.com> X-X-Sender: eww(a)wopr.wolfram.com To: Sequence Fans Mailing List <seqfan(a)ext.jussieu.fr> cc: Math-Fun mailing list <math-fun(a)CS.Arizona.EDU> Subject: primes in pi: 1, 2, 6, 38, 16208, 47577, 78073, ... Anyone curious about prime number searches may be (slightly) interested to learn that the concatenation of the first 78073 digits of pi (including the initial 3) forms a probable prime. For the complete list of known indices of primes in the digits of pi (as listed above), see http://www.research.att.com/~njas/sequences/A060421. I believe at 78073 digits, pi(78073) becomes the 65th largest known PRP (cf. http://www.primenumbers.net/prptop/prptop.php) Cheers, -Eric --------- I saw the buried cable problem in one of Gardner's puzzle columns, long long ago. I've always been curious if the solver is required to cleanup his patch cord. Rich

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[math-fun] Stupid physics question
by Henry Baker 18 Jul '06

18 Jul '06

According to Noether, energy conservation is equivalent to time-invariance & other conservation laws are equivalent to other symmetries. So here's the question: the laws of physics, as we currently understand them, imply that _information_ is conserved -- we can run physics movies backwards, so long as enough (i.e., _all_) detail is retained. This appears to be the case even in the face of black holes and quantum spookiness. So what is the corresponding symmetry?

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RE: [math-fun] Physics question
by Andy Latto 17 Jul '06

17 Jul '06

> -----Original Message----- > From: math-fun-bounces+andy.latto=pobox.com(a)mailman.xmission.com > [mailto:math-fun-bounces+andy.latto=pobox.com@mailman.xmission.com]On > Behalf Of Henry Baker > Sent: Sunday, July 16, 2006 9:23 PM > To: David Wilson > Cc: math-fun > Subject: Re: [math-fun] Physics question > > > There seems to be another analog to this system: semiconductors. > > The electrons act a little like a gas & bounce around, while the > places they leave are called "holes". However, they aren't identical: > electrons have higher mobility than holes, which is why NPN > transistors > are faster than PNP transistors. When I was very young, one of my favorite books was _Many_Moons_, by James Thurber. One of the things the Royal Astronomer (or possibly the Royal Mathematician) brags to the king that he has done is calculating the Speed of Darkness. This was just meant to be silly, but I was intrigued by it. If a room is brightly lit by a lightbulb, and you turn the bulb off, it gets dark, but not instantaneously. It will be some amount of time before the far corner of the room is dark, so there is a meaningful speed of darkness. At the time, I was proud of myself for figuring out that this speed would be exactly the speed of light. While that's true if you're only considering direct illumination, if you take into account the fact that some of the reason that the far corner is light is not through direct illumination, but via reflections off of the walls, the matter becomes more complicated. Andy Latto andy.latto(a)pobox.com >

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Re: [math-fun] Buried Cable problem
by Henry Baker 17 Jul '06

17 Jul '06

Doesn't Bell Labs have a patent on a solution to this problem or a close variant? --- If you have an unlimited amount of patch cord, why not simply bury that and lose the cable? Only one 10 mile walk is required, carrying or pushing the cable reel. On second thought, 10 miles of optical fiber doesn't weight very much, so bury that instead -- it has thousands of times the bandwidth of the 120 individual wires. At 06:55 AM 7/17/2006, Michael Kleber wrote: >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] Physics question
by David Wilson 17 Jul '06

17 Jul '06

Suppose there was a closed glass cylinder, with a thin planar membrane separating the cylinder into two chambers. The chambers contain colored water of two different colors, having the same density and no chemical reaction with one another. The water in one chamber is warmer than the water in the other chamber, but we do not know what color water is warmer. The cylinder is put into a zero-gravity environment, allowed to come to rest, and the membrane is removed, allowing the water allowed to mix. Can we tell by the flow patterns of the mixing water which color is warmer and which cooler?

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[math-fun] Calculating pi
by Jason Holt 16 Jul '06

16 Jul '06

A friend of mine recently asked me what pi is all about. This was rather remarkable, as he's failing out of high school, and people treat him like a failure. Turns out he's fascinated by physics and math. Anyway. So I described pi in terms of the experimental method of drawing circles and noticing that their circumference always seems a little more than 3 times their diameter. That led to a great discussion about experimental vs. theoretical science, and eventually he asked me how we went from experimental methods of calculating pi to computational approaches. I checked the wikipedia, but its approaches all seem to assume advanced math to prove that the approximations describe pi, and he hasn't gotten past algebra and geometry. At this point, the simplest approach I can think of is to write a program to choose random points in a circumscribed circle, then calculate the proportion of the points which fall within it. But that still has a heuristic element to it. Are there any plain arguments for the deterministic approximations? (For that matter, how about doing the same thing for the trig functions? I took the same approach for sin/cos/tan, talking about building a table of angle/ratio measurements, but I had no idea how to describe computational approaches.) -J

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Re: [math-fun] Physics question
by dasimov＠earthlink.net 16 Jul '06

16 Jul '06

David Wilson writers: << Suppose there was a closed glass cylinder, with a thin planar membrane separating the cylinder into two chambers. The chambers contain colored water of two different colors, having the same density and no chemical reaction with one another. The water in one chamber is warmer than the water in the other chamber, but we do not know what color water is warmer. The cylinder is put into a zero-gravity environment, allowed to come to rest, and the membrane is removed, allowing the water allowed to mix. Can we tell by the flow patterns of the mixing water which color is warmer and which cooler? >> The place where these assumptions don't quite correspond to reality -- in that the warmer and cooler waters have equal density -- leaves it unclear to me just exactly what ideal problem is to be answered here. Also I'm not sure how to interpret "come to rest", since having a temperature means there is molecular motion. --Dan

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[math-fun] Re: Diameter of Rubik's Cube
by Steve Witham 16 Jul '06

16 Jul '06

Since there's been such thoughtful silence on the Rubik's Cube problem, I thought I'd barge in with more amateur thoughts. First of all, I said that 6,576,625,523 (the square root of the number of Rubik's positions) x 10 bytes per position is about 66 terabytes, but it's only 66 gigabytes. But now it seems to me that the Rubik's cube is not a square-root meet- in-the-middle type of problem. That is, if you graph how many positions you can get to in a given number of moves, you get a sigmoid curve, reaching half (not the square root) of the possible positions in half the maximum number of moves. Thinking that your odds of going backwards goes up like the number of positions you've covered so far. Am I getting this wrong? or looking at something irrelevant? On the other hand, would it be useful to have just the raw numbers of positions at distances up until you ran out of disk space? Would it help to know just how many moves it takes to get to the square root? Would that number help focus the search for the actual diameter? How much effort and space can be saved by having a list of symmetries, and only storing canonical representatives of positions? Er, is that where the square root comes in? --Steve >From: "Schroeppel, Richard" <rschroe(a)sandia.gov> > >Has the diameter of the Rubik's cube puzzle been established yet? >We now have enough computing power to find & prove the distance- >to-start for any particular position, using the obvious square-root >meet-in-the-middle algorithm and a big disk. We can search for >edge positions by generating random positions and using the >exact distance algorithm to guide a hill-climb away from the >start position. But this is only heuristic. >From Steve Witham <sw(a)tiac.net> >http://en.wikipedia.org/wiki/Rubik%27s_Cube_group says that the number >of positions is > ( 8! 3^8 12! 2^12 ) / 12 = 43,252,003,274,489,856,200 >so the square root is about 6,576,625,523 > >Each position needs about 5 bytes, and I guess having two copies of the >set would make it go a lot faster, so that's 66 terabytes...by my >amateur guess. >From: Jason Holt <jason(a)lunkwill.org> > >If somebody is serious about coding up an algorithm, let's work up some >resource estimates and I can ask if we can get them for this important bit of >mathematical research. I work for a company that reportedly has servers >spread in at least 25 locations around the world.

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