April 2015 - mathfun@mailman.xmission.com

Re: [math-fun] Aristarchus of Samos, 300 BC
by Warren D Smith 19 Apr '15

19 Apr '15

Yes, RWG has a very good point that Aristarchus could have used Moon-Earth-Sun triangles of other angles than right by using other crescents than "half full." PUZZLE: And what angle would have been the best, in the sense of returning the most accuracy despite measurement errors? M | alpha | | | theta E-----------------S Another brilliant measurement was by Ole Roemer who in 1676 deduced the speed of light by observing Jupiter's moons. You need a telescope or at least good binoculars for this, plus excellent clocks, so the ancient Greeks could not have done it. A sketch of how he presumably did it is here: http://www.amnh.org/education/resources/rfl/web/essaybooks/cosmic/p_roemer.… Roemer actually never published his method because the great experts all told him he was an idiot, and him doing this was some kind of insult or something. But I think it was really fantastic, I very much doubt I ever would have thought of his technique.

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[math-fun] Aristarchus of Samos, 300 BC
by Warren D Smith 19 Apr '15

19 Apr '15

> I don't get it. How do you get any indication of the Sun's size > relative to the moon? --Aristarchus's input data were the following (from which you should be able to see how he did it). 1. He knew the size of the Earth based on two latitude measurements at two places (Alexandria & Syene) a known distance apart along a Meridian. 2. When the moon is exactly half full, the angle between sun & moon is 87 degrees. (Except that really, it is 89.85 degrees. A difficult measurement to make, since judging "half full" by eye.) 3. During a total eclipse, moon & sun are essentially the same angular diameter. (True.) 4. The Earth's shadow at the location of the moon is about twice as wide as the moon. (Actually, 3.67.) 5. The moon subtends an angle of 2 degrees. (Actually, 0.52 degrees, it is hard to understand how Aristarchus could have been so far off on this. Archimedes gave 0.5 degrees.) I don't think Aristarchus had good measuring equipment, he could have been a lot more accurate.

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[math-fun] Aristarchus of Samos, 300 BC
by Warren D Smith 19 Apr '15

19 Apr '15

I had known that Aristarchus computed a good estimate of the size of the Earth using trigonometry. But I had not known that he also computed the sizes of, and distances to, the Moon & Sun. However, he got those numerically quite wrong, due to bad input data (his mathematics was ok). Nevertheless, he did correctly realize that the Sun was much bigger than the Earth, which led him to suggest the "Copernican" idea that the Sun was central with Earth rotating around it, rather than the reverse. Wikipedia says JW Draper contends he then ordered the planets correctly by distance to the central Sun. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] packing 17 points into 9X9 array of squares cells
by Neil Sloane 18 Apr '15

18 Apr '15

To Math Fun: I have been going through old copies of the Popular Computing (Calabasas, CA) newsletter. Problem 178 asks for max number of pieces that can be placed on an nXn chess-board so that no two are closer than a knight's-move apart. See https://oeis.org/a233735_1.pdf for the problem. This can be re-stated as an error-correcting coding question: find max number of cells so that you cannot go from one to another by crossing two edges. Back in 2003 I seem to have thought that the number for a 9x9 board was 17, as in A085577. But now I can only get 16, which would mean that A085577 is wrong and the true answer is A233735, which even has a remarkably simple explicit formula. Comments, anyone? Incidentally Bill Gosper gets a nice mention on the same page where the problem is stated (see scan mentioned above). For other nice problems, mostly unsolved, from Pop. Computing and other old magazines, go to https://oeis.org/A256969 and work backwards through the A-numbers. There is lots of fun to be had here... Neil

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[math-fun] Question about Paris-Harrington Theorem
by David Wilson 18 Apr '15

18 Apr '15

Wikipedia states that The Paris-Harrington theorem is true, but unprovable, in Peano arithmetic. The article goes on to intimate that PH can be proved in second-order arithmetic (the argument seems to be that second-order arithmetic can be used to prove PH from the infinite Ramsey theorem. A proof is given for the latter, which I assume uses Peano Arithmetic, am I right?) At any rate, I would think that if PH were true but unprovable in PA, then either PH or ¬PH would be consistent with PA. This would mean that we have two flavors of arithmetic: PA + PH, which is consistent with second-order arithmetic, and PA + ¬PH, which is not. So when the text says that PH is true but unprovable in PA, Im hearing true in second-order arithmetic but unprovable in PA, with that tacit understanding that arithmetic subsumes second-order arithmetic? Does this simply reveal a bias toward second-order arithmetic over other flavors of arithmetic such as PA + ¬PH? Or is my argument above wrong? Is it somehow possible that PA + ¬PH is contradictory, yet PH is unprovable in PA?

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[math-fun] What's the proof?
by David Wilson 17 Apr '15

17 Apr '15

x*cos(x/2)*cos(x/4)*cos(x/8)*. = sin(x).

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[math-fun] Mensa Correctional Facility, reply to Fred Lunnon
by Warren D Smith 17 Apr '15

17 Apr '15

> From: Fred Lunnon <fred.lunnon(a)gmail.com> > I cannot understand how this idea is supposed to work. > Suppose we have 16 coins, heads = 1, tails = 0, > numbered 0..15, and the prisoners agree coins 0..6 > to communicate the Hamming(7, 4) code as shown at > http://en.wikipedia.org/wiki/Hamming%287,4%29 > The warder selects coin 0 , then turns all coins heads. > > How is prisoner A to convert this 1111111 to a message > (correctable to) 0000000, using only a single turn? > > WFL --no, prisoner A merely converts it to a message WITHIN THE 000..00 COSET of the hamming code, i.e. due to the fact that 0000=0000 in your example, here just within the hamming code itself. In your example, prisoner A actually can just do nothing, I think. My idea is that prisoner A's message to B is encoded in the answer to "which coset of the Hamming code is the (after modification by A) bitstring in?" I believe Goucher's solution of the problem is equivalent to mine, by the way, but taking advantage of the so-called "syndrome" coset-identifier elegant trick which only works for Hamming codes in a specific format, but does not work for general error correcting codes. My general idea will solve this whole class of problems provided you have a suitable error correcting code which can tile the full space of codewords via disjoint translates (aka cosets). (To solve my "*extended* mensan problem" is, however, more involved, or anyhow you in general would need to know more about the coset "graph structure." It might be that, for this extended problem, merely looking up codes in tables of good codes like a moron, is not going to be good enough.) -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Mensa Correctional Facility + Steganography
by Warren D Smith 16 Apr '15

16 Apr '15

I'd asked whether my solution involving cosets of error correcting codes, had any use. Here is a use, sort of. Knuth in some exercise asked "if in a table of random numbers, there is a misprint -- does that matter, i.e. can it stop the numbers from being random?" He then presented a "proof" that the answer is "yes." Well, the Mensan problem is another such, rather more direct, "proof." This also yields a way, I suppose, to do "steganography." (And the bitstring we begin with can be either random or not, steganography still works.) If you wanted to make a harder version of the Mensan problem, you could make one that depends on hairier error correcting codes (and knowledge of their properties) than the Hamming codes. Two example good versions: A. suppose the Warden had 23 coins (12+11, 2-row array), and prisoner #1 had to flip at most 3 coins to deliver an 11-bit message to prisoner #2. B. suppose the Warden had 15 coins (7 and 8 in a 2-row array), and prisoner #1 had to flip at most 2 coins to deliver a 7-bit message to prisoner #2. For an even harder version of the problem, which also seems perhaps even more relevant to steganography... change the rules as follows: ENHANCED_MENSAN_PROBLEM(N,J,K,L): 1. prisoners #1 & #2 agree on a protocol. 2. prisoner #2 leaves. 3. Warden chooses N-bit string. 4. prisoner #1 flips up to J bits. 5. prisoner #1 leaves. 6. Warden flips up to K bits in attempt to ruin it. 7. prisoner #2 re-enters, observes pattern, and deduces message from prisoner #1, which could have been up to L possible messages. Anyone see how to solve that one (& good parameter sets N,J,K,L to suggest)? This is sort of an adversarial steganography model.

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Re: [math-fun] Mensa Correctional Facility
by Warren D Smith 16 Apr '15

16 Apr '15

If there is a binary error correcting code on N=2^J bits such that (1) exactly N cosets of it constitute the whole set of 2^N possible bitstrings (2) any N-bit word is distance <=1 away from a codeword then you can solve the problem by changing a bit to move it into coset X, thus communicating the value of X to the other prisoner. The questioners asked about J=4 and J=6 and a single-error-correcting Hamming code should do the job. Actually I do better. If N= 2^J - 1 (for example N=127, J=7), then Hamming codes have cardinality=2^(N-J-1) hence we can by my method provide J+1 bits of information to the other prisoner by flipping only at most one coin. But only J bits of info [in fact only log2(2^J - 1) bits] actually are needed. Is there an actual use for this (not just in a silly puzzle)?

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[math-fun] Mensa Correctional Facility
by Veit Elser 16 Apr '15

16 Apr '15

There’s a growing class of puzzles I call Mensa Correctional Facility puzzles. In these puzzles there is always a warden who challenges prisoners with math/logic puzzles and grants them freedom when they succeed in solving them. Does anyone know the origins of the following one: In this puzzle the warden challenges a pair of Mensa inmates. The first inmate is shown to the warden’s room and the warden proceeds to place identical coins, each one either heads up or tails up, onto the 64 squares of a checkerboard. He then points to one of the coins and declares it to be the “freedom coin”. The inmate watching this is then invited to flip one of the coins to help his partner identify the freedom coin. Then, without allowing the inmates to communicate with each other, the second inmate is lead into the room, shown the coins, and after a little head scratching points to the freedom coin — he is, after all, a Mensa inmate — and the prisoners are set free. How did they do it? -Veit

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