June 2012 - mathfun@mailman.xmission.com

[math-fun] Incircle center location
by Henry Baker 26 Jun '12

26 Jun '12

I think I finally came up with a simple formula in complex numbers for the location of the incenter of a triangle in the complex plane: If a,b,c are the complex numbers which represent the _vectors_ of the triangle sides (a+b+c=0), if R is the (real) circumradius (R=|a||b||c|/(4 Area)), if r is the (real) inradius, if O is the (complex) circumcenter, and if I is the (complex) incenter, then I = O + d (d is a complex number which is the vector from O to I) = O + R*(d/R) = O + R*(a/|a|+b/|b|+c/|c|)/i (i = sqrt(-1)) = O - i*R*(a/|a|+b/|b|+c/|c|) = O - i*R*(abar + bbar + cbar) (abar, bbar, cbar are unit vectors along a,b,c) So we add up the _unit vectors_ along a,b,c, multiply by the circumradius & take a right turn. Note also that d*d' = |d|^2 = R^2 - 2*r*R (Euler's triangle theorem.) Proving Euler's theorem in this fashion required some very deep conversations with Maxima to deal with intermediate expression "swell"; if you're not careful, some of the intermediate expressions can grow to thousands of terms, with no hope of factoring on a laptop.

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[math-fun] recording sought....insert your favorite zero math content warning here
by Thane Plambeck 26 Jun '12

26 Jun '12

Some time ago I had a link to an audio recording made by a math professor at a US (or possibly Canadian) university. He had a Russian accent as I recall. In the recording, he says things like "I love your Calculus problem set....Your homework was excellent....Your algebra solution set was wonderful...." The recording goes on for perhaps one minute or so. He made the recording after receiving student complaints that he offered no "positive feedback" in his instruction, and told class members to simply listen to the recording whenever they felt the need to do so. I'd like to find the recording and the person again. Googling hasn't turned up much for me. Does anyone remember this? -- Thane Plambeck tplambeck(a)gmail.com http://counterwave.com/

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Re: [math-fun] The final Google Turing puzzle: what is 11010110110101101011011...
by Ray Tayek 26 Jun '12

26 Jun '12

At 12:22 PM 6/23/2012, you wrote: >... > > On Fri, Jun 22, 2012 at 9:40 PM, Robert Munafo <mrob27(a)gmail.com> wrote: > >> By step 49720, it has written: > >> 11010110110101101011 ... > > > > It looks remarkably like this sequence that grows like > Fibonacci's rabbits: ... http://blog.jgc.org/2012/06/what-final-turing-machine-google-doodle.html --- co-chair http://ocjug.org/

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Re: [math-fun] Vector puzzle
by Warren Smith 25 Jun '12

25 Jun '12

I don't think any inductive proof of Kleber's ilk is going to crack this ultra-annoying Asimov vector puzzle, because in the n=3 example with 8 vectors I gave, the minimum subset with 0-sum has cardinality=5, which note is greater than 2^(n-1). Ditto for the 4 vector example with n=2 and min subset cardinality=3. You can permute the columns and change the sign of any column... a fact which seems to have no use. There may be a solution via horrible OVERKILL: Rephrase it as an 0-1 integer programming problem, then show the linear program automatically has an 0-1 integer solution. (This would be equivalent to Asimov's desires.) It is however probably more interesting to go in the other direction: Use whatever Asimov's proof technique is, to discover a new class of linear programs which automatically have 0-1 integer solutions.

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Re: [math-fun] Vector puzzle
by Dan Asimov 25 Jun '12

25 Jun '12

Since no one has solved the cute vector puzzle I heard, I'll repeat it here: << For some integer n > 0 you have all 2^n vectors of dimension n whose entries are +1 or -1. Of course the sum of all 2^n of these vectors is the 0 vector. Then your three-year-old child changes some of the entries of some of these vectors to 0. Show that there is still a nonempty subset of the new set of 2^n vectors that sums to the 0 vector. >> Since there's been a lot of traffic on math-fun lately, maybe it's too early to post the solution, but I will soon. --Dan ________________________________________________________________________________________ It goes without saying that .

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[math-fun] Interesting dimensional problem
by Henry Baker 25 Jun '12

25 Jun '12

The FIRST Robotics competition this year involved a basketball shooting robot. However, for various reasons, the basketball chosen for the competition was a "size 4" (grade-school size) basketball rather than a "size 7" (college/NBA) size basketball. The hoop/rim, however, was a standard 18" inner diameter rim with the center 15" from the backboard. I'm interested in what court dimensions would make this size 4 ball & standard size rim "equivalent" in difficulty with a standard size 7 ball and standard size rim. Here are the standard court dimensions: 15' from the free throw line to the projection of the backboard onto the floor; the top of the rim is exactly 10' from the floor. The standard size 7 basketball is 29.5-30" in circumference & weighs 22oz. --- The size 4 basketball has a circumference of 25" and a weight of 11.2oz. --- In order to make shooting a standard basket with a smaller/lighter ball "equivalent" in difficulty, how high above the floor should the rim be, and how far away from the free-throw line should the backboard be located? The tallest part of the robot may not exceed 84", so we can assume that the robots are shooting from approximately the same height as a tall adult human. --- My intuition is that air resistance may be negligible for this problem, but perhaps not.

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[math-fun] Coin-tossing
by Guy Haworth 25 Jun '12

25 Jun '12

Maybe I've misunderstood the question, but I'm wondering if JP is asking for something that cannot exist. Suppose a 'coin', simulated for example by a random-number generator on a computer, has a probability 'p' of coming up Heads. Then the probability of getting 'X' Heads out of 'n' is defined by a binomial formula involving 'p', 'X' and 'n'. However, if I observe 'X' heads occurring in 'n' tosses of the coin, I have a different piece of information - in fact less information. Even if X/n = p, I don't know that p is the probability of coming up Heads. So, in terms of knowledge, probability and statistics, I'm 'in a different place'. The case of getting HH on two coin tosses (with probability p^2) shows this. The case of getting H on one coin toss shows this too. If I could derive a binomial formula for the number of Heads in n' tosses of the coin, it would be equivalent to knowing what the probability of Heads is on each toss. But I don't know this, so I can't derive a binomial formula from observing a sequence of coin-tosses. Guy

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[math-fun] Life probability of halting/settling
by Warren Smith 24 Jun '12

24 Jun '12

> I'd like to compute the first bits of the "halting probability" of > Conway's life; since CAs don't halt, I'd like to compute the > probability that it evolves to a state that's easily detectable by > Golly, like all empty or a still life or cyclic. (Also, I know the > number is uncomputable, but that doesn't preclude a finite computable > prefix.) Since any translation or reflection of a pattern has the > same outcome as the original, I'd like to choose a measure on the set > of cells such that computing the measure of all the translations and > reflections of a pattern is easy. > Any suggestions? > > --but isn't this probability zero? >I forgot to say that I'm considering only finitely many "on" cells in the initial pattern (and therefore in its descendants). For example, number the cells starting from some arbitrary origin outwards in a spiral, let s be some computable real number greater than 1, and let the measure of a pattern p be 2^{-s|p|} where |p| is the largest index of an "on" cell in the initial pattern. Since there are nonempty finite patterns that evolve into empty ones, the measure is nonzero. --oh. Well, in that case, your answer will depend on your choice of s, and hence be kind of arbitrary and ugly and pointless to compute. (?) --- But anyhow, generate a random uniform (from your measure) start-pattern, run golly, and use Floyd cycle finding algorithm to see if it has settled into a cycle (a), and use special code to detect whether some glider or spaceship has "escaped" (b). Prob(a) is a lower bound on your probability Pstay, Prob(b) is a lower bound on 1-Pstay. So you can compute confidence intervals for both lower & upper bounds. Or: can you? In order to compute Prob(b) you need to KNOW whether some glider or spaceship has escaped. How do you know that? If we knew that nothing could travel diagonally faster than a glider, or knew nothing could travel horizontally faster than a spaceship, then maybe we could prove escape. However, I am unaware of any such proof. I.e. it might not have really escaped -- something faster will come along later to catch and kill it. In view of this, I do not see a good way to prove good upper bounds, I only see how to prove lower bounds? Well... if you can detect a situation with escaped gliders and the rest is periodic, then for THAT we can prove escape, thus getting upper bounds.

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[math-fun] Propp's coin tossing puzzle
by Warren Smith 24 Jun '12

24 Jun '12

>Jim Propp: Let me phrase the puzzle in a more (math-)fun way. A friend of yours has a biased coin, but she won't tell you what the bias p is ("Hey, I'm not THAT good a friend!"). She tosses the coin n times and tells you how many times it came up heads (call it X), and then she gives you a positive integer n'<n. Your job is to use n, n', and X, along with some supplemental randomness, to generate a number X' that is statistically indistinguishable from the (random) number of heads you would get if you tossed your friend's coin n' times. Note that you do not have access to your friend's coin; that is to say, you don't know its bias. Nonetheless, you can effectively simulate n' tosses of your friend's coin if she does you the favor of tossing it n times, where n'<n, and telling you how many heads she observed. Note that X' need not be independent of X. All we require is that X' is governed by the binomial distribution Binomial(p,n') provided that X is governed by the binomial distribution Binomial(p,n). I stress that the point of the puzzle is to find a way to do this that doesn't require knowledge of p. By the way, what's the quickest way to see that you CAN'T accomplish the above if n' is bigger than n? --re the final sentence, if n=1 it seems clear you cannot simulate n'=1000 tosses just based on knowing whether that 1 toss came up heads. You need continuum information (value of p) to do so well in the limit of large n', but you have only 1 bit of information (heads/tails). --re the simulation problem when n'<n, proceed thus. Regard the n old tosses (of which X are heads) as a string of n bits, X of them 1-bits. Randomly permute the bits. Now, as your output, generate the number of 1-bits among the first n' bits in your permuted bitstring. It seems to me that if X is a genuine sample from the binomial(n,p) distribution, then your output must be a genuine sample from the binomial(n',p) distribution. (By the way, I never looked at Propp's original puzzle since I had no idea what a "Galton board" was. I figured it was something to do with genetics. But googling tells me that guess was totally wrong.)

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Re: [math-fun] Propp's coin tossing puzzle
by Dan Asimov 24 Jun '12

24 Jun '12

Jim, what is your definition of "statistically indistinguishable" ? --Dan ________________________________________________________________________________________ It goes without saying that .

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