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Re: [math-fun] Schroedinger's daughter
by David Gale 21 Jun '06

21 Jun '06

The following clarified for me both the case of Mary and Bartholomew. There are two ways to pick a girl from a population of two child families. 1. Pick a family at random. If it contains a girl the odds are two to one that she has a brother. 2. Pick a girl at random. Then the odds are even that she has a brother, (because there are (roughly) the same number of girls with brothers as there are girls with sisters). Or is this is the classical rather than the quantum case. dg PS. Are there really twice as many boy-girl families as girl-girl or boy-boy? It shouldn't be too hard to check. Maybe certain couples are biologically more likely to produce one sex over the other. At 03:30 PM 6/20/2006, you wrote: >OK, I thought that I understood this, but here's a variation: > >A friend writes a girl's name on a piece of paper and puts it (unopened) >in your pocket. A random parent of two children comes up to you and says, >"I have a daughter named Mary." At this point, the probability that >Mary's sibling is a girl is 1/3 (I think). However, if you look at the >piece of paper, and it says "Mary", the probability changes to 1/2 >(approximately). > >Is this correct? > >Bill C. >_______________________________________________ >math-fun mailing list >math-fun(a)mailman.xmission.com >http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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RE: [math-fun] Schroedinger's daughter
by Cordwell, William R 21 Jun '06

21 Jun '06

I thought about this some more. If a family has two daughters, one of whom is named Mary, there's only a 50-50 chance that the parent will say "I have a daughter named Mary", so the families with two girls have the same chance of saying this as the families with one boy and one girl. Thus, it stays 1/3, I think. Bill C. -----Original Message----- From: math-fun-bounces+cordwell=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+cordwell=sandia.gov@mailman.xmission.com]On Behalf Of David Gale Sent: Wednesday, June 21, 2006 7:27 AM To: math-fun Subject: Re: [math-fun] Schroedinger's daughter The following clarified for me both the case of Mary and Bartholomew. There are two ways to pick a girl from a population of two child families. 1. Pick a family at random. If it contains a girl the odds are two to one that she has a brother. 2. Pick a girl at random. Then the odds are even that she has a brother, (because there are (roughly) the same number of girls with brothers as there are girls with sisters). Or is this is the classical rather than the quantum case. dg PS. Are there really twice as many boy-girl families as girl-girl or boy-boy? It shouldn't be too hard to check. Maybe certain couples are biologically more likely to produce one sex over the other. At 03:30 PM 6/20/2006, you wrote: >OK, I thought that I understood this, but here's a variation: > >A friend writes a girl's name on a piece of paper and puts it (unopened) >in your pocket. A random parent of two children comes up to you and says, >"I have a daughter named Mary." At this point, the probability that >Mary's sibling is a girl is 1/3 (I think). However, if you look at the >piece of paper, and it says "Mary", the probability changes to 1/2 >(approximately). > >Is this correct? > >Bill C. >_______________________________________________ >math-fun mailing list >math-fun(a)mailman.xmission.com >http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun _______________________________________________ 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] x^n + y^n = z^n + w^n
by Christian Boyer 20 Jun '06

20 Jun '06

We know integer solutions of x^n + y^n = z^n + w^n, for n <= 4. For example, for n=4: http://www.research.att.com/~njas/sequences/A018786 But it seems that we do not know any nontrivial solution for n > 4. Am I right? Blair Kelly found no solution for x^5 + y^5 = z^5 + w^5 < N = 1.02 x 10^26 (Guy's UPINT, 3rd edition, p.210). Do you know some other results for n > 4? Christian.

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[math-fun] 3d funsters (and some others)
by Thane Plambeck 20 Jun '06

20 Jun '06

Taken by Bruce Oberg Richard Guy: http://www.flickr.com/photos/23046609@N00/171170984/in/set-7215759417170596… Neil Sloane: http://www.flickr.com/photos/23046609@N00/171171346/in/set-7215759417170596… David Singmaster: http://www.flickr.com/photos/23046609@N00/171174311/in/set-7215759417170596… -- Thane Plambeck http://www.plambeck.org/ehome.htm

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[math-fun] Chaitin's "Meta Math!"
by David Wilson 19 Jun '06

19 Jun '06

While pursuing my deli job, I sometimes have to wait at Borders bookstore for my wife to pick me up from work. I have managed to read Chaitin's "Meta Math!" and I found it to be rather intriguing.

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Re: [math-fun] x^n + y^n = z^n + w^n
by dasimov＠earthlink.net 19 Jun '06

19 Jun '06

Was I the only one who received Richard's e-mail with an attachment? (It's labeled Unknown 166.) I definitely will not open it. But I thought the mailman.xmission.com mailer filtered out attachments, no? --Dan

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[math-fun] Re: More ZipfHuffing
by Henry Baker 19 Jun '06

19 Jun '06

Hi Steve & David: I duplicated Steve's results with my own version of Huffman encoding, so I agree with his results 100%. David's suggestion is slightly inferior to pure Huffman. Huffman gives outstanding efficiency for these Zipf distributions -- e.g., 99.67% efficient for the first 16,384 elements of the series 1/i (i=1 16384). So it's a little hard to argue for the need for a more efficient encoding. The reason for this exceptional efficiency seems to be that Zipf is a very smooth distribution. Here its "long tail" actually helps, since any bad encoding of the most frequent 2 or 3 symbols won't affect the overall efficiency very much. However, even the simplest encoding (14 bits) is 71% efficient, and appears to be in the 70% range even for the first million elements of the harmonic series. Once again, the long tail helps, because the bulk of the tail is reasonbly efficiently encoded. An interesting connection between Huffman and Egyptian fractions is the inner step of Huffman: Sort the symbols into ascending order of probabilities. Take the symbols with the two smallest probabilities and "merge" them, thus adding their probabilities. Then bubble this probability up the list until it is sorted again. (This is obviously not the most efficient algorithm; I'm focussed on the probability numbers themselves.) We are obviously building up sums from Egyptian fractions (probabilities not normalized), with the topmost sum being H_n (nth harmonic number). At 12:17 AM 6/17/2006, Steve Witham wrote: >Henry, > >Thanks for an interesting question. >Lots of web pages mention Zipf and Huffman in the same sentence, >but no one seems to have asked just what you asked before. > >Dusted off and was forced to debug a Huffman-coding routine. >Here it is generating codes for various populations of thingies. >The digits indicate the number of bits to represent each thingy. >Line 16 matches the example I gave last night; my >hand-calculations were close enough. > >Line 15 seems perfect. It represents an ideal of what normally >(or at least sometimes) happens, but I don't think it ever happens >quite that way again. But, I'm not actually sure, more exploring >to do. > > --Steve > >1: 0 >2: 11 >3: 122 >4: 1233 >5: 12344 >6: 133344 >7: 1334444 >8: 22334444 >9: 223344455 >10: 2234444455 >11: 22344445555 >12: 233344445555 >13: 2333444555555 >14: 23344444555555 >15: 233444455555555 >16: 2334444555555566 >17: 23344445555556666 >18: 233444555555556666 >19: 2334445555555666666 >20: 23344455555566666666 >21: 233444555556666666666 >22: 2334455555556666666666 >23: 23344555555666666666666 >24: 234444555555666666666666 >25: 2344445555556666666666677 >26: 23444455555666666666666677 >27: 234444555556666666666667777 >28: 2344455555556666666666667777 >29: 23444555555566666666666777777 >30: 234445555556666666666666777777 >31: 2344455555566666666666677777777 >32: 23444555555666666666667777777777 >33: 234445555556666666666777777777777 >34: 2344455555666666666666777777777777 >35: 23444555556666666666677777777777777 >36: 234445555566666666667777777777777777 >37: 2344455555666666666777777777777777777 >38: 23444555566666666666777777777777777777 >39: 234445555666666666677777777777777777777 >40: 2344555555666666666677777777777777777777 >41: 23445555556666666666777777777777777777788 >42: 234455555566666666677777777777777777777788 >43: 2344555555666666666777777777777777777778888 >44: 23445555566666666666777777777777777777778888 >45: 234455555666666666667777777777777777777888888 >46: 2344555556666666666777777777777777777777888888 >47: 23445555566666666667777777777777777777788888888 >48: 234455555666666666677777777777777777778888888888 >49: 2344555556666666666777777777777777777888888888888 >50: 23445555566666666677777777777777777777888888888888 >51: 234455555666666666777777777777777777788888888888888 >52: 2344555556666666667777777777777777778888888888888888 >53: 23445555566666666677777777777777777888888888888888888 >54: 234455555666666667777777777777777777888888888888888888 >55: 2344555556666666677777777777777777788888888888888888888 >56: 23445555666666666677777777777777777788888888888888888888 >57: 234455556666666666777777777777777778888888888888888888888 >58: 2344555566666666677777777777777777778888888888888888888888 >59: 23445555666666666777777777777777777888888888888888888888888 >60: 234455556666666667777777777777777788888888888888888888888888 >61: 2344555566666666677777777777777778888888888888888888888888888 >62: 23445555666666667777777777777777778888888888888888888888888888 >63: 234455556666666677777777777777777888888888888888888888888888888 >64: 2344555566666666777777777777777788888888888888888888888888888888 >65: 23445555666666667777777777777777888888888888888888888888888888899 >66: 234455556666666677777777777777788888888888888888888888888888888899 >67: 2344555566666666777777777777777888888888888888888888888888888889999 >68: 23445555666666677777777777777777888888888888888888888888888888889999 >69: 234455556666666777777777777777778888888888888888888888888888888999999 >70: 2344555566666667777777777777777888888888888888888888888888888888999999 >71: 23445555666666677777777777777778888888888888888888888888888888899999999

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Re: [math-fun] x^n + y^n = z^n + w^n
by MCKAY＠vax2.concordia.ca 19 Jun '06

19 Jun '06

Was the diophantine x^4+y^4=z^4+w^4 not the subject ofd a paper by the (young) Dyer in ? J/ Lond Math Soc in the aerly 50's while still at school? van der Pol gave a pretty combinatorial proof of the Jacobi 4-ic identity for null thetas. It extends to full thetas, yielding the above identity. Whether this can be used for the diophantine equation, I doubt. John McKay .

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Re: [math-fun] Zipf, Harmonic series & Egyptian fractions
by Henry Baker 16 Jun '06

16 Jun '06

Correct. There's an interesting paper that draws an analogy between prime numbers and software libraries, at least as far as their statistics are concerned: "Software Libraries & Their Reuse" by Todd Veldhuizen (use Google search to find). His observations: 1. Infinitely many primes & infinitely many SW components. 2. nth prime is factor of ~1/(n ln(n)) of the integers. nth most frequently used library component is ~1/(n log(n) log+(n)). 3. Erdos-Kac predicts # factors tends to normal distribution; similar measurements indicate the same for SW components. 4. Prime Number theorem nth prime is ~log(n ln(n)) bits long; approx. ditto for SW components. Do primes have an approx. Zipf-type distribution? Alternately, should Zipf's Law be reformulated to look more like the prime distribution? -- i.e., perhaps word frequency data would match prime distribution data better than the harmonic distribution? At 07:19 PM 6/15/2006, dasimov(a)earthlink.net wrote: ><< >Zipf's Law says that the probability of the n'th most popular thingy >is proportional to 1/n. Clearly this probability exists only if the >universe of thingy's is finite. >>> > >Actually Zipf's law applies to the commest *words* used, and assuming we're >looking at N of them gives the k'th one a probability of > > (1/k) / (sum{j=1 to N} (1/j)), > >so it's normalized to 1. > >People have looked at a generalized Zipf's law, which allows a fixed exponent >where you replace all the 1/C 's with 1/C^t 's -- which gives a good fit for things >other than words with s <> 1. > >--Dan

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[math-fun] Forgot to say: Zipf, etc.
by Steve Witham 16 Jun '06

16 Jun '06

Henry, Forgot to say that, once you split a Zipf distribution into a nice 50/50 split, now you have two groups of probabilities that aren't Zipf distributions, but you still have to split them, too! That makes a method specialized for Zipf distributions seem unlikely to me. If you were to take, say, as many of the least-likely symbols as got closest to 1/2 total probability, you'd have two Zipf-*like* distributions, and so you could do that recursively, but I don't think it would give you a code as efficient as Huffman. (I can't see how that could be better than the Shannon- Fano top-down method, which can be less efficient than Huffman's bottom-up method.) The thing about Huffman codes of sorted lists (or monotonic functions) is that, once the Huffman tree is done, ranges of thingies have the same number of bits, so you can throw away the tree and code using a table something like this (for the example where 1 <= n <= 16 is the number of the thingy with probability 1/n): 1 <= n <= 1 --> .00 + ( n - 1 ) x .01 (binary) 2 <= n <= 3 --> .010 + ( n - 2 ) x .001 4 <= n <= 7 --> .1000 + ( n - 4 ) x .0001 8 <= n <= 14 --> .11000 + ( n - 8 ) x .00001 15 <= n <= 16 --> .111110 + ( n - 15 ) x .000001 (I calculated this table by hand and it may be wrong, but the correct table definitely has that form.) When I looked up the history of the Huffman vs. Shannon-Fano codes, I found http://en.wikipedia.org/wiki/Huffman_coding which talks about the history and almost everything I mentioned, including arithmetic codes and patents, except this way of representing the code in a log-sized table. --Steve

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