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[math-fun] combinatori(c)al
by Joerg Arndt 29 Nov '02

29 Nov '02

Which of the spellings is correct/usual, the one with or without the 'c'? -- p=2^q-1 prime <== q>2, cosh(2^(q-2)*log(2+sqrt(3)))%p=0 Life is hard and then you die.

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Re: [math-fun] Puzzle: Find limiting angle as dimension -> oo
by Mike Stay 29 Nov '02

29 Nov '02

asimovd(a)aol.com wrote: > << A partridge in a pear tree? > > I have to cry fowl. The puzzle was posted before Xgiving, so would not have a > Xmas reference. > > << > Let v_n be the vector in R^(n-1) defined as (1/n, 2/n,...,(n-1)/n). > Let w_n be the vector in R^(n-1) with the same coordinates in reverse order. > Find the limit as n -> oo of the angle between v_n and w_n. > (Try this without using numerical approximation or summation-of-powers > formulae.) > >> I was referring to the number of gifts in the song, which matched the dot product of the two vectors for n=12 (well, to within a power of 2): 1*12 partridges 2*11 turtle doves 3*10 french hens 4*9 calling birds 5*8 gold rings 6*7 geese 7*6 swans 8*5 maids 9*4 ladies 10*3 lords 11*2 pipers 12*1 drummers -- Mike Stay staym(a)datawest.net

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Re: [math-fun] Puzzle: Find limiting angle as dimension -> oo
by asimovd＠aol.com 29 Nov '02

29 Nov '02

<< A partridge in a pear tree? I have to cry fowl. The puzzle was posted before Xgiving, so would not have a Xmas reference. Incidentally, when I wrote << Let v_n be the vector in R^(n-1) defined as (1/n, 2/n,...,(n-1)/n). Let w_n be the vector in R^(n-1) with the same coordinates in reverse order. Find the limit as n -> oo of the angle between v_n and w_n. (Try this without using numerical approximation or summation-of-powers formulae.) >> I didn't mean to suggest the puzzle is solvable by guesswork. --Dan

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Re: [math-fun] Puzzle: Find limiting angle as dimension -> oo
by asimovd＠aol.com 28 Nov '02

28 Nov '02

<< I'll guess the limit is 90 degrees on the grounds that random vectors in high dimensional spaces tend to be orthogonal. >> Maybe so, but these aren't random! Regards, Dan

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[math-fun] Re: Dominoes
by asimovd＠aol.com 28 Nov '02

28 Nov '02

Clearly any portion of a checkered board that can be covered by dominoes must have an equal number of squares of each color, but does the converse hold? (Does identifiying opposite edges of the board to make a torus affect the answer?) AND: Would someone please recall for me a problem that was posted here several months ago, but never answered, regarding properties of some kind of balanced two-color designs on a 4x4 square. --Dan

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[math-fun] (no subject)
by Richard Guy 27 Nov '02

27 Nov '02

Calculate the natural log of (95536/e)^{95536}.

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[math-fun] Re: [math-fun] (no subject)
by guy.haworth 27 Nov '02

27 Nov '02

No, it's a $1000000 and .02972 cents question - but nice try! Btw, what fun maths books do I buy the ex Maths Graduate and now interested maths hobbyist for Xmas? What's that compendium of M.Gardner writings called? Apols: should have noted it at the time. g Accédez au courrier électronique de La Poste : www.laposte.net ; 3615 LAPOSTENET (0,13 /mn) ; tél : 08 92 68 13 50 (0,34/mn)"

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[math-fun] Question about XOR
by Helger Lipmaa 27 Nov '02

27 Nov '02

I am sure the next problem has been solved a long time ago, but it is hard to find any references. Does anybody know a solution or references? (We came up with a fast algorithm for it ourselves, but inventing wheel is not always the best thing.) Problem: Given nonnegative integers A, B, C. Find the cardinality of {(X,Y):0 <=X < A,0 <= Y < B, 0 <= (X xor Y) < C} where XOR denotes the bitwise XOR of X and Y. Helger

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[math-fun] simultaneous rational approximation
by Mike Stay 26 Nov '02

26 Nov '02

I've been looking at simultaneous rational approximation with respect to modular arithmetic. One can 'factor' any number mod n into two numbers whose logs sum to log n easily using a continued fraction: Let m be the number we're trying to 'factor', and n be the modulus. We take cf(m/n) and find a convergent p/q. Multiply m by q, reduce mod n, and you get a number p' that is (lg m - lg q) bits long. Then m = p'/q mod n. For example, let n=629287. I want to 'factor' 529287. The continued fraction for 529287/629287 = [0 1 5 3 2 2 2 2 1 3 3 1 4 1 7]. The denominators of the convergents are 1,1,6,19,44,107,258,623,... Taking 258 as my denominator we expect that p' will be (6-3)=3 digits long. 529287*258 = 767 (Yep!), so one can "factor" 529287 as 767/258 mod 629287. What if we use bcf's for doing the approximation? Let k*m be the number we're trying to factor. Using bcf's, we take bcf (k/n, m/n) and find convergents p/r, q/r. Multiply k,m by r, reduce by n, and you get p',q' that are smaller than k,m. k, m = p'/r, q'/r mod n. Experimentally, lg p',q' ~= (lg k,m - .5 lg r), so we don't do any better than with simple continued fractions. Has anyone proven that one can't do better with any method of approximation? -- Mike Stay staym(a)datawest.net

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[math-fun] Fredkin's Theory of Everything
by Eugene Salamin 25 Nov '02

25 Nov '02

About a month ago, someone posted a message about a talk Fredkin was to give concerning an updated version of his CA model of physics. It was touted as compatible with relativity and quantum mechanics, and explained the various forces of physics. Has anybody attended that talk, or can somebody report back as to just what Fredkin's theory is? __________________________________________________ Do you Yahoo!? Yahoo! Mail Plus Powerful. Affordable. Sign up now. http://mailplus.yahoo.com

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