November 2002 - mathfun@mailman.xmission.com

[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

4 4

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

27 Nov '02

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

3 2

[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)"

2 1

[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

2 1

[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

1 0

[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

1 0

[math-fun] Bernoulli totient generalization
by R. William Gosper 23 Nov '02

23 Nov '02

This may be old news disguised by notational ignorance, but heterologous to sum over divisors, ==== \ > f(d), / ==== d | n is sum over totatives, ==== \ S(n) := > f(t) . / ==== t < n (t, n) = 1 When f is a polynomial, we can express S(n) in terms of an umbral operator on Bernoulli polynomials and two multiplicative functions: Euler's totient a a-1 phi(p ) = p (p - 1), OEIS A000010, and the unnamed(?) a k U (p ) := 1 - p , OEIS <see below>. k n - 1 ==== B (n) - B ==== \ k k + 1 k + 1 \ k k s (n) := > j = ------------------ S (n) := > t k / k + 1 k / ==== ==== j = 0 t < n (t, n) = 1 2 2 n n phi(n ) 1 -- - - ------- 2 2 2 3 2 3 n U (n) n n n phi(n ) 1 2 -- - -- + - ------- + ------- 3 2 6 3 6 2 4 3 2 4 n U (n) n n n phi(n ) 1 3 -- - -- + -- ------- + -------- 4 2 4 4 4 3 5 4 3 5 n U (n) n U (n) n n n n phi(n ) 3 1 4 -- - -- + -- - -- ------- - ------- + -------- 5 2 3 30 5 30 3 2 4 6 5 4 2 6 n U (n) 5 n U (n) n n 5 n n phi(n ) 3 1 5 -- - -- + ---- - -- ------- - -------- + ---------- 6 2 12 12 6 12 12 3 5 7 6 5 3 7 n U (n) n U (n) n U (n) n n n n n phi(n ) 5 3 1 6 -- - -- + -- - -- + -- ------- + ------- - -------- + -------- 7 2 2 6 42 7 42 6 2 2 4 6 8 7 6 4 2 8 n U (n) 7 n U (n) 7 n U (n) n n 7 n 7 n n phi(n ) 5 3 1 7 -- - -- + ---- - ---- + -- ------- + -------- - ---------- + ---------- 8 2 12 24 12 8 12 24 12 etc. All this needs proving, of course. S_1(n) is OEIS A023896. S_2(n), S_3(n), and S_4(n) are A053818 through A053820. Unsigned U_1(n) and U_3(n) are A023900 and A063453. --rwg

1 1

[math-fun] little trinklets, twice a handfull
by wouter meeussen 23 Nov '02

23 Nov '02

all have equal Absolute values for z=(-1)^(1/15), but they are devoid of any (known) symmetry. the first set of ten: z+z^5+z^6+z^10+z^12+z^14+z^17+z^22, z+z^4+z^9+z^10+z^13+z^15+z^17+z^22, z+z^3+z^5+z^9+z^10+z^14+z^15+z^17+z^22, z+z^4+z^6+z^10+z^12+z^13+z^17+z^18+z^22, z+z^2+z^6+z^7+z^9+z^11+z^18+z^23, z+z^2+z^4+z^6+z^10+z^11+z^13+z^18+z^23, z+z^2+z^7+z^9+z^10+z^12+z^14+z^19+z^23, z+z^2+z^3+z^6+z^8+z^10+z^14+z^19+z^24, z+z^3+z^6+z^10+z^11+z^12+z^14+z^19+z^24, z+z^5+z^6+z^7+z^10+z^13+z^15+z^19+z^24 the second set: z+z^3+z^6+z^8+z^9+z^13+z^14+z^17, z+z^2+z^4+z^6+z^7+z^10+z^13+z^18, z+z^2+z^4+z^6+z^7+z^11+z^12+z^15+z^18, z+z^3+z^5+z^6+z^8+z^9+z^14+z^19, z+z^2+z^5+z^7+z^8+z^10+z^13+z^15+z^19, z+z^3+z^6+z^7+z^9+z^12+z^14+z^15+z^19, z+z^2+z^4+z^6+z^7+z^8+z^12+z^15+z^20, z+z^4+z^6+z^7+z^8+z^11+z^12+z^17+z^20, z+z^5+z^6+z^8+z^9+z^11+z^12+z^17+z^22, z+z^5+z^6+z^7+z^9+z^10+z^12+z^17+z^23 sets of bicolored glass beads were once fashionable as gifts (?) Wouter Meeussen wouter.meeussen(a)pandora.be

1 0

Re: [math-fun] Multiply-gendered humanitarians
by Henry Baker 22 Nov '02

22 Nov '02

Heck, you could have at least shown how to do Bingo in N dimensions! Or fractal dimensions... That's what really distinguishes mathematicians from engineers... At 02:03 AM 11/22/02 -0500, Ken Roberts wrote: >Hi Everyone, > >Drifting a bit in topic header - about Bingo cards... > >About 11 years ago a printing company hired me to help >them develop some Bingo card "perms" (permutations) as >they are called. Those are selections of numbers which >can then be fed thru the printing process to produce >cards. Some people with a mathematical bent make some >of their income via the design of Bingo perms. > >I've never played Bingo but understand that there are >special games - normal Bingo is 5 in a row, then there >is a need-both-diagonals variant, a large C variant >(need top and bottom rows and left column), and large >box variant (top and bottom, and left and right). > >Bingo cards are often printed in triplets, that is >three cards which use all 75 numbers. As you know >col B has numbers 1-15, col I has 16-30 etc. It is >considered "bad" to have all evens or all odds in >a row or column, so you have fewer choices for each. >It is also considered "good" for distinct cards in >a perm to differ by at least 4 numbers from one >another. There are not really septillions which >are sufficiently different from one another to be >acceptable for use in a Bingo parlour. > >There's more but that may give you the flavour of >the production of actual Bingo cards. > >I didn't stay with the assignment very long. The >math ran ahead of the programming. I showed that >one of their desires, for a perm which was maximal >in a certain sense, could not be achieved - and so >they decided that their existing ad hoc perms from >random generate-test-discard programs were maybe >not so bad after all! > >Actually, I didn't have to "show" anything except >to myself. A simple statement, in a definitive >tone of voice, that such and such had been proved >to be impossible was sufficient to satisfy the >client. A three sentence execution of the golden >goose. > >Bottom line: If you want to be a mathemagician, >add some fuzz and mystery to your pronouncements. > >Ken R.

2 1

[math-fun] Steven Krantz's review of A New Kind of Science
by Thane Plambeck 22 Nov '02

22 Nov '02

Just *too* amusing. This is going to appear in the AMS Bulletin in Jan 2003, apparently. http://www.math.wustl.edu/~sk/wolfram.pdf Thane Plambeck 650 321 4884 office 650 323 4928 fax http://www.qxmail.com/home.htm

6 7