September 2005 - mathfun@mailman.xmission.com

RE: [math-fun] Multiplicative Magic Squares
by Christian Boyer 29 Sep '05

29 Sep '05

Sorry Michael, I have not completely read one of your previous messages, you mentioned already the same square: > (p,q,pq; pp,qq,r) [4,4,1] 6480 -----Message d'origine----- De : Christian Boyer [mailto:cboyer@club-internet.fr] Envoyé : lundi 26 septembre 2005 16:14 À : 'math-fun' Objet : RE: [math-fun] Multiplicative Magic Squares >From: "Michael Kleber" <michael.kleber(a)gmail.com> > >So abcdABCD makes a pretty small fraction of all multiplicative magic >squares, as the magic product (and the exponents on each prime in its >factorization) get large. Here is an interesting family of 4x4 multiplicative squares, using only three variables a, b, c: 1 abbb bc aaa ac aab ab bb aaab c abb b bbb a aa abc Magic product = a^4 * b^4 * c, for the 4 rows, 4 columns, 2 diagonals and 2 broken diagonals. And also for other sets of four cells (the corners for example, the 4 numbers of each quarter, ...). Apply a=2 and b=3, and you will get one of the various possible examples with magic product 6480. 6480 is the second smallest product after 5040, as listed in my previous email. Christian.

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RE: [math-fun] another kind of median (trying again)
by dasimov＠earthlink.net 28 Sep '05

28 Sep '05

<< The picture was supposed to show a rectangle of width 1 and height 2 whose bottom is centered at x=1, and to the right of it, a rectangle of width 1 and height 1 whose bottom is centered at x=2. The base of the first rectangle goes from x=1/2 to x=3/2, and the base of the second rectangle goes from x=3/2 to x=5/2. The total area under the histogram is (1)(2)+(1)(1) = 3. The area to the left of the line x=5/4 is (5/4-1/2)(2) = 3/2, which is half of the total area. So x=5/4 is the "median". >> After getting an explanatory e-mail from John Robertson, I thought I posted a "now I get it" post to math-fun, but it apparently didn't get through. (Your original message was perfectly clear, and my attempt to reproduce your calculation neglected to take account of the +1/2 poisitioning of the bars, even though I implied that positioning myself.) But: For very small samples, the median doesn't generally convey much information. For a large sample, however, this "bar-graph" median can be unduly influenced by outlying data -- much as an average can. But the idea of the median, as I see it, is to characterize the middle of the sorted data without much influence on the size of data points not close to the middle. Or am I still misunderstanding something? --Dan

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Re: [math-fun] Sex, Lies and Government Conducted Surveys
by dasimov＠earthlink.net 28 Sep '05

28 Sep '05

Jim writes: << Here's another kind of median I could imagine statisticians using: Suppose our data are the numbers 1, 1, and 2. The bar graph is ___ | | | |___ | | | |___|___| 1 2 We could then choose the median to be the value m such that the line x=m bisects the area inside the histogram, which (if I'm not mistaken) is m=1.25. >> I'm proably not understanding the bar graph; it doesn't seem very clear to me even in an equal-width font. But I'd expect the bar graph for data 1, 1, 2 would be a bar of width 1 and height 2 around x = 1, adjacent to a bar of width 1 and height 1 around x = 2. Which would mean that the vertical line bisecting the area would lie at the solution to 2 - 2(1-x) = 2(1-x) + 1, giving x = 3/4. --Dan

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Re: [math-fun] Sex, Lies and Government Conducted Surveys
by dasimov＠earthlink.net 28 Sep '05

28 Sep '05

David Gale writes: << Mike Greenwald asks, how a median can be 3.8. Good question! . . . Anyhow, someone please help me out on those fractional medians >> In my fairly extensive interaction with statisticians + stat textbooks, all definitions I've seen of a median of a finite sample of N real-valued data are identical: If N odd, median = middle value; if N even, median = average of the 2 middle values. BUT -- there is a more sophisticated definition that may exist: If N is even, and the two middle values are distinct but at least one of them occurs with multiplicity > 1, one might prefer to use the "asimov median", defined as follows: Assume the data sorted by size: x(1) <= x(2) <= . . . <= x(N). Let J be the smallest interval of indices that is symmetrical about the real number N/2, such that J contains a) all repetitions of x((N+1)/2) (if N is odd), or b) all repetitions of both data x(N/2) and x(N/2 + 1) (if N is even). Now define the "weighted median" as the average of all values whose indices lie in J. Thus if N = 200 and the sorted data looks like this: sort #: . . . 94 95 96 97 98 99 100 101 102 103 104 105 106 . . . datum: . . . 2 3 3 3 4 4 4 4 4 4 4 4 5 . . . then one may wish to say the asimov median is the average of data whose indices k satisfy 100.5 - 5 <= k <= 100.5 + 5, i.e., 96 <= k <= 105, so we get a.m. = (3+3+4+4+4+4+4+4+4+4)/5 = 3.8 . --Dan

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CORRECTION: Re: [math-fun] Sex, Lies and Government Conducted Surveys
by dasimov＠earthlink.net 28 Sep '05

28 Sep '05

CORRECTIONS: I stupidly rushed the "asimov median" post into print with a number of errors. Here is the corrected definition -- sorry. Correct definition: ------------------------------------------------------------------------------------------------ Assume the data sorted by size: x(1) <= x(2) <= . . . <= x(N). Let J be the smallest [interval of indices] that is symmetrical about the real number (N+1)/2, such that J contains a) all repetitions of x((N+1)/2) if N is odd, or b) all repetitions of both data x(N/2) and x(N/2 + 1) if N is even. Now define the "asimov median" as the average of all values whose indices lie in J. Thus if N = 200 and the sorted data looks like this: sort #: . . . 94 95 96 97 98 99 100 101 102 103 104 105 106 . . . datum: . . . 2 3 3 3 4 4 4 4 4 4 4 4 5 . . . then one may wish to say the asimov median is the average of data whose indices k satisfy 100.5 - 5 <= k <= 100.5 + 5, i.e., 96 <= k <= 105, so we get a.m. = (3+3+4+4+4+4+4+4+4+4)/10 = 3.8 . --Dan

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[math-fun] Sex, Lies and Government Conducted Surveys
by David Gale 28 Sep '05

28 Sep '05

On September 15 the government released a report "Sexual Behavior and Selected Health Measures" reporting on a survey "conducted by the Centers for Disease Control and Prevention's [sic] (CDC) National Center for Health Statistics (NCHS)" http://www.cdc.gov/nchs/data/ad/ad362.pdf The objective of the survey is to get information on public health concerns, fertility, sexually transmitted disease, etc. However I bring it up here because it contains the most glaring examples I have yet seen of published inconsistency. It's the well worn story that males have many more "sexual partners" than females. Interested people should look at the report. For example, Tables 10 and 11 of the survey show that the median number of partners "in lifetime" for males over forty is 8 while that for females is 3.8. To immediately recognize the inconsistency imagine the same survey with but with the words sexual partners replaced by spouses. For more of the same look at Figure 6 on page 6. More inconsistency: when men and women are asked for the number of partners over the past 12 months, Tables 1 and 2, the numbers come out the same, within epsilon. . As I said, this is an old story. See the following from New York Review of Books ten years ago by Richard Lewontin. http://www.nybooks.com/articles/article-preview?article_id=1923 The authors are aware of the problem and suggest explanations on pages 12 and 13 some of which are, well, amusing. "Some researchers have suggested that some of this difference is due to a small percentage of men who report very large numbers of partners," Another "explanation" is that women my have partners over age 44 who were not polled in the survey (this would of course skew it in the other direction). Is this an example of innumeracy in high places? Lewontin writes in http://www.nybooks.com/articles/1871 "The willingness of both social and natural scientists to recognize contradictions in their findings, and then to ignore them when it is convenient, is a serious disease of inquiry." David The authors

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RE: [math-fun] Multiplicative Magic Squares
by Torgerson, Mark D 27 Sep '05

27 Sep '05

Michael Kleber wrote >Q: Is it always possible to partition the integers from 2 to 2n-1 >into two sets of size n-1, such that the products of two elements, >one from each set, all distinct integers >=2n? >Seems like the answer is "yes with high probability", but can someone >change that to just "yes"? Looks like "no" is the real answer Suppose so... Let S,T be the two sets, with 2 in S. Let 2<a<n. It must be that a isin S. Given a, then 2a<=2n-1 and hence in one of the sets. If a isin T then we can make 2a two different ways. The first with 1*2a (no matter where 2a lives) and 2*a since 2 and a are not in the same set. So S={1,2,3,...,n-1,r} for some r>=n. Now for n of any size, there are a couple of instances where 2a and 2a+1 are in S and (because we don't know r) at least one case where 2(2a) and 2(2a+1) are in T. This gives two ways to get 4a(2a+1). The cheese ball way to get the job done is to set S={1,2,...,n} and T={1,p1,...,p(n-1)} where the pi are the n -1 smallest primes bigger than n. Mark

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RE: [math-fun] Multiplicative Magic Squares
by Lee Sallows 27 Sep '05

27 Sep '05

>I promise to explain my "secret" algorithm (but there is no great secret...) >when it will be proved enough good and optimized to successfully answer to >the 5x5 question. >Christian Hello folks, I don't know how Christian does it, but to take 3x3 as an example, I'd start with a generalization of all 3x3 multiplicative magic squares, such as: z*x z*x^2*y^2 z*y z*y^2 z*x*y z^x^2 z*x^2*y z z*x*y^2 The magic product is then x^3*y^3*z^3 = 216 when z = 1, x = 2, y = 3, the smallest integer assignmments. Stepwise incrementing of these values while skipping those results showing repeated entries will then yield squares with the remaining products on Christian's list. Ditto for 4x4. But I must say, 5x5 looks tricky. Lee

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RE: [math-fun] Multiplicative Magic Squares
by dasimov＠earthlink.net 27 Sep '05

27 Sep '05

I love the idea of multiplicative magic squares. Using n^2 unknown but distinct vectors in R^n that are indexed by (i,j) for 1 <= i,j <= n, it's the same as finding nonnegative integer coordinates for these vectors such that all vectors (i, j0) for 1 <= i <= n sum to the same as all vectors (i0,j) for 1 <= j <= n (for all choices of 1 <= i0,j0 <= n). (Or did someone already mention this?) I'm interested in how magic these can be. I find the old rule of having all rows & columns plus the 2 diagonals sum to the same sum as rather dorky: the diagonals are kind of random. 1) Better here would be all rows & columns and broken diagonals (i.e., diagonals on the torus) multiply to the same product. 2) Best of all would be for simply all affine lines on Z_n x Z_n multiply to the same product. (This is when you repeat the same step (K,L) on the torus and first come back to your starting point with exactly n steps. (There are 4x4 additive magic squares with this property; I'm not sure if there are for any other n.) Are 1) or 2) possible for a multiplicative magic square (MMS)? (And is there a better name for an MMS than multiplicative magic square?) --Dan

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RE: [math-fun] Announcement: Atlas of Natural Numbers
by dasimov＠earthlink.net 27 Sep '05

27 Sep '05

David Wilson wrote: << In the near future, I intend to begin serious development on my Atlas of Natural Numbers, a web site devoted to collecting and organizing interesting information about individual natural numbers. >> This sounds great. Can you answer a question I've had for some time? I'd like to know the smallest positive integer that cannot be described with fewer than fifteen English words. --Dan

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