June 2006 - mathfun@mailman.xmission.com

Re: [math-fun] One thousand paintings
by Henry Baker 04 Jun '06

04 Jun '06

There's always quaternionic art: E.g., http://www.bluestarfolly.com/art/quaternion.html http://mzlabs.com/gart/g4.html At 10:10 AM 6/4/2006, dasimov(a)earthlink.net wrote: ><< >Kerry Mitchell wrote: > >Here's your chance to own your favorite number, unless of course, someone already does. >http://www.onethousandpaintings.com/home/ >>> > >This number scam is at least far more artistic than the classic ('50s? '60s?) painting that >is completely black -- and it's all the same black, with indiscernible brushtrokes. Now that's >*real* art! > >But maybe we math-fun members could make a huge profit if we corner the market in >Gaussian integer paintings!!! Anyone know an average print shop? > >--Dan

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Re: [math-fun] One thousand paintings
by Henry Baker 04 Jun '06

04 Jun '06

Didn't PT Barnum say it best? There's a sucker born every minute. (Maybe not... http://en.wikipedia.org/wiki/There_is_a_sucker_born_every_minute) You can also buy a single pixel on the web: http://milliondollarhomepage.com/ Or patent your favorite number? http://www.theregister.co.uk/2003/11/26/first_integer_patented/ http://www.daen.dk/archives/000221.html At 09:44 PM 6/2/2006, Kerry Mitchell wrote: >Here's your chance to own your favorite number, unless of course, someone >already does. > >http://www.onethousandpaintings.com/home/ >Kerry >-- >lkmitch(a)gmail.com >www.fractalus.com/kerry

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[math-fun] One thousand paintings
by Kerry Mitchell 03 Jun '06

03 Jun '06

Here's your chance to own your favorite number, unless of course, someone already does. http://www.onethousandpaintings.com/home/ Kerry -- lkmitch(a)gmail.com www.fractalus.com/kerry

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[math-fun] coded messages
by Russ Cox 02 Jun '06

02 Jun '06

These two coded messages have been making the rounds. http://www.homelandstupidity.us/2006/05/31/the-212-796-0735-mystery/ http://www.homelandstupidity.us/2006/06/01/another-phone-numbers-station-41… 013 056 051 012 079 046 065 010 093 000 082 039 013 094 069 012 078 108 017 028 017 069 022 073 038 014 017 015 015 073 004 020 068 012 013 125 100 054 004 091 014 013 015 086 022 096 081 066 002 082 055 070 002 000 000 022 083 029 008 022 012 004 071 013 065 027 094 019 029 014 022 008 002 011 083 073 003 026 019 000 007 000 000 086 061 078 002 021 085 006 013 069 006 079 012 015 024 007 006 016 017 069 095 000 017 024 005 014 024 009 087 022 067 089 074 010 010 086 078 013 024 004 016 027 073 013 015 006 093 069 112 020 084 000 000 021 003 070 031 076 049 065 023 027 067 000 007 016 012 017

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[math-fun] Unusual "CDR" (REST) recursion
by R. William Gosper 01 Jun '06

01 Jun '06

We wish to count the teeth in a gear train represented as an ordered list of ordered triples. Each triple represents a gear cluster, and consists of an identifier, an ordered list of integers (the "numerator" gears), and an equally long list of "denominator" gears. If the last n numerator gears coincide with the last n of some later cluster, then the last n+1 denominator gears coincide also, and these two clusters share those gears, and we want to count their teeth only once. (defun teeth (train) (loop for clusters on train sum (+ (toot clusters #'second) (toot clusters #'third)))) (defun toot (clusters accessor &aux (gears (funcall accessor (first clusters)))) (if (or (null gears) (some #'(lambda (cluster) (loop for cogs on (funcall accessor cluster) when (equal gears cogs) return 't)) (rest clusters))) 0 (+ (first gears) (toot clusters #'(lambda (cluster) (cdr (funcall accessor cluster))))))) toot counts the numerator or denominator teeth of a cluster, depending on the accessor. When it finds an unshared gear, it adds it to the count of the rest of its list. But the list per se is not a transmitted argument on which to recurse. So we put the CDR in accessor! Termination is not running out of clusters, but rather running out of gears in the first cluster. --rwg

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Re: [math-fun] Bridge probability puzzle
by dasimov＠earthlink.net 31 May '06

31 May '06

<< What is the probability there is some pair of suits your bridge hand does not include, and your partner's hand does not include the other pair? >> My answer to this agrees with Gareth's and then Joshua's post -- as a fraction in lowest terms, p/q. My strategy was this: (how many ways you & your partner can each have hands in which exactly two suits appear) + (how many ways you can have one complete suit, & partner's hand shows exactly 2 suits) + (vice versa) + (the 12 ways each of you can have one full suit) Here's the Mathematica session: ------------------------------------------------------------------------------------------------------------ In[1]:= c = 26!/13!^2 Out[1]= 10400600 In[2]:= h22 = 4!((c-2)/2)^2 Out[2]= 649034632545624 In[3]:= h12 = 4*3!*(c-2)/2 Out[3]= 124807176 In[4]:= h21 = h12 Out[4]= 124807176 In[5]:= total = h22 + h12+h21 + 12 Out[5]= 649034882159988 In[6]:= npairhands = 52!/(13!^2 26!) Out[6]= 5157850293780050462400 In[7]:= prob = total/npairhands 54086240179999 Out[7]= --------------------- 429820857815004205200 Out[8]= 1.2583437773342733107186438969809670960643871973497 * 10^-7 [to 50 decimal places]. -------------------------------------------------------------------------------------------- Is this solution now in doubt? --Dan

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