http://qs1969.pair.com?node_id=496412


in reply to Re: Latent Japh
in thread Latent Japh

How did you know that's what I did?!!  :-)

In a related vein -- you could use such a method to generate "Just another Perl hacker" using the digits of pi.  Taking 2 digits at a time to represent a single ascii value (offset by <space> = ascii 32) , all you have to do is search for the correct digit sequence.  Write a program to calculate pi to that number of places, convert the next 48 digits into 24 ascii chars, and you're done!  (You probably won't get a lot of votes for it, though, as it might take a few years to run!)

Seriously though, I'm thinking of submitting the program I used to generate Latent Japh.  I need to clean it up a little first, and add some more user options (it's amazing how much the output can change with just a little tweaking), but I'll publish it once I've gotten it cleaned up.

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Re^3: Latent Japh
by Anonymous Monk on Oct 01, 2005 at 22:42 UTC
    I'm not a mathemagician - but I'm guessing that since pi's digits are an infinite nonrepeating sequence, then it must hold true that any other finite sequence you ever wanted to see exists somewhere in the digits of pi (possibly unfathomly deep in the numbers of pi though). So in theory, you could just fine the digit position at which the ascii codes for the sentence "Just another Perl hacker" occur naturally in order in the digits of pi, and calc down to there and then print them out.
[reply]

      Google found me a website that allows people to search the first 200 million digits of PI for any sequence of digits (up to 120 digits). If you convert the phrase, "Just another Perl hacker,\n" to its ordinal values you get "74117115116329711011111610410111432801011141083210497991071011144410 " (all delimiters have been removed). Searching for that string of digits in the first 200 million digits of PI comes back with no matches.

      The same website also shows a breakdown of probability of finding digit strings of certain lengths:

      DigitsProbability
      1-5100%
      6Nearly 100%
      799.995%
      863%
      99.5%
      100.995%
      110.09995%

      The ASCII ordinal value string for "Just another Perl hacker,\n" is 68 digits long. I don't remember from stats classes how to predict the probability of finding a 68 digit sequence in a "random" sequence of 200 million digits, but I know the probability is extremely low. And as the search results show, "Just another Perl hacker,\n" isn't found in the first 200 million digits of PI.

      As a matter of fact, the ordinal values of the ascii string "Just" also do NOT appear within the first 200 million digits of PI. And that's only an eleven digit pattern! A given 68 digit pattern might not even show up in the first 200 billion digits of PI. ...but then again it might. ;)


      Dave

[reply]
        JAPH appears, though: The string 74658072 was found at position 83,607,151 counting from the first digit after the decimal point. The 3. is not counted.
[reply]
      I'm not a mathemagician - but I'm guessing that since pi's digits are an infinite nonrepeating sequence, then it must hold true that any other finite sequence you ever wanted to see exists somewhere in the digits of pi

      No, it is not true that "it must". In fact it needs not, i.e. a number's digits being an infinite nonrepeating sequence is not an sufficient condition for it to include any given finite subsequence. Check the definition of normal number (which is itself slightly stronger than the above, involving a requirement on the limiting frequency) e.g. here. However it is indeed postulated that pi is normal, but needless to say it's extremely difficult to prove such a claim.

      In this vein you guys may also be interested in the miraculous Bailey-Borwein-Plouffe formula which gives somewhat unexpectedly (and slightly simplifying) the n-th hexadecimal digit of pi independently of the previous ones, which makes it particularly suitable for distribuited computing...

[reply]
      Yes, that's what I was getting at.  I think you said it better, though!
[reply]
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