in reply to
Latent Japh
Okay, that tears it... I'm going to find one of those sites that converts pictures to ASCII art, and just start putting #!/usr/bin/perl in front of the results, and see what executes.
Re^2: Latent Japh by liverpole (Monsignor) on Sep 30, 2005 at 13:27 UTC 
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.
 [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 (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] 

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

Latent Japh my ass.
I've definitely seen this before. Do the words "prank flash" *pop up* in mind? It should be *screaming* at you and *biting* you hard. But then again, some of us get a *heart attack* at that idea.
(Halloween joke: The thing to remember about pumpkins is that *things* can be stuck in their eyes and mouths, and that they are *soft* and *squishy* unless hollowed out...{shifty eyes}...NOT THAT I'D KNOW ANYHTING!)
 [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:
Digits  Probability 
15  100% 
6  Nearly 100% 
7  99.995% 
8  63% 
9  9.5% 
10  0.995% 
11  0.09995% 
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. ;)
 [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 BaileyBorweinPlouffe formula which gives somewhat unexpectedly (and slightly simplifying) the nth hexadecimal digit of pi independently of the previous ones, which makes it particularly suitable for distribuited computing...
 [reply] 

