If i took out a portion of pi's decimal value, and handed it to you, could you figure out where it was positioned?

Linear time, probably.

>If pi runs on forever, yet never starts to cycle, then any finite string MUST turn up sooner or later

Really? Can you actually say that, though? Put your intuition aside, because we certainly can't assume it. I can think of a number which does not 'cycle' and does not produce any finite 'string' of digits. An irrational number which only contains 0's and 1's in its decimal representation, and does not 'cycle' any certain sequences of digits, for example

I'm not saying you're wrong, but I am saying it sounds like your claim could use some backing up that I personally don't know exists or not (but would be interested in seeing if it did)

is it possible to prove then that pi does not generate that irrational number which only contains 0's and 1's in its decimal representation somewhere in its expression??

Just from the fact that it is irrational, it doesn't have to exhaust all combinations of digits.
1.10100100010001000001000001.... is irrational aswell, but you won't claim it will show 123 at some point.

en.m.wikipedia.org/wiki/Normal_number

int find_pi (string portion){
int index = 0;
while(true){
loop:
if(pi_digit(index++)==str2num(portion[0])){
for(int strindx = 1; strindx < portion.length(); ++strindx, ++index){
if(pi_digit(index) != str2num(portion[strindx])){
index -= strindx-1;
goto loop:
}
}//end for
return index-portion.length();
}//end if
}//end while
}//end find_pi

>If pi runs on forever, yet never starts to cycle, then any finite string MUST turn up sooner or later.

nope

0.101001000100001000001... never repeats but contains only 0s and 1s, and not even every possible string of 0s and 1s (like 11). Pi might contain every string though, I dunno.

I get the feeling you don't do a lot of mathematics

I can't prove that, because as far as I'm aware, there's pretty little we can work with when proving what digits pi does or doesn't contain. I wanted to know if there was some actual literature on this (a juicy proof for it) that I didn't know about. Or, you know, if you were just a layman commenting on this based on guesswork alone

Assume your string is N digits long.
We start checking each group of N digits. (We advance by 1 each time, not N)
The odds that it's NOT what we're seeking is 1 in 10^N. If N was, say, 6 that's 0.999999 Call that Z.
After examining A such groups, the chances of never hitting is Z^A. As soon as that drops below 0.5 the odds are better than even that we've found what we're looking for. By the time it drops to 0.1 the odds are 9 out of 10 that we've succeeded. And so on. You can get any confidence limit you like (short of 100%) if you keep going.

In the example I posted there's a "10" and the first line, a "11" in the second line, etc.
Any finite-string consisting of only "1" and "0" is bound to turn up eventually (the calculation is only a little more complicated than that given above) so, NO, it's not possible to prove what you asked.

"Not cycling" means that you'll eventually find any string you like. You've just made the task easier. Rather than looking for a particular 6 digit string (which is 1 in a million) , any of 2^6 strings in a million will do.

"6" is, of course, arbitrary. It could be a million.
If you're read Sagan's "Contact" you'll know that they find an odd sequence within pi. The significance isn't that it's there. It HAS to be there, somewhere. But it turns up long long before they expected it.

>For example, Chaitin's constant is normal (and uncomputable). It is widely believed that the (computable) numbers √2, π, and e are normal, but a proof remains elusive.
...huh.

Touché. Your example is similar to part of Gödel's incompleteness proof.
OK. I can't prove EVERY sequence will turn up. I can only say that, thus far, pi has passed every test for randomness. But that 's not a proof, of course.

Let's try this again
>"Not cycling" means that you'll eventually find any string you like

Why? Assume we have an irrational number that doesn't cycle, and prove why a given sequence of n digits MUST exist in the decimal representation of that number. Not "probably" exists, show that it HAS to, or else your assertion is completely useless

not cyclic doesn't force every digit to appear. It is not impossible for a number to exist that does not have a 9 in its decimal value but be irrational. If your sequence then contains a 9, tough luck.

I already conceded I couldn't prove it. You don't have to hammer.
Experimentally, pi passes the tests of "normality", but that's insufficient.

14159

149597870700

>if i took out a portion of pi's decimal value, and handed it to you, could you figure out where it was positioned?

Theoretically impossible to solve, since Pi is an irrational number of infinite length, it could fit anywhere.

>find first ocurrence
>done

problem doesn't state positioning has to be unique

use one of those wacky series that converge to pi and ask every term if they know the digits
does it make me clear myself?

>problem doesn't state positioning has to be unique

problem implied a unique position.

theoretically yes, but my computer will need some time.

there are algorithms to get the digits of pi to any precision, so i just compare your data to the sequences i get one after the other....

>Is pi a normal number?
Yes: no unique position exists given finite length sequence. Guaranteed duplicates of the same exact sequence exist.
No: I dunno sonny boi, might be impossibru