7 ms·
>In reality the digits of pi are biased, so finding a particular number of N digits is even less likely. This is actually not known to be true. In fact, it's c
by wging 13y ago
>In reality the digits of pi are biased, so finding a particular number of N digits is even less likely.
This is actually not known to be true. In fact, it's conjectured that pi is normal[1]--meaning that we should expect every sequence of digits of a given length to be as likely as any other, no matter what base we're using.
So if you know the position at which the string you're interested in appears, you can just 'compress' by giving the location in the binary expansion of pi at which your message appears, and the length of your message in bits.
Where this fails, of course, is that you're likely to have to search a long time to find your message in pi, and you're also probably unlikely to be able to express your starting offset in fewer bits than it would take to write your full message.
[1] http://en.wikipedia.org/wiki/Normal_number http://en.wikipedia.org/wiki/Normal_number
- sillysaurus2 13y agoSo pi containing 3.141592653579 is equally likely as pi containing 1415926535797, which is equally likely to contain ABCDEFGHIJKL for every random ABCDEFGHIJKL? Fascinating. Completely impossible to take advantage of, too, but fascinating.
- dbaupp 13y agoIf the conjecture of being normal is true, yes.
- weavie 13y agoHmm.. it would also mean that somewhere in pi there is a sequence of 1 trillion consecutive zeros.
- judk 13y agohttp://en.m.wikipedia.org/wiki/Feynman_point http://en.m.wikipedia.org/wiki/Feynman_point
- p4bl0 13y ago> Completely impossible to take advantage of That depends on your goals, to do geeky things it's pretty cool :). I know my birth date (in the "ddmmyy" format) appears at the 262768th decimal of π. By the way this example validates part of the answer of sillysaurus2 to mvleming: I need 6 digits to encode 6 other. Not that good a compression.
- wging 13y agoIf pi is normal, and the two strings are of equal length, then yes... not only that, but if pi is normal it's likely that each appears infinitely often. By the way, sorry to point this out, but after 535 pi actually goes 5358979...
- Someone 13y ago"if pi is normal it's likely that each appears infinitely often." Not likely, _certain_. If there is a sequence S that appears a finite number of times, there is a N such that as never appears after digit N. That, in turn, means that, in the limit of L => infinity, the probability that S appears in the first L digits is zero.
- wging 13y agoOh. Right. Whoops. Thanks.
- BuildTheRobots 13y ago> So if you know the position at which the string you're interested in appears, you can just 'compress' by giving the location in the binary expansion of pi at which your message appears, and the length of your message in bits. It's a bit tongue in cheek, but this is exactly what piFS[1] does. [1] https://github.com/philipl/pifs https://github.com/philipl/pifs