If pi is truly infinite, then it contains all the works of Shakespeare, every version of Windows, and this comment I’m typing right now.
That’s not how it’s works. Being “infinite” is not enough, the number 1.110100100010000… is “infinite”, without repeating patterns and dosen’t have other digits that 1 or 0.
to be fair, though, 1 and 0 are just binary representations of values, same as decimal and hexadecimal. within your example, we’d absolutely find the entire works of shakespeare encoded in ascii, unicode, and lcd pixel format with each letter arranged in 3x5 grids.
Doesn’t, the binary pattern 10101010 dosen’t exists on that number, for example.
Actually, there’d only be single pixels past digit 225 in the last example, if I understand you correctly.
If we can choose encoding, we can “cheat” by effectively embedding whatever we want to find in the encoding. The existence of every substring in a one of a set of ordinary encodings might not even be a weaker property than a fixed encoding, though, because infinities can be like that.
Still not enough, or at least pi is not known to have this property. You need the number to be “normal” (or a slightly weaker property) which turns out to be hard to prove about most numbers.
If it’s infinite without repeating patterns then it just contain all patterns, no? Eh i guess that’s not how that works, is it? Half of all patterns is still infinity.
No. 1011001110001111… (One 1, one 0, two 1s, two zeros…) Doesn’t contain repeating patterns. It also doesn’t contain any patterns with ‘2’ in it.
But pi is believed to be normal. https://en.m.wikipedia.org/wiki/Normal_number
So it should contain all finite patterns an infinite number of times.
Not, the example I gave have infinite decimals who doesn’t repeat and don’t contain any patterns.
What people think about when said that pi contain all patters, is in normal numbers. Pi is believed to be normal, but haven’t been proven yet.
An easy example of a number who contains “all patterns” is 0.12345678910111213…
In some encoding scheme, those digits can represent something other than binary digits. If we consider your string of digits to truly be infinite, some substring somewhere will be meaningful.
There’s no way the copyright office is actually going to approve this right?
According to Dr. Calibri, there’s a 99.9999% chance they will approve it :)
Including relevant XKCD as demanded by internet law: https://xkcd.com/10/
“I may be a staunch atheist,” said Richard Stallman, creator of the GNU + Linux operating system and self-proclaimed architect of the modern world, “but any decent analysis in comparative religion would conclude that the universe is a copyleft creation, thereby pi should automatically fall under the terms of the GNUv3 license.”
Lol, he would actually say that
Is there an algorithm or number such that we could basically pirate data from it by saying “start digit 9,031,643,679 with length 5,345,109 is an MP4 of Shrek”? Something that we could calculate in a day or less?
The short answer is no, and even if we could, the digit index you’d start at would have a larger binary representation than the actual data you were trying to encode.
An example I found: the string of digits 0123456789 occurs at position 17387594880. In this case, it took 11 digits to describe where to find a 10-digit number.
So I think such an algorithm would technically work, but your “start digit” would be so large it would use more data than just sending the raw file data. Not to mention the impossible amount of computing power needed.
What if instead we utilized an algorithm, some code, that would ultimately generate the file? I could imagine a program that generates a number which ultimately is more dense than the program. For example, if we just-so-happened to need a million digits of Pi the program would be shorter than the number. Is there a way to tailor an algorithm to collapse down to any number? As an example, what if we needed a million digits of Pi but the last 10 digits need to be all 9s?
Could we already do this by leveraging the Library of Babel?
Genuinely asking, I’m not really sure.