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Comment by mamon | Hacker News Reader

I might be ignorant here, but wouldn't it suffice to prove that digits of Pi are random? Just check whether the distribution is uniform, the same way you would validate pseudorandom sequence generator for a programming language. Once you proved that Pi digits are random the normality hypotesis becomes trivial: as number of known Pi digits approaches infinity the probablity of any finite sequence to appear approaches 1.

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That's sort of the whole point of normal numbers, and pi is not known to belong to this class.

https://en.wikipedia.org/wiki/Normal_number

Quoting:

> It is widely believed that the (computable) numbers √2, π, and e are normal, but a proof remains elusive.

Here's a brief look into the distribution of the digits of pi if you're interested: https://blogs.sas.com/content/iml/2015/03/12/digits-of-pi.ht...

Summary: It's "seems" quite random, but has yet to proven as such.

You can’t just check all the digits. There are an infinite number of them and you don’t know if they stop looking random at some point after you stop checking.

Digits of Pi are not random, instead they are uniformly distributed or "pseudo-random".

Not all sequences are uniformly distributed, thus it very unlikely that we will be able to find long pieces of non-uniformly distributed sequences in uniformly distributed sequence.

I’m sorry, but I don’t think you quite understand what randomness, or pseudorandomness, means. In a truly random sequence, the probability of any given finite sequence appearing in it is 1.

AFAIK, nobody proved that Pi digits sequence is normal. See https://www.quora.com/How-random-are-the-digits-in-pi . So it quite possible for some finite sequences to be omitted from the Pi digits sequence.

The distribution wouldn't even need to be uniform, right? As long as one or more digits never stopped appearing at any point, the infinite sequence would include every possible sequence of numerals.

I'm 100% sure that Pi doesn't include e.g. 0.(3) sequence, or e sequence, or many many other sequences, or even just million of zeroes in the row.

Why would you be sure it doesn't include a million zeroes in a row? If the digits are random, as appears likely, it definitely will include a million zeroes in a row. The problem is even if you could turn every atom in the universe into a supercomputer, and somehow get them all cooperatively working on calculating the digits, you'd still never get to the million zeroes sequence in a million universe lifetimes. But they're still out there somewhere.

Because geometry of a circle will be skewed if Pi will contains millions zeroes in a row.

Pi is not a random number even if it digit sequence looks as pseudo-random sequence.

One would expect a repdigit sequence of length k to appear once in a 10^k normal sequence. So in the 10^1000000 expansion of pi, one should expect a million zeroes to appear in a row once. You can check this to a limited extent here: http://www.angio.net/pi/. It only goes up to 200 million (2 x 10^8) digits, but sure enough 00000000 happens to appear twice. (Nonetheless, we don't actually know if pi is normal.)

Thanks for this, very interesting. My earlier intuition is more than confirmed; If you turn every atom in the universe into a supercomputer, get them all working together generating digits of pi, for a million universe life times, you only get 10^80 (atoms in universe approx) * 10^10 (10 billion digits per sec each say) * 10^11 (years in a universe life time roughly) * 10^8 (seconds in a year roughly) * 10^6 (a million universe lifetimes) = 10^115 (ish) digits. In fact that is an incomparably smaller number that 10^1000000 digits, only good for a single run of 115 zeroes in a row.

Basically there's an awful lot of room between zero and mathematical infinity, and normal large numbers that we experience or even try to reason about in everyday life don't and can't begin to stretch even a little way towards the ultimate mathematical limits.

We can use BPP formula and sampling to try to find 100 zeroes in a row.

I cannot reply to the reply from v_lisivka (for some reason) but I don't know why you'd think a temporary run of decimal digits in pi would have any effect on the geometry of a circle. It all comes down to whether pi is Normal or not. Repeating the quote above, it is thought likely but not proven at this stage;

Quoting:

> It is widely believed that the (computable) numbers √2, π, and e are normal, but a proof remains elusive.

My rough understanding is that Pi is representation of certain curve: circle. If we skew sequence too far to one side or another, then curve will lost it shape.

I hope it is becoming apparent that you need more than a rough understanding of one of the properties of pi to contribute usefully to the discussion.

Normality also includes the distribution of the substrings in the definition, it's not just "every finite string of numbers compares somewhere"

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