Can all books be found somewhere within the number Pi? | Hacker News Reader

Can all books be found somewhere within the number Pi?

jakubkotowski.com

31 pointsjkbyc29 comments

The idea of encoding a book as a string is incoherent once we look at it with intellectual rigour. How are images in illuminated manuscripts or graphic novels encoded? What does "all books" even mean? Are we talking about each individual artefact past present and future or some idealized representation of each: e.g. is my current copy of Catcher in the Rye encoded separately from the copy I was assigned in 10th grade English class?

It's all a matter of interpretation.We can just as easily choose a different arbitrary encoding and claim to have found all the books in π. There's no need to make things complicated. We are free to pick any interpretation we want once we are claiming that numbers represent books. Let:

  B = {b1, b2,...bn}

Such that it contains the set of all books. And let:

  def Find-books(num)
      if 3.14 < num < 3.15
      then return B
      else return "all the books not found"

The article assumes that there is some natural way of encoding books. But digits of π are not Unicode or Ascii characters. Though we can interpret a digit or string of digits as such, that encoding is arbitrary not a property of the natural or mathematical world.

This objection is silly. Of course if you define an encoding that maps '3' to 'the contents of all books', then you've trivially found a way to encode it in pi. But I think you're aware that the intention of the question is to find an encoding that has comparable complexity to the text of all books.

Specifically, the property we want is that if some substring of the digits of pi encodes 'all books', then you should be able to extract certain books out of substrings of that block of pi. I don't know if there's a technical term for that, but it's clear that we're imagining encoding successive paragraphs of these books in successive blocks of digits in pi.

The thing about normality is that if a number is normal, it will contain all books not only for a given encoding of a book, but for any given encoding of any book. That is true only for normal sequences, of which pi is assumed to be an example.

Nitpick: This is NOT true for only normal sequences. See disjunctive sequences: http://en.wikipedia.org/wiki/Disjunctive_sequence .

pi may contain all numbers as substrings and yet not be normal.

Exactly. So we can choose an encoding that is computationally expensive or one which is cheap. The essence of mathematical thinking is to use replace complexity with simplicity: a person who insists on calculating the sum of positive integers less than n with n - 1 operations is not to be admired for their mathematical insight [their dogged determination may be another matter].

Some encodings are more useful than others while being equivalent. All encoding requires an arbitrary interpretation. Each is equally valid...finding "books" in Pi is a flight of fantasy.

Yes, but the number which indexes the position in PI where your book starts will probably be longer than the book itself.

More precisely: Let's assume that pi is absolutely normal and we use a 256 letters alphabet (ascii, ¿latin-1?).

I'm almost sure that if we have a "book" with N characters, in average, the number of (decimal) digits of the index of that string in pi is N * log(256) / log(10). (I'm too lazy to write the proof now, so perhaps I'm making a mistake.) This is (esencially) equivalent to that the expected position is 256^N. (But I'm taking averages willy-nilly.)

If this is correct, the position increase exponentially with the book length, but the numbers of digits in the position increase linearly.

Theoretically, everything can. And there's a FUSE implementation: https://github.com/philipl/pifs

Theoretically? Citation needed. I believe this is only true if pi is normal, and while there's a lot of conjecture that pi is normal and no evidence that suggests it isn't, there is no proof.

Normal or not normal? If a work can be found in pi, then one would think a Finite State Gambler (FSG) would succeed on the portion of pi equivalent to the work. But FSGs cannot succeed on normal sequences, meaning this sequence would be non-normal.

Of course this begs the question of whether a number normal over an infinite sequence can contain sub-sequences that are non-normal - sequences of 100 tails will occur in sufficiently large completely normal (randomized) sequences of random coin tosses, e.g.

A normal sequence includes all non-normal sequences, by definition of normality.

As for the citation, there is this: http://en.wikipedia.org/wiki/Disjunctive_sequence

If this is correct, then you are correct, the conjecture would hold if pi were normal. But then wouldn't an FSG be able to succeed on the sub-sequence? Of is that acceptable, provided that the FSG fails on the overall sequence?

Normality is sufficient, but I do not think it is necessary. Normality requires that if the number is represented in any base b then each each of the b possible digits occure with density 1/b. All that is needed for every work to appear is that each digit occur at least once. They do not have to occure with equal density.

Nope, normality is necessary.

>>All that is needed for every work to appear is that each digit occur at least once.

Counter example would be a=0,123456789010011000111000011110000011111....

Normality is not necessary. Counter-counter example is the digits of pi in groups of 1, 2, 3... with strings of zeros in between:

   3.0140015900026530000....

this number is obviously nor normal since 0 occurs with greater frequency than any other digit. But if pi is normal, then this sequence contains all finite sequences too.

Each base 10 digit occurs at least once in your number. It's not obvious to me that each base b digit occurs at least once for all bases b in your number.

Exactly. But I've been searching Pi for the blueprints of a Klingon cloaking device for the past 5 years and have yet to turn up anything.

The question is: is pi a 'normal' number? It's an open question. Note that it's not particularly difficult or illuminating to generate a normal number, example: 0.(binary digits of 1)(binary digits of 2)...(binary digits of n)...

http://en.m.wikipedia.org/wiki/Normal_number

That number is binary to base 2, but not necessarily to, say, base 10. Making an absolutely normal number (that is, a number that is normal to any base) is more involved[0]. It has only been recently shown that this can be done with polynomial time complexity[1]. My licentiate thesis was an implementation of the second algorithm.

[0]http://www.sciencedirect.com/science/article/pii/S0304397501... [1]http://www.dc.uba.ar/people/profesores/becher/poly.pdf

Disclaimer: I don't know what I'm talking about.

My dim understanding of the issues leads me to consider a conflict. Pi is more or less considered to be more or less random, or some flavor of random, notwithstanding known patterns of Pi. "Random," to me, sounds a lot like "unorganized."

A book is definitely organized. A larger book is highly organized (entropically speaking). So while you probably can find the same sequence of words in a two-word or ten-word or other small book in Pi, at some point you get a book that's too highly organized to appear in Pi.

However, Pi is also infinite, so it's infinitely possible to find any sequence. (This sounds really hand-wavy to me).

But since Pi is infinite, then isn't it also infinitely unorganized?

A book from start to finish could be considered to be 'highly organised'. However, if the book you find is only an infinitely small subset of pi, then it should have no bearing on whether you call pi itself 'highly organised'.

Infinity is very large, and even with something as simple as flipping a coin, you could probably find arbitrary patterns in it if you looked hard enough. 1000 heads in a row? Alternating between heads and tails for 100 flips?

http://www.wired.com/2012/12/what-does-randomness-look-like/

> "Random," to me, sounds a lot like "unorganized." A book is definitely organized.

Search for "microstate vs macrostate" and keep checking links until you hit an explanation that you like.

Alternatively, let me give it a go: Enumerate all 10-character strings. The string "Hi, there!" appears once, just as often as "l9.gn;omeh" (which also appears once). We call these "microstates". However, if we label 10-char strings green if they look like valid English and red if they don't, then clearly most of our table is going to be colored red. Red and Green are "macrostates". Unlike microstates, red and green do not have equal probability if we choose an element of the enumeration at random.

    p(red)>>p(green)>>p("Hi, there!")=p("l9.gn;omeh")

OK, now you were probably picturing a "entropy trace" computed over windows of digits of pi, like you would get out of binwalk. ''Entropy'' is a macrostate, not a microstate. Even though all sequences of digits within a window are equally likely, if you choose a sequence at random it will probably (!!!!) have high ''entropy''. But possibly not. Having high entropy is like being labeled "red" in the 10-character enumeration table: if you pick an entry at random, you're probably going to hit red, but if you keep doing it then eventually you will hit green, and if you do it even longer then eventually you will hit "Hi, there!". Just like if you keep looking at digits of pi, eventually you'll find a block of them where the ''entropy'' is low. In fact, if pi has the properties that Mathematicians conjecture it does, you'll find infinitely many such blocks of anomalously low ''entropy''. If you keep looking long enough, you'll find one with your book in it. These anomalies aren't due to pi being less than perfectly random, they are due to the definition ''entropy'' being statistical in nature. It's only "right" about randomness most of the time.

> However, Pi is also infinite, so it's infinitely possible to find any sequence. (This sounds really hand-wavy to me).

Here is the more precise mathematical analog from the article:

> PI is believed to be a normal number and therefore all possible finite sequences of characters appear equally often in it

it's still an open question, but the trouble doesn't lie in the definitions, it lies in finding absolute proof, which they haven't done yet.

That's actually pretty good, thanks.

Take a book (which is "organised") and loss-lessly compress it. The compression uses that organisation to be able to save space. The compressed file is now "unorganised".

Maybe it could be more efficient to go the other way round - instead of searching in the digits of pi kind of invert the BBP formula [1] and try to calculate the position of what you are looking for. But because the BBP formula involves rounding this is certainly not a simple task and I have really no idea if anything could be gained.

[1] http://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93...

I'm not sure the problem is invertible like that: https://researchspace.auckland.ac.nz/handle/2292/3516

There may be no shortcut.

Theoretically, you can also find all books in a random N-long number as well. Just keep on generating.

At which digit of Pi can I download the code used by the author of the blog post?

It's been argued about a lot - with far brighter people unable to come to agreement than the author of this blog.

The reason the linked article in itself is quite worthless is because it trivializes the question from philosophy of mathematics to oh pi is random let's calculate probability - but obviously that has nothing to do with the real problem.

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