Tuesday, July 26, 2011

Some Mathematical Assistance, Please!

If there's anyone with the basic skills to help me out here, I'd much appreciate it. The rest can just enjoy the dizzying infinites implied more intuitively. We are, of course, talking about Borges.

I've been thinking about his Library of Babel again. Recall that each book has 410 pages and that every possible book is in the libary. That means that not only is every possible page (every combination of 40 lines of roughly 80 characters) but every possible combination of pages in the libary. Just as two books may differ in only a single A being a B on a particular page, so two books may differ only in page 7 being after page 8 rather than before it in one of them (I take it even the numbering may wrong in one book and right in another). That much is easy to understand.

But there is also another small possible difference between books that the library realizes. Two books may be identical except that one of them has its own title while the other is, say, the second of a four-volume set, while yet another version of the very same book may be the seventh of eighteen-volume set, etc., etc. The difference may appear only on the spine and the title page of the book. Or only on the spine. And two books may differ only by one of them containing the typographical error of being called "Volume 3" on the spine and "Volume 2" on the title page. So, while we can stick to the dogma that the library contains a finite number of books (all the possible combinations), it quickly becomes an enormous number. By putting some limits on the amount of writing on the spine (Borges doesn't say) we can calculate exactly how many books there will be.

That's not the math question I'm posing, but feel free to answer it if you want. I'm thinking of something Willard Quine once pointed out. Since all possible books in all possible sequences are part of the defintition of the library. The number of books becomes dizzying. But this, Quine notes, is also true of the amount of pages. The problem would arise even the pages were, not 410 pages long, but, say, 205, or 80, or 40, or even 1. We'd still need to arrange those pages in every possible order.

So Quine follows the logic out. What if we only had every possible line in the library? What if we only had every possile letter? What if we only had a 0 and a 1. This effectively erases the problem that the Library poses, doesn't it? The library is as meaningless an image as a zero and a one one two separate pieces of paper. The thought experiment is simply: arrange them in every possible way but under a particular set of contraints. And these constraints are simply arbitrary. "Every possible book" might just be two pieces of paper, one black, one white, until answer questions like: how many lines per page, how many characters per line, how many pages per book? And if we don't answer those questions at all, "every possible book" is an infinite among since there is an indefinite number of pages to work with, i.e., an infinite one.

Anyway, here's the math question. If we bring the library down from every possible book of 410 pages to every possible page, how much smaller would it be? I've been thinking 410410 times smaller. But is that right?


Michael Peverett said...

I'm not a mathematician, but I don't think your figure is right. All the differences you mention come down to being differences of one or more characters. (As opposed to differences in material, ink, font, temperature, location...) So in principle they are numberable, because a character can be treated as a unit. But the page number (410) has no intrinsic significance in this calculation, it is only the number of characters that is significant. The increase in possible books generated by increase in book size would be C to the power of N , where N = the size increase, measured in characters, and C = the number of possible characters in the character set.
e.g, if the size of the book increased from one page to one-page-plus-two-characters, and if the character set was (absurdly, but for the sake of argument) limited to A-Z and 0-9, then the total number of possible books would be 36 to the power of two (= 1,296) times more than whatever the previous total was.

Thomas said...

As I understand it, the amount of possible different pages is fixed. 40 lines. 80 characters per line. 25 different characters. Each space on the page might also be left blank. That is why I think a 1-page book stands in some definite relation to a 410 page book. But it does not seem to me that the 1 page book only contains 410 times fewer possibilities (i.e., that the 410 page book contains 410 times the possibilities of the 1 page book.) this is because it's not just that there are 409 more pages that each have to realize the possibilities that the first must. They can also be put in any order.

Michael Peverett said...

OK, I'll do some sums, based on the figures you just quoted.
Characters per page = 3,200.
410 pages = 1,312,000 characters.
Total number of books in the library, given a character set of 26, is thus 26 to the power of 1,312,000 (according to my formula "C to the power of N").
Which is (more or less) 410 to the power of 700,000.
It's a long time since I read the story, so I might have forgotten some constraints. As for the mind-bending question of page order, it seems best to treat this (mathematically) as transformations in character-content rather than page-sequence. i.e. the first page is always considered to be the first page, regardless of what page it claims to be or whether the same text had, in some other book, appeared on page 252. Well, back to work.
Thanks for the entertainment!

Thomas said...

Wouldn't the problem become much more manageable if we imagined different constraints on the book. Keep in mind that I only want to know how much smaller a library of all pages would be than a library of all possible 410-page books (regardless of how big the latter might be.)

Imagine much less astronomical constraints: instead of 40x80 characters, how about just 1x1, i.e. 1 character per page. And instead 26 different symbols, just 0 and 1 (with no blank pages).

It should be easy to figure out how many 410-page books it would take to realize all the options. But the library of one-page books is simpler still: there would be only two (one with a '0' on it and one with with a '1').

I guess my question is whether (or my assumption is that) the relative size of the library of single pages and the library of n-page books can be captured in a simple function of n, regardless of the informational complexity of each page.