[math-fun] The Library of Babel
Is the Library of Babel a sphere whose consummate center is any hexagon, and whose circumference is inaccessible.?
From Wikipedia; "Borges's narrator describes how his universe consists of an enormous expanse of interlocking hexagonal rooms, each of which contains the bare necessities for human survival—and four walls of bookshelves. Though the order and content of the books is random and apparently completely meaningless, the inhabitants believe that the books contain every possible ordering of just a few basic characters (letters, spaces and punctuation marks). Though the majority of the books in this universe are pure gibberish, the library also must contain, somewhere, every coherent book ever written, or that might ever be written, and every possible permutation or slightly erroneous version of every one of those books. The narrator notes that the library must contain all useful information, including predictions of the future, biographies of any person, and translations of every book in all languages. Conversely, for many of the texts some language could be devised that would make it readable with any of a vast number of different contents. Despite — indeed, because of — this glut of information, all books are totally useless to the reader, leaving the librarians in a state of suicidal despair. This leads some librarians to superstitions and cult-like behaviour, such as the "Purifiers", who arbitrarily destroy books they deem nonsense as they scour through the library seeking the "Crimson Hexagon" and its illustrated, magical books. Another is the belief that since all books exist in the library, somewhere one of the books must be a perfect index of the library's contents; some even believe that a messianic figure known as the "Man of the Book" has read it, and they travel through the library seeking him."
A fulltext copy of the story is here;http://www.daimi.au.dk/~doina/literature/Jorge_Luis_Borges_-_The_Library_of_...
From the third paragraph of the story: "Each book contains 410 pages; each page, 40 lines; each line, about 80 black letters." That makes 410 x 40 x 80 = 1,312,000 characters. The fifth paragraph tells us that "there are 25 orthographic symbols" including spaces and punctuation. So the library contains 25^1,312,000 books.
"Five shelves correspond to each one of the walls of each hexagon; each shelf contains thirty-two books of a uniform format;" "Twenty shelves - five long shelves per side - cover all sides except two;" So there are 640 books in a hexagon. Borges states what he calls "the classic dictum: The Library is a sphere whose consummate center is any hexagon, and whose circumference is inaccessible." However the total number of books and the hexagons are not commensurate, the classic dictum is not strictly speaking true. To show this, it suffices to divide the number of books in the library by the number of books in a hexagon, ignore the large quotient and focus on the remainder. Using mathematica (and continued fractions - could have done it without them) I got a remainder of 385, that is there is at least one hexagon with only 385 books - 12 shelves of 32 books and one book left over. Put that book on its own shelf and there are seven empty shelves left. Nice numbers The phrase "The library is a sphere whose exact center is any hexagon and whose circumference is unattainable" seems to rule out the existence of a uniquely special hexagon. But such a hexagon must exist, and it could be designated as the spacial origin ie the center. The remainder can be calculated with continued fractions. In mathematica ' ContinuedFraction[(5^2623999)/128] The result was a 182165 digit number followed by 5 terms of the continued fraction expansion of the remainder ie {305631237.... ..... 339715957641, 1, 1, 1, 1, 25} then convert the continued fraction terms into a fraction; FromContinuedFraction[{0,1, 1, 1, 1, 25}] = 77/128 the result; 77/128 (of a hexagon) 640 * 77/128 = 385 books I wrote to Bloch, William Goldbloom author of The Unimaginable Mathematics of Borges’ Library of Babel (2008) and he did the same calculation using mod arithmetic in Maple; irem(25^1312000,640); or 25&^1312000 mod 640; Both of them quickly return 385. The mod calc was straighforward, I still find it strange doing arithmetic on such large numbers, I wonder what the practical size limits are for large numbers in software like maple and mathematica? Some thoughts; It seems to me that the 385 book hexagon complicates Borges story in some interesting ways. On one hand you can say its nothing special - plenty of hexagons are missing books - from the book destruction by Purifier librarians, so what makes this hexagon unique? How do we even know that the 'missing' 255 book slots are not spread across the whole library? Just an empty slot on a shelf in 255 different disparate locations (or any possible combination) that are to all intents and purposes inaccessible. One can say Borges was not a mathematician so he did not embed the remainder hexagon in the story. He picked the dimensions of the books and hexagon shelving from the municipal library where he worked when writing the story. On the other hand the stupendous scale of the library focusses attention on the design of its essential elements ie the books, hexagons and librarians. The numerical relationship between numbers of books and numbers of hexagons is a central design element of the library, how can it not be deemed significant? Why not explore how this new information sits with the existing ideas, themes and concepts found in the text? Readers bring new interpretations and the meaning of a text shifts with time, details of a texts creation and the authors intent are not the whole story. And Borges may have known his choice of numbers would produce a remainder, without knowing what it was (though I doubt this). The Crimson Hexagon is very interesting when considered in relation to the 385 book hexagon, ".. the Crimson Hexagon --- books smaller than natural books, books omnipotent, illustrated and magical" the 385 book hexagon could segue into the Crimson Hexagon, if the books are illustrated they are not part of the 80 char 40 line 410 page book collection, and need separate shelving. If they are all magical then perhaps they could act as a catalog, providing an alternative to considering the library as its own catalogue. The notion of a first book and a last book is also interesting. The first book if done lexigraphically might be all aaaaaaaa... and the last book all spaces (or whatever the 25th symbol is). The 385 hexagon could accentuate its uniqueness by containing the first book or last book or both. Perhaps not, books seem randomly distributed around the library. There is also The Book (of the Book-Man) ""On some shelf in some hexagon, it was argued, there must exist a book that is the cipher and perfect compendium of all other books" If there are 385 books in a hexagon and they are stacked together, the 385th book has to go on its own shelf. This particular book is a good candidate for The Book, given the uniqueness. Then there is the circular book, though on reflection bringing that in to this 385 hexagon is pushing it, as even the narrator discounts the idea of a circular book. The empty space in the 385 book hexagon is also another angle. There is space for 255 more books. Books that are invisible, books that use different symbols, books that are yet to be written ... 255 is a Mersenne number, 255 is a repeating digit in base 2 .... of course there are many ways of looking at the library, Wikipedia mentions some of the philosphical implications; "There are numerous philosophical implications within the idea of the infinite library. One of these is that the librarians may have been horribly mistaken about the nature of some of the books of nonsense. Some of these are assuredly copies of other books, some written in a substitution cipher, others phonetically, some in made-up languages, etc. Every book in the library is intelligible, if one decodes it right. Any book in the library can be decoded into any other book in the library, using a third book in the library as a One-time pad. This lends itself to the philosophical idea proposed by Immanuel Kant, that by defining rules for the universe, we create rules of the universe. Because the librarians assumed that the books of nonsense were exactly that, they may have tossed away several copies of Directions to the crimson hexagon from where you are now standing, simply because it was written in a cipher. Additionally, because there are by definition all books, there are certainly also books of lies and falsehoods. For each copy of the catalog to the library, there will be many copies of false codices, claiming some false books to be true and some true books to be false. In short, any room in the library could be the crimson hexagon. Hidden in the gibberish of the library, there are works beyond human capacity to write, simply by definition that it contains all possible books, of which these are a possibility. The library cannot be damaged by the destruction of any of its books because even though a single book is unique, there are also similar books differing by a single letter. The library is a temptation, because it offers these gems of enlightenment, and buries them in deception. One can consider any text as being pulled from the library by the act of the author defining the search letter by letter until they reach a text close enough to the one they intended to write. The text already existed theoretically, but had to be found by the act of the author's imagination.[7] Another implication is an argument against certain proofs of the existence of God, as it is carried out by David Hume using the thought experiment of a similar library of books generated not by human mind, but by nature. [8] Stuart Anderson www.squaring.net
On Tue, Mar 13, 2012 at 6:19 AM, Stuart Anderson <stuart.errol.anderson@gmail.com> wrote:
Is the Library of Babel a sphere whose consummate center is any hexagon, and whose circumference is inaccessible.?
Using mathematica (and continued fractions - could have done it without them) I got a remainder of 385, that is there is at least one hexagon with only 385 books - 12 shelves of 32 books and one book left over. Put that book on its own shelf and there are seven empty shelves left. Nice numbers
The remainder can be calculated with continued fractions. In mathematica ' ContinuedFraction[(5^2623999)/128] The result was a 182165 digit number followed by 5 terms of the continued fraction expansion of the remainder ie {305631237.... ..... 339715957641, 1, 1, 1, 1, 25}
then convert the continued fraction terms into a fraction; FromContinuedFraction[{0,1, 1, 1, 1, 25}] = 77/128 the result; 77/128 (of a hexagon) 640 * 77/128 = 385 books
I wrote to Bloch, William Goldbloom author of The Unimaginable Mathematics of Borges’ Library of Babel (2008) and he did the same calculation using mod arithmetic in Maple; irem(25^1312000,640); or 25&^1312000 mod 640;
Both of them quickly return 385.
Maple? Mathematica? Continued Fractions? This is something you can calculate in your head in under 2 minutes. Use the Chinese Remainder Theorem. 25^1312000 = 0 mod 5. 25^1312000 = 1 mod 128, since 1312000 is a multiple of 32. (you might think that the exponent of (Z/Z128)* was 64, since it has 64 elements. But since (Z/8Z)* has exponent 2 (3^2 = 9 = 1 mod 8 is the only case you need to check, since the other elements are 1, -1, and -3), (Z/128Z) has exponent 32. (To see that 1312000 is a multiple of 32, you just need to verify that 1312 is a multiple of 4, which follows from the fact that 12 and 100 are multiples of 4). So to compute 25^1312000 mod 640, we just need to find a number mod 640 that is 0 mod 5 and 1 mod 128. 128 = 3 mod 5, and 3 * 3 + 1 = 10 = 0 mod 5, so the solution is (128 * 3) + 1 = 385. Andy
* Andy Latto <andy.latto@pobox.com> [Mar 13. 2012 19:47]:
[...]
Maple? Mathematica? Continued Fractions? This is something you can calculate in your head in under 2 minutes. Use the Chinese Remainder Theorem.
25^1312000 = 0 mod 5.
25^1312000 = 1 mod 128, since 1312000 is a multiple of 32. (you might think that the exponent of (Z/Z128)* was 64, since it has 64 elements. But since (Z/8Z)* has exponent 2 (3^2 = 9 = 1 mod 8 is the only case you need to check, since the other elements are 1, -1, and -3), (Z/128Z) has exponent 32.
(To see that 1312000 is a multiple of 32, you just need to verify that 1312 is a multiple of 4, which follows from the fact that 12 and 100 are multiples of 4).
So to compute 25^1312000 mod 640, we just need to find a number mod 640 that is 0 mod 5 and 1 mod 128. 128 = 3 mod 5, and 3 * 3 + 1 = 10 = 0 mod 5, so the solution is (128 * 3) + 1 = 385.
Andy
And the winner is: brain.
participants (3)
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Andy Latto -
Joerg Arndt -
Stuart Anderson