Maximum amount of Hexes?

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This topic contains 7 replies, has 2 voices, and was last updated by  David Coleman 3 weeks, 3 days ago.

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  • #107 Reply


    While browsing the library I asked my self: “How many Hexes there will be if the library is complete?”

    I would calculate it myself but the nubersystem the library is using isn’t base36 or anything else so…


    #115 Reply

    Jonathan Basile

    Each hexagon contains a very small number of books relative to the total number of possibilities. A complete library would have 29^1312000 books, and each hexagon holds only 640 books. so the total number of possible hexagons would be right around 29^1311998.

    Since we use 36 characters (numbers and lower case letters) in the names of hexagons, a full library would require all possibilities of up to 1,232,835 letters and numbers to distinguish them.

    I’m actually close to completing a second version of the library which is much more expansive. It won’t have all possible books, but it will have all possible pages (29^3200), and the hexagon names have become quite a hassle. They’re reaching up to about 3200 characters now. I’ll post some more information about the new version once it’s ready.

    #130 Reply


    To relate to the story, you might make the names of hexes defined by paths, rather than individually — i.e. a hex could be known by the path it takes to get there from some other hex, with specific names for each side (which you already have). This would also add the element of distance, in that some hexes would be “harder” to reach than others…

    #131 Reply


    It would also be nice if you could browse the hexes visually, like you can browse the books. Ie. from the given hex I can go “left” or “right to reach a different hex.

    #133 Reply

    Jonathan Basile

    Hi Liam and Someone! Thank you for sharing these ideas.

    Borges’ librarian-narrator does make a somewhat cryptic reference to numbered “circuits” of hexagons (some of my thoughts about this, and the architecture in general, can be read here)- but I wouldn’t want to set these down artificially. It sounds to me as though, out of habit and familiarity, the librarians dwelling in the labyrinth-library recognize and distinguish the paths they take through the hexagons. Perhaps with time visitors to the site will do the same. And I hope they share their discoveries here.

    #1304 Reply


    Hi. So I would just like to say that there are about 10^2,000,000 books in the library and 1050 books PER hexagon (5 shelves on 6 shelves with 35 books per shelf = 5*6*35 = 1050). This means there are (10^2,000,000)/1050 amounts of hexes. Just saying.

    #1347 Reply

    Jonathan Basile

    The library described by Borges in “The Library of Babel” has hexagons with four walls of books per hexagon, and two walls that are open as passages to adjacent hexagons (I wrote more about this here). The number 35 comes from a mistranslation of the story by Andrew Hurley – each shelf has 32 books on it. So there are 640 books per hexagon.

    Borges’ library uses a 25-character set, so there are 25^1,312,000 possible books. The librarians claim these books never repeat, but of course it would be impossible for them to verify that empirically. In our library it is possible to know (based on the parameters of the algorithms being used) that there are about 10^5000 unique books, at that once they are cycled through they repeat in the same order – The Order.

    #35366 Reply

    David Coleman

    Pondering the length of the hexagon address, I came upon a confusing result.
    The total number of pages represented in the library is 29^(80 characters x 40 lines) or 29^3200
    the number of books per hexagon is 640 and number of pages per book is 410 so we have 262,400 pages per hexagon so the number of hexagons required is 29^3200 divided by 262,400. The hexagon address would then be this number represented in base 36 representing the 26 letters and 10 digits. The number of these base 36 digits is the address length, best shown in logarithms:

    (3200*log(29) – log(262400)) / log(36) = 3003.436

    So the length is 3,004 characters for the hexagon address.

    Question is: Why are the hexagon addresses 3,254 characters in length? 250 more than necessary.

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