does this library really contains all the books ?

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This topic contains 5 replies, has 1 voice, and was last updated by  Ray 1 year, 3 months ago.

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  • #9363 Reply


    hello, if there is 29^(3200*410) different books but we can browse only by chosing a 3260 character chain string, the different permutation of a unique input is 3260*4*5*32 (4 for the walls, 5 for the shelves, 32 for the tomes)

    am i right until here ?

    soo this library only contains a very small fraction of the potential books or am i missing something ?


    #10408 Reply


    The library doesn’t actually contain anything. I think all that is going on is some sort of encryption. If you look close, when you search for something the so-called “location” ends up being a pretty big chunk of random looking text. Usually around 3200 characters or so. I’m guessing this very large number representing the “Hexagon location” is actually your search string (padded with random text if less than 3200 characters) encrypted. And somehow the remainder of the location (wall, shelf, volume, page) is used as part of a key or seed for the encryption algorithm. Thus by providing all the details of the “location” in the library, you are actually plugging in your ciphertext with decryption key into the algorithm and viola! You’ve “found” your text in the library. Clever trick. I actually fell for the “library” story until I figured out there is NO WAY every bit of text ever written or ever will be is in there. The creator even admits this. This magical and mystical “algorithm” is just a simple encryption engine. Nothing more.

    #10410 Reply


    I work out the math to be:
    29^(3260) = total “chambers” multiplied by (4*5*32*410) “pages” per chamber comes to about
    2.62×10^4767 total “pages” in the library. That’s like writing out every possible permutation of a 25.6Kb encryption key. That’s why “browsing” is such a futile endeavor. This site is interesting in that it offers a totally unique viewpoint on how encryption actually works.

    #10411 Reply


    That’s 29^(3260) = 2.62×10^4767 chambers.
    Each chamber has 4*5*32 = 640 books.
    Total books in the library = 1.67×10^4770.
    410 pages per book = 6.86×10^4772 total pages in the library.

    #10412 Reply


    Another correction.
    It’s 36^(3260) = 3.52×10^5073 chambers.
    …times 640 books per chamber = 2.25×10^5076 books.
    …times 410 pages per book = 9.23×10^5078 pages.
    Gotta pay attention to the details!

    #13165 Reply


    No, not all books are contained. Every page, however, is (including duplicates, as there’s actually too much space.

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