does this library really contains all the books ?
This topic contains 5 replies, has 1 voice, and was last updated by Ray 3 weeks, 5 days ago.
October 20, 2016 at 11:03 pm #9363
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 ?
RegardsDecember 12, 2016 at 4:08 pm #10408
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.December 12, 2016 at 4:39 pm #10410
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.December 12, 2016 at 4:51 pm #10411
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.December 12, 2016 at 5:27 pm #10412
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!March 1, 2017 at 8:22 pm #13165
No, not all books are contained. Every page, however, is (including duplicates, as there’s actually too much space.