I had a message early in the week from Janos with another idea about decoding the Riddle.
His first message was terse and a little cryptic:
The 5 sentences encode 5 hex digits that have to be XORed (or maybe used in another operation) with the URLs digits. The 10 nobelties refer to the 10 digits of the first 5 bytes of the URL, i.e. to 83.5c.0f.80/2b. This means that the last two digits (2b) should be XORed with same number as the 5th sentence has 2 ranks in it (lowest ones btw). And how does a user's URL usually look like? Yes, it starts with a '~'. and guess what, 2b XOR 7e = 55!
i.e. my theory works for the last sentence. So, what we need to find out is what other digits the other sentences encode. In fact, at this point you could run a brute force search for all possible combinations, which is not that hard, since we can have at most 16k (=16^4) combinations.
On the other hand, we could try to find out how the 5th sentence leads to the number 5 as this would suggest us what numbers the sentences should give out.
As I understood him, this is the process he proposes:
There are ten "users" (nobles etc - the fifth column in the matrix), and, if you take the four URL digit groups in hex, plus the "+" from the final part of the URL taken as the ASCI number, you get a string of ten digits (83 5C 0F 80 2B) .
In a second message, Janos explained that there is a one-to-one correspondence between each nybble and each noble or user. (Nybble and Noble sound a little similar: hmmm...) Thus, the first "8" corresponds to "royalty", the "3" corresponds to "duke", the "5" to "marquis" and so on.
Next, as I undertsood him, the example he uses is the final hex digit, 2B, that corresponds to the two users in the fifth line, the squire and the gentleman.
The process of decoding, at least for this hex digit, is to XOR it with the number "5" twice, as there are two users and they are in the fifth sentence. Thus you get 2B XOR 55 = 7E , which is the ASCII code for " ~ ", a very plausible start to any URL username. Neat.
I wasn't sure whether there should be another way to get the digit "5" twice, other than from the fact we are dealing with the fifth sentence, and there are two users in it.
Janos explained in his second message that there is a precise mapping between the ranks and the digits. The sentance that has the rank in it should be applied to the digit that this rank refers to. Since, he says, the fifth sentence has the two lowest ranks in it, we have to "apply" (XOR, in this case) the number determined by this sentence to the last two digits of the ten, ie 2B.
He then went on to discuss two further questions:
1. How does the fifth sentence determine the number 5?
2. What numbers do the other sentences determine? (most likely not 5)
My final question to him was what to do with the remaining "ORC" part of the URL. He feels that this should be left the same, so that the final part of the URL would be "~ORC". Which looks natural.
So the exercise becomes one of determining the "numbers" to apply to each hex nybble in the ten digit sequence, and how to apply it to the hex bytes. Presumably XOR is the preferred process.
If we take the sentence number as providing the source of the informatin to apply to the hex byte, and recognising that the third sentence supplies three numbers (3, 3, 3) , while the first supplies only one (1) , which is the simplest possible solution to this process, we get the following number sequence from the sentences: (1,2,2,3,3,3,4,4,5,5) , which, if we combine the nybbles into into hex bytes, gives us a sequence (12, 23, 33, 44, 55) .
If we XOR these with the corresponding hex bytes in the URL, we get:12 XOR 83= 91
or a sequence (in decimal) 145 127 60 196 ~
This suggests the URL 18.104.22.168/~ORC according to Janos' ideas, and using the simplest numbers from the Riddle sentences, and the XOR process. (No, the URL doesn't answer).
This is a very ingenious hypothesis, founded mainly on what seems like a reasonable idea, and one that generates a plausible " ~ " in the user address.
Worth further exploration, I think, though some might argue it doesn't quite pass the "simplicity" test. Interesting , nonetheless.