# UPDATED

Until more hints are provided, the only thing I can really supply is a dump of all of the cyclic rotations of all of the co-prime sampling intervals of the ciphertext (and, separately, the plaintext). Each are `123*80 = 9840` lines. I hope that is useful to someone. I wrote the code in Haskell. The files are each 1MB, so I am splitting them up into several parts to fit it all on pastebin.

### Cipher

Part 1: http://pastebin.com/mD2DuL5c

Part 2: http://pastebin.com/BPCNWwa3

Part 3: http://pastebin.com/XW61r97s

### Plain

Part 1: http://pastebin.com/E2ef2bRL

Part 2: http://pastebin.com/mPhx0iLC

Part 3: http://pastebin.com/TCKhwtn6

# EDIT

I have made no changes to the dumps above, but I just thought a bit more about the nature of this problem and came to a number of impressions/conclusions, which are not exactly rigorous but have some basis.

From the information provided, both in the post and in its comments, I have arrived at several conclusions:

## Automorphism

Because the decryption does not rely on the orientation of the cyclical ciphertext or the starting point, we can consider this problem from a modular-arithmetic standpoint. We also know that all of the characters in the plaintext $p$ are preserved in the ciphertext $c$, merely subjected to a rearrangement. We can therefore treat the cipher as an automorphism, because it is a mapping of $\mathbb{N}_m \rightarrow \mathbb{N}_m$, where $m$ is the length of the message. We will therefore consider $E$ to be the encryption function, and $D$ to be the decryption function, where $D(E(p)) = p$. We also know that, for an arbitrary cyclic shift $C$, we have $E(C(p)) = C(E(p))$ and $D(C(c)) = C(D(c))$. We thus conclude that $E$ and $D$ are pure, stateless functions, meaning that they produce the same result for the same output independent of any history. We also know that $E$ and $D$ depend only on the positions of letters, and not on their values. For,if this were the case, the size of the domain of $E$ would be $m*L$, where the alphabet of the messages is of size $L$, while the range would still be $m$, as the letters themselves are merely transposed and not transformed; by the Pigeon-Hole principle, there must therefore be multiple index-value pairs that map to the same index of the ciphertext, which we know not to be the case. We therefore consider $e$ instead of $E$, which is the function mapping an index in plaintext to the ciphertext index when $E$ is applied, which we state as $$p_i = c_{e(i)}\;\forall\;1 \le i \le m$$

We therefore only need to determine the mapping of $e$, which we know must be linear if $E(C(P)) = C(E(P)) \implies e(i+x) = e(i)+x$. After I create a table of the positions of each letter in ciphertext and plaintext, I will post the results of that.

# Results

(I know someone else has posted these, but this is in a slightly different form, and it doesn't hurt)


    [3,7,12,17,29,34,37,44,48,51,56,60,65,69,73,84,88,95,100,104,108,115,118] -> [3,11,12,13,21,24,27,29,44,47,53,64,69,72,73,75,85,87,91,99,113,119,121]
    [43] -> [2]
    [99] -> [59]
    [28,122] -> [82,107]
    [5,10,32,41,63,71,81,102] -> [1,22,28,43,65,86,90,118]
    [45] -> [10]
    [4,18,70,89,101] -> [33,51,70,89,100]
    [11,52,76,83] -> [14,39,45,94]
    [9,22,25,40,54,62,77,87,93,97,114] -> [7,16,26,40,49,61,66,79,98,108,115]
    [109] -> [74]
    [38,111] -> [9,78]
    [14,21,31,86,92] -> [0,54,63,102,114]
    [15,19,35,49,90,110,116] -> [5,20,60,81,84,106,112]
    [61] -> [37]
    [6,57,64,72,75,82,103] -> [31,38,46,52,58,77,103]
    [1,53,58,67,106,119] -> [30,36,41,83,93,101]
    [20,91] -> [67,116]
    [8,23,39,78,94,113] -> [19,55,71,88,110,122]
    [16,36,55,79] -> [17,35,57,117]
    [13,24,27,30,33,42,47,50,59,80,85,98,117,121] -> [4,6,15,18,23,32,34,42,62,68,76,96,104,120]
    [2,46,68,74,107,112,120] -> [25,48,56,92,95,105,109]
    [26] -> [8]
    [0,66,96,105] -> [50,80,97,111]