(22-05-2025, 08:17 AM)kckluge wrote: You are not allowed to view links. Register or Login to view.The question wasn't whether Brumbaugh was wrong -- it's clear he was -- it was a specific technical question about how to compute unicity distance for such a cipher. Even with multiple options for individual words, there almost certainly has to be some text length such that only one set of possible options for the words forms a coherent text.
(Edit: the below is wrong, I'll add a correction in a later post.)
Given there is no answer so far, I'd say that one can just naively apply the Shannon formula, as shown in the Wikipedia article. That's what I would usually do when I have no idea what I'm doing. So,
U = Hk/D
Here Hk is the key entropy. Given no specific key constraints, this is the binary logarithm of the number of ways to group Latin alphabet (+ one extra sequence like "us" to make it split evenly) into sets of three. D is the information density in bits per character. English, as per the same article, has about 1.5 bits per character, but we reduce the source set by the factor of 3 by merging letters into groups. Without much thinking (I'm not sure I can get to the exact solution here is reasonable time
), it's likely this should increase the unicity distance, so let's just divide by 3 and say we have 0.5 bits per character.Hk for splitting 27 items into 9 labelled groups of 3 items is (I delegated this to Claude, so won't vouch for this): lb( (27! / 3!) ** 9 ), which is roughly 70.
U = 70 / 0.5 = 140
So, you need about 140 characters of the ciphertext to identify the key for an English plaintext using this encoding scheme. I'd say this looks reasonable.