I am new to this site and joined the community when this question caught my eye.
Did the person who solved the puzzle use computer program of some kind to deduce the answer? I don't think it is possible to come up with the answer in any other way.
$\begingroup$
$\endgroup$
2
-
$\begingroup$ Did you read the last spoiler of that answer? $\endgroup$– MordechaiMar 26, 2017 at 20:07
-
$\begingroup$ Part of solving puzzles is being attentive to EVERYTHING, since you don't know up front the value it gives. In this case, it offered up a plain explanation of how the solution was derived. TL;DR; summaries are not useful. READ EVERYTHING. $\endgroup$– wbogaczMar 26, 2017 at 20:45
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
The answer given by the person who solved the puzzle explains how they found it. I don't see any reason to disbelieve them.
There were a few hints in the question; the first indicates that the ciphertext is longer than the plaintext (suggesting looking at groups of more than one letter), the second observes that there are a lot of As and Bs (suggesting looking at where they occur), and the third suggests treating letters A..Z as corresponding to numbers 0..25 somehow.
And then, as that answer says:
All three paragraphs have length divisible by 3, suggesting that each 3-letter sequence represents a single unit.
Doing frequency analysis for all 3-letter sequences showed that the two most frequent sequences were UQB (20,16,1) and PLB (15,11,1) - both ended with B(1) and had a difference of 4 (E) between the first two letters, suggesting I was on to something.
The rest was just plugging the values generated by the above algorithm into the 3-letter sequences ending with B, getting much of the letters and easily figuring out what the rule for sequencing ending with A is.
answered Mar 26, 2017 at 20:09
-
$\begingroup$ "...each 3-letter sequence represents a single unit...." could you please explain? $\endgroup$ Mar 26, 2017 at 20:13
-
$\begingroup$ The solver noticed that the paragraphs all had numbers of letters that would let you divide them into groups of 3, and guessed that each group of 3 corresponded to a single letter in the original plaintext. $\endgroup$– Gareth McCaughan ♦Mar 26, 2017 at 20:26