got a reply from a nice Maths chap
My gut feeling tells me it is a formal cipher: which means that they've
removed the spaces from an encrypted text and artificially grouped it into
blocks of four. However, I can't tell if it's a monoalphabetic substitution.
I've run a frequency count for you and got the following results:
Single letter counts
13: a k
12: j s
11: f n
10:
9: c e
8: m r
7: h i l o
6: d v
5: q t u w y
4: p
3: x z
2: g
1: b
Digraphs (double letters)
6: vn
5: kj
4: ir mc
3: ak js sd ls am na jf oq eu po oi
2: fk se fa aj jk ca al ry iw kd fo qi we cx xz re ut
Trigraphs (three letters)
2: kjs jsd faj als jfo foq oqi cxz poi oir ire reu eut kjf vna
You can see frequency lists in my FAQ:
letter frequency FAQ
However, I doubtful looking at this if it is as simple as a substitution
cipher; I would certainly expect "the" and "th" to appear much more often in
such a long piece of text, and as no trigraph appears more than twice it
makes me doubtful.
If I get a chance I will try out other things. I can tell you that I've
tried things like: adding the alphabetical position of each letter in the
blocks of four and converting this to a letter. For example, akjs:
1+11+10+19=41, then reduce to 1-26: 41-26=15, which is the letter O. But
nothing has worked! )c:
I still suspect it is some type of substitution, but with some clever twist.
It could, of course, be an elaborate hoax.