Uranium235
Auto-translated
Okay, this cipher is a bit more complex.
We assume that 1 symbol corresponds to 1 letter, especially since we already have spaces and punctuation marks. Well, almost – the last word is quite long, so it might be several words.
The most interesting part is the sequence "77a." There aren't many three-letter words where the first two letters are the same, and it comes after a comma.
Mostly, these are abbreviations like ООН, ФФА. The dictionary also says that there exist ААВ, ААЛ, ААХ, but I doubt they are used here.
Another possibility is that a space is also a symbol, not just a word separator (and the comma too, which would make things even more complicated).
Anyway, I haven't made any progress with this approach.
Another approach, remembering "The Gold Bug," is to count the number of symbols – the more frequently used letters should appear more often, give or take. Let's use Excel. Hmm, a simple count of the symbols shows that there are over 50, so the theory that each symbol is a letter is unlikely.
But I'll still calculate the number of repetitions, since I've already started.
7 4
Q 4
W 4
е 4
3 3
4 3
# 3
R 3
S 3
Z 3
р 3
, 2
0 2
1 2
2 2
@ 2
$ 2
F 2
N 2
Δ 2
Θ 2
Ω 2
а 2
б 2
в 2
з 2
к 2
о 2
т 2
5 1
6 1
8 1
9 1
^ 1
. 1
& 1
D 1
G 1
I 1
L 1
№ 1
U 1
V 1
Y 1
Σ 1
Ψ 1
г 1
д 1
и 1
н 1
Ө 1
у 1
Ӌ 1
ы 1
Maybe there are uppercase and lowercase letters, but I'm not sure I can distinguish them at this point. So, let's try another approach – for example, by analogy, but I can't figure out the system yet. For example, if we understand "6" as an analogue of "b," then why do we need "b" itself?
Or maybe one letter is encoded with one or more symbols, in which case counting the number of symbols doesn't really help.
Of course, it could also be a simple transition from one encoding to another, but again, there are over 50 symbols, and they don't repeat very often.
The combination of symbols Ω3бS#еΘ@з repeats in positions 1 and 3 (if we consider the space as a word separator); this is clearly not a coincidence.
We assume that 1 symbol corresponds to 1 letter, especially since we already have spaces and punctuation marks. Well, almost – the last word is quite long, so it might be several words.
The most interesting part is the sequence "77a." There aren't many three-letter words where the first two letters are the same, and it comes after a comma.
Mostly, these are abbreviations like ООН, ФФА. The dictionary also says that there exist ААВ, ААЛ, ААХ, but I doubt they are used here.
Another possibility is that a space is also a symbol, not just a word separator (and the comma too, which would make things even more complicated).
Anyway, I haven't made any progress with this approach.
Another approach, remembering "The Gold Bug," is to count the number of symbols – the more frequently used letters should appear more often, give or take. Let's use Excel. Hmm, a simple count of the symbols shows that there are over 50, so the theory that each symbol is a letter is unlikely.
But I'll still calculate the number of repetitions, since I've already started.
7 4
Q 4
W 4
е 4
3 3
4 3
# 3
R 3
S 3
Z 3
р 3
, 2
0 2
1 2
2 2
@ 2
$ 2
F 2
N 2
Δ 2
Θ 2
Ω 2
а 2
б 2
в 2
з 2
к 2
о 2
т 2
5 1
6 1
8 1
9 1
^ 1
. 1
& 1
D 1
G 1
I 1
L 1
№ 1
U 1
V 1
Y 1
Σ 1
Ψ 1
г 1
д 1
и 1
н 1
Ө 1
у 1
Ӌ 1
ы 1
Maybe there are uppercase and lowercase letters, but I'm not sure I can distinguish them at this point. So, let's try another approach – for example, by analogy, but I can't figure out the system yet. For example, if we understand "6" as an analogue of "b," then why do we need "b" itself?
Or maybe one letter is encoded with one or more symbols, in which case counting the number of symbols doesn't really help.
Of course, it could also be a simple transition from one encoding to another, but again, there are over 50 symbols, and they don't repeat very often.
The combination of symbols Ω3бS#еΘ@з repeats in positions 1 and 3 (if we consider the space as a word separator); this is clearly not a coincidence.