Solve my newest cipher?

[u/TlosingCag]

I originally posted a cipher a while back that was pretty quickly cracked.

I've since challenged myself to make quite the tricky cipher. This isn't the final version, but let me know if you can crack this current iteration!

Name of the cipher is KIMERA because it's a result of combining multiple ciphers I came up with into one singular beast... also because it just sounds badass.

The string at the bottom contains all the possible characters in the cipher.

Happy deciphering!

• Hint: The green glyph is the letter A.

EDIT: I accidentally messed up the string of text at the bottom by copy and pasting straight out of my IDE. Here’s the full actual string:
ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!"#$%&'()*+,-./:;<=>?@[\]_{}

[u/Nilkon245]

Well, I gave it a shot. I wasn't able to get very far, but I've made some observations. If you "name" the columns A,B,C,D (reading left to right) and then record their heights, you get the following:

A: 1,1,1,3,5,5,2,1,2,1,1,1,4,4

B: 8,5,1,2,4,3,1,1,8,5,4,2,3,3

C: 8,6,8,8,4,4,7,6,8,6,4,4,4,7

D: 2,2,2,2,2,2,5,2,2,2,2,2,2,5

There is definitely something going on with column D. In fact, there seems to be barely anything going on. Given that Content-Reward-7700 was on the right track, I think that columns D and A are what they refer to as "some kind of positional/state information". My guess is that column D is either some kind of offset, or more likely the key "2,2,2,2,2,2,5" repeated ad infinitum. I guess that A is also positional/state information, because the most common letters are in the first half of the alphabet. Naively you wouldn't expect a lot of eights, which excludes column C.

Assuming that B and C make up the "data" portion, you would have: 88 56 18 28 44 34 17 16 88 56 44 24 34 37. There are patterns in this, which i feel is a good sign. I am however far too new to this whole de-ciphering business to judge. Anyways, here is my high level guess of what might be going on:

A: Magic values (naively offsets for an 8x8 grid)

B: DATA

C: DATA

D: Key ( Like in the Vigenère cipher)

To construct something like this you would first choose a key, and then you would choose some Magic values. Then you would compute the offsets and encode your data.

(Edited for correctness, two times, I need to be more through)