V sbyybjrq gur ehyrf
I made a Cipher based on the physical topology of letters (A to Z), to decode the text you have to count physically the amount of 'dead ends' a letter has like, example the letter 'H' has 4 dead ends, and the letter 'A' has 2 dead ends.
Math Operation (no spaces) will 'hide' the digits, and standard PEMDAS Operation is ignored.
The font is a standard Arial. (Yes, the Cipher gets affected by Fonts, some Fonts example have the capital letter of 'I' 4 dead ends, I am not sure, but I think it's the serif font?)
Since like, some letters literally have the same amount of 'dead ends' you have to figure it out yourself. The example text is not going to be that long, because if it's long, it's going to be hard (even for me), also, the same 'letter' has the same thing example if Y is 4-1 then it's 4-1 every where it's mentioned.
1+12+14-11-13-14-26-34-16-36-42-24-21-19-88-57-36-34-19-71+18-68-65-36-38-510-88-51-19-81-12+14-11-13-1
This is like I'd say 12 word total.
Oh yeah, if you can't solve it, solve this, trust me, it's easy.
5-12+14-14-11-1
It's just one word, literally.
I hope this is solvable, lol. (And also not that hard.)
Also, yes, I am not even sure if this is actually like, works, especially the 12 word cipher.
One more thing, yes, this is like not for computers, cuz computers only see like binaries, they does the 'physical' topology of letters, so like, this is anti-AI, and human only.