Roman numerals to Voynicheesy.
1. Convert plaintext to numbers using number values of the letters[1] then summing them (some words will have the same number)
2. Express the sum of the letters in a word, in Roman numerals e.g DCCLVII, in the non-subtractive form i.e VIIII not IX
3. Using multiple rules; including spaces, substitute groups of Roman numeral letters for groups of EVA-letters e.g ('IIII' --> 'in')
Example.
Some of the substitution rules:
'_CC' --> '_qo'
'CC' --> 'ch'
'C' --> 'k'
'XX' --> 'e'
'I_' --> 'm_'
Some letter values:
t = 116
h = 104
e = 101
c = 99
a = 97
PLAINTEXT = "the cat sat on the mat"
the :: 321 :: [100, 100, 100, 10, 10, 1] 321 ['C', 'C', 'C', 'X', 'X', 'I']
cat :: 312 :: [100, 100, 100, 10, 1, 1] 312 ['C', 'C', 'C', 'X', 'I', 'I']
... etc ...
INTERMEDIATE = "CCCXXI CCCXII CCCXXVIII CCXXI CCCXXI CCCXXII"
CIPHERTEXT = "chkem qoshol qokeriin qoem qokem qokeol"
[1] Alphabetic numeral system : You are not allowed to view links.
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Numbers can be also be assigned arbitrarily or methodically e.g using a Polybius square.
For this post, i used the the number representing the unicode code of a specified character.
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This is a non-deterministic cipher, some words have the same number.
Also, i don't know (idk) if you can reverse the substitution rules for the letter-groups.
There is no way to derive the EVA-letters,
if this method was used, why, for example, substitute 'XX' for 'e', is it a stylistic choice ?
is there some underlying relation between vms glyphs, that can be seen as latin abbreviations, and the roman numerals they substitute for ?
With more work the rules could be changed, expanded upon, to render the ciphertext more like voynichese, but idk how close you could get.
Is it possible that with some small rule changes the voynichese dialects could be recreated ?
Is it possible that the substitution rules could be performed by a volvelle ?
I have not persued the idea further than this because, well, its kinda interesting but idk.
Voynichese checklist:
Entropy - check
Pseudo-repetition - check
Binomially distributed word lengths - hmm, with English plaintext its close, but using the same rules with italian plaintext , not so much.
Other - ?