Strong's "peculiar system of a double reversed arithmetic progression of
a multiple alphabet" is a puzzling description, but GC recently
(Feb 2010) explained it as ""double reversed
arithmetic progression" as defined by the string
1-3-5-7-9-7-5-3-1-4-7-4" (although I think the sequence given is an
example, rather than the definition). The number of alphabets is "a
handful".
If we suppose that the cipher is indeed constructed like this, then can
we crack it computationally?
First we need to make some assumptions. Let's generously assume that the
number of alphabets is 10. Let's then assume that these alphabets are
rotated through in a sequence that is 17 long (the number 17 is picked
since it crops up as a feature of the VMs text in many places). Let's
not assume that the sequence is double, or reversed, or anything else:
it's just a sequence of alphabet numbers. Let's assume that each alphabet
contains 21 characters: abcdefghilmnopqrstuvx
We then take a sample of VMs text (I chose the first "paragraph" of f1v)
h1s9 1o8am oe oek1c9 1ay Fax ap 9kcc9 1ay oy o19 81o eho89 oho8ay 1o89 8o H9 HoH9 29 8h2ii9 K9 hok1o89 8ae 8oe 1ohco 8aiy 8ap so1c9 1o ho89
and, equipped with a large dictionary of Latin words, we start to build
a possible cipher. To do this, we start by looking at the first VMs word
"h1s9", and pick a Latin word of the same length, at random: "acri". With
this pair we can start to construct the cipher table:
Voynich o 9 e 1 8 a h y c k 2 i K s m H p F x g & Alphabet 0 . . . . . . a . . . . . . . . . . . . . . Alphabet 1 . . . c . . . . . . . . . . . . . . . . . Alphabet 2 . . . . . . . . . . . . . r . . . . . . . Alphabet 3 . i . . . . . . . . . . . . . . . . . . . Alphabet 4 . . . . . . . . . . . . . . . . . . . . . Alphabet 5 . . . . . . . . . . . . . . . . . . . . . Alphabet 6 . . . . . . . . . . . . . . . . . . . . . Alphabet 7 . . . . . . . . . . . . . . . . . . . . . Alphabet 8 . . . . . . . . . . . . . . . . . . . . . Alphabet 9 . . . . . . . . . . . . . . . . . . . . .
We continue with the next word: "1o8am" and a random Latin word of the
same length: "paveo", and update the table:
Voynich o 9 e 1 8 a h y c k 2 i K s m H p F x g & Alphabet 0 . . . . . . a . . . . . . . . . . . . . . Alphabet 1 . . . c . . . . . . . . . . . . . . . . . Alphabet 2 . . . . . . . . . . . . . r . . . . . . . Alphabet 3 . i . . . . . . . . . . . . . . . . . . . Alphabet 4 . . . p . . . . . . . . . . . . . . . . . Alphabet 5 a . . . . . . . . . . . . . . . . . . . . Alphabet 6 . . . . v . . . . . . . . . . . . . . . . Alphabet 7 . . . . . e . . . . . . . . . . . . . . . Alphabet 8 . . . . . . . . . . . . . . o . . . . . . Alphabet 9 . . . . . . . . . . . . . . . . . . . . .
The next word is "eo" and the random Latin word is "do". Now
the Latin letter "o" has to be placed under the Voynich "o"
column in Alphabet 0:
Voynich o 9 e 1 8 a h y c k 2 i K s m H p F x g & Alphabet 0 . . o . . . a . . . . . . . . . . . . . . Alphabet 1 . . . c . . . . . . . . . . . . . . . . . Alphabet 2 . . . . . . . . . . . . . r . . . . . . . Alphabet 3 . i . . . . . . . . . . . . . . . . . . . Alphabet 4 . . . p . . . . . . . . . . . . . . . . . Alphabet 5 a . . . . . . . . . . . . . . . . . . . . Alphabet 6 . . . . v . . . . . . . . . . . . . . . . Alphabet 7 . . . . . e . . . . . . . . . . . . . . . Alphabet 8 . . . . . . . . . . . . . . o . . . . . . Alphabet 9 d . . . . . . . . . . . . . . . . . . . .
We continue in this vein, picking random Latin words to match
the VMs words, and attempting to place them into the cipher.
This starts off easily, but rapidly becomes impossible, with
the Latin words chosen: when we come to place a letter
into the required column at the current alphabet in the sequence,
we find that the position is already occupied by a different
letter, or that the alphabet already contains that letter but
in a different column.
In such cases we try to select a different Latin word to see if it
will fit. If we exhaust all possible Latin words, then we backtrack
to the beginning, and start afresh with a new sequence and new
choices.
Most of the time, this algorithm doesn't get further than a few
words into the text before failing. Occasionally it gets quite
a long way. Of course, the search space of possible Latin word
combinations is staggering ...
This is one of the more interesting attempts at deciphering f2v:
Voynich o 9 e 1 8 a h y c k 2 i K s m H p F x g & Alphabet 0 l . t a s . b o f . n . . . . . . . . . . Alphabet 1 m i . o l b . . r . . t . . . g . . . . . Alphabet 2 f a u e . . . s . . . i . n . . . . . . . Alphabet 3 . o . v f . i . . n u . . . . . . e . . . Alphabet 4 . a . h d i . . . . . . o . . . . . . . . Alphabet 5 e s . i t o . . . . . . . . . . . . a . . Alphabet 6 u . . . r s c . . . . . . . . . . . . . . Alphabet 7 u . i . n b . s r . . . . . . m t . . . . Alphabet 8 e i . . . d u . . c . . . . a . . . . . . Alphabet 9 s . . u . . . . . n . . . . . a . . . . . Sequence vals = 0 1 2 3 4 5 6 7 8 9 0 1 2 3 5 7 8
h1s9 1o8am oe oek1c9 1ay Fax ap 9kcc9 1ay oy o19 81o eho89 oho8ay 1o89 8o H9 HoH9 29 8h2ii9 K9 hok1o89 8ae 8oe 1ohco 8aiy 8ap so1c9 1o ho89
bono herba st muniri abs eia st infra vos eo meo diu iussi fiendo offa tu mi alga us nuntio os cuculla