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VMs: VMS: the "Mario cipher" again. Last of it I hope.



Unexpected results coming up...

I have already called "Mario" that Papal cipher that uses
a 5x5 grid to produce a stream of digits. Let me distinguish
between that "Mario proper" and my "20-letter Mario" by
calling the original one "Mario-10" and mine "Mario-20".

I took my THES book review and enciphered it into both
(I reduced the English alphabet to 25 by replacing every 
'z' with 's'):

   Plaintext    Mario-10    Mario-20
h0  4.70044     3.32193     4.32193
h1  4.15998     3.21802     4.21734
h2  3.53698     2.38840     2.28870
h3  2.65030     2.08113     1.98587

Next, I took Galileo's dedication to Cosimo de Medici of his
Sidereus Nuncius. Here is an excerpt:

Praeclarvm sane atqve hvmanitatis plenvm eorvm fvit 
institvtvm, qvi excellentivm virtvte virorvm res 
praeclare gestas ab invidia tvtari, eorvmqve immortalitate 
digna nomina ab oblivione atqve interitv vindicare,
conati svnt.

Here are the results:

   Plaintext    Mario-10    Mario-20
h0  4.45943     3.32193     4.32193
h1  3.91803     3.25482     4.25126
h2  3.37095     2.26053     2.15267
h3  2.67525     1.93378     1.84085

We have there very Voynich-like figures, except for
h1 which, in both cases, is very close to h0.

The surprising thing is how Mario-20 _increases_ h1.
Well, with hindsight and thinking about it a bit,
it is not so surprising after all.

Now, let me imagine myself the author of the VMS.

I want to use a "Mario" cipher, but I certainly do
not want to look up a 5x5 grid for every letter 
I encipher. The easy solution is to draw up
a single-substitution cipher consisting entirely
of digraphs, the first member of which is drawn
from a subset of the alphabet, the second from
another subset of the alphabet, the two having
no intersection, e.g.:

a   cf
b   lo
c   ca
d   ci
e   cm
f   li
g   cj

and so on. Call the first letter of each
digraph its "initial", the second its "final"

Once I have memorized this alphabet (not difficult),
I can directly  "write in Mario". Example:

Plaintext: dad

1. d => ci
2. a => cf oh! same initial, omit it: => f
3. d => ci     same initial, omit it: => i

cipher: cifi

Plaintext: deaf

1. d => ci
2. e => cm... oh! same initial, so omit it: => m
3. a => cf        same initial, so omit it: => f
4. l => li  new initial, so keep it: => li

cipher: cimfli


Using this system two sets of five letters each are
sufficient, but I  my cipher to be more secure,
without complicating the encipherment process.

Easy: I assign two shapes to each letter, one
"straight", the other "curly", and I make up
this rule: straight after straight, curly after
curly. So for instance:

"straight" version:  iiiv
"curly" version:     cccu

Now doesn't that remind you of Voynichese?

I can break the rule any time I like: that 
will not impair the decipherment.

I can insert nulls any time I like by writing
several initials in succession: only the last
one is functional.

Finally, there is no compelling reason for my
diagraphs to fit in a 5x5 grid, nor for each 
and every letter to have two forms. As long
as the set of initials does not intersect
the set of finals, all is well: the cipher works.

So, we have here a cipher which reduces the
entropy of the plaintext and exhibits the
same "shape harmony" as the VMS.

Do I believe this to be the solution?

Yes and no.

1. Yes, because it fits nicely.

2. No, because it is very easy to crack and so,
if it was the solution, the VMS would have been 
deciphered long ago. 

Twenty years ago I found a solution to the 
4-colour problem.  Very simple and elegant. 
But I never bothered because I reasoned that, 
if it was a valid solution, it would have been 
discovered long ago. So there just had to be 
a flaw somewhere in my proof.

In the same manner, I cannot bring myself to 
seriously believe that the "Extended Mario Cipher" 
is the solution to the VMS. It might be, but
I doubt it.



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