Hello Simon

(What's happened to everyone else?)

Well after reading the link you sent and breaking it down to a simple form
and from what I understand:


For any cell in the matrix. I'm not going to go into all the specific
indexing of each cell, as it is the same for each anyway.

P1 = plaintext message 1
P2 = plaintext message 2
P3 = plaintext message 3

C x P1 = e1

C x P2 = e2

C x P3 = e3


As P1,2 and 3 are not the same value, there is going to be a problem reverse
calculating the C

Really how I see this working is like that code which uses the 26 x 26
square of alphabet subsitution (cipher (some french word)).

You would have to look for the Crib and determine one of the code's letters
then you would have a chance at doing it.

Otherwise I can't see how you would do it.




On 18-Nov-04, you wrote:

>
> Hi all,
> Ive been set the challenge by one of my fellow Maths
> Ungraduates of cracking a code that I claimed was easy
> to break.
> Basically the Plaintext is put into a 3*3 matrix (P),
> and the Codematrix (C), remains constant and is
> invertable.
> I have three Coded matrices with the same C.
> Any ideas?



Regards
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Andrew Bruno
abruno@...