MHB Hill Cipher Attack: Eve Can Crack Alice's Message

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Eve can exploit her knowledge of one entry of the Hill cipher matrix and the fact that Alice's message consists solely of 'A's to potentially decipher the message. The discussion highlights that the matrix must be invertible, requiring the determinant condition to hold true. Participants express skepticism about the problem's completeness, suggesting it may only apply to a limited alphabet like 'A' and 'B' for a conclusive proof. The consensus is that with a full 26-letter alphabet, the problem lacks sufficient information to demonstrate that Eve can definitively crack the code. Overall, the discussion underscores the complexities of the Hill cipher and the necessity of clarity in problem statements.
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Suppose that we are in the situation that Alice is using a Hill cipher consisting of a $2 \times 2$ matrix $M$ to send her message, which is $100$ ‘A’s. If Eve intercepts this message and knows that plaintext contained only one letter, and she also knows anyone of the entries of the matrix $M$, then prove that Eve can use this information to find the plaintext and the complete key.

So I tried writing a matrix $M$ as $M = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$ and assuming that Eve saw messages $c_1$ and $c_2$, I got

$\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} \begin{bmatrix} x \\ x \\ \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \\ \end{bmatrix}$ which implies $x \begin{bmatrix} a + b \\ c + d \\ \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \\ \end{bmatrix}$ where $x$ is a letter that Eve wants to find.

But I don't know how to conclude from this that she can do it. I tried assuming that, for example, she knows the entry $a$ or any other and proceed but it led me nowhere.If you could help or give me some hint, I would appreciate it very much.
 
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Hi Mathick,

We also have that the matrix must be invertible.
That is, $ad-bc\not\equiv 0 \pmod{\text n}$, where $n$ is the number of letters.

Either way, it seems to me that the problem statement is incomplete.
Is it possible that we only use the letters 'A' and 'B'?
Because then we can indeed prove what is requested.
 
Klaas van Aarsen said:
Hi Mathick,

We also have that the matrix must be invertible.
That is, $ad-bc\not\equiv 0 \pmod{\text n}$, where $n$ is the number of letters.

Either way, it seems to me that the problem statement is incomplete.
Is it possible that we only use the letters 'A' and 'B'?
Because then we can indeed prove the what is requested.

Hi, thanks for your reply!

That is what I thought - that this problem is incomplete. It says that we use a full alphabet (26 letters) and I also came to the conclusion that the statement can't be proven. I will try to find out where the typo is and come back.
 
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