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

  • Thread starter Thread starter Mathick
  • Start date Start date
  • Tags Tags
    Hill
Mathick
Messages
23
Reaction score
0
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.
 
Mathematics news on Phys.org
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.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Thread 'Imaginary Pythagorus'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...

Similar threads

Back
Top