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Well, I went and solved the puzzle using brute force and Maple. I won't spoil the actual answer, but here are some tips that ought to make the process a bit more quicker. Solving the linear system modulo 2 gives you the parity of the second and third letters of the unknown plaintext. Note that all vowels in the English alphabet map to even numbers using ...

There are 18 plaintext and ciphertext letters $p_j$ and $c_j$, $0\le j<18$ (with $j<6$ for the "first plaintext"), all of which are known except $p_7..p_{17}$. Let $M=\pmatrix{m_{0,0}&m_{0,1}&m_{0,2}\\m_{1,0}&m_{1,1}&m_{1,2}\\m_{2,0}&m_{2,1}&m_{2,2}}$ be the key matrix (unknown, except that it is invertible). We have 18 linear ...

Your reasoning is correct. However, there is still some information to be exploited: as you know already, the matrix used to encrypt (the secret key) need to be invertible in order to allow decryption. Since you basically know (say) the first column and the second column of the secret key from your plaintext/ciphertext, you derive from the invertibility of ...

Sure. Assuming that you're using the encoding $A = 0$, $B = 1$, etc., just choose your plaintext messages to be the one-block strings: $$ BA \dots A \\ AB \dots A \\ \vdots \\ AA \dots B $$ The encryptions of these strings will then directly give you the columns of your key matrix.

This can be broken. The exact nature of the attack will depend what modulus you use for the Hill cipher: are you working modulo a prime number, or working modulo 26? Working modulo a prime $p$ A simple attack, with no fancy mathematics needed. One simple attack is to start by requesting the encryption of the 26 messages AAAA, BBBB, CCCC, DDDD, ..., ZZZZ. ...

One time pad is definitely both easy to do and has perfect secrecy, but key management is a pain and can compromise security. Basically a Vigenère cipher with a key as long as the the message should be secure, because different keys can create ALL possible messages with equal probabilities. Again, it's a one time pad, so no KPA, CPA, or CCA security. ...

Well, I'll assume that we'll use the same mapping between letters and integers both to translate the plaintext into integers (to be matrix multipled), and the integers (after the matrix multiply) back into ciphertext. And, we don't know that mapping, the key matrix $K$, and possibly the value of $n$. If so, the obvious place to start is to attempt to solve ...