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3

Yes. Remember that, in a Vigenère cipher, the $n$-th ciphertext letter is calculated by adding the $n$-th plaintext letter and the $n$-th key letter (where the key is repeated as many times as necessary to make it as long as the plaintext) modulo 26 (for the standard English alphabet), i.e.: $$c_n \equiv p_n + k_n \mod 26 \tag1$$ (Here, I'll assume the ...

1

I've thought quite a lot about this, and I think in general the answer is no, it would not be a good idea to use a KCV for those kind of situations. Using a hash or even better a MAC (using the key as MAC Key) would be a much better idea if a KCV is required. Instead of zero's, it would be much better to use a block of bytes that is not likely to be of use ...

0

With this cipher, it's pretty easy to retrieve at least 1 key that is consistent with 2 pairs of plaintext,ciphertext . (Other ciphers are better or worse at making it nearly impossible to recover even 1 key consistent with the given plaintext,ciphertext). With this cipher, it is not possible to fully retrieve the key from only 2 known pairs of ...

0

There is a straightforward brute force method. Take for example the lowest 8 bits of everything and check for valid values of $K_1$ and $K_2$, mod $2^8$. You will need about $2^{16}$ checks to get the lower 8 bits of $K_1$ and $K_2$. Proceed then to values mod $2^{16}$, as you know the lower 8 bits of $K_1$ and $K_2$, only bits 8..15 of these must be ...

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