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You can find the part of the key for which you know the plaintext bits by simply applying XOR over the plaintext bits and ciphertext bits at the positions concerned. In the XOR cipher the bits of the plaintext, key and ciphertext are only related to the bits in the same position. That means you won't get any information about the bits that are not in the ...


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You should first try to determine the possible lengths of the key. Then, it would depends if your plaintext bit is bigger or shorter than the possible key length: If it's longer, you just have to map the part of the ciphertext that are coherent with your plaintext (at n and n+KeyLength, same delta between the two ciphertext characters and the two ...


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The triple DES (3DES) block cipher works by essentially running the block through DES three times. Triple DES is also known as "DES EDE" (encrypt-decrypt-encrypt) and under the name given by the standard document: "TDEA". The TDEA algorithm is described in FIPS NIST Special Publication 800-67 Revision 1 where paragraph 3.2 describes the TDEA Keying Options. ...


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It looks fine; whether you use the secret $S_0, S_1$ as the HMAC key, or whether you use the random value $r$ as the HMAC key; if $t' = t$, it implies that either $S_0 = S_1$, or we found a collision in the underlying hash function. I would personally suggest you use $S_0, S_1$ as the key. With HMAC, it doesn't really matter; however if we extend this to ...


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Assuming the Alphabet is only 26 values, you have: 26! or 403,291,461,126,606,000,000,000,000 ways of rearranging it. This is not key space for a single substitution alpha. You would need to be able to create, or at least have the means of creating, n random Alphas for the number to be relevant. An encryption of 3 characters (say abc) over a random 26 ...



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