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How does a small block size reduce the key space? Let $b$ be the block bit size of a block cipher. There are $2^b$ possible block values, for a plaintext, and for a ciphertext. For a fixed key, there is a single ciphertext corresponding to any given plaintext, and vice-versa. A key and the block cipher thus implement a bijection on the set of $n=2^b$ ...


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It really depends on the block mode that you are going to use. If you want to use something like CBC, have a look into PKCS#7. If you were to use CTR mode, then you do not require any padding, as the input will always be in chunks of 16 bytes, and you use as much of it as you require.


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JoJoTheCodeDude's answer is correct. I just want to add some detail which may clear up some possible misunderstandings and followup questions. AES is a block cipher. Contrary to the name Advanced Encryption Standard, it's not actually useful for practical encryption of anything. That's because, like any block cipher, it can safely encrypt exactly one block ...


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I cannot see why such values are required for AES-192 and AES-256. That's because you have a correct understanding of the situation, and the answer manual is wrong. Your analysis in "what I tried" is correct; there's not a whole lot I can add to it...


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There are three major options available to you, depending on the available preferences and guarantees, ordered in descending order of generality: Format-Preserving Encryption is a class of encryption schemes that can accept arbitrary input spaces and then produce ciphertext from that input space as output. A classical example would be encryption of credit ...


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A few things have gone wrong: You've reversed the initial state, but continued to shift from left to right. If you do reverse then the shift is from right to left. You've confused the output bit (the oldest bit in the register) with the feedback bit (the new value that is introduced after shifting) The feedback rule for the polynomial $x^4+x^3+1$ is $s_{i+4}...


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