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I apologize if this is a dumb question but I'm trying to understand how the final block of partial plaintext is XORed using only n number of bytes in the final keystream block.

How do we determine the number of bytes in the keystream that are required to produce a final block of the same size (128 bit) as the others?

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  • $\begingroup$ They can all handle any number of bits. Why do you have a concept of a block of partial plaintext here? Are you using a padding scheme, not mentioned in the question? $\endgroup$ – Squeamish Ossifrage Apr 17 at 16:00
  • $\begingroup$ Hi @SqueamishOssifrage I was reading Wikipedia and it says "The last partial block of plaintext is XORed with the first few bytes of the last keystream block, producing a final ciphertext block that is the same size as the final partial plaintext block." and I'm looking for a more in-depth explanation of how this works. $\endgroup$ – onetruekey Apr 17 at 16:12
  • $\begingroup$ You just take as many bits as you need? For, e.g., AES-CTR to encrypt an 87-bit message, you use AES to generate a 128-bit block and then take the first 87 bits to xor with your plaintext, and discard the remaining 41. $\endgroup$ – Squeamish Ossifrage Apr 17 at 16:14
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Suppose a block is $b$ bits long, say $b = 128$, and the message is $\ell$ bits long, say $\ell = 187$. Then the final plaintext block is $n = \ell - \lfloor\ell/b\rfloor \cdot b$ bits long. So use $n$ bits of the final keystream block; in this example, $n = 59$.

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