Timeline for repeating-key xor and hamming distance
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Nov 10, 2014 at 4:41 | comment | added | D.W. |
Hi @GabeHollombe, for new questions, I recommend you post a new question. But the short answer is: yes. If you guessed the key length correctly, you're looking at $\text{wt}(X \oplus K \oplus Y \oplus K) = \text{wt}(X \oplus Y)$, which is 2-3 bits. If you guessed it incorrectly, you're looking at $\text{wt}(X \oplus K \oplus Y \oplus K')$, which is about 3 bits (here all of $X,Y,K,K'$ are independently distributed English ASCII letters). ASCII lowercase letters are 0x61 to 0x7A, so the xor of four of those is close to uniform on its low 6 bits, and thus has 3 bits set on average.
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| Nov 10, 2014 at 1:49 | comment | added | Gabe Hollombe | All of this makes sense to me (thanks, @D.W.), but I'm stuck on one issue: If the plaintext consists of only ASCII letters, and the key consists of only ASCII letters, then will comparing the bitwise Hamming distance of ciphertext bites still help? All ASCII letters share the same last two bits (01), so I don't understand how we'll end up with an average bitwise Hamming distance of 4 when doing these comparisons. I'm sure I'm missing something, I just don't know what it is. Can you point me in the right direction? | |
| S Apr 25, 2013 at 14:54 | history | suggested | Dilip Sarwate | CC BY-SA 3.0 |
included clarification that the length of $X$ here is the length in bytes, not the length in bits.
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| Apr 25, 2013 at 11:21 | review | Suggested edits | |||
| S Apr 25, 2013 at 14:54 | |||||
| Apr 25, 2013 at 6:49 | history | answered | D.W. | CC BY-SA 3.0 |