Timeline for How does one attack a two-time pad (i.e. one time pad with key reuse)?
Current License: CC BY-SA 3.0
17 events
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| Jan 17, 2022 at 4:33 | comment | added | jw013 |
With so many (10) extra ciphertexts, techniques like crib dragging or statistical analysis are not required. The only assumption you need is that the plain text consists of mostly lower case letters ([a-z]) and spaces. When a space (0x20) is xor'd with a lower case letter it becomes upper case. Pick any $C$ and xor it with every other $C$. For any position $i$ where most of the results are capital letters you can guess that $M[i]$ is a space. After doing this for a few more $C$'s you can guess the rest like on "Wheel of Fortune".
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| Oct 25, 2017 at 1:57 | review | Suggested edits | |||
| Oct 25, 2017 at 6:44 | |||||
| Apr 13, 2017 at 12:48 | history | edited | CommunityBot |
replaced http://crypto.stackexchange.com/ with https://crypto.stackexchange.com/
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| Jan 31, 2016 at 10:22 | answer | added | Yehuda Lindell | timeline score: 10 | |
| Jun 25, 2013 at 14:09 | history | protected | CommunityBot | ||
| S Apr 27, 2013 at 20:09 | history | post merged (destination) | |||
| Jun 18, 2012 at 16:37 | vote | accept | SRKX | ||
| S Apr 27, 2013 at 20:09 | |||||
| Jun 16, 2012 at 17:54 | answer | added | Ilmari Karonen | timeline score: 33 | |
| Jun 16, 2012 at 11:56 | comment | added | Maeher | @DilipSarwate: If $m_1=m_2$ then the key has still only been used once. Therefore it is implicitly clear that $m_1\neq m_2$ | |
| Jun 16, 2012 at 11:35 | comment | added | Dilip Sarwate | Does "the class" include a disclaimer that the method needs to have $m_1 \neq m_2$ in order for the method to work? | |
| Apr 3, 2012 at 16:56 | history | edited | Paŭlo Ebermann | CC BY-SA 3.0 |
formatting, tags,
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| Apr 2, 2012 at 14:38 | history | tweeted | twitter.com/#!/StackCrypto/status/186824813678051329 | ||
| Apr 1, 2012 at 20:23 | vote | accept | Samer Makary | ||
| Apr 1, 2012 at 20:21 | comment | added | Samer Makary | Yes, your totally right. I was mistaken with that part, but it is my very first trial in cryptography. I got it now | |
| Apr 1, 2012 at 1:35 | comment | added | B-Con | If string1[i] == string2[i], that only means that C2[i] == C3[i] (and thus M2[i] == M3[i]). You can check that fact without doing any of those XOR operations. And XORing string1 by C2 gives you back C1 XOR "space", which, at a glance, doesn't seem to be related to key[i]. I haven't looked at that EC assignment for the class in any detail yet, but I think you're supposed to use the fact that XORing by "space" only changes the case of the original character. You might use that fact to analyze the XOR or two plaintext messages, which you obtain by XORing their two respective ciphertexts. | |
| Apr 1, 2012 at 1:34 | answer | added | poncho | timeline score: 83 | |
| Mar 31, 2012 at 22:29 | history | asked | Samer Makary | CC BY-SA 3.0 |