Timeline for Shortcuts / practicality of brute forcing block cipher (AES) + ECB with known plaintext [duplicate]
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Mar 12, 2019 at 9:54 | history | closed |
Squeamish Ossifrage Maeher kelalaka Maarten Bodewes♦ |
Duplicate of Is it possible to obtain AES-128 key from a known ciphertext-plaintext pair? | |
| Mar 11, 2019 at 22:05 | review | Close votes | |||
| Mar 12, 2019 at 9:54 | |||||
| Sep 27, 2012 at 0:02 | history | tweeted | twitter.com/#!/StackCrypto/status/251109601121009664 | ||
| Sep 19, 2012 at 15:23 | comment | added | Konrads | You're right, there's two tihngs: a) plain ol' brute force and then b) there are some areas of the system where I don't know the plaintext but I would imagine that it uses same encryption methods | |
| Sep 19, 2012 at 15:14 | vote | accept | Konrads | ||
| Sep 19, 2012 at 13:49 | answer | added | Ilmari Karonen | timeline score: 4 | |
| Sep 19, 2012 at 13:03 | comment | added | mikeazo | Is there a padding oracle present? That is very different from a brute-force attack. Padding oracle will get you the plaintext, but not the encryption key. Brute force would get you the encryption key (and the plaintext but you already know that). Statistical analysis will work with ECB but is typically only practical in rare cases and will again only get you the plaintext, not the key. | |
| Sep 19, 2012 at 12:53 | review | First posts | |||
| Oct 5, 2012 at 9:46 | |||||
| Sep 19, 2012 at 12:49 | history | asked | Konrads | CC BY-SA 3.0 |