Timeline for Known text attack on Hash function (SHA 256 or SHA512)
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Apr 23, 2021 at 19:09 | history | edited | kelalaka | CC BY-SA 4.0 |
added 126 characters in body
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| Apr 23, 2021 at 18:48 | vote | accept | Sami | ||
| Apr 23, 2021 at 18:04 | comment | added | kelalaka | @Sami the precious moment of the LEA is that the attacker can execute a signature without learning the key. The easiest mitigation is use SHA3, which naturally has resistance, use Blake2 which uses the HAIFA. Your modification is secret suffix is secure. Why don't you use HMAC or KMAC? | |
| Apr 23, 2021 at 17:47 | comment | added | Sami | I have a very naive question. For length extension attacks, the format is H(secret || message). But I change the format, is it still vulnerable? for example: A and B are secrets and C is the message. Then, I concatenate (A || C || B) and generate the hash H(A || C || B), will it still be vulnerable to length extension attack? If so, how? | |
| Apr 23, 2021 at 16:49 | history | edited | kelalaka | CC BY-SA 4.0 |
polish
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| Apr 22, 2021 at 22:04 | comment | added | kelalaka | @Gilles'SO-stopbeingevil' By considering that knowing C is a subset length extension. Then, if one can extract the information from $H(A||B||C)$ by knowing only the $C$, then take any $H(X)$ and extend it, since we know the extended part, we can extract the $X$. Not a great argument, I know. | |
| Apr 22, 2021 at 21:53 | comment | added | Gilles 'SO- stop being evil' | I don't understand how length extension attacks come into play. We know a length extension attack on SHA-256: If you know $H(A||B||C)$ then there is one specific $D$ for which you also know $H(A||B||C||D)$. So what? How does that matter towards finding $A||B$? | |
| Apr 22, 2021 at 19:19 | history | answered | kelalaka | CC BY-SA 4.0 |