Timeline for How many trials does it take to break HMAC-MD5?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Nov 1, 2019 at 4:25 | history | edited | Squeamish Ossifrage | CC BY-SA 4.0 |
Fix spacing.
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| Feb 27, 2019 at 19:10 | history | edited | Squeamish Ossifrage | CC BY-SA 4.0 |
Use a 256-bit (32-byte) key in the illustrative code.
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| Feb 27, 2019 at 16:59 | comment | added | Squeamish Ossifrage | @SEJPM Done, by reference to crypto.stackexchange.com/a/67341. | |
| Feb 27, 2019 at 16:59 | history | edited | Squeamish Ossifrage | CC BY-SA 4.0 |
Suggest 256-bit keys, not 512-bit keys and definitely not >512-bit keys.
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| May 24, 2018 at 5:20 | history | edited | Squeamish Ossifrage | CC BY-SA 4.0 |
Cite a fun application for which HMAC-MD5 falls flat on its face.
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| May 24, 2018 at 1:11 | history | edited | Squeamish Ossifrage | CC BY-SA 4.0 |
Point out why the additional cost of collision search doesn't figure into the advantage figure.
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| May 24, 2018 at 0:57 | history | edited | Squeamish Ossifrage | CC BY-SA 4.0 |
Summarize the security conjecture.
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| May 24, 2018 at 0:47 | history | edited | Squeamish Ossifrage | CC BY-SA 4.0 |
This answer was completely wrong. What were you thinking upvoting it, crypto.SE hivemind? Seriously!
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| May 21, 2018 at 15:37 | vote | accept | ladybug | ||
| May 21, 2018 at 14:30 | comment | added | SEJPM | You may want to mention the trivial HMAC "collision" that comes from $k=\operatorname{MD5}(\text{"a very long string, that is longer than one MD5 block, at least I hope that."})$ and $k'=\text{"a very long string, that is longer than one MD5 block, at least I hope that."}$ which should result in $\operatorname{HMAC-MD5}_k(m)=\operatorname{HMAC-MD5}_{k'}(m)$. | |
| May 21, 2018 at 14:03 | history | answered | Squeamish Ossifrage | CC BY-SA 4.0 |