Is $H(k \operatorname\| H(k \operatorname\| m))$ with double length $k$ like NMAC?

Asked Dec 06 '16 at 01:45

Active Dec 08 '16 at 05:49

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I know that NMAC is mathematically proved with two different keys with length $L$.

But, if I choose to implement a MAC this way: $$H(k \operatorname\| H(k \operatorname\| msg))$$ with a key $k$ of length $2L$, am I vulnerable to some attacks?

edited Dec 06 '16 at 03:35

Ilmari Karonen

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asked Dec 06 '16 at 01:45

Abmen

Is $H(k \operatorname\| H(k \operatorname\| m))$​ with double length $k$ like NMAC?

0 Answers0

Is $H(k \operatorname\| H(k \operatorname\| m))$ with double length $k$ like NMAC?