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kelalaka
kelalaka
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Suppose $H$ is a hash function; why is $H(k\mathbin\|H(k\mathbin\|m))$$$H(k\mathbin\|H(k\mathbin\|m))$$ not secure?

See this HMAC definition. In there, indeed two keys are used and the mac algorithm is $H(k_1\mathbin\|H(k_2\mathbin\|m))$ .$$H(k_1\mathbin\|H(k_2\mathbin\|m)).$$ Why don't we use $H(k\mathbin\|H(k\mathbin\|m))$,$$H(k\mathbin\|H(k\mathbin\|m)),$$ which has only one key?

Suppose $H$ is a hash function; why is $H(k\mathbin\|H(k\mathbin\|m))$ not secure?

See this HMAC definition. In there, indeed two keys are used and the mac algorithm is $H(k_1\mathbin\|H(k_2\mathbin\|m))$ . Why don't we use $H(k\mathbin\|H(k\mathbin\|m))$, which has only one key?

Suppose $H$ is a hash function; why is $$H(k\mathbin\|H(k\mathbin\|m))$$ not secure?

See this HMAC definition. In there, indeed two keys are used and the mac algorithm is $$H(k_1\mathbin\|H(k_2\mathbin\|m)).$$ Why don't we use $$H(k\mathbin\|H(k\mathbin\|m)),$$ which has only one key?

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forest
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Suppose $H$ is a hash function; why is $H(k||H(k||m))$$H(k\mathbin\|H(k\mathbin\|m))$ not secure?

See this HMAC definition. In there, indeed two keys are used and the mac algorithm is $H(k_1 ||H(k_2 ||m))$$H(k_1\mathbin\|H(k_2\mathbin\|m))$ . Why don't we use $H(k||H(k||m))$$H(k\mathbin\|H(k\mathbin\|m))$, which has only one key?

Suppose $H$ is a hash function; why is $H(k||H(k||m))$ not secure?

See this HMAC definition. In there, indeed two keys are used and the mac algorithm is $H(k_1 ||H(k_2 ||m))$ . Why don't we use $H(k||H(k||m))$, which has only one key?

Suppose $H$ is a hash function; why is $H(k\mathbin\|H(k\mathbin\|m))$ not secure?

See this HMAC definition. In there, indeed two keys are used and the mac algorithm is $H(k_1\mathbin\|H(k_2\mathbin\|m))$ . Why don't we use $H(k\mathbin\|H(k\mathbin\|m))$, which has only one key?

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otus
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Why does HMAC definition by using only one keyuse two different keys?

Why does HMAC definition by using only one keyuse two different keys?

Suppose $H$ is a hash function; why is $H(k||H(k||m))$ not secure?

See this HMAC definition. In there, indeed two keys are used and the mac algorithm is $H(k_1 ||H(k_2 ||m))$ . Why don't we use $H(k||H(k||m))$, which has only one key?

HMAC definition by using only one key

Suppose $H$ is a hash function; why is $H(k||H(k||m))$ not secure?

See this HMAC definition. In there, indeed two keys are used and the mac algorithm is $H(k_1 ||H(k_2 ||m))$ . Why don't we use $H(k||H(k||m))$, which has only one key?

Why does HMAC use two different keys?

Suppose $H$ is a hash function; why is $H(k||H(k||m))$ not secure?

See this HMAC definition. In there, indeed two keys are used and the mac algorithm is $H(k_1 ||H(k_2 ||m))$ . Why don't we use $H(k||H(k||m))$, which has only one key?

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