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I understand how for hash functions which are vulnerable to length extension attacks (such as SHA1 and SHA2) it is safer to use a HMAC construction.

What I don't understand is, how or why is $\operatorname{HMAC\_SHA256}_\mathrm{key}(\mathrm{message})$ safer (in terms of resistant against certain attacks) than $\operatorname{SHA256}(\mathrm{key}_1 \mathbin\| \mathrm{message} \mathbin\| \mathrm{key}_2)$, assuming that all key strings are sufficient in length and entropy?

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    $\begingroup$ HMAC_SHA256(message,key) has a security proof; we do not have a ready-made one for SHA256(key1+message+key2). That's quite an argument. That said, for reasons similar to HMAC_SHA256, SHA256(key1+message+key2) intuitively seems quite strong: there's a key1 initially, making collision hard; then a final key2, further increasing security. However the lack of alignment to block boundary in SHA256(key1+message+key2) makes it quite hard to devise a proof. $\endgroup$
    – fgrieu ♦
    Mar 20, 2014 at 20:19
  • $\begingroup$ @fgrieu : $\:$ (I realize it's been over a year, but) One could consider SHA256(key1+message+padtoblockboundary+key2) instead. $\;\;\;\;$ $\endgroup$
    – user991
    Apr 6, 2015 at 21:03

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The construction you are proposing is called the "envelope" or "sandwich" MAC, it predates HMAC, and it is in fact secure—provided the key and message are appropriately padded. That is,

$$ \text{SHA256}(k \parallel m \parallel 1 \parallel 0^{b - 1 - (|m| \bmod b)} \parallel k) $$

is secure, as long as $k$ is the underlying hash function's block length $b$ ($b = 512$ in the case of SHA256), or is padded to that effect. Note that the padding after the message is crucial: without making sure the second key is in its own block the security of the envelope MAC is lower than that of HMAC. This was proved by Yasuda, and later Koblitz and Menezes (§6) argue that there is no sound theoretical reason to prefer HMAC over (correctly padded) envelope-MAC, beyond the former's more established nature.

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  • $\begingroup$ For all messages m, if |m| is not a multiple of b then your formula will give $\:$ m || 0 $\:$ the same MAC as m. $\;\;\;\;$ $\endgroup$
    – user991
    May 6, 2015 at 23:33
  • $\begingroup$ Right, I got the message padding wrong. Fixed now. $\endgroup$
    – Samuel Neves
    May 6, 2015 at 23:35
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    $\begingroup$ Being the envelope MAC was the point! A single key is enough to be secure, though two independent keys are also fine. $\endgroup$
    – Samuel Neves
    May 6, 2015 at 23:53
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    $\begingroup$ Correct; that applies to the version without the $10^t$ padding. Yasuda notes: "We note that it is the lack of appropriate filling between the message M and the last key K, rather than the usage of a single key, that contributes to this key recovery attack." $\endgroup$
    – Samuel Neves
    May 6, 2015 at 23:59
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    $\begingroup$ @chux Yes, same key. Check Figure 1 in page 4 of 2013/248 for a graphical representation of it. $\endgroup$
    – Samuel Neves
    Jun 30, 2016 at 7:09
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Some research brought up this paper On the Security of Two MAC Algorithms (Preneel and Oorschot, 1995).

The authors state that it's possible to significantly reduce the claimed security so that the security is about the same as collision resistance instead of preimage resistance. The details can be read there and in the references.

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