# How is HMAC(message,key) more secure than Hash(key1+message+key2)

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?

• 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. – fgrieu Mar 20 '14 at 20:19
• @fgrieu : $\:$ (I realize it's been over a year, but) One could consider SHA256(key1+message+padtoblockboundary+key2) instead. $\;\;\;\;$ – user991 Apr 6 '15 at 21:03

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.

• For all messages m, if |m| is not a multiple of b then your formula will give $\:$ m || 0 $\:$ the same MAC as m. $\;\;\;\;$ – user991 May 6 '15 at 23:33
• Right, I got the message padding wrong. Fixed now. – Samuel Neves May 6 '15 at 23:35
• Should that be parsed as $\: (10)^{-|m| \operatorname{mod} b} \:$ or $\;\;\; 1 \: || \: 0^{-|m| \operatorname{mod} b} \:\:\:\:$? $\;\;\;\;\;\;\;\;$ – user991 May 6 '15 at 23:38
• 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." – Samuel Neves May 6 '15 at 23:59
• @chux Yes, same key. Check Figure 1 in page 4 of 2013/248 for a graphical representation of it. – Samuel Neves Jun 30 '16 at 7:09

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.