In the book Information Security Principles and Practice, the author explains why HMAC should be used for integrity with a symmetric key with the message, to prevent a MiTM to replace both the message and its MAC.

He explains why it is not secure to prepend the key to the message, with notation h(K, M). I don't understand how the attack works. Here's the paragraph:

Suppose we choose to compute an HMAC as h(K,M). Most cryptographic hashes hash the message in blocks—for MD5, SHA-1, and Tiger, the block size is 512 bits. As a result, if M = {B1, B2), where each Bi is 512 bits, then h(M) = F(F(A, B1), B2) = F(h(B1), B2) (5.2)
for some function F, where A is a fixed initial constant.
For example, in the Tiger hash, the function F consists of the outer rounds illustrated in Figure 5.2, with each Bi corresponding to a 512-bit block of input and A
corresponding to the 192-bit initial value (a,b,c).

If Trudy chooses M' so that M' = (M, X), Trudy might be able to use equation (5.2) to find
h(K, M') from h(K, M) without knowing K since, for K, M, and X
of the appropriate size, h(K, M') = h(K, M, X) = F(h(K, M),X), where the function F is known.

The last paragraph is the attack.

I don't understand, how does the attack work? How can Trudy find h(K, M') from h(K, M) without knowing K?

  • 1
    $\begingroup$ This is the length extension attack. Also, the hash functions pads the message before hashing. So one needs to consider {B1,B2} as padded message. $\endgroup$ – kelalaka Apr 10 at 13:03
  • $\begingroup$ Could you also edit the quoted part to make it clear? the vvv's and 5,2 has no usage here. You can also add a picture for the figure. $\endgroup$ – kelalaka Apr 10 at 13:07
  • $\begingroup$ @kelalaka edited; there is no picture $\endgroup$ – kewiro5 Apr 11 at 14:17

This is a basic length extension attack. For hash functions like SHA-1 and SHA-2 the final state of hashing $h(K \| M)$ block by block is the output. If you hash $h(K \| M \| X)$ then you simply continue hashing from the given state (i.e. the last output).

In practice though the final block must contain a valid padding and length, which is added to $K \| M$. So basically you can create a valid hash for $K \| M \| P \| LE \| X$ where $P$ is the padding and $LE$ is the length encoding. You cannot just create any message that starts with $M$.

| improve this answer | |
  • 1
    $\begingroup$ Length extension attacks are KMPLEX :P $\endgroup$ – Maarten Bodewes Apr 13 at 2:42

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