2
$\begingroup$

I just can't figure out why on earth the $IV$ in Merkle–Damgård has to be fixed (that's what the Katz-Lindell book says)? Because even if you choose it randomly from say $\{{0,1}\}^n$ then the resulting construction still remains collision-resistant, or not?

$\endgroup$
  • 1
    $\begingroup$ One reason is convenience: a variable IV would need to be transmitted, perhaps authenticated.. Another reason is explained here following with the IV replaced by a constant of unknown origin $\endgroup$ – fgrieu Aug 23 '15 at 14:27
  • $\begingroup$ @fgrieu If you'll add your comment as an answer, I'll accept it. Thank you btw $\endgroup$ – pa5h1nh0 Aug 23 '15 at 23:28
7
$\begingroup$

One reason is convenience: if the IV was variable, it would have to be transmitted to a party verifying the hash, and we'd have to deal with its integrity.


Another reason is that it avoids an attack in a Merkle–Damgård variant with padding reduced to appending a single 1 bit and as many zeros as necessary to end the block. With such scheme and common compression functions, like the ones in MD5, SHA-1, and the SHA-2 familly, ability to choose the IV implies ability to chose a secret prefix that does not change the hash.

Notice that the compression function $C$ used in these hashes is of the form $$(X,K)\rightarrow C(X,K)=X\tilde+ E(X,K)$$ where $X$ is the current state (initially the IV), $K$ is the block to hash, $\tilde+$ is a variant of addition (with a few carries suppressed), and $X\rightarrow E(X,K)$ is a reversible block cipher with key $K$.

Who defined an $IV$ that's not a nothing-up-my-sleeves number could have chosen a secret one-block $K$, computed $IV = E^{-1}(0,K)$, which insures $IV=C(IV,K)$. That allows insertion of $K$ at the beginning of any message, without changing the hash.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.