# The Merkle–Damgård Transform

# The Merkle–Damgård Transform#

Let (Gen, \(h\) ) be a compression function for inputs of length \(n+n^{\prime} \geq 2 n\) with output length \(n\). Fix \(\ell \leq n^{\prime}\) and \(I V \in\{0,1\}^{n}\).

Construct hash function (Gen, \(H\) ) as follows:

Gen: remains unchanged.

\(H\) : on input a key \(s\) and a string \(x \in\{0,1\}^{*}\) of length \(L<2^{\ell}\), do:

Append a 1 to \(x\), followed by enough zeros so that the length of the resulting string is \(\ell\) less than a multiple of \(n^{\prime}\). Then append \(L\), encoded as an \(\ell\)-bit string. Parse the resulting string as the sequence of \(n^{\prime}\)-bit blocks \(x_{1}, \ldots, x_{B}\).

Set \(z_{0}:=I V\).

For \(i=1, \ldots, B\), compute \(z_{i}:=h^{s}\left(z_{i-1} \| x_{i}\right)\).

Output \(z_{B}\).

If \((\mathrm{Gen}, h)\) is collision resistant, then so is \((\mathrm{Gen}, H)\)

Proof. Let \(\mathcal{A}\) be a probabilistic polynomial-time adversary and \(\varepsilon\) a function such that \(\mathcal{A}\) succeeds in the \(\mathrm{Hash}\)-\(\mathrm{coll}\) experiment with (Gen, \(H\) ) with probability \(\varepsilon\). We construct an adversary \(\mathcal{A}^{\prime}\) that succeeds in the \(\mathrm{Hash}\)-\(\mathrm{coll}\) experiment with (Gen,\(h)\) with probability \(\varepsilon\).

Upon receiving \(s\), \(\mathcal{A}^{\prime}\) invokes \(\mathcal{A}\) upon \(s\) and receives back a pair \(x, x^{\prime} .\)

We show that for any \(s\), a collision in \(H^{s}\) yields a collision in \(h^{s}\). Let \(x\) and \(x^{\prime}\) be two different strings of length \(L\) and \(L^{\prime}\), respectively, such that \(H^{s}(x)=H^{s}\left(x^{\prime}\right)\). Let \(x_{1}, \ldots, x_{B}\) be the \(B\) blocks of the padded \(x\), and let \(x_{1}^{\prime}, \ldots, x_{B^{\prime}}^{\prime}\) be the \(B^{\prime}\) blocks of the padded \(x^{\prime}\). Let \(z_{0}, z_{1}, \ldots, z_{B}\) (resp., \(z_{0}^{\prime}, z_{1}^{\prime}, \ldots, z_{B^{\prime}}^{\prime}\) ) be the intermediate results during computation of \(H^{s}(x)\) (resp., \(\left.H^{s}\left(x^{\prime}\right)\right)\). There are two cases to consider:

Case 1: \(L \neq L^{\prime}\). In this case, the last step of the computation of \(H^{s}(x)\) is \(z_{B}:=h^{s}\left(z_{B-1} \| x_{B}\right)\), and the last step of the computation of \(H^{s}\left(x^{\prime}\right)\) is \(z_{B^{\prime}}^{\prime}:=h^{s}\left(z_{B^{\prime}-1}^{\prime} \| x_{B^{\prime}}^{\prime}\right)\). Since \(H^{s}(x)=H^{s}\left(x^{\prime}\right)\) we have \(h^{s}\left(z_{B-1} \| x_{B}\right)=\) \(h^{s}\left(z_{B^{\prime}-1}^{\prime} \| x_{B^{\prime}}^{\prime}\right)\). However, \(L \neq L^{\prime}\) and so \(x_{B} \neq x_{B^{\prime}}^{\prime}\). (Recall that the last \(\ell\) bits of \(x_{B}\) encode \(L\), and the last \(\ell\) bits of \(x_{B^{\prime}}^{\prime}\) encode \(L^{\prime}\).) Thus, \(z_{B-1} \| x_{B}\) and \(z_{B^{\prime}-1}^{\prime} \| x_{B^{\prime}}^{\prime}\) are a collision with respect to \(h^{s}\).

Case 2: \(L=L^{\prime}\). This means that \(B=B^{\prime}\). Let \(I_{i} \stackrel{\text { def }}{=} z_{i-1} \| x_{i}\) denote the \(i\) th input to \(h^{s}\) during computation of \(H^{s}(x)\), and define \(I_{B+1} \stackrel{\text { def }}{=} z_{B}\). Define \(I_{1}^{\prime}, \ldots, I_{B+1}^{\prime}\) analogously with respect to \(x^{\prime}\). Let \(N\) be the largest index for which \(I_{N} \neq I_{N}^{\prime}\). Since \(|x|=\left|x^{\prime}\right|\) but \(x \neq x^{\prime}\), there is an \(i\) with \(x_{i} \neq x_{i}^{\prime}\) and so such an \(N\) certainly exists. Because

we have \(N \leq B\). By maximality of \(N\), we have \(I_{N+1}=I_{N+1}^{\prime}\) and in particular \(z_{N}=z_{N}^{\prime}\). But this means that \(I_{N}, I_{N}^{\prime}\) collide under \(h^{s}\).

If \(H^{s}(x) \neq H^{s}\left(x^{\prime}\right)\) then \(\mathcal{A}^{\prime}\) just halts.

Otherwise,

IF \(L \neq L^{\prime}\) then \(\mathcal{A}^{\prime}\) outputs \(z_{B} \| L\) and \(z_{B^{\prime}}^{\prime} \| L^{\prime}\) and halts.

IF \(L=L^{\prime}\), then \(\mathcal{A}^{\prime}\) works backwards from the last block to the first to find the maximal \(N\).

Then, \(\mathcal{A}^{\prime}\) outputs \(I_N\) and \(I^{\prime}_N\) and halts.

Note that \(\mathcal{A}^{\prime}\) can compute all the \(z\)-values by computing \(H\) and storing all the intermediate values. So whenever \(\mathcal{A}\) finds a collision, \(\mathcal{A}^{\prime}\) also finds a collision. Thus, \(\mathcal{A}^{\prime}\) succeeds in \(\mathrm{Hash}\)-\(\mathrm{coll}\) for (Gen, \(\left.h\right)\) with probability \(\varepsilon\), implying that \(\varepsilon(n)\) is negligible. Thus, (Gen, \(H)\) is collision-resistant as required.