11.2: Merkle-Damgård Construction

Building a hash function, especially one that accepts inputs of arbitrary length, seems like a challenging task. In this section, we’ll see one approach for constructing hash functions, called the Merkle-Damgård construction.

Instead of a full-fledged hash function, imagine that we had a collision-resistant function whose inputs were of a single fixed length, but longer than its outputs. In other words, \(h:\{\theta, 1\}^{n+t} \rightarrow\{0,1\}^{n}\), where \(t>0\). We call such an \(h\) a compression function. This is not compression in the usual sense of the word - we are not concerned about recovering the input from the output. We call it a compression function because it "compresses" its input by \(t\) bits (analogous to how a pseudorandom generator "stretches" its input by some amount).

The following construction is one way to build a full-fledged hash function (supporting inputs of arbitrary length) out of such a compression function:

Construction \(11.2\) (Merkle-Damgård)

Let \(h:\{0,1\}^{n+t} \rightarrow\{0,1\}^{n}\) be a compression function. Then the Merkle-Damgård transformation of \(h\) is \(M D_{h}:\{0,1\}^{*} \rightarrow\{0,1\}^{n}\), where:

Figure \(\PageIndex{1}\): Copy and Paste Caption here. (Copyright; author via source)

The idea of the Merkle-Damgård construction is to split the input \(x\) into blocks of size \(t\). The end of the string is filled out with \(\vartheta s\) if necessary. A final block called the "padding block" is added, which encodes the (original) length of \(x\) in binary.

Example

Suppose we have a compression function \(h:\{0,1\}^{48} \rightarrow\{0,1\}^{32}\), so that \(t=16\). We build a Merkle-Damgård hash function out of this compression function and wish to compute the hash of the following 5-byte (40-bit) string:

Figure \(\PageIndex{1}\): Copy and Paste Caption here. (Copyright; author via source)

We must first padx appropriately \((M D P A D(x))\) :

• Since \(x\) is not a multiple of \(t=16\) bits, we need to add 8 bits to make it so.
• Since \(|x|=40\), we need to add an extra 16-bit block that encodes the number 40 in \(\operatorname{binary}(101000)\)

After this padding, and splitting the result into blocks of length 16 , we have the following:

Figure \(\PageIndex{1}\): Copy and Paste Caption here. (Copyright; author via source)

The final hash of \(x\) is computed as follows:

Figure \(\PageIndex{1}\): Copy and Paste Caption here. (Copyright; author via source)

We are presenting a simplified version, in which \(\mathrm{MD}_{h}\) accepts inputs whose maximum length is \(2^{t}-1\) bits (the length of the input must fit into \(t\) bits). By using multiple padding blocks (when necessary) and a suitable encoding of the original string length, the construction can be made to accommodate inputs of arbitrary length (see the exercises).

The value \(y_{0}\) is called the initialization vector (IV), and it is a hard-coded part of the algorithm.

As discussed above, we will not be making provable security claims using the librarystyle definitions. However, we can justify the Merkle-Damgård construction with the following claim:

Claim 11.3

Suppose \(h\) is a compression function and \(M D_{h}\) is the Merkle-Damgård construction applied to \(h\). Given a collision \(x, x^{\prime}\) in \(M D_{h}\), it is easy to find a collision in \(h\). In other words, if it is hard to find a collision in \(h\), then it must also be hard to find a collision in \(M D_{h}\).

Proof

Suppose that \(x, x^{\prime}\) are a collision under \(M_{h}\). Define the values \(x_{1}, \ldots, x_{k+1}\) and \(y_{1}, \ldots, y_{k+1}\) as in the computation of \(M D_{h}(x)\). Similarly, define \(x_{1}^{\prime}, \ldots, x_{k^{\prime}+1}^{\prime}\) and \(y_{1}^{\prime}, \ldots, y_{k^{\prime}+1}^{\prime}\) as in the computation of \(\operatorname{MD}_{h}\left(x^{\prime}\right)\). Note that, in general, \(k\) may not equal \(k^{\prime}\)

Recall that: \[\begin{gathered} \operatorname{MD}_{h}(x)=y_{k+1}=h\left(y_{k} \| x_{k+1}\right) \\ \mathrm{MD}_{h}\left(x^{\prime}\right)=y_{k^{\prime}+1}^{\prime}=h\left(y_{k^{\prime}}^{\prime} \| x_{k^{\prime}+1}^{\prime}\right) \end{gathered}\] Since we are assuming \(\operatorname{MD}_{h}(x)=\operatorname{MD}_{h}\left(x^{\prime}\right)\), we have \(y_{k+1}=y_{k^{\prime}+1}^{\prime} .\) We consider two cases:

Case 1: If \(|x| \neq\left|x^{\prime}\right|\), then the padding blocks \(x_{k+1}\) and \(x_{k^{\prime}+1}^{\prime}\) which encode \(|x|\) and \(\left|x^{\prime}\right|\) are not equal. Hence we have \(y_{k}\left\|x_{k+1} \neq y_{k^{\prime}}^{\prime}\right\| x_{k^{\prime}+1}^{\prime}\), so \(y_{k} \| x_{k+1}\) and \(y_{k^{\prime}}^{\prime} \| x_{k^{\prime}+1}^{\prime}\) are a collision under \(h\) and we are done.

Case 2: If \(|x|=\left|x^{\prime}\right|\), then \(x\) and \(x^{\prime}\) are broken into the same number of blocks, so \(k=k^{\prime}\). Let us work backwards from the final step in the computations of \(M D D_{h}(x)\) and \(M_{h}\left(x^{\prime}\right)\).

We know that: \[\begin{aligned} &y_{k+1}=h\left(y_{k} \| x_{k+1}\right) \\ &= \\ &y_{k+1}^{\prime}=h\left(y_{k}^{\prime} \| x_{k+1}^{\prime}\right) \end{aligned}\] If \(y_{k} \| x_{k+1}\) and \(y_{k}^{\prime} \| x_{k+1}^{\prime}\) are not equal, then they are a collision under \(h\) and we are done. Otherwise, we can apply the same logic again to \(y_{k}\) and \(y_{k}^{\prime}\), which are equal by our assumption.

More generally, if \(y_{i}=y_{i}^{\prime}\), then either \(y_{i-1} \| x_{i}\) and \(y_{i-1}^{\prime} \| x_{i}^{\prime}\) are a collision under \(h\) (and we say we are "lucky"), or else \(y_{i-1}=y_{i-1}^{\prime}\) (and we say we are "unlucky"). We start with the premise that \(y_{k}=y_{k}^{\prime}\). Can we ever get "unlucky" every time, and not encounter a collision when propagating this logic back through the computations of \(\mathrm{MD}_{h}(x)\) and \(\mathrm{MD}_{h}\left(x^{\prime}\right)\) ? The answer is no, because encountering the unlucky case every time would imply that \(x_{i}=x_{i}^{\prime}\) for all \(i\). That is, \(x=x^{\prime}\). But this contradicts our original assumption that \(x \neq x^{\prime}\). Hence we must encounter some "lucky" case and therefore a collision in \(h\).