GenericAttacksHash

1 Preimage attacks

The resistance of a hash function to collision and (second) preimage attacks depends in the first place on the length $n$ of the hash value. Regardless of how a hash function is designed, an adversary will always be able to find preimages or second preimages after trying out about $2^n$ different messages. In case an adversary is given $2^k$ distinct target hashes, preimages can be found after trying about $2^{n-k}$ different messages~\cite[Chapter 2, pages 12-13]{Merkle79SecrecyAuthenticationAnd} %, chapter 2, pages 12-13}.

2 Collision attacks

As independently observed by Merkle~\cite{conf/crypto/Merkle89} and Yuval~\cite{journal/cryptologia/Yuval79} in 1979, finding collisions requires a much smaller number of trials: about $2^{n/2}$, as described subsequently in Sect.~\ref{sec:birth}. As a result, hash functions producing less than 160 bits of output are currently considered inherently insecure. Moreover, if the internal structure of a particular hash function allows collisions or preimages to be found more efficiently than what could be expected based on its hash length, then the function is considered to be broken.

2.1 The birthday attack

The birthday attack (also calledsquare-root attack) is a generic attack which considers a hash function as black box. Therefore, a birthday attack is successful for every hash function. For any message $m$ we can compute the $n$-bit hash value $y = h(m)$. Since at least a fraction $2^{-n}$ of the pairs $(m,m^*)$ satisfies $h(m) = h(m^*)$, one can expect to find a colliding message pair after trying about $2^n$ arbitrary message pairs, \cf~\eqref{eqn:collision}. Nevertheless, it follows from the birthday paradox that one can check $2^n$ pairs with only $2^{n/2}$ evaluations of $h$. A birthday attack works as follows: \begin{enumerate}

 \item Pick any message $m$ and compute $h(m)$.
 \item Update list $L$ %Check if $h(m)$ is in the list $L$.
   \begin{itemize}
       \item if $(h(m),m)$ is already in $L$, a colliding message pair has been found.
       \item else save the pair $(h(m),m)$ in the list $L$ and go back to step 1.
   \end{itemize}

\end{enumerate} From the birthday paradox we know that we can expect to find a matching entry, after performing about $2^{n/2}$ hash evaluations. Note that in a birthday attack an attacker has full control over the messages. Hence, as pointed out by Yuval~\cite{}, this method enables an attacker to construct meaningful collisions.

3 Attacks in the quantum setting

Results of quantum complexity theorists as well as newly invented algorithms suggest that even with the power of hypothetic quantum computers applied against commonly used hash functions, no exponential improvement over classical computers is possible. Here we briefly discuss hash function related aspects of this work.

One of the celebrated results in quantum algorithms is Grover's~\cite{Grover1996AFastQuantum} from 1996. %(building up on pioneering work of Deutsch, etc..~\cite{}): The search for a particular element in an unordered database of size $r$ takes at most $O(r^{1/2})$, an actual algorithm is provided that uses xxx memory. Matching lower bounds exist for this problem as well~\cite{boyer98tight,zalka99GroversQuantumSearching}. This algorithm is of wide interest as it does not rely on a particular structure of the elements in the search space.

This result has already direct implications on hash functions: There, the search for a preimage or a second preimage is at most as hard as a search in an unordered database, hence security against these types of generic attacks is lowered from $2^{n}$ to $2^{n/2}$ in the quantum setting.

How about collision attacks in the quantum setting? Here, the fact that many collisions exist and the problem is to find a single one indeed leads to (both asymptotically and for commonly used finite dimensions\todo{better wording than finite dim. needed!}) faster algorithms. An actual quantum algorithm for the collision problem is due to Brassard, H{\o}yer, and Tapp~\cite{DBLP:conf/latin/BrassardHT98} from 1997. This combination of Grover's algorithm with the birthday effect yields a runtime of $2^{n/3}$ for a hash function with $n$ bit output size. The algorithm requires $\Theta(n^{1/3}$log $n)$ classical bits of memory. To the best of the author's knowledge, no time/memory tradeoffs are known. Is this the best one can do? Nontrivial lower bounds for the collision problem were an open problem for some time. Only in 2001, Aaronson~\cite{DBLP:conf/stoc/Aaronson02} proved a query complexity of $\Omega(n^{1/5})$. Subsequent work of Shi improved this bound to $\Omega(n^{1/4})$ and finally to $\Omega(n^{1/3})$(\cite{DBLP:conf/focs/Shi02,DBLP:journals/jacm/AaronsonS04}). As this constitutes a tight bound, indeed one can not do better than Brassard~\etal. That is, given our current axiomatic assumptions about the nature of quantum mechanics.

The bottom line here is as follows. Ignoring practical problems with implementations of large, stable quantum computers, and still requiring a (what now became standard) 128-bit security level gives rise to the following minimal output sizes for hash functions. If only oneway-ness but not collision resistance is required then 256-bits would be enough, for collision resistance at least 384 bits are needed. Of course, these blackbox results might not be the end of the story, dedicated quantum cryptanalytic techniques have not been considered yet. Fast quantum algorithms to compute median and mean values~\cite{DBLP:conf/stoc/Grover98}, and other basic building blocks seem to be an interesting starting point.