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I'm struggling to give a formal proof that $CRH \implies OWF$ using the definition below.

Intuitively, I see why a $CRH$ would be "hard to predict" and might be used as a $PRF$, but I'm unable to give a formal proof.


$\mathcal{H} = \{\mathcal{H}_n = \{h:\{0,1\}^* \rightarrow \{0,1\}^n \}\}$ is a $CRH$ if it can be efficiently sampled and $$\forall_{A;PPT} \exists_N \forall_{n>N} \Pr_{h \rightarrow H_n} [A(1^n, h) = (x, x') \land x \neq x' \land h(x) = h(x')] = neg(n)$$

Proving the existence of $PRG$ or $PRF$s would suffice.

Related but answered question here.

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Collision-resistance implies one-wayness. Proof by reduction: if you have an algorithm $A$ which is able to invert $H: b^* \rightarrow b^n$, we can build an algorithm $B$ which is able to find a collision (with almost the same time and probability). Here $b$ means $\{0,1\}$.

Let's consider only input strings of some length $l\ge n$. Notice that $H$ splits the set $b^l$ into the following $2^n$classes: $$ \{C_h = \{x: x\in b^l \land H(x)=h \}\}_{h\in b^n}. $$ On expectation (if function behaves randomly), each class consists of $k = 2^{l-n}$ elements. But for our concrete function some of the classes could be empty, while some of them have more elements than others.

Algorithm $B$ behaves as follows:

  1. Picks random $x\in b^l$ and evaluates $h = H(x)$.
  2. Runs $A$ on input $h$: $x' = A(h)$.
  3. If $x \neq x'$, returns a collision $(x, x')$. Goes to step 1 otherwise.

Easy to see that probability of success on step 3 is $1 - \frac{1}{k}$ (which is quite a lot, and you can increase $k$ or just make several rounds of algorithm to achieve any probability you want). The explanation is: when you're picking $x$ on first step, you're picking one of ${C_h}$ with probability proportional to a number of elements in the class: $Pr[C_h] = \frac{|C_h|}{2^l}$. Then, on the step 3, probability of failure, i.e. $x = x'$, is $1/|C_h|$, taking into account uniformly random choosing of $x$ and the fact that $A$ "doesn't know" which element $x$ from the class $C_h$ you selected (you've given him just $h$). So, the common probability of failure is calculated as: $$ Pr[\texttt{"failure"}] = \sum_{h\in b^n} {Pr[C_h] \cdot Pr[\texttt{"failure on C_h}"]} = \sum_{h\in b^n} {\frac{|C_h|}{2^l} \cdot \frac{1}{|C_h|}} = 2^n\cdot \frac{1}{2^l}. $$

Remark that we do not require any random properties of $A$ nor $H$. We achieve this probability estimate just using randomness of $B$ (uniformly random choosing of $x$ on step 1).

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