Suppose $f$ is a probabilistic one-way function. Then my question is, does there exist a construction of a deterministic one-way function $g$ based on $f$?

Or is it possible that probabilistic one-way functions exist but deterministic one-way functions do not?

EDIT: Probabilistic one-way functions are defined in definition 2.2.2 here.

  • $\begingroup$ What's your definition of a probabilistic one-way function? $\endgroup$
    – Maeher
    Nov 1, 2020 at 19:08
  • $\begingroup$ @Maeher The probabilistic part means that for any given input f doesn’t always give the same output, but instead randomly selects from a set of possible outputs. The one-wayness part means that for any probabilistic polynomial time adversary A, if A is given f(x) where x is a randomly chosen n-bit string, then the probability that A outputs an x’ such that f(x) = f(x’) is less than or equal to a negligible function of n. $\endgroup$ Nov 1, 2020 at 19:16
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    $\begingroup$ This doesn't seem like a useful definition. What about $f$ that ignores its input and returns a random $\kappa$-bit string? Is that "probabilistically one way"? $\endgroup$
    – Mikero
    Nov 1, 2020 at 19:43
  • $\begingroup$ No, because then any string x’ would be a valid inversion. Maybe I should rephrase as, “the probability that A outputs an x’ such that f(x) is a possible output of f when applied to input x’ is less than or equal to a negligible function of n”. $\endgroup$ Nov 1, 2020 at 19:46
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    $\begingroup$ In either case it would appear that you can construct a deterministic one-way function by considering the randomness as part of the input. If this function were not one-way, you could find $(x',r')$, such that $F(x',r')=F(x,r)$. This would also allow you to break the one-wayness of the probabilistic function. $\endgroup$
    – Maeher
    Nov 1, 2020 at 20:33

1 Answer 1


Let's look at the definition in the linked thesis:

Definition 2.2.2 (probabilistic one-way function). A probabilistic function, $F$ (with randomness domain $R_n$), with a corresponding deterministic verifier, $V_F$ , is called one-way with respect to a well-spread distribution, $\mathbb{X}$, if for any PPT, $A$: $$\Pr\bigl[x \gets X_n, r \gets R_n, V_F\bigl(A(F(x, r)), F(x, r)\bigr) = 1\bigr] < \mu(n).¹$$ $F$ is called one-way if it is one-way with respect to the uniform distribution.

I will assume, since you did not specify anything about the distribution, that we are looking at the uniform distribution.

This definition is a bit problematic since it does not contain any kind of correctness guarantee for the verifier. In particular, take any function $F$ and define $V_F$ to be the constant $0$ function. According to the definition above the pair $(F,V_F)$ is one-way. Clearly this is not what was meant.

Definition 2.5.1 defines efficient verifiability, though this definition doesn't quite fit for a one-way function as defined above since it talks about an ensemble of keyed functions. However, in the same spirit, I will assume that Definition 2.2.2 meant to require the following from $F$:

Definition (Effcient Verification). A function, $F$ satisfies efficient verification if there exists a deterministic polynomial time algorithm, $V_F$, such that: $$\forall x \in X_n, r \in R_n, V_F(x, F(x, r)) = 1.$$

If that is the case then the following holds.

Theorem Let $F : X_n \to Y_n$ be a probabilistic one-way function with randomness domain $R_n$. Then the deterministic function $G : X_n \times R_n \to Y_n$ defined as $G(x,r)=F(x,r)$ is a deterministic one-way function.

Proof. Let $A$ denote an arbitrary PPT algorithm such that $$\Pr[(x,r) \gets X_n\times R_n, G(A(G(x,r))) = G(x,r)] =\epsilon(n).$$ Note, that since the distribution over $X_n$ is uniform it holds that \begin{align} \epsilon(n) = &\Pr[(x,r) \gets X_n\times R_n, G(A(G(x,r))) = G(x,r)]\\ =&\Pr[x \gets X_n, r\gets R_n, G(A(G(x,r))) = G(x,r)]\\ =&\Pr[x \gets X_n, r\gets R_n, F(A(F(x,r))) = F(x,r)] \end{align} where the last equality follows by the definition of $G$.

Now, consider the PPT algorithm $B$ that upon input $y$, executes $(x',r')\gets A(y)$ and outputs $x'$. By the definition of efficient verifiability, for all $x,x'\in X_n$, $r',r \in R_n$ it must hold that $$F(x',r') = F(x,r) \implies V_F(x',F(x,r))=1.$$ Thus, \begin{align} \epsilon(n) = &\Pr[x \gets X_n, r\gets R_n, F(A(F(x,r))) = F(x,r)]\\ \leq & \Pr[x \gets X_n, r\gets R_n, V_F(B(F(x,r)),F(x,r))=1] \leq \mu(n), \end{align} where the last inequality follows from the assumption that $F$ is a probabilistic one-way function.

Since the above holds for arbitrary PPT $A$, the theorem statement follows. $\quad\quad\Box$

¹Where $\mu$ denotes an unspecified negligible function.


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