The $n$-strong Diffie Hellman assumption state that given the subset $\{g, g^s,\cdots,g^{s^n}\} \subseteq \mathbb{G}$ in a cyclic group $\mathbb{G}$ of prime order $p$, a PPT algorithm cannot output $g^{\frac{1}{s+\alpha}}$ for any $\alpha \in \mathbb{F}_p$ except with negligible probability.

Does it somehow imply that no PPT algorithm can output an irreducible polynomial $f(X)\in \mathbb{F}_p[X]$ and the element $g^{\frac{1}{f(s)}}$? Or does that entail a strictly stronger assumption?

  • $\begingroup$ I guess it is weaker than $n$-SDH asumption. The $g^{f(s)}$ can be expressed by the subset {${g^{s^0},...,g^{s^n}}$}, thus the problem is: Given the $g^{f(s)}$, no PPT algorithm can output $g^{1/f(s)}$. $\endgroup$
    – X.H. Yue
    Sep 13, 2021 at 12:35
  • $\begingroup$ Actually, I meant for any irreducible $f(X)$ rather than a prescribed $f(X)$. Since linear polynomials are an example of irreducible polynomials, I think this would be a stronger assumption that $n$-SDH. But I am not sure whether it can be reduced to the $n$-SDH $\endgroup$ Sep 13, 2021 at 16:08

1 Answer 1


If I understand your quantifies (for any given irreducible $f(x)$, there does not exist such an algorithm), then it’s a stronger assumption and one that is unlikely to be true as $n$ grows. First, note that if we write $x_i$ for $g^{s_i}$ then the degree (at most) $n$ polynomial $\sum c_is^i$ gives $$g^{\sum c_is^i}=\prod x_i^{c_i}$$ as easily calculable.

Now for any polynomial $h(x)$ of degree at most $n$ with no roots mod $p$, let $f(x)$ be a solution to $$f(x)h(x)\equiv 1\pmod {p, x^p-x}$$ Then $$g^{1/f(s)}=g^{h(s)}$$ and so can be calculated easily. As $n$ grows the number of possible $h(x)$ grows and soon we will be guaranteed that one of our $f(x)$ is irreducible.

  • $\begingroup$ Good point. I should have specified that the degree of $f(X)$ is bounded. What if we add the condition that $\deg(f(X))≤n$ or is polynomial in $n$? Also, the PPT algorithm won't be able to compute the coefficients of $f(X)$ since it's degree would be $\geq p-\deg(h(X))$. $\endgroup$ Sep 14, 2021 at 0:32

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