In the wikipedia article on Time Complexity it is written that:

The exponential time hypothesis (ETH) is that 3SAT, the satisfiability problem of Boolean formulas in conjunctive normal form with, at most, three literals per clause and with n variables, cannot be solved in time $2^{o(n)}$. More precisely, the hypothesis is that there is some absolute constant $c>0$ such that 3SAT cannot be decided in time $2^{cn}$ by any deterministic Turing machine.

We have $2^{o(n)} = \bigcap_{c > 0} O(2^{cn})$.

Proof: For the inclusion $2^{o(n)} \subseteq \bigcap_{c > 0} O(2^{cn})$. Let $f \in o(n)$, then we know that for each $c > 0$ we can find an $N_c$ such that for $n > N_c$ we have $f(n) < cn$, or $2^{f(n)} < 2^{cn}$ which gives this inclusion. For the inclusion $\bigcap_{c > 0} O(2^{cn}) \subseteq 2^{o(f)}$. Let be some function $g : \mathbb N \to \mathbb N \in \bigcap_{c > 0} O(2^{cn})$, i.e., for each $c > 0$ we have some $N_c$ such that for $n > N_c$ we have $g(n) < 2^{cn}$. Taking logarithms this gives $\log g(n) < cn$. So we have $\log g(n) \in o(n)$.

But ETH is about algorithms, so if we can find for each $c > 0$ some algorithm running in time $O(2^{cn})$, this does not imply that we have a single algorithm running in the intersection of those times, i.e., in $2^{o(n)}$ by the above. So there is still the possiblity that we cannot solve a problem in $2^{o(n)}$, but we can solve it by giving for each $c > 0$ some algorithm $A_c$ that solves it in $O(2^{cn})$. So claiming that something is not solvable in $2^{o(n)}$ is actually a stronger claim than saying we have a $c > 0$ such that the problem cannot be solved in $O(2^{cn})$.

But is there any example of such a problem, i.e. a problem such that for each $c > 0$ we have an algorithm (deterministic TM) running in time $O(2^{cn})$, but we cannot give an algorithm running in time $2^{o(n)}$?

  • 3
    $\begingroup$ This is not exactly what you are asking for but in one of my paper (arxiv.org/abs/1703.01928, Section 4) with Yann Strozecki, we prove that if ETH holds then a hierarchy is strict. We do this by contradiction by constructing for some $c < 1$, a $2^{c^i n}$ time algorithm for SAT for every $i$. So we disprove ETH if the hierarchy is not strict but not by constructing a $2^{o(n)}$ algorithm for SAT. $\endgroup$
    – holf
    Feb 22, 2019 at 16:13

1 Answer 1


Yes, an example is:
Accept $0^k 1^m$ iff $k>0$ and the $k$th Turing machine halts in less than $2^{m/k}$ steps on the empty input. Other strings are rejected. See question Does $∩_{ε>0} \mathrm{DTIME}(O(n^{2+ε})) = \mathrm{DTIME}(n^{2+o(1)})$? for the proof.

I expect that there is a relativization barrier against proving equivalence of ETH with its weakening using $2^{o(n)}$, and that there are oracles relative to which (note that this is not ETH) for every $ε>0$ every relativized NP-complete problem can be solved in time $O(2^{n^ε})$ but not in time $2^{n^{o(1)}}$.


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