Consider the following decision problem

Input: A monotone CNF $\Phi$ and a monotone DNF $\Psi$.

Question: Is $\Phi \to \Psi$ a tautology?

Definitely you can solve this problem in $O(2^n \cdot \mathrm{poly}(l))$-time, where $n$ is the number of variables in $\Phi \to \Psi$ and $l$ is the length of input. On the other hand, this problem is coNP-complete. Moreover, a reduction which establishes coNP-completeness also shows that, unless SETH fails, there is no $O(2^{(1/2 - \varepsilon)n} \mathrm{poly}(l))$-time algorithm for this problem (this holds for any positive $\varepsilon$). Here is this reduction. Let $A$ be a (non-monotone) CNF and let $x$ be its variable. Replace every positive occurence of $x$ by $y$ and every negative occurence of $x$ by $z$. Do the same for every variable. Let the resulting monotone CNF be $\Phi$. It is easy to see that $A$ is satisfiable iff $\Phi \to yz \lor \ldots $ is not a tautology. This reduction blows up the number of variables by a factor of 2, which implies $2^{n/2}$ (SETH-based) lower bound mentioned above.

So there is a gap between $2^{n/2}$ and $2^n$-time. My question is whether any better algorithm or better reduction from SETH is known?

Just two remarks seemingly related to the problem:

  • a reverse problem of whether a monotone DNF implies a monotone CNF is trivially solvable in polynomial time.

  • interestingly, the problem of deciding whether $\Phi$ and $\Psi$ compute the same function can be solved in quasi-polynomial time due to Fredman and Khachiyan (On the Complexity of Dualization of Monotone Disjunctive Normal Forms, Journal of Algorithms 21 (1996), no. 3, pp. 618–628, doi: 10.1006/jagm.1996.0062)


1 Answer 1


Assuming SETH, the problem is not solvable in time $O\bigl(2^{(1-\epsilon)n}\mathrm{poly}(l)\bigr)$ for any $\epsilon>0$.

First, let me show that this is true for the more general problem where $\Phi$ and $\Psi$ may be arbitrary monotone formulas. In this case, there is a poly-time ctt reduction from TAUT to the problem that preserves the number of variables. Let $T^n_t(x_0,\dots,x_{n-1})$ denote the threshold function $$T^n_t(x_0,\dots,x_{n-1})=1\iff\bigl|\{i<n:x_i=1\}\bigr|\ge t.$$ Using the Ajtai–Komlós–Szemerédi sorting network, $T^n_t$ can be written by a polynomial-size monotone formula, constructible in time $\mathrm{poly}(n)$.

Given a Boolean formula $\phi(x_0,\dots,x_{n-1})$, we may use De Morgan rules to write it in the form $$\phi'(x_0,\dots,x_{n-1},\neg x_0,\dots,\neg x_{n-1}),$$ where $\phi'$ is monotone. Then $\phi(x_0,\dots,x_{n-1})$ is a tautology if and only if the monotone implications $$T^n_t(x_0,\dots,x_{n-1})\to\phi'(x_0,\dots,x_{n-1},N_0,\dots,N_{n-1})$$ are valid for every $t\le n$, where $$N_i=T^{n-1}_t(x_0,\dots,x_{i-1},x_{i+1},\dots,x_{n-1}).$$

For the left-to-right implication, let $e$ be an assignment satisfying $T^n_t$, i.e., with at least $t$ ones. There exists $e'\le e$ with exactly $t$ ones. Then $e'\models N_i\leftrightarrow\neg x_i$, thus $e'\models\phi$ implies $e'\models\phi'(x_0,\dots,x_{n-1},N_0,\dots,N_{n-1})$. Since this is a monotone formula, we also have $e\models\phi'(x_0,\dots,x_{n-1},N_0,\dots,N_{n-1})$. The right-to-left implication is similar.

Now, let me return to the original problem. I will show the following: if the problem is solvable in time $2^{\delta n}\mathrm{poly}(l)$, then for any $k$, $k$-DNF-TAUT (or dually, $k$-SAT) is solvable in time $2^{\delta n+O(\sqrt{kn\log n})}\mathrm{poly}(l)$. This implies $\delta\ge1$ if SETH holds.

So, assume we are given a $k$-DNF $$\phi=\bigvee_{i<l}\Bigl(\bigwedge_{j\in A_i}x_j\land\bigwedge_{j\in B_i}\neg x_j\Bigr),$$ where $|A_i|+|B_i|\le k$ for each $i$. We split the $n$ variables into $n'=n/b$ blocks of size $b\approx\sqrt{k^{-1}n\log n}$ each. By the same argument as above, $\phi$ is a tautology if and only if the implications $$\tag{$*$}\bigwedge_{u<n'}T^b_{t_u}(x_{bu},\dots,x_{b(u+1)-1})\to \bigvee_{i<l}\Bigl(\bigwedge_{j\in A_i}x_j\land\bigwedge_{j\in B_i}N_j\Bigr)$$ are valid for every $n'$-tuple $t_0,\dots,t_{n'-1}\in[0,b]$, where for any $j=bu+j'$, $0\le j'<b$, we define $$N_j=T^{b-1}_{t_u}(x_{bu},\dots,x_{bu+j'-1},x_{bu+j'+1},\dots,x_{b(u+1)-1}).$$ We can write $T^b_{t}$ as a monotone CNF of size $O(2^b)$, hence the LHS of $(*)$ is a monotone CNF of size $O(n2^b)$. On the right-hand side, we may write $N_j$ as a monotone DNF of size $O(2^b)$. Thus, using distributivity, each disjunct of the RHS can be written as a monotone DNF of size $O(2^{kb})$, and the whole RHS is a DNF of size $O(l2^{kb})$. It follows that $(*)$ is an instance of our problem of size $O(l2^{O(kb)})$ in $n$ variables. By assumption, we may check its validity in time $O(2^{\delta n+O(kb)}l^{O(1)})$. We repeat this check for all $b^{n'}$ choices of $\vec t$, thus the total time is $$O\bigl((b+1)^{n/b}2^{\delta n+O(kb)}l^{O(1)}\bigr)=O\bigl(2^{\delta n+O(\sqrt{kn\log n})}l^{O(1)}\bigr)$$ as claimed.

We get a tighter connection with the (S)ETH by considering the bounded-width version of the problem: for any $k\ge3$, let $k$-MonImp denote the restriction of the problem where $\Phi$ is a $k$-CNF, and $\Psi$ is a $k$-DNF. The (S)ETH concerns the constants $$\begin{align*} s_k&=\inf\{\delta:k\text-\mathrm{SAT}\in\mathrm{DTIME}(2^{\delta n})\},\\ s_\infty&=\sup\{s_k:k\ge3\}. \end{align*}$$ Likewise, let us define $$\begin{align*} s'_k&=\inf\{\delta:k\text-\mathrm{MonImp}\in\mathrm{DTIME}(2^{\delta n})\},\\ s'_\infty&=\sup\{s'_k:k\ge3\}. \end{align*}$$ Clearly, $$s'_3\le s'_4\le\dots\le s'_\infty\le1$$ as in the SAT case. We also have $$s'_k\le s_k,$$ and the double-variable reduction in the question shows $$s_k\le2s'_k.$$ Now, if we apply the construction above with constant block-size $b$, we obtain $$s_k\le s'_{bk}+\frac{\log(b+1)}b,$$ hence $$s_\infty=s'_\infty.$$ In particular, SETH is equivalent to $s'_\infty=1$, and ETH is equivalent to $s'_k>0$ for all $k\ge3$.

  • $\begingroup$ Thank you for your answer! I'm curious whether it is possible to make $\Phi$ and $\Psi$ constant-depth in this construction? Namely, I'm not aware whether subexponential-size constant-depth monotone Boolean formulas (or even non-monotone circuits) are known for $T^n_k$ (in particular for Majority)? Of course there is a $2^{n^{\Omega(1/d)}}$ lower bound for depth-$d$, but, say, $2^{\sqrt{n}}$ size would be OK. $\endgroup$ Sep 6, 2018 at 16:14
  • $\begingroup$ $T^n_k$, and in general anything computable by polynomial-size formulas (i.e., in NC^1), has depth-$d$ circuits of size $2^{n^{O(1/d)}}$. See e.g. cstheory.stackexchange.com/q/14700 . I will have to think if you can make them monotone, but it sounds plausible. $\endgroup$ Sep 6, 2018 at 16:52
  • $\begingroup$ OK. First, the generic construction works fine in the monotone setting: if a function has poly-size monotone formulas, it has depth-$d$ monotone circuits of size $2^{n^{O(1/d)}}\mathrm{poly}(n)$ for any $d\ge2$. Second, for $T^n_k$ specifically, it is easy to construct monotone depth-$3$ circuits of size $2^{O(\sqrt{n\log n})}$ by splitting the input into blocks of size $\Theta(\sqrt{n\log n})$. $\endgroup$ Sep 6, 2018 at 17:28
  • $\begingroup$ Actually, pushing this idea a little bit more, it does provide an answer to the original question: assuming SETH, the lower bound holds already for $\Phi$ monotone CNF and $\Psi$ monotone DNF. I will write it up later. $\endgroup$ Sep 6, 2018 at 18:05
  • $\begingroup$ I would guess that you can divide all the variables into about $\sqrt{n}$ blocks $x^1, \ldots x^{\sqrt{n}}$ and then write $T^{\sqrt{n}}_{k_1}(x^1)\land \ldots \land T^{\sqrt{n}}_{k_\sqrt{n}}(x^{\sqrt{n}}) \to \phi^\prime$ for every $k_1 + \ldots + k_{\sqrt{n}} \le n$. You can use $2^{\sqrt{n}}$-size CNF for every threshold function. But then on a right hand side you will have not DNF but a depth-3 formula... $\endgroup$ Sep 6, 2018 at 18:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.