Let $\phi$ be negative monotone 2-CNF on $n$ variables and $n^{3/2}$ clauses.

What is the complexity of finding satisfying assignment with maximum number of ones $k$?

Alternatively let $G$ be a graph of order $n$ and $n^{3/2}$ edges. $G$ is dense.

What is the complexity of finding $k$-independent set of $G$?

Each variables $x_i$ is in about $\sqrt{n}$ clauses (alternatively the degrees of vertices of $G$ are about $\sqrt{n}$).

Can we get fixed parameter tractable algorithm with parameter $k$?

Can we increase the exponent $3/2$ to get polynomial solutions?

If necessary assume $n$ is square to get rid of the fractions.

Added later

Third question:

Let $\frac12 \le C < n$ and $d=Cn$. Let $G$ be $d$-regular graph. It has $C n^2/2$ edges and each vertex has $d$ neighbors.

In CNF notation there are $C n^2/2$ clauses and each variable is in $d$ clauses.

Can we find MIS in $G$ in subexponential or polynomial time?

More details about this construction are on Mathoverlfow


1 Answer 1


In both questions the answer is the same for any exponent $1 < \alpha < 2$.

For the first question, we can use a degeneracy based algorithm to find an independent set of size $\Omega(\frac{n}{n^{\alpha-1}}) = \Omega(n^{2-\alpha})$ in polynomial time. Therefore we get an FPT algorithm for every constant $\alpha$ simply by outputting YES if $k < c n^{2-\alpha}$ for some constant $c$, and otherwise solving the problem with bruteforce in $2^n = 2^{O(k^{1/(2-\alpha)})}$ time.

For the second question, we can prove that the problem is NP-hard for all $1 < \alpha < 2$ by reducing from maximum independent set problem in 3-regular graphs, which is NP-hard [1]. For some constant $\beta$ we make $n^\beta$ copies of each vertex, making the set of copies a clique for each vertex. Now the new graph has exactly the same maximum independent set as the old graph. It is $4 n^\beta - 1$ regular and has $n^{\beta+1}$ vertices, so we can choose $\beta \approx \frac{\alpha-1}{2-\alpha}$ to make it work for any $1 < \alpha < 2$.

[1] Garey, Michael R., David S. Johnson, and Larry Stockmeyer. "Some simplified NP-complete problems." STOC 1974.

  • $\begingroup$ Thanks. I don't understand replacing with cliques. Would you please give the transformed $K_4$ which is of independence number $1$? $\endgroup$
    – joro
    Commented Dec 6, 2020 at 11:24
  • $\begingroup$ The reduction replaces each vertex $v$ with a vertex set $X_v$ which is a clique and contains $n^{\beta}$ vertices. The neighbors of each $u \in X_v$ are $N(u) = X_v \setminus \{u\} \bigcup_{w \in N(v)} X_w$. So the graph $K_4$ transforms into $K_{4 n^{\beta}}$. $\endgroup$
    – Laakeri
    Commented Dec 6, 2020 at 16:22
  • $\begingroup$ I added question with exponent 2 and degree()>= n/2 and a Mathoverflow link for another construction. $\endgroup$
    – joro
    Commented Dec 8, 2020 at 13:37
  • $\begingroup$ I think it's against the rules to add more questions, but maybe I can try to answer if you write out what the question is (not just the definition of the graph). If the question if whether you can obtain a polynomial (or even subexponential) algorithm in this case I would bet the answer is NO. $\endgroup$
    – Laakeri
    Commented Dec 8, 2020 at 19:39
  • $\begingroup$ By the way, for explaining your heuristic ideas and experiments, I would recommend the paper "Determining the stability number of a graph" of Chvatal (SICOMP 1977, preprint in apps.dtic.mil/dtic/tr/fulltext/u2/a038864.pdf). $\endgroup$
    – Laakeri
    Commented Dec 8, 2020 at 20:04

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