An Improved Approximation Bound for Minimum Weight Dominating Set on Graphs of Bounded Arboricity

Abstract

We consider the minimum weighted dominating set problem on graphs having bounded arboricity. We show that the natural LP has an integrality gap of at most one more than the arboricity of our graph via a primal-dual algorithm. This is nearly the best approximation possible, as Bansal and Umboh have shown it is NP-hard to approximate dominating sets to within one less than the arboricity of the graph.

Supported by University of Waterloo.

Appendix 1.A Example of Bansal and Umboh Algorithm Yielding \(2a-1\) Approximation

The following example shows the Bansal and Umboh Algorithm returning a solution $2a-1-o(1)$ times the optimum. Let $a \in \mathbb {N}$ and $n=k(a-1)+1$. Construct graph G with vertex set $\{ v_1,v_2,..,v_{k(a-1)+1} \}$ and edge set $\{ v_i v_j: \ 0 \le i-j \le a-1 \} \cup \{ v_i v_j: \ 0 \le i+k(a-1)+1-j \le a-1 \}$. One can see that $|E(G(U))| \le a(|U|-1)$ for any $U \subset V$ and hence G has arboricity at most a. For $i > k(a-1)+1$ and $i<0$, define $v_i = v_{i \pmod {k(a-1)+1} }.$

Consider the Bansal and Umboh algorithm on G with unit weights. We claim $x= \frac{1}{2a-1} \mathbb {1}$ is the optimal LP solution for the a-MWDS LP on G with unit weights. Note that the LP cost $\sum _{v \in V(G) } x_v$ of x is $\frac{k(a-1)+1}{2a-1}$. Adding up $\sum _{i=-a+1}^{a-1} v_{i+j} \ge 1$ for $j=1,2,.., k(a-1)+1$ we get

$$ (2a-1) \sum _{v \in V(G) } x_v \ge k(a-1)+1 $$

so $\sum _{v \in V(G) } x_v \ge \frac{k(a-1)+1}{2a-1}$ and that equality can only hold if $\sum _{i=-a+1}^{a-1} v_{i+j} = 1$ for all i. It thus follows the unique optimal solution satisfies $x_{v_i}=x_{v_j} \ \ \forall i,j \in [k(a-1)+1]$ and is thus $\frac{1}{2a-1} \mathbb {1}$.

The algorithm of Bansal and Umboh would thus take every vertex of G, which has a cost of $k(a-1)+1$. However, the vertices $\{ v_{(2a-1)i}: 1 \le i \le \lceil \frac{k(a-1)+1}{2a-1} \rceil \}$ are a solution of cost $\lceil \frac{k(a-1)+1}{2a-11} \rceil \approx \frac{k(a-1)+1}{2a-1}$ for large k. Hence the solution returned by the algorithm is $2a-1-o(1)$ times the optimum.