Stabbing Line Segments with Disks: Complexity and Approximation Algorithms

Abstract

Computational complexity and approximation algorithms are reported for a problem of stabbing a set of straight line segments with the least cardinality set of disks of fixed radii $r>0$ where the set of segments forms a straight line drawing $G=(V,E)$ of a planar graph without edge crossings. Close geometric problems arise in network security applications. We give strong NP-hardness of the problem for edge sets of Delaunay triangulations, Gabriel graphs and other subgraphs (which are often used in network design) for $r\in [d_{\min },\eta d_{\max }]$ and some constant $\eta$ where $d_{\max }$ and $d_{\min }$ are Euclidean lengths of the longest and shortest graph edges respectively. Fast $O(|E|\log |E|)$-time O(1)-approximation algorithm is proposed within the class of straight line drawings of planar graphs for which the inequality $r\ge \eta d_{\max }$ holds uniformly for some constant $\eta >0,$ i.e. when lengths of edges of G are uniformly bounded from above by some linear function of r.

This work was supported by Russian Science Foundation, project 14-11-00109.

A Proof of the Lemma 1

Proof

Let $u=(x,y),\,v=(x_1,y_1)$ and $w=(x_2,y_2)$ be distinct points of X. Consider an arbitrary radius r circle (out of two circles) which passes through v and w, and denote its center by O. A lower bound is obtained below for the distance $\pi =\pi (u;v,w)$ from that circle to the point $u\notin C(v,w).$

Let $\varDelta =|v-w|_2,$ $\lambda =\sqrt{r^2-\frac{ \varDelta ^2}{ 4}}$, $a=(u-v,u-w)$ and $b=(u-v,(v-w)^{\perp })$, where $(v-w)^{\perp }=\pm (y_1-y_2,-x_1+x_2)$. The distance $\pi >0$ can be written in the form:

$$\begin{aligned} \pi =\pi (u;v,w)=\left| \left| \frac{v+w}{2}-\lambda \frac{(v-w)^{\perp }}{|v-w|_2}-u\right| _2-r\right| =\left| \frac{a+\frac{ {2\lambda b}}{ \varDelta }}{\sqrt{a+\frac{ {2\lambda b}}{ \varDelta }+r^2}+r}\right| . \end{aligned}$$

Without loss of generality it is assumed that u is in the 2r radius disk centered at O. Indeed, we get $\pi \ge r\ge \frac{1}{r}$ otherwise. Let us bound denominator of the fraction $\pi ,$ taking into account that $\varDelta \le 2r,$ $|u-v|_2\le |u-O|_2+|O-v|_2\le 3r$ and $|b|/\varDelta \le 3r:$

$$\begin{aligned} \sqrt{a+\frac{2\lambda b}{\varDelta }+r^2}+r\le 5r. \end{aligned}$$

As points of X have integer coordinates, a and b are integers. For $\varDelta ^2=4r^2$ we get $\pi \ge \frac{ 1}{ 5r}$. When $\varDelta ^2\le 4r^2-1,$ it is enough to prove the inequality

$$\begin{aligned} \left| a+\frac{2\lambda b}{\varDelta }\right| \ge \frac{1}{96r^4}. \end{aligned}$$

(1)

Indeed, again, combining this bound with the aforementioned upper bound for denominator of the fraction $\pi ,$ we get $\pi \ge \frac{1}{480r^5}$.

For integer $\frac{2\lambda b}{\varDelta }$ the left-hand side of the inequality (1) is at least 1. Thus, it remains for us to prove the inequality (1) for the case where $\frac{2\lambda b}{\varDelta }\notin \mathbb {Z}$. Suppose that $q=\left\{ \left| \frac{2\lambda b}{\varDelta }\right| \right\} >0$ and $k=\left[ \left| \frac{2\lambda b}{\varDelta }\right| \right] ,$ where $\{\cdot \}$ and $[\cdot ]$ denote fractional and integer part of real number respectively. In fact, the term $\min \{q,1-q\}$ can be bounded from below. Let us start estimating with q. First, it is assumed that $\gamma =\frac{ 4r^2b^2}{ \varDelta ^2}\in ~\mathbb {Z}.$ We have $k^2<\frac{ 4\lambda ^2b^2}{ \varDelta ^2}<(k+1)^2.$ As $q>0,$ we get $q\ge \left\{ \sqrt{k^2+1}\right\} .$ Due to concavity of the square root we have

$$\begin{aligned} \left\{ \sqrt{k^2+1}\right\} =\left\{ \sqrt{\frac{2k\cdot k^2}{2k+1}+ \frac{(k+1)^2}{2k+1}}\right\} \ge \left\{ k+\frac{1}{2k+1}\right\} =\frac{1}{2k+1} \end{aligned}$$

$$\begin{aligned} \ge \frac{1}{\frac{ 4\lambda |b|}{ \varDelta }+1} \ge \frac{ 1}{ 13r^2}. \end{aligned}$$

Now the case is considered where $\gamma \notin \mathbb {Z}$. As $2kq+q^2\ge \{2kq+q^2\}=\{\gamma \}$, we have that

$$\begin{aligned} q\ge \sqrt{k^2+\{\gamma \}}-k\ge \frac{\{\gamma \}}{\sqrt{k^2+\{\gamma \}}+k} \ge \frac{\frac{ 1}{ \varDelta ^2}}{\frac{ 4r|b|}{ \varDelta }} \ge \frac{1}{12r^2\varDelta ^2}\ge \frac{1}{48r^4}. \end{aligned}$$

Let us get a lower bound for $1-q$. Again, assume that $\gamma \in \mathbb {Z}$. Arguing analogously, we arrive at the bound

$$\begin{aligned} 2k(1-q)+(1-q)^2\ge \{(k+1-q)^2\} =\left\{ (k+1)^2-\frac{4\lambda ^2b^2}{\varDelta ^2}-2q(1-q)\right\} \ge \frac{1}{2}. \end{aligned}$$

Resolving the quadratic inequality with respect to $1-q$, we get:

$$\begin{aligned} 1-q\ge \sqrt{k^2+\frac{1}{2}}-k =\frac{\frac{1}{2}}{\sqrt{k^2+\frac{1}{2}}+k} \ge \frac{1}{\frac{ 8r|b|}{ \varDelta }}\ge \frac{1}{24r^2}. \end{aligned}$$

Now let $\gamma \notin \mathbb {Z}$. Let us consider the subcase, where $\{\gamma \}+2q(1-q)>1$. We get

$$\begin{aligned} \left\{ (k+1)^2-\frac{4\lambda ^2b^2}{\varDelta ^2}-2q(1-q)\right\} \ge 1-\{\gamma \}\ge \frac{1}{\varDelta ^2}. \end{aligned}$$

Resolving the corresponding inequality with respect to $1-q$, we arrive at the analogous lower bound $1-q\ge \frac{1}{48r^4}$.

Now we are to address the case where $\{\gamma \}+2q(1-q)<1$. Obviously,

$$\begin{aligned} \left\{ (k+1)^2-\frac{4\lambda ^2b^2}{\varDelta ^2}-2q(1-q)\right\} =1-\{\gamma \}-2q(1-q). \end{aligned}$$

For $1-q<\frac{ 1}{ 4\varDelta ^2}$ we have $1-\{\gamma \}-2q(1-q) \ge \frac{1}{\varDelta ^2}-\frac{1}{2\varDelta ^2} =\frac{ 1}{ 2\varDelta ^2}.$ Arguing analogously, we obtain the following bound $1-q\ge \frac{ 1}{ 96r^4}$; otherwise, we get $1-q\ge \frac{ 1}{ 4\varDelta ^2}\ge \frac{ 1}{ 16r^2}$.

For $\{\gamma \}+2q(1-q)=1$ we have:

$$\begin{aligned} 1-q=\frac{1-\{\gamma \}}{ 2q} \ge \frac{1-\{\gamma \}}{2}\ge \frac{1}{2\varDelta ^2}\ge \frac{1}{8r^2}. \end{aligned}$$

Finally, we arrive at the claimed bound:

$$\begin{aligned} \min \{q,1-q\}\ge \frac{1}{96r^4}. \end{aligned}$$