Approximating Minimum Dominating Set on String Graphs

Abstract

A string graph is an intersection graph of simple curves on the plane. For $k\ge 0$, $B_k$-VPG graphs are intersection graphs of simple rectilinear curves having at most k cusps (bends). It is well-known that any string graph is a $B_k$-VPG graph for some value of k. For $k\ge 0$, unit $B_k$-VPG graphs are intersection graphs of simple rectilinear curves having at most k cusps (bends) and each segment of the curve being unit length. Any string graph is a unit-$B_k$-VPG graph for some value of k.

In this article, we show that the Minimum Dominating Set (MDS) problem for unit $B_k$-VPG graphs is NP-Hard for all $k \ge 1$ and provide an $O(k^4)$-approximation algorithm for all $k\ge 0$. Furthermore, we also provide an 8-approximation for the MDS problem for the vertically-stabbed L-graphs, intersection graphs of L-paths intersecting a common vertical line. The same problem is known to be APX-Hard (MFCS, 2018). As a by-product of our proof, we obtained a 2-approximation algorithm for the stabbing segment with rays (SSR) problem introduced and studied by Katz et al. (Comput. Geom. 2005).

1 Introduction

A string graph is a graph with simple curves on a plane as vertices, and two vertices are adjacent if they intersect. String graphs are important as it contains all intersection graphs of connected sets in $\mathbb {R}^2$. Asinowski et al. [1] introduced the concept of $B_k$-VPG graphs to initiate a systematic study of string graphs and its subclasses. A path is a simple rectilinear curve and a k-bend path is a path having k cusps (bend). The $B_k$-VPG graphs are intersection graphs of k-bend paths. Any string graph has a $B_k$-VPG representation for some k [1]. A unit k-bend path is a k-bend path with each segments being of unit length. The unit $B_k$-VPG graphs are intersection graphs of unit k-bend paths. Any string graph has a unit $B_{k'}$-VPG representation for some $k'$.

A dominating set of a graph $G=(V,E)$ is a subset D of vertices V such that each vertex in $V\setminus D$ is adjacent to some vertex in D. The Minimum Dominating Set (MDS) problem is to find a minimum cardinality dominating set of a graph G. It is not possible to approximate the MDS problem on string graphs with n vertices to within $(1-\alpha )\ln n$ unless $NP\subseteq DTIME(n^{O(\log \log n)})$ [5]. Researchers have studied the MDS problem extensively on important graph classes like planar graphs, permutation graphs, cocomparability graphs all of which are subclasses of string graphs [1, 13, 14, 16] and hence unit $B_k$-VPG graphs for some k. The MDS problem admits constant factor approximation algorithms on intersection graphs of unit disks, rectangles with bounded aspect ratio, r-regular polygons etc. all of which are also subclasses of string graphs [10, 12]. Mehrabi [18] gave constant factor approximation algorithms for the MDS problem on restricted subclasses of $B_1$-VPG graphs. However, for $k\ge 1$, it is not known if there is an f(k)-approximation algorithm for the MDS problem on $B_k$-VPG graphs. We show that the MDS problem remains NP-Hard on unit $B_k$-VPG graphs with $k\ge 1$.

Theorem 1

It is NP-Hard to solve the MDS problem on unit $B_k$-VPG graphs with $k\ge 1$.

We achieved a constant factor approximation algorithm for the MDS problem on unit $B_k$-VPG graphs for a given unit $B_k$-VPG representation and fixed k.

Theorem 2

Given a unit $B_k$-VPG representation of a graph G with n vertices, there is an $O(k^2n^5)$-time $O(k^4)$-approximation algorithm to solve the MDS problem on G.

The MDS problem remains difficult in restricted families of $B_1$-VPG graphs. An L-path is a 1-bend path having the shape ‘L’. A vertically-stabbed-L-representation of a graph is a collection of L-paths along with a vertical line intersecting all the paths such that each path in the collection represents a vertex of the graph and two paths intersect if and only if the vertices they represent are adjacent in the graph. A graph is a vertically-stabbed-L graph if it has a vertically-stabbed-L-representation. Bandyapadhyay et al. [2] proved APX-hardness for the MDS problem on vertically-stabbed-L graphs. The class of vertically-stabbed-L graphs was introduced by McGuinness [17] and it contains many important graph classes like interval graphs, outerplanar graphs, permutation graphs, circle graphs as subclasses. Researchers have studied the MDS problem on these classes of graphs [4, 7,8,9, 11]. An $\epsilon$-net based algorithm of Mehrabi [18] gives an O(1)-approximation algorithm for the MDS problem on vertically-stabbed-L-graphs. The specific value of the constant (which is at least 32) was not reported by the author. In this paper, we prove the following.

Theorem 3

Given a vertically-stabbed-L-representation of a graph G with n vertices, there is an $O(n^5)$-time 8-approximation algorithm to solve the MDS problem on G.

To prove Theorems 2 and 3, we needed to prove a lemma about the stabbing segment with rays (SSR) problem introduced by Katz et al. [15]. In this problem, the inputs are a set of disjoint leftward-directed horizontal rays and a set of disjoint vertical segments. The objective is to select a minimum number of leftward-directed horizontal rays that intersect all vertical segments. Throughout this article, we let $\mathcal {SSR}(R,V)$ denote an SSR instance where R is a given set of disjoint leftward-directed horizontal rays and V is a given set of disjoint vertical segments. We observe the following lemma.

Lemma 1

Let R be a set of leftward-directed horizontal rays and V be a set of vertical segments. The cost of an optimal solution for the Integer Linear Program (ILP) of $\mathcal {SSR}(R, V)$ is at most twice the cost of an optimal solution for the relaxed linear program (LP) of $\mathcal {SSR}(R, V)$.

As a consequence of the above Lemma 1, we have a subquadratic 2-approximation algorithm for the SSR problem.

Theorem 4

There is an $O((n+m)\log (n+m))$-time 2-approximation algorithm for SSR problem where n and m are the number of rays and segments, respectively.

We prove both Lemma 1 and Theorem 4 in Sect. 2 and prove Theorem 3 in Sect. 3. In Sect. 4, we prove Theorems 1 and 2.

2 Proof of Lemma 1 and Theorem 4

In this section, we represent a leftward-directed horizontal ray by simply a ray and a vertical segment by a segment in short. Let R be a set of disjoint rays and V be a set of disjoint vertical segments. We assume each segment intersects at least one ray in R and no two segments in V has the same x-coordinate.

To prove Lemma 1, first we present an iterative algorithm consisting of three main steps. The first step is to include all rays $r\in R$ in heuristic solution S whenever some segments in V intersect precisely a single ray r in that iterative step. In the next step, delete all segments intersecting any ray in S from V. In the final step, find a ray in $R\setminus S$ whose x-coordinate of the right endpoint is the smallest among all rays in $R\setminus S$ and delete it from R (when there are multiple such rays, choose anyone arbitrarily). We repeat the above three steps until V is empty. The above algorithm takes $O((|R|+|V|)\log (|R|+|V|))$ time (using segment trees [3]) and outputs a set S of rays such that all segments in V intersect at least one ray in S.

We describe the above algorithm formally in Algorithm 1. Below we introduce some notations used to describe the algorithm. We assign token $T_r=\{r\}$ for each $r\in R$ initially. For $i\ge 1$, let $R_i,V_i,S_i$ be the set of rays, the set of segments and the heuristic solution constructed by this Algorithm 1, respectively at the end of $i^{th}$ iteration. A ray $r\in R_i$ is critical if there is a segment $v\in V_i$ such that r is the only ray in $R_i$ that intersects v. We describe a discharging technique below.

Let D be a subset of R. A ray $r\in D$ lies in between two rays $r',r''\in D$ if the y-coordinate of r lies in between those of $r',r''$. A ray $r\in D$ lies just above (resp. just below) a ray $r'\in D$ if y-coordinate of r is greater (resp. smaller) than that of $r'$ and no other ray lies in between $r,r'$ in D. Two rays $r,r'\in D$ are neighbours of each other if r lies just above or below $r'$.

Discharging Method: Let $r\in R_{i-1}\setminus S_{i}$ be a ray whose x-coordinate of the right endpoint is the smallest. The phrase “r discharges the token to its neighbours” in the $i^{th}$ iteration means the following operations in the given order.

• (i) Let $r'$ lie just above r and $r''$ lie just below r in $R_{i-1}\setminus S_i$. For all $x\in T_r$ (x and r not necessarily distinct) do the following. If there is a segment in $V_{i}$ that intersects $x,r'$ and r then assign $T_{r'}=T_{r'}\cup \{x\}$ and if there is a segment in $V_{i}$ that intersects $x,r''$ and r then $T_{r''}=T_{r''}\cup \{x\}$.

• (ii) Make $T_r=\emptyset$ after performing the above step.

We have the following observation.

Observation A. For some $v\in V_k, k\ge 1,$ if some ray $r\in R_0$ intersects v, then either $r\in R_k$ or there exists some ray $r'\in R_k$ such that $r\in T_{r'}$.

Proof

Assume $r\notin R_k$. Let ${<}r_1,r_2,\ldots ,r_k{>}$ be a sorted order of the rays such that for $i<j$, $r_i$ discharged the token to the neighbours before $r_j$. Due to step 5 of SSR-algorithm, $X={<}r_1,r_2,\ldots ,r_k{>}$ is an increasing sequence based on the x-coordinate of their right endpoint. Observe that, whenever a ray $r_i\in X$ discharged its token to its neighbours in the $i^{th}$ iteration, all the vertical segments in $V_i$ intersected by $r_i$ also intersects one of the immediate neighbours of $r_i$. Again as $v\in V_k$, v is not intersected by critical ray within k iteration. Hence the result follows.   $\square$

Lemma 2

For a ray r, there are at most two tokens containing r.

Proof

If r never discharged its token to its neighbours, then the statement is true. Let r discharged the token to its neighbours at iteration i. Note that, r discharged tokens to at most two of its neighbours. Since r gets deleted after the discharging step, the rays whose token contain r become neighbours of each other.

Let j be the minimum integer with $i<j$ such that at the end of $(j-1)^{th}$ iteration, there is a ray $p\in R_{j-1}$ which is critical and $r\in T_p$. Note that iteration of SSR-Algorithm may stop before encountering such events. However, within iteration i to $j-1$, there may exist some rays which discharged their tokens containing r due to step 5 of SSR-Algorithm.

To prove the lemma, we use induction to show that there are at most two tokens containing r in any iteration from i upto $j-1$. Consider some k, $i<k<j$, such that $x_1$, $x_2\in R_{k-1}$ be only two rays where $r\in T_{x_1}$ and $r\in T_{x_2}$. Notice that, $x_1$ and $x_2$ are neighbours of each other and without loss of generality assume $x_1$ lies just above $x_2$ in $V_{k-1}$. Assume $x_1$ discharged its token at $k^{th}$ iteration. If there exists a neighbour of $x_1$ (say $x_3$) which is different from $x_2$, then due to the discharging step of $k^{th}$ iteration, $x_1$ passes the token to its neighbours (i.e $x_2$ and $x_3$) and gets deleted from $R_{k-1}$ to create $R_{k}$. If $x_3$ does not exist, then $x_1$ shall pass the token only to $x_2$. Therefore $x_2$ becomes the top-most ray among those rays in $R_k$ which intersect some segment intersecting r.

Moreover, if x was the only ray in $R_{k-1}$ such that $r\in T_x$, then x was the top-most (or bottom-most) ray among those rays in $R_{k-1}$ which intersect some segment intersecting r. Therefore, at the end of $k^{th}$ iteration there is exactly one ray $x'\in R_{k}$ such that $r\in T_{x'}$ and $x'$ must be the top-most (resp. bottom-most) ray among those rays in $R_k$ which intersect some segment intersecting r.

Hence we conclude that for each k with $i\le k<j$, there is at most two rays $r', r''\in R_{k}$ such that $r\in T_{r'}\,\cap \,T_{r''}$ and they are neighbours. If there is exactly one ray $r'''\in R_{k}$ such that $r\in T_{r'''}$ then $r'''$ must be the top-most or bottom-most ray among those rays in $R_k$ which intersect some segment intersecting r.

In iteration j, ray p is critical and $r\in T_p$ and p is put in heuristic solution. If p is the only ray whose token contained r, only $T_p$ will contain r after the termination of Algorithm 1. Let $r',p\in R_{j-1}$ be the rays whose token contained r. They must be neighbours. Without loss of generality assume that p lies just above $r'$. If both $r',p$ are selected in $S_j$, then there is nothing to prove. Now consider the set A of segments in $V_{j}$ that intersects r but not p. Note that, no ray above p intersects any segment in A. Hence $r'$ becomes the only ray in next iterative step whose token contains r and $r'$ turns to be bottom most ray among those rays in $R_{j-1}$ which intersect some segment intersecting r. Now consider any iteration $k>j$. By similar arguments as above, there would be at most one ray in $R_k$ that contains the token r. Hence the lemma follows.   $\square$

For a segment $v\in V$, let $N(v)\subseteq R$ be the set of rays that intersect v. Let $r\in S$ be a ray, i be the minimum integer such that $r\in S_i$. There must exist a segment $\nu _r\in V_{i-1}$ such that r is the only ray in $R_{i-1}$ that intersects $\nu _r$ and all rays in $N(\nu _r)\setminus \{r\}$ must have passed the token to its neighbours. So, for each ray $r\in S$, there exists a segment $\nu _r$ such that for all $x\in N(\nu _r)\setminus \{r\}$ we have $T_{x}=\emptyset$. We call $\nu _r$ a critical segment with respect to r.

Observation B. For a ray $r\in S$ let $\nu _r$ be a critical segment with respect to r. Then $N(\nu _r)\subseteq T_r$.

Proof

Consider any arbitrary but fixed deleted ray $y\in N(\nu _r)\setminus \{r\}$ which was deleted at some $j^{th}$ iteration. By Observation A, there exists a ray $y'\in R_j$ such that $y'$ intersects v and $y\in T_{y'}$. Applying the above argument for all rays in $N(\nu _r)\setminus \{r\}$, we have the proof.   $\square$

Lemma 3

If S is the set returned by SSR-algorithm with rays R and segments V, then $|S|\le 2|OPT|$, where OPT is an optimum solution of $\mathcal {SSR}(R,V)$.

Proof

Let R be the set of rays and V be the set of segments with $|R|=n, |V|=m$. Consider the ILP formulation Q of $\mathcal {SSR}(R,V)$. For each ray $r\in R$, let $x_r\in \{0,1\}$ denote the variable corresponding to r. Objective is to minimize $\sum \limits _{r\in R} x_r$ with constraints $\sum \limits _{r\in N(v)} x_r \ge 1$ for all $v\in V$. Let the corresponding relaxed LP formulation be $Q_l$.

Let $\mathbf Q _l=\{x_r\}_{r\in R}$ be an optimal solution of $Q_l$. Consider SSR-algorithm. Here, define $y_r=1$ if $r\in S$, $y_r=0$ if $r\notin S$ and $\mathbf Q '=\{y_r\}_{r\in R}$, obtained by the algorithm. This is a feasible solution of Q as SSR-algorithm terminates only when no segments are left in $V_i$. Now we fix any arbitrary $r\in S$ and $\nu _r$ be a critical segment with respect to r. Then due to Observation B, we know that for all $z\in N(\nu _r)\setminus \{r\}$ we have $T_{z}=\emptyset$ and $N(\nu _r)\subseteq T_r$. Therefore, for the constraint corresponding to $\nu _r$ in $Q_l$, we have that

$$\sum _{z\in N(\nu _r)} y_z=1\le \sum _{z\in N(\nu _r)} x_z \le \sum _{z\in T_r} x_z \left[ \text {since }N(\nu _r)\subseteq T_r\text { by Observation B}\right] $$

Therefore, from above argument and from Lemma 2 we conclude that

$$|S|=\sum _{r\in S} y_r=\sum _{r\in S} \sum _{z\in N(\nu _r)} y_z\le \sum _{r\in S} \sum _{z\in T_r} x_z \le 2\sum _{z\in R} x_z \le 2|OPT|.$$

Hence we have the proof.   $\square$

The proofs of Lemma 1 and Theorem 4 follows directly from the proof of Lemma 3.

3 Proof of Theorem 3

In this section, we shall give an $O(n^5)$-time 8-approximation algorithm to solve the MDS problem on vertically-stabbed-L graphs. In the rest of the paper, OPT(Q) and $OPT(Q_l)$ denote the cost of the optimum solution of an ILP Q and LP $Q_l$ respectively.

Overview of the Algorithm: First, we solve the relaxed LP formulation of the ILP of the MDS problem on the input vertically-stabbed-L graph G and create two subproblems. We shall show that one of those two subproblems is equivalent to the SSR problem and the other is equivalent to a Stabbing Rays with Segments problem (defined below) introduced by Katz et al. [15]. We solve these two subproblems individually.

In the Stabbing Rays with Segments (SRS) problem, the input is a set R of disjoint leftward-directed horizontal rays and a set V of disjoint vertical segments. The objective is to select a minimum cardinality subset of V that intersects all rays in R. We shall propose a heuristic algorithm that gives a feasible solution, and its cost is at most twice the cost of the optimum solution of the ILP of SRS.

2-Approximation Algorithm for SRS Problem: With each segment $v\in V$, we associate a token $T_v$ which is a subset of V. Initialise $T_v=\emptyset$ for each $v\in V$. Let $r_i$ be the ray whose right-endpoint, $(x_i , y_i)$, has the smallest x-coordinate. We assume without loss of generality that x- and y-coordinates of the endpoints of the rays are all distinct. Assuming that there is a feasible solution to the SRS instance, there must exist a segment of V that intersects $r_i$. Let $N(r_i)\subseteq V$ be the set of segments that intersect $r_i$. Let $v_{top}$ (resp. $v_{bot}$) be a segment in $N(r_i)$ whose top endpoint is top-most (resp., bottom endpoint is bottom-most); it may be that $v_{top}=v_{bot}$. We add both $v_{top}$ and $v_{bot}$ to our heuristic solution set S. Also we set $T_{v_{top}}=T_{v_{bot}}=N(r_i)$. We remove from R all of the rays that intersect $v_{top}$ or $v_{bot}$, delete all segments in $N(r_i)$ and then repeat the above steps untill $R=\emptyset$. Observe that for each ray r, there is a segment $v\in S$ that intersects r. Also observe that for each segment $v\in V$, there are at most two tokens such that both of them contains v.

Lemma 4

Let Q be the ILP of the SRS instance with a set of rays R and set of segments V as input and $Q_l$ be the corresponding relaxed LP. Then $OPT(Q)\le 2\cdot OPT(Q_l)$.

Proof

Let $\mathbf X =\{x_v\}_{v\in V}$ be an optimal solution of $Q_l$ where $x_v$ denotes the value of the variable in $Q_l$ corresponding to $v\in V$. Let S be the solution returned by the above algorithm with R, V as input. Now define for each $v\in V$, $y_v=1$ if $v\in S$, $y_v=0$ if $v\notin S$ and let $\mathbf Y =\{y_v\}_{v\in V}$. Observe that $\mathbf Y$ is a feasible solution of Q. For each $z\in S$, there is a ray $r_i$ such that $T_z=N(r_i)$. Therefore, $y_z= 1 \le \displaystyle \sum \limits _{v\in N(r_i)} x_v = \displaystyle \sum \limits _{v\in T_{z}} x_v$.

As a segment v is contained in at most two tokens, using the above inequality we have

$$|S|= \displaystyle \sum \limits _{v\in S} y_v\le \displaystyle \sum \limits _{v\in S} \displaystyle \sum \limits _{v'\in T_v} x_{v'}\le 2\displaystyle \sum \limits _{v'\in V} x_{v'}= 2\cdot OPT(Q_l)$$

Hence the result follows.   $\square$

Now we describe our approximation algorithm for MDS problem on vertically-stabbed-L graphs. Let $\mathcal {R}=\{\textsc {L}_u\}_{u\in V}$ be a vertically-stabbed-L-representation of a graph $G=(V,E)$. Without loss of generality, we assume that (i) the vertical line $x=0$ intersects all the L-paths in $\mathcal {R}$ and the x-coordinate of the corner point of each L-path in $\mathcal {R}$ is strictly less than 0, and (ii) whenever two distinct L-paths intersect in $\mathcal {R}$, they intersect at exactly one point.

For a vertex $u\in V$, let N[u] denote the closed neighbourhood of u in G, $H_u=\{c\in N[u]:\textsc {L}_c~\text {intersects the horizontal segment of}~\textsc {L}_u\}$ and let $V_u$ denote the set $N(u)\setminus H_u$. Based on these we have the following ILP (say Q) of the problem of finding a minimum dominating set of G.

Let $Q_l$ be the the relaxed LP formulation of Q and $\mathbf Q _l=\{x_v:v\in V\}$ be an optimal solution of $Q_l$. Now we define the following sets.

$$A_1=\left\{ u\in V:\displaystyle \sum \limits _{v\in H_u} x_v \ge \frac{1}{2} \right\} , A_2=\left\{ u\in V:\displaystyle \sum \limits _{v\in V_u} x_v \ge \frac{1}{2} \right\} $$

$$H=\bigcup \limits _{u\in A_1} H_u,V=\bigcup \limits _{u\in A_2} V_u$$

Based on these, we consider the following two integer programs $Q'$ and $Q''$.

Let $Q'_l$ and $Q''_l$ be the relaxed LP of $Q'$ and $Q''$ respectively. Clearly, the solutions of $Q'$ and $Q''$ gives a feasible solution for Q. Hence $OPT(Q)\le OPT(Q') + OPT(Q'')$. For each $x_v\in \mathbf Q _l$, define $y_v=\min \{1,2x_v\}$ and define $\mathbf Y _l=\{y_v\}_{x_v\in \mathbf Q _l}$. Notice that $\mathbf Y _l$ gives a solution to $Q'_l$ and $Q''_l$. Therefore, $OPT(Q'_l) + OPT(Q''_l)\le 4\cdot OPT(Q_l)$. We have the following lemma.

Lemma 5

$OPT(Q')\le 2\cdot OPT(Q'_l)$ and $OPT(Q'')\le 2\cdot OPT(Q''_l)$.

Proof

Note that for each vertex $u\in A_1$, $H_u$ is non-empty and for each $v\in H_u$, $L_v$ intersects the horizontal segment of $L_u$. Let R be the set of horizontal segments of the L-paths representing the vertices in $A_1$ and S be the set of vertical segments of the L-paths representing the vertices in H. Since all horizontal segments in R intersect the vertical line $x=0$ and the x-coordinates of the vertical segments in S is strictly less than 0, we can consider the horizontal segments in R as rightward directed rays. Hence, solving $Q'$ is equivalent to solving the ILP, say $\mathcal {E}$, of the problem of finding a minimum cardinality subset of vertical segments S that intersects all rays in the set R of rightward-directed rays. Hence solving $\mathcal {E}$ is equivalent to solving an SRS instance with R and S as input. By Lemma 4, we have that

$$OPT(Q')=OPT(\mathcal {E})\le 2\cdot OPT(\mathcal {E}_l)\le 2\cdot OPT(Q'_l)$$

where $\mathcal {E}_l$ is the relaxed LP of $\mathcal {E}$. Hence we have proof of the first part.

For the second part, using similar arguments as above, we can show that solving $Q''$ is equivalent to solving an SSR instance. Hence, by Lemma 1, we have that $OPT(Q'')\le 2\cdot OPT(Q''_l)$. Hence the proof follows.   $\square$

Proof of Theorem 3: Lemma 5 implies that solving $Q'$ (resp. $Q''$) is equivalent to solving SRS (resp. SSR) problem instance. Let A be the union of the solutions returned by 2-approximation algorithm for SRS problem and SSR-algorithm, used to solve $Q'$ and $Q''$ respectively. Hence,

$$|A|\le 2(OPT(Q'_l)+OPT(Q''_l)) \le 8\cdot OPT(Q_l) \le 8\cdot OPT(Q)$$

Since $Q_l$ consists of n variables where $n=|V|$, solving $Q_l$ takes $O(n^5)$ time [19]. Solving both the SSR and SRS instances takes a total of $O(n\log n)$ time and therefore the total running time of the algorithm is $O(n^5)$.

4 Proof of Theorem 1 and Theorem 2

First we prove the NP-hardness for the MDS problem on unit $B_1$-VPG graphs. The (h, w)-grid is the undirected graph G with vertex set $\{(x,y):x,y\in \mathbb {Z}, 1\le x\le h, 1\le y\le w\}$ and edge set $\{(u,v)(x,y):|u-x|+|v-y|=1\}$. A graph G is a grid graph if G is an induced subgraph of (h, w)-grid for some positive integers h, w.

Proof of Theorem 1: We shall reduce the NP-complete MDS problem on grid graphs [6] to the MDS problem on unit $B_1$-VPG graphs. It is sufficient to show that for any positive integers h, w the (h, w)-grid has a unit $B_1$-VPG representation. Let $G=(V,E)$ be a (h, w)-grid and $\epsilon =\frac{1}{hw}$. For each $(x,y)\in V$ consider the unit L-path $\textsc {L}_{(x,y)}$ such that top endpoint of vertical segment of $\textsc {L}_{(x,y)}$ is $(x,y-\epsilon (x-1))$. The set $\mathcal {R}=\{\textsc {L}_{(x,y)}:(x,y)\in V\}$ is a unit $B_1$-VPG representation of G. This completes the proof.

To prove Theorem 2 we will need Lemma 6. For sake of clarity, first we shall prove Theorem 2 assuming Lemma 6 to be true. The proof of Lemma 6 uses Lemma 1, but it is omitted due to space constraints.

Lemma 6

Let $S_1$ and $S_2$ be sets of orthogonal unit length segments. Let $\mathcal {C}$ be the ILP of the problem of finding a minimum cardinality subset D of $S_2$ such that every segment in $S_1$ intersects some segment in D. There is an $O(n^5)$-time algorithm to compute a set $D'$ which gives a solution of $\mathcal {C}$ and $|D'|\le 18\cdot OPT(\mathcal {C}_l)$ where $n=|S_1\cup S_2|$ and $\mathcal {C}_l$ is the relaxed LP of $\mathcal {C}$.

Completion of Proof of Theorem 2 Assuming Lemma 6: Let $\mathcal {R}$ be a unit $B_k$-VPG representation of a unit $B_k$-VPG graph $G=(V,E)$. We shall assume that every vertex of G has a self loop (this does not contradict the intersection model as every rectilinear path intersects itself). For a vertex $v\in V$, let P(v) denote the path in $\mathcal {R}$ that corresponds to v. For a vertex $v\in V$, N(v) and N[v] denote the open neighbourhood and closed neighbourhood of v, respectively. Throughout this section, we assume that the segments of each path $P\in \mathcal {R}$ are numbered consecutively starting from the leftmost segment by $1,2,\ldots ,t$ where $t (\le k+1)$ is the number of segments in P.

Define $\phi :E\rightarrow \mathbb {N}\times \mathbb {N}$ such that for an edge uv, $\phi (uv)=(i,j)$ if and only if the $i^{th}$ segment of P(u) intersects the $j^{th}$ segment of P(v), and for all $1\le a<i$ and $1\le b<j$, the $a^{th}$ segment of P(u) does not intersect the $b^{th}$ segment of P(v).

For a vertex $u\in V$, let $X_u(i,j)=\{v\in N[u] :\phi (uv)=(i,j)\}$. For distinct pairs (i, j) and $(i',j')$ the sets $X_u(i,j)$ and $X_u(i',j')$ are disjoint. Let $\mathcal {K}$ denote the set $\{1,2,\ldots ,k+1\}\times \{1,2,\ldots ,k+1\}$. Based on these we have the following ILP of the MDS problem on G.

First step of our algorithm is to solve the relaxed LP formulation (say $Q_l$) of Q. Let $\mathbf {Q}_l=\{x_v:v\in V \}$ be an optimal solution of $Q_l$. For each vertex $u\in V$, there is a pair $(i,j)\in \mathcal {K}$ such that $\sum \limits _{v\in X_u(i,j)} x_v \ge \frac{1}{(k+1)^2}$. For each pair $(i,j)\in \mathcal {K}$, define

$$A(i,j)=\left\{ u\in V:\sum \limits _{v\in X_u(i,j)} x_v \ge \frac{1}{(k+1)^2}\right\} , B(i,j)=\displaystyle \bigcup \limits _{u\in A(i,j)} X_{u}(i,j)$$

Based on these, we have the following ILP for each pair $(i,j)\in \mathcal {K}$.

For each pair $(i,j)\in \mathcal {K}$, let $Q_l(i,j)$ be the relaxed LP of Q(i, j). We have the following

$$OPT(Q)\le \displaystyle \sum \limits _{(i,j)\in \mathcal {K}} OPT(Q(i,j))$$

For each $x_v\in \mathbf Q _l$, define $y_v=\min \{1,x_v(k+1)^2\}$ and define $\mathbf Y _l=\{y_v\}_{x_v\in \mathbf Q _l}$. Clearly, $\mathbf Y _l$ gives a solution to $Q_l(i,j)$ for each $(i,j)\in \mathcal {K}$. Moreover,

$$\displaystyle \sum \limits _{(i,j)\in \mathcal {K}} OPT(Q_l(i,j)) \le (k+1)^4\cdot OPT(Q_l)$$

Now we have the following lemma.

Lemma 7

For each pair $(i,j)\in \mathcal {K}$, there is a solution D(i, j) for Q(i, j) such that $|D(i,j)|\le 18\cdot OPT(Q_l(i,j))$.

Proof

For any $(i,j)\in \mathcal {K}$, solving Q(i, j) is equivalent to finding a minimum cardinality subset D of B(i, j) such that each vertex $u\in A(i,j)$ has a neighbour in $D\cap X_u(i,j)$. Notice that, for each $u\in A(i,j)$ the set $X_u(i,j)$ is non-empty and each $v\in X_u(i,j)$, the $i^{th}$ segment of P(u) intersects the $j^{th}$ segment of P(v). Let $S=\{i^{th}\text { segment of } P(u):u\in A(i,j)\}, T=\{j^{th}\text { segment of } P(v):v\in B(i,j)\}$.

Solving Q(i, j) is equivalent to the problem of finding a minimum cardinality subset D of T such that every segment in S intersect at least one segment in D. Moreover, every segment in $S\cup T$ have unit length. Hence by Lemma 6, we have a solution (say D(i, j)) for Q(i, j) such that $|D(i,j)|\le 18\cdot OPT(Q(i,j))$.   $\square$

For each pair $(i,j)\in \mathcal {K}$, due to Lemma 7, we have a solution D(i, j) of Q(i, j) such that $|D(i,j)|\le 18\cdot OPT(Q_l(i,j))$ in polynomial time. Let D be the union of D(i, j)’s for all $(i,j)\in \mathcal {K}$. We have that

$$\begin{aligned} \begin{aligned} |D|&= \displaystyle \sum \limits _{(i,j)\in \mathcal {K}} |D(i,j)| \\&\le \displaystyle \sum \limits _{(i,j)\in \mathcal {K}} 18\cdot OPT(Q_l(i,j)) \\&\le 18\cdot (k+1)^4\cdot OPT(Q_l)\le 18\cdot (k+1)^4\cdot OPT(Q) \end{aligned} \end{aligned}$$

This completes the proof of Theorem 2.

5 Conclusion

We initiated the study of approximating MDS of string graphs in terms of their unit $B_k$-VPG representation. It is unlikely that there is $o(\log k)$-approximaiton algorithm for MDS problem on $B_k$-VPG graphs. This naturally leads to the following question.

Question 1

Is there an $O(\log k)$-approximation algorithm for the MDS problem on $B_k$-VPG graphs or unit $B_k$-VPG graphs?