The Geodetic Hull Number is Hard for Chordal Graphs

TL;DR

This paper proves that determining the geodetic hull number is computationally hard for chordal graphs, highlighting a significant complexity challenge in this graph class.

Contribution

It establishes the NP-hardness of computing the geodetic hull number specifically for chordal graphs, a previously unresolved problem.

Findings

Geodetic hull number is NP-hard for chordal graphs

Complexity results impact algorithms for related graph problems

Highlights difficulty in exact computation for this graph class

Abstract

We show the hardness of the geodetic hull number for chordal graphs.

Equations

$$B$$

$$Z$$

$$S=\bigcup _{i\in [n]}\left\{x_i^{\prime 1},x_i^{\prime 2},{\bar{x}}_i^{\prime \prime }\right\}\cup \bigcup _{i\in [n]:x_{i}truein\mathcal{S}}\left\{x_i\right\}\cup \bigcup _{i\in [n]:x_{i}falsein\mathcal{S}}\left\{{\bar{x}}_i\right\}.$$

$$\bigcup _{i\in [n]:x_{i}truein\mathcal{S}}\left\{x_i\right\}\cup \bigcup _{i\in [n]:x_{i}falsein\mathcal{S}}\left\{{\bar{x}}_i\right\}.$$

$$V_j=\{c_j\}\cup \bigcup _{i\in [n]:j=j_i^{(1)}}\left\{x_i,x_i^{\prime },x_i^1\right\}\cup \bigcup _{i\in [n]:j=j_i^{(2)}}\left\{x_i,x_i^{\prime },x_i^2\right\}\cup \bigcup _{i\in [n]:j=j_i^{(3)}}\left\{{\bar{x}}_i,{\bar{x}}_i^{\prime }\right\}$$

Full text

1 Laboratoire d’Informatique, de Robotique et de Microélectronique de Montpellier

Université de Montpellier, Montpellier, France, stephane.bessy@lirmm.fr

2 Departamento de Ciência da Computação, Instituto de Matemática

Universidade Federal do Rio de Janeiro, Rio de Janeiro, Brazil, mitre@dcc.ufrj.br

3 Institute of Optimization and Operations Research

Ulm University, Ulm, Germany, $\{$lucia.penso,dieter.rautenbach$\}$@uni-ulm.de

[TABLE]

1 Introduction

One of the most well studied convexity notions for graphs is the shortest path convexity or geodetic convexity, where a set $X$ of vertices of a graph $G$ is considered convex if no vertex outside of $S$ lies on a shortest path between two vertices inside of $S$. Defining the convex hull of a set $S$ of vertices as the smallest convex set containing $S$, a natural parameter of $G$ is its hull number $h(G)$ [7], which is the minimum order of a set of vertices whose convex hull is the entire vertex set of $G$. The hull number is NP-hard for bipartite graphs [2], partial cubes [1], and $P_{9}$-free graphs [5], but it can be computed in polynomial time for cographs [4], $(q,q-4)$-graphs [2], $\{{\rm paw},P_{5}\}$-free graphs [3, 5], and distance-hereditary graphs [9]. Bounds on the hull number are given in [2, 6, 7].

In [9] Kanté and Nourine present a polynomial time algorithm for the computation of the hull number of chordal graphs. Unfortunately, their correctness proof contains a gap described in detail at the end of the present paper. As our main result we show that computing the hull number of a chordal graph is NP-hard, which most likely rules out the existence of a polynomial time algorithm.

Before we proceed to our results, we collect some notation and terminology. We consider finite, simple, and undirected graphs. A graph $G$ has vertex set $V(G)$ and edge set $E(G)$. A graph $G$ is chordal if it does not contain an induced cycle of order at least $4$. A clique in $G$ is the vertex set of a complete subgraph of $G$. A vertex of a graph $G$ is simplicial in $G$ if its neighborhood is a clique. The distance ${\rm dist}_{G}(u,v)$ between two vertices $u$ and $v$ in $G$ is the minimum number of edges of a path in $G$ between $u$ and $v$. The diameter ${\rm diam}(G)$ of $G$ is the maximum distance between any two vertices of $G$. The eccentricity $e_{G}(u)$ of a vertex $u$ of $G$ is the maximum distance between $u$ and any other vertex of $G$. For a positive integer $k$, let $[k]$ be the set of the positive integers at most $k$.

Let $G$ be a graph, and let $S$ be a set of vertices of $G$. The interval $I_{G}(S)$ of $S$ in $G$ is the set of all vertices of $G$ that lie on shortest paths in $G$ between vertices from $S$. Note that $S\subseteq I_{G}(S)$, and that $S$ is convex in $G$ if $I_{G}(S)=S$. The set $S$ is concave in $G$ if $V(G)\setminus S$ is convex. Note that $S$ is concave if and only if $S\cap I_{G}(\{v,w\})=\emptyset$ for every two vertices $v$ and $w$ in $V(G)\setminus S$. The hull $H_{G}(S)$ of $S$ in $G$, defined as the smallest convex set in $G$ that contains $S$, equals the intersection of all convex sets that contain $S$. The set $S$ is a hull set if $H_{G}(S)=V(G)$, and the hull number $h(G)$ of $G$ [5, 7] is the smallest order of a hull set of $G$.

2 Result

We immediately proceed to our main result.

Theorem 2.1**.**

For a given chordal graph $G$, and a given integer $k$, it is NP-complete to decide whether the hull number $h(G)$ of $G$ is at most $k$.

Proof.

Since the hull of a set of vertices of $G$ can be computed in polynomial time, the considered decision problem belongs to NP. In order to prove NP-completeness, we describe a polynomial reduction from a restricted version of Satisfiability. Therefore, let ${\cal C}$ be an instance of Satisfiability consisting of $m$ clauses $C_{1},\ldots,C_{m}$ over $n$ boolean variables $x_{1},\ldots,x_{n}$ such that every clause in ${\cal C}$ contains at most three literals, and, for every variable $x_{i}$, there are exactly two clauses in ${\cal C}$, say $C_{j_{i}^{(1)}}$ and $C_{j_{i}^{(2)}}$, that contain the literal $x_{i}$, and exactly one clause in ${\cal C}$, say $C_{j_{i}^{(3)}}$, that contains the literal $\bar{x}_{i}$, and these three clauses are distinct. Using a polynomial reduction from [LO1] [8], it has been shown in [5] that Satisfiability restricted to such instances is still NP-complete.

Let the graph $G$ be constructed as follows starting with the empty graph:

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For every $j\in[m]$, add a vertex $c_{j}$.

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For every $i\in[n]$, add three $y_{i},\bar{y}_{i},$ and $z_{i}$.

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Add edges such that $B\cup Z$ is a clique, where

[TABLE]

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For every $i\in[n]$, add $9$ vertices and $25$ edges to obtain the subgraph indicated in Figure 1.

Note that ${\rm dist}_{G}(x_{i},\bar{x}_{i}^{\prime})={\rm dist}_{G}(\bar{x}_{i},x^{\prime 1}_{i})=3$ for every $i$ in $[n]$. Since every vertex of $G$ has a neighbor in the clique $B\cup Z$, the diameter of $G$ is $3$. Furthermore, since no vertex is universal, all vertices in $B\cup Z$ have eccentricity $2$.

Let $k=4n$.

Note that the order of $G$ is $12n+m$.

It remains to show that $G$ is chordal, and that $\cal C$ is satisfiable if and only if $h(G)\leq k$.

In order to show that $G$ is chordal, we indicate a perfect elimination ordering, which is a linear ordering $v_{1},\ldots,v_{12n+m}$ of its vertices such that $v_{i}$ is simplicial in $G-\{v_{1},\ldots v_{i-1}\}$ for every $i$ in $[12n+m]$. Such an ordering is obtained by

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starting with the vertices $x^{\prime 1}_{i},x^{\prime 2}_{i},$ and $\bar{x}^{\prime\prime}_{i}$ for all $i\in[n]$ (in any order),

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continuing with the vertices $x^{1}_{i},x^{2}_{i},$ and $\bar{x}^{\prime}_{i}$ for all $i\in[n]$,

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continuing with the vertices $x^{\prime}_{i}$ for all $i\in[n]$,

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continuing with the vertices $x_{i}$ and $\bar{x}_{i}$ for all $i\in[n]$, and

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ending with the vertices in the clique $B\cup Z$.

Now, let $\cal S$ be a satisfying truth assignment for $\cal C$.

Let

[TABLE]

Clearly, $|S|=k=4n$. For every $i$ in $[n]$, we have $\{z_{i},\bar{y}_{i}\}\subseteq I_{G}(\{x_{i},\bar{x}^{\prime\prime}_{i}\})$, $\{z_{i},y_{i}\}\subseteq I_{G}(\{\bar{x}_{i},x^{\prime 1}_{i}\})$, $y_{i}\in I_{G}(\{\bar{y}_{i},x^{\prime 1}_{i}\})$, and $\bar{y}_{i}\in I_{G}(\{y_{i},\bar{x}^{\prime\prime}_{i}\})$, which implies $\{z_{i},y_{i},\bar{y}_{i}\}\subseteq H_{G}(S)$. Since ${\cal S}$ is a satisfying truth assignment, for every $j$ in $[m]$, there is a neighbor, say $v$, of $c_{j}$ in

[TABLE]

If $v\in\bigcup\limits_{i\in[n]:\,\,x_{i}\,\,true\,\,in\,\,{\cal S}}\left\{x_{i}\right\}$, then $c_{j}\in I_{G}(\{v,\bar{x}^{\prime\prime}_{i}\})$, otherwise $c_{j}\in I_{G}(\{v,x^{\prime 1}_{i}\})$. Hence, $B\cup Z\subseteq H_{G}(S)$.

Now, for some $i$ in $[n]$, let $c_{j}$, $c_{k}$, and $c_{\ell}$ be the neighbors in $B\setminus\{y_{i},\bar{y}_{i}\}$ of $x^{1}_{i}$, $x^{2}_{i}$, and $\bar{x}^{\prime}_{i}$, respectively, similarly as in Figure 1. We have $x^{1}_{i}\in I_{G}(\{x^{\prime 1}_{i},c_{j}\})$, $x^{2}_{i}\in I_{G}(\{x^{\prime 2}_{i},c_{k}\})$, $x_{i}^{\prime}\in I_{G}(\{x^{1}_{i},x^{2}_{i}\})$, $\bar{x}^{\prime}_{i}\in I_{G}(\{\bar{x}^{\prime\prime}_{i},c_{\ell}\})$, $x_{i}\in I_{G}(\{x_{i}^{\prime},z_{i}\})$, and $\bar{x}_{i}\in I_{G}(\{\bar{x}^{\prime}_{i},z_{i}\})$.

Altogether, we obtain that $S$ is a hull set of $G$ of order $4n$.

Finally, let $S$ be a hull set of $G$ of order at most $4n$.

Claim 1**.**

For every $i\in[n]$, the set $\{x_{i},z_{i},\bar{x}_{i}\}$ is concave.

Proof of Claim 1: For a contradiction, suppose that some vertex in $S^{\prime}=\{x_{i},z_{i},\bar{x}_{i}\}$ lies on a shortest path $P$ in $G$ between two vertices $v$ and $w$ in $V(G)\setminus S^{\prime}$. Since the diameter of $G$ is $3$, the path $P$ contains at most $2$ vertices of $S^{\prime}$. Since the neighbors outside of $S^{\prime}$ of each vertex in $S^{\prime}$ form a clique, the path $P$ contains exactly $2$ adjacent vertices of $S^{\prime}$, that is, either $P=vx_{i}z_{i}w$ or $P=v\bar{x}_{i}z_{i}w$. In both cases, the vertex $w$ has eccentricity at least $3$. However, every neighbor $w$ of $z_{i}$ outside $S^{\prime}$ belongs to $B\cup Z$, and thus, has eccentricity $2$, a contradiction. $\Box$

Claim 2**.**

For every $j\in[m]$, the set

[TABLE]

is concave.

Proof of Claim 2: First, suppose that $C_{j}$ contains the positive literal $x_{i}$. By symmetry, we may assume that $j=j_{i}^{(1)}$ and $j_{i}^{(2)}=k$ for some $k$ in $[m]\setminus\{j\}$.

First, suppose that some shortest path $P$ between two vertices $v$ and $w$ in $\bar{V}_{j}=V(G)\setminus V_{j}$ contains $x_{i}$. Choosing $P$ of minimum length, it follows that $v$ and $w$ are the only vertices of $P$ in $\bar{V}_{j}$. Since the diameter of $G$ is $3$, the length of $P$ is at most $3$, and we may assume that $v$ is a neighbor of $x_{i}$, which implies $v\in\{z_{i},c_{k},y_{i}\}$. Since $\{z_{i},c_{k},y_{i}\}$ is a clique, the vertex $w$ is not a neighbor of $x_{i}$, and $P$ contains exactly one vertex $u$ of $V_{j}$ different of $x_{i}$, which implies $P=vx_{i}uw$ and $u\in\{x^{\prime}_{i},c_{j}\}$. Suppose that $u=x^{\prime}_{i}$. This implies $w\in\{x^{2}_{i},c_{k},y_{i}\}$. Since $c_{k},y_{i}\in N_{G}(x_{i})$, we obtain $w=x^{2}_{i}$ and $v=z_{i}$. However, ${\rm dist}_{G}(z_{i},x^{2}_{i})=2$, which is a contradiction. Hence, $u=c_{j}$ and $w\in B\cup Z$. However, every vertex in $B\cup Z$ has eccentricity $2$, which is a contradiction. Hence, no shortest path between two vertices in $\bar{V}_{j}$ contains $x_{i}$.

Next, suppose that some shortest path $P$ between two vertices $v$ and $w$ in $\bar{V}_{j}$ contains $x_{i}^{\prime}$. Similarly as above, we may assume that $v$ and $w$ are the only vertices of $P$ in $\bar{V}_{j}$, the length of $P$ is at most $3$, and $v$ is a neighbor of $x_{i}^{\prime}$, which implies $v\in\{x^{2}_{i},y_{i},c_{k}\}$. Since $\{x^{2}_{i},y_{i},c_{k}\}$ is a clique, the path $P$ contains exactly one vertex $u$ of $V_{j}$ different of $x_{i}^{\prime}$, which implies $P=vx^{\prime}_{i}uw$ and $u\in\{x_{i}^{1},c_{j}\}$, where we use that $P$ does not contain $x_{i}$. Suppose that $u=x_{i}^{1}$. This implies $w\in\{x^{\prime 1}_{i},y_{i}\}$. Since $y_{i}\in N_{G}(x^{\prime}_{i})$, we obtain $w=x^{\prime 1}_{i}$ and $v=x^{2}_{i}$. However, ${\rm dist}_{G}(x^{2}_{i},x^{\prime 1}_{i})=2$, which is a contradiction. Hence, $u=c_{j}$ and $w\in B\cup Z$. However, every vertex in $B\cup Z$ has eccentricity $2$, which is a contradiction. Hence, no shortest path between two vertices in $\bar{V}_{j}$ contains $x^{\prime}_{i}$.

Next, suppose that some shortest path $P$ between two vertices $v$ and $w$ in $\bar{V}_{j}$ contains $x_{i}^{1}$. Similarly as above, we may assume that $v$ and $w$ are the only vertices of $P$ in $\bar{V}_{j}$, the length of $P$ is at most $3$, and $v$ is a neighbor of $x_{i}^{1}$, which implies $v\in\{x^{\prime 1}_{i},y_{i}\}$. Since $\{x^{\prime 1}_{i},y_{i}\}$ is a clique, the path $P$ contains exactly one vertex $u$ of $V_{j}$ different of $x_{i}^{1}$, which implies $P=vx_{i}^{1}c_{j}w$ and $w\in B\cup Z$, where we use that $P$ does not contain $x^{\prime}_{i}$. However, every vertex in $B\cup Z$ has eccentricity $2$, which is a contradiction. Hence, no shortest path between two vertices in $\bar{V}_{j}$ contains $x_{i}^{1}$.

Next, suppose that $C_{j}$ contains the negative literal $\bar{x}_{i}$, that is, $j=j_{i}^{(3)}$.

First, suppose that some shortest path $P$ between two vertices $v$ and $w$ in $\bar{V}_{j}$ contains $\bar{x}_{i}$. Similarly as above, we may assume that $v$ and $w$ are the only vertices of $P$ in $\bar{V}_{j}$, the length of $P$ is at most $3$, and $v$ is a neighbor of $\bar{x}_{i}$, which implies $v\in\{z_{i},\bar{y}_{i}\}$. Since $\{z_{i},\bar{y}_{i}\}$ is a clique, the vertex $w$ is not a neighbor of $\bar{x}_{i}$, and $P$ contains exactly one vertex $u$ of $V_{j}$ different of $\bar{x}_{i}$, which implies $P=v\bar{x}_{i}uw$ and $u\in\{\bar{x}^{\prime}_{i},c_{j}\}$. Suppose that $u=\bar{x}^{\prime}_{i}$. This implies $w\in\{\bar{x}^{\prime\prime}_{i},\bar{y}_{i}\}$. Since $\bar{y}_{i}\in N_{G}(\bar{x}_{i})$, we obtain $v=z_{i}$ and $w=\bar{x}^{\prime\prime}_{i}$. However, ${\rm dist}_{G}(z_{i},\bar{x}^{\prime\prime}_{i})=2$, which is a contradiction. Hence, $u=c_{j}$ and $w\in B\cup Z$. However, every vertex in $B\cup Z$ has eccentricity $2$, which is a contradiction. Hence, no shortest path between two vertices in $\bar{V}_{j}$ contains $\bar{x}_{i}$.

Next, suppose that some shortest path $P$ between two vertices $v$ and $w$ in $\bar{V}_{j}$ contains $\bar{x}_{i}^{\prime}$. Similarly as above, we may assume that $v$ and $w$ are the only vertices of $P$ in $\bar{V}_{j}$, the length of $P$ is at most $3$, and $v$ is a neighbor of $\bar{x}_{i}^{\prime}$, which implies $v\in\{\bar{x}_{i}^{\prime\prime},\bar{y}_{i}\}$. Since $\{\bar{x}_{i}^{\prime\prime},\bar{y}_{i}\}$ is a clique, the path $P$ contains exactly one vertex $u$ of $V_{j}$ different of $\bar{x}_{i}^{\prime}$, which implies $P=v\bar{x}_{i}^{\prime}c_{j}w$ and $w\in B\cup Z$, where we use that $P$ does not contain $\bar{x}_{i}$. However, every vertex in $B\cup Z$ has eccentricity $2$, which is a contradiction. Hence, no shortest path between two vertices in $\bar{V}_{j}$ contains $\bar{x}_{i}^{\prime}$.

Finally, since the neighbors of $c_{j}$ outside of $V_{j}$ form a clique, no shortest path between two vertices in $\bar{V}_{j}$ contains $c_{j}$, which completes the proof of the claim. $\Box$

Note that all $3n$ simplicial vertices in $\bigcup\limits_{i\in[n]}\left\{x^{\prime 1}_{i},x^{\prime 2}_{i},\bar{x}^{\prime\prime}_{i}\right\}$ belong to $S$.

Since $S$ contains at most $n$ non-simplicial vertices, Claim 1 implies that, for every $i$ in $[n]$, the set $S$ contains exactly one of the three vertices in $\{x_{i},z_{i},\bar{x}_{i}\}$, and that these are the only non-simplicial vertices in $S$. Now, Claim 2 implies that, for every $j$ in $[m]$, there is some $i\in[n]$ such that

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either $C_{j}$ contains the literal $x_{i}$ and the vertex $x_{i}$ belongs to $S$

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or $C_{j}$ contains the literal $\bar{x}_{i}$ and the vertex $\bar{x}_{i}$ belongs to $S$.

Therefore, setting the variable $x_{i}$ to true if and only if the vertex $x_{i}$ belongs to $S$ yields a satisfying truth assignment ${\cal S}$ for ${\cal C}$, which completes the proof. ∎

As pointed out in the introduction, the correctness proof in [9] contains a gap. In lines 14 and 15 on page 322 of [9] it says

“At iteration $i+1$, the vertex $x_{i+1}$ is a simplicial vertex in $G_{i+1}$. We first claim that there exists no functional dependency of the form $zt\to x_{i+1}$ in $\Sigma$.”

Consider applying the algorithm from [9] to the graph in Figure 2. In iteration 1, it would decide to add $x_{1}$ to $K$. In iteration 2, it would decide not to add $x_{2}$ to $K$, because of $t\to x_{2}$. Furthermore, because of $t\to x_{2}$ and $z,x_{2}\to x_{3}$, it would replace $z,x_{2}\to x_{3}$ within $\Sigma$ with $z,t\to x_{3}$. Therefore, in iteration 3, $\Sigma$ would actually contain $z,t\to x_{3}$, contrary to the claim cited above.