Power Edge Set and Zero Forcing Set Remain Difficult in Cubic Graphs

Abstract

This paper presents new complexity and non-approximation results concerning two color propagation problems, namely Power Edge Set and Zero Forcing Set. We focus on cubic graphs, exploiting their structural properties to improve and refine previous results. We also give hardness results for parameterized precolored versions of these problems, and a polynomial-time algorithm for Zero Forcing Set in proper interval graphs.

1 Introduction

Motivation. In power networks, synchrophasors are time-synchronized numbers that represent both the magnitude and phase angle of the sine waves on network links. A Phasor Measurement Unit (PMU) is an expensive measuring device used to continuously collect the voltage and phase angle of a single station and the electrical lines connected to it. The problem of minimizing the number of PMUs to place on a network for complete network monitoring is an important challenge for operators and has gained a considerable attention over the past decade [4, 7, 8, 12, 13, 15, 17, 19, 21, 22, 25]. The problem is known as Power Dominating Set  [25] and we state it below. We model the network as a graph $G = (V, E)$ with $|V|=:n$ and $|E|=:m$. We denote the set of vertices and edges of G by V(G) and E(G), respectively. We let $N_G(v)$ denote the set of neighbors of $v\in V$ in G and $d_G(v)=|N_G (v)|$ its degree in G. Further, we let $N_G[v]=N_G(v)\cup \{v\}$ denote the closed neighborhood of v in G, and we let G[W] denote the subgraph of G induced by vertices $W\subseteq V(G)$. The problem is described through monitoring of nodes of the network, corresponding to monitoring vertices V(G) by PMUs, propagated using the following rules.

• $\textsc {Rule}~R_1$ $^*$: A vertex v of G on which a PMU is placed will be called a monitored vertex, and all its neighbors vertices $N_G(v)$ automatically become monitored.

• $\textsc {Rule}~R_2$: if all but one neighbor of a monitored vertex are monitored, then this unmonitored vertex will become monitored as well.

Letting $\Gamma _P(G)$ denote the minimum number of PMUs to place on vertices to obtain a full monitoring of the network (using $\textsc {Rule}~R_2$), the decision version of the problem is described as follows:

• Power Dominating Set (PDS)

• Input: a graph $G=(V,E)$ and some

• Question: Is $\Gamma _P(G)\le k$?

Power Dominating Set is $\mathcal {NP}$-complete in general graphs [15]. A large amount of literature is devoted to this problem, describing a wide range of approaches, either exact such as integer linear programming [12] or branch-and-cut [21], or heuristic, such as greedy algorithms [17], approximations [4] or genetic algorithms [19]. The problem has also been shown to be polynomial-time solvable on grids [7], but $\mathcal {NP}$-complete in unit-disk graphs [22].

In this paper, we consider two variants of the problem, called Power Edge Set  (PES) [23, 24] and Zero Forcing Set (ZFS) [3], which respectively consist in placing PMUs on the links, and reducing the monitoring range of a PMU placed on a node. This leads us to replace in each of these problems as follows ($\textsc {Rule}~R_2$ remains unchanged):

• $\textsc {Rule}~R_1$ (PES): two endpoints of an edge bearing a PMU are monitored.

• $\textsc {Rule}~R_1$ (ZFS): only the node bearing a PMU is monitored.

We let pes(G) and zfs(G) denote the minimum number of PMUs to place on the edges, resp. nodes, of G to entirely monitor G. Both PES and ZFS can be seen as a problem of color propagation with colors 0 (white) and 1 (black), respectively designating the states not monitored and monitored of a vertex of G. As input to PES or ZFS, we will consider a connected graph $G = (V, E)$. For each vertex $v\in V$, let c(v) be the color assigned to v (we abbreviate $\bigcup _{v\in X}c(v)=:c(X)$). Before placing the PMUs, we have $c(V)=\{0\}$ and the aim is to obtain $c(V)=\{1\}$ using $\textsc {Rule}~R_1$ and $\textsc {Rule}~R_2$ while minimizing the number of PMUs. See Figs. 1 and 2 for detailed examples illustrating the differences between PES and ZFS.

Fig. 1.

PMU propagation on PES problem: before any PMU placement, all vertices are white (a). A PMU on $\{b,c\}$ induces $c(b)=c(c)=1$ (black) by $\textsc {Rule}~R_1$ (b). By applying $\textsc {Rule}~R_2$ on b, we obtain $c(a)=1$ (c). Then $\textsc {Rule}~R_2$ on a induces $c(d)=1$ (d), and $\textsc {Rule}~R_2$ on c or d induces $c(e)=1$ (e). A second PMU is required to obtain a complete coloring. Placing a PMU on $\{e,f\}$ gives us $c(f)=1$ by $\textsc {Rule}~R_1$ (f). Finally, $\textsc {Rule}~R_2$ on e induces $c(g)=1$ (g). The set of edges where PMUs have been placed is $S= \{bc,ef\}$, giving (b, c, a, d, e, f, g) as a valid order for G.

Fig. 2.

PMU propagation on ZFS problem: before any PMU placement, all vertices are white (a). Placing one PMU on $\{b\}$ allows to monitor it. (b). Placing a second PMU on $\{c\}$ allows to monitor it (c), and now we can apply $\textsc {Rule}~R_2$ on b, to obtain $c(a)=1$ (d). Then $\textsc {Rule}~R_2$ on a induces $c(d)=1$ (e), and $\textsc {Rule}~R_2$ on c or d induces $c(e)=1$ (f). A third PMU is required to obtain a complete coloring. For example, placing a PMU on f (g) allows to apply $\textsc {Rule}~R_2$ on e to obtain $c(g)=1$ (h).

Previous work. Assigning a minimum number of PMUs to monitor the whole network is known to be $\mathcal {NP}$-hard in general for both PES and ZFS. For the former, some complexity results and a lower bound on approximation of $1.12-\epsilon$ with $\epsilon >0$ have been shown by Toubaline et al. [23], who also present a linear-time algorithm on trees by reduction to Path Cover. Poirion et al. [20] propose a linear program with binary variables indexed by the necessary iterations using propagation rules. Recently, inapproximability results have been proposed on planar or bipartite graphs [5]. In this work, we develop hardness results on complexity and approximation for special cases of Power Edge Set and Zero Forcing Set.

Preliminaries. In the following, we will consider a total order $\sigma$ of vertices of a graph G as a sequence $(v_1,v_2,\ldots )$ such that $v_i$ occurs before $v_j$ in the sequence if and only if $v_i<_\sigma v_j$.

Definition 1

(valid order). Let $G=(V,E)$ be a graph, let $S\subseteq E$ (resp. $S\subseteq V$), and let $\sigma$ be a total order of V, such that for each $v\in V(G)$, there is an edge incident to v in S (resp. $v\in S$) or there is a vertex $u\in N_G(v)$ which verifies $N_G[u]\le _\sigma v$. Then, $<_\sigma$ is called valid for S.

Given a graph $G=(V,E)$, any set $S\subseteq V$ (or $S\subseteq E$) such that repeated application of $\textsc {Rule}~R_1$ (ZFS) (or $\textsc {Rule}~R_1$ (PES)) and $\textsc {Rule}~R_2$ leads to G being completely monitored is called a zero forcing set (or power edge set). Using Definition 1, we can formally define the propagation process in G. For instance, in Fig. 1, a valid order for $S=\{bc, ef\}$ is (b, c, a, d, e, f, g).

Observation 1

Let $G=(V,E)$ be a graph and let $S\subseteq E$ (resp. $S\subseteq V$). Then, S is a power edge set (res. a zero forcing set) if and only if there is a valid order $\sigma$ on G, with respect to S.

Note that, for a graph $G=(V,E)$, any set $S\subseteq E$ is a power edge set if and only if $\bigcup _{e\in S}e$ is a zero forcing set for G. It is therefore a natural and unambiguous to also call such an edge set zero forcing set.

Finally, we call a vertex v propagating to $x\in N_G(v)$ if $c(x) = 0$ and for all $y \in N_G[v]\setminus \{x\}$, we have $c(y)=1$. Note that each maximal clique of G can contain at most one propagating vertex.

Lemma 1

Let $G=(V,E)$ be a graph, let S be a zero forcing set of G, and let $\mathcal {C}:=\{C_1,\ldots ,C_c\}$ be a set of maximal cliques in G covering E. Then $|V\setminus S|\le c$.

Proof

Let $\sigma$ be a valid order for S. We show that each $C_i$ contains at most one edge uv such that $v\notin S$ and $N_G[u]\le _\sigma v$. Since $\mathcal {C}$ covers E, this implies $|V\setminus S|\le |\mathcal {C}|=c$. Let $C\in \mathcal {C}$ and let C contain an edge uv such that $N_G[u]\le _\sigma v$ and $v\notin S$. Then, $C\subseteq N_G[u]$, implying $C\le _\sigma v$. Thus, v is the last vertex of C with respect to $\sigma$ and this vertex is unique.    $\square$

Contribution. The next section is devoted to the NP-completeness for cubic graph for Power Edge Set. We show that Power Edge Set and Zero Forcing Set are W[2]-hard parameterized by the size of the solution in Sect. 3. Section 4 is mainly dedicated to inapproximability and we show that there is an $\frac{n}{2}$-approximation for Power Edge Set. In the last section, we propose a linear polynomial-time algorithm on proper interval graph for Zero Forcing Set.

2 Computational Results

Most results presented in this section rely on reductions from graph problems using gadgets for vertices or edges of the original instance. We model the propagation process using the notion of valid order with respect to the solution set, whatever the nature of it: set of edges for PES, of vertices for ZFS.

We present new lower bounds for Power Edge Set that hold even in the very restricted case that G is cubic (i.e. all vertices in G have degree three). Previous results show that the problem is $\mathcal {NP}$-complete even if G is a subgraph of the grid with bounded degree at most three [5]. In this paper, we show the problem remains $\mathcal {NP}$-complete if G is cubic and planar. The proof is done by reduction from Vertex Cover (see below) on 3-regular, planar graphs, which is $\mathcal {NP}$-complete [11] but admits a PTAS [1], and a -approximation [2].

• 3-regular planar Vertex Cover (3RPVC).

• Input: a 3-regular planar graph $G=(V,E)$, some $k\in \mathcal {N}$.

• Question: Is there a size-k set $S\subseteq V$ covering E, i.e. $\forall _{e\in E}\;e\cap S \ne \varnothing$?

Construction 1

For a given cubic planar graph $G=(V,E)$ with n vertices, we construct a graph $G'$ as follows:

• For each $v \in V$, construct $H_v$ (see Fig. 3).

• If x is adjacent to y in G, we add exactly one of the edge between $x_0, x_1$ or $x_2$ and $y_0, y_1$ or $y_2$ to connect $H_x$ and $H_y$

Fig. 3.

The gadget $H_v$ for a vertex v.

The graph $G'$ is clearly cubic and planar and Construction 1 is applied in polynomial time. The construction is linear in n and k.

Lemma 2

The gadget $H_v$ needs at least one PMU to be fully colored: if $x_1$, $y_2$ and $z_0$ are propagating respectively to $v_1$, $v_2$ and $v_0$, then one PMU is sufficient; otherwise two PMUs are needed to fully color $H_v$.

Proof

First, if $x_1$, $y_2$ and $z_0$ are propagating respectively to $v_1$, $v_2$ and $v_0$, then, after application of $\textsc {Rule}~R_2$, $c(v_0)=c(v_1)=c(v_2)=1$. Thus this is the beginning of a valid order: $(v_0, v_1, v_2, v_3, v_5, v_4, v_6, v_7, v_{12}, v_9, v_{10})$. There is no more possible propagation, it is necessary to assign a new PMU. If we place it on the edge $v_{14}v_{16}$, the remainder of a valid order for $H_v$ is: $(v_{14}, v_{16}, v_{11}, v_8, v_{13}, v_{15}, v_{16})$.

Second, we show that $H_v$ may be colored by two PMUs in every case. If PMUs are assigned to the edges $v_{11}v_{13}$ and $v_{15}v_{16}$, we the following order is valid: $(v_{11}, v_{13}, v_{15}, v_{16}, v_{7}, v_{8}, v_{9}, v_{14}, v_4, v_6, v_{10}, v_3, v_{12}, v_3, v_{12}, v_1, v_2, v_5, v_0)$.

Third, we show that even if $x_1$ and $z_0$ are propagating to respectively $v_1$ and $v_0$, and $y_2$ is not, we need two PMUs to color $H_v$. The beginning of the propagation is given by the following order: $(v_0, v_1, v_3, v_5)$. There is no more possible propagation, therefore we have to put one more PMU. As more than two uncolored vertices remain, so we have to initiate propagation with this PMU. So the potential edges are $v_6v_{12}$, $v_4v_2$, $v_6v_9$ or $v_{10}v_{12}$ (other edges won’t start a propagation, and we need to color more than two vertices). By exhaustive search, we find that it is impossible to color $H_v$ with only one PMU on any one of these edges. We use the same kind of argument if $x_1$ and $y_2$ or $y_2$ and $z_0$ propagate.    $\square$

Theorem 1

Power Edge Set remains $\mathcal {NP}$-complete on planar cubic graphs.

Proof

Let $G'$ be the graph obtained by using Construction 1 on $G=(V,E)$, a cubic planar graph. We show that G has a size-k vertex cover iff Power Edge Set has a solution of size $n+k$ on $G'$. Clearly, Power Edge Set is in $\mathcal {NP}$.

“$\Rightarrow$”: With a size-k vertex cover S for G, we build a power edge set $S'$ for $G'$:

$$ S':=\bigcup _{v\in S}\{v_{11}v_{16}, v_{13}v_{15}\} \cup \bigcup _{v\in V\setminus S}\{v_{14}v_{16}\} $$

Then, $|S'|=n+k$ and, by Lemma 2, all vertices of $G'$ are colored by $S'$.

“$\Leftarrow$”: Suppose that $G'$ is colored with $n+k$ PMUs. By Lemma 2, there is at least one PMU on each gadget. Further, if a gadget $H_v$ is colored with a single PMU, then every $H_x$ with $x\in N_G(v)$ is colored with two PMUs inside (by Lemma 2). Then, $\{v\, |\, H_v \text { admits two PMUs}\}$ is a vertex cover for G.    $\square$

3 Parameterized Hardness

In what follows, we introduce parameterized versions of our problems and recall the notion of parameterized reduction. Using known results for Dominating Set, we deduce hardness results for Power Edge Set and Zero Forcing Set. First, we recall the parameterized Dominating Set problem. We obtain hardness results for a restricted version of our problems, when a precoloring exists on a particular set of vertices.

• Dominating Set (DS)

• Input: a graph $G=(V,E)$, some

• Question: Does G have a size-k dominating set?

• Parameter: k

• Precolored Zero Forcing Set/Precolored Power Edge Set

• Input: a graph $G=(V,E)$, a set $B\subseteq V$, some $c:V \rightarrow \{0,1\}$ with $c^{-1}(1)=B$, and an integer k

• Question: Is there a set $S'\subseteq V$ (resp. $S' \subseteq E$) of size k such that $B\cup S'$ (resp. $B\cup \bigcup _{e\in S'}e$) is a zero forcing set for G?

• Parameter: k

We prove the hardness using a parameterized reduction from Dominating Set. First, we introduce a gadget which allows to propagate a coloration, but only in one direction. It is called “check-valve”.

Definition 2

(Check-valve). A check-valve $C_{x,y}$ from x to y is a graph $G=(V,E)$, with $V = \{x,y,x_1,x_2,x_3,x_4\}$ and $E=\{xx_1$, $xx_2$, $x_1x_3$, $x_1x_4$, $x_2x_4$, $x_3y$, $x_4y\}$, with a coloring function , such that $c(x) = c(x_1) = 1$ and all other vertices are colored by 0. A check-valve $C_{x,y}$ is illustrated on Fig. 4.

Fig. 4.

The check-valve $C_{x,y}$

Observation 2

Let $C_{x,y}$ be a check-valve inserted between two vertices a and b, depicted by Fig. 4. Then:

• 1. If $c(a)=1$ then $c(b)=1$ after exhaustive application of $\textsc {Rule}~R_2$.

• 2. If $c(b)=1$, and $c(a)=0$, then c(a) is still 0 after exhaustive application of $\textsc {Rule}~R_2$, and it is necessary to add a PMU in order to have $c(a)=1$.

Construction 2

Let xy be a edge such that $c(x)=1$ and $c(y)=0$, we construct the gadget $C_{xy}$: we add vertices $x_1$, $x_2$, $x_3$ and $x_4$ and we add edges $xx_1$, $x_1x_2$, $x_3y$, $yx_4$, $x_4x_2$ et $x_2x$. Notice that xy is deleted.

Construction 3

For given $G=(V,E)$, construct $G' = (V',E')$ as follows:

• 1. For all $x \in V$, build $J_x$ depicted in Fig. 5, containing a core graph $(\{E_x,R_x, V_x,$ $x_1,x_2,x_3,x_4\},$ $\{E_xx_3, E_xx_4, E_xV_x, x_1x_3, x_2x_4, R_xx_1, R_xx_2, R_xV_x\})$ with precolored vertices $V_x$, $x_3$ and $x_4$, and outgoing check-valves: $d_G(x)$ many $C_{x_{v_i}^1, x_{v_i}^2}$ connected to $E_x$, and n many $C_{x_{s_i}^1, x_{s_i}^2}$ connected to $R_x$.

• 2. For all $v_i \in N(x)$, add edges $x^2_{v_i}R_{v_i}$ with $x^2_{v_i} \in J_x$ and $R_{v_i} \in J_{v_i}$.

• 3. For all $s_i \in V$, add edges $x^2_{s_i}V_{s_1}$ with $x^2_{s_i} \in J_x$ and $V_{s_1} \in J_{s_i}$.

Fig. 5.

The gadget $J_x$ for a vertex x. Note that for sake of clarity, some external vertices have been duplicated. Indeed, $\{R_{v_1}, \ldots R_{v_t}\}$, where $v_1$, ..., $v_t$ are neighbors of x, is included in $\{R_{s_1}, \ldots , R_{s_n}\}$.

Lemma 3

For all $x \in V$, if $c(E_x)=1$ then, after exhaustive application of $\textsc {Rule}~R_2$, $c(J_x)=1$ and $c(R_v)=1$ for all $v \in N(x)$.

Proof

If $V= \{s_1, \ldots , s_n\}$ and $N(x) = \{v_1, \ldots , v_t\}$, then the following sequence is a valid order: $(E_x, x_1, x_2 ,R_x, x^2_{s_1}, \ldots , x^2_{s_n}, x^2_{v_1}, \ldots , x^2_{v_t}, R_{v_1}, \ldots R_{v_t})$.    $\square$

Lemma 4

Let $c(R_x)=1$ for all $x\in V$. Then, after exhaustive application of $\textsc {Rule}~R_2$, $G'$ becomes fully colored.

Proof

Clearly, all vertices in $N(V_x) \setminus \{E_x\}$ are colored by $R_x$ for all $x\in V$. Then, $E_x$ is colored by $V_x$. By Lemma 3, $c(E_x)=1$ leads to $J_x$ being fully colored. As $V'=\bigcup _{z \in V}V(J_z)$, $G'$ becomes fully colored.    $\square$

Theorem 2

Precolored Zero Forcing Set and Precolored Power Edge Set are W[2]-hard wrt. the solution size k.

Proof

Let $G=(V,E)$ be a graph and let $G'$ the product of Construction 3 on G. We show that G has a size-k dominating set if and only if $G'$ has a size-k zero forcing set (power edge set).

“$\Rightarrow$”: Let S be a size-k dominating set for G. A size-k zero forcing set $S'$ for $G'$ is obtained as follows: for all $x \in S$, we place a PMU on $E_x$, (resp. $E_xx_4$). By Lemmas 3 and 4, $G'$ is fully colored after applying $\textsc {Rule}~R_2$ exhaustively.

“$\Leftarrow$”: Let $S'$ be a zero forcing set of size k for $G'$. Let S be the set of vertices $x \in V(G)$ such that $J_x$ has at least one vertex, resp. one edge, in $S'$ (for each $x,y \in V$, if there is a PMU on the edge $E_xR_y$ or $R_xV_y$ it counts as an edge of $J_x$). Suppose that S is not a dominating set for G. So, there is some $y \in V$ such that no $u \in V(J_y)$ is in $S'$ and no $v\in V(J_x)$ is in $S'$ for any $x \in N(y)$. (for PES, there is some $y \in V$ such that no $u_1u_2\in E(J_y)$ is in $S'$ and no $v_1v_2\in E(J_x)$ is in $S'$ for any $x \in N(y)$). Since $J_y$ is fully colored, this coloration comes from a vertex (resp. an edge) outside of $J_y$. Four cases have to be considered:

Case 1: There is some $v_i \in N(y)$ such that $c(y^2_{v_i})=1$ for $y^2_{v_i} \in J_y$ and this coloration comes from $R_{v_i} \in J_{v_i}$. By Observation 2, we have $c(E_y)=1$ only if at least one PMU is assigned on the check-valve.

Case 2: There is some $s_i \in V$ such that $c(y^2_{s_i})=1, \in J_y$, and this coloration comes from $V_{s_i} \in J_{s_i}$. By Observation 2, for $R_y$ to be colored, at least one PMU has to be assigned to the check-valve.

Case 3: $V_y$ be a propagator. But then, $S'$ is not zero forcing since $c(E_y)=0$ and $c(R_y)=0$ and they are in N(y).

Case 4: There is some $v_i \in N(y)$ such that a coloration happens on $R_y \in J_y$ from $E_{v_i} \in J_{v_i}$. Then, either there is some $t \in J_{v_i}\cap S$ and so S is a dominating set, or no PMU is assigned on $J_{v_i}$, but we already know that $E_{v_i}$ cannot be colored (see Case 1). Consequently, if $c(E_{v_i}) = 1$ then $c(w)=1$ for some $w \in J_{v_i}$ contradicting S not being a dominating set.

Thus S is a dominating set of G. Further, Construction 3 can be carried out in polynomial time and $|S|=|S'|$, yielding the desired result.    $\square$

4 Non-approximation

In this section, we will show that the reductions presented in the proofs of Theorems 1 and 2 are L-reductions.

But above all, it is clear it exists a $\frac{n}{2}$-approximation; it is sufficient to put one PMU incident to each vertex (at most n), and the lower bound for optimal solution is at least two PMUs (in cubic graph) so we obtain a $\frac{n}{2}$-approximation.

Theorem 3

Power Edge Set is $\frac{n}{2}$-approximable

For the first, by construction, we have $OPT(I')=OPT(I)+n$. Let S be a solution to I, suppose that $n>3|S|$. By the pigeon hole principle, there is a vertex which cover at least four edges, which is impossible because the degree of each vertex is three, so $n \le 3|S|$. Thus $OPT(I') \le 4OPT(I)$.

Moreover, by construction, we have

$$\begin{aligned} val(g(S') \le val(S') - n \le val(S') - OPT(I')+OPT(I) \end{aligned}$$

Thus, we construct an L-reduction with $\alpha _1 = 4$ and $\alpha _2 = 1$.

Assuming $\mathcal {P} \ne \mathcal {NP}$, Vertex Cover is hard to approximate to a factor 1.36 [6] and [9], thus yielding the desired result:

$$\begin{aligned} |S'|&\ge |g(S')| + OPT(I') - OPT(I) \\&\ge 1.36 OPT(I) + OPT(I') - OPT(I) \\&\ge 1.09 OPT(I') \end{aligned}$$

   $\square$

Corollary 1

Under $\mathcal {P} \ne \mathcal {NP}$, Power Edge Set on cubic graph cannot be approximated to within a factor better than 1.09.

Assuming, Vertex Cover is hard to approximate to a factor $2 - \epsilon$ [16] and [9], thus yielding the desired result:

   $\square$

Corollary 2

Under $\mathcal {UGC}$, Power Edge Set on cubic graph cannot be approximated to within a factor better than .

Previous results mainly show that Power Edge Set do not admit a PTAS algorithm, even on cubic graphs.

For the second, we got $OPT(I)=OPT(I')$, so clearly it is a S-reduction. Dominating Set do not admit a polynomial time approximation algorithm with ratio O(log n) ([18]), so Precolored Power Edge Set and Precolored Zero Forcing Set do not too.

Corollary 3

Under $NP \ne DTIME(n^{polylogn})$, Precolored Power Edge Set and Precolored Zero Forcing Set do not admit a polynomial time approximation algorithm with ratio O(log n).

Fig. 6.

An interval graph (a), with its interval representation (b), a perfect path decomposition of this graph (c) and its bag partition according to Definition 4 (d).

5 ZFS on Proper Interval Graphs

Preliminaries. A graph G is an interval graph if it is the intersection graph of a family of intervals on the real line. Each interval is represented by a vertex of G and an intersection between two intervals is represented by an edge between the corresponding vertices (see Fig. 6). G is called proper interval if it has an interval representation in which no interval is properly contained in another.

In the following, we use perfect path decompositions to solve Power Edge Set on proper interval graphs.

Definition 3

A path decomposition $\mathcal {D}$ of a graph $G=(V,E)$ is a sequence $(X_i)_{i=1\ldots \ell }$ of subsets of V (called bags), verifying the following properties:

• (a) for each $xy \in E$, there is some $X_i$ with $x,y \in X_i$ (each edge is in a bag),

• (b) for $i \le j \le k$, $X_i \cap X_k \subseteq X_j$ (bags containing any $v\in V$ are consecutive).

$\mathcal {D}$ is called perfect if the number of bags and their sizes are minimal under (a) and (b). The pathwidth of $\mathcal {D}$ is the size of the largest $X_i$ minus one.

Lemma 5

If G is connected, then $X_i \cap X_{i-1} \ne \varnothing$ for all $i>1$.

Proof

Towards a contradiction, assume that $X_i \cap X_{i-1} = \varnothing$. Then, by Definition 3(b), $A:=\bigcup _{1 \le k \le i-1} X_k$ and $B:=\bigcup _{i \le l} X_l$ are disjoint. Since G is connected, there is an edge xy between A and B, but no bag contains both x and y, contradicting Definition 3(a).    $\square$

Lemma 6

Let G be an interval graph. A perfect path decomposition $\mathcal {D}$ of G can be computed in linear time and

each bag of $\mathcal {D}$ is a maximal clique in G.

Proof

Being an interval graph, G admits a linear order of its maximal cliques such that, for each vertex v, all maximal cliques containing v are consecutive [10] and this order can be computed in $O(n+m)$ time [14]. Such a “clique path” naturally corresponds to a perfect path decomposition and we know that vertices of each bags induce maximal cliques. In a clique path, the size and the number of bags are minimal.    $\square$

Now we can present our algorithm, using previous results:

The Algorithm.

In the following, G is a connected proper interval graph and $\mathcal {D}= (X_1,...,X_\ell )$ is a perfect path decomposition of G. We show that it is possible to apply $\textsc {Rule}~R_2$ once per maximal clique $X_i$ in interval graphs. The central concept is a partition of the bags of $\mathcal {D}$ into four sets.

Definition 4

(Bag partition, see Fig. 6). Let $X_i$ be a bag in a perfect path decomposition of an interval graph.

• IO (Inside Only) is the set $X_i\setminus (X_{i-1}\cup X_{i+1})$.

• LO (Left Only) is the set $X_i\cap X_{i-1}\setminus X_{i+1}$.

• RO (Right Only) is the set $X_i\cap X_{i+1}\setminus X_{i-1}$.

• LR (Left Right), contains all remaining vertices of $X_i$.

Note that $RO(X_i)$ and $RO(X_j)$ are disjoint for $i\ne j$. Further, since G is proper interval, $RO(X_i)\ne \varnothing$ for all $i<\ell$. Our algorithm will simply choose any vertex of $RO(X_i)\cup IO(X_i)$ for all i. This can clearly be done in linear time and we show that it is correct and optimal.

Lemma 7

Let G be a connected interval graph and let $\mathcal {D}=(X_1,\ldots ,X_\ell )$ be a perfect path decomposition of G. Let $\overline{S}$ be a set intersecting each $RO(X_i)$ for all $1\le i <\ell$ in exactly one vertex and intersecting $IO(X_\ell )$ in exactly one vertex. Then, $S:=V\setminus \overline{S}$ is an optimal zero forcing set for G.

Proof

For each i, let $x_i$ be the $i^{\text {th}}$ vertex of $\overline{S}$, that is, $\overline{S}\cap (RO(X_i)\cup IO(X_i))=\{x_i\}$ for each $X_i\in \mathcal {D}$. We show that the order $\sigma$ consisting of S in any order followed by $(x_1,\ldots ,x_\ell )$ is valid for S. To this end, let $1\le j<\ell$. Note that $IO(X_j)\cup LO(X_j)=((X_j\setminus X_{j+1})\cap X_{j-1})\cup ((X_j\setminus X_{j+1})\setminus X_{j+1})=X_j\setminus X_{j+1}$. Thus, there is some $u\in IO(X_j)\cup LO(X_j)$ as otherwise, $X_j \subseteq X_{j+1}$ contradicting $\mathcal {D}$ being perfect. Towards a contradiction, assume $N_G[u]\not \le _\sigma x_j$, that is, there is some $v\in N_G[u]$ with $x_j<_\sigma v$. By construction of $\sigma$, there is a $k>j$ such that $v=x_k$. By construction of $\overline{S}$, we have $x_k\in IO(X_k)\cup RO(X_k)$, implying $x_k\notin X_{k-1}$ by definition of RO and IO. Further, since $ux_k$ is an edge of G, there is a bag $X_i$ containing both u and $x_k$ and, since $u\in IO(X_j)\cup LO(X_j)$ we know $i\le j$. But then, $x_k$ occurs in $X_i$, not in $X_{k-1}$ but again in $X_k$, contradicting $\mathcal {D}$ being perfect. It remains to treat $x_\ell$, but since $x_\ell$ is the last vertex of $\sigma$, $N_G[u]\le _\sigma x_\ell$ for all $u\in N_G(x_\ell )$.

Finally, optimality of S is implied by Lemma 1 as $|\overline{S}|=|\mathcal {D}|$.    $\square$

Theorem 4

Zero Forcing Set is solvable in $O(n+m)$ time on proper interval graphs.

Proof

We know that our algorithm is exact, compute the path decomposition PD can done in linear time (Lemma 6) and partitioning its vertices is easy. So there is a polynomial time algorithm for Zero Forcing Set in proper interval graph.   $\square$

6 Conclusion and Perspectives

In this article, we investigated Power Edge Set and Zero Forcing Set from the point of view of computational complexity. We obtained a series of negative results, refining the previous hardness results and excluding certain exact algorithms. On the positive side, we give a linear-time algorithm in case the input is a proper interval graph and a naive approximation algorithm. There is a big gap between positive and negative result in approximation so further research will be focused on developing efficient polynomial-time approximation algorithms, as well as considering more special cases and structural parameterizations.