# On Efficient Domination for Some Classes of H-Free Bipartite Graphs

A vertex set D in a finite undirected graph G is an efficient dominating set (e.d.s. for short) of G if every vertex of G is dominated by exactly one vertex of D. The Efficient Domination (ED) problem, which asks for the existence of an e.d.s. in G, is known to be NP-complete even for very restricted graph classes such as for 2P_3-free chordal graphs while it is solvable in polynomial time for P_6-free chordal graphs (and even for P_6-free graphs). On the other hand, ED is solvable in linear time for 2P_2-free and even for P_5-free graphs, and thus for P_5-free bipartite graphs. Lu and Tang showed that ED is NP-complete for chordal bipartite graphs and for planar bipartite graphs;actually, ED is -complete even for planar bipartite graphs with vertex degree at most 3. Thus, ED is NP-complete for K_1,4-free bipartite graphs. For K_1,3-free bipartite graphs, however, ED is solvable in polynomial time: For classes of bounded clique-width, ED is solvable in polynomial time. Dabrowski and Paulusma published a dichotomy for clique-width of H-free bipartite graphs. For instance, clique-width of S_1,2,3-free bipartite graphs is bounded (which includes K_1,3-free bipartite graphs). We show that ED is NP-complete for K_3,3-free bipartite graphs. Moreover, we show that (weighted) ED can be solved in polynomial time for H-free bipartite graphs when H is P_7 or S_1,2,4 or ℓ P_4 for fixed ℓ, and similarly for P_9-free bipartite graphs with vertex degree at most 3.

## Authors

• 7 publications
• 8 publications
• ### Finding Efficient Domination for P_8-Free Bipartite Graphs in Polynomial Time

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• ### Eliminating Odd Cycles by Removing a Matching

We study the problem of determining whether a given graph G=(V, E) admit...
10/21/2017 ∙ by Carlos V. G. C. Lima, et al. ∙ 0

• ### Parameterized Complexity of Finding Subgraphs with Hereditary Properties on Hereditary Graph Classes

We investigate the parameterized complexity of finding subgraphs with he...
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• ### Linear-time Algorithms for Eliminating Claws in Graphs

Since many NP-complete graph problems have been shown polynomial-time so...
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• ### Characterizing and decomposing classes of threshold, split, and bipartite graphs via 1-Sperner hypergraphs

A hypergraph H is said to be 1-Sperner if for every two hyperedges the s...
05/09/2018 ∙ by Endre Boros, et al. ∙ 0

• ### Finding Efficient Domination for S_1,1,5-Free Bipartite Graphs in Polynomial Time

A vertex set D in a finite undirected graph G is an efficient dominating...
10/29/2020 ∙ by Andreas Brandstädt, et al. ∙ 0

• ### Intersection models and forbidden pattern characterizations for 2-thin and proper 2-thin graphs

The thinness of a graph is a width parameter that generalizes some prope...
04/08/2021 ∙ by Flavia Bonomo-Braberman, et al. ∙ 0

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## 1 Introduction

Let be a finite undirected graph. A vertex dominates itself and its neighbors. A vertex subset is an efficient dominating set (e.d.s. for short) of if every vertex of is dominated by exactly one vertex in ; for any e.d.s.  of , for every (where denotes the closed neighborhood of ). Note that not every graph has an e.d.s.; the Efficient Dominating Set (ED) problem asks for the existence of an e.d.s. in a given graph .

The Exact Cover Problem asks for a subset of a set family over a ground set, say , containing every vertex in exactly once. As shown by Karp [22], this problem is -complete even for set families containing only -element subsets of (see problem X3C [SP2] in [21]).

Clearly, ED is the Exact Cover problem for the closed neighborhood hypergraph of . The notion of efficient domination was introduced by Biggs [5] under the name perfect code.

In [3, 4], it was shown that the ED problem is -complete. Moreover, Lu and Tang [25] showed that ED is -complete for chordal bipartite graphs (i.e., hole-free bipartite graphs). Thus, for every , ED is -complete for -free bipartite graphs.

Moreover, ED is -complete for planar bipartite graphs [25] and even for planar bipartite graphs of maximum degree 3 [14] and girth at least for every fixed [27]. Thus, ED is -complete for -free bipartite graphs and for -free bipartite graphs.

In [11], it is shown that WED is solvable in polynomial time for interval bigraphs, and convex bipartite graphs are a subclass of them (and of chordal bipartite graphs). Moreover, Lu and Tang [25] showed that weighted ED is solvable in linear time for bipartite permutation graphs (which is a subclass of convex bipartite graphs). It is well known (see e.g. [12, 23]) that is a bipartite permutation graph if and only if is AT-free bipartite if and only if is (,hole)-free bipartite (see Figure 1). Thus, while ED is -complete for -free bipartite graphs (since and contain and contains ), we will show that WED is solvable in polynomial time for -free (and more generally, for -free) bipartite graphs.

In this paper, we will also consider the following weighted version of the ED problem:

Weighted Efficient Domination (WED) Instance: A graph , vertex weights . Task: Find an e.d.s. of minimum finite total weight, or determine that contains no such e.d.s.

In [8], it is shown that one can extend polynomial time algorithms for ED to such algorithms for WED.

For a set of graphs, a graph is called -free if contains no induced subgraph isomorphic to a member of . In particular, we say that is -free if is -free. Let denote the disjoint union of graphs and , and for , let denote the disjoint union of copies of . For , let denote the chordless path with vertices, and let denote the complete graph with vertices (clearly, ). For , let denote the chordless cycle with vertices.

For indices , let denote the graph with vertices , , such that the subgraph induced by forms a , the subgraph induced by forms a , and the subgraph induced by forms a , and there are no other edges in . Thus, claw is , chair is , and is isomorphic to e.g. .

For a vertex , denotes its (open) neighborhood, and denotes its closed neighborhood. A vertex sees the vertices in and misses all the others. The non-neighborhood of a vertex is . For , and .

We say that for a vertex set , a vertex has a join (resp., co-join) to if (resp., ). Join (resp., co-join) of to is denoted by (resp., ). Correspondingly, for vertex sets with , denotes for all and denotes for all . A vertex contacts if has a neighbor in . For vertex sets with , contacts if there is a vertex in contacting .

If but has neither a join nor a co-join to , then we say that distinguishes . A set of at least two vertices of a graph is called homogeneous if and every vertex outside is either adjacent to all vertices in , or to no vertex in . Obviously, is homogeneous in if and only if is homogeneous in the complement graph . A graph is prime if it contains no homogeneous set. In [9, 10, 14], it is shown that the WED problem can be reduced to prime graphs.

It is well known that for a graph class with bounded clique-width, ED can be solved in polynomial time [16]. Thus we only consider ED on -free bipartite graphs for which the clique-width is unbounded. In [18], the clique-width of all classes of

-free bipartite graphs is classified. For example, while ED is

-complete for claw-free graphs (even for ()-free perfect graphs [24]), the clique-width of claw-free bipartite graphs is bounded.

###### Theorem 1 ([18]).

The clique-width of -free bipartite graphs is bounded if and only if one of the following cases appears:

1. , ;

2. ;

3. ;

4. ;

5. .

For graph , the square of has the same vertex set , and two vertices , , are adjacent in if and only if . Let , i.e., is the subset of vertices which have distance 2 to , and correspondingly for . The WED problem on can be reduced to Maximum Weight Independent Set (MWIS) on (see e.g. [11, 8, 14, 26] and the survey in [7]).

## 2 WED in polynomial time for some H-free bipartite graphs

### 2.1 A general approach

A vertex is forced if for every e.d.s.  of ; is excluded if for every e.d.s.  of . Analogously, if we assume that for a vertex then is -forced if for every e.d.s.  of with , and is -excluded if for every e.d.s.  of with . Similarly, is -forced if for every e.d.s.  of with , and correspondingly, is -excluded if for such e.d.s. .

In this manuscript, we solve some cases in polynomial time by the following approach: Assume that has an e.d.s. . Then for any vertex , . Assume that for a vertex , and let , , denote the distance levels of in (in particular, ). Since is bipartite, every is independent. By the e.d.s. property, we have:

 D∩(N1∪N2)=∅. (1)
 If for some x∈N2,N(x)∩N3=∅ then G has% no e.d.s. D with v∈D. (2)
 If for y∈N3,N(y)∩N4=∅ then y is v−forced. (3)

Let such that , and let , . By the e.d.s. property, we have:

 If N(yi)∩N4=∅,i=1,2, then G has no e.d.s% . D with v∈D. (4)

Thus, for finding an e.d.s.  with , we can assume:

 For every y∈N3,N(y)∩N4≠∅. (5)

More generally, for , let us write

• , and

• ,

i.e., is a partition of .

For , let and with and . Clearly, for every pair , , we have:

 x1,x2,y1,y2,z1,z2 induce a 2P3. (6)
###### Claim 2.1.

Assume that for . Then:

• If at least three vertices in , say , have private neighbors in then there is an in .

• If all of have a common neighbor in then there is an in .

• If neither nor appears then there is an or in .

Proof. : Let be the private neighbors of , . Then , (with center ) induce an .

: Let be a common neighbor of , . Then , (with center ) induce an .

: Assume that not all of have private neighbors in and there is no common neighbor of in (since neither nor appears). Without loss of generality, let be a common neighbor of such that , and let with .

If and then (with center ) induce an , and similarly, if and then (with center ) induce an .

Finally, if and then (with center ) induce an . Thus, Claim 2.1 is shown.

Let

 Gi:=G[{v}∪N1∪…∪Ni]

and assume that

 Di:=D∩({v}∪N1∪…∪Ni)

is known as an e.d.s. for . In particular, . Note that for , additionally dominates parts of .

Let

 N′i:=Ni∖(Di∪N(Di)).

By the e.d.s. property, for possible -candidates from , we have to exclude from (recall the notion of as in the Introduction), i.e.,

 Wi+1:=Ni+1∖(N(Di)∪N2(Di)).

The collection of possible -candidates in has to dominate .

Thus, for constructing from , for a possible subset such that dominates (following the e.d.s. property), we have .

Finally, if is the last distance level of then (if the e.d.s. property is correct).

The stepwise construction of from is possible e.g. when candidates are forced: Recall that is an e.d.s. for . Similarly as for -forced vertices, a vertex is -forced if for every e.d.s.  of with , being an e.d.s. for .

As in (3) and (5), a non-excluded vertex (with respect to ) with neighbor in is -forced if . Thus we can assume that , i.e., .

If the number of distance levels of is unbounded, it leads to a polynomial time algorithm for WED when e.g.  has polynomially many subsets in and for each , , the candidates for are -forced. This can be done e.g. for -free bipartite graphs (see section 2.3).

One of the helpful arguments is that in some cases, has only a fixed number of distance levels; for instance, this is the case for -free bipartite graphs (e.g., for -free bipartite graphs, we have ) as well as, more generally, for -free bipartite graphs (but the complexity of ED is still open for -free bipartite graphs and for -free bipartite graphs).

A more general case is when for all , every has at most one neighbor in ; for instance, this is the case for -free bipartite graphs (for which the complexity of ED is still open).

Another more general case is when an e.d.s.  in has only polynomially many subsets in , . If has a fixed number of distance levels , then, starting with , for to , we can produce a polynomial number of . This can be done for -free bipartite graphs and for -free bipartite graphs (see section 2.2).

If the number of distance levels of is not fixed then a helpful property would be that for all distance levels , from a fixed number , there is at most one neighbor of in , and the number of e.d.s. for is polynomial. This can be done for -free bipartite graphs (see section 2.3).

### 2.2 WED in polynomial time for H-free bipartite graphs when H=p7 or H=ℓP4

Recall that ED is -complete for -free graphs but polynomial for -free graphs (see e.g. [15]). Moreover, for -free bipartite graphs , for every , and thus, is unique (if it is really an e.d.s.) and thus, WED can be solved in linear time for -free bipartite graphs; actually, WED is done in linear time for -free graphs [15].

The subsequent lemma implies further polynomial cases for WED:

###### Lemma 1 ([9, 10]).

If WED is solvable in polynomial time for -free graphs then WED is solvable in polynomial time for -free graphs.

This clearly implies the corresponding fact for -free graphs. By Theorem 1, the clique-width of -free bipartite graphs is unbounded.

Recall that for graph , the distance between two vertices of is the number of edges in a shortest path between and in .

###### Theorem 2 ([2]).

Every connected -free graph admits a vertex such that for every .

###### Theorem 3.

For -free bipartite graphs, WED is solvable in time .

Proof. Let be a connected -free bipartite graph. Recall that , , denote the distance levels of in ; since is bipartite, is independent for every . By Theorem 2, there is a vertex whose distance levels , , are empty. Moreover, for an e.d.s.  of , either or there is a neighbor of such that . Thus, since is -free, for every , .

Thus, we can check for every whether is part of an e.d.s.  of . By (1), for , we have . By (3), if for , then is -forced. Thus, we can reduce the graph and from now on, by (5) assume that for every , . If there are two such vertices (with neighbors , and , , ) then, since induce a in , would contain a . Thus, after the reduction, ; let . Then again, since , every is -forced.

The algorithm has the following steps (according to section 2.1):

• Find a vertex in with .

• For each , check whether there is an e.d.s.  of with , as follows:

• Add all vertices with to the initial and reduce correspondingly. If has a contradiction to the e.d.s. properties then has no e.d.s.  with .

• For the reduced graph , check for any vertex with whether is an e.d.s. of .

Clearly, this can be done in time . Thus, Theorem 3 is shown. ∎

###### Theorem 4.

For -free bipartite graphs, WED is solvable in polynomial time for every fixed .

Proof. Let be an e.d.s. of . We can assume that there is a vertex with : If for all then we can check whether the set of leaves in forms an e.d.s.; by the modular decomposition property, we can assume that no two leaves have a common neighbor in . Thus, let ; since is bipartite, induce a in .

Since is an induced subgraph of , it follows from Theorem 3 that WED is solvable in polynomial time for -free bipartite graphs.

As a next step, let , and again let , , be the distance levels of with . Then obviously, since otherwise, there is a in .

By (3) and (5), assume that for every , .

According to (6), we first claim that : Otherwise, if there are then for the neighbors and of , , induce a in .

After reducing by for , we can again assume that every vertex which is nonadjacent to and is -forced. Thus we can assume that for every remaining vertex , .

Next we claim that : Otherwise, if there are then for the neighbors and of , , induce a in .

The same principle holds for , . Thus, contains at most 8 vertices which are not -forced, and thus, we can solve WED in polynomial time.

For every fixed , the same principle can be done which again leads to WED in polynomial time for -free bipartite graphs. ∎

For the more general case of -free bipartite graphs, we again show that WED is polynomial.

###### Theorem 5 ([19, 20]).

The family of independent sets of every -free graph of vertices has members and can be computed in time . The family of independent sets of every -free graph, for any fixed , has polynomially many members and can be computed in polynomial time.

###### Theorem 6.

For -free bipartite graphs, for any fixed , WED is solvable in polynomial time.

Proof. The proof is similar to the approach in Section 2.1. In particular we have:

• Since is -free, we have for every .

• For possible e.d.s.  with , has polynomially many members and can be computed in polynomial time since subgraph is -free (cf. Theorem 5): If contains with -edges then by construction and since is prime, for , every pair leads to a -subgraph (with vertices , where have a common neighbor and there is a vertex distinguishing and ), and have a pairwise co-join.

Then, starting with one of the possible subsets , it can be continued for the fixed number of remaining distance levels as in the approach in Section 2.1. Thus, Theorem 6 is shown. ∎

###### Corollary 1.

For every -free bipartite graph, for any fixed , the e.d.s. family contains polynomially many members and can be computed in polynomial time.

Recall that ED is -complete for bipartite graphs of vertex degree at most 3 [14] and girth at least for every fixed [27]. If the degree of all vertices in is at most 3 then we can show:

###### Theorem 7.

For -free bipartite graphs with vertex degree at most , WED is solvable in polynomial time.

Proof. Let be a -free bipartite graph with vertex degree at most . Again, by Theorem 2 and by the e.d.s. property, when checking whether is part of an e.d.s. of , we can assume that its distance levels , , are empty. By (3) and (5), we can assume that every vertex in has a neighbor in . We first show:

###### Claim 2.2.

.

Proof. Suppose to the contrary that ; let , and let , be neighbors of in and let , be neighbors of in . Clearly, by the e.d.s. property and by (6), induce a in . Let , , be a common neighbor of and . By the degree bound 3, there is no common neighbor of , and .

First assume that for two of , there is a common neighbor in ; without loss of generality, let and for . Then, by the degree bound 3, , and thus, there is a distinct neighbor with . Since do not induce a in , we have , and since do not induce a in , we have , which is a contradiction to the degree bound 3. Thus, there is no common neighbor in of two of the vertices ; each of has its private neighbor , . But then induce a in , which is a contradiction. Thus .

Thus, for every pair , we can check whether there is an e.d.s.  of with by reducing the graph correspondingly; let . Again, we can assume that all vertices in have a neighbor in since otherwise, such vertices are -forced by the assumption that . By similar arguments as for Claim 2.2, we can show that and finally, for vertices , is forced. Thus, Theorem 7 is shown. ∎

By the degree bound 3, it is obvious that a bipartite graph with induced subgraph has no e.d.s. Moreover, for a with degree 3 vertices and , these two vertices are excluded. What is the complexity of ED for -free bipartite graphs with vertex degree at most 3?

### 2.3 WED for S2,2,4-free bipartite graphs in polynomial time

In this section, we generalize the WED approach for -free bipartite graphs. Recall that the clique-width of -free bipartite graphs is bounded and the clique-width of -free bipartite graphs as well as of -free bipartite graphs is unbounded.

As usual, we check for every whether is part of an e.d.s.  of . Let , , denote the distance levels of in ; since is bipartite, every is an independent vertex subset. Recall by (1) that , and by (5), for every , , i.e., subsequently we consider only -candidates in which are not -forced.

The following is a general approach which will be used for -free bipartite graphs, , and for -free bipartite graphs:

Recall that and assume that , is an e.d.s. for . Moreover, recall and . Clearly, if for , there is no neighbor of in then there is no such e.d.s. , and if then the corresponding neighbor of in is -forced. Now assume that for .

###### Claim 2.3.

If for , is -free bipartite and for , , then for the -vertex which dominates , we have for every , .

Proof. Let be the -vertex which dominates , and let , . Let be any neighbor of in . Since , has to be dominated by a neighbor , and since for a shortest path , , between and , the subgraph induced by vertices (with center ) do not contain an induced , we have . Thus, , and Claim 2.3 is shown.

###### Theorem 8.

For -free bipartite graphs, WED is solvable in time .

Proof. Let be an -free bipartite graph. Recall that, as in section 2.1, for , we denote , , , and .

By the e.d.s. property, the collection of possible -candidates from has to dominate . Thus, for constructing from , for a possible subset such that dominates , we have .

First let us see how many subsets of are candidates for , i.e., for .

###### Claim 2.4.

For any , the remaining -vertices in are -forced.

Proof. Let us fix any possible vertex , and let , , , and ; let and , and let . By construction and by the e.d.s. property, the remaining -vertices in are in and their neighbors are in .

Then let us fix any . Since we assumed that , vertex has to be dominated by some vertex in .

Clearly, if there is no neighbor of in then there is no such e.d.s. , and if then the corresponding neighbor of in is -forced. Now assume that