# Finding Efficient Domination for P_8-Free Bipartite Graphs in Polynomial Time

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 -complete for P_7-free graphs, and even for very restricted H-free bipartite graph classes such as for K_1,4-free bipartite graphs as well as for C_4-free bipartite graphs while it is solvable in polynomial time for P_7-free bipartite graphs as well as for S_2,2,4-free bipartite graphs. Here we show that ED can be solved in polynomial time for P_8-free bipartite graphs.

## Authors

• 7 publications
• 8 publications
• ### Finding Efficient Domination for S_1,3,3-Free Bipartite Graphs in Polynomial Time

A vertex set D in a finite undirected graph G is an efficient dominating...
08/10/2020 ∙ by Andreas Brandstädt, et al. ∙ 0

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

A vertex set D in a finite undirected graph G is an efficient dominatin...
05/30/2018 ∙ by Andreas Brandstädt, et al. ∙ 0

• ### Biclique Graphs of K_3-free Graphs and Bipartite Graphs

A biclique of a graph is a maximal complete bipartite subgraph. The bicl...
05/29/2020 ∙ by Marina Groshaus, et al. ∙ 0

• ### A polynomial time approximation schema for maximum k-vertex cover in bipartite graphs

The paper presents a polynomial time approximation schema for the edge-w...
09/18/2019 ∙ by Vangelis Th. Paschos, et al. ∙ 0

• ### When Maximum Stable Set can be solved in FPT time

Maximum Independent Set (MIS for short) is in general graphs the paradig...
09/18/2019 ∙ by Édouard Bonnet, et al. ∙ 0

• ### Dichotomizing k-vertex-critical H-free graphs for H of order four

For k ≥ 3, we prove (i) there is a finite number of k-vertex-critical (P...
06/30/2020 ∙ by Ben Cameron, et al. ∙ 0

• ### Vertex deletion into bipartite permutation graphs

A permutation graph can be defined as an intersection graph of segments ...
10/22/2020 ∙ by Łukasz Bożyk, 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 for short) problem asks for the existence of an e.d.s. in a given graph . The notion of efficient domination was introduced by Biggs [3] under the name perfect code.

The Exact Cover Problem asks for a subset of a set family over a ground set, say , containing every vertex in exactly once, i.e., forms a partition of . As shown by Karp [12], this problem is -complete even for set families containing only -element subsets of (see problem X3C [SP2] in [11]).

Clearly, ED is the Exact Cover problem for the closed neighborhood hypergraph of , i.e., if is an e.d.s. of then forms a partition of (we call it the e.d.s. property). In particular, the distance between any pair of distinct -vertices is at least 3. In [1, 2], it was shown that the ED problem is -complete.

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 , , (and center ) 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 .

In [10, 16, 17], it was shown that ED is -complete for -free chordal unipolar graphs and thus, in general, for -free graphs. In [8], ED is solvable in polynomial time for -free graphs (which leads to a dichotomy).

A bipartite graph is chordal bipartite if is -free for every . Lu and Tang [14] 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 [14] and even for planar bipartite graphs of maximum degree 3 [7] and girth at least for every fixed [15]. Thus, ED is -complete for -free bipartite graphs and for -free bipartite graphs.

In [4], it is shown that one can extend polynomial time algorithms for Efficient Domination to such algorithms for weighted Efficient Domination. Thus, from now on, we focus on the unweighted ED problem.

In [5], it is shown that ED is solvable in polynomial time for AT-free graphs. Moreover, in [5], it is shown that ED 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 [14] showed that 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. [6, 13]) 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 ), in [9] we have shown that ED is solvable in polynomial time for -free (and more generally, for -free) bipartite graphs.

In [9] we have shown that ED is solvable in polynomial time for -free bipartite graphs as well as for -free bipartite graphs. Now in this manuscript, we show:

###### Theorem 1.

For -free bipartite graphs, the ED problem is solvable in polynomial time.

## 2 Preliminaries

Recall that is a -free bipartite graph, i.e., every vertex in is black, and every vertex in is white.

A vertex contacts if . A vertex contacts if has a neighbor in . For a subset , a vertex has a join to , say , if contacts every vertex in . Moreover, has a co-join to , say , if does not contact any vertex in .

A vertex is forced if for every e.d.s.  of ; is excluded if for every e.d.s.  of . For example, if are leaves in and is the neighbor of then are excluded and is forced.

By a forced vertex, can be reduced to as follows:

###### Claim 2.1.

If is forced then has an e.d.s.  with if and only if the reduced graph has an e.d.s.  such that all vertices in are excluded in .

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 . For checking whether has an e.d.s.  with , we can clearly reduce by forced vertices as well as by -forced vertices when we assume that :

###### Claim 2.2.

If we assume that and is -forced then has an e.d.s.  with if and only if the reduced graph has an e.d.s.  with such that all vertices in are -excluded in .

Similarly, for , is -forced if for every e.d.s.  of with , and correspondingly, is -excluded if for such e.d.s. , and can be reduced by the same principle.

Clearly, for every connected component of , the e.d.s. problem is independently solvable. Thus, we can assume that is connected.

Let denote the distance between and in . By the e.d.s. property, the distance between two -vertices is at least 3. Moreover, we have:

###### Claim 2.3.

If is connected then:

• For every and , or .

• For every or , or .

Proof. : By the e.d.s. property and since is bipartite, for and , . In particular, , .

If then induce a in , which is a contradiction. Thus, or .

: Without loss of generality, assume that . Then by the e.d.s. property and since is bipartite, . In particular, , . If If then there is a in , which is a contradiction. Thus, or . ∎

Without loss of generality, assume that , , with , .

First, let (for example, ), i.e., , and let , , be the distance levels of . Since is -free bipartite, we have . By the e.d.s. property, we have:

 D∩(N1∪N2)=∅ (1)
 Every vertex u∈N2 must have a D-neighbor in N3 (2)

(else there is no such e.d.s.  in ).

 Every vertex u∈N2 has at least two neighbors in N3 (3)

(else the only neighbor of is -forced and can be reduced, i.e., ).

 Every vertex v∈N3 has a neighbor in N3∪N4 (4)

(else is -forced and can be reduced, i.e., ).

###### Claim 2.4.

Every is no endpoint of a , , with and .

Proof. Let . Since is -free bipartite, is no endpoint of any with and .

If is an endpoint of a with and then, since by (1), , has no -neighbor (else there is a with -neighbor of in ) and there is no such e.d.s. Thus, is an endpoint of a , , with and .

If is an endpoint of a with and then must have a -neighbor , and since is -free, by (4), has no further neighbor (else there is a in ). Thus, is -forced and can be reduced, i.e., .

Moreover, if has two such neighbors then assume that , and must have a -neighbor . But then induce a in , which is a contradiction. Thus, there is no such e.d.s.

Thus, is no endpoint of such a , , and Claim 2.4 is shown. ∎

In the next section, we assume:

## 3 When every vertex in N2 is a P5-endpoint whose remaining vertices are in N0∪N1

### 3.1 General remarks

###### Lemma 1.

If every vertex in is an endpoint of a whose remaining vertices are in then we have:

• Every vertex in is no endpoint of any whose remaining vertices are in , .

• , is independent, and every edge in does not contact .

Proof. : Suppose to the contrary that is the endpoint of a with , , . Recall that is the endpoint of a whose remaining vertices are in . But then and the with endpoint induce a in , which is a contradiction. Thus, is no endpoint of any whose remaining vertices are in , .

: By , every is no endpoint of any whose remaining vertices are in , .

If , say , , with , and , then is the endpoint of such a , which is impossible by . Thus, .

Analogously, if is not independent, say with , then let be a neighbor of . Clearly, since is bipartite. Recall that there is a neighbor of . But then is the endpoint of such a , which is impossible by . Thus, is independent.

Finally, if there is an edge with which contacts , say without loss of generality, with , then let for . But then is the endpoint of such a , which is impossible by . Thus, every edge in does not contact .

Now, Lemma 1 is shown. ∎

By Lemma 1 , we have:

###### Corollary 1.

For every and , we have .

Since by Lemma 1 , is independent and every edge in does not contact , we have:

###### Corollary 2.

For every and , .

###### Claim 3.1.

If and then is -excluded.

Proof. By Corollary 2, we have or . Clearly, and have the same color (either black or white) since .

First assume that and . Suppose to the contrary that . Then by the e.d.s. property, and , but now, has no -neighbor in (recall that ), i.e., has no such e.d.s. Thus, is -excluded.

Next assume that and . Suppose to the contrary that . Then by the e.d.s. property, and , but now, has no -neighbor in (recall that by Lemma 1 , and is independent), i.e., has no such e.d.s. Then again, is -excluded. ∎

###### Corollary 3.

The following statements hold:

• If and then is -forced, and can be reduced, i.e., .

• If and , , and there are two such , , , then there is no such e.d.s.

Proof. : Recall that by Claim 3.1, are -excluded and must have a -neighbor. Clearly, do not have any -neighbor in since and by Lemma 1 , every edge in does not contact . Thus, is the only possible -neighbor of , i.e., is -forced, and can be reduced, i.e., .

: Assume and , . Recall again that by Claim 3.1, are -excluded and must have a -neighbor.

Without loss of generality, assume that . Then either or is -excluded and has no -neighbor. Thus, there is no such e.d.s. ∎

###### Corollary 4.

The following statements hold:

• If with and such that , , then and are -forced, and can be reduced, i.e., .

• If and , , , with , , , then there is no such e.d.s.

Proof. : Recall that by Claim 3.1, are -excluded and must have a -neighbor. Then and are -forced, i.e., .

: Recall that by Claim 3.1, are -excluded and must have a -neighbor. Then the only possible -neighbors are and but then and do not have any -neighbor, and there is no such e.d.s. ∎

In particular, if are leaves with a common neighbor in , say with , , then is -forced, and can be reduced. It can also lead to a contradiction, i.e., if for a vertex , are leaves and with , , and , , then and are -forced, which is a contradiction since and .

Recall that, as in (1), , as well as in (2) and (3), for every , (else has no -neighbor if or the only neighbor of in is -forced). It can also lead to a contradiction, i.e., if has two neighbors which are leaves then and are -forced, which is a contradiction since and . Moreover, if has two neighbors such that is a leaf then is -forced, and can be reduced, i.e., .

Recall Corollary 4. Assume that there is an edge with and , , . If , , and , , then . Analogously, if , , and , , then . But then do not have any -neighbor, i.e., there is no such e.d.s.

Thus, assume that at most one of has such neighbors in .

Recall that for , , say , , and must have a -neighbor in , say without loss of generality, . Then for every , , and contacts (else there is no e.d.s.  with –recall Claim 3.1). Then we have:

###### Claim 3.2.

If for , , , and then for every , , say , and vertex is -forced.

Proof. Without loss of generality, assume that . Since , we have , and must have a -neighbor in . Suppose to the contrary that , say , and without loss of generality let . Then and must have a -neighbor. Recall that by Lemma 1 , , is independent, and every edge in does not contact . Then either or .

First assume that . Then must have a -neighbor in , say with . But then by the e.d.s. property, , i.e., induce a , which is impossible by Lemma 1 .

Next assume that . Then again must have a -neighbor in , say with . But then again , i.e., induce a , which is impossible by Lemma 1 .

Thus , say , and is -forced. ∎

Thus can be reduced by such -forced vertices; let . This can also lead to the case that has no such e.d.s. in :

If contacts only and then has no -neighbor, and there is no such e.d.s.

If with then there is no such e.d.s. in .

Clearly, if two vertices in have a common neighbor in then by Claim 3.2, there is no such e.d.s. in .

Let be a component in , say . By Lemma 1 , either is independent or .

###### Claim 3.3.

Let which contacts . If then , and if then .

Proof. Without loss of generality, let , i.e., is black. Let with . Let induce a with and . By Lemma 1 , do not induce a , i.e., , and in general, it leads to . Analogously, if then . ∎

###### Claim 3.4.

There is no in with , , and , .

Proof. Suppose to the contrary that with , , and , , induce a in . Recall that by (4), has a neighbor in (else is -forced and can be reduced, i.e., ); let with .

Since by Lemma 1 , is no endpoint of a , we have , and analogously, (else induce a ), and (else induce a ). But then induce a in , which is a contradiction. ∎

### 3.2 Components Q in G[N2∪N3∪N4] with independent V(Q)∩N3

Let be a component in , and assume that is independent. Recall that by (1), and by (4), for with