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., holefree 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 ATfree 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 nonneighborhood of a vertex is . For , and .
We say that for a vertex set , a vertex has a join (resp., cojoin) to if (resp., ). Join (resp., cojoin) 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 cojoin 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 cliquewidth, ED can be solved in polynomial time [16]. Thus we only consider ED on free bipartite graphs for which the cliquewidth is unbounded. In [18], the cliquewidth of all classes of
free bipartite graphs is classified. For example, while ED is
complete for clawfree graphs (even for ()free perfect graphs [24]), the cliquewidth of clawfree bipartite graphs is bounded.Theorem 1 ([18]).
The cliquewidth of free bipartite graphs is bounded if and only if one of the following cases appears:

, ;

;

;

;

.
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 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:
(1) 
(2) 
(3) 
Let such that , and let , . By the e.d.s. property, we have:
(4) 
Thus, for finding an e.d.s. with , we can assume:
(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:
(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
and assume that
is known as an e.d.s. for . In particular, . Note that for , additionally dominates parts of .
Let
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.,
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 nonexcluded 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 free bipartite graphs when or
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 cliquewidth 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 .
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 cojoin.
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 free bipartite graphs in polynomial time
In this section, we generalize the WED approach for free bipartite graphs. Recall that the cliquewidth of free bipartite graphs is bounded and the cliquewidth 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
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