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., holefree 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 ATfree 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 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 ), 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 cojoin 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:
(1) 
(2) 
(else there is no such e.d.s. in ).
(3) 
(else the only neighbor of is forced and can be reduced, i.e., ).
(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 is a endpoint whose remaining vertices are in
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. ∎
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