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 . The notion of efficient domination was introduced by Biggs [2] under the name perfect code. The Efficient Domination (ED) problem asks for the existence of an e.d.s. in a given graph (note that not every graph has an e.d.s.)
A set of edges in a graph is an efficient edge dominating set (e.e.d.s. for short) of if and only if it is an e.d.s. in its line graph . The Efficient Edge Domination (EED) problem asks for the existence of an e.e.d.s. in a given graph . Thus, the EED problem for a graph corresponds to the ED problem for its line graph . Note that not every graph has an e.e.d.s. An efficient edge dominating set is also called dominating induced matching (d.i.m. for short), and the EED problem is called the Dominating Induced Matching (DIM) problem in various papers (see e.g. [3, 4, 5, 6, 7, 9, 11, 12]); subsequently, we will use this notation instead of EED.
In [10], it was shown that the DIM problem is complete; see also [3, 9, 13, 14]. However, for various graph classes, DIM is solvable in polynomial time. For mentioning some examples, we need the following notions:
Let denote the chordless path with vertices, say , and edges , ; we also denote it as .
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 ; is called the center of . Thus, claw is , and is isomorphic to .
For a set of graphs, a graph is called free if no induced subgraph of is contained in . If , say , then instead of free, is called free.
The following results are known:
Theorem 1.
2 Definitions and Basic Properties
2.1 Basic notions
Let be a finite undirected graph without loops and multiple edges. Let or denote its vertex set and or its edge set; let and . For , let denote the open neighborhood of , and let denote the closed neighborhood of . If , we also say that and see each other, and if , we say that and miss each other. A vertex set is independent in if for every pair of vertices , . A vertex set is a clique in if for every pair of vertices , , . For let and .
For , let denote the subgraph of induced by vertex set . Clearly is an edge in exactly when and ; thus, can simply be denoted by (if understandable).
For and , , we say that ( and miss each other) if there is no edge between and , and and see each other if there is at least one edge between and . If a vertex has a neighbor then contacts . If every vertex in sees every vertex in , we denote it by . For , we simply denote by , and correspondingly by . If for , , we say that is isolated in . For graphs , with disjoint vertex sets, denotes the disjoint union of , , and for , denotes the disjoint union of copies of . For example, is the disjoint union of two edges.
As already mentioned, a chordless path , , has vertices, say , and edges , ; the length of is . We also denote it as .
A chordless cycle , , has vertices, say , and edges , , and ; the length of is .
Let , , denote the clique with vertices. Let or diamond be the graph with four vertices, say , such that forms a and ; its midedge is the edge .
A butterfly has five vertices, say, , such that induce a with edges and (the peripheral edges of the butterfly), and .
We often consider an edge to be a set of two vertices; then it makes sense to say, for example, and , for an edge . For two vertices , let denote the distance between and in , i.e., the length of a shortest path between and in . The distance between a vertex and an edge is the length of a shortest path between and , i.e., . The distance between two edges is the length of a shortest path between and , i.e., . In particular, this means that if and only if .
An edge subset is an induced matching if the pairwise distance between its members is at least 2, that is, is isomorphic to for . Obviously, if is a d.i.m. then is an induced matching.
Clearly, has a d.i.m. if and only if every connected component of has a d.i.m.; from now on, connected components are mentioned as components.
Note that if has a d.i.m. , and denotes the vertex set of then is an independent set, say , i.e.,
(1) 
From now on, all vertices in are colored white and all vertices in are colored black. According to [11], we also use the following notions: A partial blackwhite coloring of is feasible if the set of white vertices is an independent set in and every black vertex has at most one black neighbor. A complete blackwhite coloring of is feasible if the set of white vertices is an independent set in and every black vertex has exactly one black neighbor. Clearly, is a d.i.m. of if and only if the black vertices and the white vertices form a complete feasible coloring of .
2.2 Reduction steps, forbidden subgraphs, forced edges, and excluded edges
Various papers on this topic introduced and applied some forcing rules for reducing the graph to a subgraph such that has a d.i.m. if and only if has a d.i.m., based on the condition that for a d.i.m. , has the partition such that all vertices in are black and all vertices in are white (recall (1)).
A vertex is forced to be black if for every d.i.m. of , . Analogously, a vertex is forced to be white if for every d.i.m. of , .
Clearly, if and if are forced to be black, then is contained in every (possible) d.i.m. of .
An edge is a forced edge of if for every d.i.m. of , . Analogously, an edge is an excluded edge of if for every d.i.m. of , .
For the correctness of the reduction steps, we have to argue that has a d.i.m. if and only if the reduced graph has one (provided that no contradiction arises in the vertex coloring, i.e., it is feasible).
Then let us introduce two reduction steps which will be applied later.
Vertex Reduction. Let . If is forced to be white, then

color black all neighbors of , and

remove from .
Let be the reduced subgraph. Clearly, Vertex Reduction is correct, i.e., has a d.i.m. if and only if has a d.i.m.
Edge Reduction. Let . If and are forced to be black, then

color white all neighbors of and of (other than and ), and

remove and (and the edges containing or ) from .
Again, clearly, Edge Reduction is correct, i.e., has a d.i.m. if and only if the reduced subgraph has a d.i.m.
The subsequent notions and observations lead to some possible reductions (some of them are mentioned e.g. in [3, 4, 5]).
Observation 1 ([3, 4, 5]).
Let be a d.i.m. of .

contains at least one edge of every odd cycle
in , , and exactly one edge of every odd cycle , , in . 
No edge of any can be in .

For each either exactly two or none of its edges are in .
Proof. See e.g. Observation 2 in [4].
In what follows, we will also refer to Observation 1 (with respect to ) as to the triangleproperty, and to Observation 1 as to the property.
Since by Observation 1 , every triangle contains exactly one edge, and the pairwise distance of edges is at least 2, we have:
Corollary 1.
If has a d.i.m. then is free.
Assumption 1. From now on, by Corollary 1, we assume that the input graph is free (else it has no d.i.m.).
Clearly, it can be checked (directly) in polynomial time whether the input graph is free.
By Observation 1 with respect to and the distance property, we have the following:
Observation 2.
The midedge of any diamond in and the two peripheral edges of any induced butterfly are forced edges of .
Assumption 2. From now on, by Observation 2, we assume that the input graph is (diamond,butterfly)free.
In particular, we can apply the Edge Reduction to each midedge of any induced diamond and to each peripheral edge of any induced butterfly; that can be done in polynomial time.
2.3 The distance levels of an edge in a
Based on [5], we first describe some general structure properties for the distance levels of an edge in a d.i.m. of . Since is , diamond, butterfly)free, we have:
Observation 3.
For every vertex of , is the disjoint union of isolated vertices and at most one edge. Moreover, for every edge , there is at most one common neighbor of and .
Since it is trivial to check whether has a d.i.m. with exactly one edge, from now on we can assume that . Since is connected and butterflyfree, we have:
Observation 4.
If then there is an edge in which is contained in a of .
Proof. Let and assume that is not part of an induced of . Since is connected and , , and since we assume that is not part of an induced of and is  and diamondfree, there is exactly one neighbor of , namely a common neighbor, say of and . Again, since , has a neighbor , and since is  and diamondfree, induce a paw. Clearly, the edge is excluded and has to be dominated by a second edge, say but now, since is butterflyfree, . Thus, induce a in , and Observation 4 is shown. ∎
Recall [5] for Observation 4. Let be an edge for which there is a vertex such that induce a with edge . By the assumption that , we have that and are black, and it could lead to a feasible coloring (if no contradiction arises).
Let and for , let
denote the distance levels of . We consider a partition of into , , with respect to the edge (under the assumption that ). Clearly, by Observation 4 and since is free, we have:
(2) 
Recall that by (1), is a partition of where is the set of black vertices and is the set of white vertices which is independent.
Since we assume that (and is an edge in a ), clearly, and thus:
(3) 
Moreover, no edge between and is in . Since and all neighbors of vertices in are in , we have:
(4) 
Let denote the set of edges with and let denote the set of isolated vertices in ; is a partition of . Obviously:
(5) 
If for , an edge is contained in every d.i.m. of with , we say that is an forced edge, and analogously, if an edge is contained in no d.i.m. of with , we say that is excluded. The Edge Reduction for forced edges can also be applied for forced edges (then, in the unsuccessful case, has no d.i.m. containing ), and correspondingly for forced white vertices (resulting from the black color of and ), the Vertex Reduction can be applied.
Obviously, by (5), we have:
(6) 
Thus, from now on, after applying the Edge Reduction for edges, we can assume that , i.e., . For every , let denote the mate of (i.e., ). Let denote the set of edges with one endpoint in (and the other endpoint in ). Obviously, by (5) and the distance condition for a d.i.m. , the following holds:
(7) 
As a consequence of (7) and the fact that every triangle contains exactly one edge (recall Observation 1 ), we have:
(8) 
This means that for the edge , the Edge Reduction can be applied, and from now on, we can assume that there is no such triangle with and , i.e., for every edge in :
(9) 
According to and the assumption that (recall ), let:

,

, , and

.
By definition, is the set of private neighbors of in (note that ), is a partition of , and is a partition of .
Lemma 1 ([5]).
The following statements hold:

For all , .

For all , is the disjoint union of vertices and at most one edge.

is bipartite.

, i.e., is an independent subset of white vertices.

If a vertex sees two vertices in , , , then is an forced edge.
Proof. : Holds by definition of and by the distance condition of a d.i.m. .
: Holds by Observation 3.
: If , i.e., sees at least two vertices then clearly, , and thus, is an independent subset (recall that is an independent set).
: Suppose that sees and in . If then would induce a diamond in . Thus, and now, induce a in ; by Observation 1 , no edge in the is in , and by (7), the only possible edge for dominating is , i.e., . ∎
By Lemma 1 and the Vertex Reduction for the white vertices of , we can assume:

, i.e., .
By Lemma 1 , we can assume:

For , , every vertex has at most one neighbor in .
In particular, if for some , , then there is no d.i.m. of with , and if , say , then is an forced edge. Thus, we can assume:

For every , .
Let us say that a vertex , , is an outvertex of if it is adjacent to some vertex of with , or it is adjacent to some vertex of , and is an invertex of otherwise.
For finding a d.i.m. with , one can remove all but one invertices; that can be done in polynomial time. In particular, if there is an edge between two invertices , , then either or is black, and thus, is completely colored. Thus, let us assume:

For every , has at most one invertex.
Lemma 2.
Assume that has a d.i.m. with . Then:

For every , there are at most two edges between and .

If there are two edges between and , say and for and , , , then every vertex in is white.
Proof. : Suppose to the contrary that there are three edges between and , say , , and for , . Then is black if and only if is white, is black if and only if is white, and is black if and only if is white. Without loss of generality, assume that is black, and is white. Then is white, and is black, but now, and are white, which is a contradiction.
: Let , , be two such edges between and . By Lemma 1 , , and . Then again, or is black as well as or is black, and thus, every other vertex in or is white.
Thus Lemma 2 is shown. ∎
By Lemma 2 , we can assume:

For , , there are at most two edges between and .
Recall that . If there is an edge in , say with and there is a third vertex then either or is black, and thus is forced to be white, and by the Vertex Reduction and by Lemma 2 , we can assume:

If there is an edge in then . Analogously, if there are two edges between and then and .
Then let us introduce the following forcing rules (which are correct). Since no edge in is in (recall (7)), we have:

All neighbors of a black vertex in must be colored white, and all neighbors of a white vertex in must be colored black.
Moreover, we have:

Every , , should contain exactly one vertex which is black. Thus, if is black then all the remaining vertices in must be colored white.

If all but one vertices of , , are white and the final vertex is not yet colored, then must be colored black.
Since no edge between and is in (recall (7)), we have:

For every edge with and , is white if and only if is black and vice versa.
Subsequently, for checking if has a d.i.m. with , we consider the cases and . In particular, we have the following property:
Observation 5.
If for then is an endpoint of an induced , say with vertices such that and with edges , , , , . Analogously, if then is an endpoint of a corresponding induced .
Proof. If then clearly there is such a . Thus, assume that . Then and . Recall that induce a . If then induce a . Thus assume that . Let be a neighbor of . Now, if then induce a , and if but then induce a . Analogously, if then is an endpoint of an induced (which could be part of the above). Thus, Observation 5 is shown. ∎
Recall that the distance between two vertices in graph , is the number of edges in a shortest path in between and .
Theorem 2 ([1]).
Every connected free graph admits a vertex such that for every .
We call such a vertex a central vertex. Theorem 2 implies that every connected free graph admits a vertex such that for every . Then let be a connected free graph and let be a central vertex such that for every . Clearly, for every edge , for any . Moreover, since is (,diamond,butterfly)free (and is no triangle, else the DIM problem is trivial), is contained in an induced of .
If one could check, for any edge , whether there is a d.i.m. of with , then one could conclude: Either has a d.i.m. with , or has no d.i.m. with ; in particular, in the latter case, if none of the edges is in a d.i.m. then is white and one can apply the Vertex Reduction to and in particular remove .
Then let us introduce the following recursive algorithm which formalizes the approach we will adopt to check if has a d.i.m.
Algorithm DIM()
Input. A connected  and (,diamond,butterfly)free graph .
Output. A d.i.m. of or the proof that has no d.i.m.

Compute a central vertex, say , of such that for every .

For each edge of [contained in a of ] do:

compute the distance levels with respect to and apply the reduction steps as shown above: if no contradiction arose and if assumptions (A1)(A6) hold, then go to Step (B.2), else take another edge with ;

check if has a d.i.m. with ; if , then return it, and STOP;


Apply the Vertex Reduction to [and in particular remove ]; let denote the resulting graph, where the neighbors of in are colored by black; if is disconnected, then execute Algorithm DIM() for each connected component of ; otherwise, go to Step (B), with .

Return “ has no d.i.m.” and STOP. ∎
Then, by the above, Algorithm DIM() is correct and can be executed in polynomial time as soon as Step (B.2) can be so.
Then in what follows let us try to show that Step (B.2) can be solved in polynomial time, with the agreement that is (,diamond,butterfly)free and enjoys assumptions (A1)(A6): in particular recall that for .
Thus we consider the cases and . Let and .
3 The Case
In this section, we show that for the case (i.e., ), one can check in polynomial time whether has a d.i.m. with ; we consider the feasible colorings for . Recall that for every edge , and are black, for , every vertex in is white, and all , , are black, , and recall assumptions (A1)(A6) and rules (R1)(R4). In particular, , i.e., .
Clearly, in the case , all the components of can be independently colored. Every component with at most three vertices has a polynomial number of feasible colorings. Thus, we can focus on components with at least four vertices.
A in is isolated in if it is not part of a in .
Claim 1.
If every in component in is isolated then has at most three vertices.
Proof. Suppose to the contrary that has at least four vertices, say with contacts between and , , say , , and . But then induce a , which is a contradiction. Thus, Claim 1 is shown. ∎
From now on, we can assume that there is at least one with contact between and in .
Claim 2.
For any ’s and in such that are not in the ’s of , there is an edge between and .
Proof. Suppose to the contrary that there is no such edge between the ’s and in . Since is diamondfree, are in at least two ’s; assume that . Then, by Lemma 1 and since , , we have ; let , i.e., . Then either or ; without loss of generality, let . Analogously, since are not in , assume that and .
Let be any induced path in between and through . Then the subgraph of induced by , , and contains an induced , which is a contradiction. Thus, Claim 2 is shown. ∎
For a in with and , vertex is a special endpoint of in .
Claim 3.
There is no in with special endpoint .
Proof. Suppose to the contrary that is a in with special endpoint and . But then by Observation 5, vertex is the midpoint of a , which is a contradiction. Thus, Claim 3 is shown. ∎
Claim 4.
If is a in with exactly one vertex and , , , (possibly or ) then and are forced to be black, i.e., and are forced edges, and thus, and are completely colored.
Proof. Suppose to the contrary that there is a in (possibly or ) such that is white. Then and are black, which implies that is white, and thus, is black, which is a contradiction. Analogously, if is white then it leads to the same contradiction. Thus, Claim 4 is shown. ∎
After the Edge Reduction step, we can assume that there is no such in , i.e., every in has either two vertices of or none of it.
Claim 5.
If is a in then has exactly two vertices in , say , and then is forced to be black, i.e., is an forced edge.
Proof. Let be a in . Recall that by Lemma 1 , there is no in . Thus, , and clearly, .
If there is exactly one vertex in a in , say with then
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