Finding Efficient Domination for S 1,3,3 -Free Bipartite Graphs in Polynomial Time
aa r X i v : . [ c s . D M ] A ug Finding Efficient Domination for S , , -FreeBipartite Graphs in Polynomial Time Andreas Brandst¨adt
Institut f¨ur Informatik, Universit¨at Rostock, D-18051 Rostock, Germany [email protected]
Raffaele Mosca
Dipartimento di Economia, Universit´a degli Studi “G. D’Annunzio”, Pescara 65121, Italy [email protected]
August 11, 2020
Abstract
A vertex set D in a finite undirected graph G is an efficient dominatingset ( e.d.s. for short) of G if every vertex of G is dominated by exactly onevertex of D . The Efficient Domination (ED) problem, which asks for theexistence of an e.d.s. in G , is NP -complete for various H -free bipartite graphs,e.g., Lu and Tang showed that ED is NP -complete for chordal bipartite graphsand for planar bipartite graphs; actually, ED is NP -complete even for planarbipartite graphs with vertex degree at most 3 and girth at least g for everyfixed g . Thus, ED is NP -complete for K , -free bipartite graphs and for C -freebipartite graphs.In this paper, we show that ED can be solved in polynomial time for S , , -free bipartite graphs. Keywords : Efficient domination; S , , -free bipartite graphs. Let G = ( V, E ) be a finite undirected graph. A vertex v dominates itself and itsneighbors. A vertex subset D ⊆ V is an efficient dominating set ( e.d.s. for short)of G if every vertex of G is dominated by exactly one vertex in D ; for any e.d.s. D of G , | D ∩ N [ v ] | = 1 for every v ∈ V (where N [ v ] denotes the closed neighborhoodof v ). 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 G .In [1, 2], it was shown that the ED problem is NP -complete, and it is NP -completefor P k -free graphs, k ≥
7. Moreover, ED is NP -complete for claw-free graphs (even1or ( K , , K − e )-free perfect graphs [8]). However, in [4], we have shown that EDis solvable in polynomial time for P -free graphs which leads to a dichotomy of EDfor H -free graphs.Lu and Tang [9] showed that ED is NP -complete for chordal bipartite graphs(i.e., hole-free bipartite graphs). Thus, for every k ≥
3, ED is NP -complete for C k -free bipartite graphs. Moreover, ED is NP -complete for planar bipartite graphs[9] and even for planar bipartite graphs of maximum degree 3 [3] and girth at least g for every fixed g [10]. Thus, ED is NP -complete for K , -free bipartite graphs andfor C -free bipartite graphs.It is well known that for every graph class with bounded clique-width, ED can besolved in polynomial time [6]; for instance, the clique-width of claw-free bipartite,i.e., K , -free bipartite graphs is bounded. Dabrowski and Paulusma [7] publisheda dichotomy for clique-width of H -free bipartite graphs. For instance, the clique-width of S , , -free bipartite graphs is bounded (which includes K , -free bipartitegraphs). However, the clique-width of S , , -free bipartite graphs is unbounded.In [5], we solved ED in polynomial time for P -free bipartite graphs, for ℓP -free bipartite graphs for fixed ℓ , for S , , -free bipartite graphs as well as for P -freebipartite graphs with degree at most 3, but we had some open problems: What isthe complexity of ED for − P k -free bipartite graphs, k ≥ − S , , -free bipartite graphs, − S , , -free bipartite graphs, − S , ,k -free bipartite graphs for k ≥ − chordal bipartite graphs with vertex degree at most 3?In this manuscript, we will show: Theorem 1.
For S , , -free bipartite graphs, ED can be solved in polynomial time. Let G = ( X, Y, E ) with V ( G ) = X ∪ Y , be a bipartite graph, say, every vertex in X is black, and every vertex in Y is white. For subsets U, W ∈ V ( G ), the join U (cid:13) W denotes uw ∈ E for every u ∈ U and every w ∈ W . Moreover, the cojoin U (cid:13) W denotes uw / ∈ E for every u ∈ U and every w ∈ W . In particular, for u ∈ V ( G ), u (cid:13) W if uw ∈ E for every w ∈ W and u (cid:13) W if uw / ∈ E for every w ∈ W .Let P i , i ≥
1, denote the chordless path with i vertices, and in bipartite graphs,let C i , i ≥
2, denote the chordless cycle with 2 i vertices. Recall that a subgraph 2 P i ,2 ≥
3, has two P i ’s P , P ′ without any contact between P and P ′ , i.e., P (cid:13) P ′ . Forexample, if ( u , v , w ), ( u , v , w ) induce a 2 P in G then { u , v , w } (cid:13){ u , v , w } .Analogously, subgraph 3 P i , i ≥
3, has three P i ’s P , P ′ , P ′′ without any contactbetween P , P ′ and P ′′ , i.e., P (cid:13) P ′ , P (cid:13) P ′′ , and P ′ (cid:13) P ′′ .For indices i, j, k ≥
0, let S i,j,k denote the graph with vertices u, x , . . . , x i , y , . . . , y j , z , . . . , z k (and midpoint u ) such that the subgraph induced by u, x , . . . , x i forms a P i +1 ( u, x , . . . , x i ), the subgraph induced by u, y , . . . , y j forms a P j +1 ( u, y , . . . , y j ), and the subgraph induced by u, z , . . . , z k forms a P k +1 ( u, z , . . . , z k ),and there are no other edges in S i,j,k . Thus, claw is S , , , chair is S , , , and P k isisomorphic to S , ,k − .Clearly, an S , , S in G has eight vertices, say S = ( r , s, r , r , r , r , r , r ) suchthat ( r , . . . , r ) induce a P with r i r i +1 ∈ E , 1 ≤ i ≤ s contacts only r , i.e., r s ∈ E and r i s / ∈ E for i ∈ { , , , , , } , and r is the midpoint of S , , S . Thebipartite graph G is S , , -free if there is no induced S , , in G .Clearly, in general, S , , -free bipartite graphs are not P k -free for any k ≥
1; forexample, every bipartite graph G = P k , k ≥
1, is S , , -free.Let D be a possible e.d.s. of G . For example, if G = P , say ( r ), then D = { r } ,if G = P , say ( r , r ), then D = { r } or D = { r } , if G = P , say ( r , r , r ),then D = { r } , if G = P , say ( r , r , r , r ), then D = { r , r } etc. For instance, if G = P , say ( r , . . . , r ), then D = { r , r , r , r } . In general, for every G = P k , k ≥ G is S , , -free bipartite and has an e.d.s. which can be solved in polynomialtime.Another example is G = C k , k ≥
2; every such G = C k is S , , -free bipartite.Clearly, G = C has no e.d.s., G = C has three e.d.s., G = C has no e.d.s., G = C has no e.d.s., G = C has three e.d.s., etc. In general, G = C k , k ≥
2, has an e.d.s.if and only if k = 3 i , i ≥ u ∈ V ( G ) is forced if u ∈ D for every e.d.s. D of G ; u is excluded if u / ∈ D for every e.d.s. D of G . For example, if x , x ∈ X are leaves in G and y isthe neighbor of x , x then x , x are excluded and y is forced.By a forced vertex, G can be reduced to G ′ as follows: Claim 2.1. If u is forced then G has an e.d.s. D with u ∈ D if and only if thereduced graph G ′ = G \ N [ u ] has an e.d.s. D ′ = D \ { u } such that all vertices in N ( u ) are excluded in G ′ . Analogously, if we assume that v ∈ D ∩ V ( G ) then u ∈ V ( G ) is v -forced if u ∈ D for every e.d.s. D of G with v ∈ D , and u is v -excluded if u / ∈ D for every e.d.s. D of G with v ∈ D . For checking whether G has an e.d.s. D with v ∈ D , we canclearly reduce G by forced vertices as well as by v -forced vertices when we assumethat v ∈ D : Claim 2.2. If v ∈ D and u is v -forced then G has an e.d.s. D with v ∈ D if andonly if the reduced graph G ′ = G \ N [ u ] has an e.d.s. D ′ = D \ { u } with v ∈ D ′ suchthat all vertices in N ( u ) are v -excluded in G ′ . k ≥ u ∈ V ( G ) is ( v , . . . , v k ) -forced if u ∈ D for every e.d.s. D of G with v , . . . , v k ∈ D , and correspondingly, u ∈ V ( G ) is ( v , . . . , v k ) -excluded if u / ∈ D for such e.d.s. D , and G can be reduced by the same principle.Clearly, for every connected component of G , the e.d.s. problem is independentlysolvable. Thus, we can assume that G is connected. For every vertex v ∈ V ( G ),either v ∈ D or | N ( v ) ∩ D | = 1.Without loss of generality, assume that there is a D -vertex d ∈ D ∩ X , i.e., d isblack. Let N i , i ≥
1, be the distance levels of d . By the e.d.s. property, we have: D ∩ ( N ∪ N ) = ∅ (1)If there is a vertex x ∈ N with N ( x ) ∩ N = ∅ then G has no e.d.s. D with d ∈ D .In particular, if N = ∅ then G has no e.d.s. D with d ∈ D . Thus, we can assume:For every vertex x ∈ N , N ( x ) ∩ N = ∅ . (2)Next we have:If for x ∈ N , N ( x ) ∩ N = { y } then y is d -forced . (3)If y ∈ N with N ( y ) ∩ N = ∅ then y is d -forced . (4)By the e.d.s. property, it can lead to a contradiction: If two d -forced vertices y , y ∈ N (e.g. with N ( y i ) ∩ N = ∅ , i ∈ { , } ), have a common neighbor in N ∪ N thenthere is no e.d.s. D with d ∈ D . Thus assume that every such pair of d -forcedvertices y , y has no common neighbor.In particular, if N = ∅ then every y ∈ N is d -forced, i.e., N ⊂ D with d ∈ D ,and it leads to a simple e.d.s. solution in polynomial time or a contradiction if twovertices in N have a common neighbor in N . Thus, we can assume: N = ∅ . (5) Claim 2.3.
For i ≥ and j ∈ { , } , let u j ∈ N i , v j ∈ N i +1 , w j ∈ N i +2 such that ( u , v , w ) , ( u , v , w ) induce a P in G . Then we have: ( i ) N ( u ) ∩ N ( u ) ∩ N i − = ∅ . ( ii ) If v , v ∈ D then N ( w ) ∩ N ( w ) ∩ N i +3 = ∅ . Proof. ( i ): Suppose to the contrary that u and u have a common neighbor r ∈ N i − . Recall that i ≥ r has a neighbor s ∈ N i − , say rs ∈ E (possibly s = d if i = 2). But then ( r, s, u , v , w , u , v , w ) (with midpoint r ) induce an S , , in G , which is a contradiction. Thus, N ( u ) ∩ N ( u ) ∩ N i − = ∅ .( ii ): If v , v ∈ D and w , w have a common neighbor in N i +3 , say rw ∈ E , rw ∈ E with r ∈ N i +3 , then r must have a D -neighbor s ∈ D ∩ ( N i +2 ∪ N i +4 ). Butthen ( r, s, w , v , u , w , v , u ) (with midpoint r ) induce an S , , in G , which is acontradiction. Thus, N ( w ) ∩ N ( w ) ∩ N i +3 = ∅ , and Claim 2.3 is shown.4 orollary 1. If y , y ∈ D ∩ N with N ( y i ) ∩ N = ∅ , i ∈ { , } , then for every x i ∈ N ( y i ) ∩ N , i ∈ { , } , there is no common neighbor of x and x in N . Proof.
Let x i y i ∈ E for x i ∈ N , i ∈ { , } , and let y i z i ∈ E for z i ∈ N , i ∈ { , } .By the e.d.s. property, ( x , y , z ) and ( x , y , z ) induce a 2 P in G [ N ∪ N ∪ N ].By Claim 2.3 ( i ), x and x have no common neighbor in N . Claim 2.4.
There are at most two D -vertices y , y ∈ D ∩ N with N ( y i ) ∩ N = ∅ , i ∈ { , } . Proof.
Suppose to the contrary that there are three D -vertices y , y , y ∈ D ∩ N with N ( y i ) ∩ N = ∅ , 1 ≤ i ≤
3. Recall that by the e.d.s. property, thereis a 3 P ( x , y , z ), ( x , y , z ), ( x , y , z ) in G [ N ∪ N ∪ N ]. Recall that byCorollary 1, there is no common neighbor of x i , x j , i = j , in N ; let u i x i ∈ E for u i ∈ N , i ∈ { , , } . But then ( d, u , u , x , y , u , x , y ) (with midpoint d )induce an S , , in G , which is a contradiction (in this case, there is even an S , , ( d, u , x , y , z , u , x , y , z , u , x , y , z ) with midpoint d in G ). Thus, Claim 2.4 isshown. Claim 2.5.
If there is a d -forced vertex y ∈ D ∩ N with N ( y ) ∩ N = ∅ then thereis at most one D -vertex y ′ ∈ D ∩ N with N ( y ′ ) ∩ N = ∅ . Proof.
Let y ∈ D ∩ N with N ( y ) ∩ N = ∅ and xy ∈ E with x ∈ N . Recallthat by (4), y is d -forced. Suppose to the contrary that there are two D -vertices r , s ∈ D ∩ N with contact to N . By the e.d.s. property, ( r , r , r ) and ( s , s , s ), r i , s i ∈ N i , 2 ≤ i ≤
4, induce a 2 P in G . By Claim 2.3 ( i ), r and s do not have anycommon neighbor in N , i.e., r r ∈ E and s s ∈ E for r , s ∈ N and r s / ∈ E , s r / ∈ E . First assume that xr ∈ E . Clearly, by the e.d.s. property, xr / ∈ E and xs / ∈ E . Since ( r , x, r , r , r , d, s , s ) (with midpoint r ) do not induce an S , , in G , we have xs ∈ E . But then ( x, y, r , r , r , s , s , s ) (with midpoint x )induce an S , , in G , which is a contradiction. Thus, xr / ∈ E and analogously, xs / ∈ E . Now assume that ux ∈ E for u ∈ N . Since ( d, u, r , r , r , s , s , s ) (withmidpoint d ) do not induce an S , , in G , we have ur ∈ E or us ∈ E . Recall thatby Claim 2.3 ( i ), r and s do not have any common neighbor in N . Thus either ur / ∈ E or us / ∈ E . Without loss of generality, assume that ur ∈ E and us / ∈ E .But then ( u, x, r , r , r , d, s , s ) (with midpoint u ) induce an S , , in G , which is acontradiction. Thus, there is at most one D -vertex y ′ ∈ D ∩ N with N ( y ′ ) ∩ N = ∅ ,and Claim 2.5 is shown. Claim 2.6.
There is at most one D -vertex in D ∩ N with contact to N . Proof.
Suppose to the contrary that there are two D -vertices r , s ∈ D ∩ N withcontact to N , say ( r , r , r ) and ( s , s , s ), r i , s i ∈ N i , 3 ≤ i ≤
5, induce a 2 P in G .By Claim 2.3 ( i ), r and s do not have any common neighbor in N , i.e., r r ∈ E and s s ∈ E for r , s ∈ N and r s / ∈ E , s r / ∈ E . Then again, ( r , r , r ) and( s , s , s ), r i , s i ∈ N i , 2 ≤ i ≤
4, induce a 2 P in G . Analogously, r and s do not5ave any common neighbor in N , i.e., r r ∈ E and s s ∈ E for r , s ∈ N and r s / ∈ E , s r / ∈ E . Then again, ( r , r , r ) and ( s , s , s ), r i , s i ∈ N i , 1 ≤ i ≤ P in G .Since by the e.d.s. property, r / ∈ D , r must have a D -neighbor in N , say r y ∈ E for y ∈ D ∩ N . But then ( r , y, r , r , r , r , d, s ) (with midpoint r )induce an S , , in G , which is a contradiction.Thus, there is at most one D -vertex in D ∩ N with contact to N , and Claim 2.6is shown. Claim 2.7. If N = ∅ then the e.d.s. problem can be solved in polynomial time. Proof.
Recall (3), (4) for d -forced vertices in N , and recall (5). Case . There are no d -forced vertices in D ∩ N .Then by Claim 2.4, there are at most two D -vertices y , y ∈ D ∩ N with N ( y i ) ∩ N = ∅ , i ∈ { , } . Moreover, every y ∈ N \ D , y = y , y , must have a D -neighborin N \ ( N ( y ) ∪ N ( y )) which is ( d, y , y )-forced or it leads to a contradiction, e.g., ifthere is a vertex y ∈ N \ D , y = y , y , which has no neighbor in N \ ( N ( y ) ∪ N ( y )),or there are z , z ∈ N \ ( N ( y ) ∪ N ( y )) with common neighbor in N , or if thereis a vertex y ∈ N \ D , y = y , y , with z , z ∈ N ( y ) ∩ ( N \ ( N ( y ) ∪ N ( y ))). Case . There exist d -forced vertices in D ∩ N .Let D be the subset of d -forced vertices in D ∩ N . Case . . The d -forced vertices in D contact every vertex in N , i.e., D (cid:13) N .Then by the e.d.s. property, for every y ∈ N \ D , y / ∈ D and y must have a D -neighbor in D ∩ N .For example, if N = { u } , N = { x , x } , N = { y , y , y } , and N = { z } , suchthat ( u, x , y , x ) induce a C in G , and du ∈ E , x y ∈ E , x y ∈ E , y z ∈ E , arethe only other edges in G then y , y are d -forced and z is ( d, y , y )-forced whichleads to an e.d.s. D = { d, y , y , z } in G .Recall that by Claim 2.4, there are at most two such d -forced vertices y , y ∈ D with N ( y i ) ∩ N = ∅ , i ∈ { , } . If every y ∈ N \ D has exactly one neighbor z ∈ N and z does not contact the d -forced vertices y , y ∈ D , then z is ( d, y , y )-forced,and it is easy to solve the e.d.s. problem in polynomial time.Moreover, if y ∈ N \ D has two neighbors in N which do not contact the d -forced vertices y , y ∈ D , then it leads to a contradiction. Finally, if for y ∈ N \ D ,every neighbor of y in N contacts one of the d -forced vertices y , y ∈ D , then itleads to a contradiction, and there is no such e.d.s. D with d ∈ D . Case . . The d -forced vertices in D do not contact every vertex in N . Case . . . At least one of the d -forced vertices in D does not contact N .Then by Claim 2.5, there is at most one D -vertex y ′ ∈ D ∩ N with contact to N . Let N ′ be the part of N which has no contact to d -forced vertices in D or to6 ′ . Moreover, every y ∈ N \ D must have a D -neighbor in N ′ . Then N ′ ⊂ D whichis a solution or it leads to a contradiction. Case . . . Every d -forced vertex in D contacts N .Then again by Claim 2.4, there are at most two D -vertices y , y ∈ D ∪ D ∩ N with N ( y i ) ∩ N = ∅ , i ∈ { , } (recall Case 1). Moreover, every y ∈ N \ D , y = y , y , must have a D -neighbor in N \ ( N ( y ) ∪ N ( y )) which is ( d, y , y )-forced or it leads to a contradiction, e.g., if there is a vertex y = y , y , which hasno neighbor in N \ ( N ( y ) ∪ N ( y )), or there are z , z ∈ N \ ( N ( y ) ∪ N ( y ))with common neighbor in N , or if there is a vertex y ∈ N \ D , y = y , y , with z , z ∈ N ( y ) ∩ ( N \ ( N ( y ) ∪ N ( y ))).Finally, Claim 2.7 is shown.Thus, from now on, we can assume: N = ∅ . (6) Corollary 2. If N = ∅ but N = ∅ then the e.d.s. problem can again be solved inpolynomial time. Proof.
Recall Claim 2.6 that there is at most one D -vertex z ∈ D ∩ N with N ( z ) ∩ N = ∅ . If z (cid:13) N then the e.d.s. problem can be solved as in Claim 2.7.However, if N \ N ( z ) = ∅ and since N = ∅ then N \ N ( z ) ⊂ D , and by Claim 2.7,the e.d.s. problem can be solved in polynomial time or it leads to a contradiction.Thus, from now on, we can assume: N = ∅ . (7) d ∈ D and N = ∅ Recall d ∈ D , N i are the distance levels of d and N = ∅ , say P = ( d, r , r , . . . , r )is a P in G with r i ∈ N i , 1 ≤ i ≤ P ( s , d, r , . . . , r ) with s , r ∈ N and r i ∈ N i , ≤ i ≤ Claim 3.1.
If there is a non-neighbor s ∈ N of r , i.e., s r / ∈ E then N ( r ) ∩ N = { r } , and r is d -forced.Proof. Suppose to the contrary that | N ( r ) ∩ N | ≥
2, say r , r ′ ∈ N ( r ) ∩ N . Since( r , r ′ , r , d, s , r , r , r ) (with midpoint r ) do not induce an S , , in G , we have r ′ r ∈ E , and in general, r (cid:13) N ( r ) ∩ N . Clearly, r must have a D -neighbor in N ,say without loss of generality, r ∈ D . Recall that D ∩ N = ∅ . Then r ′ must have a D -neighbor r ′ ∈ D ∩ N . Since ( r ′ , r ′ , r , r , d, r , r , r ) (with midpoint r ′ ) do not7nduce an S , , in G , we have r ′ r ∈ E . But then ( r , r , r , d, s , r ′ , r ′ , r ) (withmidpoint r ) induce an S , , in G , which is a contradiction. Thus, N ( r ) ∩ N = { r } , r is d -forced, and Claim 3.1 is shown.Recall (4): If y ∈ N with N ( y ) ∩ N = ∅ then y is d -forced. Case A.
There exist d -forced vertices y j ∈ D ∩ N with N ( y j ) ∩ N = ∅ , 1 ≤ j ≤ k .Recall Claim 2.5 that in this case, there is at most one D -vertex in D ∩ N withcontact to N . Then by Claim 3.1, r is the only D -vertex in N with contact to N , and r is d -forced. Let N ′ := N ∩ ( N ( y ) ∪ . . . ∪ N ( y k )) and N ′′ := N \ N ′ .Then G can be reduced to G ′ := G \ ( N [ d ] ∪ N [ y ] ∪ . . . ∪ N [ y k ]), i.e., G hasan e.d.s. with d, y , . . . , y k ∈ D if and only if G ′ has an e.d.s. with such a d -forcedvertex r ∈ D ∩ N , and the distance levels N i ( r ), i ≥
1, of r in G ′ are lessthan the distance levels of d in G : Recall that N i = N i ( d ). Then for example, N ( r ) = N ′′ ∪ N ( d ) and N i ( r ) = N i +3 ( d ), i ≥ Case B.
There is no d -forced vertex y ∈ D ∩ N with N ( y ) ∩ N = ∅ .Recall that by Claim 3.1, r is d -forced. Case B.1.
First assume that r (cid:13) N :Then G can be reduced to G ′ := G \ N [ d ], i.e., G has an e.d.s. with d ∈ D if andonly if G ′ has an e.d.s. with such a d -forced vertex r ∈ D ∩ N , and the distancelevels of r in G ′ are less than the distance levels of d in G . Case B.2.
Now assume that r has no join to N :Then there is an s ∈ N with s r / ∈ E and there is a D -neighbor s ∈ D ∩ N for s . By Case B, N ( s ) ∩ N = ∅ , say s s ∈ E for a neighbor s ∈ N . Recallthat ( r , r , r ), ( s , s , s ) induce a 2 P , and by Claim 2.3 ( i ) (i.e., by Corollary 1), r and s do not have any common neighbor in N , say s s ∈ E . Moreover, byClaim 2.3 ( ii ), r and s do not have any common neighbor in N , i.e., s r / ∈ E .Recall that by Claim 2.4, there are at most two D -vertices in N , and r is already d -forced. For a second D -vertex s ∈ D ∩ N , G can be reduced to G ′ := G \ N [ d ],i.e., G has an e.d.s. with d ∈ D if and only if G ′ has an e.d.s. with such vertices r , s ∈ D ∩ N . In particular, if N ( s ) ∩ N = { s } then by (3), s is d -forced,and moreover, if N ( s ) ∩ N = ∅ , say s ∈ N and s ∈ N with P ( s , . . . , s )then ( r , d, s , . . . , s ) is a P in G , and by Claim 3.1, s is d -forced. Then G can bereduced to G ′ := G \ N [ d ].For example, if G = ( r , . . . , r , d, s , . . . , s ) induce a C then G has an e.d.s. D with d ∈ D if and only if G ′ := G \ N [ d ] has an e.d.s. D ′ = { r , r , s } .Now assume that the second D -neighbor s ∈ D ∩ N is not yet d -forced, i.e., | N ( s ) ∩ N | ≥ N ( s ) ∩ N = ∅ . Assume that s , s ′ ∈ N ( s ) ∩ N ; without lossof generality, assume that s ∈ D and thus, s ′ must have a D -neighbor s ′ ∈ D ∩ N .Recall that r s / ∈ E .First we claim s ′ (cid:13) N : 8f s ′ has a neighbor s ′ ∈ N then ( s , s , s ′ , s ′ , s ′ , s , d, r ) (with midpoint s )induce an S , , in G , which is a contradiction. Thus, s ′ (cid:13) N .Moreover, we claim s (cid:13) N :Suppose to the contrary that s s ∈ E for s ∈ N . Since ( s , s ′ , s , s , s , s , d, r )(with midpoint s ) do not induce an S , , in G , we have s ′ s ∈ E . Since s musthave a D -neighbor t ∈ D ∩ ( N ∪ N ) and since N ( s ) ∩ N = ∅ (else s is d -forced),we have t ∈ D ∩ N . But then ( s ′ , s ′ , s , s , t, s , s , d ) (with midpoint s ′ ) induce an S , , in G , which is a contradiction. Thus, s (cid:13) N , and G ′ is disconnected.If G ′ is disconnected then for G ′ , the e.d.s. problem can be solved independentlyfor the connected components of G ′ . For example, if G = ( r , . . . , r , d, s , . . . , s )induce a P then G has an e.d.s. D with d ∈ D if and only if G ′ := G \ N [ d ] = G \ { d, r , s } has an e.d.s. for its components, namely r , r ∈ D for component G ′ = ( r , . . . , r ) and s , s ∈ D for component G ′ = ( s , . . . , s ).Now assume that G ′ is connected. Then N ( r , s ) := { r , s } in G ′ and N i ( r , s ), i ≥
1, are the distance levels of r , s in G ′ . If N ( r , s ) = ∅ then it can be solvedas in the previous cases for d ∈ D and N = ∅ . Now assume that N ( r , s ) = ∅ and ( r , r , . . . , r ) induce a P in G ′ as well as ( s , s , . . . , s ) induce a P in G ′ with r i , s i ∈ N i ( d ), 2 ≤ i ≤
9. Then by Claim 3.1, r and s are ( r , s )-forced and G ′ can be reduced as in the previous case, i.e., G ′′ := G ′ \ ( N [ r ] ∪ N [ s ]) and G ′ hasan e.d.s. if and only if G ′′ has an e.d.s., and so on. P ( s , d, r , . . . , r ) with s , r ∈ N and r i ∈ N i , ≤ i ≤ Now assume that r (cid:13) N for every P -endpoint r ∈ N with P ( r , r , r , r , r ), r i ∈ N i , 2 ≤ i ≤ N = ∅ and let ( d, r , . . . , r ) be a P with r i ∈ N i , 1 ≤ i ≤ Claim 3.2. If ( d, r , . . . , r ) is a P with r i ∈ N i , ≤ i ≤ , then r ∈ D ∩ N or r ∈ D ∩ N .Proof. Suppose to the contrary that r / ∈ D and r / ∈ D . Then r must have a D -neighbor s ∈ D ∩ N , say r s ∈ E . Since ( r , s, r , r , r , r , r , d ) (with midpoint r ) do not induce an S , , , we have sr ∈ E , and r / ∈ D . Recall that r must havea D -neighbor t ∈ D ∩ ( N ∪ N ).First assume that t ∈ D ∩ N . Since ( r , t, r , r , r , r , r , r ) (with midpoint r )do not induce an S , , , we have r t ∈ E . But then ( r , s, r , r , r , r , t, r ) (withmidpoint r ) induce an S , , , which is a contradiction. Thus, t / ∈ D ∩ N , i.e., t ∈ D ∩ N .Since ( r , t, r , r , r , r , r , r ) (with midpoint r ) do not induce an S , , , we have r t ∈ E . But then ( r , s, r , t, r , r , r , d ) (with midpoint r ) induce an S , , , whichis a contradiction. Thus, r ∈ D ∩ N or r ∈ D ∩ N , and Claim 3.2 is shown.9f r ∈ D then ( r , . . . , r ) is a P in G as in Section 3.1, and the e.d.s. problemcan be solved as in Section 3.1. Now assume that r / ∈ D , and by Claim 3.2, r ∈ D .Then ( r , . . . , r ) is a P in G as in Section 3.1, and the e.d.s. problem can be solvedas in Section 3.1.From now on, assume that N = ∅ . Claim 3.3. If r ∈ N contacts N then r / ∈ D , i.e., r is d -excluded.Proof. Recall that r ∈ N has a neighbor r ∈ N , say ( d, r , . . . , r ) induce a P in G with r i ∈ N i , 1 ≤ i ≤
6. Suppose to the contrary that r ∈ D . Then r / ∈ D and r must have a D -neighbor s ∈ D ∩ N (recall that D ∩ N = ∅ ). Butthen ( r , s, r , r , d, r , r , r ) (with midpoint r ) induce an S , , in G , which is acontradiction. Thus, r / ∈ D , i.e., r is d -excluded, and Claim 3.3 is shown. Corollary 3. If r ∈ N contacts N then r / ∈ D , i.e., r is d -excluded.Proof. Let N = ∅ and ( d, r , . . . , r ) induce a P with r i ∈ N i , 1 ≤ i ≤ r ∈ D , and recall that r is d -excluded. Then r musthave a D -neighbor s ∈ D ∩ ( N ∪ N ). First assume that s ∈ D ∩ N . But then( r , s, r , r , r , r , r , r ) (with midpoint r ) induce an S , , , which is a contradiction.Thus s ∈ D ∩ N . Since ( r , s, r , r , r , r , r , r ) (with midpoint r ) do not inducean S , , , we have r s ∈ E , and r must have a D -neighbor t ∈ D ∩ ( N ∪ N ). If t ∈ D ∩ N then ( r , t, r , r , r , r , r , r ) (with midpoint r ) induce an S , , , whichis a contradiction. Thus t ∈ D ∩ N . Since ( r , t, r , r , r , r , r , r ) (with midpoint r ) do not induce an S , , , we have r t ∈ E . But then ( r , s, r , r , r , r , t, r ) (withmidpoint r ) induce an S , , , which is a contradiction. Thus, r is d -excluded, andCorollary 3 is shown.If N = ∅ and N = ∅ then recall that by Corollary 3, every vertex in N has no D -neighbor in N , and thus, every r ∈ N is d -forced, such that (unless acontradiction arises) G can be reduced to N = ∅ by Claim 2.2.Now assume that N = ∅ . Recall that by (7) and in this section, N = ∅ . Claim 3.4.
There is at most one D -vertex in D ∩ N with contact to N .Proof. Suppose to the contrary that there are two such D -vertices in D ∩ N withcontact to N , say ( r , r , r ), ( s , s , s ) induce a 2 P in G with r i , s i ∈ N i , 2 ≤ i ≤ i ), r and s do not have any common neighbor in N , say r r ∈ E and s s ∈ E for r , s ∈ N with r s / ∈ E and s r / ∈ E . Let t ∈ N and t ∈ N with t t ∈ E . Since in Section 3.2, there is no P ( s , d, r , . . . , r , t , t ),we have r t / ∈ E , and analogously, since there is no P ( r , d, s , . . . , s , t , t ), wehave s t / ∈ E . Let t t ∈ E for t ∈ N , t = r , s .Analogously, since there is no such P with last-but-one vertex d , we have r t / ∈ E and s t / ∈ E ; let t t ∈ E for t ∈ N , t = r , s .Analogously, since there is no such P with last-but-one vertex d , we have r t / ∈ E and s t / ∈ E ; let t t ∈ E for t ∈ N , t = r , s .10hen ( t , . . . , t ) is a P with t i ∈ N i , 2 ≤ i ≤
6. Recall that t (cid:13) N , i.e., t r ∈ E and t s ∈ E .Since ( t , t , r , r , r , s , s , s ) (with midpoint t ) do not induce an S , , , wehave t r ∈ E or t s ∈ E ; by the e.d.s. property, assume without loss of generalitythat t r ∈ E and t s / ∈ E . But then ( t , r , t , t , t , s , s , s ) (with midpoint t )induce an S , , , which is a contradiction.Thus, Claim 3.4 is shown. Corollary 4.
There is at most one D -vertex in D ∩ N with contact to N , and ingeneral, there is at most one D -vertex in D ∩ N i , i ≥ , with contact to N i +1 .Proof. Suppose to the contrary that there are two such D -vertices in D ∩ N withcontact to N , say ( r , r , r ), ( s , s , s ) induce a 2 P in G with r i , s i ∈ N i , 5 ≤ i ≤ i ), r and s do not have any common neighbor in N , say r r ∈ E and s s ∈ E for r , s ∈ N . Then ( r , r , r , r ), ( s , s , s , s ) induce a 2 P in G with r i , s i ∈ N i , 4 ≤ i ≤
7. Analogously, by Claim 2.3 ( i ), r and s do not haveany common neighbor in N , say r r ∈ E and s s ∈ E for r , s ∈ N , r and s do not have any common neighbor in N , say r r ∈ E and s s ∈ E for r , s ∈ N ,as well as r and s do not have any common neighbor in N , say r r ∈ E and s s ∈ E for r , s ∈ N . But then there is a P ( s , d, r , . . . , r ) with s , r ∈ N and r i ∈ N i , 2 ≤ i ≤
6, which is a contradiction in Section 3.2. Thus, there is atmost one D -vertex in D ∩ N with contact to N , and in general, there is at mostone D -vertex in D ∩ N i , i ≥
6, with contact to N i +1 .If N = ∅ and of course, N = ∅ , then recall that by Claim 3.3, every N -neighbor of every r ∈ N is d -excluded, and thus, every r ∈ N is d -forced, and(unless a contradiction arises) G can be reduced to N = ∅ by Claim 2.2, i.e., thee.d.s. problem can be solved in polynomial time.Now assume that N = ∅ and N = ∅ .Recall that by Corollary 4, there is at most one D -vertex in D ∩ N with contactto N . First assume that there is exactly one r ∈ D ∩ N with contact to N . If r (cid:13) N then as in the case N = ∅ , the e.d.s. problem can be solved in polynomialtime. Now assume that N \ N ( r ) = ∅ . Then N \ N ( r ) ⊂ D and again, the e.d.s.problem can be solved in polynomial time. Finally, if no r ∈ D ∩ N contacts N then N ⊂ D . In every such case (unless a contradiction arises), G can be reducedto N = ∅ by Claim 2.2.Finally assume that N = ∅ and N = ∅ .Recall that by Corollary 4, there is at most one D -vertex in D ∩ N with contactto N . Now the e.d.s. problem can be solved in polynomial time as in the case N = ∅ and N = ∅ .Finally, the e.d.s. problem can be done in polynomial time in every cases ofSection 3 (in general, N k = ∅ , k ≥ Acknowledgment.
The second author would like to witness that he just tries topray a lot and is not able to do anything without that - ad laudem Domini.11 eferences [1] D.W. Bange, A.E. Barkauskas, and P.J. Slater, Efficient dominating sets in graphs, in: R.D.Ringeisen and F.S. Roberts, eds., Applications of Discrete Math. (SIAM, Philadelphia, 1988)189-199.[2] D.W. Bange, A.E. Barkauskas, L.H. Host, and P.J. Slater, Generalized domination and effi-cient domination in graphs,
Discrete Math.
159 (1996) 1-11.[3] A. Brandst¨adt, M. Milaniˇc, and R. Nevries, New polynomial cases of the weighted efficientdomination problem, extended abstract in: Conference Proceedings of MFCS 2013, LNCS8087, 195-206. Full version: CoRR arXiv:1304.6255, 2013.[4] A. Brandst¨adt and R. Mosca, Weighted efficient domination for P -free and P -free graphs,extended abstract in: Proceedings of WG 2016, P. Heggernes, ed., LNCS 9941, pp. 38-49,2016. Full version: SIAM J. Discrete Math.
Vol. 30, No. 4 (2016) 2288-2303.[5] A. Brandst¨adt and R. Mosca, On efficient domination for some classes of H -free bipartitegraphs, Discrete Applied Math.
270 (2019) 58-67.[6] B. Courcelle, J.A. Makowsky, and U. Rotics, Linear time solvable optimization problems ongraphs of bounded clique-width,
Theory of Computing Systems (2000) 125-150.[7] K. Dabrowski and D. Paulusma, Classifying the clique-width of H -free bipartite graphs, Dis-crete Applied Math.
200 (2016) 43-51.[8] C.L. Lu and C.Y. Tang, Solving the weighted efficient edge domination problem on bipartitepermutation graphs,
Discrete Applied Math.
87 (1998) 203-211.[9] C.L. Lu and C.Y. Tang, Weighted efficient domination problem on some perfect graphs,
Discrete Applied Math.
117 (2002) 163-182.[10] R. Nevries, Efficient Domination and Polarity, Ph.D. Thesis, University of Rostock, 2014.117 (2002) 163-182.[10] R. Nevries, Efficient Domination and Polarity, Ph.D. Thesis, University of Rostock, 2014.