Finding Efficient Domination for P_8-Free Bipartite Graphs in Polynomial Time
aa r X i v : . [ c s . D M ] J a n Finding Efficient Domination for P -Free BipartiteGraphs 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]
January 7, 2021
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 known to be NP -complete for P -free graphs,and even for very restricted H -free bipartite graph classes such as for K , -freebipartite graphs as well as for C -free bipartite graphs while it is solvable inpolynomial time for P -free bipartite graphs as well as for S , , -free bipartitegraphs. Here we show that ED can be solved in polynomial time for P -freebipartite graphs. Keywords : Efficient domination; P -free bipartite graphs; polynomial-time algorithm. 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 . Thenotion of efficient domination was introduced by Biggs [3] under the name perfectcode . 1he Exact Cover Problem asks for a subset F ′ of a set family F over a groundset, say V , containing every vertex in V exactly once, i.e., F ′ forms a partition of V .As shown by Karp [12], this problem is NP -complete even for set families containingonly 3-element subsets of V (see problem X3C [SP2] in [11]).Clearly, ED is the Exact Cover problem for the closed neighborhood hypergraphof G , i.e., if D = { d , . . . , d k } is an e.d.s. of G then N [ d ] ∪ . . . ∪ N [ d k ] forms apartition of V (we call it the e.d.s. property ). In particular, the distance betweenany pair of distinct D -vertices d i , d j is at least 3. In [1, 2], it was shown that theED problem is NP -complete.For a set F of graphs, a graph G is called F -free if G contains no inducedsubgraph isomorphic to a member of F ; in particular, we say that G is H -free if G is { H } -free. Let H + H denote the disjoint union of graphs H and H , and for k ≥
2, let kH denote the disjoint union of k copies of H . For i ≥
1, let P i denote thechordless path with i vertices, and let K i denote the complete graph with i vertices(clearly, P = K ). For i ≥
4, let C i denote the chordless cycle with i vertices.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 center 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 − .In [10, 16, 17], it was shown that ED is NP -complete for 2 P -free chordal unipolargraphs and thus, in general, for P -free graphs. In [8], ED is solvable in polynomialtime for P -free graphs (which leads to a dichotomy).A bipartite graph G is chordal bipartite if G is C k -free for every k ≥
3. Luand Tang [14] 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 -freebipartite graphs. Moreover, ED is NP -complete for planar bipartite graphs [14] andeven for planar bipartite graphs of maximum degree 3 [7] and girth at least g forevery fixed g [15]. Thus, ED is NP -complete for K , -free bipartite graphs and for C -free bipartite graphs.In [4], it is shown that one can extend polynomial time algorithms for EfficientDomination to such algorithms for weighted Efficient Domination. Thus, from nowon, 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 bi-graphs, and convex bipartite graphs are a subclass of them (and of chordal bipartitegraphs). Moreover, Lu and Tang [14] showed that ED is solvable in linear time forbipartite permutation graphs (which is a subclass of convex bipartite graphs).It is well known (see e.g. [6, 13]) that G is a bipartite permutation graph ifand only if G is AT-free bipartite if and only if G is ( H , H , H ,hole)-free bipartite2igure 1: Forbidden induced subgraphs H = S , , , H , H for bipartite permutationgraphs.(see Figure 1). Thus, while ED is NP -complete for ( H , H )-free bipartite graphs(since H and H contain C and H contains K , ), in [9] we have shown thatED is solvable in polynomial time for S , , -free (and more generally, for S , , -free)bipartite graphs.In [9] we have shown that ED is solvable in polynomial time for P -free bipartitegraphs as well as for S , , -free bipartite graphs. Now in this manuscript, we show: Theorem 1.
For P -free bipartite graphs, the ED problem is solvable in polynomialtime. Recall that G = ( X, Y, E ) is a P -free bipartite graph, i.e., every vertex in X isblack, and every vertex in Y is white.A vertex u contacts v if uv ∈ E . A vertex u / ∈ W contacts W if u has a neighborin W . For a subset U ⊆ X ∪ Y , a vertex v / ∈ U has a join to U , say v ○ U , if v contacts every vertex in U . Moreover, v has a co-join to U , say v ○ U , if v does notcontact any vertex in U .A vertex u ∈ V 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 is theneighbor 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 for a vertex v ∈ V = X ∪ Y then u ∈ V 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 with3 ∈ D , we can clearly reduce G by forced vertices as well as by v -forced verticeswhen we assume that v ∈ D : Claim 2.2.
If we assume that v ∈ D and u is v -forced then G has an e.d.s. D with v ∈ D if and only if the reduced graph G ′ = G \ N [ u ] has an e.d.s. D ′ = D \ { u } with v ∈ D ′ such that all vertices in N ( u ) are v -excluded in G ′ . Similarly, for k ≥ u ∈ V 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 is ( v , . . . , v k ) -excluded if u / ∈ D forsuch 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.Let dist G ( u, v ) denote the distance between u and v in G . By the e.d.s. property,the distance between two D -vertices is at least 3. Moreover, we have: Claim 2.3. If G = ( X, Y, E ) is connected then: ( i ) For every x ∈ D ∩ X and y ∈ D ∩ Y , dist G ( x, y ) = 3 or dist G ( x, y ) = 5 . ( ii ) For every v, v ′ ∈ D ∩ X or v, v ′ ∈ D ∩ Y , dist G ( v, v ′ ) = 4 or dist G ( v, v ′ ) = 6 . Proof. ( i ): By the e.d.s. property and since G is bipartite, for x ∈ D ∩ X and y ∈ D ∩ Y , dist G ( x, y ) ≥
3. In particular, dist G ( x, y ) ≥ k + 1, k ≥ dist G ( x, y ) = 7 then ( x, y , x , y , x , y , x , y ) induce a P in G , which is acontradiction. Thus, dist G ( x, y ) = 3 or dist G ( x, y ) = 5.( ii ): Without loss of generality, assume that v, v ′ ∈ D ∩ X . Then by the e.d.s.property and since G is bipartite, dist G ( v, v ′ ) ≥
4. In particular, dist G ( x, y ) ≥ k , k ≥
2. If If dist G ( v, v ′ ) = 8 then there is a P in G , which is a contradiction. Thus, dist G ( v, v ′ ) = 4 or dist G ( v, v ′ ) = 6.Without loss of generality, assume that d , . . . , d k ∈ D , k ≥
1, with dist G ( d i , d j ) ≥ i = j .First, let D basis := { d , . . . , d k } (for example, k ∈ { , , } ), i.e., N := D basis ,and let N i , i ≥
1, be the distance levels of D basis . Since G is P -free bipartite, wehave N = ∅ . By the e.d.s. property, we have: D ∩ ( N ∪ N ) = ∅ (1)Every vertex u ∈ N must have a D -neighbor in N (2)(else there is no such e.d.s. D in G ).Every vertex u ∈ N has at least two neighbors in N (3)4else the only neighbor v ∈ N of u ∈ N is D basis -forced and G can be reduced, i.e., D basis := D basis ∪ { v } ).Every vertex v ∈ N has a neighbor in N ∪ N (4)(else v is D basis -forced and G can be reduced, i.e., D basis := D basis ∪ { v } ). Claim 2.4.
Every u ∈ N is no endpoint of a P k P = ( w , . . . , w k − , u ) , k ≥ , with w , . . . , w k − ∈ N ∪ N and w k − ∈ N . Proof.
Let u ∈ N . Since G is P -free bipartite, u is no endpoint of any P P = ( w , . . . , w , u ) with w , . . . , w ∈ N ∪ N and w ∈ N .If u is an endpoint of a P P = ( w , . . . , w , u ) with w , . . . , w ∈ N ∪ N and w ∈ N then, since by (1), D ∩ ( N ∪ N ) = ∅ , u has no D -neighbor (else thereis a P P = ( w , . . . , w , u, v ) with D -neighbor v ∈ D ∩ N of u in G ) and there isno such e.d.s. Thus, u is an endpoint of a P k , k ≤ P = ( w , . . . , w k − , u ) with w , . . . , w k − ∈ N ∪ N and w k − ∈ N .If u ∈ N is an endpoint of a P P = ( w , . . . , w , u ) with w , . . . , w ∈ N ∪ N and w ∈ N then u must have a D -neighbor v ∈ D ∩ N , and since G is P -free, by(4), v has no further neighbor w ∈ N ∪ N (else there is a P ( w , . . . , w , u, v, w )in G ). Thus, v is D basis -forced and G can be reduced, i.e., D basis := D basis ∪ { v } .Moreover, if u has two such neighbors v, v ′ ∈ N then assume that v ′ / ∈ D ∩ N ,and v ′ must have a D -neighbor w ′ ∈ D ∩ ( N ∪ N ). But then ( w , . . . , w , u, v ′ , w ′ )induce a P in G , which is a contradiction. Thus, there is no such e.d.s.Thus, u is no endpoint of such a P k P = ( w , . . . , w k − , u ), k ≥
6, and Claim 2.4is shown.In the next section, we assume: N is a P -endpoint whoseremaining vertices are in N ∪ N Lemma 1.
If every vertex in N is an endpoint of a P whose remaining verticesare in N ∪ N then we have: ( i ) Every vertex in N is no endpoint of any P whose remaining vertices are in N i , i ≥ . ( ii ) N = ∅ , N is independent, and every edge in N does not contact N . roof. ( i ): Suppose to the contrary that u ∈ N is the endpoint of a P ( u, u , u , u )with u ∈ N , u ∈ N ∪ N , u ∈ N ∪ N ∪ N . Recall that u is the endpoint ofa P P whose remaining vertices are in N ∪ N . But then u , u , u and the P P with endpoint u induce a P in G , which is a contradiction. Thus, u is no endpointof any P whose remaining vertices are in N i , i ≥ ii ): By ( i ), every u ∈ N is no endpoint of any P whose remaining vertices are in N i , i ≥ N = ∅ , say u i ∈ N i , 2 ≤ i ≤
5, with u u ∈ E , u u ∈ E and u u ∈ E , then u is the endpoint of such a P ( u , u , u , u ), which is impossible by ( i ). Thus, N = ∅ .Analogously, if N is not independent, say u v ∈ E with u , v ∈ N , then let u ∈ N be a neighbor of u . Clearly, u v / ∈ E since G is bipartite. Recall that thereis a neighbor u ∈ N of u . But then u is the endpoint of such a P ( u , u , u , v ),which is impossible by ( i ). Thus, N is independent.Finally, if there is an edge u v ∈ E with u , v ∈ N which contacts N , saywithout loss of generality, v v ∈ E with v ∈ N , then let u u ∈ E for u ∈ N .But then u is the endpoint of such a P ( u , u , v , v ), which is impossible by ( i ).Thus, every edge in N does not contact N .Now, Lemma 1 is shown.By Lemma 1 ( i ), we have: Corollary 1.
For every u ∈ N and w ∈ N ∩ N ( u ) , we have N ( w ) ∩ N ⊆ N ( u ) ∩ N . Since by Lemma 1 ( ii ), N is independent and every edge in N does not contact N , we have: Corollary 2.
For every v ∈ N and w ∈ N , N ( v ) ∩ N ( w ) = ∅ . Claim 3.1. If v , v ∈ N ∪ N and N ( v ) ∩ ( N ∪ N ) ⊆ N ( v ) ∩ ( N ∪ N ) then v is D basis -excluded. Proof.
By Corollary 2, we have v , v ∈ N or v , v ∈ N . Clearly, v and v havethe same color (either black or white) since N ( v ) ∩ ( N ∪ N ) ⊆ N ( v ) ∩ ( N ∪ N ).First assume that v , v ∈ N and N ( v ) ∩ N ⊆ N ( v ) ∩ N . Suppose to thecontrary that v ∈ D . Then by the e.d.s. property, N ( v ) ∩ D = ∅ and v / ∈ D , butnow, v has no D -neighbor in G (recall that D ∩ N = ∅ ), i.e., G has no such e.d.s.Thus, v is D basis -excluded.Next assume that v , v ∈ N and N ( v ) ∩ N ⊆ N ( v ) ∩ N . Suppose to thecontrary that v ∈ D . Then by the e.d.s. property, N ( v ) ∩ D = ∅ and v / ∈ D , butnow, v has no D -neighbor in G (recall that by Lemma 1 ( ii ), N = ∅ and N isindependent), i.e., G has no such e.d.s. Then again, v is D basis -excluded.6 orollary 3. The following statements hold: ( i ) If v , v ∈ N and N ( v ) ∩ N = N ( v ) ∩ N = { w } then w is D basis -forced,and G can be reduced, i.e., D basis := D basis ∪ { w } . ( ii ) If v , v ∈ N and N ( v ) ∩ N = N ( v ) ∩ N = { w , . . . , w ℓ } , ℓ ≥ , and thereare two such N ( w i ) ∩ N = N ( w j ) ∩ N = { v , v } , ≤ i, j ≤ ℓ , i = j , thenthere is no such e.d.s. Proof. ( i ): Recall that by Claim 3.1, v , v are D basis -excluded and must have a D -neighbor. Clearly, v , v do not have any D -neighbor in N since N ( v ) ∩ N = N ( v ) ∩ N = { w } and by Lemma 1 ( ii ), every edge in N does not contact N .Thus, w ∈ D is the only possible D -neighbor of v , v , i.e., w is D basis -forced, and G can be reduced, i.e., D basis := D basis ∪ { w } .( ii ): Assume v , v ∈ N and N ( v ) ∩ N = N ( v ) ∩ N = { w , . . . , w ℓ } , ℓ ≥
2. Recallagain that by Claim 3.1, v , v are D basis -excluded and must have a D -neighbor.Without loss of generality, assume that N ( w ) ∩ N = N ( w ) ∩ N = { v , v } .Then either w or w is D basis -excluded and has no D -neighbor. Thus, there is nosuch e.d.s. Corollary 4.
The following statements hold: ( i ) If v , v ∈ N with N ( v ) ∩ N = { w } and N ( v ) ∩ N = { w , w } such that N ( w ) ∩ N = { v , v } , N ( w ) ∩ N = { v } , then v and w are D basis -forced,and G can be reduced, i.e., D basis := D basis ∪ { v , w } . ( ii ) If v , v , v ∈ N and N ( v ) ∩ N = { w } , N ( v ) ∩ N = { w , w } , N ( v ) ∩ N = { w , w , w } , with N ( w ) ∩ N = { v , v , v } , N ( w ) ∩ N = { v , v } , N ( w ) ∩ N = { v } , then there is no such e.d.s. Proof. ( i ): Recall that by Claim 3.1, v , w are D basis -excluded and must have a D -neighbor. Then v and w are D basis -forced, i.e., D basis := D basis ∪ { v , w } .( ii ): Recall that by Claim 3.1, v , v , w , w are D basis -excluded and must have a D -neighbor. Then the only possible D -neighbors are v and w but then v and w do not have any D -neighbor, and there is no such e.d.s.In particular, if t , t ∈ N are leaves with a common neighbor in N , say v ∈ N with vt i ∈ E , i ∈ { , } , then v is D basis -forced, and G can be reduced. It can alsolead to a contradiction, i.e., if for a vertex u ∈ N , t , t , t , t ∈ N ∩ N ( u ) areleaves and v , v ∈ N ∩ N ( u ) with v t i ∈ E , i ∈ { , } , and v t j ∈ E , j ∈ { , } ,then v and v are D basis -forced, which is a contradiction since uv ∈ E and uv ∈ E .Recall that, as in (1), D ∩ ( N ∪ N ) = ∅ , as well as in (2) and (3), for every u ∈ N , | N ( u ) ∩ N | ≥ u has no D -neighbor if N ( u ) ∩ N = ∅ or the onlyneighbor of u in N is D basis -forced). It can also lead to a contradiction, i.e., if u hastwo neighbors v , v ∈ N which are leaves then v and v are D basis -forced, which7s a contradiction since uv ∈ E and uv ∈ E . Moreover, if u has two neighbors v , v ∈ N such that v is a leaf then v is D basis -forced, and G can be reduced, i.e., D basis := D basis ∪ { v } .Recall Corollary 4. Assume that there is an edge xy ∈ E with x, y ∈ N and u x ∈ E , u y ∈ E , u , u ∈ N . If u v ∈ E , u v ∈ E , v , v ∈ N and N ( v ) ∩ N = { w } , N ( v ) ∩ N = { w , w } , then v ∈ D . Analogously, if u v ∈ E , u v ∈ E , v , v ∈ N and N ( v ) ∩ N = { w } , N ( v ) ∩ N = { w , w } , then v ∈ D . But then x, y do not have any D -neighbor, i.e., there is no such e.d.s.Thus, assume that at most one of u , u has such neighbors in N .Recall that for u ∈ N , | N ( u ) ∩ N | ≥
2, say N ( u ) ∩ N = { v , . . . , v r } , r ≥ u must have a D -neighbor in N ( u ) ∩ N , say without loss of generality, v ∈ D .Then for every i ≥ v i / ∈ D , and v i contacts ( N ∪ N ) \ N ( v ) (else there is noe.d.s. D with D basis ∪ { v } ⊂ D –recall Claim 3.1). Then we have: Claim 3.2.
If for u ∈ N , N ( u ) ∩ N = { v , . . . , v r } , r ≥ , and v ∈ D then forevery i ≥ , | N ( v i ) ∩ (( N ∪ N ) \ N ( v )) | = 1 , say N ( v i ) ∩ (( N ∪ N ) \ N ( v )) = { w i } ,and vertex w i is D basis ∪ { v } -forced. Proof.
Without loss of generality, assume that i = 2. Since v ∈ D , we have v / ∈ D ,and v must have a D -neighbor in ( N ∪ N ) \ N ( v ). Suppose to the contrary that | N ( v ) ∩ ( N ∪ N ) \ N ( v )) | ≥
2, say w , w ∈ N ( v ) ∩ ( N ∪ N ) \ N ( v ), and withoutloss of generality let w ∈ D . Then w / ∈ D and w must have a D -neighbor. Recallthat by Lemma 1 ( ii ), N = ∅ , N is independent, and every edge in N does notcontact N . Then either w , w ∈ N or w , w ∈ N .First assume that w , w ∈ N . Then w must have a D -neighbor in N , say w v ∈ E with v ∈ D ∩ N . But then by the e.d.s. property, uv / ∈ E , i.e.,( u, v , w , v ) induce a P , which is impossible by Lemma 1 ( i ).Next assume that w , w ∈ N . Then again w must have a D -neighbor in N ,say w v ∈ E with v ∈ D ∩ N . But then again uv / ∈ E , i.e., ( u, v , w , v ) inducea P , which is impossible by Lemma 1 ( i ).Thus | N ( v ) ∩ ( N ∪ N ) \ N ( v )) | = 1, say N ( v ) ∩ ( N ∪ N ) \ N ( v )) = { w } ,and w is D basis ∪ { v } -forced.Thus G can be reduced by such D basis ∪ { v } -forced vertices; let D basis := D basis ∪{ v } . This can also lead to the case that G has no such e.d.s. in G :If v contacts only ( N ∪ N ) ∩ N ( v ) and v ∈ D then v has no D -neighbor,and there is no such e.d.s.If N ∩ N ( u ) = { w } with w ○ N ∩ N ( u ) then there is no such e.d.s. in G .Clearly, if two vertices in ( N ( u ) \ N ( v )) ∩ N have a common neighbor in N then by Claim 3.2, there is no such e.d.s. in G .Let Q be a component in G [ N ∪ N ], say Q = ( X Q , Y Q , E Q ). By Lemma 1 ( ii ),either V ( Q ) ∩ N is independent or V ( Q ) ∩ N = ∅ .8 laim 3.3. Let u ∈ N which contacts Q . If u ∈ N ∩ X then u ○ Y Q ∩ N , and if u ∈ N ∩ Y then u ○ X Q ∩ N . Proof.
Without loss of generality, let u ∈ N ∩ X , i.e., u is black. Let v ∈ Y Q ∩ N with uv ∈ E . Let ( v , w , v ) induce a P with w ∈ X Q ∩ ( N ∪ N ) and v ∈ Y Q ∩ N .By Lemma 1 ( i ), ( u, v , w , v ) do not induce a P , i.e., uv ∈ E , and in general, itleads to u ○ Y Q ∩ N . Analogously, if u ∈ N ∩ Y then u ○ X Q ∩ N . Claim 3.4.
There is no P ( v , u , v , u , v , u , v ) in G with u i ∈ N , ≤ i ≤ ,and v i ∈ N , ≤ i ≤ . Proof.
Suppose to the contrary that P = ( v , u , v , u , v , u , v ) with u i ∈ N ,1 ≤ i ≤
3, and v i ∈ N , 1 ≤ i ≤
4, induce a P in G . Recall that by (4), v has a neighbor in N ∪ N (else v is D basis -forced and G can be reduced, i.e., D basis := D basis ∪ { v } ); let w ∈ N ∪ N with v w ∈ E .Since by Lemma 1 ( i ), u is no endpoint of a P ( u , v , w, v ), we have v w / ∈ E ,and analogously, v w / ∈ E (else ( u , v , w, v ) induce a P ), and v w / ∈ E (else( u , v , w, v ) induce a P ). But then ( w, v , u , v , u , v , u , v ) induce a P in G ,which is a contradiction. Q in G [ N ∪ N ∪ N ] with independent V ( Q ) ∩ N Let Q be a component in G [ N ∪ N ∪ N ], and assume that V ( Q ) ∩ N is independent.Recall that by (1), D ∩ N = ∅ and by (4), for v ∈ N ∩ V ( Q ) with N ( v ) ∩ N = ∅ , v is D basis -forced (which also leads to a contradiction when two D basis -forced vertices v , v have a common neighbor in N ). Clearly, G can be reduced by such D basis -forced vertices in N which are in the next D basis . Now assume that every vertex in V ( Q ) ∩ N has a neighbor in N . Moreover, recall that by Lemma 1 ( ii ), N = ∅ and N is independent.For every component Q in G [ N ∪ N ∪ N ], the e.d.s. problem for G can be doneindependently. Lemma 2.
If for component Q in G [ N ∪ N ∪ N ] , V ( Q ) ∩ N is independent thenthe e.d.s. problem for Q can be done in polynomial time. Proof.
Let N X := N ∩ X and N Y := N ∩ Y . Case 1. V ( Q ) ∩ N X = ∅ or V ( Q ) ∩ N Y = ∅ :Without loss of generality, assume that V ( Q ) ∩ N Y = ∅ and V ( Q ) ∩ N X = ∅ . Case 1.1.
There is a u ∈ N X with u ○ V ( Q ) ∩ N :Then N ( u ) ∩ N = V ( Q ) ∩ N , and u has a D -neighbor in V ( Q ) ∩ N . ByClaim 3.2, the e.d.s. problem for Q can be done in polynomial time.9 ase 1.2. There are exactly two u , u ∈ N X with ( N ( u ) ∪ N ( u )) ∩ N = V ( Q ) ∩ N :Then there are at most two D -neighbors for u , u , and by Claim 3.2, the e.d.s.problem for Q can be done in polynomial time. Case 1.3.
There are at least three neighbors u , . . . , u i ∈ N X , i ≥
3, of Q with( N ( u ) ∪ . . . ∪ N ( u i )) ∩ N = V ( Q ) ∩ N :Let u , u in N X with special neighbors v , v ∈ V ( Q ) ∩ N induce a 2 P , i.e., u v ∈ E , u v ∈ E , and u v / ∈ E , u v / ∈ E . Recall that v and v have neighbors in N (else v or v is D basis -forced, and G can be reduced). Thus assume that v w ∈ E and v w ∈ E for w , w ∈ N . Recall that by Lemma 1 ( i ), u is no endpoint ofany P whose remaining vertices are in N ∪ N , i.e., v , v do not have any commonneighbor w ∈ N (else ( u , v , w, v ) induce a P , which is a contradiction), i.e., w = w , v w / ∈ E and v w / ∈ E .We first claim that there is no path in N ∪ N between v and v :Since G is P -free, the distance between v and v in Q is at most 6. Recall that v , v do not have any common neighbor w ∈ N .If there is a path ( v , w , v, w , v ) with w , w ∈ N and v ∈ N then u v / ∈ E (else ( u , v, w , v ) induce a P , which is a contradiction), and analogously, u v / ∈ E .But now, ( u , v , w , v ) induce a P , which is a contradiction.Moreover, if there is a path ( v , w , v, w, v ′ , w , v ) with w , w, w ∈ N and v, v ′ ∈ N then u v ′ / ∈ E (else ( u , v ′ , w , v ) induce a P , which is a contradiction), andthus, u v / ∈ E (else ( u , v, w, v ′ ) induce a P , which is a contradiction), but now,( u , v , w , v ) induce a P , which is a contradiction.Thus, there is no path in N ∪ N between v and v .Since v , v ∈ V ( Q ), there is a path in N X ∪ N between v and v . Recall thatby Claim 3.4, there is no P in N ∪ N with endpoints v , v . Thus, there is a P ( v , u , v, u , v ) in ( N X ∪ N ) ∩ V ( Q ), i.e., u and u have a common neighbor v ∈ N ∩ V ( Q ).Let u , u , u ∈ V ( Q ) ∩ N X with special neighbors v i ∈ V ( Q ) ∩ N for u i ,1 ≤ i ≤
3. If there is no common neighbor of u , u , u in V ( Q ) ∩ N , say u v ∈ E and u v ∈ E but u v / ∈ E , as well as u v ′ ∈ E and u v ′ ∈ E but u v ′ / ∈ E , v, v ′ ∈ V ( Q ) ∩ N , then ( v , u , v, u , v ′ , u , v ) induce a P , which is a contradictionby Claim 3.4. Thus, in general, all u i ∈ V ( Q ) ∩ N X have a common neighbor v ∈ V ( Q ) ∩ N . Then either v ∈ D , and by Claim 3.2, the e.d.s. problem for Q canbe done in polynomial time, or v / ∈ D , vw ∈ E for w ∈ V ( Q ) ∩ N , and every u i hasa special D -neighbor which can also be done in polynomial time.Analogously, the e.d.s. problem for Q can be done in polynomial time when V ( Q ) ∩ N Y = ∅ and V ( Q ) ∩ N X = ∅ . Case 2. V ( Q ) ∩ N X = ∅ and V ( Q ) ∩ N Y = ∅ :10ince N ∩ V ( Q ) is independent, and N is independent, the only contact between V ( Q ) ∩ N X and V ( Q ) ∩ N Y is in N , and the e.d.s. problem can be done independentlyfor V ( Q ) ∩ N X and for V ( Q ) ∩ N Y .Then the e.d.s. problem can be done in polynomial time for Q as in Case 1, andLemma 2 is shown. Q in G [ N ∪ N ∪ N ] with N -edges A component K in G [ N ] is nontrivial if K has an edge (otherwise, | V ( K ) ∩ N | = 1and K is trivial in G [ N ]). Claim 3.5.
For every nontrivial component K = ( X K , Y K , E K ) in N , we have: ( i ) N ( K ) ∩ N = ∅ . ( ii ) For every edge xy ∈ E in K and its neighbors u x , u y in N , we have u x ○ X K and u y ○ Y K . ( iii ) If there is a P in K then | D ∩ X K | = 1 , say D ∩ X K = { x } , as well as | D ∩ Y K | = 1 , say D ∩ Y K = { y } , and x ○ Y K as well as y ○ X K . In particular,if K is P -free then | D ∩ V ( K ) | = 1 , say D ∩ V ( K ) = { v } and v ○ V ( K ) \ { v } . Proof. ( i ): Recall that by Lemma 1 ( ii ), every edge in N does not contact N .Thus, N ( K ) ∩ N = ∅ .( ii ): If K is an edge, i.e., P , in N , say V ( K ) = { x, y } , i.e., X K = { x } and Y K = { y } then for the only edge xy ∈ E in K and its neighbors u x , u y in N , u x x ∈ E , u y y ∈ E ,we have u x ○ X K and u y ○ Y K .Now assume that every P in K is part of a P , say ( x, y, x ′ ) in K . Recall u x x ∈ E and u y y ∈ E for u x , u y ∈ N . Since by Lemma 1 ( i ), u x is no endpoint ofa P ( u x , x, y, x ′ ), we have u x x ′ ∈ E , and in general, u x ○ X K . Analogously, for a P ( y, x, y ′ ), we have u y y ′ ∈ E , and in general, u y ○ Y K .( iii ): Let P = ( x, y, x ′ , y ′ ) with x, x ′ ∈ X K , y, y ′ ∈ Y K be a P in K . Recall that D ∩ N = ∅ and by Claim 3.5 ( i ), every edge in K has no neighbor in N . Thus, ifthere is no D -vertex in K then there is no such e.d.s. Moreover, if D ∩ X K = ∅ or D ∩ Y K = ∅ then there is no such e.d.s. In particular, if D ∩ X K = ∅ then y or y ′ does not have any D -neighbor (if y ∈ D then y ′ / ∈ D and vice versa). Analogously,if D ∩ Y K = ∅ then x or x ′ does not have any D -neighbor (if x ∈ D then x ′ / ∈ D andvice versa).Now assume that | D ∩ X K | ≥ | D ∩ Y K | ≥ D ∩ V ( K ) ≥
3, i.e., either | D ∩ X K | ≥ | D ∩ Y K | ≥
2. Without loss of generality, say | D ∩ X K | ≥
2, i.e., x , x ∈ X K ∩ D ; let u x x ∈ E for u x ∈ N . Recall that by Claim 3.5 ( ii ), u x ○ X K . But then it leadsto a contradiction by the e.d.s. property. Thus, if there is a P P = ( x, y, x ′ , y ′ ) in K | D ∩ X K | = 1 as well as | D ∩ Y K | = 1, say D ∩ X K = { x } and D ∩ Y K = { y ′ } .Since D ∩ N = ∅ and N ( K ) ∩ N = ∅ , we have x ○ Y K as well as y ′ ○ X K (else thereis no e.d.s. in K ).In particular, if K is P -free then we have that | D ∩ V ( K ) | = 1, say D ∩ V ( K ) = { v } and v ○ V ( K ) \ { v } . Corollary 5.
If there is a P P = ( x , y , x , y , x , y , x ) in K then there are two D -vertices x, y ∈ D ∩ N with x ○ { y , y , y } and y ○ { x , x , x , x } . Analogously,if there is a P P = ( y , x , y , x , y , x , y ) in K then there are two D -vertices x, y ∈ D ∩ N with x ○ { y , y , y , y } and y ○ { x , x , x } . Proof.
For the P P = ( x , y , x , y , x , y , x ) in N , recall that D ∩ N = ∅ and N ( P ) ∩ N = ∅ . Let K = ( X K , Y K , E K ) be the component in G [ N ] containing P . By Claim 3.5 ( iii ), we have | D ∩ X K | = 1, say D ∩ X K = { x } , as well as | D ∩ Y K | = 1, say D ∩ Y K = { y } , and x ○ Y K as well as y ○ X K . Thus, x i / ∈ D ,1 ≤ i ≤
4, and y j / ∈ D , 1 ≤ j ≤ x ○ { y , y , y } as well as y ○ { x , x , x , x } . Analogously, the same holds for a P P = ( y , x , y , x , y , x , y ) in G [ N ].By Claim 3.5 ( iii ), it can also lead to a contradiction. For example, if thenontrivial component K in N is a C , i.e., K is P -free, then there is no such e.d.s.:Let K = ( x , y , x , y ) be a C in N . If x ∈ D then x / ∈ D and x has no D -neighbor, which is a contradiction, and similarly for x ∈ D , y ∈ D or y ∈ D .Moreover, if the nontrivial component K in N is exactly a P then there is nosuch e.d.s.: Let K = ( x , y , x , y , x , y , x ) be such a P . Then by Claim 3.5 ( iii ), | D ∩ X K | = 1, as well as | D ∩ Y K | = 1 but then there is no such e.d.s.Then K can be reduced: • If there is a P in K and by Claim 3.5 ( iii ), D ∩ X K = { x } and D ∩ Y K = { y } then reduce K to a P ( x, y ′ , x ′ , y ). • If there is no P in K and by Claim 3.5 ( iii ), D ∩ V ( K ) = { v } then reduce K to a P ( v, w ). Corollary 6.
The following statements hold: ( i ) For u, u ′ ∈ N , there is no P ( x , y ) , ( x , y ) , ( x , y ) and no P + P ( x , y , x , y ) , ( x , y ) , in ( N ( u ) ∪ N ( u ′ )) ∩ N . ( ii ) For u , u , u ∈ N , there is no P ( x , y ) , ( x , y ) , ( x , y ) , ( x , y ) , andthere is no P + 2 P ( x , y , x , y ) , ( x , y ) , ( x , y ) , with y , y ∈ N ( u ) ∩ N , x , x , x , x ∈ N ( u ) ∩ N , y , y ∈ N ( u ) ∩ N . Proof. ( i ): Let u ∈ N with uy i ∈ E and u ′ ∈ N with u ′ x i ∈ E , 1 ≤ i ≤
3. First suppose to the contrary that there is a 3 P ( x , y ), ( x , y ), ( x , y ) in( N ( u ) ∪ N ( u ′ )) ∩ N . Recall that by Lemma 1 ( ii ), the P ’s in N do not contact12 . Assume without loss of generality that y ∈ D and x ∈ D . But then x , y donot have any D -neighbor, which is a contradiction by the e.d.s. property.Next suppose to the contrary that there is a P + P ( x , y , x , y ), ( x , y ) in( N ( u ) ∪ N ( u ′ )) ∩ N . Assume without loss of generality that x ∈ D and y ∈ D .But then x , y do not have any D -neighbor, which is a contradiction by the e.d.s.property.( ii ): First suppose to the contrary that there is a 4 P ( x , y ), ( x , y ), ( x , y ),( x , y ), with y , y ∈ N ( u ) ∩ N , x , x , x , x ∈ N ( u ) ∩ N , y , y ∈ N ( u ) ∩ N .Without loss of generality, assume that x ∈ D and y ∈ D . But then x , x / ∈ D and y , y ∈ D , which is a contradiction by the e.d.s. property.Next suppose to the contrary that there is a P + 2 P ( x , y , x , y ), ( x , y ),( x , y ), with y , y ∈ N ( u ) ∩ N , x , x , x , x ∈ N ( u ) ∩ N , y , y ∈ N ( u ) ∩ N .Recall that in every P ( x , y , x , y ) in N , there are two D -vertices, say x ∈ D with x u ∈ E , and y ∈ D with y u ∈ E . But then x , x / ∈ D and y , y ∈ D with y u ∈ E , y u ∈ E , which is a contradiction by the e.d.s. property. Claim 3.6.
Let K i , ≤ i ≤ ℓ , be nontrivial components in G [ N ] . The followingstatements hold: ( i ) For K i , K j , i = j , with P in K i and K j , there is no common N -neighbor of K i , K j , and there is no P P in N with distance between P and K i as wellas distance between P and K j . ( ii ) For K i , K j , i = j , with P in K i and with common N -neighbor of K i , K j , the D -vertices in K j are forced by the D -vertices in K i or there is a contradiction. Proof. ( i ): Suppose to the contrary that there is a common neighbor u ∈ N of K , K with P ’s in K and in K . Then by Claim 3.5 ( iii ), there is a P ( x , y , x , y )in K with x , y ∈ D and a P ( x , y , x , y ) in K with x , y ∈ D . Without lossof generality, let u ∈ X ∩ N with uy ∈ E . Then by Claim 3.5 ( ii ), u ○ Y K , andthus uy ∈ E , i.e., u contacts two D -vertices, which is a contradiction by the e.d.s.property. Analogously, there is no common N -neighbor u ∈ Y ∩ N of K , K .If there is a P P = ( x, y ) in N with distance 2 between P and K as well asdistance 2 between P and K then x / ∈ D and y / ∈ D but then there is no such e.d.s.( ii ): Without loss of generality, let i = 1, i.e., K is a nontrivial component in G [ N ]with P in K . By Claim 3.5 ( iii ), there are two D -vertices in V ( K ), say withoutloss of generality, y ∈ D ∩ Y K . Let u ∈ X ∩ N be an N -neighbor of the D -vertex y in K , and assume that u contacts K , say uy ′ ∈ E for y ′ ∈ Y K . Then by the e.d.s.property, y ′ / ∈ D , and by Claim 3.5 ( ii ), u ○ Y K and u ○ Y K . Thus, D ∩ Y K = ∅ ,and K is P -free, i.e., by Claim 3.5 ( iii ), K has only one D -vertex in X K which isthe only neighbor of y ′ in K (else there is a contradiction if there is a P ( x, y ′ , x ′ )in K ). Analogously, we have the same case if u ∈ Y ∩ N is an N -neighbor of a D -vertex x ∈ D ∩ X K . 13learly, the e.d.s. problem can be done independently for the components in G [ N ∪ N ∪ N ]. Now let Q be a component in G [ N ∪ N ∪ N ] with nontrivialcomponents K , . . . , K ℓ , ℓ ≥
1, in G [ N ]. Clearly, in any nontrivial component K i in G [ N ], there is a D -vertex (recall Claim 3.5 ( iii )). Then every vertex in N haseither a D -neighbor in a P in N or in no P in N .If ℓ = 1, there is only one nontrivial component K in Q , and the remainingparts of Q are partial components Q ′ with independent V ( Q ′ ) ∩ N .If K has a P then | D ∩ X K | = 1, say x ∈ D ∩ X K , as well as | D ∩ Y K | = 1, say y ∈ D ∩ Y K , and the D -vertices in Q ′ are D basis ∪ { x } -forced or D basis ∪ { y } -forced(recall Claim 3.2).Now assume that there is no P in K but a P in K , say xy with x ∈ D , andthere is a u ∈ N with ux ∈ E . Then for the partial components Q ′ with independent V ( Q ′ ) ∩ N and with contact to u , the D -vertices in Q ′ are D basis ∪ { x } -forced (recallClaim 3.2). For the partial components Q ′ with independent V ( Q ′ ) ∩ N and withoutcontact to u , it can be done independently.Now assume that ℓ ≥ Q in G [ N ∪ N ] with nontrivial compo-nents K , . . . , K ℓ . Clearly, V ( Q ) ∩ N has enough possible edges in N since G is P -free.Recall that by Claim 3.4, there is no P ( v , u , v , u , v , u , v ) with u , u , u ∈ N and v , v , v , v ∈ N . Thus, for the distance between K i and K j in component Q , there are at most two such u i , u j ∈ N with the same color.If there is a common neighbor u ∈ N of K i and K j and uv ∈ E for v ∈ D ∩ K i as well as uv ∈ E for v ∈ K j then v is D basis ∪ { v } -excluded, and v must havea D -neighbor in K j . Recall that by Lemma 1 ( i ), u is no endpoint of a P whoseremaining vertices are in N . If v has at least two neighbors in K j , say v w ∈ E with w ∈ D and v w ′ ∈ E then w ′ must have a D -neighbor in K j , say w ′ v ′ ∈ E for v ′ ∈ D , but now, u is the endpoint of a P ( u, v , w ′ , v ′ ), which is a contradiction.Thus, v has only one neighbor in K j , and its neighbor in K j is D basis ∪ { v } -forced(recall Claim 3.2).If there is no common N -neighbor between K i and K j , say x , y ∈ K with x y ∈ E and x , y ∈ K with x y ∈ E then assume that there are u , u ∈ N with u x ∈ E and u does not contact K as well as u x ∈ E and u does not contact K .Since there is no such P with three N -vertices of the same color (recall Claim 3.4), u and u have a common neighbor x ∈ N . Now P = ( y , x , u , x, u , x , y ) inducea P in G .First assume that for the P -midpoint x ∈ N of P in Q , x ∈ D : Then x / ∈ D and x / ∈ D which implies that y ∈ D and y ∈ D (which are D basis ∪ { x } -forced).Moreover, every other D -neighbor in this component is D basis ∪ { x } -forced (recallClaim 3.2) and it leads to an e.d.s. or a contradiction. For example, if there is a P in K or in K then x / ∈ D . 14f P = ( y , x , u , x, u , x , y ) induce a P in G [ N ∪ N ] with x ∈ D and thereis a next P P ′ = ( x , y , u ′ , y, u , y , x ) in G [ N ∪ N ] then y / ∈ D since y ∈ D ,and a D -neighbor w ∈ D ∩ N of y is D basis ∪ { x } -forced. Clearly, there are contactsbetween u , u and u ′ , u since G is P -free. Moreover, the next subcomponent with u , y , x etc. can be done independently.If x / ∈ D then u , u must have D -neighbors in N , say u i w i ∈ E with w i ∈ D ∩ N , i ∈ { , } , (possibly w = x or w = x ), and there is a D -neighbor w ∈ D ∩ N of x , which is D basis ∪ { w } -forced. In this case, the e.d.s. problem can be doneindependently for subcomponents Q i with u i , x i , y i , i ∈ { , } , etc.Next assume that no such P -midpoint in Q is in D ∩ N . Then for K and K ,and in general, for every K i in Q , the e.d.s. problem can be solved independently,i.e., we have: Corollary 7.
For every component in G [ N ∪ N ∪ N ] in this section, there is apolynomial time solution for finding an e.d.s. or a contradiction. D -vertices x, y ∈ D Recall that the e.d.s. problem can be done independently for every connected com-ponent in G = ( X, Y, E ); assume that G = ( X, Y, E ) is a connected component.For an e.d.s. D in the P -free bipartite graph G = ( X, Y, E ), let | D | ≥ D ∩ X = ∅ or D ∩ Y = ∅ then G is not connected. Thus, | D ∩ X | ≥ | D ∩ Y | ≥ | D ∩ X | = 1, say D ∩ X = { x } . Then every y ∈ Y \ N ( x )has no D -neighbor and D = { x } ∪ ( Y \ N ( x )) or there is no e.d.s. in G , which is atrivial e.d.s. solution. Analogously, if | D ∩ Y | = 1 then it is a trivial e.d.s. solution.From now on, assume that | D ∩ X | ≥ | D ∩ Y | ≥ x ∈ D ∩ X and y ∈ D ∩ Y , dist G ( x, y ) = 3or dist G ( x, y ) = 5, and for every v, v ′ ∈ D ∩ X or v, v ′ ∈ D ∩ Y , dist G ( v, v ′ ) = 4 or dist G ( v, v ′ ) = 6.Assume that x ∈ D ∩ X with xy ∈ E and y x ∈ E . Then x must have a D -neighbor y ∈ D ∩ Y , and dist G ( x, y ) = 3.If | N ( x ) | = 1 and | N ( y ) | = 1, say ( x, y , x , y ) induce a P with leaves x, y , then x , y are excluded, x, y are forced, and G can be reduced. Thus, from now on,assume that either | N ( x ) | ≥ | N ( y ) | ≥ G withdistance 3); let N ( x ) = { y , . . . , y i } and N ( y ) = { x , . . . , x j } , i ≥ j ≥
2, andlet x y ∈ E (possibly there could be more edges between N ( x ) and N ( y )).First assume that ( x, y , x , y ) induce a P , x , x ∈ N ( y ), y , y ∈ N ( x ) and dist G ( x, x ) = 4 as well as dist G ( y, y ) = 4. Then x and y are midpoints of a P ,i.e., ( y , x, y , x ) induce a P , ( y , x , y, x ) induce a P .15et N = D basis , i.e., x, y ∈ D basis and every ( x, y )-forced vertex is in D basis .Moreover, N i , i ≥
1, are the distance levels of D basis . Recall that by (1), D ∩ ( N ∪ N ) = ∅ . Claim 4.1.
The following statements hold: ( i ) For every u ∈ N which contacts N ( x ) or N ( y ) , u is an endpoint of a P whoseremaining vertices are in N ∪ N . ( ii ) For every u ∈ N with u ○ ( N ( x ) ∪ N ( y )) , u is an endpoint of a P whoseremaining vertices are in N ∪ N . ( iii ) N = ∅ , N is independent, and every edge in N does not contact N . Proof. ( i ): Without loss of generality, assume that u ∈ N ∩ X . Since u isblack, u ○ N ( y ) but u contacts N ( x ). If uy ∈ E then ( u, y , x , y ) induce a P with endpoint u . If uy / ∈ E and uy ′ ∈ E for y ′ ∈ N ( x ) and y ′ x / ∈ E then( u, y ′ , x, y , x , y ) induce a P with endpoint u , which is impossible by Claim 2.4.Thus, y ′ x ∈ E and then ( u, y ′ , x, y ) induce a P with endpoint u .Analogously, it can be done if u ∈ N ∩ Y .( ii ): Without loss of generality, assume that u ∈ N ∩ X with u ○ ( N ( x ) ∪ N ( y )).Let uv ∈ E for some v ∈ N ( w ), w ∈ D basis , w = x .First assume that v contacts N ( y ). If vx ∈ E then ( u, v, x , y , x ) induce a P with endpoint u . If vx / ∈ E but vx ′ ∈ E for some x ′ ∈ N ( y ) then ( u, v, x ′ , y, x )induce a P with endpoint u .Now assume that v ○ N ( y ) for every neighbor v ∈ N of u . Recall that G isconnected and by Claim 2.3, dist G ( w, y ) = 3 or dist G ( w, y ) = 5.First assume that dist G ( w, y ) = 3; since v ○ N ( y ), there is a P ( y, x ′ , v ′ , w )(possibly x ′ = x ) with v ′ = v . Since v ′ contacts N ( y ), we have uv ′ / ∈ E , and then( u, v, w, v ′ , x ′ , y ) induce a P with endpoint u , which is impossible by Claim 2.4.Now assume that dist G ( w, y ) = 3 and dist G ( w, y ) = 5, say ( y, x ′ , v ′ , x ′′ , v ′′ , w ) issuch a P . Since v ′ contacts N ( y ), we have uv ′ / ∈ E . If uv ′′ ∈ E then ( u, v ′′ , x ′′ , v ′ , x ′ , y )induce a P with endpoint u , which is impossible by Claim 2.4. Now assume that uv ′ / ∈ E and uv ′′ / ∈ E . Recall that u contacts N ( w ), say uv ∈ E for v ∈ N ( w ).If vx ′′ / ∈ E then ( u, v, w, v ′′ , x ′′ , v ′ ) induce a P with endpoint u , which is impos-sible by Claim 2.4. Thus, vx ′′ ∈ E . Recall v ○ N ( y ), i.e., vx ′ / ∈ E and again,( u, v, x ′′ , v ′ , x ′ , y ) induce a P with endpoint u , which is impossible by Claim 2.4.Analogously, it can be done if u ∈ N ∩ Y .( iii ): By ( i ) and since G is P -free, u is not the endpoint of a P ( u, v , v , v , v ) with uv ∈ E , v v ∈ E , v v ∈ E , v v ∈ E , and v ∈ N , v ∈ N ∪ N , v ∈ N ∪ N ∪ N , v ∈ N ∪ N ∪ N ∪ N . In particular, u is not the endpoint of a P ( u, v , v , v , v )with v i ∈ N i , 3 ≤ i ≤
6. Thus, N = ∅ . 16nalogously, u is not the endpoint of a P ( u, v , v , v , v ) with v ∈ N , v ∈ N ,and v , v ∈ N . Thus, N is independent.Finally, u is not the endpoint of a P ( u, v , v , v , v ) with v , v ∈ N and v ∈ N . Thus, every edge in N does not contact N . D -vertices in N Recall that x, y ∈ D basis , N = D basis , N i , i ≥
1, are the distance levels of D basis ,and D ∩ ( N ∪ N ) = ∅ . If N contains black and white vertices, say u ∈ N ∩ X , u ∈ N ∩ Y , then there are at least two white and black D -neighbors of u , u ∈ N in N , say v ∈ D ∩ N ∩ Y with u v ∈ E and v ∈ D ∩ N ∩ X with u v ∈ E (elsethere is no such e.d.s. with x, y ∈ D ).If such a v ∈ N has only neighbors in N , i.e., v ○ N ∪ N , then by (4), v ∈ D is D basis -forced, and G can be reduced, i.e., v ∈ D basis . Thus assume that every D -vertex in N has a neighbor in N ∪ N .Assume that for v i ∈ D ∩ N , i ∈ { , } , ( u i , v i , w i ) induce a P with u i ∈ N and w i ∈ N ∪ N . Clearly, w i ○ N , i ∈ { , } , and u ○ N ( D basis ∩ Y ) as well as u ○ N ( D basis ∩ X ).Clearly, v , v are not yet D basis -forced, but assume that D basis := D basis ∪{ v , v } , N = D basis , and N ′ i , i ≥
1, are the new distance levels of the new D basis . Clearly,again D ∩ ( N ′ ∪ N ′ ) = ∅ . Lemma 3.
For every u ∈ N ′ , u is an endpoint of a P whose remaining verticesare in N ′ ∪ N ′ . Proof.
Recall that v , v ∈ D basis , v is white and v is black. Moreover, recall that( u , v , w ) is a P and w ○ N ( D basis \ { v , v } ) as well as ( u , v , w ) is a P and w ○ N ( D basis \ { v , v } ).Assume without loss of generality that u ∈ N ′ ∩ X , i.e., u is black. Then u doesnot contact N ( v ) for every v ∈ D basis ∩ Y ; in particular, u ○ ( N ( y ) ∪ N ( v )). Clearly, u contacts N ( r ) for some r ∈ ( D basis \ { v , v } ) ∩ X (possibly r = x ).If u ○ N ( x ) then recall that by Claim 4.1 ( ii ), u is an endpoint of a P whoseremaining vertices are in N ′ ∪ N ′ . Thus assume that u contacts N ( x ), i.e., u ∈ N for the previous distance level N . Case . u and u have a common neighbor z ∈ N ∩ Y :If z ∈ N then z must have a D -neighbor r ∈ N (possibly r = x ), and then z ∈ N ′ . Since w z / ∈ E (recall w ∈ N ∪ N ), then v ∈ N ′ , u , w , z ∈ N ′ , and( u, z, u , v , w ) induce a P with endpoint u . Case . u and u do not have any common neighbor in N ∩ Y :Let uz ∈ E , uz ′ / ∈ E and u z ′ ∈ E , u z / ∈ E for z, z ′ ∈ N , z = z ′ . Since u ∈ N ′ and v ∈ N ′ , uv / ∈ E . If z and z ′ have a common neighbor r ∈ N , say z, z ′ ∈ N ( r )17or r ∈ N (possibly r = x ) then ( u, z, r, z ′ , u , v , w ) induce a P with endpoint u ,which is impossible by Claim 2.4. Analogously, if z and z ′ have a common neighbor x ′ ∈ N then ( u, z, x ′ , z ′ , u , v , w ) induce a P with endpoint u , which is impossibleby Claim 2.4.Now assume that z and z ′ do not have any common neighbor in N ∪ N , say z ∈ N ( r ) (possibly r = x ), z ′ ∈ N ( r ′ ) for r, r ′ ∈ N , r = r ′ . Recall that G isconnected and by Claim 2.3, dist G ( r, r ′ ) = 4 or dist G ( r, r ′ ) = 6.First assume that dist G ( r, r ′ ) = 4, say ( r, y ′ , x ′ , y ′′ , r ′ ) induce a P . Recall that uy ′′ / ∈ E since N -neighbors of u, u do not have any common D -neighbor in N . If uy ′ / ∈ E and uz ∈ E for z ∈ N ( r ) then, since uy ′′ / ∈ E , ( u, z, r, y ′ , x ′ , y ′′ , r ′ ) inducea P with endpoint u if x ′ z / ∈ E , which is impossible by Claim 2.4. Thus, x ′ z ∈ E and then ( u, z, x ′ , y ′′ , r ′ ) induce a P with endpoint u . If uy ′ ∈ E and x ′ v ∈ E then( u, y ′ , x ′ , v , w ) induce a P with endpoint u .Next assume that dist G ( r, r ′ ) = 6, say ( r, y ′ , x ′ , y ′′ , x ′′ , y ′′′ , r ′ ) induce a P ; recallthat uy ′′′ / ∈ E . If uy ′′ ∈ E then ( u, y ′′ , x ′′ , y ′′′ , r ′ ) induce a P with endpoint u . Nowassume that uy ′′ / ∈ E . If uy ′ ∈ E then ( u, y ′ , x ′ , y ′′ , x ′′ , y ′′′ , r ′ ) induce a P with end-point u , which is impossible by Claim 2.4. Thus, uy ′ / ∈ E and uz ∈ E with z ∈ N ( r ).Since dist G ( r, r ′ ) = 6, we have zx ′′ / ∈ E . Since by Claim 2.4, ( u, z, r, y ′ , x ′ , y ′′ , x ′′ )does not induce a P with endpoint u , zx ′ ∈ E , and then ( u, z, x ′ , y ′′ , x ′′ , y ′′′ , r ′ )induce a P with endpoint u , which is impossible by Claim 2.4.Thus, in general, u is an endpoint of a P whose remaining vertices are in N ′ ∪ N ′ .Analogously, the same can be shown when u ∈ N ′ ∩ Y , i.e., u is white, andLemma 3 is shown.Similarly, Lemma 3 can also be shown if | N ( x ) | ≥ | N ( y ) | = 1 or | N ( x ) | = 1and | N ( y ) | ≥ D -vertices in N . D -vertices in N have the same color Recall that by Claim 4.1 ( ii ), N = ∅ . Without loss of generality, assume that all D -vertices in N are black, i.e., D ∩ N ⊂ X and D ∩ N ∩ Y = ∅ . Then N ⊂ Y (else for a black vertex in N , there is no white D -neighbor in N , and there is nosuch e.d.s. D with x, y ∈ D basis ), N ⊂ X , N ⊂ Y , and N ⊂ X , i.e., every N i ,2 ≤ i ≤
5, is independent. Clearly, N ( x ) is white and thus, N ( x ) ○ N . N = ∅ If for every r ∈ N , r ∈ D , i.e., N ⊂ D , then G can be reduced, i.e., D basis := D basis ∪ N , and N ′ = ∅ for the new distance level N ′ .18ow assume that there is a r ∈ N with r / ∈ D . Then r must have a D -neighbor r ∈ D ∩ N . Let ( r , r , r , r , r , r ) be a P with r i ∈ N i , 0 ≤ i ≤ r = y ). Then r must have a D -neighbor s ∈ D ∩ N .If s ○ N then s is D basis -forced or it leads to a contradiction if r has two suchneighbors s , s ′ with s ○ N , s ′ ○ N . Now assume that s s ∈ E with s ∈ N .Then either ( r , r , r , s , s , r , r , r ) induce an S , , with midpoint u if s r / ∈ E and s r / ∈ E , or s r ∈ E or s r ∈ E ; we call it an S ∗ , , .Now assume that D basis := D basis ∪ { s , r } . Again, N ′ = D basis and N ′ i , i ≥ D basis . Lemma 4.
Each vertex in N ′ is the endpoint of a P whose remaining vertices arein N ′ ∪ N ′ . Proof.
Let u ∈ N ′ . Clearly, u has a neighbor v ∈ N ′ . Recall that for r , s , r ∈ N ′ , r is white, s is black and r is white. Case . u ∈ N ′ ∩ X , i.e., u is black. Then u ○ ( N ( r ) ∪ N ( r )). Case . . u contacts N ( s ):If ur ∈ E then ( u, r , r , r , r ) induce a P with endpoint u . If ur / ∈ E and us ∈ E then ( u, s , s , r , r , r , r ) induce a P with endpoint u , which isimpossible by Claim 2.4. Now assume that ur / ∈ E , us / ∈ E , and us ∈ E for s ∈ N ( s ), s = r , s . If sr / ∈ E then ( u, s, s , r , r , r ) induce a P with endpoint u , which is impossible by Claim 2.4. Now assume that sr ∈ E , i.e., s ∈ N andrecall that r ∈ N for the previous distance levels N , N . Then sr / ∈ E , andsince ( u, s, r , r , r , r , r ) does not induce a P with endpoint u , we have sr ∈ E .Finally, ( u, s, r , r , r ) induce a P with endpoint u . Case . . u ○ N ( s ):Let uv ∈ E for v ∈ N ′ , v / ∈ N ( s ). Clearly, vw ∈ E for w ∈ N ′ . Since u is black, v is white and w is black. Recall that G is connected and by Claim 2.3, dist G ( w, s ) = 4 or dist G ( w, s ) = 6.First assume that dist G ( w, s ) = 4, i.e., ( s , y ′ , x ′ , y ′′ , w ) induce a P . Clearly, uy ′ / ∈ E , and if uy ′′ ∈ E then ( u, y ′′ , x ′ , y ′ , s ) induce a P with endpoint u . Nowassume that uy ′′ / ∈ E and recall uv ∈ E for v ∈ N ( w ). Since ( u, v, w, y ′′ , x ′ , y ′ , s )does not induce a P with endpoint u , we have vx ′ ∈ E , and then ( u, v, x ′ , y ′ , s )induce a P with endpoint u .Next assume that dist G ( w, s ) = 6, i.e., ( s , y ′ , x ′ , y ′′ , x ′′ , y ′′′ , w ) induce a P .Again, uy ′ / ∈ E , and if uy ′′ ∈ E then ( u, y ′′ , x ′ , y ′ , s ) induce a P with endpoint u . Now assume that uy ′′ / ∈ E . Since ( u, y ′′′ , x ′′ , y ′′ , x ′ , y ′ , s ) does not induce a P with endpoint u , we have uy ′′′ / ∈ E , and recall uv ∈ E for v ∈ N ( w ). If vx ′ ∈ E then ( u, v, x ′ , y ′ , s ) induce a P with endpoint u . Now assume that vx ′ / ∈ E . Since( u, v, x ′′ , y ′′ , x ′ , y ′ , s ) does not induce a P with endpoint u , we have vx ′′ / ∈ E . Butthen ( u, v, w, y ′′′ , x ′′ , y ′′ , x ′ , y ′ ) induce a P , which is a contradiction.19hus, every u ∈ N ′ ∩ X is the endpoint of a P whose remaining vertices are in N ′ ∪ N ′ . Case . u ∈ N ′ ∩ Y , i.e., u is white. Then u ○ N ( s ). Case . . u contacts N ( r ) ∪ N ( r ):First assume that u contacts N ( r ). If ur ∈ E then ( u, r , r , s , s ) induce a P with endpoint u . Now assume that ur / ∈ E and ur ′ ∈ E for r ′ ∈ N ( r ). Clearly, r ′ s / ∈ E . Since ( u, r ′ , r , r , r , s , s ) does not induce a P with endpoint u , wehave r ′ r ∈ E and then again ( u, r ′ , r , s , s ) induce a P with endpoint u .Now assume that u ○ N ( r ) and u contacts N ( r ). If ur ∈ E then ( u, r , r , r , r )induce a P with endpoint u . If ur / ∈ E and ur ∈ E then ( u, r , r , r , r , r , r )induce a P with endpoint u , which is impossible. Now assume ur / ∈ E , ur / ∈ E and ur ∈ E , r ∈ N ( r ), r = r , r .Since ( u, r, r , r , r , r , r ) does not induce a P with endpoint u , we have rr ∈ E and then ( u, r, r , r , r ) induce a P with endpoint u . Case . . u ○ ( N ( r ) ∪ N ( r )):Let uv ∈ E for v ∈ N ′ , v / ∈ N ( r ) ∪ N ( r ). Clearly, vw ∈ E for w ∈ N ′ . Since u is white, v is black and w is white. Recall that G is connected and by Claim 2.3, dist G ( w, r ) = 4 or dist G ( w, r ) = 6.First assume that dist G ( w, r ) = 4, say ( w, x ′ , y ′ , x ′′ , r ) (possibly x ′′ = r ) inducea P . If ux ′ / ∈ E and ux ′′ / ∈ E then, since ( u, v, w, x ′ , y ′ , x ′′ , r ) does not induce a P with endpoint u , we have vy ′ ∈ E and then ( u, v, y ′ , x ′′ , r ) induce a P withendpoint u . If ux ′ / ∈ E and ux ′′ ∈ E then ( u, x ′′ , y ′ , x ′ , w ) induce a P with endpoint u . Analogously, if ux ′ ∈ E and ux ′′ / ∈ E then ( u, x ′ , y ′ , x ′′ , r ) induce a P withendpoint u .Now assume that ux ′ ∈ E and ux ′′ ∈ E . If ur ∈ E then ( u, r , r , s , s ) inducea P with endpoint u . If ur / ∈ E then, since ( u, x ′′ , r , r , r , s , s ) does not inducea P with endpoint u , we have x ′′ r ∈ E (clearly, x ′′ s / ∈ E since x ′′ ∈ N ( r )), andthen ( u, x ′′ , r , s , s ) induce a P with endpoint u .Next assume that dist G ( w, r ) = 6, say ( w, x ′ , y ′ , x ′′ , y ′′ , x ′′′ , r ) (possibly x ′′′ = r )induce a P . Again, if ur ∈ E then ( u, r , r , s , s ) induce a P with endpoint u .Thus assume that ur / ∈ E . If ux ′′′ ∈ E then, since ( u, x ′′′ , r , r , r , s , s ) does notinduce a P with endpoint u , we have x ′′′ r ∈ E (clearly, x ′′′ s / ∈ E since x ′′ ∈ N ( r )),and then ( u, x ′′′ , r , s , s ) induce a P with endpoint u .Now assume that ux ′′′ / ∈ E . If ux ′′ ∈ E then ( u, x ′′ , y ′′ , x ′′′ , r ) induce a P withendpoint u . Now assume that ux ′′ / ∈ E and ux ′′′ / ∈ E . Since ( u, x ′ , y ′ , x ′′ , y ′′ , x ′′′ , r )does not induce a P with endpoint u , we have ux ′ / ∈ E . Recall that uv ∈ E with v ∈ N ( w ). If vy ′′ ∈ E then ( u, v, y ′′ , x ′′′ , r ) induce a P with endpoint u . Thusassume that vy ′′ / ∈ E . Since ( u, v, w, x ′ , y ′ , x ′′ , y ′′ ) does not induce a P with endpoint u , we have vy ′ ∈ E . But then ( u, v, y ′ , x ′′ , y ′′ , x ′′′ , r ) induce a P with endpoint u ,which is impossible. 20hus, Lemma 4 is shown.Then the e.d.s problem can be solved in polynomial time as in Section 3. N = ∅ Case 1. | D ∩ N | = 1, say without loss of generality, D ∩ N ∩ X = { v } :Clearly, v ○ N (else there is no such e.d.s. D with x, y ∈ D ). Then every vertexin N \ { v } is the endpoint of a P whose remaining vertices are in N ∪ N ∪ N .Now we can assume that v ∈ D basis , i.e., D basis := D basis ∪ { v } , and N ′ i , i ≥ D basis . Then N ′ = N \ { v } (recall N = ∅ ), every u ∈ N ′ is the endpoint of a P whose remaining vertices are in N ′ ∪ N ′ , and thee.d.s. problem in Case 1 with D ∩ N = { v } can be solved in polynomial time as inSection 3.Thus, for every possible v ∈ D ∩ N with v ○ N , the e.d.s. problem can be donein polynomial time. Case 2. | D ∩ N | ≥ N = ∅ , N ∪ N is white and N is black.Let v , v ∈ D ∩ N with P ( u i , v i , w i ), and u i ∈ N , w i ∈ N , i ∈ { , } . Clearly,by the e.d.s. property, ( u i , v i , w i ) and ( u j , v j , w j ), i = j , induce a 2 P in G . Since G is P -free bipartite, u and u must have a common neighbor in N .If D ∩ N = ∅ then clearly, N ⊂ D , which leads to a polynomial time solutionfor an e.d.s. or a contradiction. Now assume that D ∩ N = ∅ . If there is only one D -vertex w ∈ D ∩ N then all vertices in N \ N ( w ) are D basis ∪ { w } -forced, whichleads to a polynomial time solution for an e.d.s. or a contradiction.Now assume that | D ∩ N | ≥
2, say w, w ′ ∈ D ∩ N , and vw ∈ E , v ′ w ′ ∈ E for v, v ′ ∈ N . Thus, for every w ∈ N , N ( w ) ⊆ N .Recall that by (3), for u i ∈ N , i ∈ { , } , | N ( u i ) ∩ N | ≥
2, say v i , v ′ i ∈ N ( u i ) ∩ N with v i ∈ D and v ′ i / ∈ D . Then v ′ i must have a D -neighbor w ′ i ∈ D ∩ N (possibly w ′ = w ′ ).Let D basis := D basis ∪ { v , w ′ , v } , let N ′ := D basis and let N ′ i , i ≥
1, be the newdistance levels of the new D basis . Lemma 5.
For every u ∈ N ′ , u is an endpoint of a P whose remaining verticesare in N ′ ∪ N ′ . Proof.
Let u ∈ N ′ ; clearly, u has a neighbor in N ′ . Recall that x, v are black and y, w ′ are white. Recall that G is connected and by Claim 2.4, u is no endpoint of a P or P whose remaining vertices are in N ′ ∪ N ′ . Case . u ∈ N ′ ∩ X , i.e., u is black. 21ecall that for the previous distance levels, N ∪ N is white and N is black.Then for black u ∈ N ′ ∩ X , u ∈ N and u ○ N ( x ).Let ur ∈ E for r ∈ N ′ and rs ∈ E for s ∈ D basis (possibly r ∈ N ( v ), i.e., s = v )but r / ∈ N ( x ) since u ○ N ( x ), i.e., s = x .First assume that r contacts N ( y ). If rx ∈ E then ( u, r, x , y , x ) induce a P with endpoint u . Analogously, if rx / ∈ E and rx ′ ∈ E for x ′ ∈ N ( y ) then( u, r, x ′ , y, x ) induce a P with endpoint u .Now assume that r ○ N ( y ). Recall that by Claim 2.3, either dist G ( s, x ) = 4 or dist G ( s, x ) = 6.First assume that dist G ( s, x ) = 4, say ( x, y ′ , x ′ , y ′′ , s ) induce a P in G . If uy ′′ ∈ E then, since u ○ N ( x ), i.e., uy ′ / ∈ E , ( u, y ′′ , x ′ , y ′ , x ) induce a P with endpoint u .Now assume that uy ′′ / ∈ E ; recall that ur ∈ E with rs ∈ E . Since by Claim 2.4,( u, r, s, y ′′ , x ′ , y ′ , x ) does not induce a P with endpoint u , we have rx ′ ∈ E , and then( u, r, x ′ , y ′ , x ) induce a P with endpoint u .Next assume that dist G ( s, x ) = 6, say ( x, y ′ , x ′ , y ′′ , x ′′ , y ′′′ , s ) induce a P in G .Recall that uy ′ / ∈ E . If uy ′′ ∈ E then ( u, y ′′ , x ′ , y ′ , x ) induce a P with endpoint u . If uy ′′ / ∈ E then by Claim 2.4, ( u, y ′′′ , x ′′ , y ′′ , x ′ , y ′ , x ) does not induce a P with endpoint u , and thus, uy ′′′ / ∈ E ; recall ur ∈ E with rs ∈ E . Since ( u, r, s, y ′′′ , x ′′ , y ′′ , x ′ , y ′ )does not induce a P , we have rx ′′ ∈ E or rx ′ ∈ E . Since ( u, r, x ′′ , y ′′ , x ′ , y ′ , x ) doesnot induce a P with endpoint u , we have rx ′ ∈ E , and then ( u, r, x ′ , y ′ , x ) induce a P with endpoint u . Case . u ∈ N ′ ∩ Y , i.e., u is white.Then u ○ ( N ( x ) ∪ N ( v )). Recall that for x, y ∈ N , N is black and N ∪ N iswhite, i.e., u / ∈ N and u ∈ N ∪ N . Case . . u ∈ N :Then u ○ N ( y ). Recall that G is connected and u has a black neighbor in N ′ ,say ur ∈ E for r ∈ N ′ , i.e., r / ∈ N ( y ). Let rs ∈ E for white s ∈ N ′ , s = y . Recallthat by Claim 2.3, dist G ( s, y ) = 4 or dist G ( s, y ) = 6.First assume that dist G ( s, y ) = 4, say ( s, x ′ , y ′ , x ′′ , y ) induce a P in G ; recallthat ux ′′ / ∈ E . If ux ′ ∈ E then ( u, x ′ , y ′ , x ′′ , y ) induce a P with endpoint u . Nowassume that ux ′ / ∈ E ; recall ur ∈ E for r ∈ N ( s ). Since ( u, r, s, x ′ , y ′ , x ′′ , y ) does notinduce a P with endpoint u , we have ry ′ ∈ E and then ( u, r, y ′ , x ′′ , y ) induce a P with endpoint u .Next assume that dist G ( s, y ) = 6, say ( s, x ′ , y ′ , x ′′ , y ′′ , x ′′′ , y ) induce a P in G ;recall that ux ′′′ / ∈ E . If ux ′′ ∈ E then ( u, x ′′ , y ′′ , x ′′′ , y ) induce a P with endpoint u .Now assume that ux ′′ / ∈ E . Since ( u, x ′ , y ′ , x ′′ , y ′′ , x ′′′ , y ) does not induce a P withendpoint u , we have ux ′ / ∈ E ; recall ur ∈ E for r ∈ N ( s ). Since ( u, r, s, x ′ , y ′ , x ′′ , y )does not induce a P with endpoint u , we have ry ′ ∈ E or ry ′′ ∈ E . If ry ′ ∈ E and22 y ′′ / ∈ E then ( u, r, y ′ , x ′′ , y ′′ , x ′′′ , y ) induce a P with endpoint u , which is impossibleby Claim 2.4. Thus, ry ′′ ∈ E , and then ( u, r, y ′′ , x ′′′ , y ) induce a P with endpoint u . Case . . u ∈ N :First assume that u / ∈ N ( v ), say u ∈ N ( v i ), i ≥
2. Since G is P -free, u and u have a common neighbor x ′ ∈ N (possibly x ′ ∈ N ( y )), x ′ w / ∈ E , and since uv / ∈ E , ( u, x ′ , u , v , w ) induce a P with endpoint u .Now assume that u ∈ N ( v ). Recall that v , v ∈ D ∩ N . Then u and u havea common neighbor x ′ ∈ N , and ( u, x ′ , u , v , w ) induce a P with endpoint u .Thus, Lemma 5 is shown.Then the e.d.s problem can be done in polynomial time as in Section 3.Finally, Theorem 1 is shown. Open problem : What is the complexity of ED for P k -free bipartite graphs, 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.
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