Finding Efficient Domination for S 1,1,5 -Free Bipartite Graphs in Polynomial Time
aa r X i v : . [ c s . D M ] O c t 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]
November 2, 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 .1n [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 (evenfor ( K , , K − e )-free perfect graphs [10]). 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 [11] 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[11] and even for planar bipartite graphs of maximum degree 3 [3] and girth at least g for every fixed g [12]. 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 [8]; for instance, the clique-width of claw-free bipartite,i.e., K , -free bipartite graphs is bounded. Dabrowski and Paulusma [9] 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 [6], we showed already that for S , , -free bipartite graphs, ED can be solvedin polynomial time. 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 ), V ( G ) = X ∪ Y , be an S , , -free bipartite graph with black X and white Y .We say that for a vertex set U ⊆ V ( G ), 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 contacts no vertex in U . 2orrespondingly, for vertex sets U, W ⊆ V ( G ) with U ∩ W = ∅ , U ○ W denotes u ○ W for all u ∈ U , and U ○ W denotes u ○ W for all u ∈ U .A vertex u / ∈ W contacts W if u has a neighbor in W . For vertex sets U, W with U ∩ W = ∅ , U contacts W if there is a vertex in U contacting W .Let dist G ( x, y ) denote the distance between x and y in G .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 with v ∈ 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 component of G , the e.d.s. problem is independently solvable.Thus, we can assume that G is connected. Since G is S , , -free bipartite, G is not P k -free for every k ≥ D -vertex is midpoint of a P in G then every midpoint of P is excluded,and every endpoint of P in G is forced or there is no e.d.s. in G if two endpoints of P have a common neighbor. In this case, the e.d.s. problem can be easily done inpolynomial time.Thus, from now on, assume that there exists a d ∈ D which is midpoint of a P . ( v , . . . , v ) is a P or C in G with v , v ∈ D In this section, we assume that for ( v , . . . , v ), v i v i +1 ∈ E , 1 ≤ i ≤
5, with v ∈ D and v ∈ D , either( i ) P = ( v , . . . , v ) is a P in G , or( ii ) C = ( v , . . . , v ) is a C in G with v v ∈ E .3e call P a 2 D − P and C a 2 D − C . Clearly, G contains polynomially manyinduced P and C . Then let us set D basis := { v , v } and N := D basis . Without lossof generality, assume that v , v , v ∈ X and v , v , v ∈ Y . Let N i := { v ∈ X ∪ Y :min { dist G ( v, v ) , dist G ( v, v ) } = i } , i ≥
1, be the distance levels with respect to { v , v } . By the e.d.s. property, we have: D ∩ ( N ∪ N ) = ∅ . (1)If there is a vertex u ∈ N with N ( u ) ∩ N = ∅ then G has no e.d.s. D with v , v ∈ D . Thus, we assume:Every vertex u ∈ N has a neighbor in N . (2)If there is a vertex u ∈ N with | N ( u ) ∩ N | = 1, say N ( u ) ∩ N = { w } then w is ( v , v )-forced (or it can also lead to an S , , if w has two neighbors in N ∪ N ).Then one can update D basis := D basis ∪ { w } and redefine the distance levels withrespect to D basis , i.e., N i := { v ∈ X ∪ Y : min { dist ( v, d ′ ) : d ′ ∈ D basis } = i } , i ≥ u ∈ N has at least two neighbors in N . (3)If there is a vertex w ∈ N with w ○ N ∪ N then w is ( v , v )-forced (or it canlead to a contradiction if there are two such w, w ′ with common neighbor in N and w ○ N ∪ N as well as w ′ ○ N ∪ N ). Then one can update D basis := D basis ∪ { w } and redefine the distance levels with respect to D basis as above. Thus, we assume:Every vertex w ∈ N has a neighbor in N ∪ N . (4)By an iterative application of the above operations, one has finally: Claim 3.1.
There is a set D basis , with { v , v } ⊆ D basis , such that: ( i ) every vertex in D basis \ { v , v } is ( v , v ) -forced; ( ii ) G [ D basis ∪ N ( D basis )] is connected; ( iii ) assumptions (1) − (2) − (3) − (4) hold for the distance levels N i with respect to D basis , namely, N i := { v ∈ X ∪ Y : min { dist ( v, d ′ ) : d ′ ∈ D basis } = i } , i ≥ ,and N = D basis .In particular, D basis can be computed in polynomial time. In what follows let D basis be just the set of Claim 3.1 [i.e. let us assume that sucha set has been computed] and let N i , i ≥
1, be the corresponding distance levels.
Claim 3.2. If u ∈ N ∩ X contacts y ∈ N ( x ) for a ( v , v ) -forced vertex x ∈ N ∩ X then u contacts N ( v ) . Analogously, if u ∈ N ∩ Y contacts x ∈ N ( y ) for a ( v , v ) -forced vertex y ∈ N ∩ Y then u contacts N ( v ) . roof. Without loss of generality, assume that u ∈ N ∩ X contacts y ∈ N ( x ) fora ( v , v )-forced vertex x ∈ N ∩ X . Since it is a component with N [ v ] and N [ v ]and y is white, we assume without loss of generality that y contacts N ( v ). Clearly, u is black and u does not contact N ( v ). Moreover, u must have a D -neighbor in D ∩ N , by (3), u has at least two neighbors r, r ′ ∈ N , say r ∈ D ∩ N , and by (4), r has a neighbor s ∈ N ∪ N .Suppose to the contrary that u does not contact N ( v ).First assume that yv / ∈ E and yv / ∈ E , say yv ∈ E with v ∈ N ( v ), v = v , v . Then ( v , v , v , v, y, u, r, s ) (with midpoint v ) induce an S , , , which is acontradiction. Thus, yv ∈ E or yv ∈ E . Case 1. P = ( v , . . . , v ) is a 2 D − P .Since ( u, r, r ′ , y, v , v , v , v ) (with midpoint u ) do not induce an S , , , we have yv / ∈ E . Thus, yv ∈ E . But then ( u, r, r ′ , y, v , v , v , v ) (with midpoint u ) inducean S , , , which is a contradiction. Thus, in this case, u contacts N ( v ). Case 2. C = ( v , . . . , v ) is a 2 D − C with v v ∈ E .Recall that yv ∈ E or yv ∈ E . Since ( u, r, r ′ , y, v , v , v , v ) (with mid-point u ) do not induce an S , , , we have yv / ∈ E and thus, yv ∈ E . But then( u, r, r ′ , y, v , v , v , v ) (with midpoint u ) induce an S , , , which is a contradiction.Thus, again, in this case, u contacts N ( v ).Analogously, if white u ∈ N contacts x ∈ N ( y ) for a ( v , v )-forced vertex y ∈ Y then u contacts N ( v ).Thus, Claim 3.2 is shown. Claim 3.3.
Let P = ( v , . . . , v ) be a D - P with v , v ∈ D . ( i ) If u ∈ N contacts N ( v ) and uv ∈ E then uv ∈ E . Moreover, if uv / ∈ E then for every v ∈ N ( v ) \ { v , v } with uv ∈ E , vv ∈ E . ( ii ) If u ∈ N contacts N ( v ) and uv ∈ E then uv ∈ E . Moreover, if uv / ∈ E then for every v ∈ N ( v ) \ { v , v } with uv ∈ E , vv ∈ E . Proof. ( i ): Recall that by (3), | N ( u ) ∩ N | ≥
2, say w, w ′ ∈ N ( u ) ∩ N . Assume that u ∈ N contacts N ( v ). First assume that uv ∈ E . Since ( u, w, w ′ , v , v , v , v , v )(with midpoint u ) do not induce an S , , , we have uv ∈ E .Now assume uv / ∈ E . Then uv / ∈ E , i.e., uv ∈ E for v ∈ N ( v ) \ { v , v } .Since ( u, w, w ′ , v, v , v , v , v ) (with midpoint u ) do not induce an S , , , we have vv ∈ E . Thus, ( i ) is shown, and analogously, ( ii ) can be shown in the same way.Thus, Claim 3.3 is shown. Claim 3.4.
Let C = ( v , . . . , v ) be a D - C with v , v ∈ D . ( i ) If u ∈ N contacts N ( v ) then either uv ∈ E and uv ∈ E or uv / ∈ E and uv / ∈ E , and for every v ∈ N ( v ) \ { v , v } with uv ∈ E , we have vv ∈ E and vv ∈ E . ii ) If u ∈ N contacts N ( v ) then either uv ∈ E and uv ∈ E or uv / ∈ E and uv / ∈ E and for every v ∈ N ( v ) \ { v , v } with uv ∈ E , we have vv ∈ E and vv ∈ E . Proof. ( i ): Recall that by (3), | N ( u ) ∩ N | ≥
2, say w, w ′ ∈ N ( u ) ∩ N . Assume that u ∈ N contacts N ( v ). First assume that uv ∈ E . Since ( u, w, w ′ , v , v , v , v , v )(with midpoint u ) do not induce an S , , , we have uv ∈ E . Analogously, if uv ∈ E and since ( u, w, w ′ , v , v , v , v , v ) (with midpoint u ) do not induce an S , , , we have uv ∈ E .Now assume that uv / ∈ E and uv / ∈ E . Then uv ∈ E for v ∈ N ( v ) \ { v , v } .Since ( u, w, w ′ , v, v , v , v , v ) (with midpoint u ) do not induce an S , , , we have vv ∈ E . Analogously, since ( u, w, w ′ , v, v , v , v , v ) (with midpoint u ) do notinduce an S , , , we have vv ∈ E . Thus, ( i ) is shown, and analogously, ( ii ) can beshown in the same way. Thus, Claim 3.4 is shown. Definition 1.
Let R = ( r , r , r , r ) , r i r i +1 ∈ E , ≤ i ≤ , be a P with endpoint r ∈ N and r ∈ N , r ∈ N ∪ N , r ∈ N ∪ N ∪ N . Claim 3.5.
Let R = ( r , r , r , r ) be such a P as in Definition . ( i ) If r contacts N ( v ) then r v ∈ E or r v ∈ E . ( ii ) If r contacts N ( v ) then r v ∈ E or r v ∈ E . Proof. ( i ): Suppose to the contrary that r v / ∈ E and r v / ∈ E , i.e., r v ∈ E , v = v , v . But then ( v , v , v , v, r , r , r , r ) (with midpoint v ) induce an S , , ,which is a contradiction. Thus, ( i ) is shown, and analogously, ( ii ) can be shown inthe same way. Thus, Claim 3.5 is shown. Corollary 1.
Let ( v , . . . , v ) with v , v ∈ D be a D − P . If R = ( r , r , r , r ) issuch a P as in Definition then r v ∈ E or r v ∈ E . Proof.
Without loss of generality, assume that r ∈ N is white and r contacts N ( v ) (recall Claim 3.2). By Claim 3.5 ( i ), r v ∈ E or r v ∈ E . Recall that( v , . . . , v ) with v , v ∈ D is a 2 D − P , and assume that r v ∈ E . Then byClaim 3.3 ( i ), r v ∈ E . Analogously, if r is black then r v ∈ E . Thus, Corollary 1is shown. Corollary 2.
Let ( v , . . . , v ) with v , v ∈ D be a D − C . If R = ( r , r , r , r ) issuch a P as in Definition then r v ∈ E and r v ∈ E or r v ∈ E and r v ∈ E . Proof.
Without loss of generality, assume that r ∈ N is white and r contacts N ( v ) (recall Claim 3.2). Recall that by Claim 3.4, either r v ∈ E and r v ∈ E or r v / ∈ E and r v / ∈ E . By Claim 3.5 ( i ), r v ∈ E or r v ∈ E . Thus, r v ∈ E and r v ∈ E . Analogously, if r is black then r v ∈ E and r v ∈ E . Thus, Corollary 2is shown. 6 laim 3.6. Let R = ( r , r , r , r ) be such a P as in Definition . Then the r -neighbor r ∈ N has exactly one neighbor in N ∪ N , namely r ∈ N ∪ N . Proof.
Without loss of generality, assume that r ∈ N ∩ Y , i.e., r is white. Recallthat by Claim 3.5, r v ∈ E or r v ∈ E . Then by Claims 3.3 and 3.4, r v ∈ E .Suppose to the contrary that r has at least two neighbors r , r ′ ∈ N ∪ N .Recall that ( v , . . . , v ) is a 2 D − P or a 2 D − C with v , v ∈ D . But then( r , r , r ′ , r , v , v , v , v ) (with midpoint r ) induce an S , , , which is a contradic-tion. Thus, r ∈ N ∩ X has exactly one neighbor in N ∪ N , and analogously, r ∈ N ∩ Y has exactly one neighbor in N ∪ N . Thus, Claim 3.6 is shown. Corollary 3.
For every such P R = ( r , r , r , r ) as in Definition , we have r / ∈ D and r is ( v , v ) -forced. Proof.
Assume that there is such a P R = ( r , r , r , r ). Without loss of generality,assume that r is white. Recall that by Corollaries 1 and 2, r v ∈ E .Suppose to the contrary that r ∈ D . Then by the e.d.s. property, r / ∈ D and r must have a D -neighbor s ∈ D ∩ ( N ∪ N ∪ N ∪ N ). But then( v , v , v , r , r , r , r , s ) (with midpoint v ) induce an S , , , which is a contradic-tion. Thus, r / ∈ D and must have a D -neighbor in D ∩ ( N ∪ N ). Recall that byClaim 3.6, the only neighbor of r in N ∪ N is r . Thus, r is ( v , v )-forced. Thus,Corollary 3 is shown.Thus, G can be reduced, i.e., D basis := D basis ∪ { r } , and then we have: Corollary 4.
There is no such P R = ( r , r , r , r ) as in Definition . Claim 3.7. N = ∅ , N is independent, and for every edge ( r , s ) with r , s ∈ N , s ○ N and r ○ N . Proof.
Recall that by Corollaries 3 and 4 (and after the reduction), there is no such P R = ( r , r , r , r ). Then N = ∅ and N is independent.Assume that there is an edge r s ∈ E with r , s ∈ N and r r ∈ E with r ∈ N . If s s ∈ E for s ∈ N then ( r , r , s , s ) induce a P , which is impossible.Thus, s ○ N and analogously, r ○ N . Thus, Claim 3.7 is shown. Claim 3.8.
Let ( r, s, t ) be a P in N ∪ N with midpoint s . ( i ) If s ∈ N ∩ X then for us ∈ E with u ∈ N , uv / ∈ E and uv / ∈ E . Moreover,for uv ∈ E with v ∈ N ( v ) , v ○ N ( v ) . ( ii ) If s ∈ N ∩ Y then for us ∈ E with u ∈ N , uv / ∈ E and uv / ∈ E . Moreover,for uv ∈ E with v ∈ N ( v ) , v ○ N ( v ) . Proof. ( i ): Assume that there is a P ( r, s, t ) in N ∪ N with midpoint s ∈ N ∩ X . Since ( s, r, t, u, v , v , v , v ) (with midpoint s ) do not induce an S , , , wehave uv / ∈ E , and analogously, uv / ∈ E . Let uv ∈ E with v ∈ N ( v ), v = v , v .7ince ( s, r, t, u, v, v , v , v ) (with midpoint s ) do not induce an S , , , we have vv ∈ E , and analogously, vv ∈ E . Moreover, if v ′ ∈ N ( v ), v ′ = v , v , then, since( s, r, t, u, v, v , v , v ′ ) (with midpoint s ) do not induce an S , , , we have vv ′ ∈ E ,and v ○ N ( v ).( ii ): Analogously, the same can be done for a P ( r, s, t ) in N ∪ N with midpoint s ∈ N ∩ Y . Thus, Claim 3.8 is shown. Claim 3.9.
Let P = ( r, s, t ) in N ∪ N with midpoint s ∈ N and P ′ = ( s ′ , t ′ ) with s ′ ∈ N , t ′ ∈ N ∪ N , such that s, s ′ ∈ X or s, s ′ ∈ Y have the same color, be a P + P . Then s and s ′ have a common N -neighbor. Proof.
Without loss of generality, assume that P = ( y , x , y ) in N ∪ N withmidpoint x ∈ N ∩ X and P ′ = ( x , y ) with x ∈ N ∩ X and y ∈ N ∪ N is a P + P .Suppose to the contrary that there is no common N -neighbor of s and s ′ , say u x ∈ E and u x ∈ E for u , u ∈ N , u = u . If u and u have a commonneighbor v ∈ N ( v ) (possibly v = v or v = v ) then ( x , y , y , u , v, u , x , y ) (withmidpoint x ) induce an S , , , which is a contradiction. Thus, u and u have nocommon neighbor v ∈ N ( v ) but then it leads to an S , , , which is a contradiction.Thus, x and x have a common N -neighbor.Analogously, if there is a P + P ( x , y , x ) in N ∪ N with midpoint y ∈ N ∩ Y and ( y , x ) with y ∈ N ∩ Y and x ∈ N ∪ N then y and y have a common N -neighbor. Thus, Claim 3.9 is shown. Claim 3.10.
If there is a P ( r, s, t ) in N ∪ N with midpoint s ∈ D ∩ N ∩ X then D ∩ N ∩ X = { s } . Analogously, if there is a P ( r, s, t ) in N ∪ N with midpoint s ∈ D ∩ N ∩ Y then D ∩ N ∩ Y = { s } . Proof.
Without loss of generality, assume that there is a P ( y , x , y ) in N ∪ N with midpoint x ∈ N ∩ X . Suppose to the contrary that there is a second D -neighbor x ∈ D ∩ N . Clearly, x has a neighbor y ∈ N ∪ N (else x is ( v , v )-forced). Let u x ∈ E with u ∈ N and u x ∈ E with u ∈ N . By the e.d.s.property, u = u . If u and u have a common neighbor v ∈ N ( v ) (possibly v = v or v = v ) then ( x , y , y , u , v, u , x , y ) (with midpoint x ) induce an S , , , whichis a contradiction. Thus, u and u have no common neighbor v ∈ N ( v ) but thenit leads to an S , , , which is a contradiction.Analogously, if there is a P ( r, s, t ) in N ∪ N with midpoint s ∈ D ∩ N ∩ Y then D ∩ N ∩ Y = { s } . Thus, Claim 3.10 is shown. Corollary 5.
There is at most one P ( r, s, t ) in N ∪ N with midpoint s ∈ D ∩ N ∩ X and at most one P ( r ′ , s ′ , t ′ ) in N ∪ N with midpoint s ′ ∈ D ∩ N ∩ Y . From now on, assume that there is no such P -midpoint s ∈ D ∩ N ∩ X or s ′ ∈ D ∩ N ∩ Y . Then for every such P in N ∪ N , it leads to a P ( x , y , x , y )in N ∪ N with x , y ∈ D . 8 laim 3.11. If there is a P ( x , y , x , y ) in N with x , y ∈ D then D ∩ N = { x , y } . Proof.
Assume that there is a P ( x , y , x , y ) in N with x , y ∈ D .Suppose to the contrary that there is another D -vertex in N . Assume withoutloss of generality that x ∈ D ∩ N , and by (4), let x y ∈ E for y ∈ N ∪ N . Let u x ∈ E for u ∈ N and u x ∈ E for u ∈ N . Clearly, by the e.d.s. property, u = u . Recall that u , u have a common neighbor v ∈ N ( v ). Since ( u , x , y , x )do not induce a P (which is impossible by Corollary 4), we have u x ∈ E .If u x ∈ E then ( u , x , y , x ) induce a P which is impossible by Corollary 4.Thus, u x / ∈ E . Moreover, since ( u , x , y , x ) do not induce a P , we have y x / ∈ E . But then ( x , y , y , u , v, u , x , y ) (with midpoint x ) induce an S , , , which isa contradiction. Thus, there is no other D -vertex x ∈ D ∩ X ∩ N .Analogously, there is no other D -vertex y ∈ D ∩ Y ∩ N , and Claim 3.11 isshown. Corollary 6.
If there is a P ( x , y , x , y ) in N ∪ N with x , x ∈ N and y , y ∈ N , and x , y ∈ D then D ∩ X ∩ N = { x } . Analogously, if there isa P ( y , x , y , x ) in N ∪ N with y , y ∈ N and x , x ∈ N , and y , x ∈ D then D ∩ Y ∩ N = { y } . By the way, if for the P ( x , y , x , y ) in N with x , y ∈ D and u x ∈ E , u x ∈ E , u y ∈ E , u y ∈ E , there is a P ( x , y ) with u x ∈ E , u y ∈ E thenthere is no such e.d.s.Now assume that there is no such P in N ∪ N and there is no such P in N ∪ N with midpoint in N . Then for every P ( x, y ) in N , either x ∈ D or y ∈ D (else there is no such e.d.s. in G ). Claim 3.12.
If there are two P ’s ( x , y ) , ( x , y ) in N with u x ∈ E and u x ∈ E for u ∈ N as well as u y ∈ E and u y ∈ E for u ∈ N then either x , y ∈ D or x , y ∈ D . Moreover, there are no three such P ’s with common N -neighbors. Proof.
Without loss of generality, assume that for P ( x , y ) in N , x ∈ D . Then x / ∈ D and y ∈ D . Analogously, if y ∈ D then y / ∈ D and x ∈ D .If there is a third P ( x , y ) with common N -neighbors u , u then there is noe.d.s. Thus, Claim 3.12 is shown.By the way, if there are two P ’s ( x , y ), ( x , y ) in N with u x ∈ E and u x ∈ E for u ∈ N as well as u y ∈ E and u y ∈ E for u ∈ N and there aretwo P ’s ( x , y ), ( x , y ) in N with u x ∈ E and u x ∈ E for u ∈ N as well as u y ∈ E and u y ∈ E for u ∈ N then there is no such e.d.s.Then the set N ∪ N ∪ N can be partitioned into two subsets, namely, H and H such that H := [( N ∪ N ) ∩ X ] ∪ [ N ∩ Y ] and H := [( N ∪ N ) ∩ Y ] ∪ [ N ∩ X ].9 .1 When H = ∅ or H = ∅ Assume without loss of generality that H = ∅ . Then N ∪ N ⊂ Y and N ⊂ X ,i.e., N i are independent, 2 ≤ i ≤
4. The e.d.s. problem can be solved independentlyfor every component Q in G [ N ∪ N ∪ N ].If for Q , | D ∩ N | ≤ | D ∩ N | = k , k ≥
3. For every x i ∈ D ∩ N , 1 ≤ i ≤ k ,( u , x , y ) , . . . , ( u k , x k , y k ) induce a kP in Q . If xy i ∈ E for x ∈ N then, sincethere is no such P , we have xu i ∈ E (recall Corollary 4).Moreover, if xu i ∈ E and xu j ∈ E with u i x i ∈ E , x i ∈ D , and u j x j ∈ E , x j ∈ D , i = j , then xy i / ∈ E and xy j / ∈ E since there is no such P in Q . Claim 3.13.
If in Q , there are at least three P ( u i , x i , y i ) with x i ∈ D , ≤ i ≤ ,and x ∈ N contacts u , u then x ○ N . Proof.
Suppose to the contrary that neither x ○ N nor x ′ ○ N , i.e., xu ∈ E , xu ∈ E and xu / ∈ E , say x ′ u ∈ E , x ′ u ∈ E and x ′ u / ∈ E . Let y ∈ D ∩ N be a D -neighbor of x . Since there is no P ( u , x, y, x ′ ), x ′ y / ∈ E . But then( x, y, u , u , x ′ , u , x , y ) (with midpoint x ) induce an S , , , which is a contradiction.Thus, x ○ N , and Claim 3.13 is shown.Then for every component Q , either x ∈ D and for all such P ( u i , x i , y i ), y i ∈ D ,or x / ∈ D , y ∈ D ∩ N with xy ∈ E , and x i ∈ D , 1 ≤ i ≤ k . Thus, for everycomponent in G [ N ∪ N ∪ N ], the e.d.s. problem can be done in polynomial time. H = ∅ and H = ∅ Recall that by Claim 3.7, N = ∅ , N is independent, and every edge in N doesnot contact N . If there is no contact between components in H ∩ ( N ∪ N ) and H ∩ ( N ∪ N ) then the e.d.s. problem can be done in polynomial time as in section3.1. Now assume that there are such contacts between components in H ∩ ( N ∪ N )and H ∩ ( N ∪ N ). Since G is S , , -free bipartite, it is possible that N ∩ X ○ N ∩ Y .However, if in Q , every x ∈ N ∩ X and y ∈ N ∩ Y have no common N -neighborthen Q is no connected component. Thus, assume that for every component Q in H ∪ H , there are such common N -neighbors.Recall Claims 3.8, 3.9, 3.10 and Corollary 5 as well as Claims 3.11, 3.12 andCorollary 6.Without loss of generality, assume that x ∈ D ∩ X ∩ V ( Q ). Let ux ∈ E with u ∈ N . If u contacts any other edge ( x ′ , y ′ ) in N then by the e.d.s. property, x ′ / ∈ D and y ′ ∈ D . Moreover, if u contacts ( x ′ , y ′ ) with x ′ ∈ N and y ′ ∈ N thenanalogously, x ′ / ∈ D and y ′ ∈ D . Analogously, assume that y ∈ D ∩ Y ∩ V ( Q ), uy ∈ E with u ∈ N and u contacts any other edge ( y ′ , x ′ ) in N or in N ∪ N .Finally, if there is no such case then the component Q can be reduced for smallercomponents. 10hus, for every component Q in N ∪ N ∪ N , the e.d.s. problem can be solvedin polynomial time. D − P and no D − C in G In this section, assume that there is no such 2 D − P or 2 D − C ( w , . . . , w ) with w , w ∈ D . Recall that G is connected.If | D | = 1 in G then the e.d.s. problem is solvable in polynomial time. Nowassume that | D | ≥ G . Claim 4.1. If P = ( v , . . . , v ) induce a P in G then v / ∈ D and v / ∈ D . Proof.
Without loss of generality, assume that P = ( x , y , x , y , x , y , x ) inducea P in G . Suppose to the contrary that x ∈ D or x ∈ D , say without loss ofgenerality, x ∈ D . Then by the e.d.s. property, x , y , y , x / ∈ D and x as well as x must have a D -neighbor.Let y ∈ D with x y ∈ E . Clearly, y = y since y / ∈ D . If yx ∈ E then( y, x , y , x , y , x ) induce a 2 D − C which is impossible in this section. Thus, yx / ∈ E . Then x must have a D -neighbor y ′ ∈ D . If y ′ = y then ( y , x , y , x , y , x )induce a 2 D − P which is impossible in this section. Thus, y / ∈ D and y ′ = y .But then ( x , y , y ′ , y , x , y , x , y ) (with midpoint x ) induce an S , , , which is acontradiction. Thus, x / ∈ D and analogously, x / ∈ D , i.e., the third and fifth vertex x , x of every P P = ( x , y , x , y , x , y , x ) is not in D .Analogously, if P = ( y , x , . . . , x , y ) induce a P then y / ∈ D and y / ∈ D , andClaim 4.1 is shown. Corollary 7. If P = ( v , v , v , v , v , v , v , v ) induce a P in G then v , v , v , v / ∈ D . Lemma 1. If P = ( v , v , . . . , v , v ) induce a P in G then v ∈ D and v ∈ D . Proof.
Assume that P = ( x , y , x , y , x , y , x , y ) induce a P in G . Recall thatby Corollary 7, x , y , x , y / ∈ D . Suppose to the contrary that y / ∈ D or x / ∈ D . Case 1. y / ∈ D and x / ∈ D :Then x must have a D -neighbor y ∈ D , and y must have a D -neighbor x ∈ D .Since ( x , y, y , y , x , y , x , y ) (with midpoint x ) do not induce an S , , , wehave yx ∈ E or yx ∈ E . Analogously, since ( y , x, x , x , y , x , y , x ) (withmidpoint y ) do not induce an S , , , we have xy ∈ E or xy ∈ E . Case 1.1 xy ∈ E and yx ∈ E :Then ( x, y , x , y, x , y ) induce a 2 D − C which is impossible in this section. Case 1.2 xy ∈ E and yx ∈ E :Then ( x, y , x , y, x , y ) induce a 2 D − C which is impossible in this section.11 ase 1.3 xy ∈ E and yx ∈ E :Then ( x, y , x , y, x , y ) induce a 2 D − C which is impossible in this section.Thus, assume that xy / ∈ E and yx / ∈ E . Then xy ∈ E and yx ∈ E . Since( y , x, x , x , y, x , y , x ) (with midpoint y ) do not induce an S , , , we have yx ∈ E . But then ( y , x, y , x , y, x ) induce a 2 D − P which is impossible in this section.Thus, y ∈ D or x ∈ D . Case 2. y ∈ D and x / ∈ D :Then y must have a D -neighbor x ∈ D . Since ( y , x, x , x , y , x , y , x ) (withmidpoint y ) do not induce an S , , , we have xy ∈ E . But then ( x , y , x , y , x, y )induce a 2 D − P which is impossible in this section.Analogously, y / ∈ D and x ∈ D is impossible. Thus, y ∈ D and x ∈ D , andLemma 1 is shown.In this section, by Lemma 1, for every P P = ( x , y , . . . , x , y ) in G , we have y , x ∈ D . Moreover, since y , x / ∈ D , y and x must have a D -neighbor, say xy ∈ E with x ∈ D and yx ∈ E with y ∈ D . Then D basis := D basis ∪ { y , x } .Assume that x , x , y , y ∈ N and y , x ∈ N . Then x, y ∈ D ∩ N , and sincethere is no 2 D − P in this section, x does not have any neighbor in N ∪ N , andanalogously, y does not have any neighbor in N ∪ N . Thus, x, y are ( y , x )-forcedand D basis := D basis ∪ { y , x , x, y } , or it leads to a contradiction.Thus, G can be reduced by every P in G or it leads to a contradiction.From now on, in the reduced G , there are only P ’s in N ∪ N , and for everycomponent Q in G \ ( N ∪ N ), Q is P -free bipartite. Clearly, the e.d.s. problemcan be done independently for every such component Q .In [7], we showed that ED is solvable in polynomial time for P -free bipartitegraphs. Thus, Theorem 1 is shown. 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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