Two Minimum Dominating Sets With Minimum ... - Semantic Scholar
Two Minimum Dominating Sets With Minimum Intersection In Chordal Graphs Peter Eades
Mark Keil
y
Mirka Miller
Paul D. Manuel
Abstract
We prove that the problem of nding two minimum dominating sets (connected dominating sets or vertex clique covers) with minimum intersection is linearly solvable in interval graphs. Furthermore, the problem of deciding whether or not there exist two disjoint minimum dominating sets (connected dominating sets or vertex clique covers) is shown to be NP-hard for chordal graphs. Keywords: vertex clique cover, dominating set, chordal graph.
1 Introduction
A graph is said to be chordal if it contains no induced cycle of length 4 or greater. A graph is said to be an interval graph if it is the intersection graph of a family of intervals along the real line. A set S of vertices in a graph G is a dominating set if every vertex, not in S , is adjacent to at least one vertex in S . A dominating set S is a connected dominating set if the subgraph induced by S is connected. A subset A of V is a clique of a graph G if it induces a complete subgraph of G. A clique C is maximal if there is no clique of G which properly contains C as a subset. A vertex v and Department of Computer Science, University of Newcastle, Newcastle, Australia. 2308 y Department of Computational Science, University of Saskatchewan, Saskatoon, SK,
Canada. S7N 0W0. This work was supported nancially by the Natural Science and Engineering Research Council of Canada.
1
a maximal clique C cover each other if the vertex v is in the maximal clique C . A set of vertices which covers all the maximal cliques of G is a vertex clique cover of G. This has also been called the clique transversal of G [Tu 90, An 91]. Several `combinatorial graph problems' have been studied for various classes of graphs [Ga 79]. A combinatorial graph problem is to nd an optimum (minimum or maximum etc.) set (of vertices or edges or cliques etc.) satisfying a given combinatorial property in a graph. An example of a combinatorial problem is the vertex cover problem: nd a minimum set of vertices covering all the edges of a given graph. Here the \covering all the edges of the graph" is the combinatorial property. As a generalization, Grinstead and Slater [Gr90] suggest investigating the problem of nding two optimum sets with minimum intersection, each set satisfying the given combinatorial property. For example, the more general version of the vertex cover problem is to nd two minimum vertex covers S1 and S2 such that the cardinality of S1 \ S2 is minimum. Obviously, the problem of nding one optimum set with a given graph combinatorial property is a subproblem of the more general version which requires two optimum sets (with minimum intersection), each satisfying the graph combinatorial property. Grinstead and Slater [Gr90] have given a linear time algorithm for nding two minimum dominating sets with minimum possible intersection in a tree and later [Gr 89] extended the algorithm to series-parallel graphs. Here we consider this idea for the combinatorial problems such as the domination problem, the connected domination problem and the vertex clique cover problem. We demonstrate that the problem of locating two minimum dominating sets (connected dominating sets or vertex clique covers) with minimum intersection is linearly solvable in interval graphs. Another variation of a combinatorial graph problem is the problem of deciding whether or not there exist two disjoint optimum sets of the combinatorial problem satisfying the given combinatorial property. The corresponding problem for the vertex cover problem is to decide whether or not there exist two disjoint minimum vertex covers of a graph. Notice that this problem does not generalize the normal combinatorial problem since it is just to report the boolean answer whether or not there exist two disjoint optimum sets of the combinatorial problem. Grinstead and Slater [Gr90] have shown that the problem of deciding whether or not there exist two disjoint minimum dominating sets is NP-hard for bipartite graphs. We show that the problem 2
of deciding whether or not there exist two disjoint minimum dominating sets (connected dominating sets or vertex clique covers) is shown to be NP-hard for chordal graphs
2 Preliminaries
Gavril [Ga 74] has shown that a graph G is chordal if and only if it is the intersection graph of a family of subtrees in a tree. The intersection model for a chordal graph can always be chosen so that the nodes of the tree correspond to the maximal cliques of the original graph G. Each vertex of G then corresponds to the subtree comprised of exactly those maximal cliques to which it belongs. We call such an intersection model a clique tree of the chordal graph. G is an interval graph if and only if the maximal cliques of G can be linearly ordered such that, for every vertex v of G, the maximal cliques containing v occur consecutively [Gi 64]. This can also be explained in the clique tree model. Interval graphs are the intersection graphs of a family of subpaths of a path tree [Mo 86]. For a given interval graph G(V; E ) with V = f1; 2; : : : ; ng, it is possible to construct a path (the clique tree) T with the maximal cliques of G as its nodes and P = fpi j i 2 V and pi is a subpath of T containing all the maximal cliques having ig such that the intersection graph of P in T is G [Gi 64, Mo 86]. Note that two paths pu and pv of P intersect each other (share a common node) if and only if u is adjacent to v in G. Let PS denote fpv j v 2 S g for any subset S of V . Here onwards T denotes the clique tree of G and Ci; 0 i q denote the maximal cliques of G which are the nodes of T . Also C0; C1; C2; : : : ; Cq is the order of the nodes of T placed left to right in T . (See Figure 1). We say Ci Cj if and only if i j . (See C0
C1
C2
Cj
Ci
Cq?1 Cq
Figure 1: The nodes C0; C1; C2; : : :; Cq of T are placed left to right in T . Ci Cj if and only if i j . Figure 1). We refer to the vertices of T as nodes and use V (T ) and E (T ) to denote the nodes and the edges of the clique tree T respectively. For every vertex ni, the respective path pni is written as fCl; Cl+1; : : :; Cr g, where the 3
vertex ni lies in the maximal cliques Cl; Cl+1; : : : ; Cr and fCl; Cl+1; : : :; Cr g is a subpath of T . (See Figure 2). For a path pni = fCl; Cl+1; : : :; Cr g, Cr is
pn i C0
C1
C2
Cl
Cl+1
Cr
Cq?1 Cq
Figure 2: Clique tree T of an interval graph. pRni = Cr and pLni = Cl. called the right end of pni and denoted by pRni . Similarly, Cl is called the left end of pni and denoted by pLni . (See Figure 2). A rightmost path of P from a node Ci0 is a path p = fCl; Cl+1; : : :; Cr g of P containing Ci0 whose right end node Cr is the nearest to Cq. (See Figure 3). A leftmost path of P from a node Ci0 is a path p = fCl; Cl+1; : : :; Cr g of P containing Ci0 whose left end node Cl is the nearest to C0.
p C0
C1
C2
Cq?1 Cq
Ci0
Figure 3: p is the rightmost path from Ci0
3 Polynomial algorithm in interval graphs
3.1 Connected Domination Problem
In this section we show how to compute two minimum connected dominating sets with minimum intersection in an interval graph G. The algorithm consists of two phases. In the rst phase, a set M = fM0; M1; M2; : : :; M g of nodes of V (T ) is identi ed with the property that the i-th rightmost path of PS must extend to (or to the right of) point Mi where S is a minimum connected dominating set of G and is the cardinality of a minimum connected dominating set of G. In the second phase the two minimum connected 4
dominating sets for G with minimum intersection are computed using the set M for guidance.
3.1.1 Phase I
It is known that Lemma 3.1.1 [Ke 94, Ma 93] For a chordal graph G, let T be its clique tree and P = fpv j v 2 Vg be the collection of subtrees in T such that its intersection graph is G. A set S = fg1 ; g2 ; : : : ; g g is a connected dominating set of G if and only if PS = fpg1 ; pg2 ; : : :; pg g of P covers E (T ), the edge set of the clique tree T of G. First we will construct a minimum collection fpm1 ; pm1 ; : : : ; pm g of paths from Cq to C0 so that these paths cover E (T ) of T of G. Let pm1 be the leftmost path from Cq . Given pmi?1 , we identify pmi as the leftmost path from pLmi?1 . We continue this process until some pm reaches C0. Then we set Mi = pLm?i for i = 0; 1; 2; : : : ; ? 1;. M is Cq and M0 is C0. (See Figure pm C0
pm?1
pm2
pmi
pm1
pLm
pLm?1
pLmi
pLm2
pLm1
M0
M1
M ?i
M ?2
M ?1
Cq M
Figure 4: After Phase I of the connected domination problem, the clique tree T is marked with M = fM0 ; M1; M2 ; : : :; M g 4). By the construction of the paths pmi , the set S = fm1; m2; : : :; m g is a minimum connected dominating set of G [Ke 94, Ma 93]. Lemma 3.1.2 Let F = ffi : i = 1; 2; : : : ; g be a minimum connected dominating set of G with vertices sorted by the right end points of the paths pfi . Then pRfi Mi for i = 1; 2; : : : ; , where pRfi is the right end of pfi . 5
Proof Suppose pRfi < Mi for some i. By phase I, we know that we need a minimum of ( ? i) paths of P to cover the edges between Mi and M in T . Since pRfi < Mi, the paths pf1 ; pf2 ; : : :; pfi do not cover all the edges between M0 and Mi in T and thus the set ff1; f2; : : :; fig cannot be a subset of a minimum connected dominating set of G.
pa1
pb1 C0
M0
M1
pRb1
M ?1
pRa1
Cq
M
Figure 5: Since pRb1 M1, b1 2 B
3.1.2 Phase II
In this phase, we compute two minimum connected dominating sets A and B of G with minimum intersection as follows: We construct A and B simultaneously by traversing from C0 to Cq over T with M0; M1; : : : ; M marked on T from phase I. A path denoted by paj is included in PA and a path denoted by pbj is included in PB . We construct paj and pbj ,i = 1; 2; : : : ; in such a way that pRaj Mj and pRbj Mj in order to satisfy Lemma 3.1.2. Let pa1 be the rightmost path from M0. By Lemma 3.1.2, pRa1 M1 since it is true that every graph has at least one minimum connected dominating set. Add a1 to A. Find the next rightmost path pb1 (di erent from pa1 ) from M0. If pRb1 M1, add b1 to B (See Figure 5); otherwise b1 = a1. We continue to add vertices to A and B as follows. At stage i, let pai and pbi be the most recently selected paths for PA and PB respectively. Without loss of generality we assume that pRai pRbi . (See Figure 6). Since pRbi pRai , we rst nd a rightmost path pbi+1 from pRbi and add bi+1 to B . (Note that if pRai pRbi , then we will rst construct pai+1 from pRai ). The existence of such pbi+1 satisfying the condition that pRbi+1 Mi+1 is guaranteed by Lemma 3.1.2 since it is true that every graph has at least one minimum connected dominating set. Next nd another rightmost path pai+1 (di erent from pbi+1 ) from pRai . If pRai+1 Mi+1 , then add ai+1 to A (See Figure 7); otherwise ai+1 = bi+1. (See Figure 8). The algorithm terminates when pa and pb both reach M . Now we shall verify the correctness of the above algorithm. 6
pai C0
M0
pbi
Mi
pRbi
pRai
Cq
Mi+1
M
Figure 6: pRai pRbi
pai+1
pai C0
Mi
M0
p bi
pRbi
pRai
Mi+1 pRbi+1
Cq
pRai+1
M
pRbi+1
Cq M
pbi+1 Figure 7: Since pRai+1 Mi+1, ai+1 2 A
pai+1
pai C0
Mi
M0
p bi
pRbi
pRai
pRai+1 Mi+1
pbi+1 Figure 8: Since pRai+1 < Mi+1, ai+1 is not considered for A. 7
Theorem 3.1.1 The sets A and B constructed by the algorithm are mini-
mum connected dominating sets for G with minimum possible intersection. Proof. Lemma 3.1.2 and the construction of the two sets A and B imply that A and B are minimum connected dominating sets for G. It remains to show that A \ B is minimum. To do this we rst de ne the common vector of a pair PQ, PR of sets of paths of P as the vector whose elements contain the right end points of the paths common to both PQ and PR, sorted in increasing order. The theorem then follows from the following claim. Claim: There exist two sets PY and PZ of paths of P such that Y and Z satisfy (i) each is a minimum connected dominating set for G.
(ii) among all pairs of sets satisfying (i), jY \ Z j is minimum. (iii) among all pairs of sets satisfying (i) and (ii), PY and PZ lexicographically maximize their common vector. and such that after i stages of the algorithm fpa1 ; pa2 ; : : :; pai g PY and fpb1 ; pb2 ; : : :; pbi g PZ . Proof of the Claim: Let pyi and pzi be the i-th paths of PY and PZ respectively when the paths of PY and PZ are sorted in increasing order of right end point. We prove the claim by induction on i, the number of stages in the algorithm. After zero stages of the algorithm, A and B are empty and the claim trivially holds. As an inductive assumption, assume the claim holds after i ? 1 stages of the algorithm, that is, py1 = pa1 , py2 = pa2 , : : : , pyi?1 = pai?1 and pz1 = pb1 , pz2 = pb2 , : : : , pzi?1 = pbi?1 . Now consider the i-th elements ai and bi added to the sets A and B respectively. Without loss of generality we may assume pRai?1 pRbi?1 . We proceed by case analysis Case 1 (pyi 6= pzi ): In this case, Lemma 3.1.2 and the construction used in the algorithm imply that pai 6= pbi . If pRzi pRyi , then the construction used in the algorithm implies that pRbi pRzi and pRai pRyi and the inductive assumption assures that PY 0 = fpa1 ; pa2 ; : : :; pai ; pyi+1 ; : : :; py g and PZ0 = fpb1 ; pb2 ; : : : ; pbi ; pzi+1 ; : : :; pz g are the sets required in the claim. The remaining subcase has pRzi < pRyi . If pyi 6= pbi , then the inductive assumption and the construction used in the algorithm ensure that PY 0 = fpa1 ; pa2 ; : : :; pai ; pyi+1 ; : : : ; py g and PZ0 = fpb1 ; pb2 ; : : : ; pbi ; pzi+1 ; : : :; pz g are the sets required in the claim. If pyi = pbi , then the new sets required in the 8
claim are PY 0 = fpa1 ; pa2 ; : : :; pai ; pzi+1 ; : : :; pz g and PZ0 = fpb1 ; pb2 ; : : : ; pbi ; pyi+1 ; : : :; py g. Case 2 (pyi = pzi ): Here if pai = pbi , then the construction and the inductive assumption assure that PY = fpa1 ; pa2 ; : : :; pai?1 ; pbi ; pyi+1 ; : : :; py g and PZ = fpb1 ; pb2 ; : : :; pbi?1 ; pbi ; pzi+1 ; : : :; pz g are the two sets required in the claim. The remaining possibility has pai 6= pbi for this case. We will now show this can not occur. Assume to contrary that it can occur and let i0 be the smallest integer greater than or equal to i such that pRbi0 pRai0 . For i j < i0 the construction used in the algorithm imply the following properties: (a) paj 6= pbj (b) pRyj pRbj < pRaj and pRzj pRbj < pRaj Consider the two sets PY 0 = fpa1 ; pa2 ; : : :; pai?1 ; pai ; : : :; pai0 ?1 ; pbi0 ; pyi0 +1 ; : : :; py g and PZ0 = fpb1 ; pb2 ; : : : ; pbi?1 ; pbi ; : : : ; pbi0 ?1 ; pbi0 ; pzi0 +1 ; : : :; pz g. The inductive assumption implies that PY 0 and PZ0 satisfy properties (i) and (ii) of the claim; And properties (a) and (b) above imply that PY 0 and PZ0 have a lexicographically larger common vector than PY and PZ . This contradicts the inductive assumption. Hence the subcase pai 6= pbi cannot occur.
3.1.3 Time complexity
In the rst phase, the algorithm constructs the set M = fM0; M1; M2; : : : ; M g by traversing from Cq to C0. During the traversal, each path is visited at most once. Hence the rst phase runs in linear time. In the second phase, the two minimum connected dominating sets A and B with minimum intersection are computed. The algorithm traverses from C0 to Cq visiting each path at most once. Thus the second phase also runs in linear time.
3.2 Domination Problem
We now consider the problem of computing two minimum dominating sets with minimum intersection in interval graphs. For any path pni = fCl; Cl+1; : : : ; Crg of P on T , de ne the right head of pni , denoted by hR(pni ), as the node Ck0 of T , where k0 = min fk j p0 = fCj ; Cj+1; : : : ; Ck g such that j r + 1g p0 2P 9
(See Figure 9). In a similar way, one can de ne the left head of pni , denoted by hL(pni ), as the node Cj0 , where 0 = fC ; C ; : : : ; C g such that k l ? 1g j 0 = max f j j p j j +1 k 0 p 2P
p0 p0
pn i C0
Cl
p0
Cr Cr+1
Ck0
Cq
Figure 9: hR(pni ) = Ck0 in the domination problem The algorithm for the domination problem is very similar to the algorithm for the connected domination problem. The only di erence between the two algorithms is that the right end (left end) of a path in the connected domination algorithm is replaced by the right head (left head) of the path in the domination algorithm.
3.2.1 Phase I A set S = fg1; g2; : : :; g g is a dominating set of an interval graph G if and only if every path of P , not in PS = fpg1 ; pg2 ; : : :; pg g, is intersected by at
least one path of PS . Let denote the cardinality of a minimum dominating set of G. We construct such a minimum collection of paths from Cq to C0 as follows: Let pm1 be the leftmost path from Cq . Then select a leftmost path from hL(pm1 ) and call it pm2 . Given pmi?1 , we identify pmi as the leftmost path from hL(pmi?1 ). (See Figure 10). We continue this process until some pm reaches C0. The above constructed collection S = fm1; m2; : : : ; m g is a minimum dominating set of G [Ke 94, Ma 93]. Then we set Mi = hL(pm ?i ) for i = 0; 1; 2; : : : ; ? 1. M is Cq and M0 is C0. (See Figure 11). The following lemma will be used in the proof of Lemma 3.2.2. Lemma 3.2.1 Let D = fd1; d2; : : : ; d g be a minimum dominating set of G such that the corresponding paths of D are ordered from Cq to C0 (that is, 10
pm ?1
pm C0
hL (pm ) hL (pm ?1 )
pmi hL (pmi?1 )
hL (pmi )
pLm ?1
pm1 hL (pm1 ) pLm1
pLmi?1
pLmi
Cq
Figure 10: pmi is the leftmost path from hL(pmi?1 ). C0
hL (pm )
hL (pm ?1 )
hL (pm ?i )
hL (pm1 )
M0
M1
Mi
M ?1
Cq
M
Figure 11: After Phase I of the domination problem, the clique tree T is marked with M = fM0; M1; M2 ; : : : ; M g the path pd1 originates at Cq , then follows pd2 and nally pd terminates at C0). Then hL(pdi ) hL(pmi ) for i = 1; 2; : : : ; , where hL (pmi ) = M ?i and hL(pdi ) is the left head of pdi . (See Figure 12).
Proof The proof follows from the greedy construction of fpm1 ; pm2 ; : : : ; pm g [Ke 94, Ma 93]. 2 C0
hL (pm ) M0
hL(pmi ) hL (pdi ) M ?i
hL (pm1 ) hL(pd1 ) M ?1
Cq
M
Figure 12: hL (pdi ) hL(pmi )
Lemma 3.2.2 Let F = ff1; f2; : : :; f g be a minimum dominating set of G with vertices sorted by the right end points of the paths pfi . Then
pRfi Mi for i = 1; 2; : : : ; , where pRfi is the right end of pfi .
11
Proof. Suppose pRfi < Mi for some i. While traversing from Cq to C0 in
phase I of this section, there exists a path p0 lying between pLm ?i and Mi and not intersecting the path pm ?i . (See Figure 13 and Figure 9). Since pRfi < Mi, the paths fpf1 ; pf2 ; : : : ; pfi g do not intersect p0. By our construction, the paths fpm1 ; pm2 ; : : :; pm ?i g between Mi and M do not intersect pm ?i+1
pm C0
M0
pf1
pfi
p0 Mi
pm1
pm ?i pLm ?i
Mi+1
Cq M
Figure 13: Here pRfi < Mi and pfi does not intersect p0.
p0. fpm1 ; pm2 ; : : : ; pm ?i g are the most optimal paths between Mi and M by Lemma 3.2.1. Thus pfi must intersect p0 if ff1; f2; : : : ; fig is a subset of a minimum dominating set of G. Since pfi do not intersect p0, ff1; f2; : : :; fig cannot be a subset of any minimum dominating set of G. This is a contradiction. 2
3.2.2 Phase II
We give a sketch of the algorithm to compute two minimum dominating sets A and B for G with minimum intersection. Given pai and pbi , the most recently selected paths for PA and PB respectively, we assume without loss of generality that pRai pRbi . Now nd a rightmost path pbi+1 from hR(pbi ) and add bi+1 to B . The existence of such pbi+1 with the condition that pRbi+1 Mi+1 is guaranteed by Lemma 3.2.2 since it is true that every graph has at least one minimum dominating set. Then nd the next rightmost path pai+1 (di erent from pbi+1 ) from hR(pai ). If pRai+1 Mi+1 , add ai+1 to A; otherwise ai+1 = bi+1. The algorithm terminates at Cq. The proof of correctness is similar to the proof for the connected domination algorithm. Furthermore, it is obvious that the algorithm runs in linear time. 12
3.3 Vertex clique cover problem
Lemma 3.3.1 [Ke 94, Ma 93] For a chordal graph G, let T be its clique tree and P = fpv j v 2 Vg be the collection of subtrees in T such that its intersection graph is G. A set S = fg1 ; g2 ; : : : ; g g is a vertex clique cover of G if and only if PS = fpg1 ; pg2 ; : : :; pg g of P covers V (T ), the vertex set of
the clique tree T of G. For this problem, for any path pni = fCl; Cl+1; : : :; Cr g of P , we de ne the right head of pni , denoted by hR(pni ), as Cr+1 and the left head of pni , denoted by hL(pni ), as Cl?1. The algorithm for nding two minimum vertex clique covers with minimum intersection thus proceeds exactly as the algorithm for nding two minimum dominating sets with minimum intersection.
x1 x3 uij
x5
x6
uij
x4 x2 Figure 14: The H subgraph of G with designated vertices uij and uij .
4 NP-hard in chordal graphs In this section we show that the problem of deciding whether or not there exist two disjoint minimum total dominating sets is NP-hard for bipartite 13
graphs. We transform this problem to the corresponding connected domination problem on chordal graphs and prove that the problem of deciding whether or not there exist two disjoint minimum connected dominating sets is NP-hard for chordal graphs. Using this result, we prove that the problem of deciding whether or not there exist two disjoint minimum dominating sets (or vertex clique covers) is NP-hard for chordal graphs.
4.1 Disjoint Total Domination Problem in bipartite graphs is NP-hard
A dominating set S is a total dominating set if the subgraph induced by S has no isolates. Grinstead and Slater [Gr 90] have shown that the problem of deciding whether or not there exist two disjoint minimum dominating sets is NP-hard for bipartite graphs. We modify their construction to prove that the problem of deciding whether or not there exist two disjoint minimum total dominating sets is NP-hard for bipartite graphs. We use the same notation and terminology of [Gr 90]. We construct a bipartite graph G from a set U = fu1; u2; : : : ; ung of variables and a collection C = fc1 ^ c2 ^ : : : ^ cmg of clauses over U . Here ci = (si1 _ si2 _ si3) and each sij is uh or uh for some 1 h n. For 1 i m and 1 h n, uih is uh and uih is uh . If sij is uh, then sij is uih. If sij is uh , then sij is uih. For example, if ci = (u2 _ u4 _ u5), then si1 = ui2 = u2, si2 = ui4 = u4, si3 = ui5 = u5. For every sij , 1 i m, 1 j 3, we construct a subgraph H which forms a building block for G. Notice that the H subgraph is always referred to uih , where sij = uih. The H subgraph is given in Figure 14. The graph G will contain 3m copies of the graph H . We need to note that H is bipartite with vertices labeled uij and uij in the same set of the bipartition and the only vertices in H adjacent to other vertices of G will be uij and uij . fx1; x2; x5; x6g and fuij ; uij ; x3; x4g form the bipartition of H . Each copy of H in G will be called an H -subgraph with designated vertices uij and uij . Any minimum total dominating set of H must contain at least one of uij and uij . Any minimum total dominating set of H contains exactly two vertices. The possible pairs are fuij ; x5g, fuij ; x6g, fuij ; x5g and fuij ; x6g. If H has two disjoint minimum total dominating sets D1 and D2, then there are two possibilities. If D1 contains fuij ; x5g, then D2 will contain fuij ; x6g. If D1 contains fuij ; x6g, then D2 will contain fuij ; x5g. 14
Most importantly for each uij , if one minimum total dominating set of H contains uij , the other minimum total dominating set of H must contain uij . For each clause ci = (si1 _ si2 _ si3), we construct a graph Gi as follows. (See Figure 15). Suppose si1 = ua or ua, si2 = ub or ub, and si3 = uc or uc.
b1i
ui1
H
b2i
ui1 ui2
b3i
H
ui2 ui5
b4i
H
ui5
Figure 15: A subgraph Gi of G for ci = (u1 _ u2 _ u5). Let Gi contain three copies of H with designated vertices uia and uia, uib and uib, and uic and uic. Each of b1i and b2i is connected to uia if si1 = ua, to uia if si1 = ua, to uib if si2 = ub, to uib if si2 = ub, to uic if si3 = uc, and to uic if si3 = uc. Each of b3i and b4i is connected to the three designated vertices not adjacent to b1i and b2i. Let G be the graph containing disjoint copies of G1; G2; : : : ; Gm to which we add the following vertices and edges. For each occurance of a uh or uh with 1 h n in distinct clauses ci and cj , add four vertices of degree 2 as follows. Assume sir is uh or uh and assume sjt is uh or uh where 1 i < j m and 1 r; t 3. Let two of the four vertices be adjacent to uih and to ujh , and let the other two be adjacent to uih and to ujh. The graph G is illustrated in Figure 16 for C = (u1 _ u2 _ u5) ^ (u2 _ u3 _ u4). We derive the following information from G. G is a bipartite graph. Any minimum total dominating set of G must contain at least one of uij and uij . Any minimum total dominating set of G contains exactly two vertices of H . The possible pairs are fuij ; x5g, fuij ; x6g, fuij ; x5g and fuij ; x6g. If G has two disjoint minimum total dominating sets D1 and D2, then there are two possibilities in H . If D1 contains fuij ; x5g, then D2 will contain fuij ; x6g. If D1 contains fuij ; x6g, then D2 will contain fuij ; x5g. Most importantly, 15
u11
b11
b21
H u11
u12
u23
x
u22
H u22
b31
b41
H u12
u15
H u15
H u23
u24
H u24
y
b12 b22 b32 b42 Figure 16: Graph G for C = (u1 _ u2 _ u5) ^ (u2 _ u3 _ u4).
16
if one minimum total dominating set of G contains uij , the other minimum total dominating set of G must contain uij . The NOT-ALL-EQUAL 3SAT problem [Ga 79] is a 3SAT problem with an additional condition that each clause has at least one true literal and at least one false literal. A NOT-ALL-EQUAL 3SAT truth assignment is a truth assignment for U such that each clause in C has at least one true literal and at least one false literal. The NOT-ALL-EQUAL 3SAT problem is NP-complete [Ga 79]. Now we will prove that G has two disjoint minimum total dominating sets if and only if U has a NOT-ALL-EQUAL 3SAT truth assignment for C . Suppose there is a NOT-ALL-EQUAL 3SAT truth assignment for C . Let S1 consist of those ui in U that receive the value true and S2 = U ? S1. Construct two vertex subsets D1 and D2 of V (G) as follows. For each uij in G if uj 2 S1 then add uij and x5 of H to D1 and add uij and x6 of H to D2, and if uj 2 S2, then add uij and x5 of H to D2 and add uij and x6 of H to D1. (For example, consider C = (u1 _ u2 _ u5) ^ (u2 _ u3 _ u4). One NOTALL-EQUAL 3SAT truth assignment is S1 = fu1; u4g. Then D1 contains the six darkened vertices u11; u12; u15; u22; u23 and u24 with one additional vertex from each H . D2 contains the six undarkened vertices u11; u12; u15; u22; u23 and u24 with one additional vertex from each H ). First we prove that D1 is a dominating set of G. Since D1 contains either uij and x5 or uij and x6, it dominates the remaining vertices of H . Each clause has one true literal and one false literal. The vertex of D1 corresponding to the true literal dominates b1i and b2i, and the vertex of D1 corresponding to the false literal dominates b3i and b4i. For each occurance of a uh or uh in distinct clauses ci and cj , four vertices of degree 2 were added to G. If uik and ujk are vertices of G with i 6= j , then both are in D1 or both are in D2. Thus the four vertices of degree 2 are dominated by D1 . In the same way, we can prove that D2 is also a dominating set of G. It is easy to verify that D1 and D2 are total dominating sets of G. As we mentioned earlier, every minimum total dominating set of G must contain at least two vertices of each H . Since D1 and D2 contain exactly two vertices of each H and both are total dominating sets of G, D1 and D2 are minimum total dominating sets of G. Also D1 and D2 are disjoint. Hence, G has two disjoint minimum total dominating sets. To prove the converse, suppose G has two disjoint minimum total dominating sets, say D1 and D2. As we noted earlier, for each uij of G, either D1 or D2 contains uij and a corresponding vertex x5 (or x6) from its subgraph 17
H , and the other contains uij and a corresponding vertex x6 (or x5) from the same subgraph H . Without loss of generality, let us assume that one of D1 and D2 contains uij and x5, and the other contains uij and x6. Suppose uik and ujk are vertices of G with i 6= j (for example, u12 and u22 in Figure 16). To see that uik and ujk can not both be in D1 or both be in D2, note that if uik and ujk are in D1 ( and so uik and ujk are in D2 ), then D1 must also contain the two vertices x and y of degree 2 which are adjacent to uik and ujk (for example, vertices x and y in Figure 16). But the set D1 [ fuik g n fx; yg is also a total dominating set which contradicts the minimality of D1. Consequently, if uik and ujk are vertices of G with i 6= j , then both are in D1 or both are in D2. Therefore, the following is a well de ned truth assignment for U . For each uk 2 U , nd a uik in G, and let uk be true if uik 2 D1 and false if uik 2 D2. It remains only to show that each clause ci has at least one true literal and at least one false literal. If not, we can assume clause ci has three true literals. Then each of b3i and b4i is not dominated by D1, so b3i and b4i are in D1. Let z be one of the vertices adjacent to b3i and b4i. Then z is some uil or uil. We can verify that D1 [ fzg n fb3i; b4ig is also a total dominating set. This contradicts the minimality of D1. It follows that U has a NOT-ALL-EQUAL 3SAT truth assignment for C . Since the NOT-ALL-EQUAL 3SAT problem is NP-complete, we can state that Theorem 4.1.1 The problem of deciding whether or not there exist two disjoint minimum total dominating sets is NP-hard for bipartite graphs.
4.2 Disjoint Connected Domination Problem in chordal graphs is NP-hard
A vertex v of G is simplicial if the set of vertices adjacent to v in G is completely connected. Now we transform the problem of deciding whether or not there exist two disjoint minimum total dominating sets in bipartite graphs to the problem of deciding whether or not there exist two disjoint minimum connected dominating sets in chordal graphs. Let G(V1 [ V2; E ) be any bipartite graph. Let V1 = fvj j j = 1; 2; : : : ; ng, V2 = fui j i = 1; 2; : : : ; mg. We construct a chordal graph G0(V 0; E 0) from G(V1 [ V2 ; E ) as follows: The vertex set V 0 = fvj ; vj0 j vj 2 V1 ; j = 1; 2; : : : ; ng [ fui; u0i j ui 2 18
v1
u1
v2
u2
v3
Figure 17: A bipartite graph G(V1 [ V2 ; E )
V2; i = 1; 2; : : : ; mg. The edge set E 0 of G0 is de ned as follows. (vj ; u0i), (vj0 ; ui) 2 E 0 if (vj ; ui) 2 E . Also any two vertices of V1 [ V2 form an edge in E 0 so that V1 [ V2 induces a completely connected subgraph in G0. (See Figure 17 and Figure 18.) v1 u01 u02
u1
v2 u2
v3
v10 v20 v30
Figure 18: The corresponding chordal graph G0(V 0; E 0)
Claim 4.2.1 If a subset S of V 0 is a minimum connected dominating set of G0, then S V1 [ V2. Proof of the Claim Suppose S has some u0i of V 0. Since S is connected, S has some vj of V1 which is adjacent to u0i in G0. If u0i and vj are adjacent in G0, then N [u0i] N [vj ], where N [v] = fvg [ fu j u is adjacent to vg. That is, any u0i or vj0 is a simplicial vertex of G0. Thus S n fu0ig is still a connected dominating set of G0 which contradicts the minimality of S . The same arguments hold if S has any vj0 of V 0. Hence S V1 [ V2. 2 Claim 4.2.2 A set S of vertices is a minimum total dominating set of G if and only if S is a minimum connected dominating set of G0 . 19
Proof of the Claim Let and be the cardinality of a minimum total
dominating set of G and a minimum connected dominating set of G0 respectively. Suppose S is a minimum total dominating set of G. We rst prove that S is a dominating set of G0. It is true that S V1 [ V2. Since V1 [ V2 induces a completely connected subgraph of G0, S dominates V1 [ V2 in G0. Let u0i be an arbitrary vertex of G0. Since S is a total dominating set of G, the vertex ui of G is dominated by some vertex of S in G and thus u0i is dominated by the same vertex of S in G0. In the same way, the vertices of the form vj0 in G0 are dominated by S . So S is a dominating set of G0. Since S is a subset of the completely connected subgraph V1 [ V2 in G0, S is connected and hence S is a connected dominating set of G0. Therefore, . Suppose S is a minimum connected dominating set of G0. By Claim 4.2.1, S V1 [ V2. We rst prove that S is a total dominating set of G. Let ui be an arbitrary vertex of G. Since S is a dominating set of G0 and S V1 [ V2, u0i is dominated by some vertex vj of S in G0. Since u0i and vj are adjacent in G0, ui and vj are adjacent in G. Thus the vertices of V2 are dominated by S in G. In the same way, we can show that the vertices of V1 are dominated by S in G. Hence S is a total dominating set of G. Therefore . Hence the claim. 2 It is easy to observe from Claim 4.2.2 that Claim 4.2.3 S1 and S2 are disjoint minimum total dominating sets of G if and only if S1 and S2 are disjoint minimum connected dominating sets of G0 . Since the problem of deciding whether or not there exist two disjoint minimum total dominating sets is NP-hard for bipartite graphs, we can state that Theorem 4.2.1 The problem of deciding whether or not there exist two disjoint minimum connected dominating sets is NP-hard for chordal graphs. Note that the above arguments do not immediately follow for the domination and vertex clique cover problems. We use Theorem 4.2.1 to prove that the problem of deciding whether or not there exist two disjoint minimum dominating sets (or vertex clique covers) is NP-hard for chordal graphs. 20
4.3 Disjoint Domination Problem in chordal graphs is NP-hard
We transform the connected domination problem on a chordal graph G to the domination problem (or vertex clique cover problem) on a similar chordal graph G0. To de ne G0, we construct a clique tree T 0 from the clique tree T of G and a family P 0 of subtrees in T 0. For every edge (Ci; Cj ) of T , we introduce two new nodes between the nodes Ci and Cj to construct T 0. Let Cij and Cji be the new nodes introduced to the edge (Ci; Cj ) of T where Cij is the node close to Ci and Cji is the node close to Cj . Each edge (Ci; Cj ) of a subtree pv of P in T is replaced by the path segment fCi; Cij ; Cji; Cj g to construct the corresponding subtree p0v of P 0 in T 0. For each edge (Ci; Cj ) of T , let pvij and pvji be the subtrees containing the single node Cij and Cji respectively. (See Figure 19). These subtrees are added to the family P 0 of subtrees in T 0. Thus to de ne G0, the family P = fpv j v 2 V g of Ci
Cj
Ci
pvij
pvji
Cij
Cji
Cj
T0
T
Figure 19: subtrees in the intersection model for G is augmented to P 0 = fp0v j pv 2 P and v 2 V g [ fpvij , pvji j (Ci; Cj ) is an edge in Tg. The intersection graph of the subtrees of P 0 in T 0 is G0. The vertex set of G0 is V 0 = V [ U where V denotes the set of vertices corresponding to the subtrees fp0v j pv 2 P and v 2 V g and U denotes the set of vertices corresponding to the subtrees fpvij , pvji j (Ci; Cj ) is an edge in Tg. In other words, V 0 = V [ fvij ; vji j (Ci; Cj ) is an edge in T g. When G is a chordal graph, G0 is also a chordal graph. Let PS0 denote fp0v 2 P 0 j v 2 S g. The following lemmas are useful to study the relation between the connected domination problem and domination problem in chordal graphs. Claim 4.3.1 If S is a minimum dominating set (or vertex clique cover) of G0, then S V . 21
Proof of the Claim Any vertex vij or vji of U is a simplicial vertex of G0
since it is in exactly one maximal clique of G0. Suppose S contains some vertex of U , say vij . Then S does not contain any vertices adjacent to vij . Otherwise we will arrive at a contradiction to the minimality of S by removing vij from S . We observe that the set of vertices adjacent to vij is the same as the set of vertices adjacent to vji in G0. Since S contains vij and S does not contain any vertices adjacent to vij , S must have vji. Otherwise it will contradict the fact that S is a dominating set of G0. Thus S has both vij and vji of G0. Since G0 is a connected graph, every edge of its clique tree T 0 is covered by a subtree of P 0. Let p0u be the subtree of P 0 which covers the edge (Ci; Cj ). Then the vertex u of G0 is adjacent to both vij and vji. Replace vij and vji of S by u to form a set S 1. Since S 1 is still a dominating set of G0, it contradicts the fact that S is a minimum dominating set of G0. Thus S V. 2 Claim 4.3.2 S is a minimum connected dominating set of G if and only if S is a minimum dominating set (or vertex clique cover) of G0. Proof of the Claim Let and be the cardinality of a minimum connected dominating set of G and a minimum dominating set of G0 respectively. Suppose S is a minimum connected dominating set of G. By Lemma 3.1.1, PS covers all the edges of T . That is, PS0 covers all the edges of T 0 and hence the subtrees of PS0 intersect all other subtrees of P 0. Note that two subtrees pu and pv of P intersect each other if and only if u is adjacent to v in G. Thus S dominates the vertices of V n S of G0 and the vertices of U of G0. Hence S is a dominating set of G0. Therefore, . Suppose S is a minimum dominating set of G0. By Claim 4.3.1, S V . Let e = (Ci; Cj ) be an edge of T . Since S dominates U , the subtrees pvij and pvji are intersected by a subtree p0u of PS0 in T 0. Thus the subtree p0u contains the path segment fCi; Cij ; Cji; Cj g of T 0. Therefore pu of PS contains the edge (Ci; Cj ) of T . This shows that the edges of T are covered by PS . By Lemma 3.1.1, S is a connected dominating set of G. Therefore, . The corresponding result for the vertex clique cover can be derived in the same way using Lemma 3.3.1. 2 Claim 4.3.3 S1 and S2 are disjoint minimum connected dominating sets of G if and only if S1 and S2 are disjoint minimum dominating sets (or vertex clique covers) of G0 . 22
As a consequence of Claim 4.3.3 and Theorem 4.2.1, we conclude the following result. Theorem 4.3.1 The problem of deciding whether or not there exist two disjoint minimum dominating sets (or vertex clique covers) is NP-hard for chordal graphs.
References [An 91] T. Andreae, M. Schughart and Zs. Tuza, Clique-transversal sets of line graphs and complements of line graphs, Discrete Math. 88 (1991), pp 11-20. [Ga 79] M. R. Garey and D. S. Johnson, Computers and intractability: A guide to the theory of NP-completeness, W. H. Freeman and Co, 1979. [Ga 74] F. Gavril, The intersection graphs of subtrees are exactly the chordal graphs, J. Comb. Theory B, 16 (1974), pp 47-56. [Gi 64] P. C. Gilmore and A. J. Ho man, A characterization of comparability graphs and of interval graphs, Canad. J. Math. 16 (1964), pp 539-548. [Gr 89] D. L. Grinstead, Algorithmic templates and multiset problems in graphs, Ph.D. Thesis, Univ. of Alabama in Huntsville, 1989. [Gr 90] D. L. Grinstead and P. J. Slater, On minimum dominating sets with minimum intersection, Discrete Mathematics, 86 (1990), pp 239-254. [Ke 94] J. M. Keil, R. C. Laskar and P. D. Manuel, The vertex clique cover problem and some related problems in chordal graphs, SIAM conference on Discrete Algorithms, San Francisco, 1994. [Ma 93] P. D. Manuel, Vertex Clique Cover Problems and Domination Problems in Chordal Graphs, M.Sc. Thesis, Dept. of Comp. Science, Univ. of Saskatchewan, Canada, August 1993. 23
[Mo 86] C. L. Monma and V. K. Wei, Intersection graphs of paths in a tree, J. Comb. Theory B, 41 (1986), pp 141-181. [Tu 90] Zs. Tuza, Covering all cliques of a graph, Discrete Math. 86 (1990), pp 117-126.
24
Mark Keil
y
Mirka Miller
Paul D. Manuel
Abstract
We prove that the problem of nding two minimum dominating sets (connected dominating sets or vertex clique covers) with minimum intersection is linearly solvable in interval graphs. Furthermore, the problem of deciding whether or not there exist two disjoint minimum dominating sets (connected dominating sets or vertex clique covers) is shown to be NP-hard for chordal graphs. Keywords: vertex clique cover, dominating set, chordal graph.
1 Introduction
A graph is said to be chordal if it contains no induced cycle of length 4 or greater. A graph is said to be an interval graph if it is the intersection graph of a family of intervals along the real line. A set S of vertices in a graph G is a dominating set if every vertex, not in S , is adjacent to at least one vertex in S . A dominating set S is a connected dominating set if the subgraph induced by S is connected. A subset A of V is a clique of a graph G if it induces a complete subgraph of G. A clique C is maximal if there is no clique of G which properly contains C as a subset. A vertex v and Department of Computer Science, University of Newcastle, Newcastle, Australia. 2308 y Department of Computational Science, University of Saskatchewan, Saskatoon, SK,
Canada. S7N 0W0. This work was supported nancially by the Natural Science and Engineering Research Council of Canada.
1
a maximal clique C cover each other if the vertex v is in the maximal clique C . A set of vertices which covers all the maximal cliques of G is a vertex clique cover of G. This has also been called the clique transversal of G [Tu 90, An 91]. Several `combinatorial graph problems' have been studied for various classes of graphs [Ga 79]. A combinatorial graph problem is to nd an optimum (minimum or maximum etc.) set (of vertices or edges or cliques etc.) satisfying a given combinatorial property in a graph. An example of a combinatorial problem is the vertex cover problem: nd a minimum set of vertices covering all the edges of a given graph. Here the \covering all the edges of the graph" is the combinatorial property. As a generalization, Grinstead and Slater [Gr90] suggest investigating the problem of nding two optimum sets with minimum intersection, each set satisfying the given combinatorial property. For example, the more general version of the vertex cover problem is to nd two minimum vertex covers S1 and S2 such that the cardinality of S1 \ S2 is minimum. Obviously, the problem of nding one optimum set with a given graph combinatorial property is a subproblem of the more general version which requires two optimum sets (with minimum intersection), each satisfying the graph combinatorial property. Grinstead and Slater [Gr90] have given a linear time algorithm for nding two minimum dominating sets with minimum possible intersection in a tree and later [Gr 89] extended the algorithm to series-parallel graphs. Here we consider this idea for the combinatorial problems such as the domination problem, the connected domination problem and the vertex clique cover problem. We demonstrate that the problem of locating two minimum dominating sets (connected dominating sets or vertex clique covers) with minimum intersection is linearly solvable in interval graphs. Another variation of a combinatorial graph problem is the problem of deciding whether or not there exist two disjoint optimum sets of the combinatorial problem satisfying the given combinatorial property. The corresponding problem for the vertex cover problem is to decide whether or not there exist two disjoint minimum vertex covers of a graph. Notice that this problem does not generalize the normal combinatorial problem since it is just to report the boolean answer whether or not there exist two disjoint optimum sets of the combinatorial problem. Grinstead and Slater [Gr90] have shown that the problem of deciding whether or not there exist two disjoint minimum dominating sets is NP-hard for bipartite graphs. We show that the problem 2
of deciding whether or not there exist two disjoint minimum dominating sets (connected dominating sets or vertex clique covers) is shown to be NP-hard for chordal graphs
2 Preliminaries
Gavril [Ga 74] has shown that a graph G is chordal if and only if it is the intersection graph of a family of subtrees in a tree. The intersection model for a chordal graph can always be chosen so that the nodes of the tree correspond to the maximal cliques of the original graph G. Each vertex of G then corresponds to the subtree comprised of exactly those maximal cliques to which it belongs. We call such an intersection model a clique tree of the chordal graph. G is an interval graph if and only if the maximal cliques of G can be linearly ordered such that, for every vertex v of G, the maximal cliques containing v occur consecutively [Gi 64]. This can also be explained in the clique tree model. Interval graphs are the intersection graphs of a family of subpaths of a path tree [Mo 86]. For a given interval graph G(V; E ) with V = f1; 2; : : : ; ng, it is possible to construct a path (the clique tree) T with the maximal cliques of G as its nodes and P = fpi j i 2 V and pi is a subpath of T containing all the maximal cliques having ig such that the intersection graph of P in T is G [Gi 64, Mo 86]. Note that two paths pu and pv of P intersect each other (share a common node) if and only if u is adjacent to v in G. Let PS denote fpv j v 2 S g for any subset S of V . Here onwards T denotes the clique tree of G and Ci; 0 i q denote the maximal cliques of G which are the nodes of T . Also C0; C1; C2; : : : ; Cq is the order of the nodes of T placed left to right in T . (See Figure 1). We say Ci Cj if and only if i j . (See C0
C1
C2
Cj
Ci
Cq?1 Cq
Figure 1: The nodes C0; C1; C2; : : :; Cq of T are placed left to right in T . Ci Cj if and only if i j . Figure 1). We refer to the vertices of T as nodes and use V (T ) and E (T ) to denote the nodes and the edges of the clique tree T respectively. For every vertex ni, the respective path pni is written as fCl; Cl+1; : : :; Cr g, where the 3
vertex ni lies in the maximal cliques Cl; Cl+1; : : : ; Cr and fCl; Cl+1; : : :; Cr g is a subpath of T . (See Figure 2). For a path pni = fCl; Cl+1; : : :; Cr g, Cr is
pn i C0
C1
C2
Cl
Cl+1
Cr
Cq?1 Cq
Figure 2: Clique tree T of an interval graph. pRni = Cr and pLni = Cl. called the right end of pni and denoted by pRni . Similarly, Cl is called the left end of pni and denoted by pLni . (See Figure 2). A rightmost path of P from a node Ci0 is a path p = fCl; Cl+1; : : :; Cr g of P containing Ci0 whose right end node Cr is the nearest to Cq. (See Figure 3). A leftmost path of P from a node Ci0 is a path p = fCl; Cl+1; : : :; Cr g of P containing Ci0 whose left end node Cl is the nearest to C0.
p C0
C1
C2
Cq?1 Cq
Ci0
Figure 3: p is the rightmost path from Ci0
3 Polynomial algorithm in interval graphs
3.1 Connected Domination Problem
In this section we show how to compute two minimum connected dominating sets with minimum intersection in an interval graph G. The algorithm consists of two phases. In the rst phase, a set M = fM0; M1; M2; : : :; M g of nodes of V (T ) is identi ed with the property that the i-th rightmost path of PS must extend to (or to the right of) point Mi where S is a minimum connected dominating set of G and is the cardinality of a minimum connected dominating set of G. In the second phase the two minimum connected 4
dominating sets for G with minimum intersection are computed using the set M for guidance.
3.1.1 Phase I
It is known that Lemma 3.1.1 [Ke 94, Ma 93] For a chordal graph G, let T be its clique tree and P = fpv j v 2 Vg be the collection of subtrees in T such that its intersection graph is G. A set S = fg1 ; g2 ; : : : ; g g is a connected dominating set of G if and only if PS = fpg1 ; pg2 ; : : :; pg g of P covers E (T ), the edge set of the clique tree T of G. First we will construct a minimum collection fpm1 ; pm1 ; : : : ; pm g of paths from Cq to C0 so that these paths cover E (T ) of T of G. Let pm1 be the leftmost path from Cq . Given pmi?1 , we identify pmi as the leftmost path from pLmi?1 . We continue this process until some pm reaches C0. Then we set Mi = pLm?i for i = 0; 1; 2; : : : ; ? 1;. M is Cq and M0 is C0. (See Figure pm C0
pm?1
pm2
pmi
pm1
pLm
pLm?1
pLmi
pLm2
pLm1
M0
M1
M ?i
M ?2
M ?1
Cq M
Figure 4: After Phase I of the connected domination problem, the clique tree T is marked with M = fM0 ; M1; M2 ; : : :; M g 4). By the construction of the paths pmi , the set S = fm1; m2; : : :; m g is a minimum connected dominating set of G [Ke 94, Ma 93]. Lemma 3.1.2 Let F = ffi : i = 1; 2; : : : ; g be a minimum connected dominating set of G with vertices sorted by the right end points of the paths pfi . Then pRfi Mi for i = 1; 2; : : : ; , where pRfi is the right end of pfi . 5
Proof Suppose pRfi < Mi for some i. By phase I, we know that we need a minimum of ( ? i) paths of P to cover the edges between Mi and M in T . Since pRfi < Mi, the paths pf1 ; pf2 ; : : :; pfi do not cover all the edges between M0 and Mi in T and thus the set ff1; f2; : : :; fig cannot be a subset of a minimum connected dominating set of G.
pa1
pb1 C0
M0
M1
pRb1
M ?1
pRa1
Cq
M
Figure 5: Since pRb1 M1, b1 2 B
3.1.2 Phase II
In this phase, we compute two minimum connected dominating sets A and B of G with minimum intersection as follows: We construct A and B simultaneously by traversing from C0 to Cq over T with M0; M1; : : : ; M marked on T from phase I. A path denoted by paj is included in PA and a path denoted by pbj is included in PB . We construct paj and pbj ,i = 1; 2; : : : ; in such a way that pRaj Mj and pRbj Mj in order to satisfy Lemma 3.1.2. Let pa1 be the rightmost path from M0. By Lemma 3.1.2, pRa1 M1 since it is true that every graph has at least one minimum connected dominating set. Add a1 to A. Find the next rightmost path pb1 (di erent from pa1 ) from M0. If pRb1 M1, add b1 to B (See Figure 5); otherwise b1 = a1. We continue to add vertices to A and B as follows. At stage i, let pai and pbi be the most recently selected paths for PA and PB respectively. Without loss of generality we assume that pRai pRbi . (See Figure 6). Since pRbi pRai , we rst nd a rightmost path pbi+1 from pRbi and add bi+1 to B . (Note that if pRai pRbi , then we will rst construct pai+1 from pRai ). The existence of such pbi+1 satisfying the condition that pRbi+1 Mi+1 is guaranteed by Lemma 3.1.2 since it is true that every graph has at least one minimum connected dominating set. Next nd another rightmost path pai+1 (di erent from pbi+1 ) from pRai . If pRai+1 Mi+1 , then add ai+1 to A (See Figure 7); otherwise ai+1 = bi+1. (See Figure 8). The algorithm terminates when pa and pb both reach M . Now we shall verify the correctness of the above algorithm. 6
pai C0
M0
pbi
Mi
pRbi
pRai
Cq
Mi+1
M
Figure 6: pRai pRbi
pai+1
pai C0
Mi
M0
p bi
pRbi
pRai
Mi+1 pRbi+1
Cq
pRai+1
M
pRbi+1
Cq M
pbi+1 Figure 7: Since pRai+1 Mi+1, ai+1 2 A
pai+1
pai C0
Mi
M0
p bi
pRbi
pRai
pRai+1 Mi+1
pbi+1 Figure 8: Since pRai+1 < Mi+1, ai+1 is not considered for A. 7
Theorem 3.1.1 The sets A and B constructed by the algorithm are mini-
mum connected dominating sets for G with minimum possible intersection. Proof. Lemma 3.1.2 and the construction of the two sets A and B imply that A and B are minimum connected dominating sets for G. It remains to show that A \ B is minimum. To do this we rst de ne the common vector of a pair PQ, PR of sets of paths of P as the vector whose elements contain the right end points of the paths common to both PQ and PR, sorted in increasing order. The theorem then follows from the following claim. Claim: There exist two sets PY and PZ of paths of P such that Y and Z satisfy (i) each is a minimum connected dominating set for G.
(ii) among all pairs of sets satisfying (i), jY \ Z j is minimum. (iii) among all pairs of sets satisfying (i) and (ii), PY and PZ lexicographically maximize their common vector. and such that after i stages of the algorithm fpa1 ; pa2 ; : : :; pai g PY and fpb1 ; pb2 ; : : :; pbi g PZ . Proof of the Claim: Let pyi and pzi be the i-th paths of PY and PZ respectively when the paths of PY and PZ are sorted in increasing order of right end point. We prove the claim by induction on i, the number of stages in the algorithm. After zero stages of the algorithm, A and B are empty and the claim trivially holds. As an inductive assumption, assume the claim holds after i ? 1 stages of the algorithm, that is, py1 = pa1 , py2 = pa2 , : : : , pyi?1 = pai?1 and pz1 = pb1 , pz2 = pb2 , : : : , pzi?1 = pbi?1 . Now consider the i-th elements ai and bi added to the sets A and B respectively. Without loss of generality we may assume pRai?1 pRbi?1 . We proceed by case analysis Case 1 (pyi 6= pzi ): In this case, Lemma 3.1.2 and the construction used in the algorithm imply that pai 6= pbi . If pRzi pRyi , then the construction used in the algorithm implies that pRbi pRzi and pRai pRyi and the inductive assumption assures that PY 0 = fpa1 ; pa2 ; : : :; pai ; pyi+1 ; : : :; py g and PZ0 = fpb1 ; pb2 ; : : : ; pbi ; pzi+1 ; : : :; pz g are the sets required in the claim. The remaining subcase has pRzi < pRyi . If pyi 6= pbi , then the inductive assumption and the construction used in the algorithm ensure that PY 0 = fpa1 ; pa2 ; : : :; pai ; pyi+1 ; : : : ; py g and PZ0 = fpb1 ; pb2 ; : : : ; pbi ; pzi+1 ; : : :; pz g are the sets required in the claim. If pyi = pbi , then the new sets required in the 8
claim are PY 0 = fpa1 ; pa2 ; : : :; pai ; pzi+1 ; : : :; pz g and PZ0 = fpb1 ; pb2 ; : : : ; pbi ; pyi+1 ; : : :; py g. Case 2 (pyi = pzi ): Here if pai = pbi , then the construction and the inductive assumption assure that PY = fpa1 ; pa2 ; : : :; pai?1 ; pbi ; pyi+1 ; : : :; py g and PZ = fpb1 ; pb2 ; : : :; pbi?1 ; pbi ; pzi+1 ; : : :; pz g are the two sets required in the claim. The remaining possibility has pai 6= pbi for this case. We will now show this can not occur. Assume to contrary that it can occur and let i0 be the smallest integer greater than or equal to i such that pRbi0 pRai0 . For i j < i0 the construction used in the algorithm imply the following properties: (a) paj 6= pbj (b) pRyj pRbj < pRaj and pRzj pRbj < pRaj Consider the two sets PY 0 = fpa1 ; pa2 ; : : :; pai?1 ; pai ; : : :; pai0 ?1 ; pbi0 ; pyi0 +1 ; : : :; py g and PZ0 = fpb1 ; pb2 ; : : : ; pbi?1 ; pbi ; : : : ; pbi0 ?1 ; pbi0 ; pzi0 +1 ; : : :; pz g. The inductive assumption implies that PY 0 and PZ0 satisfy properties (i) and (ii) of the claim; And properties (a) and (b) above imply that PY 0 and PZ0 have a lexicographically larger common vector than PY and PZ . This contradicts the inductive assumption. Hence the subcase pai 6= pbi cannot occur.
3.1.3 Time complexity
In the rst phase, the algorithm constructs the set M = fM0; M1; M2; : : : ; M g by traversing from Cq to C0. During the traversal, each path is visited at most once. Hence the rst phase runs in linear time. In the second phase, the two minimum connected dominating sets A and B with minimum intersection are computed. The algorithm traverses from C0 to Cq visiting each path at most once. Thus the second phase also runs in linear time.
3.2 Domination Problem
We now consider the problem of computing two minimum dominating sets with minimum intersection in interval graphs. For any path pni = fCl; Cl+1; : : : ; Crg of P on T , de ne the right head of pni , denoted by hR(pni ), as the node Ck0 of T , where k0 = min fk j p0 = fCj ; Cj+1; : : : ; Ck g such that j r + 1g p0 2P 9
(See Figure 9). In a similar way, one can de ne the left head of pni , denoted by hL(pni ), as the node Cj0 , where 0 = fC ; C ; : : : ; C g such that k l ? 1g j 0 = max f j j p j j +1 k 0 p 2P
p0 p0
pn i C0
Cl
p0
Cr Cr+1
Ck0
Cq
Figure 9: hR(pni ) = Ck0 in the domination problem The algorithm for the domination problem is very similar to the algorithm for the connected domination problem. The only di erence between the two algorithms is that the right end (left end) of a path in the connected domination algorithm is replaced by the right head (left head) of the path in the domination algorithm.
3.2.1 Phase I A set S = fg1; g2; : : :; g g is a dominating set of an interval graph G if and only if every path of P , not in PS = fpg1 ; pg2 ; : : :; pg g, is intersected by at
least one path of PS . Let denote the cardinality of a minimum dominating set of G. We construct such a minimum collection of paths from Cq to C0 as follows: Let pm1 be the leftmost path from Cq . Then select a leftmost path from hL(pm1 ) and call it pm2 . Given pmi?1 , we identify pmi as the leftmost path from hL(pmi?1 ). (See Figure 10). We continue this process until some pm reaches C0. The above constructed collection S = fm1; m2; : : : ; m g is a minimum dominating set of G [Ke 94, Ma 93]. Then we set Mi = hL(pm ?i ) for i = 0; 1; 2; : : : ; ? 1. M is Cq and M0 is C0. (See Figure 11). The following lemma will be used in the proof of Lemma 3.2.2. Lemma 3.2.1 Let D = fd1; d2; : : : ; d g be a minimum dominating set of G such that the corresponding paths of D are ordered from Cq to C0 (that is, 10
pm ?1
pm C0
hL (pm ) hL (pm ?1 )
pmi hL (pmi?1 )
hL (pmi )
pLm ?1
pm1 hL (pm1 ) pLm1
pLmi?1
pLmi
Cq
Figure 10: pmi is the leftmost path from hL(pmi?1 ). C0
hL (pm )
hL (pm ?1 )
hL (pm ?i )
hL (pm1 )
M0
M1
Mi
M ?1
Cq
M
Figure 11: After Phase I of the domination problem, the clique tree T is marked with M = fM0; M1; M2 ; : : : ; M g the path pd1 originates at Cq , then follows pd2 and nally pd terminates at C0). Then hL(pdi ) hL(pmi ) for i = 1; 2; : : : ; , where hL (pmi ) = M ?i and hL(pdi ) is the left head of pdi . (See Figure 12).
Proof The proof follows from the greedy construction of fpm1 ; pm2 ; : : : ; pm g [Ke 94, Ma 93]. 2 C0
hL (pm ) M0
hL(pmi ) hL (pdi ) M ?i
hL (pm1 ) hL(pd1 ) M ?1
Cq
M
Figure 12: hL (pdi ) hL(pmi )
Lemma 3.2.2 Let F = ff1; f2; : : :; f g be a minimum dominating set of G with vertices sorted by the right end points of the paths pfi . Then
pRfi Mi for i = 1; 2; : : : ; , where pRfi is the right end of pfi .
11
Proof. Suppose pRfi < Mi for some i. While traversing from Cq to C0 in
phase I of this section, there exists a path p0 lying between pLm ?i and Mi and not intersecting the path pm ?i . (See Figure 13 and Figure 9). Since pRfi < Mi, the paths fpf1 ; pf2 ; : : : ; pfi g do not intersect p0. By our construction, the paths fpm1 ; pm2 ; : : :; pm ?i g between Mi and M do not intersect pm ?i+1
pm C0
M0
pf1
pfi
p0 Mi
pm1
pm ?i pLm ?i
Mi+1
Cq M
Figure 13: Here pRfi < Mi and pfi does not intersect p0.
p0. fpm1 ; pm2 ; : : : ; pm ?i g are the most optimal paths between Mi and M by Lemma 3.2.1. Thus pfi must intersect p0 if ff1; f2; : : : ; fig is a subset of a minimum dominating set of G. Since pfi do not intersect p0, ff1; f2; : : :; fig cannot be a subset of any minimum dominating set of G. This is a contradiction. 2
3.2.2 Phase II
We give a sketch of the algorithm to compute two minimum dominating sets A and B for G with minimum intersection. Given pai and pbi , the most recently selected paths for PA and PB respectively, we assume without loss of generality that pRai pRbi . Now nd a rightmost path pbi+1 from hR(pbi ) and add bi+1 to B . The existence of such pbi+1 with the condition that pRbi+1 Mi+1 is guaranteed by Lemma 3.2.2 since it is true that every graph has at least one minimum dominating set. Then nd the next rightmost path pai+1 (di erent from pbi+1 ) from hR(pai ). If pRai+1 Mi+1 , add ai+1 to A; otherwise ai+1 = bi+1. The algorithm terminates at Cq. The proof of correctness is similar to the proof for the connected domination algorithm. Furthermore, it is obvious that the algorithm runs in linear time. 12
3.3 Vertex clique cover problem
Lemma 3.3.1 [Ke 94, Ma 93] For a chordal graph G, let T be its clique tree and P = fpv j v 2 Vg be the collection of subtrees in T such that its intersection graph is G. A set S = fg1 ; g2 ; : : : ; g g is a vertex clique cover of G if and only if PS = fpg1 ; pg2 ; : : :; pg g of P covers V (T ), the vertex set of
the clique tree T of G. For this problem, for any path pni = fCl; Cl+1; : : :; Cr g of P , we de ne the right head of pni , denoted by hR(pni ), as Cr+1 and the left head of pni , denoted by hL(pni ), as Cl?1. The algorithm for nding two minimum vertex clique covers with minimum intersection thus proceeds exactly as the algorithm for nding two minimum dominating sets with minimum intersection.
x1 x3 uij
x5
x6
uij
x4 x2 Figure 14: The H subgraph of G with designated vertices uij and uij .
4 NP-hard in chordal graphs In this section we show that the problem of deciding whether or not there exist two disjoint minimum total dominating sets is NP-hard for bipartite 13
graphs. We transform this problem to the corresponding connected domination problem on chordal graphs and prove that the problem of deciding whether or not there exist two disjoint minimum connected dominating sets is NP-hard for chordal graphs. Using this result, we prove that the problem of deciding whether or not there exist two disjoint minimum dominating sets (or vertex clique covers) is NP-hard for chordal graphs.
4.1 Disjoint Total Domination Problem in bipartite graphs is NP-hard
A dominating set S is a total dominating set if the subgraph induced by S has no isolates. Grinstead and Slater [Gr 90] have shown that the problem of deciding whether or not there exist two disjoint minimum dominating sets is NP-hard for bipartite graphs. We modify their construction to prove that the problem of deciding whether or not there exist two disjoint minimum total dominating sets is NP-hard for bipartite graphs. We use the same notation and terminology of [Gr 90]. We construct a bipartite graph G from a set U = fu1; u2; : : : ; ung of variables and a collection C = fc1 ^ c2 ^ : : : ^ cmg of clauses over U . Here ci = (si1 _ si2 _ si3) and each sij is uh or uh for some 1 h n. For 1 i m and 1 h n, uih is uh and uih is uh . If sij is uh, then sij is uih. If sij is uh , then sij is uih. For example, if ci = (u2 _ u4 _ u5), then si1 = ui2 = u2, si2 = ui4 = u4, si3 = ui5 = u5. For every sij , 1 i m, 1 j 3, we construct a subgraph H which forms a building block for G. Notice that the H subgraph is always referred to uih , where sij = uih. The H subgraph is given in Figure 14. The graph G will contain 3m copies of the graph H . We need to note that H is bipartite with vertices labeled uij and uij in the same set of the bipartition and the only vertices in H adjacent to other vertices of G will be uij and uij . fx1; x2; x5; x6g and fuij ; uij ; x3; x4g form the bipartition of H . Each copy of H in G will be called an H -subgraph with designated vertices uij and uij . Any minimum total dominating set of H must contain at least one of uij and uij . Any minimum total dominating set of H contains exactly two vertices. The possible pairs are fuij ; x5g, fuij ; x6g, fuij ; x5g and fuij ; x6g. If H has two disjoint minimum total dominating sets D1 and D2, then there are two possibilities. If D1 contains fuij ; x5g, then D2 will contain fuij ; x6g. If D1 contains fuij ; x6g, then D2 will contain fuij ; x5g. 14
Most importantly for each uij , if one minimum total dominating set of H contains uij , the other minimum total dominating set of H must contain uij . For each clause ci = (si1 _ si2 _ si3), we construct a graph Gi as follows. (See Figure 15). Suppose si1 = ua or ua, si2 = ub or ub, and si3 = uc or uc.
b1i
ui1
H
b2i
ui1 ui2
b3i
H
ui2 ui5
b4i
H
ui5
Figure 15: A subgraph Gi of G for ci = (u1 _ u2 _ u5). Let Gi contain three copies of H with designated vertices uia and uia, uib and uib, and uic and uic. Each of b1i and b2i is connected to uia if si1 = ua, to uia if si1 = ua, to uib if si2 = ub, to uib if si2 = ub, to uic if si3 = uc, and to uic if si3 = uc. Each of b3i and b4i is connected to the three designated vertices not adjacent to b1i and b2i. Let G be the graph containing disjoint copies of G1; G2; : : : ; Gm to which we add the following vertices and edges. For each occurance of a uh or uh with 1 h n in distinct clauses ci and cj , add four vertices of degree 2 as follows. Assume sir is uh or uh and assume sjt is uh or uh where 1 i < j m and 1 r; t 3. Let two of the four vertices be adjacent to uih and to ujh , and let the other two be adjacent to uih and to ujh. The graph G is illustrated in Figure 16 for C = (u1 _ u2 _ u5) ^ (u2 _ u3 _ u4). We derive the following information from G. G is a bipartite graph. Any minimum total dominating set of G must contain at least one of uij and uij . Any minimum total dominating set of G contains exactly two vertices of H . The possible pairs are fuij ; x5g, fuij ; x6g, fuij ; x5g and fuij ; x6g. If G has two disjoint minimum total dominating sets D1 and D2, then there are two possibilities in H . If D1 contains fuij ; x5g, then D2 will contain fuij ; x6g. If D1 contains fuij ; x6g, then D2 will contain fuij ; x5g. Most importantly, 15
u11
b11
b21
H u11
u12
u23
x
u22
H u22
b31
b41
H u12
u15
H u15
H u23
u24
H u24
y
b12 b22 b32 b42 Figure 16: Graph G for C = (u1 _ u2 _ u5) ^ (u2 _ u3 _ u4).
16
if one minimum total dominating set of G contains uij , the other minimum total dominating set of G must contain uij . The NOT-ALL-EQUAL 3SAT problem [Ga 79] is a 3SAT problem with an additional condition that each clause has at least one true literal and at least one false literal. A NOT-ALL-EQUAL 3SAT truth assignment is a truth assignment for U such that each clause in C has at least one true literal and at least one false literal. The NOT-ALL-EQUAL 3SAT problem is NP-complete [Ga 79]. Now we will prove that G has two disjoint minimum total dominating sets if and only if U has a NOT-ALL-EQUAL 3SAT truth assignment for C . Suppose there is a NOT-ALL-EQUAL 3SAT truth assignment for C . Let S1 consist of those ui in U that receive the value true and S2 = U ? S1. Construct two vertex subsets D1 and D2 of V (G) as follows. For each uij in G if uj 2 S1 then add uij and x5 of H to D1 and add uij and x6 of H to D2, and if uj 2 S2, then add uij and x5 of H to D2 and add uij and x6 of H to D1. (For example, consider C = (u1 _ u2 _ u5) ^ (u2 _ u3 _ u4). One NOTALL-EQUAL 3SAT truth assignment is S1 = fu1; u4g. Then D1 contains the six darkened vertices u11; u12; u15; u22; u23 and u24 with one additional vertex from each H . D2 contains the six undarkened vertices u11; u12; u15; u22; u23 and u24 with one additional vertex from each H ). First we prove that D1 is a dominating set of G. Since D1 contains either uij and x5 or uij and x6, it dominates the remaining vertices of H . Each clause has one true literal and one false literal. The vertex of D1 corresponding to the true literal dominates b1i and b2i, and the vertex of D1 corresponding to the false literal dominates b3i and b4i. For each occurance of a uh or uh in distinct clauses ci and cj , four vertices of degree 2 were added to G. If uik and ujk are vertices of G with i 6= j , then both are in D1 or both are in D2. Thus the four vertices of degree 2 are dominated by D1 . In the same way, we can prove that D2 is also a dominating set of G. It is easy to verify that D1 and D2 are total dominating sets of G. As we mentioned earlier, every minimum total dominating set of G must contain at least two vertices of each H . Since D1 and D2 contain exactly two vertices of each H and both are total dominating sets of G, D1 and D2 are minimum total dominating sets of G. Also D1 and D2 are disjoint. Hence, G has two disjoint minimum total dominating sets. To prove the converse, suppose G has two disjoint minimum total dominating sets, say D1 and D2. As we noted earlier, for each uij of G, either D1 or D2 contains uij and a corresponding vertex x5 (or x6) from its subgraph 17
H , and the other contains uij and a corresponding vertex x6 (or x5) from the same subgraph H . Without loss of generality, let us assume that one of D1 and D2 contains uij and x5, and the other contains uij and x6. Suppose uik and ujk are vertices of G with i 6= j (for example, u12 and u22 in Figure 16). To see that uik and ujk can not both be in D1 or both be in D2, note that if uik and ujk are in D1 ( and so uik and ujk are in D2 ), then D1 must also contain the two vertices x and y of degree 2 which are adjacent to uik and ujk (for example, vertices x and y in Figure 16). But the set D1 [ fuik g n fx; yg is also a total dominating set which contradicts the minimality of D1. Consequently, if uik and ujk are vertices of G with i 6= j , then both are in D1 or both are in D2. Therefore, the following is a well de ned truth assignment for U . For each uk 2 U , nd a uik in G, and let uk be true if uik 2 D1 and false if uik 2 D2. It remains only to show that each clause ci has at least one true literal and at least one false literal. If not, we can assume clause ci has three true literals. Then each of b3i and b4i is not dominated by D1, so b3i and b4i are in D1. Let z be one of the vertices adjacent to b3i and b4i. Then z is some uil or uil. We can verify that D1 [ fzg n fb3i; b4ig is also a total dominating set. This contradicts the minimality of D1. It follows that U has a NOT-ALL-EQUAL 3SAT truth assignment for C . Since the NOT-ALL-EQUAL 3SAT problem is NP-complete, we can state that Theorem 4.1.1 The problem of deciding whether or not there exist two disjoint minimum total dominating sets is NP-hard for bipartite graphs.
4.2 Disjoint Connected Domination Problem in chordal graphs is NP-hard
A vertex v of G is simplicial if the set of vertices adjacent to v in G is completely connected. Now we transform the problem of deciding whether or not there exist two disjoint minimum total dominating sets in bipartite graphs to the problem of deciding whether or not there exist two disjoint minimum connected dominating sets in chordal graphs. Let G(V1 [ V2; E ) be any bipartite graph. Let V1 = fvj j j = 1; 2; : : : ; ng, V2 = fui j i = 1; 2; : : : ; mg. We construct a chordal graph G0(V 0; E 0) from G(V1 [ V2 ; E ) as follows: The vertex set V 0 = fvj ; vj0 j vj 2 V1 ; j = 1; 2; : : : ; ng [ fui; u0i j ui 2 18
v1
u1
v2
u2
v3
Figure 17: A bipartite graph G(V1 [ V2 ; E )
V2; i = 1; 2; : : : ; mg. The edge set E 0 of G0 is de ned as follows. (vj ; u0i), (vj0 ; ui) 2 E 0 if (vj ; ui) 2 E . Also any two vertices of V1 [ V2 form an edge in E 0 so that V1 [ V2 induces a completely connected subgraph in G0. (See Figure 17 and Figure 18.) v1 u01 u02
u1
v2 u2
v3
v10 v20 v30
Figure 18: The corresponding chordal graph G0(V 0; E 0)
Claim 4.2.1 If a subset S of V 0 is a minimum connected dominating set of G0, then S V1 [ V2. Proof of the Claim Suppose S has some u0i of V 0. Since S is connected, S has some vj of V1 which is adjacent to u0i in G0. If u0i and vj are adjacent in G0, then N [u0i] N [vj ], where N [v] = fvg [ fu j u is adjacent to vg. That is, any u0i or vj0 is a simplicial vertex of G0. Thus S n fu0ig is still a connected dominating set of G0 which contradicts the minimality of S . The same arguments hold if S has any vj0 of V 0. Hence S V1 [ V2. 2 Claim 4.2.2 A set S of vertices is a minimum total dominating set of G if and only if S is a minimum connected dominating set of G0 . 19
Proof of the Claim Let and be the cardinality of a minimum total
dominating set of G and a minimum connected dominating set of G0 respectively. Suppose S is a minimum total dominating set of G. We rst prove that S is a dominating set of G0. It is true that S V1 [ V2. Since V1 [ V2 induces a completely connected subgraph of G0, S dominates V1 [ V2 in G0. Let u0i be an arbitrary vertex of G0. Since S is a total dominating set of G, the vertex ui of G is dominated by some vertex of S in G and thus u0i is dominated by the same vertex of S in G0. In the same way, the vertices of the form vj0 in G0 are dominated by S . So S is a dominating set of G0. Since S is a subset of the completely connected subgraph V1 [ V2 in G0, S is connected and hence S is a connected dominating set of G0. Therefore, . Suppose S is a minimum connected dominating set of G0. By Claim 4.2.1, S V1 [ V2. We rst prove that S is a total dominating set of G. Let ui be an arbitrary vertex of G. Since S is a dominating set of G0 and S V1 [ V2, u0i is dominated by some vertex vj of S in G0. Since u0i and vj are adjacent in G0, ui and vj are adjacent in G. Thus the vertices of V2 are dominated by S in G. In the same way, we can show that the vertices of V1 are dominated by S in G. Hence S is a total dominating set of G. Therefore . Hence the claim. 2 It is easy to observe from Claim 4.2.2 that Claim 4.2.3 S1 and S2 are disjoint minimum total dominating sets of G if and only if S1 and S2 are disjoint minimum connected dominating sets of G0 . Since the problem of deciding whether or not there exist two disjoint minimum total dominating sets is NP-hard for bipartite graphs, we can state that Theorem 4.2.1 The problem of deciding whether or not there exist two disjoint minimum connected dominating sets is NP-hard for chordal graphs. Note that the above arguments do not immediately follow for the domination and vertex clique cover problems. We use Theorem 4.2.1 to prove that the problem of deciding whether or not there exist two disjoint minimum dominating sets (or vertex clique covers) is NP-hard for chordal graphs. 20
4.3 Disjoint Domination Problem in chordal graphs is NP-hard
We transform the connected domination problem on a chordal graph G to the domination problem (or vertex clique cover problem) on a similar chordal graph G0. To de ne G0, we construct a clique tree T 0 from the clique tree T of G and a family P 0 of subtrees in T 0. For every edge (Ci; Cj ) of T , we introduce two new nodes between the nodes Ci and Cj to construct T 0. Let Cij and Cji be the new nodes introduced to the edge (Ci; Cj ) of T where Cij is the node close to Ci and Cji is the node close to Cj . Each edge (Ci; Cj ) of a subtree pv of P in T is replaced by the path segment fCi; Cij ; Cji; Cj g to construct the corresponding subtree p0v of P 0 in T 0. For each edge (Ci; Cj ) of T , let pvij and pvji be the subtrees containing the single node Cij and Cji respectively. (See Figure 19). These subtrees are added to the family P 0 of subtrees in T 0. Thus to de ne G0, the family P = fpv j v 2 V g of Ci
Cj
Ci
pvij
pvji
Cij
Cji
Cj
T0
T
Figure 19: subtrees in the intersection model for G is augmented to P 0 = fp0v j pv 2 P and v 2 V g [ fpvij , pvji j (Ci; Cj ) is an edge in Tg. The intersection graph of the subtrees of P 0 in T 0 is G0. The vertex set of G0 is V 0 = V [ U where V denotes the set of vertices corresponding to the subtrees fp0v j pv 2 P and v 2 V g and U denotes the set of vertices corresponding to the subtrees fpvij , pvji j (Ci; Cj ) is an edge in Tg. In other words, V 0 = V [ fvij ; vji j (Ci; Cj ) is an edge in T g. When G is a chordal graph, G0 is also a chordal graph. Let PS0 denote fp0v 2 P 0 j v 2 S g. The following lemmas are useful to study the relation between the connected domination problem and domination problem in chordal graphs. Claim 4.3.1 If S is a minimum dominating set (or vertex clique cover) of G0, then S V . 21
Proof of the Claim Any vertex vij or vji of U is a simplicial vertex of G0
since it is in exactly one maximal clique of G0. Suppose S contains some vertex of U , say vij . Then S does not contain any vertices adjacent to vij . Otherwise we will arrive at a contradiction to the minimality of S by removing vij from S . We observe that the set of vertices adjacent to vij is the same as the set of vertices adjacent to vji in G0. Since S contains vij and S does not contain any vertices adjacent to vij , S must have vji. Otherwise it will contradict the fact that S is a dominating set of G0. Thus S has both vij and vji of G0. Since G0 is a connected graph, every edge of its clique tree T 0 is covered by a subtree of P 0. Let p0u be the subtree of P 0 which covers the edge (Ci; Cj ). Then the vertex u of G0 is adjacent to both vij and vji. Replace vij and vji of S by u to form a set S 1. Since S 1 is still a dominating set of G0, it contradicts the fact that S is a minimum dominating set of G0. Thus S V. 2 Claim 4.3.2 S is a minimum connected dominating set of G if and only if S is a minimum dominating set (or vertex clique cover) of G0. Proof of the Claim Let and be the cardinality of a minimum connected dominating set of G and a minimum dominating set of G0 respectively. Suppose S is a minimum connected dominating set of G. By Lemma 3.1.1, PS covers all the edges of T . That is, PS0 covers all the edges of T 0 and hence the subtrees of PS0 intersect all other subtrees of P 0. Note that two subtrees pu and pv of P intersect each other if and only if u is adjacent to v in G. Thus S dominates the vertices of V n S of G0 and the vertices of U of G0. Hence S is a dominating set of G0. Therefore, . Suppose S is a minimum dominating set of G0. By Claim 4.3.1, S V . Let e = (Ci; Cj ) be an edge of T . Since S dominates U , the subtrees pvij and pvji are intersected by a subtree p0u of PS0 in T 0. Thus the subtree p0u contains the path segment fCi; Cij ; Cji; Cj g of T 0. Therefore pu of PS contains the edge (Ci; Cj ) of T . This shows that the edges of T are covered by PS . By Lemma 3.1.1, S is a connected dominating set of G. Therefore, . The corresponding result for the vertex clique cover can be derived in the same way using Lemma 3.3.1. 2 Claim 4.3.3 S1 and S2 are disjoint minimum connected dominating sets of G if and only if S1 and S2 are disjoint minimum dominating sets (or vertex clique covers) of G0 . 22
As a consequence of Claim 4.3.3 and Theorem 4.2.1, we conclude the following result. Theorem 4.3.1 The problem of deciding whether or not there exist two disjoint minimum dominating sets (or vertex clique covers) is NP-hard for chordal graphs.
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