This section first demonstrates Garey and Johnson’s proof that the domination problem is NP-complete. The proof is adapted for split graphs, which are special chordal graphs. A split graph is a graph whose vertex set is the disjoint union of a clique C and a stable set S. Notice that a split graph is chordal as the ordering with the vertices in S first and the vertices in C next gives a perfect elimination ordering.
Theorem 5.2 The domination problem is NP-complete for split graphs.
Proof. The proof is given by transforming the vertex cover problem in general graphs to the domination problem in split graphs. Given a graph G = (V, E), construct the graph G′= (V′, E′) with
vertex set V′ = V ∪ E and
edge set E′= {v1v2: v16= v2in V } ∪ {ve : v ∈ e}.
Notice that G′ is a split graph whose vertex set V′ is the disjoint union of the clique V and the stable set E.
If G has a vertex cover C of size at most k, then C is a dominating set of G′ of size at most k, by the definition of G′. On the other hand, suppose G′ has a
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Figure 9: A transformation to a split graph.
dominating set D of size at most k. If D contains any e ∈ E, say e = uv, then replace e with u to get a new dominating set of size at most k. In this way, assume that D is a subset of V . It is then clear that D is a vertex cover of G of size at most k.
Since the vertex cover problem is NP-complete, the domination problem is also NP-complete for split graphs.
Note that the dominating set of G′ in the proof above in fact induces a connected subgraph.
Corollary 5.3 The total and the connected domination problems are NP-complete for split graphs.
In fact, the proof can be modified to get
Theorem 5.4 The domination problem is NP-complete for bipartite graphs.
Proof. The vertex cover problem in general graphs is transformed to the dom-ination problem in bipartite graphs as follows. Given a graph G = (V, E), construct the graph G′= (V′, E′) with
vertex set V′ = {x, y} ∪ V ∪ E and
edge set E′= {xy} ∪ {yv : v ∈ V } ∪ {ve : v ∈ e}.
Notice that G′ is a bipartite graph whose vertex set V′ is the disjoint union of two stable sets {x} ∪ V and {y} ∪ E.
If G has a vertex cover C of size at most k, then {y} ∪ C is a dominating set of G′ of size at most k + 1. On the other hand, suppose G′ has a dominating set D of size at most k + 1. Since NG′[x] = {x, y}, D must contain x or y. One may assume that D contains y but not x, as (D \ {x}) ∪ {y} is also a dominating set of size at most k + 1. Since y ∈ D, if D contains any e ∈ E, say e = uv, one can replace e with u to get a new dominating set of size at most k + 1. In this way, one may assume that D \ {y} is a subset of V . It is then clear that D \ {y}
is a vertex cover of G of size at most k.
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Figure 10: A transformation to a bipartite graph.
Since the vertex cover problem is NP-complete, the domination problem is NP-complete for bipartite graphs.
Corollary 5.5 The total and the connected domination problems are NP-complete for bipartite graphs.
There are many other NP-complete results for variations of domination, see [13, 16, 26, 68, 78, 82, 98, 116, 128, 166, 203, 204] and [101, 133, 134]. Most of the proofs are more or less similar to the above two. Only very few are proved by using different methods. As an example, Booth and Johnson [26] proved that the domination problem is NP-complete for undirected path graphs, which is another subclass of chordal graphs, by reducing the 3-dimensional matching problem to it.
Theorem 5.6 The domination problem is NP-complete for undirected path graphs.
Proof. Consider an instance of the 3-dimensional matching problem, in which there are three disjoint sets W , X and Y each of size q and a subset
M = {mi= (wr, xs, yt) : wr∈ W, xs∈ X and yt∈ Y for 1 ≤ i ≤ p}
of W × X × Y having size p. The problem is to find a subset M′ of M having size exactly q such that each wr ∈ W , xs ∈ X and yt ∈ Y occurs in precisely one triple of M′.
Given an instance of the 3-dimensional matching problem, construct a tree T having 6p+3q+1 vertices from which an undirected path graph G is obtained.
The vertices of the tree, which are represented by sets, are explained below.
For each triple mi ∈ M there are six vertices depend only upon the triple itself and not upon the elements within the triple:
{Ai, Bi, Ci, Di} {Ai, Bi, Di, Fi}
{Ci, Di, Gi} {Ai, Bi, Ei} {Ai, Ei, Hi}
{Bi, Ei, Ii} for 1 ≤ i ≤ p.
These six vertices form the subtree of T corresponding to mi, which is illustrated in Figure 11. Next, there is a vertex for each element of W , X and Y that depends upon the triples of M to which each respective element belongs:
{Rr} ∪ {Ai: wr∈ mi} for wr∈ W , {Ss} ∪ {Bi: xs∈ mi} for xs∈ X, {Tt} ∪ {Ci: yt∈ mi} for yt∈ Y .
Finally, {Ai, Bi, Ci : 1 ≤ i ≤ p} is the last vertex of T . The arrangement of these vertices in the tree T is shown in Figure 11. This then results in an undirected path graph G with vertex set
{Ai, Bi, Ci, Di, Ei, Fi, Gi, Hi, Ii: 1 ≤ i ≤ p} ∪ {Rj, Sj, Tj: 1 ≤ j ≤ q}
of size 9p + 3q, where the undirected path in T corresponding to a vertex v of G consists of those vertices (sets) containing v in the tree T .
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{Ai, Ei, Hi} {Bi, Ei, Ii} for mi∈ M
{Rr}S
{Ai: wr ∈ mi} {Bi: xs∈ mi} {Ss}S
{Tt}S {Ci: yt∈ mi}
for yt∈ Y for xs∈ X
for wr∈ W
tree T
. . . . . . . .
. . . .
Figure 11: A transformation to an undirected path graph.
The theorem then follows from the claim that G has a dominating set of size 2p + q if and only if the 3-dimensional matching problem has a solution.
Suppose D is a dominating set of G of size 2p + q. Observe that for any i, the only way to dominate the vertex set {Ai, Bi, Ci, Ci, Di, Ei, Fi, Gi, Hi, Ii} corresponding to mi with two vertices is to choose Di and Ei, and that any larger dominating set might just as well consist of Ai, Bi and Ci, since none of the other possible vertices dominate any vertex outside of the set. Consequently,
D consists of Ai, Bi and Ci for t mi’s, and Di and Ei for p − t other mi’s, and at least max{3(q − t), 0} Rr, Ss, Tt. Then,
2p + q = |D| ≥ 3t + 2(p − t) + 3(q − t) = 2p + 3q − 2t
and so t ≥ q. Picking q triples mi for which Ai, Bi and Ci are in D form a matching M′ of size q.
Conversely, suppose the 3-dimensional matching problem has a solution M′ of size q. Let
D = {Ai, Bi, Ci: mi∈ M′} ∪ {Di, Ei: mi∈ M \ M′}.
It is straightforward to check that D is a dominating set of G of size 3q + 2(p − q) = 2p + q.