On Disjoint Maximal Independent Sets in Graphs
Oliver Schaudt
Institut für Informatik
Universität zu Köln
Weyertal 80, 50931 Cologne, Germany
[email protected]
February 26, 2014
Abstract
An old problem in graph theory is to characterize the graphs that
admit two disjoint maximal independent sets.
In this paper, we characterize the class of graphs for which every induced subgraph admits two disjoint maximal independent sets. Moreover,
we state a polynomial-time algorithm which, for an arbitrary input graph
G, either constructs two disjoint maximal independent sets or puts out an
induced subgraph of G from the list of forbidden induced subgraphs.
As a second result, we show the following: the computation of two
disjoint maximal independent sets of minimum total size, NP-hard for
general graphs, can be done in linear time on trees.
keywords: independent set, dominating set, forbidden induced subgraph
MSC: 05C69, 05C85, 05C75, 05C05
1
Introduction
An independent set in a graph is a vertex subset whose vertices are mutually
non-adjacent. Independence is one of the most fundamental concepts in graph
theory. Its structural and algorithmical properties are studied in numerous
scientific papers. Usually one is interested in maximal independent sets, that
is, independent sets such that no further vertex can be added without violating
the property of independence. One of the important properties of maximal
independent sets is that they are dominating sets, i.e., every vertex outside of
the set has at least one neighbor in the set.
In many applications of graph theory, one is interested in a robustness against
vertex failure. For example, consider an application where a maximal independent set is used, say to serve a network modelled by a graph. When the vertices
of the independent set have to be reloaded or replaced, one needs a second maximal independent set at hand (disjoint to the first set) to be able to still serve
the network. From this consideration it is a natural question whether a given
graph has two maximal independent sets that are disjoint. Fig. 1 shows a graph
with two disjoint maximal independent sets.
One of the first to consider the problem of the existence of two disjoint
maximal independent sets was Berge. From there on it was studied by several
1
Figure 1: Two disjoint maximal independent sets (in black resp. grey).
Figure 2: The coronas of the chordless cycles on 3 and on 5 vertices.
graph theorists, among others were Erdös et al. [1], Payan [2], and Cockayne
et al. [3]. Very recently, the problem was rediscovered by Hedetniemi et al. [4],
Henning et al. [5], and Raczek et al. [6].
From the viewpoint of computational complexity, the problem is interesting
for its difficulty. In fact, Henning et al. [5] proved that the decision problem
whether there are two disjoint maximal independent sets in a graph is NPcomplete. Moreover, Raczek et al. [6] announced that the problem remains NPcomplete even when restricted to graphs of maximum degree four. Although
the concept of disjoint maximal independent sets is studied in the literature
by several authors, many questions remain open and some of them seem quite
central.
Since the decision problem is NP-complete, Hedetniemi et al. [4] asked for
sufficient conditions for the existence of two disjoint maximal independent sets.
Obviously, every bipartite graph without isolated vertices has two disjoint maximal independent sets. In a sequence of papers it was discovered that regular
graphs of low or high degree always admit two disjoint maximal independent
sets (see [1] for a brief history). Apparently, there are no further sufficient
conditions known by now.
Our main result in this paper is a sufficient condition in terms of forbidden
induced subgraphs. These forbidden subgraphs are defined as follows.
For a graph G we define its corona to be the graph obtained from G by
attaching a pendant vertex to every vertex of G. For example, Fig. 2 shows the
coronas of the chordless cycles of length 3 and 5. In the following, a cycle of
odd length is called an odd cycle. As observed by Raczek et al. [6], coronas of
chordless odd cycles do not admit two disjoint maximal independent sets.
Observation 1 (Raczek et al. [6]). Any corona of a chordless odd cycle does
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not admit two disjoint maximal independent sets.
1.1
Our contribution
We call a graph isolate-free if it does not have isolated vertices. In the following,
we will often restrict our attention to isolate-free graphs. Note that this is not a
real restriction, since a graph with an isolated vertex can never have two disjoint
maximal independent sets.
We show that the absence of coronas of chordless odd cycles is sufficient for
the existence of two disjoint maximal independent sets. This yields the following
characterization.
Theorem 1. Let G be a graph. The following assertions are equivalent.
1. Every isolate-free induced subgraph of G has two disjoint maximal independent sets.
2. G does not contain the corona of a chordless odd cycle as an induced
subgraph.
In fact, we obtain Theorem 1 from the following stronger result. For a graph
G and a vertex set X ⊆ V (G) let G[X] denote the subgraph induced by X.
Theorem 2. There is a polynomial time algorithm that computes, for a given
isolate-free graph G, two disjoint maximal independent sets of G or finds a set
Z ⊆ V such that G[Z] is the corona of a chordless odd cycle.
Note that the algorithm mentioned in the statement of Theorem 2 is robust
in the sense that it can deal with arbitrary isolate-free input graphs. This
seems to be an important feature to us, as it is apparently not known whether
the recognition of graphs not containing the corona of a chordless odd cycle as
induced subgraph can be done in polynomial time.
The proof of Theorems 1 and 2, and some consequences, are given in Section 2. Let us mention that the main idea leading to the two theorems above is
to translate the problem to a problem of total dominating sets.
We then turn to the optimization aspect of the problem. Hedetniemi et al.
[4] define the parameter ii(G) for a graph G to be the least k such that there
are two disjoint maximal independent sets I and J of G with |I| + |J| ≤ k.
Of course, ii(G) is only defined for graphs G that admit two disjoint maximal
independent sets. This parameter was also studied by Henning et al. [5], who
prove that it is NP-complete to decide whether ii(G) ≤ k for given G and k.
Given this hardness result, it is interesting to know for which classes of
graphs ii(G) is computable in polynomial time. In Section 3 we show that, for
trees, ii(G) equals the minimum size of a 2-tuple dominating set. Using the fact
that a minimum 2-tuple dominating set is computed in linear time for trees (cf.
Liao et al. [7]), we derive our next result.
Theorem 3. There is a linear time algorithm to construct, for every given tree
on at least 2 vertices, two disjoint maximal independent sets of total size ii(G).
In particular, ii(G) can be computed in linear time in the class of trees.
We close the paper with a short discussion, giving some questions for future
research on this topic.
3
2
The proof of Theorem 2
In order to give the proof of our main results, we need to recall the definition of a
total dominating set. This is a dominating set whose induced subgraph is isolatefree. The results of this section are based on the following observation. It gives
a characterization of the graphs that admit two disjoint maximal independent
sets, involving total dominating sets.
Lemma 1. Let G be a graph.
1. If G has two disjoint maximal independent sets, say X and Y , then T =
X ∪ Y is a total dominating set of G. Moreover, G[T ] is bipartite.
2. If T is a total dominating set which is maximal with respect to the property
that G[T ] is bipartite, every bipartition (X, Y ) of G[T ] has the property
that X and Y are two disjoint maximal independent sets.
Proof. Let G = (V, E) be a graph.
To 1: First we assume that G has two disjoint maximal independent sets, say
X and Y . Let T = X ∪Y . By maximality, every vertex of V \X has a neighor in
X and the analogous statement holds for Y . Thus T is a dominating set. Since
X and Y are disjoint, every member of X has a neighbor in Y and vice versa.
Thus T does not have isolated vertices and is therefore a total dominating set.
Furthermore, (X, Y ) is a bipartition of G[T ].
To 2: Now let G have a total dominating set T such that G[T ] is bipartite.
We assume that T is maximal with the property that G[T ] is bipartite. Let
(X, Y ) be any bipartition of G[T ]. Suppose that X is not an independent
dominating set, i.e., there is a vertex v ∈ V \ X such that v does not have a
neighbor in X. Since T is a total dominating set, v must have a neighbor in Y .
Thus T ′ = T ∪ {v} is a total dominating set and (X ∪ {v}, Y ) is a bipartition
of G[T ′ ]. This contradicts the choice of T . By a symmetric argumentation we
see that both X and Y are independent dominating sets.
In particular, a graph G has two disjoint maximal independent sets if and
only if G has a total dominating set T such that G[T ] is bipartite. We can now
give the proof of Theorem 2.
Proof of Theorem 2. We now explain Algorithm 1 and thereby observe its correctness. For this, let G = (V, E) be an isolate-free graph to which the algorithm
is applied.
Algorithm 1 works in two phases.
In the first phase, it tries to find a total dominating set that induces a
bipartite subgraph of G. For this, it maintains a vertex subset T which is a
total dominating set throughout the algorithm. Initially, T = V (line 1).
In every iteration (lines 2-16), the algorithm searches for a chordless odd
cycle in G[T ]. If there is no such cycle, the first phase of the algorithm is
finished and it proceeds to the second phase described below. If there is such a
cycle, say with vertex set C, the algorithm tries to delete a vertex of C from T .
If this is not possible, the corona of a chordless odd cycle is found.
Iteratively, for all u ∈ C the set Pu of vertices not dominated by T \ {u}
is computed (line 5). In each iteration, when for a vertex u ∈ C the set Pu is
computed, the algorithm distinguishes two cases.
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Algorithm 1 Computing two disjoint maximal independent sets
Require: An isolate-free graph G = (V, E).
Ensure: Either a pair (X, Y ) of two disjoint maximal independent sets or a set
Z such that G[Z] is a corona of a chordless odd cycle.
1: T ← V
2: while G[T ] is not bipartite do
3:
compute a set C ⊆ T such that G[C] is a chordless odd cycle
4:
for all u ∈ C do
5:
compute the set Pu of vertices that do not have a neighbor among
T \ {u}
6:
if Pu = ∅ then
7:
T ← T \ {u}
8:
go to step 2
9:
else
10:
choose pu ∈ Pu arbitrary
11:
T ← T ∪ {pu }
12:
end if
13:
end for
14:
Z ← {u, pu : u ∈ C}
15:
return Z
16: end while
17: for all v ∈ V \ T do
18:
if G[T ∪ {v}] is bipartite then
19:
T ← T ∪ {v}
20:
end if
21: end for
22: compute a bipartition (X, Y ) of G[T ]
23: return (X, Y )
5
Case 1. If Pu 6= ∅, the algorithm chooses an arbitrary member pu ∈ Pu and
adds it to T . Clearly that does not create new odd cycles in G[T ]. Now, the set
Pu′ is computed for the next vertex u′ ∈ C.
Case 2. If Pu = ∅, T \ {u} is a total dominating set of T . The algorithm
removes u from T and stops the loop. In this way, the odd cycle G[C] is removed
from G[T ] while no new odd cycle is created in G[T ]. Indeed, the number of
vertices in T that belong to an odd cycle of G[T ] decreased. The algorithm
re-iterates and searches for an odd cycle in G[T ] again.
If for every vertex u ∈ C Case 1 is applied, the algorithm gives out the set Z
defined by Z = {u, pu : u ∈ C} and stops. By construction, G[Z] is the corona
of the chordless odd cycle G[C].
As stated above, the number of vertices in T that belong to an odd cycle of
G[T ] decreases everytime Case 2 is applied. Thus Case 2 is, during the whole
procedure, applied at most |V | − 2 times. Hence, after at most |V | − 2 iterations
of the first phase the algorithm necessarily either finds the corona of a chordless
odd cycle and stops or reaches the second phase.
In the second phase, starting in line 17, the subgraph induced by the total
dominating set T is bipartite. Now the algorithm expands T such that it becomes maximal with respect the property that G[T ] is bipartite. This is done in
at most |V | − 2 steps by iteratively checking for every v ∈ V \ T whether it can
be added to T without violating the bipartiteness of G[T ] (lines 17-21). After
the expansion, a bipartition (X, Y ) of G[T ] is given out by the algorithm (lines
22 and 23). By Lemma 1, X and Y are two disjoint maximal independent sets.
The running time of the algorithm is easily seen to be polynomial.
As an immediate consequence of Theorem 2 we obtain the following.
Corollary 1. Let G be an isolate-free graph that does not contain a corona of
a chordless odd cycle as induced subgraph. Then G has two disjoint maximal
independent sets and such two sets can be computed in polynomial time.
Using Corollary 1, we can now prove Theorem 1.
Proof of Theorem 1. Connecting Corollary 1 and Observation 1, we see that
the coronas of chordless odd cycles are the minimal isolate-free graphs, with
respect to the induced subgraph relation, that do not admit two disjoint maximal
independent sets. This proves Theorem 1.
Let us now discuss special cases of Theorem 1. The corona of a triangle
is usually called the net. Recall that a graph G is called perfect if, in every
induced subgraph of G, the chromatic number equals the size of the largest
clique. Since chordless odd cycles of length at least five are not perfect, we
observe the following.
Corollary 2. Every isolate-free net-free perfect graph has two disjoint maximal
independent sets.
Under a complementary view, Theorem 1 states the following. A central vertex of a graph is a vertex that is adjacent to all other vertices of the graph. Note
that a graph with a central vertex cannot have two disjoint maximal cliques.
Corollary 3. For every graph G, the following assertions are equivalent.
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Figure 3: The 3-sun.
1. Every induced subgraph of G that does not have a central vertex has two
disjoint maximal cliques.
2. G does not contain the complement of a corona of a chordless odd cycle
as induced subgraph.
The complement of a net is usually called a 3-sun or the Hajós graph (see
Fig. 3). The complementary statement to Corollary 2 is formulated as follows.
Corollary 4. Every 3-sun-free perfect graph without central vertex has two
disjoint maximal cliques.
3
Proof of Theorem 3
We now come to the computational complexity of the parameter ii(G) in the
class of trees. For this, we need to introduce the following notions.
A 2-tuple dominating set of a graph G is a set X ⊆ V such that |N [v]∩X| ≥ 2
for all v ∈ V . The 2-tuple domination number of a graph G, denoted γ×2 (G),
is defined as the minimal size of a 2-tuple dominating set of G. This parameter
is defined for every isolate-free graph.
Note that in general ii(G) ≥ γ×2 (G) holds, since the union of two disjoint
maximal independent sets is a 2-tuple dominating set. As our next result shows,
both parameters coincide on the class of trees.
Lemma 2. Let G be a tree on at least 2 vertices. Then, ii(G) = γ×2 (G).
Proof. Let G be a tree on at least 2 vertices. It is clear that ii(G) ≥ γ×2 (G).
To see ii(G) ≤ γ×2 (G), let X be a 2-tuple dominating set of G. We show
how to find a partition of X into two disjoint maximal independent sets, I and
J. For this, we consider I and J to be initially empty. Let G be rooted at any
vertex x ∈ X. Put x into I.
After this initial step, we proceed with the following iteration. If a vertex,
say u, has been considered, do the following for every child w of u:
If w ∈ I, put all children of w that belong to X into J. Conversely, if w ∈ J,
put all children of w that belong to X into I. If w ∈
/ X and u ∈ I, put all
children of w that belong to X into J. Conversely, if w ∈
/ X and u ∈ J, put
all children of w that belong to X into I. Finally, if w ∈
/ X and u ∈
/ X, choose
any child of w that belongs to X and put it into I. All other children of w that
belong to X are put into J.
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Figure 4: A 2-tuple dominating set and the corresponding two disjoint maximal
independent sets.
We close the proof by observing the following facts which hold after the
procedure is finished. Every vertex that does not belong to I or J has at least
one neighbor in I and one in J. Moreover, every vertex in I has a neighbor in
J and vice versa. Finally, I and J are independent sets.
All in all, ii(G) = γ×2 (G).
An example of a tree with a 2-tuple dominating set and two disjoint maximal
independent sets is displayed in Fig. 4.
Note that the proof of Lemma 2 yields an efficient algorithm that, given a
2-tuple dominating set X, constructs two disjoint maximal independendent sets
of total size |X|. Moreover, this algorithm can be organized to run in linear
time, by applying the iterations in a depth-first-search manner.
Liao et al. [7] present a linear-time algorithm to compute a minimum 2-tuple
dominating set in the class of trees. Using this result and Lemma 2, we derive
Theorem 3.
4
Discussion
In this paper, we characterized the maximal hereditary class of graphs that admit two disjoint maximal independent sets. Moreover, we have given a polynomial time algorithm to compute, given any isolate-free graph, either two disjoint
maximal independent sets or one of the forbidden induced subgraphs. In addition, we have shown how two disjoint maximal independent sets of minimum
total size can be computed for trees.
These results we obtained using the close connection of disjoint maximal
independent sets to dominating sets: total dominating sets in the case of the
first two results, and 2-tuple domination in case of the last result.
Future research on our topic could involve the following questions. Recall
that in general ii(G) ≥ γ×2 (G) holds. It would be interesting to know the
largest hereditary graph class for which indeed ii(G) = γ×2 (G) holds in every
(isolate-free) induced subgraph. A small graph violating this equality is C4 ,
since ii(C4 ) = 2 > γ×2 (C4 ) = 3.
Another question is which graphs admit k disjoint maximal independent
sets. Here, one could again aim for sufficient condtions in terms of forbidden
induced subgraphs. The assumption of a graph to be isolate-free is necessary
for the existence of two disjoint maximal independent sets. In the case of k
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disjoint maximal independent sets, we would suggest to replace this condition
by assuming the minimum degree to be at least k − 1.
References
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independent sets of vertices in graphs, Discrete Mathematics 42 (1982), 57–
61.
[2] C. Payan, Sur une classe de problemes de couverture, C. R. Acad. Sci. Paris
A 278 (1974), 233–235.
[3] E.J. Cockayne, S.T. Hedetniemi, Disjoint independent dominating sets in
graphs, Discrete Mathematics 15 (1976), 213–222.
[4] S.M. Hedetniemi, S.T. Hedetniemi, R.C. Laskar, L. Markus, P. Slater, Disjoint dominating sets in graphs, Manuscript 2007.
[5] M.A. Henning, C. Löwenstein, D. Rautenbach, Remarks about disjoint dominating sets, Discrete Math. 309 (2009), 6451–6458.
[6] J. Raczek, R. Janczewski, A. Malafiejska, M. Malafiejski, private communication (2011).
[7] C.-S. Liao, G.J. Chang, Algorithmic aspects of k-tuple domination in graphs,
Taiwanese Journal of Mathematics 6 (2002), 415–420.
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