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European Journal of Combinatorics 34 (2013) 609–619
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European Journal of Combinatorics
journal homepage: www.elsevier.com/locate/ejc
Three complexity results on coloring P
k
-free graphs
✩
Hajo Broersma
a,1
, Fedor V. Fomin
b
, Petr A. Golovach
b
, Daniël Paulusma
a
a
School of Engineering and Computing Sciences, Durham University, DH1 3LE Durham, United Kingdom
b
Department of Informatics, University of Bergen, PB 7803, 5020 Bergen, Norway
a r t i c l e i n f o
Article history:
Available online 24 August 2012
a b s t r a c t
We prove three complexity results on vertex coloring problems
restricted to P
k
-free graphs, i.e., graphs that do not contain a
path on k vertices as an induced subgraph. First of all, we show
that the pre-coloring extension version of 5-coloring remains
NP-complete when restricted to P
6
-free graphs. Recent results of
Hoàng et al. imply that this problem is polynomially solvable on
P
5
-free graphs. Secondly, we show that the pre-coloring extension
version of 3-coloring is polynomially solvable for P
6
-free graphs.
This implies a simpler algorithm for checking the 3-colorability
of P
6
-free graphs than the algorithm given by Randerath and
Schiermeyer. Finally, we prove that 6-coloring is NP-complete
for P
7
-free graphs. This problem was known to be polynomially
solvable for P
5
-free graphs and NP-complete for P
8
-free graphs, so
there remains one open case.
© 2012 Elsevier Ltd. All rights reserved.
1. Introduction
In this paper we consider computational complexity issues related to vertex coloring problems
restricted to P
k
-free graphs. Due to the fact that the usual ℓ-Coloring problem is NP-complete for
any fixed ℓ ≥ 3, there has been considerable interest in studying its complexity when restricted
to certain graph classes. Without doubt one of the most well-known results in this respect is that
ℓ-Coloring is polynomially solvable for perfect graphs. More information on this classic result and
✩
An extended abstract of this paper appeared in the proceedings of IWOCA 2009.
E-mail addresses: hajo.broersma@durham.ac.uk, h.j.broersma@utwente.nl (H. Broersma), fedor.fomin@ii.uib.no
(F.V. Fomin), petr.golovach@ii.uib.no (P.A. Golovach), daniel.paulusma@durham.ac.uk (D. Paulusma).
1
Current address: Faculty of EEMCS, University of Twente, P.O. Box 217, 7500 AE Enschede, The Netherlands. Tel.: +44
1913341712.
0195-6698/$ – see front matter © 2012 Elsevier Ltd. All rights reserved.
doi:10.1016/j.ejc.2011.12.008
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610 H. Broersma et al. / European Journal of Combinatorics 34 (2013) 609–619
related work on coloring problems restricted to graph classes can be found in, e.g., [11,13]. Instead
of repeating what has been written in so many papers over the years, we also refer to these surveys
for motivation and background. Here we continue the study of ℓ-Coloring and its variants for P
k
-free
graphs, a problem that has been studied in several earlier papers by different groups of researchers
(see, e.g., [2,3,6,8–10,15]). We summarize all these results in the table in Section 5.
1.1. Terminology
We refer to [1] for standard graph theory terminology and to [5] for terminology on computational
complexity.
Let G = (V , E) be a graph and k a positive integer. We say that G is P
k
-free if G does not have a path
on k vertices as an induced subgraph.
A (vertex) coloring of a graph G = (V , E) is a mapping φ : V → {1, 2, . . .} such that φ(u) = φ(v)
whenever uv ∈ E. Here φ(u) is usually referred to as the color of u in the coloring φ of G. An ℓ-coloring
of G is a mapping φ : V → {1, 2, . . . , ℓ} such that φ(u) = φ(v) whenever uv ∈ E. The problem
ℓ-Coloring asks if a given graph has an ℓ-coloring.
In list-coloring we assume that V = {v
1
, v
2
, . . . , v
n
} and that for every vertex v
i
of G there is a
list L
i
of admissible colors (a subset of the natural numbers). Given these lists, a list-coloring of G is a
coloring φ : V → {1, 2, . . .} such that φ(v
i
) ∈ L
i
for all i ∈ {1, 2, . . . , n}; we say that φ respects the
lists L
i
.
In pre-coloring extension we assume that a (possibly empty) subset W ⊆ V of G is pre-colored
with φ
W
: W → {1, 2, . . .} and the question is whether we can extend φ
W
to a coloring of G. If φ
W
is
restricted to {1, 2, . . . , ℓ} and we want to extend it to an ℓ-coloring of G, we say we deal with the pre-
coloring extension version of ℓ-Coloring. In fact, we consider a slight variation on the latter problem
which can be considered as list coloring, but which has the flavor of pre-coloring: lists have varying
sizes including some of size 1. We will slightly abuse terminology and call these problems pre-coloring
extension problems too.
1.2. Results of this paper
We prove the following three complexity results on vertex coloring problems restricted to P
k
-free
graphs.
First of all, in Section 2 we show that the pre-coloring extension version of 5-Coloring remains
NP-complete when restricted to P
6
-free graphs. Recent results of Hoàng et al. [6] imply that this
problem is polynomially solvable on P
5
-free graphs. Their algorithm for ℓ-Coloring for any fixed ℓ
is in fact a list-coloring algorithm where the lists are from the set {1, 2, . . . , ℓ}.
Secondly, in Section 3 we show that the pre-coloring extension version of 3-Coloring is
polynomially solvable for P
6
-free graphs. The 3-Coloring problem was known to be polynomially
solvable for P
6
-free graphs from a paper by Randerath and Schiermeyer [10]. Their approach is as
follows. First they note that the input graph G may be assumed to be K
4
-free, i.e., does not contain
a complete graph on four vertices as a subgraph, as otherwise it is not 3-colorable. Their algorithm
then determines if G contains a C
5
. If so, it exploits the existence of this C
5
in G in a clever way. If
not, the authors use the Strong Perfect Graph Theorem to deduce that G is perfect. This allows them
to use the polynomial time algorithm of Tucker [12] for finding a χ -coloring of a K
4
-free perfect
graph. Here χ denotes the chromatic number of a graph, i.e., the smallest ℓ such that the graph
is ℓ-colorable. We follow a different approach. First, our algorithm is independent of the Strong
Perfect Graph Theorem, and second it uses a recent characterization of P
6
-free graphs in terms of
dominating subgraphs [14]. This way we can indeed show that the pre-coloring extension version
of 3-Coloring is polynomially solvable for P
6
-free graphs, whereas the approach of Randerath and
Schiermeyer [10] does not immediately lead to this result. The reason for this lies in the second part
of their algorithm that focuses on K
4
-free perfect graphs. Already for a subclass of this class, namely the
class of bipartite graphs, Kratochvíl [8] showed that the pre-coloring extension version of 3-Coloring
is an NP-complete problem.
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H. Broersma et al. / European Journal of Combinatorics 34 (2013) 609–619 611
Finally, in Section 4 we show that 6-Coloring is NP-complete for P
7
-free graphs. This problem was
known to be polynomially solvable for P
5
-free graphs [6] and NP-complete for P
8
-free graphs [15], so
there remains one open case.
2. Pre-coloring extension of 5-coloring for P
6
-free graphs
In this section we show that the pre-coloring extension version of 5-Coloring remains NP-
complete when restricted to P
6
-free graphs. We use a reduction from Not-All-Equal 3-Satisfiability
with positive literals only which we denote as NAE 3SATPL. This NP-complete problem [5] is also
known as Hypergraph 2-Colorability and is defined as follows. Given a set X = {x
1
, x
2
, . . . , x
n
} of
logical variables, and a set C = {C
1
, C
2
, . . . , C
m
} of three-literal clauses over X in which all literals
are positive, does there exist a truth assignment for X such that each clause contains at least one true
literal and at least one false literal?
We consider an arbitrary instance I of NAE 3SATPL and define a graph G
I
and a pre-coloring on some
vertices of G
I
, and next we show that G
I
is P
6
-free and that the pre-coloring on G
I
can be extended to a
5-coloring of G
I
if and only if I has a satisfying truth assignment in which each clause contains at least
one true literal and at least one false literal.
Theorem 1. The pre-coloring extension version of 5-Coloring is NP-complete for P
6
-free graphs.
Proof. Let I be an instance of NAE 3SATPL with variables {x
1
, x
2
, . . . , x
n
} and clauses {C
1
, C
2
, . . . , C
m
}.
We define a graph G
I
corresponding to I and lists of admissible colors for its vertices based on the
following construction. We note here that the lists we introduce below are only there for convenience
to the reader; it will be clear later that all lists other than {1, 2, . . . , 5} are in fact forced by the pre-
colored vertices.
1. We introduce one new vertex for each of the clauses, and use the same labels C
1
, C
2
, . . . , C
m
for
these m vertices; we assume that for each of these vertices there is a list {1, 2, 3} of admissible
colors. We say that these vertices are of C-type and use C to denote the set of C-type vertices.
2. We introduce one new vertex for each of the variables, and use the same labels x
1
, x
2
, . . . , x
n
for
these n vertices; we assume that for each of these vertices there is a list {4, 5} of admissible colors.
We say that these vertices are of x-type and use X to denote the set of x-type vertices.
3. We join all C-type vertices to all x-type vertices to form a complete bipartite graph with |C| |X|
edges.
4. For each clause C
j
we fix an arbitrary order of its variables x
i
, x
k
, and x
r
, and we introduce three
pairs of new vertices {a
i,j
, b
i,j
}, {a
k,j
, b
k,j
}, {a
r,j
, b
r,j
}; we assume the following lists of admissible
colors for these three pairs, respectively: {{1, 4}, {2, 5}}, {{2, 4}, {3, 5}}, {{3, 4}, {1, 5}}. We say
that these vertices are of a-type and b-type, and use A and B to denote the set of a-type and b-type
vertices, respectively. We add edges between x-type and a-type vertices whenever the first index
of the a-type vertex is the same as of the x-type vertex, and similarly for the b-type vertices. We
add edges between C-type and a-type vertices whenever the second index of the a-type vertex is
the same as the index of the C-type vertex, and similarly for the b-type vertices. Hence each clause
with three variables is represented by three 4-cycles that have one C-type vertex in common.
5. For each a-type vertex we introduce a copy of a K
2,3
, as follows: for a
i,j
we add five vertices
{p
i,j,1
, . . . , p
i,j,5
}, and we add all edges between {p
i,j,1
, p
i,j,2
, p
i,j,3
} and {p
i,j,4
, p
i,j,5
}. We say that
these vertices are of p-type and use P to denote the set of p-type vertices. We add edges between
each a-vertex and the p-vertices of its corresponding K
2,3
depending on its list of admissible colors.
In particular, we join the a-vertex to the three p-vertices of its K
2,3
that have a third index which
is not in its list of admissible colors. So, if a
i,j
has list {1, 4}, we join it to p
i,j,2
, p
i,j,3
, p
i,j,5
. We use
P
1
to denote the set of all p-type vertices with the third index in {1, 2, 3} and P
1
to denote all
other p-type vertices.
6. For each b-type vertex we introduce a new copy of a K
2,3
on five vertices of q-type, in the same
way as we introduced the p-type vertices for the a-type vertices. Edges are added in a similar way,
depending on the indices and the lists. We use Q to denote the set of q-type vertices, Q
1
to denote
the set of all q-type vertices with third index in {1, 2, 3} and Q
1
to denote all other q-type vertices.
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612 H. Broersma et al. / European Journal of Combinatorics 34 (2013) 609–619
Fig. 1. The (complete bipartite) subgraph of G
I
induced by vertices of type C, p, q, x.
Fig. 2. (i) The subgraph of G
I
for clause C
1
with ordered variables x
1
, x
2
, x
3
. (ii) How a
1,1
and b
1,1
are connected to P and Q,
respectively.
7. We join all the p-type and q-type vertices with third indices 1, 2, 3 to all the p-type and q-type
vertices with third indices 4, 5 to form a complete bipartite graph with |P
1
∪ Q
1
||P
1
∪ Q
1
| edges.
8. We join all x-type vertices to all p-type and q-type vertices with third indices 1, 2, 3.
9. We join all C-type vertices to all p-type and q-type vertices with third indices 4, 5.
10. We pre-color all the p-type and q-type vertices according to their third index, so p
i,j,ℓ
will be pre-
colored with color ℓ ∈ {1, 2, . . . , 5}. Note that we can now in fact replace all lists introduced
earlier by {1, 2, . . . , 5}, since the shorter lists will be forced by the given pre-coloring.
See Figs. 1 and 2 for sketches of the ingredients in the construction of the graph G
I
; in Fig. 2 we
illustrate an example in which C
1
is a clause with ordered variables x
1
, x
2
, x
3
.
We now prove that G
I
is P
6
-free. In order to obtain a contradiction, suppose that the graph G
I
contains an induced subgraph H that is isomorphic to P
6
. We first consider the complete bipartite
subgraph with bipartition classes V
1
= C ∪ P
1
∪ Q
1
and V
2
= X ∪ P
1
∪ Q
1
.
Suppose that H contains at least four vertices from V
1
∪ V
2
. Since P
6
contains no independent
set of cardinality four, H then contains at least one vertex from each of V
1
and V
2
. This either yields
a vertex with degree at least three in H or a cycle on four vertices in H, a contradiction. Hence
|V (H) ∩ (V
1
∪ V
2
)| ≤ 3. Since A ∪ B is an independent set, we also have |V (H) ∩ (A ∪ B)| ≤ 3. Since
|V (H)| = 6, this implies that both inequalities are in fact equalities.
Let V (H) = {v
1
, v
2
, . . . , v
6
} and E(H) = {v
1
v
2
, v
2
v
3
, v
3
v
4
, v
4
v
5
, v
5
v
6
}. By symmetry, we may
assume that either {v
1
, v
3
, v
5
} ⊂ V (H) ∩ (A ∪ B) or {v
1
, v
3
, v
6
} ⊂ V (H) ∩ (A ∪ B). Noting that every
vertex of P ∪Q has at most one neighbor in A∪B, in both cases v
2
∈ C∪X. We next observe that every
vertex of A ∪ B has precisely one neighbor in C and precisely one neighbor in X. This implies that
we can neither have {v
2
, v
4
} ⊂ X nor {v
2
, v
4
} ⊂ C. Since v
2
v
4
∈ E(G
I
), we cannot have v
4
∈ C ∪ X.
This rules out the first case, and in the remaining case we may assume {v
1
, v
3
, v
6
} ⊂ V (H) ∩ (A ∪ B),
with v
2
∈ C ∪ X and v
4
∈ P ∪ Q. Since v
5
is a neighbor of v
4
while v
2
is not a neighbor of v
4
, we
find that v
5
∈ C ∪ X. Hence v
5
∈ P ∪ Q. Because v
4
v
5
is an edge and v
4
, v
5
both belong to P ∪ Q,
one of them belongs to V
1
and the other one to V
2
. However, then either v
2
v
4
or v
2
v
5
is an edge of
G
I
, because v
2
∈ C ∪ X is either adjacent to all vertices in V
1
or else to all vertices in V
2
. This is not
possible, and we conclude that G
I
is P
6
-free.
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H. Broersma et al. / European Journal of Combinatorics 34 (2013) 609–619 613
We claim that I has a truth assignment in which each clause contains at least one true and at least
one false literal if and only if the pre-coloring of G
I
can be extended to a 5-coloring of G
I
.
First suppose that I has a satisfying truth assignment in which each clause contains at least one true
and at least one false literal. We use color 4 to color the x-type vertices representing the true literals
and color 5 for the false literals. Now consider the lists assigned to the a-type and b-type vertices that
come in pairs chosen from {{1, 4}, {2, 5}}, {{2, 4}, {3, 5}}, {{3, 4}, {1, 5}}. If the adjacent x-type vertex
has color 4, color 1, 2 or 3 is forced on one of the adjacent a-type or b-type vertices, respectively, while
on the other one we can use color 5; similarly, if the adjacent x-type vertex has color 5, color 2, 3 or 1
is forced on one of the adjacent a-type or b-type vertices, respectively, while on the other one we can
use color 4. Since precisely two of the three x-type vertices of one clause gadget have the same color,
this leaves at least one of the colors 1, 2 and 3 admissible for the C-type vertex representing the clause.
By coloring the vertices associated with each clause and variable as described above, a 5-coloring of
the pre-colored graph G
I
is obtained.
Now suppose that we have a 5-coloring of the graph G
I
that respects the pre-coloring. Then
each of the x-type vertices has color 4 or 5, and each of the C-type vertices has color 1, 2 or 3.
We define a truth assignment that sets a variable to TRUE if the corresponding x-type vertex has
color 4, and to FALSE otherwise. Suppose that one of the clauses contains only true literals. Then
the three x-type vertices in the corresponding clause gadget of G
I
all have color 4. Now consider
the lists assigned to the a-type and b-type vertices of this gadget that come in pairs chosen from
{{1, 4}, {2, 5}}, {{2, 4}, {3, 5}}, {{3, 4}, {1, 5}}. Since the adjacent x-type vertices all have color 4,
colors 1, 2 and 3 are forced on three of the a-type and b-type vertices adjacent to the C-type vertex
of this gadget, a contradiction, since the C-type vertex has color 1, 2 or 3. This proves that every
clause contains at least one false literal. Analogously, every clause contains at least one true literal.
This completes the proof of Theorem 1.
3. Pre-coloring extension of 3-coloring for P
6
-free graphs
In this section we show that the pre-coloring extension version of 3-Coloring is polynomially
solvable for P
6
-free graphs. A key ingredient in our approach is the following characterization of P
6
-
free graphs [14]. Here a subgraph H of a graph G is said to be a dominating subgraph of G if every
vertex of V (G) \ V (H) has a neighbor in H.
Lemma 2 ([14]). A graph G is P
6
-free if and only if each connected induced subgraph of G on more than
one vertex contains a dominating induced cycle on six vertices or a dominating (not necessarily induced)
complete bipartite subgraph. Moreover, these dominating subgraphs can be obtained in polynomial time.
Another key ingredient in our approach is the following lemma. Its proof follows from the fact that
the decision problem in this case can be modeled and solved as a 2SAT-problem. This approach has
been introduced by Edwards [4] and is folklore now, see also [6,10].
Lemma 3 ([4]). Let G be a graph in which every vertex has a list of admissible colors of size at most 2.
Then checking if G has a list-coloring is solvable in polynomial time.
An important subroutine in our algorithm works as follows. Let G be a graph in which every vertex
has a list of admissible colors. Let U ⊆ V (G) contain all vertices that have a list consisting of exactly
one color. For every vertex u ∈ U we remove the unique color in its list from the lists of its neighbors.
Next we remove u from G. We repeat this process in the remaining graph as long as there exists a
vertex with a list of size 1. This process is called updating the graph. We note the following.
Lemma 4. A graph G with lists of admissible colors on its vertices can be updated in polynomial time. If
this results in a vertex with an empty list, then G does not have a list-coloring respecting the original lists.
We are now ready to state the main result of this section. We prove a slightly stronger statement,
namely that we can decide in polynomial time whether a P
6
-free graph, in which each vertex has a list
of admissible colors from the set {1, 2, 3}, has a coloring respecting these lists; note that a pre-coloring
corresponds to lists of size 1 on the pre-colored vertices.
