A note on k-colorability of P
5
-free graphs
Ch
´
ınh T. Ho
`
ang
∗
Marcin Kami
´
nski
†
Vadim Lozin
‡
J. Sawada
§
X. Shu
¶
Abstract
A polynomial time algorithm that determines whether or not, for a fixed k, a P
5
-free graph can
be k-colored is presented in this paper. If such a coloring exists, the algorithm will produce a valid
k-coloring.
Keywords: P
5
-free, graph coloring, dominating clique
1 Introduction
Graph coloring is among the most important and applicable graph problems. The k-colorability prob-
lem is the question of whether or not the vertices of a graph can be colored with one of k colors so
that no two adjacent vertices are assigned the same color. In general, the k-colorability problem is NP-
complete [10]. Even for planar graphs with no vertex degree exceeding 4, the problem is NP-complete
[5]. However, for other classes of graphs, like perfect graphs [8], the problem is polynomial-time solv-
able. For the following special class of perfect graphs, there are efficient polynomial time algorithms
for finding optimal colorings: chordal graphs [6], weakly chordal graphs [9], and comparability graphs
[4]. For more information on perfect graphs, see [1], [3], and [7].
Another interesting class of graphs are those that are P
t
-free, that is, graphs with no chordless paths
v
1
, v
2
, . . . , v
t
of length t − 1 as induced subgraph for some fixed t. If t = 3 or t = 4, then there exists
efficient algorithms to answer the k-colorability question (see [3]). However, it is known that CHRO-
MATIC NUMBER for P
5
-free graphs is NP-complete [11]. Thus, it is of some interest to consider the
problem of k-coloring a P
t
-free graph for some fixed k ≥ 3 and t ≥ 5. Taking this parameterization
∗
Physics and Computer Science, Wilfrid Laurier University, Canada. Research supported by NSERC. E-mail:
choang@wlu.ca
†
RUTCOR, Rutgers University, 640 Bartholomew Road, Piscataway, NJ 08854, USA. E-mail:
mkaminski@rutcor.rutgers.edu
‡
Mathematics Institute, University of Warwick, Coventry CV4 7AL UK. E-mail: V.Lozin@warwick.ac.uk
§
Computing and Information Science, University of Guelph, Canada. Research supported by NSERC. E-mail:
sawada@cis.uoguelph.ca
¶
Computing and Information Science, University of Guelph, Canada. E-mail: xshu@uoguelph.ca
1
k\t 3 4 5 6 7 8 ... 12 ...
3 O(m) O(m) O(n
α
) O(mn
α
) ? ? ? ? .. .
4 O(m) O(m) ?? ? ? ? ? NP
c
.. .
5 O(m) O(m) ?? ? ? NP
c
NP
c
NP
c
.. .
6 O(m) O(m) ?? ? ? NP
c
NP
c
NP
c
.. .
7 O(m) O(m) ?? ? ? NP
c
NP
c
NP
c
.. .
.. . .. . .. . ... ... .. . .. . . .. ... ...
Table 1: Known complexities for k-colorability of P
t
-free graphs
into account, a snapshot of the known complexities for the k-colorability problem of P
t
-free graphs is
given in Table 1. From this chart we can see that there is a polynomial algorithm for the 3-colorability
of P
6
-free graphs [12].
In this paper we focus on P
5
-free graphs. Notice that when k = 3, the colorability question for P
5
-
free graphs can be answered in polynomial time (see [13]). We obtain a theorem (Theorem 2) on the
structure of P
5
-free graphs and use it to design a polynomial-time algorithm that determines whether
a P
5
-free graph can be k-colored. If such a coloring exists, then the algorithm will yield a valid k-
coloring.
The remainder of the paper is presented as follows. In Section 2 we present relevant definitions, con-
cepts, and notations. Then in Section 3, we present our recursive polynomial-time algorithm that an-
swers the k-colorability question for P
5
-free graphs.
2 Background and Definitions
In this section we provide the necessary background and definitions used in the rest of the paper. For
starters, we assume that G = (V, E) is a simple undirected graph where |V | = n and |E| = m. If A is
a subset of V , then we let G(A) denote the subgraph of G induced by A.
DEFINITION 1 A set of vertices A is said to dominate another set B, if every vertex in B is adjacent to
at least one vertex in A.
The following structural result about P
5
-free graphs is from Bacs
´
o and Tuza [2]:
THEOREM 1 Every connected P
5
-free graph has either a dominating clique or a dominating P
3
.
DEFINITION 2 Given a graph G, an integer k and for each vertex v, a list l(v) of k colors, the k-list
coloring problem asks whether or not there is a coloring of the vertices of G such that each vertex
receives a color from its list.
2
DEFINITION 3 The restricted k-list coloring problem is the k-list coloring problem in which the lists
l(v) of colors are subsets of {1, 2, . . . , k}.
Our general approach is to take an instance of a specific coloring problem Φ for a given graph and
replace it with a polynomial number of instances φ
1
, φ
2
, φ
3
, . . . such that the answer to Φ is “yes” if
and only if there is some instance φ
k
that also answers “yes”.
For example, consider a graph with a dominating vertex u where each vertex has color list {1, 2, 3, 4, 5}.
This listing corresponds to our initial instance Φ. Now, by considering different ways to color u, the
following set of four instances will be equivalent to Φ:
φ
1
: l(u) = {1} and the remaining vertices have color lists {2, 3, 4, 5},
φ
2
: l(u) = {2} and the remaining vertices have color lists {1, 3, 4, 5},
φ
3
: l(u) = {3} and the remaining vertices have color lists {1, 2, 4, 5},
φ
4
: l(u) = {4, 5} and the remaining vertices have color lists {1, 2, 3, 4, 5}.
In general, if we recursively apply such an approach we would end up with an exponential number of
equivalent coloring instances to Φ.
3 The Algorithm
Let G be a connected P
5
-free graph. This section describes a polynomial time algorithm that decides
whether or not G is k-colorable. The algorithm is outlined in 3 steps. Step 2 requires some extra
structural analysis and is presented in more detail in the following subsection.
1. Identify and color a maximal dominating clique or a P
3
if no such clique exists (Theorem 1). This
partitions the vertices into fixed sets indexed by available colors. For example, if a P
5
-free graph
has a dominating K
3
(and no dominating K
4
) colored with {1, 2, 3} and k = 4, then the fixed
sets would be given by: S
124
, S
134
, S
234
, S
14
, S
24
, S
34
. For an illustration, see Figure 1. Note that
all the vertices in S
124
are adjacent to the vertex colored 3 and thus have color lists {1, 2, 4}. This
gives rise to our original restricted list-coloring instance Φ. Although the illustration in Figure 1
does not show it, it is possible for there to be edges between any two fixed sets.
2. Two vertices are dependent if there is an edge between them and the intersection of their color
lists is non-empty. In this step, we remove all dependencies between each pair of fixed sets.
This process, detailed in the following subsection, will create a polynomial number of coloring
instances {φ
1
, φ
2
, φ
3
, . . .} equivalent to Φ.
3. For each instance φ
i
from Step 2 the dependencies between each pair of fixed sets has been
removed which means that the vertices within each fixed set can be colored independently. Thus,
3
S
134
S
14
S
124
S
24
S
4
S
34
S
234
2 3
1
Figure 1: The fixed sets in a P
5
-free graph with a dominating K
3
where k = 4.
for each instance φ
i
we recursively see if each fixed set can be colored with the corresponding
restricted color lists (the base case is when the color lists are a single color). If one such instance
provides a valid k-coloring then return the coloring. Otherwise, the graph is not k-colorable.
As mentioned, the difficult part is reducing the dependencies between each pair of fixed sets (Step 2).
3.1 Removing the Dependencies Between Two Fixed Sets
Let S
list
denote a fixed set of vertices with color list given by list. We partition each such fixed set
into dynamic sets that each represent a unique subset of the colors in list. For example: S
123
=
P
123
∪ P
12
∪ P
13
∪ P
23
∪ P
1
∪ P
2
∪ P
3
. Initially, S
123
= P
123
and the remaining sets in the partition are
empty. However, as we start removing dependencies, these sets will dynamically change. For example,
if a vertex u is initially in P
123
and one of its neighbors gets colored 2, then u will be removed from
P
123
and added to P
13
.
Recall that our goal is to remove the dependencies between two fixed sets S
p
and S
q
. To do this,
we remove the dependencies between each pair (P, Q) where P is a dynamic subset of S
p
and Q is
a dynamic subset of S
q
. Let col(P ) and col(Q) denote the color lists for the vertices in P and Q
respectively. By visiting these pairs in order from largest to smallest with respect to |col(P )| and then
|col(Q)|, we ensure that we only need to consider each pair once. Applying this approach, the crux
of the reduction process is to remove the dependencies between a pair (P, Q) by creating at most a
polynomial number of equivalent colorings.
Now, observe that there exists a vertex v from the dominating set found in Step 1 of the algorithm that
dominates every vertex in one set, but is not adjacent to any vertex in the other. This is because P and
Q are subsets of different fixed sets. WLOG assume that v dominates Q.
4
Y
4
Y
3
Y
2
Y
1
QP
y
1
y
2
y
4
y
3
v
a
c
d
b
Figure 2: Illustration for proof of Theorem 2
THEOREM 2 Let H be a P
5
-free graph partitioned into three sets P , Q and {v} where v is adjacent to
every vertex in Q but not adjacent to any vertex in P . If we let Q
′
denote all components of H(Q) that
are adjacent to some vertex in P then one of the following must hold.
1. There exists exactly one special component C in G(P ) that contains two vertices a and b such that
a is adjacent to some component Y
1
∈ G(Q) but not adjacent to another component Y
2
∈ G(Q)
while b is adjacent to Y
2
but not Y
1
.
2. There is a vertex x that dominates every component in Q
′
, except at most one (call it T).
PROOF: Suppose that there are two unique components X
1
, X
2
∈ G(P ) with a, b ∈ X
1
and c, d ∈ X
2
and components Y
1
6= Y
2
and Y
3
6= Y
4
from G(Q) such that:
• a is adjacent to Y
1
but not adjacent to Y
2
,
• b is adjacent to Y
2
but not adjacent to Y
1
,
• c is adjacent to Y
3
but not adjacent to Y
4
,
• d is adjacent to Y
4
but not adjacent to Y
3
.
Let y
1
(respectively, y
2
, y
3
, y
4
) be a vertex in Y
1
(respectively, Y
2
, Y
3
, Y
4
) that is adjacent to a (respec-
tively, b, c, d) and not b (respectively, a, d, c). Since H is P
5
-free, there must be edges (a, b) and (c, d),
otherwise a, y
1
, v, y
2
, b or c, y
3
, v, y
4
, d would be P
5
s. An illustration of these vertices and components
is given in Figure 2.
Suppose Y
2
= Y
3
. Then b is not adjacent to y
3
, for otherwise there exists a P
5
a, b, y
3
, c, d. Now, there
exists a P
5
y
1
, a, b, y
2
, y
3
(if y
2
is adjacent to y
3
) or a P
5
a, b, y
2
, v, y
3
(if y
2
is not adjacent to y
3
). Thus,
Y
2
and Y
3
must be unique components. Similarly, we have Y
2
6= Y
4
. Now since b, y
2
, v, y
3
, c cannot be
5