A certifying algorithm for 3-colorability of P
5
-free
graphs
Daniel Bruce
?
Ch
´
ınh T. Ho
`
ang
??
Joe Sawada
? ? ?
Abstract. We provide a certifying algorithm for the problem of deciding whether
a P
5
-free graph is 3-colorable by showing there are exactly six finite graphs that
are P
5
-free and not 3-colorable and minimal with respect to this property.
1 Introduction
An algorithm is certifying if it returns with each output a simple and easily verifiable
certificate that the particular output is correct. For example, a certifying algorithm for
the bipartite graph recognition would return either a 2-coloring of the input graph prov-
ing that it is bipartite, or an odd cycle proving it is not bipartite. A certifying algorithm
for planarity would return a planar embedding or one of the two Kuratowski subgraphs.
The notion of certifying algorithm [9] was developed when researchers noticed that a
well known planarity testing program was incorrectly implemented. A certifying al-
gorithm is a desirable tool to guard against incorrect implementation of a particular
algorithm. In this paper, we give a certifying algorithm for the problem of deciding
whether a P
5
-free graph is 3-colorable. We will now discuss the background of this
problem.
A class C of graphs is called hereditary if for each graph G in C, all induced sub-
graphs of G are also in C. Every hereditary class of graphs can be described by its
forbidden induced subgraphs, i.e. the unique set of minimal graphs which do not be-
long to the class. A comprehensive survey on coloring of graphs in hereditary classes
can be found in [12]. An important line of research on colorability of graphs in heredi-
tary classes deals with P
t
-free graphs. The induced path on t vertices is called P
t
, and
a graph is called P
t
-free if it does not contain P
t
as an induced subgraph.
It is known that 4 -COLORABILITY is NP-complete for P
9
-free graphs [14] and
5-COLORABILITY is NP-complete for P
8
-free graphs [10]. And most recently it was
proved that 6-COLORABILITY is NP-complete for P
7
-free [2]. On the other hand, the
k-COLORABILITY problem can be solved in polynomial time for P
4
-free graphs (since
they are perfect). In [5] and [6], it is shown that k-COLORABILITY can be solved for
the class of P
5
-free graphs in polynomial time for every particular value of k. For
t = 6, 7, the complexity of the problem is generally unknown, except for the case
?
Computing and Information Science, University of Guelph, Canada. email:
dbruce01@uoguelph.ca
??
Physics and Computer Science, Wilfred Laurier University, Canada. Research supported by
NSERC. email: choang@wlu.ca
? ? ?
Computing and Information Science, University of Guelph, Canada. Research supported by
NSERC. email: jsawada@uoguelph.ca
of 3-COLORABILITY of P
6
-free graphs [13]. Known results on the k-COLORABILITY
problem in P
t
-free graphs are summarized in Table 1 (n is the number of vertices in the
input graph, m the number of edges, and α is matrix multiplication exponent known to
satisfy 2 ≤ α < 2.376 [3]).
k\t 3 4 5 6 7 8 9 10 11 12 . . .
3 O(m) O(m) O(n
α
) O(mn
α
) ? ? ? ? ? ? . . .
4 O(m) O(m) P ? ? ? N P
c
N P
c
N P
c
N P
c
. . .
5 O(m) O(m) P ? ? N P
c
N P
c
N P
c
N P
c
N P
c
. . .
6 O(m) O(m) P ? N P
c
N P
c
N P
c
N P
c
N P
c
N P
c
. . .
7 O(m) O(m) P ? N P
c
N P
c
N P
c
N P
c
N P
c
N P
c
. . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Table 1. Known complexities for k-colorability of P
t
-free graphs
In this paper, we study the coloring problem for the class of P
5
-free graphs. This
class has proved resistant with respect to other graph problems. For instance, P
5
-free
graphs is the unique minimal class defined by a single forbidden induced subgraph
with unknown complexity of the MAXIMUM INDEPENDENT SET and MINIMUM INDE-
PENDENT DOMINATING SET problems. Many algorithmic problems are known to be
NP-hard in the class of P
5
-free graphs, for example DOMINATING SET [7] and CHRO-
MATIC NUMBER [8]. In contrast to the NP-hardness of finding the chromatic number
of a P
5
-free graph, it is known [5] that k-COLORABILITY can be solved in this class in
polynomial time for every particular value of k. This algorithm produces a k-coloring if
one exists, but does not produce an easily verifiable certificate when such coloring does
not exist. We are interested in finding a certificate for non-k-colorability of P
5
-free
graphs. For this purpose, we start with k = 3.
Besides [5], there are several polynomial-time algorithms for 3-coloring a P
5
-free
graph ([6, 11, 14]) but none of them is a certifying algorithm. In this paper, we obtain a
certifying algorithm for 3-coloring a P
5
-free graphs by proving there are a finite number
of minimally non-3-colorable P
5
-free graphs and each of these graphs is finite.
Theorem 1.1. A P
5
-free graph is 3-colorable if and only if it does not contain any of
the six graphs in Fig. 1 as a subgraph.
It is an easy matter to verify the graphs in Fig. 1 are not 3-colorable, the rest of the paper
involves proving the other direction of the theorem. In the last Section, we will discuss
open problems arising from our work.
2 Definition and Background
Let k and t be positive integers. An MNkPt is a graph G that (i) is not k-colorable and
is P
t
-free and (ii) every proper subgraph of G is either k-colorable or has a P
t
. We will
be interested specifically in the case where k = 3 and t = 5. We will use the following
notations. Let G be a simple undirected graph. A set S of vertices of G is dominating if
v
v
v
v
w
v
v
v
2
2
3
4
5
1
2
3
1
K
5
4
4
5
1
2
W
5
4
2
v
v
z
w
x
y
x
w
u
5
v
v
v
v
1
S
v
u
v
S
3
2
v
A
x
1
v
B
1
v
v
T
u
A
u
1
u
B
u
2
x
2
3
u
2
w
u
1
u
0
v
2
v
3
u
B
v
5
v
4
v
1
v
0
u
5
u
4
Fig. 1. All 6 MN3P5s
every vertex in G − S has a neighbor in S. A k-clique is a clique on k vertices. u ∼ v
will mean vertex u is adjacent to vertex v. u v will mean vertex u is not adjacent
to vertex v. For any vertex v, N (v) denotes the set of vertices that are adjacent to v.
We write G
∼
=
H to mean G is isomorphic to H. The clique number of G, denoted by
ω(G), is the number of vertices in a largest clique of G. The chromatic number of G,
denoted by χ(G), is the smallest number of colors needed to color the vertices of G. A
hole is an induced cycle with at least four vertices, and it is odd (or even) if it has odd
(or even) length. An anti-hole is the complement of a hole. A k-hole (k-anti-hole) is a
hole (anti-hole) on k vertices. A graph G is perfect if each induced subgraph H of G
has χ(G) = ω(G).
Theorem 2.1 (The Strong Perfect Graph Theorem [4]). A graph is perfect if and
only if it does not contain an odd hole or odd anti-hole as an induced subgraph.
Let G = {K
4
, W
5
, S
1
, S
2
, T, B} be the set of graphs in Fig. 1. We will denote these
graphs in the following way.
– P
5
(v
1
v
2
v
3
v
4
v
5
) means there is a P
5
being v
1
, v
2
, v
3
, v
4
and v
5
.
– K
4
(wxyz) means {w, x, y, z} form a K
4
.
– W
5
(v
1
v
2
v
3
v
4
v
5
, w) means v
1
, v
2
, v
3
, v
4
, v
5
and w form a W
5
where v
1
v
2
v
3
v
4
v
5
form a 5-cycle and w is adjacent to every other vertex.
– S
1
(v
1
v
2
v
3
v
4
v
5
, u
2
, u
5
) means v
1
, v
2
, v
3
, v
4
, v
5
, u
2
, u
5
form an S
1
where v
1
is the
only degree 4 vertex and N(v
1
) = {u
5
, u
2
, v
5
, v
2
}. Also N (v
3
) = {v
4
, v
2
, u
2
}
and N(v
4
) = {v
3
, v
5
, u
5
}, and v
1
v
2
v
3
v
4
v
5
form a 5-cycle.
– S
2
(v
1
v
2
v
3
v
4
v
5
, w, x) means v
1
, v
2
, v
3
, v
4
, v
5
, w and x form an S
2
where N(w) =
{v
2
, v
3
, v
4
, v
5
}, N(x) = {v
1
, v
3
, v
4
} and v
1
v
2
v
3
v
4
v
5
form a 5-cycle.
– T (u
1
u
A
u
B
u
2
, v
1
v
A
v
B
v
2
, x
1
, x
2
) means a T graph is present as shown previously.
– B(w, u
0
u
1
u
2
u
3
u
4
u
5
, v
0
v
1
v
2
v
3
v
4
v
5
) means a B graph is present as shown previ-
ously.
We will rely on the following result.
Theorem 2.2 ([1]). Every connected P
5
-free graph has a dominating clique or a dom-
inating P
3
.
The following lemma is folklore.
Lemma 2.1 (The neighborhood lemma). Let G be a minimally non k-colorable graph.
If u and v are two non-adjacent vertices in G, then N(u) * N(v).
Proof. Assume N (u) ⊆ N (v). Then the graph G − v admits a k-coloring. By giving u
the color of v, we see that G is k-colorable, a contradiction. ut
The neighborhood lemma is used predominantly throughout this paper. Writing
N(v, w) → u will denote the fact that N(v) * N (w) by the neighborhood lemma
so there exists a vertex u where u ∼ v, but u w .
The following fact is well-known and easy to establish.
Fact 2.1. In a minimally non k-colorable graph every vertex has degree at least k. 2
3 Intermediate Results
In this section, we establish a number of intermediate results needed for proving the
main theorem.
Lemma 3.1. Let G be an MN3P5 graph with a 5-hole C = {v
1
, v
2
, v
3
, v
4
, v
5
} and a
vertex w adjacent to at least 4 vertices of C. Then G ∈ G.
Proof. If w is adjacent to all five vertices of C, then G clearly is isomorphic to W
5
.
Now, assume N (w) ∩ {v
1
, v
2
, v
3
, v
4
, v
5
} = {v
2
, v
3
, v
4
, v
5
}.
We have N(v
1
, w) → x.
Assume for the moment that x v
3
, v
4
. We have
x ∼ v
5
, otherwise, we have P
5
(xv
1
v
5
v
4
v
3
).
x ∼ v
2
, otherwise, we have P
5
(xv
1
v
2
v
3
v
4
).
But then G contains S
1
(v
1
v
2
v
3
v
4
v
5
, x, w). This means x ∼ v
3
or x ∼ v
4
. By sym-
metry, we may assume x ∼ v
3
. We have x ∼ v
2
or x ∼ v
4
, otherwise, G contains
P
5
(xv
1
v
2
wv
4
). If x ∼ v
2
then G properly contains S
1
(v
1
v
2
v
3
v
4
v
5
, x, w), a contradic-
tion. This means x ∼ v
4
; so G contains S
2
(v
1
v
2
v
3
v
4
v
5
, w, x) and G
∼
=
S
2
. ut
Theorem 3.1. Every MN3P5 graph different from K
4
contains a 5-hole.
Proof. Let G be an MN3P5 graph different from a K
4
. We have ω(G) ≤ 3 and χ(G) ≥
4. Thus, G is not perfect. By Theorem 2.1, G contains an odd hole or an odd anti-hole
H. H cannot be a hole of size 7 or greater because G is P
5
-free. We may assume H is
an anti-hole of length at least seven, for otherwise we are done (observe that the hole
on five vertices is self-complementary). Let v
1
, v
2
, v
3
, v
4
, v
5
, v
6
, v
7
be the cyclic order
of the hole in the complement of G. Then G properly contains S
1
(v
4
v
6
v
3
v
5
v
2
, v
1
, v
7
),
a contradiction. ut
Lemma 3.2. Let G be an MN3P5 graph that has a dominating clique {a, b, c}. Also
assume that there is a vertex v /∈ {a, b, c} adjacent to two vertices from {a, b, c}. Then
G ∈ G.
Proof. The proof is by contradiction. Suppose that G /∈ G. We may assume v is ad-
jacent to b and c. We have v a, otherwise, G contains K
4
(abcv). Through repeated
applications of the Neighborhood Lemma, we will eventually add nine vertices to G
to arrive at a contradiction. In the end, we will obtain the graph B (see Fig. 2 for the
order in which vertices are added). Each time we add a vertex we will consider its adja-
cency to the other vertices of the graph. In every case, the adjacency can be completely
determined at each step.
v
b
1
9
4
8
7
6
v
v
3
2
5
B
a
c
v
v
v
v
v
v
v
Fig. 2. The graph B obtained in the proof of Lemma 3.2
N(v, a) → v
1
.
• v
1
∼ c: since {a, b, c} is dominating, v
1
is adjacent to either b or c. Without
loss of generality, assume v
1
∼ c.
• v
1
b: otherwise, G contains K
4
(bcvv
1
).
N(v
1
, b) → v
2
.
• v
2
∼ a: assume v
2
a. We have v
2
∼ v, otherwise, G contains P
5
(v
2
v
1
vba).
Also, v
2
∼ c since {a, b, c} is a dominating set. But then, G contains K
4
(v
1
v
2
vc).
• v
2
c: otherwise, G contains W
5
(abvv
1
v
2
, c).
• v
2
∼ v: otherwise, c has four neighbors in the 5-hole v
2
abvv
1
contradicting
Lemma 3.1.
N(v
2
, c) → v
3
.
• v
3
∼ b: assume v
3
b. We have v
3
∼ a since {a, b, c} is a dominating
set. We have v
3
v
1
, otherwise, G contains S
1
(vbav
3
v
2
, c, v
1
). But then G
contains P
5
(v
3
v
2
v
1
cb).
• v
3
v; otherwise, G contains W
5
(bcv
1
v
2
v
3
, v).
• v
3
∼ v
1
: otherwise, v has four neighbors in the 5-hole v
3
bcv
1
v
2
contradict-
ing Lemma 3.1.
• v
3
a: otherwise G contains S
1
(v
3
acvv
1
, b, v
2
).