Forbidden Subgraphs Generating a Finite Set
Jun Fujisawa
Faculty of Business and Commerce
Keio University
Hiyoshi 4–1–1, Kohoku-Ku
Yokohama, Kanagawa 223–8521
JAPAN
E-mail : [email protected]
Michael D. Plummer
Department of Mathematics
Vanderbilt University
Nashville, Tennessee 37240
U.S.A
E-mail : [email protected]
and
Akira Saito∗
Department of Computer Science
Nihon University
Sakurajosui 3–25–40
Setagaya-Ku, Tokyo 156–8550
JAPAN
E-mail : [email protected]
Abstract
For a set F of connected graphs, a graph G is said to be F-free if G does not
contain any member of F as an induced subgraph. The members of F are referred
to as forbidden subgraphs. When we study the relationship between forbidden
subgraphs and a certain graph property, we often allow the existence of exceptional
graphs as long as their number is finite. However, in this type of research, if the
set of k-connected F-free graphs itself, denoted by Gk (F), is finite, then every
∗
Partially supported by Japan Society for the Promotion of Science, Grant-in-Aid for Scientific
Research (C), 22500018, 2011
1
graph in Gk (F) logically satisfies all the graph properties, except for finite number
of exceptions. If this occurs, F does not give any information about a particular
property. We think that such sets F obscure the view in the study of forbidden
subgraphs, and we want to remove them. With this motivation, we study the sets
F with finite Gk (F). We prove that if |F| ≤ 2 and Gk (F) is finite, then either
K1,2 ∈ F or F consists of a complete graph and a star. For each of the values of k,
1 ≤ k ≤ 6, we then characterize all the pairs {Kl , K1,m } such that Gk ({Kl , K1,m })
is finite. We also give a complete characterization of F with |F| ≤ 3 and finite
G2 (F).
Keywords: forbidden subgraphs, k-connected graphs, hamiltonian cycles
1
Introduction
In this paper, we consider only finite graphs. Let H be a connected graph. A graph
G is said to be H-free if G does not contain H as an induced subgraph. We also say
that H is forbidden in G. For a set of connected graphs F, G is said to be F-free if G
is F -free for every F ∈ F. The set F is often called a set of forbidden subgraphs. In
general, F can be an infinite set. For example, if F is the set of all cycles, then the set
of connected F-free graphs coincides with the set of trees, and if F is the set of all odd
cycles, then the set of F-free graphs coincides with the set of bipartite graphs.
If we choose a set F of connected graphs appropriately, F-free graphs may satisfy
a certain graph property. Sumner [17] and Las Vergnas [14] proved that a connected
K1,3 -free graph of even order has a perfect matching. Duffus, Gould and Jacobson [5]
proved that a 2-connected {K1,3 , Net}-free graph has a hamiltonian cycle, where Net
is the unique graph having degree sequence (3, 3, 3, 1, 1, 1) (see Figure 1). The latter
result stimulated the study of pairs of forbidden subgraphs which imply the existence of
a hamiltonian cycle. A number of pairs of forbidden subgraphs implying hamiltonicity
have been discovered, and finally all such pairs were determined by Bedrossian [2].
In order to state Bedrossian’s theorem and the subsequent results clearly, we introduce some notation and several definitions. For a pair of connected graphs H1 and
H2 , we write H1 ≺ H2 if H1 is an induced subgraph of H2 . For two sets F1 and F2
of connected graphs, we write F1 ≤ F2 if for each F2 ∈ F2 , there exists F1 ∈ F1 with
F1 ≺ F2 . It is easy to see that if F1 ≺ F2 , then every F1 -free graph is F2 -free (see [12]).
Let W be as depicted in Figure 1, and let Pk denote the path of order k.
2
r
r
r
·
r""
·
r
··T
r
T
TTr
bbr
Net
r
T
r
T
TT·
r
r
b
b
b
b
"r
"
"
r
"
r
·
·
·
r
r
r
Z3
W
Figure 1: Net, W and Z3
Theorem A (Bedrossian [2]) Let F1 and F2 be connected graphs of order at least
two. Then every 2-connected {F1 , F2 }-free graph is hamiltonian if and only if {F1 , F2 } ≤
{K1,3 , Net}, {F1 , F2 } ≤ {K1,3 , W } or {F1 , F2 } ≤ {K1,3 , P6 }.
Let Ck be the cycle of order k, and let Zk be the graph obtained from C3 and Pk by
joining one end of Pk and one vertex of C3 by an edge (see Figure 1). There exists a
2-connected {K1,3 , Z3 }-free graph of order nine that does not have a hamiltonian cycle.
Thus, {K1,3 , Z3 } does not belong to the set of pairs in Bedrossian’s theorem. However,
Faudree et al. [11] proved that every 2-connected {K1,3 , Z3 }-free graph of order at least
ten has a hamiltonian cycle. Therefore, allowing a finite number of exceptions, we
can say that this is also a set of forbidden subgraphs which implies the existence of a
hamiltonian cycle in the class of 2-connected graphs. Faudree and Gould [7] conducted a
research in this direction, and proved that even if we allow a finite number of exceptions,
{K1,3 , Z3 } is the only pair that is added to Bedrossian’s set.
Theorem B (Faudree and Gould [7]) Let F1 and F2 be a pair of connected graphs.
Then every 2-connected {F1 , F2 }-free graph, possibly with a finite number of exceptions,
has a hamiltonian cycle if and only if {F1 , F2 } ≤ {K1,3 , Net}, {F1 , F2 } ≤ {K1,3 , W },
{F1 , F2 } ≤ {K1,3 , P6 } or {F1 , F2 } ≤ {K1,3 , Z3 }.
Since the above result was proved, triples of forbidden subgraphs implying the existence
of a hamiltonian cycle have been sought. Some partial results have been obtained, but
the complete characterization has not been completed as yet (see [3, 8, 9, 10]).
As illustrated in Theorem B, in the research of forbidden subgraphs, we do not want
to say that a given statement fails just because of a finite number of counterexamples.
If we know that the number of counterexamples is finite, we would rather say that the
statement holds with a finite number of exceptions. However, this approach poses a new
3
problem : What happens if the set of F-free graphs itself is finite for a given set F?
If it is finite, then it logically satisfies all the graph properties with a finite number of
exceptions, for we can simply declare that all of its members are exceptions. But this
means that F does not give any information about a particular property. It introduces
“noise” into the research. We want to remove this noise and clarify the picture in the
research of forbidden subgraphs. This is the motivation for this paper.
For a set F of connected graphs, a graph is F-free if and only if all of its components
are F-free. Thus, we will restrict ourselves to connected graphs. Moreover, there is no
point in forbidding K1 . Hence we will assume that all the forbidden subgraphs have
order at least two. With these assumptions in mind, we denote by G the set of all the
finite connected graphs which have at least two vertices. For F ⊂ G, we denote by G(F)
the set of all the F-free graphs in G : G(F) = {G ∈ G : G is F-free}.
We are not the first to consider this problem. Actually, the following theorem is
proved in Diestel’s textbook [6] as an application of Ramsey numbers.
Theorem C ([6], Proposition 9.4.1) Let F be a (possibly infinite) set of connected
graphs. Then G(F) is finite if and only if {Kl , K1,m , Pn } ⊂ F for some positive integers
l, m and n.
Note that the above theorem does not say that F needs three distinct elements to make
G(F) finite. For example, K2 is a complete graph, a star and a path at the same time.
Therefore, the above theorem says that G({K2 }) is finite.
Theorem C solves the problem of forbidden subgraphs generating a finite set in the
domain of connected graphs. However, we sometimes need to discuss the same problem
in the domain of graphs of higher connectivity. For example, when we discuss the
existence of a hamiltonian cycle, we naturally assume that a graph in consideration
is 2-connected since no graph with a cutvertex has a hamiltonian cycle. Similarly,
since all the hamiltonian-connected graphs are 3-connected, it is natural to assume 3connectedness when we discuss hamiltonian-connectedness. For k ≥ 2, the set of finite
k-connected graphs is smaller than the set of connected graphs. Therefore, for a given
set F, there may be only finitely many k-connected F-free graphs even if G(F) is an
infinite set. We will actually see a number of such examples in this paper. And as the
connectivity increases, we obtain more forbidden subgraphs generating a finite set.
For a positive integer k, let Gk denote the set of all the finite k-connected graphs,
4
and for F ⊂ G, we define Gk (F) by
Gk (F) = {G ∈ Gk : G is F-free} = Gk ∩ G(F).
The main purpose of this paper is to study F such that Gk (F) is finite.
For basic graph-theoretic notation, we refer the reader to [4]. Let G and H be
vertex-disjoint graphs. Then let G ∪ H, G + H and G × H denote the union, the join
and the Cartesian product of G and H, respectively. For a positive integer n, we let nG
denote the union of n copies of G. The compliment of G is denoted by G. If G contains
a cycle, we denote by girth(G) the girth of G, which is the length of a shortest cycle in
G. For x ∈ V (G), we write NG (x) for the neighborhood of x in G. We denote by κ(G)
and λ(G) the connectivity and the edge-connectivity of G, respectively. We write L(G)
for the line graph of G.
For positive integers k and m, recall that the Ramsey number r(k, m) is defined as
the smallest integer n such that every graph of order n contains either a clique of order
k or an independent set of order m.
2
One Forbidden Subgraph
In this section, we forbid one subgraph. As a corollary of Theorem C, it is easy to see
that G({F }) is finite if and only if F = K2 . Therefore, the question in this section is
whether there is any other graph F such that Gk ({F }) is finite for sufficiently large k.
We give a negative answer to this question.
We first make a simple but useful observation.
Lemma 1 Let k be an integer with k ≥ 2 and let F ⊂ G. If Gk (F) is finite, then
{Kl , Km1 ,m2 } ⊂ F for some integer l, m1 and m2 with l ≥ 2, m2 ≥ m1 ≥ 1 and m1 ≤ k.
Proof.
Let H1 = {Kp : p ≥ k + 1}. Then H1 ⊂ Gk and H1 is infinite. Therefore,
H1 ̸⊂ Gk (F) and hence Kp ∈
/ G(F) for some p ≥ k + 1. Then Kp contains some F ∈ F as
an induced subgraph. This implies that F itself is a complete graph. Let F = Kl (l ≥ 2).
Let H2 = {Kk,p : p ≥ k}. Then H2 ⊂ Gk and H2 is infinite. Therefore, H2 ̸⊂ Gk (F) and
hence Kk,p1 ∈
/ G(F) for some p1 ≥ k. Then Kk,p1 contains some F ′ ∈ F as an induced
subgraph. Since F ′ is connected and |F ′ | ≥ 2, we can write F ′ = Km1 ,m2 for some m1
and m2 with m2 ≥ m1 ≥ 1 and m1 ≤ k.
¤
5
We can answer the question of this section by means of the following corollary of the
above lemma.
Theorem 2 Let F ∈ G and let k be an integer with k ≥ 2. Then Gk ({F }) is finite if
and only if F = K2 .
Proof.
Since Gk ({K2 }) = ∅, the sufficiency is trivial. We prove the necessity.
Suppose Gk ({F }) is finite. Then by Lemma 1, F = Km1 ,m2 for some integers m1
and m2 with m2 ≥ m1 ≥ 1 and m1 ≤ k. Also by Lemma 1, F is a complete graph. This
is possible only if m1 = m2 = 1, or F = K2 .
3
¤
Two Forbidden Subgraphs
In this section, we study pairs of forbidden subgraphs which generate a finite set.
First, we discuss redundancy in a set of forbidden subgraphs. Suppose F ⊂ G and
Gk (F) is finite. Then for every F ′ with F ⊂ F ′ ⊂ G, Gk (F ′ ) ⊂ Gk (F) and hence
Gk (F ′ ) is finite. Thus, when we study F ⊂ G with finite Gk (F), we can concentrate on
inclusion-minimal sets. This leads us to the following definition.
When Gk (F) is finite, we say that F is non-redundant if Gk (F ′ ) is infinite for every
proper subset F ′ of F. In other words, F is non-redundant if F is an inclusion-minimal
set under the property that Gk (F) is finite. We shall be interested in non-redundant
sets of forbidden subgraphs.
In the family of all non-redundant sets, the relation ≤ defined in Section 1 is a partial
order (see [12]). Moreover, for F1 , F2 ⊂ G, if F1 ≤ F2 , then Gk (F1 ) ⊂ Gk (F2 ). Thus, if
Gk (F2 ) is finite, Gk (F1 ) is also finite.
By Theorem 2, if Gk (F) is finite and K2 ∈ F, then F is non-redundant if and only if
F = {K2 }. Therefore, when we discuss non-redundant pairs F of forbidden subgraphs
generating a finite set, we may assume K2 ∈
/ F.
By Lemma 1, if Gk (F) is finite, F contains a complete bipartite graph as well as
a complete graph. We first observe that if |F| = 2, this complete bipartite graph is
actually a star.
Theorem 3 Let k be an integer with k ≥ 2 and let F ⊂ G −{K2 }. If |F| = 2 and Gk (F)
is finite, then F = {Kl , K1,m } for some integers l and m with l ≥ 3 and 1 ≤ m ≤ k.
We use the following results to prove this theorem.
6
Theorem D ([15]) For any two given positive integers g and κ with g ≥ 3, there
exists a graph G with girth g and vertex connectivity κ. In particular, for every positive
integers g and k, there exist infinitely many k-connected graphs of girth at least g.
Theorem E ([16]) For any positive integers n, k with k ≥ 2 and n ≥ k + 2, there
exists a k-regular k-connected balanced bipartite graph of order 2n. In particular, for
every integer k with k ≥ 2, there exist infinitely many k-connected k-regular bipartite
graphs.
Proof of Theorem 3.
Let F = {F1 , F2 }. By Lemma 1, we may assume F1 = Kl
and F2 = Km1 ,m2 for some integers l, m1 and m2 with l ≥ 3, m2 ≥ m1 ≥ 1 and m1 ≤ k.
Note that F1 contains C3 . Assume m1 ≥ 2. Then F2 contains a C4 as an induced
subgraph. Let H1 = {G ∈ Gk : girth(G) ≥ 5}. Since girth(F1 ) = 3 and girth(F2 ) = 4,
every graph in H1 is F-free and hence H1 ⊂ Gk (F). Then since Gk (F) is finite, H1 is
also a finite set. But this contradicts Theorem D. Hence we have m1 = 1.
Next, assume m2 ≥ k + 1. Let H2 = {G ∈ Gk : G is k-regular and bipartite} and let
H ∈ H2 . Then since H is bipartite and l ≥ 3, H is Kl -free, and since H is k-regular
and m2 ≥ k + 1, H is K1,m2 -free. Hence we have H2 ⊂ Gk (F). But this is again a
contradiction since H2 is an infinite set by Theorem E. Therefore, we have m2 ≤ k.
¤
By the above theorem, when we study pairs of forbidden subgraphs generating a
finite set, we may assume that these pairs consist of a complete graph and a star.
If l2 ≥ l1 ≥ 3 and m2 ≥ m1 ≥ 2, then {Kl1 , K1,m1 } ≤ {Kl2 , K1,m2 }. This fact and
the basic property of connectivity lead us to the following easy but useful observation.
Lemma 4 For each integers k, l and m with k ≥ 1, l ≥ 3 and m ≥ 2, the following
hold:
(1) Gk ({Kl , K1,m }) ⊂ Gk ({Kl , K1,m+1 })
(2) Gk ({Kl , K1,m }) ⊂ Gk ({Kl+1 , K1,m })
(3) Gk+1 ({Kl , K1,m }) ⊂ Gk ({Kl , K1,m })
By the above lemma, we see that there are three types of sequences in {Gk ({Kl , K1,m }) : k ≥
1, l ≥ 3, m ≥ 2}.
(1) Gk ({Kl , K1,2 }) ⊂ Gk ({Kl , K1,3 }) ⊂ Gk ({Kl , K1,4 }) ⊂ Gk ({Kl , K1,5 }) ⊂ · · ·
7
(2) Gk ({K3 , K1,m }) ⊂ Gk ({K4 , K1,m }) ⊂ Gk ({K5 , K1,m }) ⊂ Gk ({K6 , K1,m }) ⊂ · · ·
(3) G1 ({Kl , K1,m }) ⊃ G2 ({Kl , K1,m }) ⊃ G3 ({Kl , K1,m }) ⊃ G4 ({Kl , K1,m }) ⊃ · · ·
We prove that the terms in the first and the second sequences ultimately become
infinite sets, except for the case m = 2, while the terms in the third sequence become
empty at a certain stage.
Theorem 5
(1) For each positive integers k and l with l ≥ 3, there exists a positive integer m such
that Gk ({Kl , K1,m }) is an infinite set.
(2) For each positive integer k and l with l ≥ 3, Gk ({Kl , K1,2 }) is a finite set.
(3) For each positive integer k and m with m ≥ 3, there exists a positive integer l
such that Gk ({Kl , K1,m }) is an infinite set.
(4) For each integer l and m with l ≥ 3 and m ≥ 2, there exists a positive integer k
with Gk ({Kl , K1,m }) = ∅.
Before proving this theorem, we prove an easy lemma, which we use in the proof and
the subsequent arguments.
Lemma 6 Let k, l and m be integers with k ≥ 1, l ≥ 3 and m ≥ 2. Then the following
hold.
(1)
Gk ({Kl , K1,2 }) =


{Kp : k + 1 ≤ p ≤ l − 1} if l ≥ k + 2

∅
(2) If m ≥ 3, then
Gk (K3 , K1,m )
Proof.
if l ≤ k + 1.


is infinite if k ≤ m − 1

= ∅
if k ≥ m.
(1) A connected graph is K1,2 -free if and only if it is complete. Since Kl is
forbidden and the k-connectedness is required, the conclusion follows.
(2) Since m ≥ 3, by Theorem E, there exist infinitely many (m − 1)-connected (m − 1)regular bipartite graphs, all of which are {K3 , K1,m }-free. Therefore, Gm−1 ({K3 , K1,m })
is an infinite set. Then by Lemma 4 (3), Gk ({K3 , K1,m }) is infinite for each k ≤ m − 1.
8
If there exists an m-connected {K3 , K1,m }-free graph G, then take a vertex x in G.
Since G is m-connected, |NG (x)| = degG x ≥ m. then since G is K1,m -free, there exist
a pair of adjacent vertices y, z in NG (x). However, this implies that {x, y, z} induces
K3 in G, a contradiction. Therefore, Gm ({K3 , K1,m }) = ∅, and again by Lemma 4 (3),
Gk ({K3 , K1,m }) = ∅ for each k ≥ m.
¤
Now we prove Theorem 5.
Proof of Theorem 5.
(1) By Theorem C (for k = 1) and Theorem 3 (for k ≥ 2),
Gk ({Kl , K1,k+1 }) is an infinite set.
(2) It is immediate from Lemma 6 (1).
l
m
k
(3) Let l = m−1
. Let B be the set of (m − 1)-connected (m − 1)-regular bipartite
graphs. Since m − 1 ≥ 2, B is an infinite set by Theorem E. Let B ∈ B. For each
x ∈ V (B), let Bx denote a copy of Kl so that V (Bu ) ∩ V (Bv ) = ∅ if u ̸= v. Then for
each edge xy in B, we join every vertex in Bx with every vertex in By by an edge. Let
GB be the resulting graph.
Since B is a bipartite graph, for each v ∈ V (GB ), the graph induced by NGB (v) is
isomorphic to Kl−1 +(m−1)Kl , which does not have m independent vertices. Therefore,
GB is K1,m -free.
Let SB be a minimal cutset of GB . If V (Bx ) ∩ SB ̸= ∅ and V (Bx ) ̸⊂ SB , take
u ∈ V (Bx ) ∩ SB and v ∈ V (Bx ) − SB . Then by the construction, NGB (u) ∪ {u} =
NGB (v)∪{v}. Thus, since GB −SB is disconnected, GB −(SB −{u}) is also disconnected,
which contradicts the minimality of SB . Therefore, V (Bx )∩SB ̸= ∅ implies V (Bx ) ⊂ SB ,
S
and hence there exists a set of vertices S in B such that SB = x∈S V (Bx ). Then
B − S is disconnected. Since B is (m − 1)-connected, this implies |S| ≥ m − 1. Then
|SB | ≥ (m − 1)l ≥ k. therefore, GB is k-connected.
Assume GB has a subgraph H which is isomorphic to K2l+1 . Since each Bx has
exactly l vertices, there exists three distinct vertices x1 , x2 , x3 in B such that V (H) ∩
V (Bxi ) ̸= ∅ for each i, 1 ≤ i ≤ 3. However, since H is a complete graph, this implies
that {x1 , x2 , x3 } induces K3 in B. This is a contradiction since B is a bipartite graph.
Therefore, GB is K2l+1 -free.
Let H = {GB : B ∈ B}. Then H ⊂ Gk ({K2l+1 , K1,m }). Since B is an infinite set, H
is also infinite, and hence Gk ({K2l+1 , K1,m }) is an infinite set.
(4) Let k = r(l − 1, m). Suppose Gk ({Kl , K1,m }) ̸= ∅, and choose an element G ∈
9
Gk ({Kl , K1,m }). Let x ∈ V (G). Since G is k-connected, |NG (x)| ≥ k = r(l − 1, m).
Therefore, G[NG (x)] contains a set of vertices S which induces Kl−1 or Km . But this
implies that S ∪ {x} induces either Kl or K1,m , a contradiction.
¤
¡
¢
Now we look at the decreasing sequence Gk ({Kl , K1,m }) k≥1 in more detail. As we
have seen in Theorem 5 (2), if m = 2, all the elements in this sequence are finite sets.
On the other hand, Lemma 6 (2) says that if m ≥ 3 and l = 3, this sequence starts from
an infinite set, but at some stage jumps from an infinite set to the empty set. Now we
look at the sequence for (l, m) = (4, 3). In this case, we now show that the sequence
again starts from an infinite set, but reaches a finite non-empty set before it reaches the
empty set. We denote the icosahedron, the platonic graph of order 12, by Icosa. For a
graph G, the square of G is the graph obtained from G by adding an edge to every pair
of vertices that are distance two in G.
Theorem 7
(1) G4 ({K4 , K1,3 }) is an infinite set.
(2) G5 ({K4 , K1,3 }) = {Icosa}
(3) G6 ({K4 , K1,3 }) = ∅
Proof.
(1) Since the square of a cycle of order at least six is a 4-connected {K4 , K1,3 }-
free graph, G4 ({K4 , K1,3 }) is an infinite set.
(2) Let G ∈ G5 ({K4 , K1,3 }). Then no vertex in G has a set of vertices in its neighborhood
that induce K3 or K3 . Since r(3, 3) = 6, this implies that G does not have a vertex of
degree six or more, and hence G is a 5-regular 5-connected graph. Moreover, since C5
is the only graph of order five that does not have three independent vertices or a clique
of order three, the neighborhood of each vertex induces C5 .
Let u ∈ V (G) and let NG (u) = {a1 , a2 , a3 , a4 , a5 }. We may assume that a1 a2 a3 a4 a5 a1
is an induced C5 . Let NG (a1 ) = {u, a2 , a5 , b1 , b5 }. Note {b1 , b5 } ∩ {a3 , a4 } = ∅. Since
{a2 u, ua5 } ⊂ E(G), we may assume that ua5 b5 b1 a2 u is an induced C5 in G[NG (a1 )].
Then {a3 , u, a1 , b1 } ⊂ NG (a2 ). Let b2 ∈ NG (a2 )−{a3 , u, a1 , b1 }. Since both a1 a2 a3 a4 a5 a1
and ua5 b5 b1 a2 u are induced cycles, b2 ∈
/ {a4 , a5 , b5 }. Since a3 ua1 b1 is a path in G,
a3 ua1 b1 b2 a3 is an induced cycle in G. Then {a4 , u, a2 , b2 } ⊂ NG (a3 ). Let b3 ∈ NG (a3 ) −
{a4 , u, a2 , b2 }. Since a4 ua2 b2 is a path in G, a4 ua2 b2 b3 a4 is an induced C5 in G. Since
a1 a2 a3 a4 a5 a1 and a3 ua1 b1 b2 a3 are induced cycles, b3 ∈
/ {a1 , a5 , b1 }. If b3 = b5 , then
10
b2 b5 ∈ E(G). But this implies that b2 a2 a1 b5 b2 induces a cycle of order four in NG (b1 ),
a contradiction. Hence b3 ̸= b5 .
Let b4 ∈ NG (a4 ) − {a5 , u, a3 , b3 }. Since a5 ua3 b3 is a path in G[NG (a4 )], a5 ua3 b3 b4 a5
is an induced C5 . By an argument similar to the one on the previous paragraph, we have
b4 ∈
/ {a1 , a2 , b1 , b2 , b5 }. Since b5 a1 ua4 b4 is a path of order five in G[NG (a5 )], b4 b5 ∈ E(G).
Let v ∈ NG (b5 ) − {b4 , a5 , a1 , b1 }. Since b4 a5 a1 b1 is a path in G[NG (b5 )], b4 a5 a1 b1 vb4
is an induced C5 . Then since G is 5-regular, v ∈
/ {u} ∪ {ai , bi : 1 ≤ i ≤ 5}.
Since vb5 a1 a2 b2 is a path in G[NG (b1 )] and vb5 a5 a4 b3 is a path in G[NG (b4 )], {vb2 , vb3 } ⊂
E(G). Then since G is 5-regular, V (G) = {u, v} ∪ {ai , b1 : 1 ≤ i ≤ 5}, and G is a icosahedron.
(3) Since κ(Icosa) = 5, G6 ({K4 , K1,3 }) = ∅.
¤
¡
¢
As we have seen in this section, the behavior of the decreasing sequence Gk ({Kl , K1,m ) k≥1
varies, depending on the values of l and m.
4
Characterization of the Pairs for Graphs of Small Connectivity
In this section, we characterize the pairs {Kl , K1,m } such that Gk ({Kl , K1,m }) is finite
for 1 ≤ k ≤ 6.
By combining results from the previous section, we immediately obtain the following.
Theorem 8 Let l and m be integers with l ≥ 3 and m ≥ 2. Then
(1) G({Kl , K1,m }) is finite if and only if m = 2,
(2) G2 ({Kl , K1,m }) is finite if and only if m = 2,
(3) G3 ({Kl , K1,m }) is finite if and only if m = 2 or (l, m) = (3, 3), and
(4) G4 ({Kl , K1,m }) is finite if and only if m = 2 or (l, m) ∈ {(3, 3), (3, 4)}.
Proof.
(1) It immediately follows from Theorem C.
(2) Since G2 ({Kl , K1,2 }) ⊂ G({Kl , K1,2 }) by Lemma 4 (3), G2 ({Kl , K1,2 }) is a finite
set. On the other hand, by Lemma 6 (2), G2 ({K3 , K1,3 }) is an infinite set. Then for
every integer l, m with l ≥ 3 and m ≥ 3, we have G2 ({K3 , K1,3 }) ⊂ G2 ({Kl , K1,m }) by
Lemma 4 (1) and (2), and hence G2 ({Kl , K1,m }) is an infinite set.
11
(3) The finiteness of G3 ({Kl , K1,2 }) follows from Theorem 5 (2), whereas Lemma 6 (2)
yields that G3 ({K3 , K1,3 }) is finite and that G3 ({K3 , K1,4 }) is infinite. Since G4 ({K4 , K1,3 })
is an infinite set by Theorem 7 (1), G3 ({K4 , K1,3 }) is also an infinite set by Lemma 4 (3).
Since both G3 ({K3 , K1,4 }) and G3 ({K4 , K1,3 } are infinite, Lemma 4 (1) and (2) yield that
G3 ({Kl , K1,m }) is infinite for every pair of integers l and m with l ≥ 3 and m ≥ 3 unless
(l, m) = (3, 3).
(4) The finiteness of G4 ({Kl , K1,2 }) and G4 ({K3 , K1,3 }) follows from (3) and Lemma 4 (3),
and the finiteness of G4 ({K3 , K1,4 }) follows from Lemma 6 (2). Lemma 6 (2) also yields
that G4 ({K3 , K1,5 }) is infinite, and G4 ({K4 , K1,3 }) is infinite by Theorem 7 (1). Then
G4 ({Kl , K1,m }) is infinite for all the other values of (l, m) by Lemma 4 (1) and (2).
¤
In order to give the characterization for 5-connected and 6-connected graphs, we
require one lemma.
The super edge-connectivity λ′ (G) of a connected graph G is the minimum cardinality
of an edge cut F such that each component of G − F has order at least two. Note that
κ(L(G)) = λ′ (G) unless L(G) is a complete graph. Lü, Chen and Xu [13] obtained the
following lower bound of the super edge-connectivity of the Cartesian product of two
graphs.
Theorem F ([13]) Let Gi be a connected graph of order ni and edge-connectivity λi
(i = 1, 2). If n1 ≥ 3 and n2 ≥ 3, then
λ′ (G1 × G2 ) ≥ min{n1 λ2 , n2 λ1 , λ1 + 2λ2 , λ2 + 2λ1 }.
Lemma 9
(1) There exist infinitely many 6-connected {K5 , K1,3 }-free graphs.
(2) There exist infinitely many 6-connected {K4 , K1,4 }-free graphs.
Proof.
(1) For an integer p with p ≥ 3, let Gp = L(Cp × Cp ). By Theorem F,
λ′ (Cp × Cp ) ≥ 6, which implies that Gp is 6-connected. Since Gp is a line graph, Gp
is K1,3 -free, and since Cp × Cp is 4-regular, Gp is K5 -free. These facts imply {Gp : p ≥
3} ⊂ G6 ({K5 , K1,3 }), and hence G6 ({K5 , K1,3 }) is an infinite set.
(2) Let p be an integer with p ≥ 6. Consider a set of p2 vertices {vi,j : 0 ≤ i, j ≤ p − 1}.
For each i and j, 0 ≤ i, j ≤ p − 1, add edges vi,j vi,j+1 , vi,j vi+1,j and vi,j vi+1,j+1 , where
12
suffices are counted modulo p. Let Hp be the resulting graph. The neighborhood of vi,j
in Hp is {vi+1,j , vi−1,j , vi,j−1 , vi,j+1 , vi+1,j+1 , vi−1,j−1 }. Hence Hp is 6-regular. Moreover,
the subgraph of Hp induced by the neighborhood of vi,j is isomorphic to C6 . Hence Hp
is {K4 , K1,4 }-free.
We claim that Hp is 6-connected. Assume, to the contrary, Hp − S is disconnected
for some S ⊂ V (Hp ) with |S| ≤ 5. Let C t = Hp [{v0,t , v1,t , . . . , vp−1,t }] and Dt =
Hp [{vt,0 , vt,1 , . . . , Vt,p−1 }] (0 ≤ t ≤ p − 1). Then both C t and Dt are cycles of order p.
Since p ≥ 6, V (C t ) ∩ S = ∅ for some t with 0 ≤ t ≤ p − 1. Without loss of generality,
we may assume t = 0.
Since Hp − S is disconnected, there exists a vertex vi,j in Hp − S such that vi,j and
C 0 lie in different components of Hp − S. Then there does not exist a path from vi,j
to C 0 in Hp − S. Since vi,j ∈ V (Dj ) and Dj is a cycle, this implies |V (Dj ) ∩ S| ≥ 2.
Then |Dj − S| ≤ 3 and hence {vi,j+1 , vi+1,j+1 , vi,j−1 , vi−1,j−1 } ̸⊂ S. By symmetry, we
may assume {vi,j+1 , vi+1,j+1 } ̸⊂ S. Then since Dj+1 is a cycle, |V (Dj+1 ) ∩ S| ≥ 2. This
¯
¡
¢¯
implies ¯S − V (Dj ) ∪ V (Dj+1 ) ¯ ≤ 1 and hence {vi,j−1 , vi−1,j−1 } ̸⊂ S. However, this
implies |V (Dj−1 ) ∩ S| ≥ 2, and |S| ≥ 6, a contradiction. Therefore, Hp is 6-connected.
Since each Hp (p ≥ 6) is a 6-connected {K4 , K1,4 }-free graph, we have {Hp : p ≥
6} ⊂ G6 ({K4 , K1,4 }), and hence G6 ({K4 , K1,4 }) is an infinite set.
¤
We now settle the cases of 5-connected and 6-connected graphs.
Theorem 10 Let l and m be integers with l ≥ 3 and m ≥ 2. Then
(1) G5 ({Kl , K1,m }) is finite if and only if m = 2 or (l, m) ∈ {(3, 3), (3, 4), (3, 5), (4, 3)},
and
(2) G6 ({Kl , K1,m }) is finite if and only if m = 2 or (l, m) ∈ {(3, 3), (3, 4), (3, 5), (3, 6), (4, 3)}.
Proof.
The finiteness of G5 ({Kl , K1,2 }) and G6 ({Kl , K1,2 }) was proved in Theo-
rem 5 (2). Lemma 6 (2) yields that both G5 ({K3 , K1,5 }) and G6 ({K3 , K1,6 }) are finite and that both G5 ({K3 , K1,6 }) and G6 ({K3 , K1,7 }) are infinite. By Theorem 7,
G5 ({K4 , K1,3 }) = {Icosa} and G6 ({K4 , K1,3 }) = ∅. Lemma 9 (2) says G6 ({K4 , K1,4 }) is
an infinite set, which implies G6 (K4 , K1,m }) is infinite for each m ≥ 4. And Lemma 9 (1)
says that G6 ({K5 , K1,3 }) is an infinite set and hence G6 ({K5 , K1,m }) is infinite for each
m ≥ 3. But then by Lemma 4 (3), G5 ({K4 , K1,m }) is infinite for every m ≥ 4 and
G5 ({K5 , K1,m }) is infinite for every m ≥ 3.
13
¤
Currently our characterization stops at k = 6. We have no idea how to characterize
the case m ≥ 7.
5
Three Forbidden Subgraphs for 2-Connected Graphs
In this section, we study forbidden subgraphs which generate a finite set in the set of
2-connected graphs. In particular, we characterize the triples of forbidden subgraphs
that generate a finite set.
Suppose G2 (F) is finite. We know that F contains a complete graph by Lemma 1.
Also by Lemma 1, F contains either a star or a complete bipartite graph of the form
K2,m (m ≥ 2). The next observation says that it contains another type of graph if F is
finite.
Lemma 11 Let F ⊂ G. If both F and G2 (F) are finite, then Pn ∈ F for some integer
n with n ≥ 2.
Proof.
Assume F does not contain a path of any order. Let C = {Ck : k ≥ 3}. Then
C is an infinite set of 2-connected graphs. Let C ′ = C − G2 (F). Since G2 (F) is finite, C ′
is an infinite set. Let C ∈ C ′ . Then C is not F-free and hence F ≺ C for some F ∈ F.
However, since a connected induced subgraph of C is either a path or C itself, we have
F = C. This implies that C ′ ⊂ F, which contradicts the finiteness of F.
¤
Now we characterize the non-redundant sets F consisting of three forbidden subgraphs such that G2 (F) is finite. By Lemma 1, Lemma 11 and Theorem 2, we may assume that F can be written as either {Kl , K1,m , Pn } or {Kl , K2,m , Pn } for some integers
l, m and n with l ≥ 3, m ≥ 2 and n ≥ 4. But in the former case, Theorem C guarantees
that G({Kl , K1,m , Pn }) is finite. Thus, let us assume that F = {Kl , K2,m , Pn }.
We first observe that the complete graph in F is K3 .
Lemma 12 Let l, m and n be integers with l ≥ 3, m ≥ 3 and n ≥ 4. If G2 ({Kl , K2,m , Pn })
is finite, then l = 3.
Proof.
Assume l ≥ 4, and let H = {K2 + sK1 : s ≥ 1}. Then every graph in H is
2-connected and {K4 , K2,2 , P4 }-free, which implies H ⊂ G2 ({Kl , K2,m , Pn }). However,
since G2 ({Kl , K2,m , Pn }) is finite and H is infinite, this is a contradiction.
14
¤
¡
At this stage, we may thus assume F = {K3 , K2,m , Pn }. We fix m and consider
¢
{K3 , K2,m , Pn } n≥3 as an increasing sequence of n. We proceed to investigate the
threshold of n, where {K3 , K2,m , Pn } is finite, but {K3 , K2,m , Pn+1 } is infinite. It turns
out that this threshold has different values in the case m ≥ 3 and the case m = 2.
(1) m ≥ 3.
The following theorem was proved in [1].
Theorem G ([1], Lemma 7) For every integer m ≥ 3, G2 ({K3 , K2,m , P5 }) is a finite
set.
We prove that P5 in the above theorem is actually a threshold in this case.
Theorem 13 For every integer m with m ≥ 3, G2 ({K3 , K2,m , P6 }) is an infinite set.
Let s be a positive integer. Consider s + 1 pairwise disjoint copies H0 , H1 , . . . Hs of
K2 . Let V (Hi ) = {ai , bi } (0 ≤ i ≤ s). For each j with 1 ≤ j ≤ s, add edges a0 aj and
b0 bj . Let Gs be the resulting graph and let H = {Gs : s ≥ 1}. Each Gs is 2-connected
and {K3 , K2,3 , P6 }-free, and hence H ⊂ G2 ({K3 , K2,m , P6 }). Since H is an infinite set,
G2 ({K3 , K2,m , P6 }) is also an infinite set.
¤
(2) m = 2.
In contrast to the case of m ≥ 3, the following theorem has been proved. Note that
{K3 , K2,2 , P6 } = {C3 , C4 , P6 }.
Theorem H ([1], Lemma 10) Every graph in G2 ({C3 , C4 , P6 }) has order at most ten.
In particular, G2 ({C3 , C4 , P6 }) is a finite set.
We now show that P6 is the threshold in this case.
Theorem 14 G2 ({C3 , C4 , P7 }) is an infinite set.
Proof. Let Gs be the graph defined in the proof of Theorem 13, and let G′s = Gs −a0 b0 .
Let H′ = {G′s : s ≥ 1}. Then every graph in H′ is 2-connected and {C3 , C4 , P7 }-free.
Therefore, H′ ⊂ G2 ({C3 , C4 , P7 }. Since H′ is infinite, G2 ({C3 , C4 , P7 }) is also an infinite
set.
¤
15
k (connectivity)
|F| = 1
|F| = 2
|F| = 3
k=1
{K2 }
{Kl , K1,2 } (l ≥ 3)
{Kl , K1,m , Pn } (l ≥ 3, m ≥ 3, n ≥ 4)
{Kl , K1,m , Pn } (l ≥ 3, m ≥ 3, n ≥ 4)
k=2
{K2 }
{Kl , K1,2 } (l ≥ 3)
{K3 (= C3 ), K2,m , Pn } (m ≥ 2, n ≤ 5)
{K3 (= C3 ), K2,2 (= C4 ), P6 }
k=3
{K2 }
{Kl , K1,2 } (l ≥ 3)
(unknown)
{K3 , K1,3 }
k=4
{K2 }
{Kl , K1,2 } (l ≥ 3)
(unknown)
{K3 , K1,3 }, {K3 , K1,4 }
{K2 }
{Kl , K1,2 } (l ≥ 3)
{K3 , K1,m } (3 ≤ m ≤ 5)
k=5
(unknown)
{K4 , K1,3 }
{K2 }
k=6
{Kl , K1,2 } (l ≥ 3)
{K3 , K1,m } (3 ≤ m ≤ 6)
(unknown)
{K4 , K1,3 }
Table 1 : The sets F with finite Gk (F)
6
Conclusion
In this paper, we have investigated sets F of connected graphs such that the set of
k-connected F-free graphs, denoted by Gk (F), is finite. We have observed the following.
• For each positive integer k, Gk ({F }) is finite if and only if F = K2 .
• For each positive integer, if |F| = 2 and Gk (F) is finite, then F contains both a
complete graph and a star.
Next we characterized the pairs {Kl , K1,m } with finite Gk ({Kl , K1,m }) for 1 ≤ k ≤ 6.
Finally, we have studied G2 (F) and determined all the triples {F1 , F2 , F3 } such that
G2 ({F1 , F2 , F3 }) is finite. Table 1 summarizes the non-redundant sets F of connected
graphs such that Gk (F) is finite, which are determined in this paper.
We make several concluding remarks. First, we note that we have tackled the problem in terms of connectivity. This is because we aim to give useful information to
researches on forbidden subgraphs and hamiltonian properties of graphs, where they
impose assumptions on connectivity. While we believe this motivation is natural, we
16
also realize that we have not made much use of connectivity in the arguments. In many
cases, our discussion was based on degrees of vertices rather than connectivity.
The complete characterization of F with finite Gk (F) seems to be difficult for k ≥ 2.
For example, if |F| = 2, we have proved that F is a pair consisting of a complete
graph and a star. However, discussing the existence of a complete graph and a star
is equivalent to discussing the existence of a clique or a set of independent vertices in
the neighborhood of a vertex, which is the Ramsey problem. We feel that the complete
solution of the problem in this paper would involve the determination of the Ramsey
numbers.
On the other hand, we have been be able to obtain some useful information on F
with finite Gk (F). For example, We proved in Theorem 7 (2) that G5 ({K4 , K1,3 }) is
a singleton set. Except for the trivial case of m ≤ 2, as far as we know, this is the
only example in which Gk ({Kl , K1,m }) is a finite but non-empty set. Thus, we pose the
following problem to conclude this paper.
Problem. For m ≥ 3, find a triple (k, l, m) other than (5, 4, 3) such that Gk ({Kl , K1,m })
is a finite but non-empty set.
Acknowledgments
The authors are grateful to Yoshimi Egawa and Kenta Ozeki for their helpful comments.
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