On the Corr\'adi-Hajnal Theorem and a question of Dirac

0.35 MB·

by  H. A. Kierstead

ON THE CORRA´DI-HAJNAL THEOREM AND A QUESTION OF
DIRAC
H.A. KIERSTEAD, A.V. KOSTOCHKA, AND E.C. YEAGER
6 Abstract. In 1963, Corr´adi and Hajnal proved that for all k ≥ 1 and n ≥ 3k, every
1 graph G on n vertices with minimum degree δ(G) ≥ 2k contains k disjoint cycles. The
0 bound δ(G) ≥ 2k is sharp. Here we characterize those graphs with δ(G) ≥ 2k −1 that
2
contain k disjoint cycles. This answers the simple-graph case of Dirac’s 1963 question on

n the characterizationof (2k−1)-connected graphs with no k disjoint cycles.
a
EnomotoandWangrefinedthe Corr´adi-HajnalTheorem,provingthe followingOre-type
J
version: For all k ≥ 1 and n ≥ 3k, every graph G on n vertices contains k disjoint cycles,
5
provided that d(x)+d(y)≥4k−1 for all distinct nonadjacent vertices x,y. We refine this
1
furtherfork ≥3andn≥3k+1: IfGisagraphonnverticessuchthatd(x)+d(y)≥4k−3

] for all distinct nonadjacent vertices x,y, then G has k vertex-disjoint cycles if and only if
O
the independence number α(G) ≤ n−2k and G is not one of two small exceptions in the
C case k = 3. We also show how the case k = 2 follows from Lov´asz’ characterization of
. multigraphs with no two disjoint cycles.
h
t
a
m
Mathematics Subject Classification: 05C15, 05C35.
[
Keywords: Disjoint cycles, Ore-degree, graph packing, equitable coloring, minimum degree.

1
v 1. Introduction
1
9
For a graph G = (V,E), let |G| = |V|, kGk = |E|, δ(G) be the minimum degree of G, and
7
3 α(G) be the independence number of G. Let G denote the complement of G and for disjoint
0 graphs G and H, let G∨H denote G∪H together with all edges from V(G) to V(H). The
.
1 degree of a vertex v in a graph H is d (v); when H is clear, we write d(v).
H
0
In 1963, Corr´adi and Hajnal proved a conjecture of Erd˝os by showing the following:
6
1
Theorem 1.1 ([6]). Let k ∈ Z+. Every graph G with (i) |G| ≥ 3k and (ii) δ(G) ≥ 2k
:
v
contains k disjoint cycles.
i
X
Clearly, hypothesis (i) in the theorem is sharp. Hypothesis (ii) also is sharp. Indeed, if a
r
a
graph G has k disjoint cycles, then α(G) ≤ |G|−2k, since every cycle contains at least two
vertices of G−I for any independent set I. Thus H := K ∨K satisfies (i) and has
k+1 2k−1
The first two authors thank Institut Mittag-Leffler (Djursholm, Sweden) for the hospitality and creative
environment.
Department of Mathematics and Statistics, Arizona State University, Tempe, AZ 85287, USA. E-mail
address: [email protected]. Research of this author is supported in part by NSA grant H98230-12-1-0212.
DepartmentofMathematics,UniversityofIllinois,Urbana,IL,61801,USAandInstituteofMathematics,
Novosibirsk,Russia. E-mailaddress: [email protected]. Researchof this authoris supported in part
by NSF grants DMS-0965587 and DMS-1266016 and by grant 12-01-00448 of the Russian Foundation for
Basic Research.
Corresponding author. Department of Mathematics, University of Illinois, Urbana, IL, 61801, USA. E-
mail address: [email protected]. Phone: [phone]. Researchof this author is supported in part
by NSF grant DMS-1266016.
1
δ(H) = 2k−1, but does not have k disjoint cycles, because α(H) = k+1 > |H|−2k. There
are several works refining Theorem 1.1. Dirac and Erd˝os [8] showed that if a graph G has
many more vertices of degree at least 2k than vertices of degree at most 2k−2, then G has
k disjoint cycles. Dirac [7] asked:
Question 1.2. Which (2k −1)-connected graphs do not have k disjoint cycles?
He also resolved his question for k = 2 by describing all 3-connected multigraphs on at
least 4 vertices in which every two cycles intersect. It turns out that the only simple 3-
connected graphs with this property are wheels. Lova´sz [22] fully described all multigraphs
in which every two cycles intersect.
The following result inthis paper yields a full answer to Dirac’squestion for simple graphs.
Theorem 1.3. Let k ≥ 2. Every graph G with (i) |G| ≥ 3k and (ii) δ(G) ≥ 2k−1 contains
k disjoint cycles if and only if
(H3) α(G) ≤ |G|−2k, and
(H4) if k is odd and |G| = 3k, then G 6= 2K ∨K and if k = 2 then G is not a wheel.
k k
Since for every independent set I in a graph G and every v ∈ I, N(v) ⊆ V(G) − I, if
δ(G) ≥ 2k − 1 and |I| ≥ |G| − 2k + 1, then |I| = |G| − 2k + 1 and N(v) = V(G) − I for
every v ∈ I. It follows that every graph G satisfying (ii) and not satisfying (H3) contains
K and is contained in K −E(K ). The conditions of Theorem 1.3 can
2k−1,|G|−2k+1 |G| |G|−2k+1
be tested in polynomial time.
Most likely, Dirac intended his question to refer to multigraphs; indeed, his result for
k = 2 is for multigraphs. But the case of simple graphs is the most important in the
question. In [19] we heavily use the results of this paper to obtain a characterization of
(2k−1)-connected multigraphs that contain k disjoint cycles, answering Question 1.2 in full.
Studying Hamiltonianproperties ofgraphs, Oreintroduced theminimum Ore-degreeσ : If
2
G is a complete graph, then σ (G) = ∞, otherwise σ (G) := min{d(x)+d(y) : xy 6∈ E(G)}.
2 2
Enomoto [9] and Wang [24] generalized the Corr´adi-Hajnal Theorem in terms of σ :
2
Theorem 1.4 ([9],[24]). Let k ∈ Z+. Every graph G with (i) |G| ≥ 3k and
(E2) σ (G) ≥ 4k −1
2
contains k disjoint cycles.
Again H := K ∨ K shows that hypothesis (E2) of Theorem 1.4 is sharp. What
k+1 2k−1
happens if we relax (E2) to (H2): σ (G) ≥ 4k−3, but again add hypothesis (H3)? Here are
2
two interesting examples.
Example 1.5. Letk = 3andY bethegraphobtainedbytwicesubdividing oneoftheedges
1
wz ofK , i.e., replacing wz bythepathwxyz. Then|Y | = 10 = 3k+1,σ (Y ) = 9 = 4k−3,
8 1 2 1
and α(Y ) = 2 ≤ |Y |−2k. However, Y does not contain k = 3 disjoint cycles, since each
1 1 1
cycle would need to contain three vertices of the original K (see Figure 1.1(a)).
8
Example 1.6. Let k = 3. Let Q be obtained from K by replacing a vertex v and its
4,4
incident edges vw,vx,vy,vz by new vertices u,u′ and edges uu′,uw,ux,u′y,u′z; so d(u) =
3 = d(u′) and contracting uu′ in Q yields K . Now set Y := K ∨Q. Then |Y | = 10 =
4,4 2 1 2
3k +1, σ (Y ) = 9 = 4k −3, and α(Y ) = 4 ≤ |Y |−2k. However, Y does not contain
2 2 2 2 2
k = 3 disjoint cycles, since each 3-cycle contains the only vertex of K (see Figure 1.1(b)).
1
2
u v
(a) Y1 (b) Y2
Figure 1.1
Our main result is:
Theorem 1.7. Let k ∈ Z+ with k ≥ 3. Every graph G with
(H1) |G| ≥ 3k +1,
(H2) σ (G) ≥ 4k −3, and
2
(H3) α(G) ≤ |G|−2k
contains k disjoint cycles, unless k = 3 and G ∈ {Y ,Y }. Furthermore, for fixed k there is
1 2
a polynomial time algorithm that either produces k disjoint cycles or demonstrates that one
of the hypotheses fails.
Theorem 1.7 is proved in Section 2. In Section 3 we discuss the case k = 2. In Section 4
we discuss connections to equitable colorings and derive Theorem 1.3 from Theorem 1.7 and
known results.
Nowweshowexamplesdemonstratingthesharpnessofhypothesis(H2)thatσ(G) ≥ 4k−3,
then discuss some unsolved problems, and then review our notation.
Example 1.8. Let k ≥ 3, Q = K and G := K ∨(K +Q). Then |G | = 4k −2 ≥
3 k 2k−2 2k−3 k
3k + 1, δ(G ) = 2k − 2 and α(G ) = |G | −2k. If G contained k disjoint cycles, then at
k k k k
least 4k − |G | = 2 would be 3-cycles; this is impossible, since any 3-cycle in G contains
k k
an edge of Q. This construction can be extended. Let k = r +t, where k + 3 ≤ 2r ≤ 2k,
Q′ = K , and put H = G ∨ Q′. Then |H| = 4r − 2 + 2t = 2k + 2r − 2 ≥ 3k + 1,
2t r
δ(H) = 2r−2+2t = 2k−2 and α(H) = 2r−2 = |H|−2k. If H contained k disjoint cycles,
then at least 4k −|H| = 2t+2 would be 3-cycles; this is impossible, since any 3-cycle in H
contains an edge of Q or a vertex of Q′.
There are several special examples for small k. The constructions of Y and Y can be
1 2
extended tok = 4atthecost oflowering σ to4k−4. Belowisanothersmall familyofspecial
2
examples. The blow-up of G by H is denoted by G[H]; that is, V(G[H]) = V(G)×V(H)
and (x,y)(x′,y′) ∈ E(G[H]) if and only if xx′ ∈ E(G), or x = x′ and yy′ ∈ E(H).
Example 1.9. For k = 4, G := C [K ] satisfies |G| = 15 ≥ 3k + 1, δ(G) = 2k − 2 and
5 3
α(G) = 6 < |G|−2k. Since girth(G) = 4, we see that G has at most |G| < k disjoint cycles.
4
This example can be extended to k = 5,6 as follows. Let I = K and H = G∨I. Then
2k−8
|G| = 2k +7 ≥ 3k +1, δ = 2k −2 and α(G) = 6 < |G|−2k = 7. If H has k disjoint cycles
then each of the at least k−(2k−8) = 8−k cycles that do not meet I use 4 vertices of G,
and the other cycles use at least 2 vertices of G. Then 15 = |G| ≥ 2k + 2(8 − k) = 16, a
contradiction.
3
Unsolved problems. 1. For every fixed k, we know only a finite number of extremal
examples. It would be very interesting to describe all graphs G with σ (G) = 4k−4 that do
2
not have k disjoint cycles, but this most likely would need new techniques and approaches.
2. Recently, there were several results in the spirit of the Corr´adi-Hajnal Theorem giving
degree conditions on a graph G sufficient for the existence in G of k disjoint copies of such
subgraphs as chorded cycles [1, 4] and Θ-graphs [5]. It could be that our techniques can help
in similar problems.
3. One also may try to sharpen the above-mentioned theorem of Dirac and Erd˝os [8].
Notation. A bud is a vertex with degree 0 or 1. A vertex is high if it has degree at least
2k −1, and low otherwise. For vertex subsets A,B of a graph G = (V,E), let
kA,Bk := X|{uv ∈ E(G) : v ∈ B}|.
u∈A
Note A and B need not be disjoint. For example, kV,Vk = 2kGk = 2|E|. We will abuse
this notation to a certain extent. If A is a subgraph of G, we write kA,Bk for kV(A),Bk,
and if A is a set of disjoint subgraphs, we write kA,Bk for V(H),B . Similarly, for
(cid:13)SH∈A (cid:13)
(cid:13) (cid:13)
u ∈ V(G), we write ku,Bk for k{u},Bk. Formally, an edge e = uv is the set {u,v}; we often
write ke,Ak for k{u,v},Ak.
If T is a tree or a directed cycle and u,v ∈ V(T) we write uTv for the unique subpath
of T with endpoints u and v. We also extend this: if w ∈/ T, but has exactly one neighbor
u ∈ T, we write wTv for w(T +w+wu)v. Finally, if w has exactly two neighbors u,v ∈ T,
we may write wTw for the cycle wuTvw.
2. Proof of Theorem 1.7
Suppose G = (V,E) is an edge-maximal counterexample to Theorem 1.7. That is, for some
k ≥ 3, (H1)–(H3) hold, and G does not contain k disjoint cycles, but adding any edge
e ∈ E(G) to G results in a graph with k disjoint cycles. The edge e will be in precisely one
of these cycles, so G contains k − 1 disjoint cycles, and at least three additional vertices.
Choose a set C of disjoint cycles in G so that:
(O1) |C| is maximized;
(O2) subject to (O1), |C| is minimized;
PC∈C
(O3) subjectto(O1)and(O2),thelengthofalongestpathP inR := G− C ismaximized;
S
(O4) subject to (O1), (O2), and (O3), kRk is maximized.
Call such a C an optimal set. We prove in Subsection 2.1 that R is a path, and in Subsection
2.2 that |R| = 3. We develop the structure of C in Subsection 2.3. Finally, in Subsection 2.4,
these results are used to prove Theorem 1.7.
Our arguments will have the following form. We will make a series of claims about our
optimal set C, and then show that if any part of a claim fails, then we could have improved
C by replacing a sequence C ,...,C ∈ C of at most three cycles by another sequence of
1 t
cycles C′,...,C′. Naturally, this modification may also change R or P. We will express the
1 t′
contradiction by writing “C′,...,C′,R′,P′ beats C ,...,C ,R,P,” and may drop R′ and R
1 t 1 t
or P′ and P if they are not involved in the optimality criteria.
This proof implies a polynomial time algorithm. We start by adding enough extra edges—
at most 3k—to obtain from G a graph with a set C of k disjoint cycles. Then we remove the
4
extra edges in C one at a time. After removing an extra edge, we calculate a new collection
C′. This is accomplished by checking the series of claims, each in polynomial time. If a
claim fails, we calculate a better collection (again in polynomial time) and restart the check,
or discover an independent set of size greater than |G| − 2k. As there can be at most n4
improvements, corresponding to adjusting the four parameters (O1)–(O4), this process ends
in polynomial time.
We now make some simple observations. Recall that |C| = k − 1 and R is acyclic. By
(O2) and our initial remarks, |R| ≥ 3. Let a and a be the endpoints of P. (Possibly, R is
1 2
an independent set, and a = a .)
1 2
Claim 2.1. For all w ∈ V(R) and C ∈ C, if kw,Ck ≥ 2 then 3 ≤ |C| ≤ 6 − kw,Ck. In
particular, (a) kw,Ck ≤ 3, (b) if kw,Ck = 3 then |C| = 3, and (c) if |C| = 4 then the two
neighbors of w in C are nonadjacent.
−→ −→
Proof. Let C be a cyclic orientation of C. For distinct u,v ∈ N(w)∩C, the cycles wuCvw
←− −→ ←−
and wuCvw have length at least |C| by (O2). Thus 2kCk ≤ kwuCvwk + kwuCvwk =
kCk+4, so |C| ≤ 4. Similarly, if kw,Ck ≥ 3 then 3kCk ≤ kCk+6, and so |C| = 3. (cid:3)
The next claim is a simple corollary of condition (O2).
Claim 2.2. If xy ∈ E(R) and C ∈ C with |C| ≥ 4 then N(x)∩N(y)∩C = ∅.
2.1. R is a path. Suppose R is not a path. Let L be the set of buds in R; then |L| ≥ 3.
Claim 2.3. For all C ∈ C, distinct x,y,z ∈ V(C), i ∈ [2], and u ∈ V(R−P):
(a) {ux,uy,a z} * E;
i
(b) k{u,a },Ck ≤ 4;
i
(c) {a x,a y,a z,zu} * E ;
i i 3−i
(d) if k{a ,a },Ck ≥ 5 then ku,Ck = 0;
1 2
(e) k{u,a },Rk ≥ 1; in particular ka ,Rk = 1 and |P| ≥ 2;
i i
(f) 4−ku,Rk ≤ k{u,a },Ck and k{u,a },Dk = 4 for at least |C|−ku,Rk cycles D ∈ C.
i i
Proof. (a) Else ux(C −z)yu,Pa z beats C,P by (O3) (see Figure 2.1(a)).
i
(b) Else |C| = 3 by Claim 2.1. Then there are distinct p,q,r ∈ V(C) with up,uq,a r ∈ E,
i
contradicting (a).
(c) Else a x(C −z)ya ,(P −a )a zu beats C,P by (O3) (see Figure 2.1(b)).
i i i 3−i
(d) Suppose k{a ,a },Ck ≥ 5 and p ∈ N(u)∩C. By Claim 2.1, |C| = 3. Pick j ∈ [2] with
1 2
pa ∈ E, preferring ka ,Ck = 2. Then V(C)−p ⊆ N(a ), contradicting (c).
j j 3−j
(e) Since a is an end of the maximal path P, we get N(a )∩R ⊆ P; so a u ∈/ E. By (b)
i i i
(2.1) 4(k −1) ≥ k{u,a },V rRk ≥ 4k −3−k{u,a },Rk.
i i
Thus k{u,a },Rk ≥ 1. Hence G[R] has an edge, |P| ≥ 2, and ka ,Pk = ka ,Rk = 1.
i i i
(f) By (2.1) and (e),k{u,a },V rRk ≥ 4|C| − ku,Rk. Using (b), this implies the second
i
assertion, and k{u,a },Ck+4(|C|−1) ≥ 4|C|−ku,Rk implies the first assertion. (cid:3)
i
Claim 2.4. |P| ≥ 3. In particular, a a 6∈ E(G).
1 2
Proof. Suppose |P| ≤ 2. Then ku,Rk ≤ 1. As |L| ≥ 3, there is a bud c ∈ Lr{a ,a }. By
1 2
Claim 2.3(f), there exists C = z ...z z ∈ C such that k{c,a },Ck = 4 and k{c,a },Ck ≥ 3.
1 t 1 1 2
5
x z z x
y y
u a u a a
i 3−i i
(a) (b)
Figure 2.1. Claim 2.3
If kc,Ck = 3 then the edge between a and C contradicts Claim 2.3(a). If kc,Ck = 1 then
1
k{a ,a },Ck = 5, contradicting Claim 2.3(d). Therefore, we assume kc,Ck = 2 = ka ,Ck
1 2 1
and ka ,Ck ≥ 1. By Claim 2.3(a), N(a )∪N(a ) = N(c), so there exists z ∈ N(a )∩N(a )
2 1 2 i 1 2
and z ∈ N(c)−z . Then a a z a ,cz z beats C,P by (O3). (cid:3)
j i 1 2 i 1 j j±1
Claim 2.5. Let c ∈ L−a −a , C ∈ C, and i ∈ [2].
1 2
(a) ka ,Ck = 3 if and only if kc,Ck = 0, and if and only if ka ,Ck = 3.
1 2
(b) There is at most one cycle D ∈ C with ka ,Dk = 3.
i
(c) For every C ∈ C, ka ,Ck ≥ 1 and kc,Ck ≤ 2.
i
(d) If k{a ,c},Ck = 4 then ka ,Ck = 2 = kc,Ck.
i i
Proof. (a) If kc,Ck = 0 then by Claims 2.1 and 2.3(f), ka ,Ck = 3. If ka ,Ck ≥ 3 then by
i i
Claim 2.3(b), kc,Ck ≤ 1. By Claim 2.3(f), ka ,Ck ≥ 2, and by Claim 2.3(d), kc,Ck = 0.
3−i
(b) As c ∈ L, kc,Rk ≤ 1. Thus Claim 2.3(f) implies kc,Dk = 0 for at most one cycle
D ∈ C.
(c) Suppose kc,Ck = 3. By Claim 2.3(a), k{a ,a },Ck = 0. By Claims 2.4 and 2.3(d):
1 2
4k −3 ≤ k{a ,a },R∪C ∪(V −R−C)k ≤ 2+0+4(k −2) = 4k −6,
1 2
a contradiction. Thus kc,Ck ≤ 2. Thus by Claim 2.3(f), ka ,Ck ≥ 1.
i
(cid:3)
(d) Now (d) follows from (a) and (c).
Claim 2.6. R has no isolated vertices.
Proof. Suppose c ∈ L is isolated. Fix C ∈ C. By Claim 2.3(f), k{c,a },Ck = 4. By
1
Claim 2.5(d), ka ,Ck = 2 = kc,Ck; so d(c) = 2(k − 1). By Claim 2.3(a), N(a ) ∩ C =
1 1
N(c)∩C. Let w ∈ V(C)rN(c). Then d(w) ≥ 4k−3−d(c) = 2k−1 = 2|C|+1. Therefore,
either kw,Rk ≥ 1 or |N(w) ∩ D| = 3 for some D ∈ C. In the first case, c(C − w)c beats
C by (O4). In the second case, by Claim 2.5(c) there exists some x ∈ N(a ) ∩ D. Then
1
c(C −w)c,w(D−x)w beats C,D by (O3). (cid:3)
Claim 2.7. L is an independent set.
Proof. Suppose c c ∈ E(L). By Claim 2.4, c ,c ∈/ P. By Claim 2.3(f) and using k ≥ 3,
1 2 1 2
there is C ∈ C with k{a ,c },Ck = 4 and k{a ,c },Ck, k{a ,c },Ck ≥ 3. By Claim 2.5(d),
1 1 1 2 2 1
ka ,Ck = 2 = kc ,Ck; so ka ,Ck, kc ,Ck ≥ 1. By Claim 2.3(a), N(a ) ∩C,N(a )∩C ⊆
1 1 2 2 1 2
N(c )∩C. Then there are distinct x,y ∈ N(c )∩C with xa ,xa ,ya ∈ E. If xc ∈ E then
1 1 1 2 1 2
c c xc , ya Pa beats C,P by (O3). Else a Pa xa , c (C −x)c c beats C,P by (O1). (cid:3)
1 2 1 1 2 1 2 1 1 2 1
6
Claim 2.8. If |L| ≥ 3 then for some D ∈ C, kl,Ck = 2 for every C ∈ C−D and every l ∈ L.
Proof. Suppose some D ,D ∈ C and l ,l ∈ L satisfy D 6= D and kl ,D k =6 2 6= kl ,D k.
1 2 1 2 1 2 1 1 2 2
CASE 1: l ∈/ {a ,a } for some j ∈ [2]. Say j = 1. For i ∈ [2]: k{a ,l },D k 6= 4
j 1 2 i 1 1
by Claim 2.5(d); k{a ,l },D k = 4 by Claim 2.3(f); ka ,D k = 2 by Claim 2.5(d). Then
i 1 2 i 2
l ∈/ {a ,a }. By Claim 2.7, l l 6∈ E(G). Claim 2.5(c) yields the contradiction:
2 1 2 1 2
4k −3 ≤ k{l ,l },R∪D ∪D ∪(V −R−D −D )k ≤ 2+3+3+4(k −3) = 4k −4.
1 2 1 2 1 2
CASE 2: {l ,l } ⊆ {a ,a }. Let c ∈ L−l −l . As above, k{l ,c},D k =6 4, and so kc,D k =
1 2 1 2 1 2 1 1 2
2 = kl ,D k. This implies l 6= l . By Claim 2.5(a,c), kl ,D k = 1. Thus k{l ,c},D k = 4;
1 2 1 2 2 2 2 1
so kc,D k = 2, and kl ,D k = 1. With Claim 2.4, this yields the contradiction:
1 1 1
4k −3 ≤ k{l ,l },R∪D ∪D ∪(V −R−D −D )k ≤ 2+3+3+4(k −3) = 4k −4.
1 2 1 2 1 2
(cid:3)
Claim 2.9. R is a subdivided star (possibly a path).
Proof. Suppose not. Then we claim R has distinct leaves c ,d ,c ,d ∈ L such that c Rd
1 1 2 2 1 1
and c Rd are disjoint paths. Indeed, if R is disconnected then each component has two
2 2
distinct leaves by Claim 2.6. Else R is a tree. As R is not a subdivided star, it has distinct
vertices s and s with degree at least three. Deleting the edges and interior vertices of s Rs
1 2 1 2
yields disjoint trees containing all leaves of R. Let T be the tree containing s , and pick
i i
c ,d ∈ T .
i i i
By Claim 2.8, using k ≥ 3, there is a cycle C ∈ C such that kl,Ck = 2 for all l ∈ L. By
Claim 2.3(a), N(a ) ∩C = N(l) ∩ C = N(a ) ∩ C =: {w ,w } for l ∈ L − a − a . Then
1 2 1 3 1 2
replacing C in C with w c Rd w and w c Rd w yields k disjoint cycles. (cid:3)
1 1 1 1 3 2 2 3
Claim 2.10. R is a path or a star.
a a a a
1 1 1 1
p p w p w p w
r a2 r a2 r l1 r l1
d d l2 l2
(a) (b) (c) (d)
Figure 2.2. Claim 2.10
Proof. By Claim 2.9, R is a subdivided star. If R is neither a path nor a star then there are
vertices r,p,d with kr,Rk ≥ 3, kp,Rk = 2, d ∈ L−a −a and (say) pa ∈ E. Then a Rd is
1 2 1 2
disjoint from pa (see Figure 2.2(a)). By Claim 2.5(c), d(d) ≤ 1+2(k −1) = 2k −1. Then:
1
(2.2) kp,V −Rk ≥ 4k −3−kp,Rk−d(d) ≥ 4k −5−(2k −1) = 2k −4 ≥ 2.
In each of the following cases, R∪C has two disjoint cycles, contradicting (O1).
CASE 1: kp,Ck = 3 for some C ∈ C. Then |C| = 3. By Claim 2.5(a), if kd,Ck =
0 then ka ,Ck = 3 = ka ,Ck. Then for w ∈ C, wa pw and a (C − w)a are disjoint
1 2 1 2 2
cycles (see Figure 2.2(b)). Else by Claim 2.5(c), kd,Ck, ka ,Ck ∈ {1,2}. By Claim 2.3(f),
2
7
k{d,a },Ck ≥ 3, so there are l ,l ∈ {a ,d} with kl ,Ck ≥ 1 and kl ,Ck = 2; say w ∈
2 1 2 2 1 2
N(l )∩C. If l w ∈ E then wl Rl w and p(C −w)p are disjoint cycles (see Figure 2.2(c));
1 2 1 2
else l wpRl and l (C −w)l are disjoint cycles (see Figure 2.2(d)).
1 1 2 2
CASE 2: There are distinct C ,C ∈ C with kp,C k, kp,C k ≥ 1. By Claim 2.8, for some
1 2 1 2
i ∈ [2] and all c ∈ L, kc,C k = 2. Let w ∈ N(p) ∩ C . If wa ∈ E then D := wpa w is a
i i 1 1
cycle and G[(C −w)∪a Rd] contains cycle disjoint from D. Else, if w ∈ N(a )∪N(d), say
i 2 2
w ∈ N(c), then a (C −w)a and cwpRc are disjoint cycles. Else, by Claim 2.1 there exist
1 i 1
vertices u ∈ N(a )∩N(d)∩C and v ∈ N(a )∩C −u. Then ua Rdu and a v(C −u)wpa
2 i 1 i 2 1 i 1
are disjoint cycles.
CASE 3: Otherwise. Then using (2.2), kp,V −Rk = 2 = kp,Ck for some C ∈ C. In this
case, k = 3 and d(p) = 4. By (H2), d(a ),d(d) ≥ 5. Say C = {C,D}. By Claim 2.3(b),
2
k{a ,d},Dk ≤ 4. Thus,
2
k{a ,d},Ck = k{a ,d},(V −R−D)k ≥ 10−2−4 = 4.
2 2
By Claim 2.5(c, d), ka ,Ck = kd,Ck = 2 and ka ,Ck ≥ 1. Say w ∈ N(a )∩C. If wp ∈ E
2 1 1
then dRa (C −w)d contains a cycle disjoint from wa pw. Else, by Claim 2.3(a) there exists
2 1
x ∈ N(a )∩N(d)∩C. If x 6= w then xa Rdx and wa p(C −x)w are disjoint cycles. Else
2 2 1
x = w, and xa Rdx and p(C −w)p are disjoint cycles. (cid:3)
2
Lemma 2.11. R is a path.
Proof. Suppose R is not a path. Then it is a star with root r and at least three leaves, any of
which can play the role of a or a leaf in L−a −a . Thus Claim 2.5(c) implies kl,Ck ∈ {1,2}
i 1 2
for all l ∈ L and C ∈ C. By Claim 2.8 there is D ∈ C such that for all l ∈ L and C ∈ C−D,
kl,Ck = 2. By Claim 2.3(f) there is l ∈ L such that for all c ∈ L−l, kc,Dk = 2. Fix distinct
leaves l′,l′′ ∈ L−l.
Let Z = N(l′)−R and A = V r(Z ∪{r}). By the first paragraph, every C ∈ C satisfies
|Z ∩C| = 2, so |A| = |G|−2k +1. For a contradiction, we show that A is independent.
Note A∩R = L, so by Claim 2.7, A∩R is independent. By Claim 2.3(a),
(2.3) for all c ∈ L and for all C ∈ C, N(c)∩C ⊆ Z.
Therefore, kL,Ak = 0. By Claim 2.1(c), for all C ∈ C, C ∩A is independent. Suppose, for
a contradiction, A is not independent. Then there exist distinct C ,C ∈ C, v ∈ A ∩ C ,
1 2 1 1
and v ∈ A ∩ C with v v ∈ E. Subject to this choose C with kv ,C k maximum. Let
2 2 1 2 2 1 2
Z ∩C = {x ,x } and Z ∩C = {y ,y }.
1 1 2 2 1 2
CASE 1: kv ,C k ≥ 2. Choose i ∈ [2] so that kv ,C −y k ≥ 2. Then define C∗ :=
1 2 1 2 i 1
v (C −y )v , C∗ := l′x (C −v )x l′, and P∗ := y l′′rl (see Figure 2.3(a)). By (2.3), P∗ is
1 2 i 1 2 1 1 1 2 i
a path and C∗ is a cycle. Then C∗,C∗,P∗ beats C ,C ,P by (O3).
2 1 2 1 2
CASE 2: kv ,C k ≤ 1. Then for all C ∈ C, kv ,Ck ≤ 2 and kv ,C k = 1; so kv ,Ck =
1 2 1 1 2 1
kv ,C ∪(C −C )k ≤ 1+ 2(k −2) = 2k −3. By (2.3) kv ,Lk = 0 and d(l) ≤ 2k −1. By
1 2 2 1
(H2), kv ,rk = kv ,Rk = (4k − 3) −kv ,Ck −d(l) ≤ (4k −3) − (2k −3) − (2k −1) = 1,
1 1 1
and v r ∈ E. Let C∗ := l′x (C − v )x l′, C∗ := l′′y (C − v )y l′′, and P∗ := v v rl (see
1 1 1 1 1 2 2 1 2 2 2 2 1
Figure 2.3(b)). Then C∗,C∗,P∗ beats C ,C ,P by (O3). (cid:3)
1 2 1 2
2.2. |R| = 3. By Lemma 2.11, R is a path, and by Claim 2.4, |R| ≥ 3. Next we prove
|R| = 3. First, we prove a claim that will also be useful in later sections.
8
v v v v
1 2 1 2
x x y x x y y
1 2 i 1 2 1 2
l′ r l′′ l′ r l′′
l l
(a) (b)
Figure 2.3. Claim 2.10
Claim 2.12. Let C be a cycle, P = v v ...v be a path in R, and 1 < i < s. At most one of
1 2 s
the following two statements holds.
(1) (a) kx,v Pv k ≥ 1 for all x ∈ C or (b) kx,v Pv k ≥ 2 for two x ∈ C;
1 i−1 1 i−1
(2) (c) ky,v Pv k ≥ 2 for some y ∈ C or (d) N(v )∩C 6= ∅ and kv Pv ,Ck ≥ 2.
i s i i+1 s
Proof. Suppose (1) and (2) hold. If (c) holds then the disjoint graphs G[v Pv + y] and
i s
G[v Pv ∪ C − y] contain cycles. Else (d) holds, but (c) fails; say z ∈ N(v ) ∩ C and
1 i−1 i
z ∈/ N(v Pv ). If (a) holds then G[v Pv +z] and G[v Pv ∪C−z] contain cycles. If (b)
i+1 s 1 i i+1 s
holds then G[v Pv +w] and G[v Pv ∪C−w] contain cycles, where kw,v Pv k ≥ 2. (cid:3)
1 i−1 i s 1 i−1
Suppose, for a contradiction, |R| ≥ 4. Say R = a a′a′′...a′′a′a . It is possible that
1 1 1 2 2 2
a′′ ∈ {a′′,a′}, etc. Set e := a a′ = {a ,a′} and F := e ∪e .
1 2 2 i i i i i 1 2
Claim 2.13. If C ∈ C, h ∈ [2] and ke ,Ck ≥ ke ,Ck then kC,Fk ≤ 7; if kC,Fk = 7 then
h 3−h
|C| = 3, ka ,Ck = 2, ka′ ,Ck = 3, ka′′Ra ,Ck = 2, and N(a )∩C = N(e )∩C.
h h h 3−h h 3−h
Proof. We will repeatedly use Claim 2.12 to obtain a contradiction to (O1) by showing that
G[C∪R] contains two disjoint cycles. Suppose kC,Fk ≥ 7 and say h = 1. Then ke ,Ck ≥ 4.
1
There is x ∈ e with kx,Ck ≥ 2. Thus |C| ≤ 4 by Claim 2.1, and if |C| = 4 then no vertex
1
in C has two adjacent neighbors in F. Then (1) holds with v = a and v = a′, even when
1 1 i 2
|C| = 4.
If ke ,Ck = 4, as is the case when |C| = 4, then ke ,Ck ≥ 3. If |C| = 4 there is a cycle
1 2
D := yza′a y for some y,z ∈ C. As (a) holds, G[a Ra′′∪C−y−z] contains another disjoint
2 2 1 2
cycle. Thus, |C| = 3. As (c) must fail with v = a′, (a) and (c) hold for v = a′ and v = a ,
i 2 i 1 1 2
a contradiction. Then ke ,Ck ≥ 5. If ka ,Ck = 3 then (a) and (c) hold with v = a and
1 1 1 1
v = a′. Now ka ,Ck = 2, ka′,Ck = 3 and ka′′Ra ,Ck ≥ 2. If there is b ∈ P − e and
i 1 1 1 1 2 1
c ∈ N(b)∩V(C)rN(a ) then G[a′Ra +c] and G[a (C−c)a ] both contain cycles. For every
1 1 2 1 1
b ∈ R−e , N(b)∩C ⊆ N(a ). Then if ka′′Ra ,Ck ≥ 3, (c) holds for v = a and v = a′′,
1 1 1 2 1 1 1 1
contradicting that (1) holds. Now ka′′Ra ,Ck = ke ,Ck = 2 and N(a ) = N(e ). (cid:3)
1 2 1 1 2
Lemma 2.14. |R| = 3 and m := max{|C| : C ∈ C} = 4.
Proof. Let t = |{C ∈ C : kF,Ck ≤ 6}| and r = |{C ∈ C : |C| ≥ 5}|. It suffices to show r = 0
and |R| = 3: then m ≤ 4, and |V(C)| = |G|−|R| ≥ 3(k − 1)+ 1 implies some C ∈ C has
9
length 4. Choose R so that:
(P1) R has as few low vertices as possible, and subject to this,
(P2) R has a low end if possible.
Let C ∈ C. By Claim 2.13, kF,Ck ≤ 7. By Claim 2.1, if |C| ≥ 5 then ka,Ck ≤ 1 for all
a ∈ F; so kF,Ck ≤ 4. Thus r ≤ t. Hence
(2.4) 2(4k −3) ≤ kF,(V rR)∪Rk ≤ 7(k −1)−t−2r +6 ≤ 7k −t−2r −1.
Therefore, 5−k ≥ t+2r ≥ 3r ≥ 0. Since k ≥ 3, this yields 3r ≤ t+2r ≤ 2, so r = 0 and
t ≤ 2, with t = 2 only if k = 3.
CASE 1: k −t ≥ 3. That is, there exist distinct cycles C ,C ∈ C with kF,C k ≥ 7. In
1 2 i
this case, t ≤ 1: if k = 3 then C = {C ,C } and t = 0; if k > 3 then t < 2. For both
1 2
i ∈ [2], Claim 2.13 yields kF,C k = 7, |C | = 3, and there is x ∈ V(C ) with kx ,Rk = 1 and
i i i i i
ky,Rk = 3 for both y ∈ V(C −x ). Moreover, there is a unique index j = β(i) ∈ [2] with
i i
a′,C = 3. For j ∈ [2], put I := {i ∈ [2] : β(i) = j}; that is, I = {i ∈ [2] : ka′,C k = 3}.
(cid:13) j i(cid:13) j j j i
(cid:13) (cid:13)
Then V(C )−x = N(a )∩C = N(e )∩C . As x a ∈/ E, one of x ,a is high. As
i i β(i) i 3−β(i) i i β(i) i β(i)
we can switch x and a (by replacing C with a (C −x )a and R with R−a +x ),
i β(i) i β(i) i i β(i) β(i) i
we may assume a is high.
β(i)
Suppose I 6= ∅ for both j ∈ [2]; say ka′,C k = ka′,C k = 3. Then for all B ∈ C and
j 1 1 2 2
j ∈ [2], a is high, and either ka ,Bk ≤ 2 or kF,Bk ≤ 6. Since t ≤ 1, we get
j j
2k−1 ≤ d(a ) = ka ,B∪Fk+ka ,C−Bk ≤ ka ,Bk+1+2(k−2)+t ≤ 2k−2+ka ,Bk.
j j j j j
Thus N(a )∩B 6= ∅forall B ∈ C. Lety ∈ N(a )∩C . Then using Claim2.13, y ∈ N(a ),
j j 3−j j j j
and a′(C −y )a′, a′(C −y )a′, a y a y a beats C ,C by (O1).
1 1 1 1 2 2 2 2 1 1 2 2 1 1 2
Otherwise, say I = ∅. If B ∈ C with kF,Bk ≤ 6 then ke ,Bk + 2ka ,Bk ≤ kF,Bk +
1 1 2
ka ,Bk ≤ 9. Thus, using Claim 2.13,
2
2(4k −3) ≤ d(a )+d(a′)+2d(a ) = 5+ke ,Ck+2ka ,Ck ≤ 5+6(k −1−t)+9t
1 1 2 1 2
⇒ 2k ≤ 5+3t.
Since k −t ≥ 3 (by the case), we see 3(k −t) + (5 + 3t) ≥ 3(3)+ 2k and so k ≥ 4. Since
t ≤ 1, in fact k = 4 and t = 1, and equality holds throughout: say B is the unique cycle in C
with kF,Bk ≤ 6. Then ka ,Bk = ke ,Bk = 3. Using Claim 2.13, d(a )+d(a′) = ke ,Rk+
2 1 1 1 1
ke ,C−Bk+ke ,Bk = 3+4+3 = 10, andd(a ),d(a ) ≥ (4k−3)−d(a ) = 13−(1+4+3) = 5,
1 1 1 2 2
so d(a ) = d(a ) = 5. Note a and a share no neighbors: they share none in R because R
1 2 1 2
is a path, they share none in C −B by Claim 2.13, and they share no neighbor b ∈ B lest
a a′ba and a (B −b)a beat B by (O1). Thus every vertex in V −e is high.
1 1 1 2 2 1
Since ke ,Bk = 3, first suppose ka ,Bk ≥ 2, say B − b ⊆ N(a ). Then a (B − b)a ,
1 1 1 1 1
a′a′a b beat B,R by (P1) (see Figure 2.4(a)). Now suppose ka′,Bk ≥ 2, this time with
1 2 1 1
B − b ⊆ N(a′). Since d(a ) = 5 and ka ,R ∪ Bk ≤ 2, there exists c ∈ C ∈ C − B with
1 1 1
a c ∈ E(G). Now c ∈ N(a ) by Claim 2.13, so a′(B − b)a′, a′(C −c)a′, and a ca b beat
1 2 1 1 2 2 1 2
B,C, and R by (P1) (see Figure 2.4(b)).
CASE 2: k − t ≤ 2. That is, kF,Ck ≤ 6 for all but at most one C ∈ C. Then,
since 5 − k ≥ t, we get k = 3 and kF,Vk ≤ 19. Say C = {C,D}, so kF,C ∪ Dk ≥
2(4k − 3) − kF,Rk = 2(4 · 3 − 3) − 6 = 12. By Claim 2.13, kF,Ck, kF,Dk ≤ 7. Then
kF,Ck, kF,Dk ≥ 5. If |R| ≥ 5, then for the (at most two) low vertices in R, we can choose
distinct vertices in R not adjacent to them. Then kR,V −Rk ≥ 5|R|−2−kR,Rk = 3|R|.
Thus we may assume kR,Ck ≥ ⌈3|R|/2⌉ ≥ |R| + 3 ≥ 8. Let w′ ∈ C be such that q =
10