the smallest and t is the largest element of J ∪K, then also n0, n1, . . . , ns and nt, nt+1, . . . , nd are monotonic. Hence we refer to sets {s, s+ 1, . . . , t}, with s, t ∈ J∪K∪ {0, d} and s+ 1, s+ 2, . . . , t−1∈/ J∪K, as monotonic intervals. The length of such an interval is defined as t−s.
Clearly, there exists a monotonic interval of length at least |J|+|K|+1d , the average length of the monotonic intervals. The main part of this proof is devoted to improving this bound and to expressing it in terms of d and r.
For this aim we first partition J into two disjoint subsets A and B.
For 0< i < d, consider the values ofithat belong toJ. We now partition J further into two disjoint subsets, A and B, where
A={i∈J | ni−2 ≤ni−1 < ni ≥ni+1 ≥ni+2} and B =J−A.
So an element i∈J is in B if and only if ni−1 < ni ≥ni+1 and, in addition, ni−2 > ni−1 or ni+1 < ni+2.
We note that eachi∈B is an end point of a monotonic interval of length 1. Indeed, for i∈B we have ni−1 < ni ≥ni+1 and, in addition, ni−2 > ni−1, in which case i−1∈K, or ni+1 < ni+2, in which case i+ 1∈K.
Since J and K alternate, no monotonic interval has both its end points in B. Hence there exists at least|B| monotonic intervals of length 1, while the remaining |J|+|K| − |B|+ 1 monotonic intervals have length at least 1.
Ifi∈J, then 2ni > ni−1+ni+1, and thus f(i) = ni
ni−1+ni+ni+1
> 1
3 for all i∈J . We now show that
f(j−1) +f(j) +f(j+ 1)> 2
3 for all j ∈A.
Forj ∈A, leta= f(j)1 = nj−1 +nnj + nj+1
j and thus,a <3 since 1a =f(j)> 13. By nj−2 ≤nj−1 and (a−1)nj =nj−1+nj+1,
f(j−1) = nj−1
nj−2 +nj−1+nj
≥ nj−1
2nj−1+nj
= (a−1)nj−1
(2a−2)nj−1+ (a−1)nj
= (a−1)nj−1 (2a−1)nj−1+nj+1.
Similarly, by nj+2 ≤nj+1 and (a−1)nj =nj−1+nj+1, f(j + 1)≥ (a−1)nj+1
(2a−1)nj+1 + nj−1
. Hence, in total,
f(j−1) +f(j) +f(j+ 1) ≥ (a−1)nj−1
(2a−1)nj−1+nj+1
+ 1
a + (a−1)nj+1 (2a−1)nj+1+nj−1
≥ (a−1)(nj−1+nj+1) 2a(nj−1+nj+1) + 1
a
= a+ 1 2a .
Thus, by a < 3 we have f(j −1) +f(j) +f(j + 1) > 23 for all j ∈ A, as desired.
Nowr ≥r(G)>Pdi=0f(i). Since the setsJ and K alternate, J does not contain two consecutive integers. Hence {i−1, i, i+ 1} and {j} are disjoint for all i ∈ A and j ∈ B. From the definition of A it is easy to see that i+ 2 ∈/ A if i ∈ A. Hence also the sets {i−1, i, i+ 1}, i ∈ A, are disjoint.
Therefore, r > X
i∈A
(f(i−1) +f(i) +f(i+ 1)) +X
j∈B
f(j)> 2
3|A|+1 3|B|, or, equivalently,
2|A|+|B|<3r.
We have |J|+|K|+ 1 monotonic intervals of total length d. At least |B|
intervals have length 1, so the remaining |J|+|K|+ 1− |B|= |A|+|K|+ 1 intervals have average length |A|+|K|+1d−|B| . Hence there exists a monotonic interval of at least this length.
We now bound the average length in terms of d and r. Since J and K alternate, we have |K| ≤ |J|+ 1. Hence,
|A|+|K|+ 1 ≤2|J|+ 2− |B|= 2|A|+|B|+ 2.
Hence there exists a monotonic interval of length at least d− |B|
|A|+|K|+ 1 ≥ d− |B|
2|A|+|B|+ 2 ≥ d
2|A|+|B|+ 2 −1.
Let{a, a+1, a+2, . . . , b}be such an interval. By 2|A|+|B|<3r, the interval has length
b−a ≥ d
3r+ 2 −1.
We assume that ni is monotone increasing on {a, a + 1, . . . , b} (if ni is de- creasing the proof is analogous). Then f(i) = n ni
i−1+ni+ni+1 ≥ 3nni
i+1, and hence,
b−1
X
i=a+1
ni ni+1 ≤3
b−1
X
i=a+1
f(i)<3r.
Note that, for S, xi >0, the product Q
i∈I
xi is maximized subject toPi∈Ixi ≤ S if xi =S/|I|for all xi. So,
na+1 = (
b−1
Y
i=a+1
ni
ni+1)nb <( 3r
b−a−1)b−a−1nb. Hence,
1≤na+1 <3r(3r+ 2) d−6r−4
(d−6r−4)/(3r+2)
n.
Now let r be constant and let x1 = d−6r−43r+2 . If x1 ≤ 3r/e, i.e., if d ≤
3r(3r+2)
e + 6r + 4, then the inequality of the theorem is satisfied for large n. So we can assume x1 > 3r/e. Let g(x) = (3rx)x = ex(log 3r−logx). Then g0(x) = (3rx)x(log 3r−logx−1). If x > 3re, then g0(x) < 0, and hence g is decreasing. It is shown in the previous page that b−a−1 ≥ x1 and that na+1 < ng(b−a−1). This implies 1 < ng(x1). Define x0 by g(x0) = 1/n.
Thus,
x0 3r
x0
=n.
Observe that, in each case under consideration, we regard r as fixed, whereas n grows beyond all bounds if and only if x0 does. Hence, logn = x0log(x3r0), and so log logn = logx0+ log log(x3r0). Thus,
logn
log logn = x0log x3r0 logx0+ log logx3r0. Hence,
x0 =logx0+ log logx3r0 logx0−log 3r
logn log logn
,
and thus,
x0 =logx0−log 3r
logx0−log 3r +log 3r+ log logx3r0 logx3r0
logn log logn
,
Hence 1 = (3rx
0)x0n, or equivalently (x3r0)x3r0 = n3r1 , has the solution x0 = (1 +o(1))log loglognn. Sincex0 = (1 +o(1))log loglognn >3r/eand g(x0)< f g(x1),we have x1 < x0. Hence,
d−6r−4 = (3r+ 2)x1 <(3r+ 2)(1 +o(1)) logn log logn, which yields the statement of the theorem.
Chapter 4
Diameter in Minimal Claw-free Graphs
4.1 Introduction
Graphs that do not contain a star on four vertices (claw) as an induced subgraph have received much attention, especially since the publication of the excellent survey paper [24] in 1997. This class of graphs includes, among others, line graphs, interval graphs, middle graphs, inflations of graphs and graphs with independence number equal to 2. Recently, Chudnovsky and Seymour found a structural characterization of claw-free graphs; that is, they defined certain classes of “basic” claw-free graphs and then showed that all claw-free graphs can be obtained by applying certain “expansion” operations.
See [6].
Definition 3 The graph K1,n is called a star. We refer to the star K1,3 as a clawwith the vertex of degree 3 as its centre.
In [12], Dankelmann et. al considered graphs that are (edge-) minimal with respect to the property of being claw-free. This was motivated by questions about cycles in claw-free graphs, but has interest in its own right.
Definition 4 Let G be a claw-free graph without isolated vertices. If the removal of any edge of G produces a graph that is not claw-free, then G is a minimal claw-free graph, briefly denoted as an m.c.f.g..
That not every claw-free graph contains an m.c.f.g. as a subgraph may be seen, for example, by considering the line graph of K4: it is obviously claw-free but one can repeatedly remove its edges until an empty graph is obtained without creating a claw. On the other hand, the line graph of K3,3 (equivalent to the cartesian product K3×K3) is an m.c.f.g.
In [12], Dankelmann et. al examined bounds on the minimum, average and maximum degrees of an m.c.f.g. and looked at the relationship between m.c.f.g.s and line graphs. For example, a 4-regular graph is m.c.f. if and only if it is the line graph of a (K4−e)-free cubic graph.
We mention that a closely related concept, minimal line graphs, was con- sidered by Sumner [46]. A graph is a minimal line graph if it is a line graph, but removal of any edge results in a graph that is not a line graph. Sumner proved that a graph G is a minimal line graph if and only if the following four conditions hold:
(i) every edge of G lies in a triangle,
(ii) every vertex of G has degree at least 3,
(iii) if an edge e lies on a triangle whose vertices have an even degree sum, then e lies on another triangle,
(iv) each 4-clique ofGhas at least two vertices adjacent to vertices outside the 4-clique.
Condition (i) clearly holds for m.c.f.g.s, and we will see that condition (ii) also holds for m.c.f.g.s.
An example of an m.c.f.g. is the 5-regular icosahedron on 12 vertices.
Indeed, if we delete one, two or three vertices from the same triangle, then the result is still an m.c.f.g. The latter is depicted in Figure 4.1. (This is not a line graph.) An exhaustive computer search has shown that the smallest order of an m.c.f.g. is 9; apart from the above graph there are two others, namely the line graphs of the two cubic graphs of order 6.
In this chapter, we examine the diameter of m.c.f.g.s. We prove that the diameter diam(G) of an m.c.f.g. G of order n satisfies
diam(G)≤ 4
9(n−20) + 7.
Moreover, we demonstrate that this bound is best possible.
Figure 4.1: The m.c.f.g I9.