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Extremal Graph Theory
Lectured by A. Thomason
Lent Term 2013
1 The Erdos-Stone Theorem 1
2 Stability 4
3 Supersaturation 6
4 Szemeredi’s Regularity Lemma 9
5 A couple of applications 13
6 Hypergraphs 16
7 The size of a hereditary property 19
8 Containers 22
9 The Local Lemma 25
10 Tail Estimation 26
11 Martingales Inequalities 29
12 The Chromatic Number of a Random Graph 31
13 The Semi-Random Method 33
Please let me know of corrections: [email protected]
Last updated: Sat 7th May, 2016
Course description
Extremal graph theory is an umbrella title for the study of graph properties and their dependenceon the values of graph parameters. This course builds on the material introduced in the PartII Graph Theory course, in particular Turan’s theorem and the Erdos-Stone theorem, as well asdeveloping the use of randomness in combinatorial proofs. Further techniques and extensionsto hypergraphs will be discussed. It is intended to cover some reasonably large subset of thefollowing.
The Erdos-Stone theorem and stability. Supersaturation. Szemeredi’s Regularity Lemma, withapplications. The number of complete subgraphs.
Hypergraphs. Erdos’s r-partite theorem. Instability. The Fano plane. Razborov’s flag algebras.Hereditary properties and their sizes.
Probabilistic tools: the Local Lemma and concentration inequalities. The chromatic numberof a random graph. The semi-random method, large independent sets and the Erdos-Hananiproblem. Dependent random choice.
Pre-requisite Mathematics
A knowledge of the basic concepts, techniques and results of graph theory, such as that affordedby the Part II Graph Theory course.
Literature
No book covers the course but the following can be helpful.
B. Bollobas, Modern graph theory, Graduate Texts in Mathematics 184, Springer-Verlag, NewYork (1998), xiv+394 pp.
N. Alon and J. Spencer, The Probabilistic Method, Wiley, 3rd ed. (2008)
1. The Erdos-Stone TheoremLecture 1
Recall:
Turan’s theorem. If |G| = n, e(G) > tr(n) and G 6⊃ Kr+1, then G = Tr(n), the r-partiteTuran graph of order n.
Note. Tr(n) is the complete r-partite graph with class sizes ⌈n/r⌉ and ⌊n/r⌋, and tr(n) ise(Tr(n)).
This answers the extremal problem forKr+1, and there is a unique extremal graph. The structureof Tr(n) invites many proofs by induction.
In general, we are interested in ex(n, F ) for some fixed graph F , where
ex (n, F ) = max
e(G) : |G| = n, F 6⊂ G
So Turan’s theorem states that ex (n,Kr+1) = tr(n) ≈(
1− 1
r
)(
n
2
)
.
Comment. It is true that tr(n) >(
1 − 1r
)(
n2
)
. This is equivalent to the average degree being
>(
1 − 1r
)
(n − 1). But in fact the minimum degree is >(
1 − 1r
)
(n − 1), as can beseen by looking at a vertex in the largest class and noting it misses at most 1
r (n − 1)vertices as neighbours. Equality holds only if there is precisely one largest class, in whichcase the average degree is greater than the minimum anyway, so in fact we always havetr(n) >
(
1− 1r
)(
n2
)
.
For general F there might be several extremal graphs and the extremal function might be hardto evaluate exactly.
Denote by Kr(t) the complete r-partite graph with t vertices per class.
So Kr = Kr(1) and Kr(t) = Tr(rt).
Lemma 1.1. Let r > 0 be an integer and ε > 0. Then there exist d = d′(r, ε) and n1 = n1(r, ε)such that, if |G| = n > n1 and, if r > 1,
δ(G) >
(
1− 1
r+ ε
)
n ,
then G ⊃ Kr+1(t), where t = ⌊d logn⌋.
Proof. If r = 0 or ε > 1/r then the assertion is trivial. We proceed by induction on r.
By the induction hypothesis, we may assume that G has a subgraph K = Kr(T ), where
T = ⌈2t/εr⌉. (This requires only that d′(r, ε) < εr3 d
′(
r − 1, 1r(r−1)
)
.)
Now, each vertex of K sends at least(
1− 1r + ε)n− |K| edges to G−K. Let U be the set
of vertices of G−K having at least(
1− 1r + ε
2
)
|K| neighbours in K.
Writing e(G−K,K) for the number of edges between G−K and K, we have
|K|[(
1− 1
r+ ε
)
n− |K|]
6 e(G−K,K) 6 |U ||K|+(
n− |U |)
(
1− 1
r+ε
2
)
|K|
1
orεn
2− |K| 6 |U |
(
1
r− ε
2
)
which, if n1(r, ε) is large enough, implies|U |r
>εn
3.
Se we may assume that |U | > εrn
3.
Now, each vertex in U is joined to at least
(
1− 1
r+ε
2
)
|K| − (r − 1)T =
(
1− 1
r+ε
2
)
rT − (r − 1)T =εrT
2> t
vertices in each class of K, and so is joined to some Kr(t) in K. But there are only(
Tt
)r
many Kr(t) in K, and, recalling that(
nk
)
6 ( enk )k, we have
(
T
t
)r
6
(
eT
t
)tr
6
(
3e
εr
)rd logn
6εrn
3t6
|U |t
if d′(r, ε) is small and n1(r, ε) is large.
Hence there exists W ⊂ U , with |W | > t, joined to the same Kr(t) in K.
Hence Kr+1(t) ⊂ G. 2
Lemma 1.2. Let c, ε > 0. Then there exists n2 = n2(c, ε) with the following property.
Suppose that |G| = n > n2 and e(G) > (c + ε)(
n2
)
. Then G has a subgraph H such that
δ(H) > c|H | and |H | > ε1/2n.
Proof. If not, there is a sequence G = Gn ⊃ Gn−1 ⊃ Gn−2 ⊃ · · · ⊃ Gs, where s = ⌊ε1/2n⌋,such that |Gj | = j and the only vertex in Gj not in Gj−1 has degree less than cj. Then
e(Gs) > (c+ ε)
(
n
2
)
−n∑
j=s+1
cj = (c+ ε)
(
n
2
)
− c
[(
n+ 1
2
)
−(
s+ 1
2
)]
>εn2
2>
(
s
2
)
provided n2 is large enough. Contradiction. 2
Theorem 1.3 (Erdos-Stone, 1946). Let r > 0 be an integer and ε > 0. Then there existLecture 2d = d(r, ε) and n0 = n0(r, ε) such that, if |G| = n > n0 and if for r > 1 we havee(G) >
(
1− 1r + ε
)(
n2
)
, then G ⊃ Kr+1(t) where t = ⌊d logn⌋.
Proof. Provided n0 > n2
(
1 − 1r + ε
2 ,ε2
)
, we may apply Lemma 1.2 to G to obtain a subgraph
H with δ(H) >(
1− 1r + ε
2
)
|H | and |H | > ε1/2n.
Provided n0 > ε−1/2n1(r,ε2 ), we may apply Lemma 1.1 to H to obtain Kr+1(t) with
t >⌊
d1(r,ε2 ) log ε
1/2n⌋
.
Provided n0 >1ε , if we take d(r, ε) = 1
2d1(r,ε2 ), we are done. 2
As observed by Erdos and Simonovits in the mid-1960s, we can determine ex(n, F ) asymptoti-cally for every F .
2
Theorem 1.4. Let F be a fixed graph with chromatic number r = χ(F ). Then
limn→∞
ex (n, F )(
n2
) = 1− 1
r − 1
Proof. Since χ(Tr−1(n)) = r − 1, we have F 6⊂ Tr−1(n), hence
ex (n, F ) > e(Tr−1(n)) >
(
1− 1
r − 1
)(
n
2
)
.
On the other hand, given ε > 0, if |G| = n and e(G) >(
1 − 1r−1 + ε
)(
n2
)
, then G ⊃Kr(|F |) ⊃ F if n is large enough, by the Erdos-Stone theorem.
Thus for every ε > 0, we have lim supex (n, F )(
n2
) 6
(
1− 1
r − 1+ ε
)
. 2
Here’s a pretty consequence of Erdos-Stone.
Define the upper density of an infinite graph to be the supremum of densities of large finitesubgraphs:
ud(G) = limn→∞
sup
x : ∃F ⊂ G, |F | > n, e(F ) > x
(|F |2
)
Corollary 1.5. ud(G) ∈ 0, 12 , 23 , 34 , . . . ∪ 1.
Can we strengthen Erdos-Stone to obtain larger t?
Theorem 1.6 Given r ∈ N, there exists εr > 0 such that, if ε < εr, there exists n(r, ε) suchthat for all n > n(r, ε) there is a graph G of order n with e(G) >
(
1 − 1r + ε
)(
n2
)
and
G 6⊃ Kr+1(t) where t =⌊
3 lognlog(1/ε)
⌋
.
Proof. Let W be a largest vertex class of Tr(n) with |W | = w = ⌈n/r⌉. Form G by addingε(
n2
)
edges within W so that G[W ] 6⊃ K2(t) and hence G 6⊃ Kr+1(t).
To see that this addition is possible, choose edges insideW independently with probabilityp = 3εr2. Take εr = (3r2)−6. So p < 1.
Let X be the number of edges chosen and Y the number of K2(t) formed by them. Then
E(X − Y ) = EX − EY = p
(
w
2
)
− 1
2
(
w
t
)(
w − t
t
)
pt2
Now,
w2t−2pt2−1 =
[
w2pt+1]t−1
<[
w2ε56(t+1)
]t+1
<[
w2n−5/2]t−1
<1
2.
Hence E(X − Y ) >p
2
(
w
2
)
. So there is a choice with X − Y >p
2
(
w
2
)
.
Remove an edge from each K2(t) to leave at least X − Y >p
2
(
w
2
)
> ε
(
n
2
)
edges with no
K2(t). 2
This shows dependence on n is correct. Our argument gives d(r, ε) >ε
2r(r − 1)!.
3
Bollobas, Erdos, Simonovits (1976) proved that t > crlogn
log(1/ε).
Chvatal, Szemeredi (1978) proved t >logn
log(1/ε).
2. StabilityLecture 3
An extremal problem is stable if all nearly-optimal examples have the same structure.
Theorem 2.1. Let t, r > 2 be fixed and suppose that G 6⊃ Kr+1(t). If e(G) =(
1− 1r +o(1)
)(
n2
)
then
(a) there exists Tr(n) on V (G) with |E(G)E(Tr(n))| = o(n2)
(b) G contains an r-partite subgraph of size(
1− 1r + o(1)
)(
n2
)
(c) G contains an r-partite subgraph of minimum degree(
1− 1r + o(1)
)
n
Global statement. Throughout, unless otherwise stated, |G| = n.
Proof. Note that (a) ⇐⇒ (b) and (c) =⇒ (b). Note too that we may jettison o(n) verticesshould we wish to. In particular, we may assume that δ(G) >
(
1− 1r+o(1)
)
n, so (b) =⇒ (c):
for otherwise, there exist εn vertices of degree at most(
1− 1r + η)n for some small ε < 1
4η,
and their removal leaves a graph of order (1 − ε)n and size >(
1− 1r + ε2
r
)(
(1−ε)n2
)
, so byErdos-Stone we would have a Kr+1(t).
Now, G ⊃ K = Kr(s) where logn > s = s(n) → ∞. Let Ci be the ith class of K, andlet X be the set of vertices joined to at least t in every class of K. THere are
(
st
)rmany
Kr(t) in J , and since Kr+1(t) ⊂ G, we have |X | 6 t(
st
)r= o(n). Jettison X .
Let Y be the set of vertices joined to fewer than (r − 1)s − s2t + t vertices of K. Since
δ(G) >(
1 − 1r + o(1)
)
n, we have e(K,G − K) > s(r − 1 + o(1))n. Because X has beenjettisoned, we have
(n− |Y |)(
(r − 1)s+ t)
+ |Y |(
(r − 1)s− s
2t+ t)
> s(
r − 1 + o(1))
n
So |Y | = o(n). Jettison Y .
The remaining vertices are partitioned by Vi, 1 6 i 6 r, where Vi is the set of verticesjoined to fewer than t of Ci. Since δ(G) >
(
1 − 1r + o(1)
)
n, it is enough to show that
e(
G[Vi])
= o(n2) for each i, for then V1, . . ., Vr are the parts of the r-partite graph we seek.
Suppose instead that e(
G[Vi])
> εn2, say. Then G[V1] ⊃ K2(t) by Erdos-Stone with r = 1.Each vertex of Ks(t) is joined to at least s − s
2t + 1 vertices in each class Ci. Hence Ci
contains a set of s− 2t(
s2t − 1
)
> t vertices joined to all of K2(t). But then G ⊃ Kr+1(t),a contradiction. 2
Corollary 2.2. Let χ(F ) = r + 1 and let G be extremal for F – i.e., F 6⊂ G, e(G) = ex(n, F ).Then δ(G) =
(
1− 1r + o(1)
)
n.
Proof. If not, by Theorem 1.4 we have δ(G) 6(
1− 1r −ε
)
n. By Theorem 2.1(c) with t = |F |, Ghas |F | vertices x1, . . ., x|F | joined to m =
(
1− 1r + o(1)
)
n common neighbours y1, . . ., ym.
Form G∗ from G−v (where v is a vertex of lowest degree) by adding a new vertex u joined
4
to y1, . . ., ym. Then e(G∗) > e(G), so G∗ ⊃ F . This copy of F in G∗ contains u but notxi, say. But then (F − u) ∪ xi contains F in G. Contradiction. 2
Stability can sometimes be used as a bootstrapping device to obtain exact results. For example,ex(n,C5) = t2(n) for n > 6. In fact, ex (n,C2k+1) = t2(n) if n is large. This is a special case ofthe following theorem.
Theorem 2.3 (Simonovits). Let F be (r+1)-edge-critical, i.e. χ(F ) = r+1 but χ(F−e) = rfor all edges e ∈ E(F ). Then ex (n, F ) = tr(n) for large n, and Tr(n) is the unique extremalgraph.
Proof. Let G be an extremal graph of order n. Select, by Theorem 2.1(c) using t = |F |, anr-partite subgraph H of minimum degree
(
1 − 1r + o(1)
)
n. Necessarily, each part of H is
of order(
1r + o(1)
)
n. Assign the o(n) vertices of G−H to the parts of H with the fewestneighbours.
Suppose a vertex x is joined to εn vertices in its own class. Then it has εn neighboursLecture 4in each part of H . These r sets of εn vertices span
(
1 − 1r + o(1)
)(
rεn2
)
edges because
δ(H) =(
1− 1r + o(1)
)
n.
By Erdos-Stone the neighbours of x span Kr(|F |) which contains F − v for any v ∈ V (F ).Hence G ⊃ F , a contradiction.
So each vertex has only o(n) neighbours in its own part. By Corollary 2.2, we knowδ(G) =
(
1− 1r + o(1)
)
n. So each vertex of G is joined to all but o(n) vertices in each otherclass. Suppose xy is an edge inside some class. Pick a set Z of |F | vertices in this class,including x, y.
Then all but o(n) vertices of the other classes are common neighbours of Z. Hence thesecommon neighbours span Kr−1(|F |), either by Erdos-Stone or directly. But Z togetherwith Kr−1(|F |) contains F , a contradiction.
Hence G is r-partite, and since Tr(n) is the unique r-partite graph of maximum size, wehave G = Tr(n). 2
Here is another example.
Theorem 2.4. Let r, s be fixed. For large n, the unique extremal graph for sKr+1 (i.e., sdisjoint copies of Kr+1) is Ks−1 + Tr(n− s+ 1) (where ‘+’ means ‘join all to all’).
Proof. The proof is similar to that of Theorem 2.3. Proceed by induction on s. The case s = 1is Turan’s theorem.
As before, choose H and assign G − H to the classes of H . Once again, if some vertexx has εn neighbours in its own class, then the neighbours of x span some Kr(s(r + 1)).But G − x cannot contain (s − 1)Kr+1, hence e(G − x) 6 e
(
Ks−2 + Tr(n − s + 1))
, so
e(G) 6 e(
Ks−1 + Tr(n− s+ 1))
.
So equality holds in both cases, so G−x = Ks−2+Tr(n−s+1) by the induction hypothesis,so G = Ks−1 + Tr(s− n+ 1).
Hence we may suppose that each vertex is joined to o(n) in its own class. Suppose someclass contains s independent edges. Let Z be the 2s endvertices of these edges. As in theprevious proof, the common neighbours of Z spanKr−1(s(r−1)), in which caseG ⊃ sKr+1.
5
Thus no class contains s independent edges, so the jth class contains a set Aj of 2(s− 1)vertices such that every edge in the jth class meets Aj . But each vertex in Aj has o(n)neighbours in the jth class.
Hence e(G) 6 2r(s− 1)o(n) + tr(n) < e(
Ks−1 + Tr(n− s+ 1))
. 2
3. Supersaturation
Supersaturation is the study of how many copies of F must exist in G if e(G) > ex (n, F ). Thebasic theorem holds in a general context.
Recall that an ℓ-uniform hypergraph is a pair G = (V,E) where
E ⊂ V (ℓ) =
Y ⊂ V : |Y | = ℓ
.
For a class F of ℓ-uniform hypergraphs, define
ex (n,F) = max
e(G) : |G| = n,G is ℓ-uniform, G contains no F ∈ F
and
π(F) = limn→∞
ex(n,F)(
nℓ
)
Exercise. This limit exists.
Theorem 3.1.LetH be an ℓ-uniform hypergraph. Then for all ε > 0 there exists δ = δ(H, ε) > 0such that every ℓ-uniform hypergraph G with |G| = n and e(G) >
(
π(H) + ε)(
nℓ
)
contains⌊
δn|H|⌋ copies of H .
Proof. For each m-set M ∈ V (m), let G[M ] be the hypergraph induced by G on M . If thereLecture 5are η
(
nm
)
sets M with e(G[M ]) >(
π(H) + ε2
)(
mℓ
)
then
(π(H) + ε)
(
n
ℓ
)
6 e(G) =
∑
M e(
G[M ])
(
n−ℓm−ℓ
) 6η(
nm
)(
mℓ
)
+ (1− η)(
nm
) (
π(H) + ε2
) (
mℓ
)
(
n−ℓm−ℓ
)
So, assuming n > m > ℓ, we have
π(H) + ε 6 η + (1− η)(
π(H) + ε2
)
or η >
ε2
1− π(H)− ε2
> 0 .
Pick m so that ex (m,H) <(
π(H) + ε2
)(
mℓ
)
. Then H ⊂ G[M ] for each of η(
nm
)
subsets M .
So G contains > η
(
n
m
)(
n− |H |m− |H |
)−1
distinct copies of H , i.e., > η
(
n
|H |
)(
m
|H |
)−1
copies.
Pick n0 > m so that
(
n
|H |
)
>1
2
n|H|
|H |! .
If n > n0 then pick δ 61
2
1
|H |!
(
m
|H |
)−1
small enough that⌊
δn|H|⌋ works for n 6 n0. 2
6
Back to graphs.
Let kp(G) denote the number of copies of Kp in G. Ramsey’s theorem says kp(G) + kp(G) > 0,if G is large (where G is the complement of G).
Theorem 3.2 (Lorden, 1962). Let G have degree sequence d1, . . ., dn. Then
k3(G) + k3(G) =
(
n
3
)
+ (n− 2)e(G) +n∑
i=1
(
di2
)
.
Proof. The number of paths of length 2 in G and G is∑(
di
2
)
+∑(
n−1−di
2
)
.
A complete or empty vertex triple contains 3 such paths, and every other triple containsexactly 1. So
(
n
3
)
+ 2(
k3(G) + k3(G))
=∑
(
di2
)
+∑
(
n− 1− di2
)
= 2∑
(
di2
)
− 2(n− 2)e(G) + 3
(
n
3
)
.
2
Corollary 3.3 (Goodman, 1959). k3(G) + k3(G) > 124n(n− 1)(n− 5).
Proof. Let m = e(G). Then k3(G) + k3(G) >
(
n
3
)
− (n− 2)m+ n
(
2m/n
2
)
. 2
These results show that k3(G) + k3(G) depends only on the degree sequence, and the minimumdensity of monochromatic triangles is at least 1/4, as attained by a random colouring. No suchresult holds for K4 – it is known that the density can be < 1/33.
Corollary 3.4. k3(G) >m
3n(4m− n2) where n = |G|, m = e(G).
Proof. Theorem 3.2 implies that k3(G)+k3(G) =
(
n
3
)
−(n−2)m+∑
(
di2
)
, where m = e(G)
and (di) is the degree sequence of G.
But 3k3(G) 6∑(
di
2
)
. So using m =(
n2
)
−m, we have
k3(G) >
(
n
3
)
− (n− 2)m+2
3n
(
2m/n
2
)
=m
3n(4m− n2) .
2
This bound is tight only for regular graphs containing no triples with just one edge. This meansthat G is a union of complete graphs, so G is complete multipartite. Such graphs are rare: theyare Tr(n) where r | n.
Given F , let iF (G) be the number of induced subgraphs of G isomorphic to F .
So, e.g., iKp(G) = kp(G).
Theorem 3.5. Let f(G) =∑
F αF iF (G), the sum being over a finite collection of F , each ofwhich is complete multipartite, with αF ∈ R, and αF > 0 unless F is complete.
7
Then, amongst graphs G of given order, f(G) is maximized on a complete multipartitegraph. Moreover, if αK3
> 0, there are no other maxima.
Proof. We may suppose that αK3> 0, the case αK3
= 0 following by a limiting argument.
Choose a graphG of order n that maximizes f(G). Suppose G is not complete multipartite.Then G contains two non-adjacent vertices x, y whose neighbourhoods X,Y differ.
The number iF (α) contains contributions from four kinds of F : those containing, respec-tively, x but not y, y but not x, both x and y, neither x nor y. The first contributiondepends only on X , and the second only on Y . Moreover the third depends only on X ∩Yand V − (X ∪ Y ) where V = V (G), since F is complete multipartite.
This fourfold partition of iF (G) means we can write
f(G) = g(X) + g(Y ) + h(X ∩ Y, V − (X ∪ Y )) + C,
where C is independent of X and Y .
Note that h(A,B) 6 h(A′, B′) if A ⊂ A′, B ⊂ B′, because F makes no contribution to h ifLecture 6F is complete, and otherwise αF > 0. Moreover, if also B 6= B′, then h(A,B) < h(A′, B′),but αK3
> 0 and iK3(G) contributes exactly αK3
|B| to h(A,B).
We may suppose that g(X) > g(Y ) and, if g(X) 6= g(Y ), that |X | 6 |Y | and so X 6= X∪Y .Therefore
g(X) + h(X,V −X) > g(Y ) + h(X ∩ Y, V − (X ∪ Y ))
Form H from G by removing all edges y to Y and inserting all edges y to X . Then
f(H) = 2g(X) + h(X,V −X) + C > f(G).
Contradiction. 2
Note that this transformation does not increase the clique size or the chromatic number.
Let 1 6 p 6 r. For 0 6 x 6(
np
)
, let ψ(x) be the maximal convex function defined by ψ(0) = 0,
ψ(
kp(Tq(n)))
= kr(Tq(n)). (Where q = r − 1, r, r + 1, . . .)
Theorem 3.6 (Bollobas, 1976). Let G be a graph of order n. Then kr(G) > ψ(kp(G))
Proof. Let f(G) = kp(G) − ckr(G) for some real c > 0. Since ψ is convex, it suffices to provethat f is maximized on a Turan graph. So let G be a graph on which f is maximized.
kp(Tr−1) kp(Tr) kp(Tr+1)
q
q
q
q
Suppose there exists G below ψ.Consider the straight line of gradientc > 0. The intercept on the axis iskp(G)− ckr(G).
By Theorem 3.5, we know that f is maximized on a complete-partite graph, with q classesof sizes 0 < a1 6 a2 6 · · · 6 aq. We may assume that q > r, else Tr−1(n) maximizes f .
Now, f(G) = a1aqA − ca1aqB + C, where A,B,C are non-zero rationals depending ona2, a3, . . ., aq−1 and a1 + aq.
8
We may assume without loss of generality that c is irrational, so A−cB 6= 0. If A−cB < 0,then replace a1 by 0 and aq by a1 + aq to increase f . If A− cB > 0 and a1 6 aq − 2, thenreplace a1 by a1 +1 and aq by aq − 1 to increase f . Thus a1 > aq − 1 and so G is a Turangraph. 2
The exact value of min
k3(G) : |G| = n, k2(G) = m
is unknown. It is conjectured to be givenby r-partite graphs where r is minimal.
k2(T2(n)) k2(T3(n))
=n2/4 =n2/3
q
q
q
k3(T4(n))
k3(T3(n))
.
...................
.................
...............
.................................
.................. ................... ...................... .........................
..................................
................................
...............................
.............................
............................ Cor. 3.4
I think. I foolishly tried to drawthis during the lecture, and so itmight not be right. . .
The continuous envelope here for n2/4 6 m 6 n2/3 was posed as a lower bound by Fisher(1989). The whole range m 6
(
n2
)
was proved (again in the limit) by Razborov, who introducedthe method of flag algebras (2002), a lowbrow view of this being a massive generalisation ofCauchy-Schwarz, using CSP to find optimal quadratic forms.
Nikiforov (2008) did likewise for n = 4. Reiher (2012) did likewise for all r, for p = 2.
Open problem. On inducibility: what is maxilength 3 paths(G)?
4. Szemeredi’s Regularity LemmaLecture 7
A graph having the (large scale) property that its (induced) subgraphs all have roughly thesame density, the graph itself can be regarded as “pseudo-random” in a sense that can be madeprecise. Consider the following bipartite version.
Let U,W be disjoint subsets of the vertex set of some graph. The number of edges between Uand W is denoted by e(U,W ).
The density d(U,W ) ise(U,W )
|U ||W | .
Definition 4.1. Let 0 < ε < 1. The pair (U,W ) is said to be ε-uniform (or ε-regular) if forU ′ ⊂ U , W ′ ⊂W ,
∣
∣d(U ′,W ′)− d(U,W )∣
∣ < ε whenever |U ′| > ε|U |, |W ′| > ε|W |
(Clearly a lower bound on |U ′| is necessary – e.g., |U ′| = 1 is hopeless.)
An ε-uniform pair is roughly regular.
Lemma 4.2. Let (U,W ) be ε-uniform and d(U,W ) = d. Then∣
∣
u ∈ U : |Γ(u) ∩W | > (d− ε)|W |∣
∣ > (1− ε)|U |
and∣
∣
u ∈ U : |Γ(u) ∩W | < (d+ ε)|W |∣
∣ > (1− ε)|U |
9
(Recall that Γ(u) is the neighbourhood of u.)
Proof. Let X = u ∈ U : |Γ(u) ∩ W | 6 (d − ε)|W |. Then e(X,W ) 6 (d − ε)|X ||W |, sod(X,W ) 6 d−ε. By the definition of density, |X | 6 ε|U |, which proves the first inequality.
The proof of the other is similar. Alternatively, observe that in the complementary graph,(U,W ) is ε-uniform with density 1− d, and apply the first inequality. 2
Repeated applications give lots of information about graph structure. For example, given agraph H with vertices v1, . . ., vk and subsets V1, . . ., Vk of V (G) such that (Vi, Vj) is ε-uniform,we may find a copy of H in G. More is true.
Lemma 4.3. Let H be a graph with ∆(H) = d, and suppose that H has an r-colouring inwhich no colour is used more than s times. Let G be a graph containing disjoint vertexsubsets V1, . . ., Vr with |Vi| = u such that for all i, j, (Vi, Vj) is ε-uniform and d(Vi, Vj) > λ.
Suppose that (d+ 1)ε 6 λd and s 6 ⌈εu⌉. Then G contains a copy of H .
Proof. Let c : V (H) → 1, . . ., r be an r-colouring of H in which no colour is used more than stimes. Let V (H) = v1, . . ., vk. We will select vertices x1, . . ., xk in G so that xixj ∈ E(G)if vivj ∈ E(H).
We claim that for 0 6 ℓ 6 k, vertices x1, . . ., xℓ can be chosen so that xi ∈ Vc(vi) and,
for ℓ < j 6 k there is a set xℓj ⊂ Vc(vj) of candidates for xj at stage ℓ, meaning that
xiyj ∈ E(G) for every yj ∈ Xℓj and every xi ∈ N(j, ℓ) = xi : 1 6 i 6 ℓ and vivj ∈ E(H).
Moreover, |Xℓj | > (λ− ε)|N(j,ℓ)||Vc(vj)|.
The claim clearly holds for ℓ = 0 – just take X0j = Vc(vj). Proceed by induction on ℓ. In
general, for each t ∈ T = j > ℓ+ 1 : vℓ+1vj ∈ E(H), letYt =
y ∈ Xℓℓ+1 : |Γ(y) ∩Xℓ
t | 6 (λ− ε)|Xℓt |
.
Let m = N(ℓ+ 1, ℓ), and note that m+ |T | 6 d.
Since d(Yt, Xℓt ) 6 λ− ε 6 d(Vc(vℓ+1), Vc(vt))− ε and
|Xℓt | > (λ− ε)d−1|Vc(vt) > (λd−1 − (d− 1)ε)|Vc(vt)| > ε|Vc(vt)|
it follows that |Yt| 6 ε|Vc(vℓ+1)|.
Therefore,∣
∣
∣
∣
Xℓℓ+1 −
⋃
t∈T
Yt
∣
∣
∣
∣
> (λ− ε)m|Vc(vℓ+1) − (d−m)|Vc(vℓ+1)|
>(
λm −mε− (d−m)ε)
|Vc(vℓ+1)|> ⌈εu⌉ > s
At most s − 1 vertices of Xℓℓ+1 −
⋃
t∈T Yt have been used for x1, . . ., xℓ, so we may select
xℓ+1 from this set. Take Xℓ+1t −Xℓ
t ∩ Γ(xℓ+1) for t ∈ T , and Xℓ+1t = Xℓ
t for t /∈ T . 2
Corollary 4.4. Let H be a graph with vertex set v1, . . ., vk. Let 0 < λ, η < 1 satisfy kη 6 λk−1.Lecture 8Let G be a graph with vertex set V1 ∪ · · · ∪ Vk, where the Vi are disjoint sets of size y > 1.
Suppose that each pair (Vi, Vj) is η-uniform, that d(Vi, Vj) > λ if vivj ∈ E(H), and thatd(Vi, Vj) 6 1− λ if vivj /∈ E(H).
10
Then there exist vertices x1, . . ., xk (xi ∈ Vi) such that the map vi 7→ xi gives an isomor-phism between H and G[x1, . . ., xk].
Proof. Note that, by replacing the set of Vi−Vj edges by the complementary set if vivj /∈ E(H),we may assume that H is complete and d(Vi, Vj) > λ for all i, j.
The result is then immediate from Lemma 4.3 on taking ε = η, r = k, d = k − 1, ands = 1. 2
It is a remarkable fact that every graph can be partitioned into a bounded number of pieces,almost all pairs of which are ε-uniform. This result is due to Szemeredi and is used in his proofthat sets of integers with positive density contain arbitrarily long arithmetic progressions.
An equipartition of V (G) into k parts is a partition V1, . . ., Vk such that ⌊n/k⌋ 6 |Vi| 6 ⌈n/k⌉for all 1 6 i 6 k, where n = |V (G)|. The partition is ε-uniform if (Vi, Vj) is ε-uniform for all
but at most ε(
k2
)
pairs, for 1 6 i < j 6 k.
Theorem 4.5 (Szemeredi’s Regularity Lemma). Let 0 < ε < 1 and let ℓ be a naturalnumber. Then there exists L = L(ℓ, ε) such that every graph has an ε-uniform equipartitioninto m parts, for some ℓ 6 m 6 L.
Before proving the lemma, we establish two useful facts.
Lemma 4.6. Let U ′ ⊂ U and W ′ ⊂W satisfy |U ′| > (1− δ)|U | and |W ′| > (1− δ)|W |.
Then∣
∣d(U ′,W ′)− d(U,W )∣
∣ 6 2δ.
Proof. Let d = d(U,W ) and d′ = d(U ′,W ′). Then
d = d(U,W ) =e(U,W )
|U ||W | >e(U ′,W ′)
|U ||W | = d(U ′,W ′)|U ′||W ′||U ||W | > d′(1− δ)2
So d′ − d 6 d′(
1− (1− δ)2)
6 2δd′ 6 2δ.
By considering the complementary graph, (1− d′)− (1 − d) 6 2δ, i.e., d− d′ 6 2δ. 2
Lemma 4.7. Let x1, . . ., xn be real numbers such that X = 1n
n∑
i=1
xi. Let x = 1m
m∑
i=1
xi. Then
1
n
n∑
i=1
x2i > X2 +m
n−m(x−X)2 > X2 +
m
n(x−X)2
Proof. We have
1
n
n∑
i=1
x2i =1
n
m∑
i=1
x2i +1
n
n∑
i=m+1
x2i
>m
nx2 +
n−m
n
(
nX −mx
n−m
)2
= X2 +m
n−m(x−X)2
by two applications of Cauchy-Schwarz. 2
11
Proof of Theorem 4.5. Define the index ind(P ) of an equipartition P with k parts V1, . . ., Vkto be
1
k2
∑
i<j
d2(Vi, Vj) .
We shall show that, if n > k16k and 4kε5 > 100, and P is not ε-uniform, then there existsan equipartition Q into k4k parts with ind(Q) > ind(P ) + ε5/8.
This will be enough to prove the theorem. For, choose t > ℓ with 4tε5 > 100. Definef(0) = t and f(j + 1) = f(j)4f(j). Let N = f(⌈4ε−5⌉). Let L = N16N . Then if n 6 L,take an equipartition into n single vertices. Otherwise, begin with an equipartition P intot parts.
So long as the current partition into k parts is not ε-uniform, replace it with one into k4k
parts and larger index. Since ind(P ) 6 12 , replacement can occur at most ⌈4ε−5⌉ times, so
an ε-uniform partition is found with 6 L parts.
For each (Vi, Vj) that is not ε-uniform, select witness sets Xij ⊂ Vi and Xji ⊂ Vj suchthat |Xij | > ε|Xi| and |Xji| > ε|Xj|, and
∣
∣d(Xij , Xji)− d(Vi, Vj)∣
∣ > ε. For each i, the setsXij partition Vi into at most 2k−1 atoms.
Let m = ⌊n/k4k⌋ and let n = k4km + ak + b where 0 6 a < 4k and 0 6 b 6 k. Then⌊n/k⌋ = 4km+ a, so the parts of P have size 4km+ a or 4km+ a+ 1, with B parts havethe larger.
Partition each part of P into 4k sets of size m or m+ 1. Any such partition is a partitionLecture 9of G into k4k sets of sizes m or m+ 1, i.e. an equipartition.
Choose such a partition Q whose parts are as much as possible inside atoms: that is, everyatom is a union of part of Q together with at most m extra vertices. All that remains isto check that ind(Q) > ind(P ) + ε5/8.
Let the setes of Q within Vi be Vi(s), for 1 6 s 6 q = 4k. That is, Vi =⋃q
s=1 Vi(s). Notethat
∑
16s,t6q e(
Vi(s), Vj(t))
= e(Vi, Vj) and |Vi| > q|Vi(s)| mm+1 for each s, and so
1
q2
∑
16s,t6q
d(
Vi(s), Vj(t))
>
(
m
m+ 1
)2
d(Vi, Vj) .
Since n > k16k, we have
(
m
m+ 1
)2
> 1− 2
m> 1− 2
4k> 1− ε5
50.
Therefore
1
q2
∑
16s,t6q
d2(
Vi(s), Vj(t)) C.S.
>
(
1
q2
∑
16s,t6q
d(
Vi(s), Vj(t))
)2
> d2(Vi, Vj)−ε5
25
The main point is that we can improve this bound if (Vi, Vj) is not uniform. Let X∗ij be
the largest subset of Xij that is a union of parts of Q. Because we choose parts of Q asmuch as possible within atoms, and because |Vi| 6 ⌊nl⌋4km+ a > 4km, we have
∣
∣X∗ij
∣
∣ >∣
∣Xij
∣
∣− 2k−1m >∣
∣Xij
∣
∣
(
1− 2k−1m
ε|Vi|
)
>∣
∣Xij
∣
∣
(
1− 1
ε2k
)
>∣
∣Xij
∣
∣
(
1− ε
10
)
12
so by Lemma 4.6, we have∣
∣d(X∗ij , X
∗ji)− d(Xij , Xji)
∣
∣ 6ε
5.
We may assume that X∗ij =
⋃
16s6ri
Vi(s) for some number ri.
By a similar argument to that above,∣
∣
∣
∣
∣
∣
1
rirj
∑
16s6ri
∑
16t6rj
d(
Vi(s), Vj(t))
− d(X∗ij , X
∗ji)
∣
∣
∣
∣
∣
∣
6ε5
50
Recalling that |d(Xi, Xj)− d(Vi, Vj)| > ε, we see that∣
∣
∣
∣
∣
∣
1
rirj
∑
16s6ri
∑
16t6rj
d(
Vi(s), Vj(t))
− d(Vi, Vj)
∣
∣
∣
∣
∣
∣
>3ε
4
Applying Lemma 4.7 in the above, use of Cauchy-Schwarz with n = q2, m = rirj gives
1
q2
∑
16s,t6q
d2(
Vi(s), Vj(t))
> d2(Vi, Vj)−ε5
25+rirjq2
9ε2
16> d2(Vi, Vj)−
ε5
25+ε4
3
because
riq
>
(
1− 1
m
)
ri
(
m+ 1
m
)
1
q>
(
1− 1
m
) |X∗ij |
|Vi|>
(
1− 1
m
)
(
1− ε
10
) |Xij ||Vi|
>4ε
5.
Therefore,
ind(Q) =1
k2q2
∑
16i,j6k
∑
16s,t6q
d2(
Vi(s), Vj(t))
>1
k2
∑
16i,j6k
(
d2(Vi, Vj)−ε5
25
)
+1
k2ε
(
k
2
)
ε4
3
> ind(P ) +ε5
8
completing the proof. 2
The proof gives something like L = 2222
2
, with ε−5 many 2s in the tower.
Gowers (1997) showed a tower of height ε−1/16 is necessary. Indeed, define an equipartitionV1, . . ., Vk to be (ε, δ, η)-uniform if for all but η
(
k2
)
of the pairs (Vi, Vj), it holds that∣
∣d(V ′i , V
′j )−
d(Vi, Vj)∣
∣ 6 ε whenever |V ′i | > δ|Vi|.
Gowers proved that for small δ there is a graph such that every (1−δ1/16, δ, 1−20δ1/16)-uniform
partition has at least 2222
(with δ−1/16 many 2s) parts.
5. A couple of applicationsLecture 10
The simplest non-trivial case of Ramsey’s theorem asserts that there is a smallest integer R(k)such that if the edges of KR(k) are coloured red or blue then we get a monochromatic Kk. It
is known that 2k/2 6 R(k) 6 4k. The existence of R(k) implies that for any graph G there isa minimal integer r(G) such that any red/blue colouring of of Kr(G) contains a monochromaticG. Clearly r(G) 6 R(|G|).
13
Theorem 5.1. Given an integer d, there exists a number c(d) such that r(G) 6 c(d)|G| if∆(G) 6 d.
Remark. The same is conjectured to be true for every graph with e(H) 6 d|H | for all H ⊂ G.
It is known for a wider class than in the theorem, including planar graphs.
Proof. Let t = R(d+ 1). Choose ε 6 min
1t ,
12d(d+1).
Let ℓ > t2 and let L = L(ℓ, ε) be the number given by Szemeredi’s Regularity Lemma(SRL). Finally, let c = L/ε.
Let G be a gaph a maximum degree 6 d and let the edges of Kn, where n > c|G|, becoloured red/blue. Apply Szemeredi’s Lemma to R, the red subgraph of Kn, with ℓ, ε asabove.
Let H be the graph with vertex set V1, . . ., Vm, where V1, . . ., Vm is the partition ofR given by SRL. Let ViVj ∈ E(H) if (Vi, Vj) is ε-uniform. Then |H | = m > t2 ande(H) 6 ε
(
m2
)
. So H ⊃ Kt. [Or else, by Turan’s theorem, there exist integers m1, . . .,mt−1
with∑
mi = m and e(H) >∑(
mi
2
)
> (t− 1)(
m/(t−1)2
)
> ε(
m2
)
where ε 6 1t and m > t2.]
Thus we may assume that every pair (Vi, Vj), 1 6 i < j 6 t is ε-uniform. Colour the edgesof Kt green if d(Vi, Vj) >
12 , and yellow otherwise.
Since t = R(d + 1), there exists a monochromatic Kd+1, and we may assume that it isspanned by V1, . . ., Vd+1. Therefore, d(Vi, Vj) >
12 for 1 6 i < j 6 d+ 1, or d(Vi, Vj) <
12
for 1 6 i < j 6 d+ 1.
We will show that in the first case we have R ⊃ G, and that in the other case we obtainsimilarly a blue G in Kn.
Take a vertex colouring of G with d + 1 colours in which no colour is used more than|G| times. Lemma 4.3 applied with H(there) = G(here), G(there) = (subgraph of Rspanned by V1, . . ., Vd+1), u = |Vi| > n/L > c|G|/L > |G|/ε, s = |G|, r = d + 1, λ = 1
2 ,(d+ 1)ε 6 1/2d.
This completes the proof. 2
Theorem 5.1 was proved by Chvatal, Rodl, Szemeredi, Trotter in 1983. It was extended to alarger class (using a modification of Lemma 4.3) by Chen and Schelp in 1993. Graham, Rodl,
Rucinski showed in 1999 that c(d) 6 2Cd log2d (obviously without SRL).
Observe that the Erdos-Stone theorem follows from SRL and Turan’s theorem. Indeed, if e(G) >(
1− 1r + ε
)(
n2
)
, form the “reduced” graph H whose edges correspond to pairs of positive density.Then e(H) > tr(|H |) so H ⊃ Kr+1, and so G ⊃ Kr+1 by Lemma 4.3.
We make this precise, but, more interestingly, we recover stability too. We use an argument ofErdos.
Theorem 5.2 (Erdos, 1970). Let G be a Kr+1-free graph. Then there is a (complete) r-partite graph H with V (H) = V (G) and dH(v) > dG(v) for all v ∈ V (G).
Proof. By induction on r. The case r = 1 is trivial.
14
In general, let x be a vertex of maximum degree. Then G′ = G[Γ(x)] is Kr-free, soby induction there is an (r − 1)-partite graph H ′ with V (H ′) = V (G′) = Γ(x), anddH′ (v) > dG′(v) for all v ∈ Γ(x).
Form H by joining every vertex of V (G) \ Γ(x) to every vertex in Γ(x). Then H is r-partite and if v /∈ Γ(x) then dH(x) = |Γ(x)| = dG(x) > dG(v), while if v ∈ Γ(x) thendH(v) = dH′ (v) + |V (G) \ Γ(x)| > dG′(v) + |V (G) \ Γ(x)| > dG(v). 2
Theorem 5.3 (Furedi, 2010). Let G be aKr+1-free graph. Then there is a complete r-partiteLecture 11graph H with V (H) = V (G) and |E(G) \ E(H)| 6 1
2 |E(H) \E(G)|.
In particular, if e(G) > tr(|G|) − k then |E(H) E(G)| 6 3k.
Proof. Construct H as in the previous proof. The proof is via induction, the case r = 1 beingtrivial.
Now |E(G′) \E(H ′)| 6 12 |E(H ′) \E(G′)|. Let there be e edges from G inside V (G) \Γ(x),
and f edges missing from G between Γ(x) and V (G) \ Γ(x). Then |E(G) \ E(H)| =|E(G′) \E(H ′)|+ e, and |E(H) \E(G)| = |E(H ′) \E(G′)|+ f , so it suffices to show thate 6 1
2f .
For each v ∈ V (G)\Γ(x), let e(v) be the number of edges of G inside V (G)\Γ(x) meetingv, and let f(v) be the number of edges missing from G between Γ(x) and V (G) \ Γ(x)meeting v. Then e = 1
2
∑
ve(v) and f = 1
2
∑
vf(v).
But for each v, dG(x) > dG(v) = dG(x) − f(v) + e(v), so e(v) 6 f(v), and we are done.
Finally, if e(G) > tr(|G|)−k, then tr(|G|) > e(H) = e(G)−|E(G)\E(H)|+|E(H)\E(G)| >tr(G) − k + 1
2 |E(H) \ E(G)|, so |E(H) \ E(G)| 6 2k, hence |E(G) \ E(H)| 6 k, and so|E(H)E(G)| 6 3k. 2
Definition 5.4. Let G be a graph with an ε-uniform partition P into m parts V1, . . ., Vm. Letλ ∈ R. The reduced graph G(P, λ) is the graph of order M with vertex set V1, . . ., Vmand edge set ViVj : (Vi, Vj) is ε-uniform and d(Vi, Vj) 6 λ.
Lemma 5.5. Let G, P be as in this definition. If e(G) > c(|G|
2
)
then
e(G(P, λ)) 6
(
c− ε− λ− 1
m− 1
)(
m
2
)
,
provided |G| > 2m2.
Proof. Let e(G(P, λ)) = d(
m2
)
and n = |G|. Then
e(G) 6 d
(
m
2
)(
n
m+ 1
)2
+
(
m
2
)
ε
(
n
m+ 1
)2
+
(
m
2
)
λ
(
n
m+ 1
)2
+m
( nm + 1
2
)
6
(
d+ ε+ λ+1
m− 1
)(
m
2
)(
n
m+ 1
)2
6
(
d+ ε+ λ+1
m− 1
)(
n
2
)
2
Theorem 5.6. Let r, t ∈ N and let ε > 0. Then there exists n0(r, t, ε) such that, if |G| = n > n0
and G 6⊃ Kr+1(t), then G has a subgraph G′ with e(G′) > e(G)− ε(
n2
)
and G′ 6⊃ Kr+1.
15
Proof. Let λ = ε/2. Pick η so that (rt+ 1)η 6 λrt and η 6 ε/4. Let n0 = 2L(ℓ, n)2, where ℓ ispicked so that 1
ℓ−1 <ε4 . Then t 6
⌈
ε⌊n/L⌋⌉
(if not, make n0 bigger).
Take an η-uniform partition P of G into ℓ 6 m 6 L parts, as given by Theorem 4.5. IfKr+1 ⊂ G(P, λ), then by Lemma 4.3, G ⊃ Kr+1(t). Hence Kr+1 6⊂ G(P, λ).
Let G′ be the subgraph of G consisting of all edges between pairs Vi and Vj whereViVj ∈ E(G(P, λ)). Then G′ 6⊃ Kr+1, for such a subgraph would have vertices in dis-tinct parts of P .
The edges of G not in G′ are those in non-uniform pairs, pairs of density < λ, and insideparts, so
e(G)− e(G′) 6 η
(
m
2
)(
n
m+ 1
)2
+
(
m
2
)
λ
(
n
m+ 1
)2
+m
( nm + 1
2
)
6
(
η + λ+1
m− 1
)(
n
2
)
< ε
(
n
2
)
2
Theorem 5.8. Let r, t ∈ N and let ε > 0. Then there exists n0(r, t, ε) such that, if |G| = n > n0,G 6⊃ Kr+1(t) and e(G) >
(
1− 1r + ε
)(
n2
)
, then there is a complete r-partite graph T with
V (T ) = V (G) and |E(T )E(G)| 6 11ε(
n2
)
.
Remarks.
1. This is essentially Theorem 2.1.
2. Hence e(G) 6 e(T ) + 11ε(
n2
)
6(
1 − 1r + 12ε
)(
n2
)
, which is the Erdos-Stone theorem(without the logn).
Proof. By Lemma 5.6, choose G′ ⊂ G with e(G′) >(
1 − 1r − 2ε
)(
n2
)
and G′ ⊃ Kr+1. Then
e(G′) > tr(n)− 3ε(
n2
)
.
Apply Theorem 5.3 to G′ to obtain T with |E(G′)E(T )| 6 9ε(
n2
)
.
Then |E(G)E(T )| 6 11ε(
n2
)
. 2
6. HypergraphsLecture 12
Let F be an ℓ-uniform hypergraph. Define
ex (n, F ) = maxe(G) : G is ℓ-uniform, |G| = n, G 6⊃ F
and
π(F ) = limn→∞
ex(n, F )(
nℓ
)
Denote by Kℓr the complete ℓ-uniform hypergraph of order r with
(
rℓ
)
edges. The value of π(Kℓr)
is unknown for r > ℓ > 3. Turan conjectured that π(K34 ) = 5/9. Erdos offered $5000 for a
solution in his memory.
Best bounds are 16 (−1 +
√21) = 0.5971 (Giraud, 1989), 1
12 (3 +√17) = 0.5935 (Chung & Lu,
1999), 0.561555 (Razborov, using flag algebras, 2008).
16
Turan further conjectured ex (n,K35 ) =
(
n3
)
−(⌊n/2⌋
3
)
−(⌈n/2⌉
3
)
, but this is false for odd n > 9(Sidorenko, 1990s).
The situation appears unstable.
π(F ) is known only in one or two simple cases; more recently, in a couple of less trivial examples,e.g. the Fano plane. But stability holds here. (Furedi, Simonovits, Pikhurko, Sudakov, Keevadi.)
Clearly π(Kℓr) 6 1− 1
(
rℓ
) . The best known general bound is as follows.
Lemma 6.1. Let G be an ℓ-uniform hypergraph with nr copies of K lr (for r > ℓ), where
nℓ = e(G) and nℓ−1 =(
nℓ−1
)
, and n = |G|. Then provided nr−1 > 0,
nr+1
nr>
r2
(r − ℓ+ 1)(r + 1)× nr
nr−1− (ℓ− 1)(n− r) + r
(r − ℓ+ 1)(r + 1)
Proof. Let A1, . . ., Anr−1be an enumeration of theKr−1’s, and B1, . . ., Bnr enumerate theKr’s.
Let ai be the number of Kr’s containing Ai and bj be the number of Kr+1’s containingBj . Then
∑
ai = rnr and∑
bj = (r + 1)nr+1.
Let N be the number of pairs (S, T ) where S is the vertex set of a Kr, T is an r-set notKr, and |S ∩ T | = r − 1. Clearly
N =∑
ai(n− r + 1− ai) = (n− r + 1)rnr −∑
a2i 6 (n− r + 1)rnr −r2nr
nr−1
On the other hand, for each Bj we can find n− r− bj vertices x such that B ∪x doesn’tspan Kr+1. So there is an (ℓ − 1)-set Y ⊂ Bj such that Y ∪ x is not an edge.
Then each z ∈ Bj \ Y gives a pair (S, T ) = (Bj , Bi \ z ∪ x). Thus
N >∑
j
(n− r − bj)(r − ℓ+ 1) = (r − ℓ+ 1)(
(n− r)nr − (r + 1)nr+1
)
Comparing bounds on N gives the result. 2
Corollary 6.2 (de Caen, 1983). ex(n,Kℓr) 6
(
n
l
)
(
1− 1(
r−1ℓ−1
) × n− r + 1
n− ℓ+ 1
)
.
In particular, π(Kℓr) 6 1− 1
(
r−1ℓ
) .
Proof. Let G be maximal with no Kℓr. Then nℓ−1 =
(
nℓ−1
)
and ns > 0 for ℓ 6 s < r.
We show by induction that
ns
ns−1>
n− ℓ+ 1
s
(
s− 1
ℓ− 1
)
(
nℓ(
nℓ
) − 1 +n− s+ 1
n− ℓ+ 1× 1(
s−1ℓ−1
)
)
holds for ℓ 6 s < r, which proves the Corollary since nr = 0.
The case s = ℓ holds automatically since ns−1 =(
ns−1
)
.
17
Writing qs = ns/ns−1 the desired inequality is
qs >n− ℓ+ 1
s
(
s− 1
ℓ− 1
)
(
nℓ(
nℓ
) − 1
)
+n− s+ 1
s
which follows from
qs >(s− 1)2
(s− ℓ)2qs−1 −
(ℓ− 1)(n− s+ 1) + s− 1
(s− ℓ)s
(from Lemma 6.1) and induction. 2
The next theorem (Erdos, 1964) shows that ℓ-partite, ℓ-uniform hypergraphs have extremalfunction o(nℓ), i.e. π = 0.
An ℓ-uniform hypergraph H is ℓ-partite if V (H) = V1 ∪ . . . ∪ Vℓ disjoint Vi, and each edge hasone vertex in each class. The complete ℓ-partite ℓ-uniform Kℓ(t1, . . ., tℓ) has |Vi| = ti and allpossible edges.
Theorem 6.3. Let G be an ℓ-uniform hypergraph of order n and size pnℓ/ℓ!. Let t1 6 t2 6Lecture 13. . . 6 tℓ be positive integers and suppose that pt1...tℓ−1 > T 2n−1, where T =
∑
ti.
Then G contains at least 12T !p
t1...tℓnT copies of Kℓ(t1, . . ., tℓ).
Remarks.
1. Note that this quantity is similar to the expected number if the edges were chosenrandomly.
2. This implies that ex (n,Kℓ(t1, . . ., tℓ)) 6 cTnℓ−1/t1...tℓ .
Proof. Let χ : V ℓ → 0, 1 be the characteristic (indicator) function of edges.
Then∑
x1,...,xℓ
χ(x1, . . ., xl) = pnl.
Let f(t1, . . ., tℓ) =1
nT
∑
x11,...,x
t11
· · ·∑
x1ℓ ,...,x
tℓℓ
∏
xi11
· · ·∏
xiℓℓ
χ(xi11 · · ·xiℓℓ ).
Note that nT f(t1, . . ., tℓ) is the number of labelled homomorphic copies of Kℓ(t1, . . ., tℓ) inG (copies where vertices in the same class might coincide). I.e., f() is the probability thata randomly chosen collection of T vertices spans a homomorphic copy of Kℓ(t1, . . ., tℓ).Now
f(t1, . . ., tℓ) =1
nT−tℓ
∑
x11,...,x
t11
· · ·∑
xℓ−1,...,xtℓ−1
ℓ−1
∏
xi11
· · ·∏
xiℓ−1
ℓ−1
∑
x1ℓ ,...,x
tℓℓ
∏
xℓ
χ(xi11 , . . ., xiℓ−1
ℓ−1 , xiℓℓ )
=1
nT−tℓ
∑
x11,...,x
t11
· · ·∑
xℓ−1,...,xtℓ−1
ℓ−1
∏
xi11
· · ·∏
xiℓ−1
ℓ−1
(
1
n
∑
xℓ
χ(xi11 , . . ., xiℓ−1
ℓ−1 , xℓ)
)tℓ
use Jensen’s
inequality>
1
nT−tℓ
∑
x11,...,x
t11
· · ·∑
xℓ−1,...,xtℓ−1
ℓ−1
∏
xi11
· · ·∏
xiℓ−1
ℓ−1
1
n
∑
xℓ
χ(xi11 , . . ., xiℓ−1
ℓ−1 , xℓ)
tℓ
= f(t1, . . ., tℓ−1, 1)tℓ
> f(t1, . . ., tℓ−2, 1, 1)tℓ−1tℓ > · · · > f(1, . . ., 1)t1t2...tℓ = pt1...tℓ
18
The contribution to nT f(t1, . . ., tℓ) from terms where, say, xi1 = xj1 is nT−1f(t1−1, t2, .., tℓ),
so the contribution from terms where all variables are distinct, i.e. the number of labelledcopies of Kℓ(t1, . . ., tℓ) is at least
pt1...tℓnT −ℓ∑
i=1
(
ti2
)
pt1...(ti−1)...tℓnT−1> pt1...tℓnT
(
1−(
T
2
)
p−t1...tℓ−1n−1
)
>1
2pt1...tℓnT
2
7. The size of a hereditary property
A(n ℓ-uniform hyper)graph property P is a class of (ℓ-uniform hyper)graphs closed underisomorphism.
It is non-trivial if it is not the class of all graphs.
It is hereditary if it is closed under taking induced subgraphs: if G ∈ P and H is an inducedsubgraph of G, then H ∈ P . Thus P is hereditary if and only if it is closed under the removalof vertices.
P is monotone if it is closed under taking any subgraph – i.e., it is closed under the removalof edges and vertices. Thus monotone =⇒ hereditary.
Examples.
• “3-colourable” is monotone
• “planar” is monotone
• “K4-free” is monotone
• “no induced C4” is hereditary
Observe that a hereditary property P can be written P = Forb(F), where F is a class of graphsand Forb(F) is the class of graphs containing no member of F as an induced subgraph. To seethis, we could take F to be F1 = graphs not in P . More usefully, we have P = Forb(F0), whereF0 = minimal elements of F1 with respect to inclusion, i.e. F0 = F /∈ P : every proper inducedsubgraph of F is in P.
For monotone properties, it is more natural to write P = Forb∗(F) where “induced” is removedfrom the above.
Examples.
• “2-colourable” = Forb∗(odd cycles)
• “no induced C4” = Forb(C4)
Observe that, for any F , Forb(F) is hereditary, and Forb∗(F) is monotone.
How “large” is P?
Let Pn be the graphs in P with labelled vertex set [n].
19
Example: let P = Forb(K4) = Forb∗(K4). Then |Pn| > 2t3(n) = 2
(
1− 13+o(1)
)
(n2), because we cantake a Turan graph T3(n) and all its subgraphs.
We extend Definition 5.4 slightly.Lecture 14
Definition 7.1. Let G be a graph with an ε-uniform partition P into m = |P | parts V1, . . ., Vm.Let λ, µ ∈ R. The reduced graph G(P, λ, µ) is the graph of order m with vertex setV1, . . ., Vm and edge set ViVj : (Vi, Vj) is ε-uniform, λ 6 d(Vi, Vj) 6 1− µ.
Hence G(P, λ) = G(P, λ, 0).
Lemma 7.2. Let V1, . . ., Vm be a partition of [n], where n > 2m2. Let 0 < ε, λ < 14 . Let H be
a graph with vertex set V1, . . ., Vm, where e(H) = d(
m2
)
.
Then the number of graphs having vertex set [n] and having an ε-uniform partition P intoparts V1, . . ., Vm such that G(P, λ, λ) = H is at most
2(d+ε+8λ log 1λ+ 1
m−1 )(n2)
Proof. Consider a graph G of this kind. The number of possible edges of [n](2) between partscorresponding to edges of G(P, λ, λ), or to non-uniform pairs, or within parts, is (as in theproof of Lemma 5.5), at most
(
d+ ε+ 1m−1
)(
n2
)
.
Hence there are at most 2(d+ε+ 1m−1 )(
n2) ways to choose edges of G from these. In how
many ways can we choose edges of G between pairs where we must have d(Vi, Vj) 6 λ ord(Vi, Vj) > 1− λ?
Let E =(
nm + 1
)2. The number of choices between some fixed pair (Vi, Vj) is at most
2
λE∑
i=0
(
E
i
)
6 2(λE + 1)
(
E
λE
)
6
(
E
λE
)2
6
( e
λ
)2λE
6 2(8λ log 1λ )E
(using(
nk
)
6(
enk
)k).
Since E(
m2
)
6(
n2
)
, the total number of choices for G is at most
2(d+ε+ 1m−1 )(
n2) × 2(8λ log 1
λ)(n2)
2
Theorem 7.3 (Erdos, Frankl, Rodl, 1936). Let P = Forb∗(F) and let r + 1 = minχ(F ) :F ∈ F.
Then |Pn| = 2(1−1r+o(1))(n2).
Proof. Since every subgraph of Tr(n) is in P , we have |Pn| > 2(1−1r+o(1))(n2).
To prove the converse inequality, let ε > 0. We will show that, if n is large, then |Pn| 62(1−
1r+ε)(n2).
Pick F ∈ F with χ(F ) = r + 1. Let t = |F |. Thus if G ∈ P then G 6⊃ Kr+1(t). Pick λ sothat 8λ log 1
λ < ε5 . Pick 0 < η < ε
5 so that (rt + 1)η < λrt. Pick ℓ so that 1ℓ−1 <
ε5 and
tr(ℓ) <(
1− 1r + ε
5
)(
ℓ2
)
.
20
By Theorem 4.5, each G ∈ Pn has an η-uniform partition P into ℓ 6 m 6 L parts, whereL = L(ℓ, η). Since Kr+1(t) 6⊂ G, then Kr+1 6⊂ G(P, λ, λ) if n is large, by Lemma 4.3. Soe(G(P, λ, λ)) 6 tr(m) 6
(
1− 1r + ε
5
)(
m2
)
.
There are at most Ln choices for V1, . . ., Vm and at most 2(L2) choices for H such that
G(P, λ, λ) = H , and at most 2(L2) ways to specify which are the non-uniform pairs, and at
most 2(L2) ways to specify whether a non-edge has density < λ or > 1− λ.
By Lemma 7.2, the number of G ∈ Pn is at most
23(L2)Ln2(1−
1r+ ε
5+ ε
5+ ε
5+ ε
5 )(n2) < 2(1−
1r+ε)(n2)
2
How about Forb(F)? Can we find some simple hereditary property to play the role of r-colourable graphs in the previous proof?
Definition 7.4. A graph is (r, s)-colourable if its vertex set can be partitioned into r classes,s of which span complete graphs, and the others of which contains no edges.
Denote by C(r, s) the class of (r, s)-colourable graphs.
Note. C(r, 0) is the class of r-colourable graphs. C(r, r) is their complements. And C(r, s) ishereditary.
Definition. For a non-trivial property P , let r(P) = maxr : ∃sC(r, s) ⊂ P.Lecture 15
Note that r(P) is well-defined. For there exists some graph H /∈ P and H ∈ C(r, s) for allr > |H | and all s. Thus r(P) < |H |. On the other hand, if P contains arbitrarily large graphsand is hereditary, then by Ramsey’s theorem either C(1, 0) ⊂ P or C(1, 1) ⊂ P . Thus r(P) > 1.
Lemma 7.6. Let t > 3 and let R = R(t) be the Ramsey number of t. Let Kn, n > R, beedge-coloured red, blue and grey, with fewer than 1
Rt
(
n2
)
grey edges. Then there exists ared or blue Kt.
Proof. Suppose we recolour the grey edges red. Every set of R vertices contains a red or blue
Kt, so we have at least(
nR
)(
n−RR−t
)−1=(
nt
)(
Rt
)−1of these. The number of these containing
an originally grey edge is at most 1Rt
(
n2
)(
n−2t−2
)
= 1Rt
(
t2
)(
nt
)
<(
nt
)(
Rt
)−1. 2
Lemma 7.7. Let r, t ∈ N and 0 6 λ 6 14 . Then there exists n0 ∈ N and η0 > 0 such that the
following holds.
Let G be a graph with vertex set⋃r+1
i=1 Vi, there Vi being disjoint and |Vi| > n0. Suppose,for 1 6 i < j 6 r + 1, then λ 6 d(Vi, Vj) 6 1 − λ, and (Vi, Vj) is η0-uniform. Thenthere exists 0 6 s 6 r + 1 such that G contains every graph in C(r + 1, s)t as an inducedsubgraph.
Proof. Let R = R(t). Pick ε <
1Rt ,
1t+1
(
λ2
)t
. Let L = L(R, ε). Take an ε-uniform partition
Q of V1 into R 6 m 6 L parts.
Colour the edges of the reduced graph G[V1](Q, 0, 0) red or blue according to whether thedensity of the pair is 6 1
2 or < 12 . Then, by Lemma 7.6, there is a monochromatic Kt. So
we may pick subsets V1p, 1 6 p 6 t, in V1, so that (V1p, V1q) is ε-uniform, 1 6 p, q 6 t, and
21
either d(V1p, V1q) 612 for 1 6 p, q 6 t or d(V1p, V1q) >
12 for 1 6 p, q 6 t.
We call V1 a “red” class in the first case, and a “blue” class in the second case. Repeatthis on the classes V2, . . ., Vr+1. Let there be s blue classes, wlog V1, . . ., Vs.
Note that |Vip| >|Vi|L+ 1
for all i, p. Let η0 = min
ε
L+ 1,λ
2
.
Then for all U ⊂ Vip, W ⊂ Vjq with |U | > ε|Vip| > η0|Vi|, |W | > ε|Vjq | > η0|Vj |, we have|d(U,W ) − d(Vi, Vj)| < η0. In particular, |d(Vip, Vjq) − d(Vi, Vj)| < η0. Since η0 <
ε2 , we
have that (Vip, Vjq) is ε-uniform.
Moreover, λ2 6 λ − η0 6 d(Vip, Vjq) 6 1 − λ + η0 6 1 − λ
2 for all 1 6 i < j 6 r + 1,1 6 p, q 6 t.
Let H ∈ C(r + 1, s)t. Then, provided that n0 is large enough, Corollary 4.4 shows thatwe can find H between some t of the classes Vip. Indeed, H has a partition into s classesof sizes t1, . . ., ts spanning complete graphs and r − s classes of sizes ts+1, . . ., tr spanningempty graphs,
∑
ti = t. Apply Corollary 4.4 to Vip, 1 6 i 6 r + 1, 1 6 p 6 ti. All pairsare ε-uniform, with densities > λ
2 where an edge is needed, and densities 6 1− λ2 where a
non-edge is needed. 2
The following theorem was proved by Promel & Steger (1991) for P = Forb(F ), by Bollobas &Thomason and by Alekseev (1992) in general. The proof here is due to Montgomery (2011).
Theorem 7.8. Let P be a hereditary property. Then |Pn| = 2(1−1r+o(1))(n2), where r = r(P).
Proof. By definition of r(P) there exists s such that C(r, s) ⊂ P . Let H be any subgroup ofTr(n). Fill in s of the classes of H so that they are complete. This graph is in C(r, s), soin P . There are 2tr(n) choices of H , so |Pn| > 2(1−
1r+o(1))(n2).
To prove the converse inequality, let ε > 0. We show |Pn| 6 2(1−1r+ε)(n2) for all large n.
Suppose not. Then there are infinitely many n with |Pn| > 2(1−1r+ε)(n2). Pick t ∈ N. By
applying the argument of Theorem 7.8, where η is picked with η < η0, we see that thereis an infinite number of n for which some G ∈ Pn has Kr+1 ⊂ G(P, λ, λ). But then, byLemma 7.7, C(r + 1, s)t ⊂ Pn ⊂ P .
Hence for each t ∈ N, ther exists s with C(r+1, s)t ⊂ P . But there are only finitely manypossible s, so for some s, C(r+1, s)t ⊂ P for infinitely many t. But P is hereditary, so forthis s, C(r + 1, s) ⊂ P . This contradicts the definition of r(P). 2
8. ContainersLecture 16
Let H be an ℓ-uniform hypergraph. Let r = e(H). The r-uniform hypergraph G(N,H) hasvertex set [N ](ℓ) where B = v1, . . ., vr is an edge whenever B, regarded as an ℓ-graph onvertex set [N ], is isomorphic to H .
Notice that H-free ℓ-uniform hypergraphs on vertex set [N ] correspond exactly to independentsets in G(N,H) (i.e., sets I ⊂ V (G) containing no edge of G). Hence Theorem 7.3 amounts tocounting independent sets in G(N,H).
22
How many independent sets can there be in a graph? If it’s d-regular? Can be more than 2n/2
(beautiful entropy argument of Kahn, 2001, shows that the maximum is found in n2dKd,d).
How many maximal independent sets? Still many (add 1 factor to each class in previous examplegives > 2n/4).
For many applications we would need fewer. Sometimes it is enough to have instead a small setof containers.
Definition 8.1. A set of containers C for a hypergraph G is a collection of subsets C ⊂ V (G)such that, for every independent set I, there exists C ∈ C with I ⊂ C.
We would like a small collection C of containers so each C ∈ C is “not too big”.
We do this for graphs. It is not too optimistic to require |C| not too big – e.g., Kd,n−d needs acontainer with |C| > n− d.
Definition 8.2. Let G be a graph with average degree d and of order n. The degree measureof a subset S ⊂ V (G) is
µ(S) =1
nd
∑
v∈S
d(v) .
Theorem 8.3. Let G be a graph of order n and average degree d. Let ζ > 0. Then there is afunction C : P [n] → P [n] such that, for every independent set I ⊂ [n], there exists T ⊂ Iwith
(a) I ⊂ C(T )
(b) µ(T ) 6 2/ζd
(c) |T | 6 2n/ζ2d
(d) µ(C(T )) 6 12 + 2ζ + µ(T ) for all T ⊂ P [n]
Remark. This gives a collection of containers with |C| 6(
n2n/ζ2d
)
6 end log d and each container
is not big.
Proof. Order the vertices v1, . . ., vn by decreasing degree. Let m = maxj : d(vj) > ζd. GivenI, run the following algorithm to find T . Begin with sets T = ∅, Γ = ∅.
For j = 1 to m : doif vj ∈ I
if∣
∣
∣
i > j : vivj ∈ E(G) but vi /∈ Γ
∣
∣
∣ > ζd(vj), add vj to T
for i = j + 1 to n : do
if∣
∣Γ(vi) ∩ T∣
∣ > d(vi)d , add vi to Γ
Suppose instead we are given some set T . Run the next algorithm, beginning with setsD = [n], Γ = ∅.
For j = 1 to m : do
if∣
∣
∣
i > j : vivj ∈ E(G) but vi /∈ Γ
∣
∣
∣ > ζd(vj), remove vj from D
for i = j + 1 to n : do
if∣
∣Γ(vi) ∩ T ∩ v1, . . ., vj∣
∣ > d(vi)d , add vi to Γ
Observe that the set Γ constructed at each stage is the same. Observe that if T camefrom a run of the first algorithm, then I ⊂ D ∪ T where D is produced from the second
23
algorithm. Also, in this case, T ⊂ I, and each vertex of Γ has a neighbour in T , soI ∩ Γ = ∅.
Define C(T ) = (D∪T )\Γ via the second algorithm. The first algorithm shows (a) is true.
Note that when a vertex vi enters Γ, it has at most d(vi)d +1 neighbours in T , since it had
fewer than d(vi)d + 1 neighbours before. So, counting edges between T and Γ,
ζndµ(T ) = ζ∑
vj∈T
d(vj) 6∑
vi∈[n]
1 +d(vi)
d= n+ n = 2n
so (b) follows.
Since vj ∈ T means j 6 m, so d(vj) > ζd and
|T |ζd 6∑
vj∈T
d(v) = ndµ(T )
and (c) follows.
Let Y be the set of edges inside D \ Γ.Lecture 17
∑
v∈D\Γd(v) 6 2|Y |+
∑
v/∈D\Γd(v)
µ(D \ Γ) 62|Y |nd
+ 1− µ(D \ Γ)
µ(D \ Γ) 61
2+
|Y |nd
Now
|Y | =∣
∣vjvi ∈ E(G) : j < i and vj , vi ∈ D \ Γ∣
∣
6
m∑
j=1
ζd(vj) +
n∑
j=m+1
ζd 6 ζnd+ ζnd
Therefore µ(D \ Γ) 6 12 + 2ζ, so µ(C) 6 µ(D \ Γ) + µ(T ) giving (d). 2
This can be made to work for hypergraphs. Applied to G(N,H) it gives the following, where
m(H) = maxH′⊂H,e(H)>1
e(H ′)− 1
v(H ′)− ℓ
Theorem 8.4. Let ε > 0. Then there exists c > 0 and n0 such that, for N > n0, there is acollection of ℓ-uniform hypergraphs on vertex set [N ] such that
(a) every H-free ℓ-uniform hypergraph is a subgraph of some C ∈ C(b) for every C ∈ C, e(C) 6
(
π(H) + ε)(
Nℓ
)
(c) log |C| 6 cN ℓ−1/m(H) logN
Corollary 8.5. The number of H-free ℓ-graphs is 2(π(H)+ε)(Nℓ ).
24
9. The Local Lemma
Suppose we have n events A1, . . ., An and we want P(
⋂nj=1 Aj
)
> 0, i.e. it is possible that no
Aj occurs.
Suppose P(Aj) = p, 1 6 j 6 n, 0 < p < 1.
If the Ai are independent, then P(⋂
Ai) = (1 − p)n > 0.
Otherwise, if p < 1/n, then P(⋃
Ai) < np < 1, so P(⋂
Ai) > 0.
What if p > 1/n but the Ai are “somewhat independent”? For each Ai, choose Ji ⊂ [n] suchthat Ai is independent of the system Aj : j /∈ Ji. Note there is no unique minimal choice forJi.
Theorem 9.1 (Local Lemma: Erdos, Lovasz, 1975). Let A1, . . ., An and J1, . . ., Jn be asabove. Suppose there exist 0 < γi < 1 such that P(Ai) 6 γi
∏
j∈Ji(1− γj).
Then P
(
n⋂
i=1
Ai
)
>n∏
j=1
(1− γj) > 0.
Corollary 9.2. Suppose |Ji| 6 ∆ for all i and that P(Ai) 61
e(∆ + 1). Then P
(
n⋂
i=1
Ai
)
> 0.
Proof of 9.1. By induction on n, we may assume that P(
⋂n−1i=1 Ai
)
> 0, so we may condition
on this event. We shall prove (by induction) that P(
An
∣
∣
⋂n−1i=1 Ai
)
6 γn.
This implies, for any S ⊂ [n− 1] and i /∈ S, that P(
Ai
∣
∣
⋂
j∈S Aj
)
6 γi.
The theorem follows because P
(
n⋂
i=1
Ai
)
=n∏
i=1
P
(
Ai
∣
∣
⋂
j<i
Aj
)
>n∏
i=1
(1− γi). Now
P
(
An
∣
∣
⋂
j<n
Aj
)
=P
(
An ∩⋂j<n Aj
)
P
(
⋂
j /∈Jn∪nAj
) ×P
(
⋂
j /∈Jn∪nAj
)
P
(
⋂
j<n Aj
)
6
P
(
An ∩⋂j /∈JnAj
)
as above× as above
as above
=P
(
An
∣
∣
⋂
j /∈JnAj
)
P
(
⋂
j<nAj
∣
∣
⋂
j /∈Jn∪nAj
)
By definition of Jn, the top is P(An). Relabelling so that Jn = 1, 2, . . ., d, the bottom is
P
d⋂
j=1
Aj
∣
∣
n−1⋂
j=d+1
Aj
=
d∏
i=1
P
Ai
∣
∣
n−1⋂
j=i+1
Aj
>
d∏
i=1
(1− γi) =∏
j∈Jn
(1− γj)
2
Eek. Fiddly. I hope that at least some of that was transcribed correctly. . . Let me know if not.
25
Theorem 9.3. Every r-regular r-uniform hypergraph is 2-colourable if r > 9 (i.e., the verticesLecture 18can be coloured red/blue so that no edge is monochromatic).
Proof. Let the hypergraph have order n. Take a random 2-colouring. Let Ai, 1 6 i 6 nrr = n,
be the event that edge i is monochromatic. This is independent of Aj : edge j is disjointfrom edge i.
So we can take Ji = Aj : edge j meets edge i and |Ji| 6 r(r − 1), since 12i 6
1e(r(r−1)+1)
if r > 9. 2
We can use the Local Lemma to improve lower bounds for Ramsey. A simple argument gives
R(3, t) = Ω(
(
tlog t
)3/2)
.
Theorem 9.4. R(3, t) >(
1 + o(1))
( t
log t
)2
.
Proof. Fix ε > 0. Let n = (1− ε)(
tlog t
)2. Colour the edges of Kn red with probability p = 1−ε√
n
and blue otherwise.
For α ∈ [n](3), let Aα be “G[α] is red”, and for β ∈ [n](t), let Bβ be “G[β] is blue”.
We can take Jα = Aα′ , Bβ′ : |α ∩ α′| > 2, |α ∩ β′| > 2, and Jβ likewise. So
|Jα| 6 3(n− 3) +
(
n
t
)
and |Jβ | 6(
t
3
)
+
(
t
2
)
(n− t) +
(
n
t
)
By the Local Lemma, it suffices to find γ and δ such that
p3 6 γ(1− γ)3(n−3)(1− δ)(nt)
(1− p)(t2) 6 δ(1− γ)(
t3)+(
t2)(n−t)(1− δ)(
nt)
Erdos-Szekeres showed R(3, t) 6
(
s+ t− 2
s− 1
)
=
(
t+ 1
2
)
.
Ajtai-Komlos-Szemeredi (1980) and Shearer (1986) showed R(3, t) = O( t2
log t
)
.
Kim (1995) showed R(3, t) = Θ( t2
log t
)
.
10. Tail Estimation
Let X be a random variable with mean µ. Chebychev’s bound P(|X − µ| > t) 6 VarXt2 gives a
simple bound that X is far from µ, but it is weak.
E.g., if X ∼ Bin(n, 12 ), thenX−n/2√
n/4∼ N(0, 1), so P
(
|X − n2 | > εn
)
6 e−2ε2n.
P(
|X − n2 | > x
√
n/4)
∼ 1√2πe−x2
.
Chebychev gives only P(
|X − n2 | > εn
)
< 14ε2n .
26
Theorem 10.1 (Chernoff, 1952). Let I1, . . ., In be independent indicators with P(Ii = 1) =pi. Let p =
1n
∑
pi. Let S =∑
Ii. Then, for h > 0,
P(
S > (p+ h)n)
6 e−nh2/2a and P(
S 6 (p− h)n)
6 e−nh2/2b
where
a = maxα(1− α) : p 6 α 6 p+ h and b = maxβ(1− β) : p− h 6 β 6 p
Corollary 10.2. Let X ∼ Bin(n, p). Then
P(
X 6 (1− ε)pn)
6 eε2pn/2 if p 6 1
2
P(
X > (1 + ε)pn)
6 eε2pn/4 if ε 6 1
Proof of 10.2. Let h = εp. By Theorem 10.1, we obtain the first inequality. Likewise thesecond. 2
Proof of 10.1. Let z ∈ R, z > 1. Then
P(
S > γn)
= P(
zS > zγn)
6E(zS)
zγnby Markov’s inequality
= z−γnn∏
i=1
(
(1− pi) + zpi)
6 z−γn(
(1− p) + zp)n
by AM-GM
=(
(1− p)z−γ + pz(1−γ))n
= e−Hn
Put z =γ(1− p)
p(1− γ). Note γ > p, so z > 1. Write γ = p+ h.
H = (p+ h) log(
1 + hp
)
+ (q − h) log(
1− hq
)
, where q = 1− p.
By Taylor’s theorem, H =h2
2α(1− α), for some α with p 6 α 6 p+ h.
The second inequality comes from the complementary P(
S 6 (q + h)n)
. 2
Theorem 10.3. Let X be hypergeometrically distributed with parameters (n,N, pN). ThenLecture 19X =
∑ni=1 Ii for some independent indicators Ii with some probabilities pi, such that
1n
∑ni=1 pi = p.
In particular, the bounds in Theorem 10.1 and Corollary 10.2 hold for X .
Proof. PX = k is the probability that, if n balls are chosen without replacement from an urnwith N balls, pN of which are red, then k balls are red. Write R = pN .
So PX = k =
(
R
k
)(
N −R
n− k
)(
N
n
)−1
, whence
27
E
(
X
ℓ
)
=∑
k
(
R
k
)(
N −R
n− k
)(
N
n
)−1(k
ℓ
)
=
(
R
ℓ
)(
N
n
)−1∑
k
(
R− ℓ
k − ℓ
)(
N −R
n− k
)
=
(
R
ℓ
)(
N
n
)−1(N − ℓ
n− ℓ
)
=
(
R
ℓ
)(
n
ℓ
)(
N
ℓ
)−1
Therefore
E(zX) = E
∑
ℓ
(
X
ℓ
)
(z − 1)ℓ
=∑
ℓ
(
n
ℓ
)(
R
ℓ
)(
N
ℓ
)−1
(z − 1)ℓ
Noting that
(
R
ℓ
)(
N
ℓ
)−1
6 pℓ, we see that
E(zX) 6∑
ℓ
(
n
ℓ
)
pℓ(z − 1)ℓ = [q + pz]n, for z > 1,
from which we could jump in to the proof of Theorem 10.1. To get Theorem 10.3 in all itsfullness,
E(zX) =R!
N !
∑
ℓ
(
n
ℓ
)
(N − ℓ)(N − ℓ− 1) · · · (R− ℓ+ 1)(z − 1)ℓ
=R!
N !
1
xRdN−R
dxN−R
∑
ℓ
(
n
ℓ
)
xN−ℓ, where x =1
z − 1
=R!
N !
1
xRdN−R
dxN−RxN [1 + x−1]n
which has real roots. That is, E(zX), which is a polynomial of degree n, has real roots.
Hence E(zX) =∏n
i=1(αi + βiz) with αi, βi ∈ R. Since E(1X) = 1, αi + βi 6= 0 for all i, wemay write E(zX) = c
∏ni=1(qi + piz) with qi + pi = 1. Again, E(1X) = 1 means c = 1.
Moreover, qi, pi > 0 or else there exists z > 0 with E(zX) = 0, contradicting E(zX) > 0for all z > 0.
Therefore E(zX) =∏n
i=1(qi+piz) with qi+pi = 1 and qi, pi > 0. But thenX = I1+. . .+In,where there Ii are independent indicators, P(Ii = 1) = pi.
Also,∑
pi = EX = pn. 2
28
11. Martingales Inequalities
We would like bounds akin to Theorem 10.1 for variables not the sum of indicators. Thereare nowadays several inequalities available (Janson, Hoeffding, Talagrand) that give differentinformation under different conditions. We look at Hoeffding.
A filter on a space Ω is a sequence F0,F1, . . .,Fn of successively finer partitions of Ω.
A sequence of random variablesX0, X1, . . ., Xn (with Xi defined on (Ω,Fi), meaningXi constanton parts of Fi) is a martingale if E(Xi+1 | Fi) = Xi. Typically X is some random variableand Xi = E(X | Fi).
The classical example is that of a gambler in a fair (zero expectation) game. If Fi is the gameand Xi = current winnings, then E(Xi+1 | Fi) = Xi.
Let f(G) be a graph-theoretic function. Label the edges of [n](2) as 1, 2, . . ., N =(
n2
)
. Insert theedges at random with probability p one by one to obtain a random graph G ∈ G(n, p).
Let Xi = E(
f(G) | the first i edges are determined)
.
Then X0 = E(f(G)), XN = f(G) and (Xi)Ni=0 is a martingale. This is called the edge exposure
martingale. (This is mildly ambiguous – depends on the edge labelling.)
We could instead, at stage i, add all edges between vertices 1, . . ., i − 1 and vertex i. Xi =E(
f(G) | all edges inside vertices 1, . . ., i are determined)
. This is the vertex exposuremartingale and is a subsequence of (some) edge martingale.
Given a martingale, its difference sequence is Yi = Xi −Xi−1. Thus E(Yi | Fi−1) = 0.Lecture 20
We aim to show that X is concentrated near its mean if the differences |Xi−Xi−1| are bounded(e.g., when X = χ(G)).
Lemma 11.1. Let Y be a random variable with −r 6 Y 6 1 − r, where r ∈ R+ and EY = 0.
Then, for s > 0, we have
E(esY ) 6 (1− r)e−sr + rs(1−r)6 es
2/8
Proof. Since esy is a convex function of y, we have esy 6 (1− r− y)e−sr +(y+ r)es(1−r). Takeexpectations.
For the second inequality, apply Taylor’s theorem to f(s) = log(
e−sr(1− r + res))
. 2
Theorem 11.2 (Hoeffding, 1963). Let Y1, . . ., Yn be a martingale difference sequence (Xi =E(X | Fi)) with −ri 6 Yi 6 1− ri, 1 6 i 6 n. Let r = 1
n
∑ni=1 ri, with 0 6 r 6 1.
Then P(
X > EX + hn)
= P(∑n
i=1 Yi > hn)
6 e−nh2/2a
and P(
X 6 EX − hn)
= P(∑n
i=1 Yi 6 −hn)
6 e−nh2/2b
where a = maxx(1 − x) : r 6 x 6 r + h and b = maxx(1− x) : r − h 6 x 6 r.
For the equalities, recall: F0 = trivial, X0 = EX ; Fn = complete partition, Xn = X . SoX − EX = Xn −X0 =
∑
Yi.
29
Proof. We proceed in a similar way to Theorem 10.1. Thus, for s > 0,
P
(
n∑
i=1
Yi > hn
)
= P
(
es∑n
i=1Yi > eshn
)
6 E
(
es∑n
i=1Yie−shn
)
= e−shnE
(
es∑n
i=1Yi | Fn−1
)
= e−shnE
(
es∑n−1
i=1YiE
(
esYn | Fn−1
)
)
using E(A) = E(E(A | B))
6 e−shnE
(
es∑n−1
i=1Yie−srn (1− rn + rne
s))
6 e−shnn∏
i=1
e−sri (1− ri + ries) by induction
6 e−shne−srn (1− r + res)n
by AM-GM
=(
(1− r)e−s(h+r) + res(1−h−r))n
6 e−nh2/2a as before
The second inequality comes from considering the martingale (−Xi), with difference se-quence (−Yi) and using the first inequality. 2
We could estimate e−nh2/2a as in Corollary 10.2, but the following is more useful.
Theorem 11.3. Let Xi = E(X | Fi) be a martingale whose difference sequence satisfies ai 6Yi 6 ai + ci, where ai is a function on (Ω,Fi−1) and ci ∈ R, 1 6 i 6 n.
Then, for t > 0,
P(
X > EX + t)
6 e−2t2/∑n
i=1c2i
P(
X 6 EX − t)
6 e−2t2/∑n
i=1c2i
Proof. Applying Lemma 11.1 to Y = E
(
Yn
cn| Fn−1
)
we obtain
E(
esYn | Fn−1
)
= E(
ecnsY)
6 es2c2n/8
Thus, following the previous proof,
P (X > EX + t) = P
(
∑
Yi > t)
6 e−tses2
8
∑ni=1
c2i
Now take s = 4t/∑n
i=1 c2i . 2
We can think of the ci as the maximum change in Xi at stage i.
A weaker form of Theorem 11.3, known as Azuma’s inequality, gives the bound e−t2/2∑
c2i , basedon the assumption that |Yi| 6 ci. But in most applications, Yi is not symmetric about zero andso 11.3 is stronger.
A useful corollary follows.
Corollary 11.4. Let Z1, . . ., Zn be independent random variables with Zi taking values in spaceAi, 1 6 i 6 n. Suppose that f :
∏ni=1 Ai → R satisfies |f(z) − f(z′)| 6 ci whenever z, z
′
differ only in the ith coordinate. Let Z = f(Z1, . . ., Zn).
Then P (|Z − EZ| > r) 6 2e−2t2/∑
ci .
30
Proof. Define a filter on Ω =∏n
i=1Ai where Fi partitions Ω into∏i
j=1 |Aj | parts according tothe first i coordinates. Then Xi = E(Z | Fi) is a martingale where X0 = EZ, Xn = Z.The conditions on f imply that
minzXi 6 Xi−1 = E(Xi) 6 max
zXi 6 min
zXi + ci
where z runs over Ai. Hence (Yi) satisfy Theorem 11.3 with ai = minzXi −Xi−1. 2
12. The Chromatic Number of a Random GraphLecture 21
As usual, G(n, p) is the space of random graphs on vertex set [n], with edges chosen independentlyat random with probability p (we take p constant).
We say that an event holds with high probability (whp) if P(A) → 1 as n → ∞. (Strictlyspeaking, we have a sequence of events An ⊂ G(n, p).)
Martingale inequalities show that χ(G) is highly concentrated around its mean. But what is themean?
Lemma 12.1. Let 0 < p < 1 be constant. Let G ∈ G(n, p).
Then χ(G) >(
1 + o(1)) n
2 log1/q nwhp, where q = 1− p.
Proof. Let ε > 0. Let d =⌈
(2 + ε) log)1/qn⌉
+ 1. Let X be the number of independent sets ofsize d in G. Then
P
(
χ(G) 6n
d
)
6 P(X > 1) 6 EX =
(
n
d
)
(1− p)(d2) 6
[
nq(d−1)/2]d → 0 as n→ ∞
This is true for all ε > 0. 2
It is not hard to show that the greedy algorithm uses(
1 + o(1))
n/ log1/q n colours.
Chebychev’s bound shows that there exist independent sets of size 2 log1/q n whp, but not withvery high probability.
Theorem 12.2 (Bollobas, 1988). Let 0 < p < 1 be constant. Let G ∈ G(n, p).
Then χ(G) =(
1 + o(1)) n
2 log1/q nwhp, where q = 1− p.
Proof. Lemma 12.1 gives the lower bound.
Let 0 < ε < 120 , and let d =
⌈
(2 − ε) log1/q n⌉. We shall find m with n1−ε < m < n1−ε/2
such that, whp, the event
A = “every subset of m vertices contains an independent set of size d”
holds. Note that, if A holds, then χ(G) 6n
d+m since we use colour 1 on d vertices, then
colour 2 on d vertices, and so on until fewer than m vertices remain, at which point wefinish off with m more colours. Since m = o(n/d), and since the statement holds for allε > 0, the theorem follows.
31
Let Pm = P(
H ∈ G(n, p) has no independent d-set)
. Then the probability of A failing isat most
(
n
m
)
Pm 6 nmPm = em lognPm 6 e1
1+εm logmPm
So it suffices to show that Pm < e−m1+δ
for some constant δ > 0.
Let X be the number of independent d-sets in H . Then EX =
(
m
d
)
q(d2).
If m < n1−ε then EX 6
(
em
d
q2/d√q
)d
< 1.
If m > n1−ε/2 then EX >
(
m
d
qq/2√d
)d
> ndε/8> n2.
If n is large. Clearly EX increases as m increases, and moving from m− 1 to m the valueincreases by a factor m
m−d < 2 for large n. It follows that there is some value of m for
which 2m7/4 6 EX 6 4m7/4.
Define the random variables
Y = max number of independent d-sets with 6 1 vertex in common.
Z = number of independent d-sets meeting every other in 6 1 vertex.
Clearly Z 6 Y 6 X . Consider H as an element of the product space 0, 1(m2 ). The valueof Y changes by at most 1 if we change one edge. So we can apply Corollary 11.4 to Y
with ci = 1, 1 6 i 6(
m2
)
, so P(Y 6 EY − t) 6 e−2t2/(m2 ).
If we can show that EY > m7/4 then we are done, because then
Pm = P(X = 0) 6 P(Y = 0) 6 P(Y 6 EY −m7/4) 6 e−2m7/2/(m2 ) 6 e−4m3/2
Now EY > EZ, so it’s enough to show that EZ > m7/4. For 2 6 r 6 d− 1, letLecture 22
Wr = number of independent d-sets meeting another in r vertices
Then X 6 Z +W2 +W3 + . . .+Wd−1. So EZ > EX −d−1∑
r=2EWr . Then
EWr 6
(
m
d
)
q(d2)(
d
r
)(
m− d
d− r
)
q(d2)−(
r2) = EXfr
where fr =
(
d
r
)(
m− d
d− r
)
q(d2)−(
r2).
So it suffices to show thatd−1∑
r=2fr 6
12 , since EX > 2m7/4.
First,
fr 6
(
d
r
)2(m
d
)(
m
r
)−1
q(d2)−(
r2) = EX
(
d
r
)2(m
r
)−1
q−(r2)
6 EX
(
d2√q
mqr/2
)r
6 EXm−9r/10
32
for r 6 d/12. Since EX 6 m7/4, we haved/12∑
r=2fr 6 1/4.
Secondly, writing s = d− r, we have
fr =
(
d
s
)(
m− d
s
)
qsd−(s+1
2 ) 6(
dmqd−s+1
2
)s
6
(
dmq−1/2q13d/24)s
for s 6 11d/12, i.e. for r > d/12.
Now ε <1
20, so
13d
24>
13
12(1− ε) logn >
(
1 +1
40
)
logn >
(
1 +1
40
)
logm.
Sod−1∑
r=d/12
fr 6 dm−1/50 6 1/4. 2
13. The Semi-Random Method
In 1980, Ajtai-Komlos-Szemeredi proved that a triangle-free graph of order n and average degreed has an independent set of size n
d logn (as opposed to >nd by Turan).
They achieved this by choosing a small random subset of size 12 |X |, such that the conditions in
G−X − Γ(X) could be described, not much worse than in G. Repeat. This is better than
(a) removing randomly one-by-one
(b) removing a large set once
The method was used by Rodl (then Frankl-Rodl) to prove the Erdos-Hanani conjecture.
We use Chebychev’s inequality:
P(
|X − EX | > t)
= P(
(X − EX)2 > t2)
6E(
(X − EX)2)
t2=
VarX
t2
Note that E(
(X − EX)2)
= EX2 − (EX)2.
Note that if X =∑
Iα, then EX2 = E∑
α,β
IαIβ =∑
α,β
P(Iα and Iβ).
Call an r-uniform hypergraph δ-semiregular (of degree d) if
(1− δ)d 6 d(v) 6 (1 + δ)d for all but δ|G| vertices
d(v) 6 erd for all v
d(u, v) 6 δd for all u, v with u 6= v
where d(u, v) is the codegree, the number of edges containing u, v.
Lemma 13.1. For all r > 2 and 0 < ε < 1, there exists δ0 such that, if δ < δ0 and G isr-uniform δ-semiregular, then G has a subhypergraph G′ of order at most (1 − ε+ ε2)|G|that is δ1/4-semiregular, and V (G)− V (G′) is covered by at most (ε+ ε2)|G|/r edges.
33
Proof. Let G be δ-semiregular of degree d. Call the vertices not satisfying (1 − δ)d 6 d(v) 6Lecture 23(1 + δ)d bad. Let B be the set of bad vertices. Choose a set X of edges independently atrandom with probability ε/d. Let G′ be the subgraph on vertex set V (G)− V (X)−B.
We show that, with positive probability, G′ is δ1/4-semiregular and |X |+ |B| 6 (ε+ε2)n/r,where n = |G|. This will prove the lemma, because our edge cover for V (G) − V (G′) . . .(something I missed, sorry - anyone help?)
Let A be the event “|X |+ |B| 6 (ε+ ε2)n/r”B be the event “n/4 6 |G′| 6 (1 − ε+ ε2)n”C be the event “G′ is δ1/4-semiregular”
We show that A ∩ B ∩ C 6= ∅.
Note δd > 1, so we can make d, and hence also n, as large as we like, by making δ0 small.
(a) e(G) 6 1r
(
(1− δ)n(1 + δ)d+ δnerd)
6(
1 + ε2
)
ndr is δ0 is small.
By Corollary 10.2,
P
(
|X | > EX +ε2
4
n
r
)
6 P
(
|X | >(
ε+3ε2
4
)
n
r
)
6 e−ε2
16nε4r <
1
3
if δ0 is small. Hence P(A) > 2/3. (|B| 6 δn 6 ε2n/4r.)
(b) Let v be a good (i.e, not bad) vertex, and Iv be the event v ∈ V (G′).
So |G′| =∑vIv. Now
1
3<(
1− ε
d
)(1+δ)d
6 P(v ∈ G′) =(
1− ε
d
)d(v)
6
(
1− ε
d
)(1−δ)d
6 e−ε(1−δ)6 1− ε+
2
3ε2
So by Chebychev, P(B) 6 Var |G′|(ε2n/12)2
, since E|G′| 6(
1− ε+ 23ε
2)
n.
Let u, v ∈ V (G)−B, u 6= v. Then
P(Iu and Iv) =(
1− ε
d
)d(v)+d(v)−d(u,v)
6
(
1− ε
d
)−δd
P(Iu)P(Iv)
So
E(|G′|2) =∑
u,v
P(Iu and Iv) =∑
u=v
+∑
u6=v
6 E(|G′|) +(
1− ε
d
)2
E(|G′|)2
since(
1− ε
d
)−δd
6 eεδ 6 1 + 2δε. So
Var (|G′|) = E|G′|2 − (E|G′|)2 6 E|G′|+ 2δεE|G′|2 6 n+ 2δεn2
Hence P(B) 6n+ 2δεn2
(ε2n/12)2<
1
3if δ0 is small.
(c) See handout.
Theorem 13.2 (Frankl-Rodl, 1985). Given r, ε > 0, there exists δ > 0 such that if G isd-regular r-uniform of maximum codegree < δd, then G can be covered by (1 + ε)|G|/redges.
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Remark. Hence there is a set of (1 − rε)|G|/r independent edges, i.e. pairwise disjoint. Forlet W be the set of vertices in more than one cover edge. Then |G|+ |W | 6 (1 + ε)|G|, so|W | 6 ε|G|. Remove edges meeting |W |.
Proof. Choose η > 0 and k ∈ N so that1 + η
1− η+ r(1 − η + η2) < 1 + ε.
Then δ < δ0(r, η)4k will work. For Lemma 13.1 can be applied k times with r and η. The
edges selected amount to at most
(η + η2)(
1 + (1− η + η2) + (1− η + η2)2 + · · ·+ (1− η + η2)k−1)
|G| 61 + η
1− η
|G|r
and there remain 6 (1− η + η2)k|G| vertices, which can be covered by an edge apiece. 2
Define:
m(n, k, ℓ) = maximum number of k-sets in [n] such that each ℓ-set is in at most one of them.
M(n, k, ℓ) = minimum number of k-sets in [n] such that each ℓ-set is in at least one of them.
Erdos-Hanani (1963) conjectured
m(n, k, ℓ) ∼(
nℓ
)
(
kℓ
) ∼M(n, k, ℓ)
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