Lecture 10: Extremal graph theory Immustata/Slides_Lecture10_565.pdf · 2020. 10. 27. · Graphs...

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Lecture 10: Extremal graph theory I

October 8, 2020

() Lecture 10 October 8, 2020 1

What is extremal combinatorics

Extremal combinatorics deals with questions of the following form: given agraph G on n vertices, how big certain global invariants can be when Gdoes not exhibit certain local substructures? What are the graphs withoutsuch local substructures that have maximal global invariants?

The questions we will discuss: how many edges can G have if G does nothave certain types of subgraphs.

We start with the easiest such result. All graphs in these two lectures willbe simple.Proposition 1. If G is a bipartite graph on n vertices, then G can have≤ n24 edges.

Remark. This falls under the previous paradigm since G is bipartite if andonly if G does not contain any polygon of odd length.


The maximal number of edges of a bipartite graph

Proof of the proposition. If G is a bipartite graph on n vertices, then wehave disjoint subsets A and B of V (G ), with k and n − k elements, suchthat every edge of G joins a vertex in A with a vertex in B. It follows that

#E (G ) ≤ k(n − k) ≤(k + (n − k)

2

)2=

n2

4,

where the inequality is given by the inequality between the arithmetic andgeometric means.

Proof of the remark. The fact that a bipartite graph contains no polygonof odd length is clear. Suppose now that G contains no such polygon andlet’s show that G is bipartite. By treating each connected component of Gseparately, we may assume G is connected. Fix a vertex v0 of G and let Aand B be the subsets of V (G ) that can be connected to v0 via a path ofeven (respectively, odd) length. We clearly have A ∪ B = V (G ) and thehypothesis implies A ∩ B = ∅. Moreover, it implies that no two vertices inA (or B) can be neighbors in G . Hence G = G (A,B) is bipartite.


Graphs with no triangles

With a bit more work, we can prove a stronger version of the previousresult.Proposition 2 (Mantel, 1907). If G is a graph on n vertices that does not

contain any triangle, then #E (G ) ≤ n24 .

Proof. Let m = #E (G ). For every x ∈ V (G ), let dx = deg(x). Given anytwo adjacent vertices x , y ∈ V (G ), the hypothesis implies that x and yhave no common neighbors. We thus have dx + dy ≤ n. Therefore∑

{x ,y}∈E(G)

(dx + dy ) ≤ mn.

On the other hand, we have∑{x ,y}∈E(G)

(dx + dy ) =∑

x∈V (G)

d2x .


Graphs with no triangles, cont’d

Recall now that the Cauchy-Schwarz inequality says that for everya1, . . . , an and b1, . . . , bn, we have(

n∑i=1

a2i

)·

(n∑

i=1

b2i

)≥

(n∑

i=1

aibi

)2.

By taking b1 = . . . = bn = 1, we obtain

n ·

(n∑

i=1

a2i

)≥

(n∑

i=1

ai

)2.

Applying this and our inequalities, we obtain

mn2 ≥ n ·∑

x∈V (G)

d2x ≥

∑x∈V (G)

dx

2 = (2m)2 = 4m2,hence m ≤ n2/4.


Graphs with no triangles, cont’d

Remark. It follows from the proof of Proposition 2 that if G is as in theproposition and m = n

2

4 , then for every edge {x , y} in E (G ), every vertexof G is a neighbor of precisely one of x and y . If A consists of theneighbors of x and B consists of the neighbors of y , since G contains notriangles, we see that G is the complete bipartite graph on A and B. Ifa = #A and b = #B, then n = a + b and #E (G ) = ab = (a + b)2/4,hence a = b.

Remark. If n is odd and G is a graph on n vertices that does not containany triangles, then it follows from Proposition 2 that #E (G ) ≤ n2−14 . Thisis sharp, as can be seen by taking the complete bipartite graph Kk+1,k ,where k = n−12 .

Exercise. Show that in this case, too, if #E (G ) = n2−14 , then G is

isomorphic to Kk+1,k .


Graphs with no Kt+1

Next: we want to extend the result in Proposition 2, by asking what is themaximum number of edges for a graph on n vertices that does not containany Kt+1, for some t ≥ 2 (recall: Kt+1 is the complete graph on t + 1vertices).

How to construct graphs without any Kt+1: let V = V1 t . . . t Vt . If weconsider the graph G with V (G ) = V and where the edges in G are givenby all edges that connect vertices in different Vi , then G does not containany Kt+1 (for any t + 1 vertices of G , two of them must lie in the same Viby the pigeonhole principle).

Note: in this example, if di = #Vi , then #V = d1 + . . .+ dt and#E (G ) =

∑i

Graphs with no Kt+1, cont’d

Let’s prove this fact when t divides n. In this case, we have∑i

Graphs with no Kt+1, cont’d

The proof of the fact in the general setting (when t might not divide n issimilar). Note that the condition di − dj ∈ {0, 1,−1} for all i and juniquely determines the graph we constructed, up to isomorphism. Indeed,if n = tk + r , with r ∈ {0, 1, . . . , t − 1}, then after reordering the Vi , wemay assume that di = k + 1 for 1 ≤ i ≤ r and di = k for r + 1 ≤ i ≤ t.The corresponding graph is the Turán graph Tn,t .

Remark. The number of edges of the Turán graph Tn,t is equal to

n2

2

(1− 1

t

)+ lower degree terms in n. (1)

This is immediate if t divides n: in this case every Vi hasnt elements,

hence the number of edges is

1

2t · n

t·(n − n

t

)=

n2

2

(1− 1

t

).

Exercise. Prove the formula in (1) in general.() Lecture 10 October 8, 2020 9

Pál Turán

Pál Turán is another famous Hungarianmathematician, who lived 1910-1976.

Worked mostly in number theory, but alsoin analysis and graph theory. Hecollaborated extensively with Paul Erdös.

As a Jew, he could not get a university jobfor several years and was sent to labourservice at various times between 1940-44.

Peter Franks: “Mathematicians have onlypaper and pen, he doesn’t have anything incamp. So he created extremalcombinatorics for which he did not needeither.”


Turán’s theorem

Theorem (Turán). For every t ≥ 2, among all graphs on n vertices thatcontain no Kt+1, the graph Tn,t has the most edges; moreover, it is uniquewith this property.

Proof. We argue by induction on t ≥ 2. For t = 2, we have seen thatamong the graphs on n vertices that contain no triangles, the unique onewith the maximal number of edges is Kk,k (if n = 2k is even) or Kk+1,k ifn = 2k + 1 is odd. We now consider t ≥ 3 and assume the theorem knownfor t − 1.Let G be a graph on n vertices with no Kt+1 subgraphs. Choosev ∈ V (G ) such that its degree dv := degG (v) is maximal. Let Sv ⊆ V (G )be the subset consisting of the neighbors of v (hence #Sv = dv ) andTv = V (G ) r Sv .Clear: since v is a common neighbor of all vertices in Sv , the assumptionon G implies that the subgraph of G spanned by Sv contains no Ktsubgraph.


Turán’s theorem

We now modify G to get a new graph G ′ with V (G ′) = V (G ), as follows:• We keep all edges in E (G ) between the vertices in Sv .• We add edges between the vertices in Sv and the vertices in Tv .• We remove all edges in E (G ) between the vertices in Tv .

Claim. For every vertex w in V (G ), we have degG ′(w) ≥ degG (w).• This is clear if w ∈ Sv (we have only added some new edges incident tow).• If w ∈ Tv , then

degG ′(w) = #Sv = dv ≥ degG (w),

by the maximality in our choice of v .The claim, together with the formula relating the sum of the degrees withthe number of edges gives

#E (G ′) ≥ #E (G ).


Turán’s theorem

By induction hypothesis for the subgraph 〈Sv 〉 of G spanned by Sv :

#E (〈Sv 〉) ≤ #E (Tdv ,t−1).

Since G ′ contains no edges between vertices in Tv and contains all edgesbetween vertices in Sv and vertices in Tv , we conclude that

#E (G ) ≤ #E (G ′) = (#Tv ) · dv + #E (〈Sv 〉)

≤ (#Tv ) · dv + #E (Tdv ,t−1) ≤ #E (Tn,t).

Furthermore, if #E (G ) = #E (Tn,t), then we see that in G already everyvertex in Sv was adjacent to every vertex in Tv (otherwise, by the previouscomputation #E (G ′) > #E (G )). Moreover, Sv has to be isomorphic toTdv ,t−1. Since G contains no Kt+1, it follows that no vertices of Tv areadjacent in G . Hence G is constructed out of subsets V1, . . . ,Vt .Maximality in the number of edges implies that G is isomorphic to Tn,t .


The Erdös-Stone theorem

We now consider the following general problem. Given a fixed graph H,estimate for large n the number ex(n,H) consisting of the maximumnumber of edges of a graph on n vertices that does not contain any graphsisomorphic to H. The answer is provided by the Erdös-Stone theorem. Ofcourse, this is not interesting if E (H) = ∅ (in which case every graph withn ≥ #V (H) vertices contains a subgraph isomorphic to H).

Recall that the chromatic number χ(H) is the smallest number c such thatthe vertices of H can be colored with c colors such that no two adjacentvertices get the same color. Note that χ(H) = 1 if and only if #E (G ) = ∅.

Theorem (Erdös-Stone). For every (finite simple) graph H with E (G ) 6= ∅and for every � > 0, for n� 0, we have

1

2

(1− 1

χ(H)− 1− �)n2 < ex(n,H) <

1

2

(1− 1

χ(H)− 1+ �

)n2.


The Erdös-Stone theorem

Example. If H = Kt+1 (when χ(H) = t + 1), this matches the assertiongiven by Turán’s theorem, which gives

ex(n,Kt+1) =1

2

(1− 1

t

)n2 + lower order terms in n

(depending on the residue of n when divided by t).Remark. The formula in the Erdös-Stone theorem implies that ifχ(H) ≥ 3, then

limn→∞

ex(n,H)

n2=

1

2

(1− 1

χ(H)− 1

)> 0.

Hence we understand the behavior of ex(n,H) for n→∞. The situationis more delicate when χ(H) = 2 (that is, for bipartite graphs), when wecan only conclude that for every � > 0, we have ex(n,H) ≤ �n2 for n� 0.We will discuss some examples in the next lecture.


Proof of the lower bound in Erdös-Stone

This part is easy: say χ(H) = k + 1. Since the Turán graph Tn,k hasvertices in k disjoint sets, with neighbors being precisely vertices indifferent such subsets, it follows that χ(Tn,k) = k . Since χ(H) = k + 1,we deduce that Tn,k has no subgraph isomorphic to H. We thus concludethat ex(n,H) ≥ #E (Tn,k). We have seen that

#E (Tn,k) =1

2

(1− 1

k

)n2 + lower order terms in n

(with the precise formula depending on the residue of n divided by t).

It thus follows that given � > 0, we have

ex(n,H) ≥ #E (Tn,k) >1

2

(1− 1

k− �)n2 for n� 0.


A lemma for the upper bound in Erdös-Stone

Lemma. Given positive integers k and t, and 0 < � < 1/k , for everygraph G on n vertices, with n large enough (depending on k, t, and �) andwith m ≥ 12

(1− 1k + �

)n2 edges, there are disjoint subsets A1, . . . ,Ak+1

of V (G ), all of size t, such that any two vertices in two different Ai andAj are neighbors in G .Proof. We first prove the following Claim:Given any p and every �′, with 0 < �′ < �, if n is large enough, then forevery graph G on n vertices, with #E (G ) ≥ 12

(1− 1k + �

)n2, we can find

a subgraph G ′ of G on ≥ p vertices, such that for every v ∈ V (G ′), wehave

degG ′(v) ≥(

1− 1k

+ �′)·#V (G ′).

We construct G ′ by successively removing the “bad” vertices.


A lemma for the upper bound in Erdös-Stone, cont’d

More precisely, we construct a sequence of subgraphs G1,G2, . . . of G = G0as follows: if Gi has been constructed and it contains a vertex vi withdegGi (vi ) <

(1− 1k + �

′) ·#V (Gi ), then we take Gi+1 to be the subgraphof Gi spanned by V (Gi ) r {vi}; if there is no such vi , then we stop.We need to show that if n� 0, we can’t reach a subgraph G` such that#V (G`) < n

′. If this is the case, note that q := #V (G`) = n − ` < n′. Inthis case, we have #E (G`) ≤ q(q−1)2 <

n′2

2 .On the other hand, for every i ≥ 0, we have #E (Gi ) ≤

#E (Gi+1) +

(1− 1

k+ �′

)·#V (Gi ) = #E (Gi+1) +

(1− 1

k+ �′

)· (n− i).

By combining all these, we get #E (G ) ≤

n′2

2+

`−1∑i=0

(1− 1

k+ �′

)·(n−i) = n

′2

2+

(1− 1

k+ �′

)·(n(n + 1)

2− q(q + 1)

2

).



Since q is bounded above and �′ < �, this implies that for n large enough,we get

#E (G ) <1

2

(1− 1

k+ �

)n2,

a contradiction. This completes the proof of the claim.

The claim implies that from now on (after possibly replacing � by a smallervalue), we may assume that we have a graph on n vertices, such thatevery vertex has degree ≥

(1− 1k + �

)n. We show by induction on q, with

1 ≤ q ≤ k + 1, that for all t, if n� 0, then we can find disjoint subsetsA1, . . . ,Aq of V (G ), all of size t, such that any two vertices in twodifferent Ai and Aj are neighbors in G . For q = k + 1, we obtain theassertion in the lemma.



The assertion to prove is clear for q = 1, hence we may assume q ≥ 2. Weapply the induction hypothesis to get disjoint subsets A′1, . . . ,A

′q of V (G ),

all of size s = dt/�e, such that any two vertices in two different A′i and A′jare neighbors in G .

Let U = V (G ) r (A′1 ∪ . . . ∪ A′q) and consider

W = {v ∈ U | v has ≥ t neighbors in eachA′i}.

We first show that we can make #W arbitrarily large by taking n� 0.

We bound in two ways the number N of missing edges in G betweenU rW and A′1 ∪ . . . ∪ A′q. Since every vertex in U rW has < t neighborsin some A′i , it follows that

N ≥ #(UrW )·(qs−t) = (n−qs−#W )·(qs−t) ≥ (n−qs−#W )s(q−�).



On the other hand, since every vertex in G has degree ≥(

1− 1q + �)n

neighbors, we have

N ≤ #(A′1 ∪ . . . ∪ A′q) ·(

1

q− �)n = qsn

(1

q− �)

= sn(1− q�).

We now combine the two inequalities involving N to get a lower bound for#W :

(q − �)#W ≥ (n − qs)(q − �)− n(1− q�) = n(�+ 1)(q − 1)− qs(q − �).

Since q, �, and s are fixed, we see that when we make n large enough, wemay assume that #W is as large as we want.

In particular, we may and will assume that #W >(st

)q(t − 1).


Proof of the upper bound in Erdös-Stone

We now make the following construction: for every w ∈W , we choose tneighbors of w in each of the A′i and let Γw be their union.

Clearly, there are at most(st

)qsuch subsets Γw . By our condition on #W

and the pigeonhole principle, we can find distinct w1, . . . ,wt ∈W suchthat Γw1 = . . . = Γwt = Γ.

If we put Aq+1 = {w1, . . . ,wt} and let Ai = Γ ∩ A′i for 1 ≤ i ≤ q, we seethat A1, . . . ,Aq+1 all have size t, and any two vertices in two different Aiand Aj are neighbors in G .This completes the proof of the induction step for the assertion dependingof q and thus the proof of the lemma.


Proof of the upper bound in Erdös-Stone

We can now prove the upper bound in the Erdös-Stone theorem. Letk = χ(H)− 1 and t = #V (H). It follows the lemma that if n� 0 and Gis a graph on n vertices and with m ≥ 12

(1− 1k + �

)n2 edges, we have

disjoint subsets A1, . . . ,Ak+1 of V (G ) such that any two vertices in twodifferent Ai , Aj are neighbors in G .

Since χ(H) = k + 1, if we consider a coloring of H with colors 1, . . . , k + 1and if we map the vertices colored by i to distinct vertices in Ai , then wesee that G has a subgraph isomorphic to H. We thus have

ex(n,H) <1

2

(1− 1

k+ �

)n2 for n� 0.

This completes the proof of the Erdös-Stone theorem.


Reference

In the next lecture, we will consider in some examples the asymptoticbehavior of ex(n,H) for certain bipartite graphs H.

For more on extremal graph theory, see Jacob Fox’s MIT course, availableat http://math.mit.edu/˜ fox/MAT307


Lecture 10: Extremal graph theory Immustata/Slides_Lecture10_565.pdf · 2020. 10. 27. · Graphs...

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