Math 262: Topics in Combinatorial Mathematics

Lectures by Fan Chung Graham

Compiled and Edited by Robert Ellis

University of California at San Diego

Spring Quarter 2002

Contents

1 Introduction to Ramsey Theory (4/1/02) 1

1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 The party problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.3 Other Ramsey numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.4 Open problems for Ramsey numbers . . . . . . . . . . . . . . . . . . . . . . 5

2 Super Six, Part 1 (4/3/02) 6

2.1 Problem 1: Ramsey’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.2 Problem 2: Schur’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.3 Problem 3: Van der Waerden’s Theorem . . . . . . . . . . . . . . . . . . . . 8

3 Van der Waerden’s Theorem (Ron Graham, 4/8/02) 9

4 Turan Theory 15

5 More Turan Theory 16

6 Turan Theory (4/29/02) 20

i

1 Introduction to Ramsey Theory (4/1/02)

Ramsey Theory is the study of unavoidable configurations in families of graphs. A typicalproblem is to show that when the edges of a complete graph on enough vertices are 2-colored,it is impossible to avoid being able to find some specified monochromatic subgraph. In onesense, Ramsey Theory is a generalization of the Pigeonhole Principle; for instance, in 2-coloring the edges of the star S2n+1 on 2n + 1 vertices, it is impossible to avoid obtaininga monochromatic Sn+1. The seminal paper in Ramsey Theory by Erdos and Szeker is [10],which also inaugurated the related field of probabilistic combinatorics.

1.1 Definitions

An introduction to graph theory can be found in [31], and a more advanced treatment is [5].We have the usual definition of a graph as follows. For a set V , and a nonnegative integerk, the set

(Vk

)is the family of all subsets of V of size k.

Definition 1. A graph G = (V, E) consists of a set V of vertices and a set E ⊆(

V2

)of edges.

Hypergraphs are a generalization of graphs in which edges may contain any number ofvertices.

Definition 2. A hypergraph H = (V, E) consists of a set V of vertices and a set E ⊆ 2V ofhyperedges.

We may think of 2V as(

V0

)∪ · · · ∪

(V|V |

), whence the generalization of graphs on edges

consisting of 2 vertices. A hypergraph is r-uniform provided that every hyperedge consistsof r vertices, whereby E ⊆

(Vr

).

The complete graph Kn is the graph for which E =(

V2

), and the complete r-uniform

hypergraph K(r)n is the graph for which E =

(Vr

). The basic Ramsey’s Theorem is as follows.

Theorem 1 (Ramsey’s Theorem). Given l, r, k ∈ N, there exists an n0(l, r, k) so that

whenever K(r)n is edge k-colored, n > n0 guarantees the existence of a monochromatic sub-

graph K(r)l .

The original proof showed existence of n0, but the actual values of n0 for various param-eters have remained elusive for most cases.

1.2 The party problem

We illustrate Ramsey’s Theorem for the case r = 2, k = 2 and l = 3, commonly referred toas the “party problem”: given 6 people at a party, it is guaranteed that 3 of them are eithermutually acquainted or mutually unacquainted. For r = 2, we define R(k, l) as follows.

Definition 3. Given k, l ∈ N, the Ramsey number R(k, l) is the smallest n such that any2-coloring of Kn contains either a monochromatic Kk or a monochromatic Kl.

1

Thus the party problem is concerned with determining R(3, 3)

Proposition 2 (The party problem). R(3, 3) = 6.

Proof. The proof procedes by showing first R(3, 3) ≤ 6 and second R(3, 3) > 5. For theupper bound, consider the graph K6. Each vertex has degree 5, and so by switching red andblue if necessary, we may assume a particular vertex v has 3 red edges adjacent to it (seeFigure 1). Now consider the edges of the 3-clique among vertices a, b, and c. If one of theseedges is red, then that edge forms a red triangle with edges coming from v. Otherwise, noneof the edges are red, and there is a blue triangle with a, b, and c. Therefore R(3, 3) ≤ 6.

{ {{

{

PPPPPPPPPPPPPPPP

������������������

��

��

ee

eee

v

c

a

b

red

red

red

?

?

?

Figure 1: Gives an upper bound for R(3, 3) by proving the existence of a monochromatictriangle.

In order to show R(3, 3) > 5, it is sufficient to demonstrate a red-blue coloring of K5

which has no monochromatic triangles. Consider the coloring in Figure 2, where drawn edgesare red and missing edges are blue.

blue interioredges

{{ {

{{ �������

ZZ

ZZ

ZZ�

��

��

�BBBBBBB

red red

redred

red

Figure 2: Gives a lower bound for R(3, 3) by demonstrating a coloring of K5 with nomonochromatic triangle.

It is impossible to find a monochromatic 3-clique using the coloring in Figure 2. Choosingany 3 vertices, two of them must be adjacent using edges on the perimeter, and so there is ared edge. But two of them must not be adjacent using edges on the perimeter, and so thereis a blue edge. Thus no triangle is monochromatic, and R(3, 3) > 5.

2

1.3 Other Ramsey numbers

It is currently known that R(4, 4) = 18, 43 ≤ R(5, 5) ≤ 49, and 102 ≤ R(6, 6) ≤ 165. Adynamic survey of results on Ramsey numbers is maintained at [21]. The generalized Ramseynumber R(c1, . . . , ck), where k colors are used, is defined as follows.

Definition 4. The Ramsey number R(l1, . . . , lk; r) is the smallest number n such that any

hyperedge k-coloring of K(r)n with the colors ci for 1 ≤ i ≤ k forces a K

(r)li

of color ci forsome i.

When r is omitted from the above definition, the coloring is done on graphs rather thanhypergraphs; for instance, R(3, 3; 2) = R(3, 3). Furthermore, the notation Rk(l) is often usedto denote R(l, · · · , l), with l taken k times in succession. The following is due to Greenwoodand Gleason ([16]).

Proposition 3. R(3, 3, 3) = 17.

Proof. To show R(3, 3, 3) ≤ 17, isolate one vertex v0 in K17 and consider the 16 edges leavingv0. At least 6 of these edges must be of the same color; say, blue. The K6 induced by these6 vertices adjacent to v0 via blue edges either has a blue edge, or not. If so, a blue triangleis formed with v0. If not, any edge 2-coloring of K6 forces a monochromatic triangle, byTheorem 2.

To show R(3, 3, 3) ≥ 16, we must exhibit a 3-colored K16 with no monochromatic tri-angle. Again fix a vertex v0, and color an equal number of its incident edges in each color.This gives 5 edges colored, say, blue, red, and yellow. Each set of 5 red, blue, or yellowedges defines 5 neighbors of v0; denote the corresponding K5’s by Kred

5 , Kblue5 , and Kyellow

5 ,respectively. Color Kyellow

5 according to Figure 2. Color Kred5 and Kblue

5 similarly, exceptwith yellow replacing red or blue, respectively. Color the remaining edges from Kx

5 to Ky5 ,

by {red, yellow, blue} \ {x, y}. The result is a 3-coloring of K16 with no monochromatictriangles.

Other results include 51 ≤ R4(3) ≤ 64 (initial work in [6] showed R4(3) ≥ 50), R(4, 4; 3) =13, and the following 2-coloring generalization.

Proposition 4.

c1k

log k≤ R(3, k) ≤ c2

k

log k.

Theorem 5.

R(k, l) ≤ R(k − 1, l) + R(k, l − 1) ≤(

k + l − 2

k − 1

).

Proof. Let n = R(k−1, l)+R(k, l−1), which is well-defined because of Theorem 1. Considerthe complete graph Kn, fix a vertex v0, and consider the n− 1 edges incident to v0. Supposek and l correspond to the colors red and blue, respectively. Any edge {red, blue}-coloringof Kn gives either at least R(k − 1, l) red edges or at least R(k, l − 1) blue edges incidentto v0. In either case, by definition of R(k − 1, l) and R(k, l − 1), a red Kk or a blue Kl is

3

guaranteed. If there are at least R(k− 1, l) red edges incident to v0, the KR(k−1,l) defined bythe vertices adjacent to v0 along red edges have either a red Kk−1 or a blue Kl. If they havea red Kk−1, throwing in v0 creates the red Kk. The other case is similar.

The second inequality is by induction. The base cases are R(1, 1) = 2 and R(k, 1) =R(1, k) = k. Assuming by induction the inequality for R(k − 1, l) and R(k, l − 1), we have

R(k, l) ≤ R(k − 1, l) + R(k, l − 1)

≤(

k + l − 3

k − 2

)+

(k + l − 3

k − 1

)=

(k + l − 2

k − 1

).

A critical graph, usually Kn, is one for which the Ramsey number is achieved, or in otherwords, smallest graph which forces the configuration in question. In fact, the first inequalityin Theorem 5 is strict if both values R(k − 1, l) and R(k, l − 1) are even, because no Kn iscritical for n odd.

The Ramsey number R(k, k) has been studied extensively; the current lower bound is

R(k, k) ≥ k

e√

22k/2(1 + o(1)).

An easier lower bound which illustrates common techniques is the following.

Theorem 6.R(k, k) ≥ 2k/2.

Proof. Color the edges of Kn in two colors, red and blue. There are 2(n2) ways to color.

For a fixed subset S ⊆ V with |S| = k, a coloring is S-bad if the set of edges E(S) is

monochromatic. For a fixed S, there are 2(n2)−(k

2)+1 S-bad colorings. The total number ofbad colorings over all choices of S is at most(

n

k

)2(n

2)−(k2)+1,

including double counting. Then there is a good coloring if

2(n2) >

(n

k

)2(n

2)−(k2)+1, (1)

where we wish to find some n = n(k) satisfying (1). In particular, we may choose n to satisfy(n

k

)21−(k

2) < 1

nk2−k2/2 < 1

n < 2k/2.

In particular, we have shown that for n < 2k/2, there is a 2-coloring of Kn with no monochro-matic Kk, and so R(k, k) ≥ 2k/2.

4

Using Theorem 5 to bound R(k, k) above, we have

R(k, k) ≤(

2k − 2

k − 1

)≈ c22k−2

√k − 1

, (2)

by employing the Stirling approximation n! = nne−n√

2πn. Combining Theorem 6 and (2)and taking the kth root throughout yields the following.

Corollary 7. √2 ≤ (R(k, k))1/k ≤ 4.

1.4 Open problems for Ramsey numbers

Several outstanding questions remain for Ramsey numbers, including the following conjec-tures of Erdos with money attached.

Conjecture 1 (Erdos, $100). The following limit exists:

limk→∞

R(k, k)1/k.

For an additional $100, determine the value of this limit.

In general, the following bounds can be shown on R(k, n) ([7]):

c1

(n

log n

) k+12

< c2nk−1

logk−1 n(1 + o(1)). (3)

A special case is possibly strengthened by the following corollary, which would reducethe order of the gap in (3) from n5/2 to n3.

Conjecture 2 (Erdos, $250).

R(4, n) >c1n

3

logc2 n.

These and other conjectures of Erdos appear in [7].

5

2 Super Six, Part 1 (4/3/02)

This lecture begins a survey of six problems in Ramsey Theory which demonstrate circum-stances in which certain configurations are unavoidable, or complete disorder is impossible.Three problems are presented in this lecture, with the remaining three deferred to a laterlecture.

2.1 Problem 1: Ramsey’s Theorem

For the first theorem, recall Definition 4 of Ramsey numbers from Section 1. Ramsey’sTheorem is that Ramsey numbers exist.

Theorem 8 (Ramsey’s Theorem). Given c, r, and k1, . . . , kc from N, there exists an n =

n(k1, . . . , kc; r) such that any r-hyperedge c-coloring of K(r)n must contain a monochromatic

K(r)ki

for some i.

Recall from last lecture, Corollary 7 gives us that√

2 ≤ (R(k, k))1/k ≤ 4. The bounds onR(5, 5) are currently 43 ≤ R(5, 5) ≤ 49. Suppose we wish to complete the following exercise:

Exercise 1. Determine if R(5, 5) ≥ 44.

One technique to obtain lower bounds is to consider only cyclic colorings. In a cycliccoloring, the vertices [n] are arranged consecutively around the circumference of a circle, withn and 1 adjacent. The color assigned to an edge {i, j} is determined by |i−j|, correspondingto distinct chord lengths. The coloring exhibited in Figure 2 is cyclic: {i, j} is colored redwhen ch(i, j) = 1, and blue otherwise. Specifically, if a cyclic coloring on 44 vertices can bedemonstrated which has no monochromatic K5, Exercise 1 is settled in the affirmative.

The following constrained Ramsey problem uses an upper bound cited by the GuinnessBook of World Records as the largest number. Consider the k-hypercube Qk on verticesV = {0, 1}k, and build the related graph Gk with vertices V and edges E =

(V2

). Let f(4)

be the minimum k such that any 2-coloring of Gk must contain a monochromatic K4 in aplane. A restatement of this problem is available in [18, p. 54]. The best and only upperbound for f(4) is known as Graham’s number, and is a 64-times repeated tower of powers of3.

2.2 Problem 2: Schur’s Theorem

Schur’s Theorem treats partitions of the first n positive integers into t parts, and whetheror not a part contains a triple x, y, z such that x + y = z. A part which contains no suchtriple is called sum-free.

Theorem 9 (Schur’s Theorem). Given t ∈ N, there is a maximum number n = S(t) sothat there exists a partition of [n] into t classes, each of which is sum-free.

6

S(t) is known as the Schur number. Clearly, if [n] can be partitioned into t sum-freeclasses, then so can [n′], for n′ < n. Also, if [n] = S(t) and n′ > n, there does not exista partition of [n′] into t sum-free classes. Schur proved Theorem 9 in [24] by showing thatS(t) < bt!ec, and also gave the following lower bound, with the cases for equality shown in[17].

Theorem 10 (Schur number lower bound).

S(t) ≥ 1

2

(3t − 1

),

with equality for t = 1, 2, 3.

Example 1.S(2) = 4.

Consider the partition [4] = {1, 4} ∪ {2, 3}, which is sum-free, showing S(2) ≥ 4. Nowconsider any partition of [5] into to sets A and B. WLOG, 1 ∈ A, and so 2 ∈ B in order forA to be sum-free. Thus 4 ∈ A is forced, and then 3 ∈ B is forced. There is nowhere to put5 without violating the sum-free condition, so S(2) < 5.

Now consider S(3). The partition P = {{1, 4, 10, 13}, {2, 3, 11, 12}, {5, 6, 8, 9}} demon-strates that S(3) ≥ 13, as does Theorem 10. The partition P = {π1, π2, π3} induces a cycliccoloring of K14 by coloring {i, j} with color i provided that |i − j| ∈ πi. Suppose to thecontrary that a monochromatic triangle on vertices a < b < c results. Then b− a, c− a, andc − b are all in πi for some 1 ≤ i ≤ 3, and since (b − a) + (c − b) = (c − a), πi fails to besum-free, contradicting the choice of partition. This formulation gives us in general a wayto bound S(t) in terms of Rt(3).

Theorem 11 (Schur’s Ramsey Theorem).

S(t) ≤ Rt(3)− 2.

Proof. Let P = {π1, . . . , πt} be a partition of S(t) into t sum-free parts. Induce a cycliccoloring on KS(t)+1 by coloring {i, j} with color c where |i, j| ∈ πc. As described in thediscussion, the presence of a monochromatic triangle would correspond to a partition whichis not sum-free, and so Rt(3) > S(t) + 1.

In the theorem, the difference of 2 is due to one vertex being added in the cyclic coloringso that the set of chord lengths is all of [S(t)]. Then Rt(3) is the smallest value for which amonochromatic triangle is forced, and so a second vertex is added. S(4) = 44 is found in [1],and the bounds 160 ≤ S(5) ≤ 315 are due to [11]. S(6) ≥ 536 and S(7) ≥ 1680 are shownin [12].

We may relax our goals and attack variations of Schur’s Theorem, such as the follow-ing challenging exercise, which is thought to be completely open, yet tractable for smallcomputational examples.

Exercise 2 (Research question). What is the smallest subset of Z such that any partitioninto t parts has some part not sum-free?

7

2.3 Problem 3: Van der Waerden’s Theorem

(coordinate references from mathworld site) Van der Waerden’s Theorem concernspartitioning either N or the first n positive integers so as to try to avoid arithmetic progres-sions of a specified length.

Theorem 12 (van der Waerden’s Theorem). Any partition of N into r parts π1, . . . , πr

ensures the existence of arbitrarily long arithmetic progressions in some πi.

In transitioning from the infinite to the finite, we define van der Waerden’s number.

Definition 5. Given r, k ∈ N, find the smallest n = Wr(k) such that any partition of [n]into r parts guarantees a k-term arithmetic progression in some part. Wr(k) is known asvan der Waerden’s number.

The existence of Wr(k) implies well-definedness, but the proof of existence and the proofof Theorem 12 will be delayed until Section 3. Defining W (k) = W2(k), we have the following:Wr(2) = r+1, W (3) = 9, W (4) = 34, and W (5) = 178. Berlekamp (add reference) provedthe lower bound k2k ≤ W (k), and Shelah (add reference) claimed a $500 Erdos prize byshowing that W (k) is bounded above by a (check usage) primitive-recursive function asfollows. Let T2(n) be defined for all n ∈ N by the recurrence T2(n) = 2T2(n−1) for n ≥ 1 andthe initial condition T2(0) = 1. Then T2(n) can be thought of as a tower of 2’s:

T2(n) = 22···2

.

Shelah’s prize-winning bound is W (k) ≤ T2(T2(T2(T2(T2(1))))). This is interpreted as atower of 2’s, where the number of 2’s in the tower is the tower of 2’s of the next level in. Thefollowing improved bound by Gowers in (cite gowers paper) won a $1000 Erdos prize aswell as a Fields Medal.

Theorem 13 (van der Waerden’s number bound).

W (k) ≤ 22222k+9

.

R. L. Graham both proposed the following conjecture and placed a bounty on it worth$1000. (check for reference)

Conjecture 3 (Graham, $1000).

W (k) ≤ 2k2

.

Things to do:

possibly add references to Rodl, Hales-Jewett (“pure form of van der Waerden”), Graham-Leeb-Rothschild (see [15, pp. 9-10])

clean up and extend section on Graham’s number, possibly putting it after section wheretower notation is defined

8

3 Van der Waerden’s Theorem (Ron Graham, 4/8/02)

Van der Waerden’s Theorem appears in “In a conjecture of Baudet” [?], but is actually aconjecture of Schur. We state the theorem as follows.

Theorem 14 (van der Waerden’s Theorem). For all r ∈ N, all possible partitionsN = C1 ∪ · · · ∪ Cr force a k-term arithmetic progression in some Ci for all k.

A k-term arithmetic progression (k-AP for short) has the form a, a + d, . . . , a + (k − 1)dfor real numbers a and d. Theorem 14 immediately raises the question as to whether theinfinite set N is required to force a k-AP in some part, or if it suffices to consider some“long enough” subset [n] := {1, . . . , n} ⊂ N. In fact, we can transfer from the infinite to thefinite with the following Compactness Principle, which is presented in the context of van derWaerden’s Theorem, but can be extended to colorings of hypergraphs. The following proofborrows from [15, Theorem 4].

Theorem 15 (Compactness Principle (Konig infinity lemma?)). Suppose that foreach n ∈ N, there exists an r-partition of [n] for which no part contains a k-term AP. ThenN contains no k-term AP.

Proof. For each n, consider the partition of [n] into r parts as an r-coloring of [n] by thefunction χn : [n] → [r]. By the assumption of the theorem, for each n let χn define a partitionof [n], no part of which contains a k-term AP. We desire to construct a coloring χ∗ from theχn’s such that no k-term AP is monochromatic for χ∗.

Define χ∗ : N → [r] as follows. Start by letting χ∗(1) be a color which appears infinitelyoften as χn(1). Of these choices which are defined on 2, at least one color appears as χn(2)for infinitely many n’s; choose this color as χ∗(2). For each j, continue refining this set ofn’s by throwing away j−1 and defining χ∗(j) as a color which appears as χn(j) for infinitelymany n’s.

Now consider a k-term AP a1, . . . , ak. Then χ∗ agrees with χn for some n ≥ ak on [ak],and the AP is not monochromatic for χn, so it is not monochromatic for χ∗. Therefore χ∗

is a coloring of N which has no monochromatic AP’s.

The Compactness Principle can be restated in many forms; for example, every k-termAP from can be considered a hyperedge of the graph H on countably many vertices. Thisleads to the generalization that appears as [15, Theorem 4] in terms of infinite hypergraphswhere each hyperedge is on finitely many vertices. Another specific problem of this typeappears in [13], in which the unavoidable substructure is a sequence which appears as thesubsequence of another larger sequence.

Theorem 16 (Compactness Principle). Let H = (V, E) be a hypergraph where all X ∈ Eare finite (but V may be countably infinite). Suppose that, for all W ⊂ V , W finite,

χ(HW ) ≤ r.

Thenχ(H) ≤ r.

9

Theorem 15 allows us to state an equivalent finite version of van der Waerden’s Theorem,since the Compactness Principle shows that we do not require an infinite ground set in orderto force the substructures of interest, k-term AP’s.

Theorem 17 (Finite van der Waerden’s Theorem). For all positive integers k and r,there exists a W (k, r) such that all possible partitions [W (k, r)] = C1 ∪ · · · ∪ Cr forces ak-term AP in some Ci.

We may abbreviate the statement of this theorem as [W (k, r)]r→ k−AP; any partition

of {1, . . . ,W (k, r)} into r parts yields a k-term arithmetic progression.

Some base cases for W (k, r) are immediate. For all k, W (k, 1) = k, since the unique partis a k-term AP. For all r, W (2, r) = r + 1 by the Pigeonhole Principle, since at least onepart contains two elements which comprise a 2-term AP. The first nontrivial we examine isW (3, 2). In showing an upper bound for W (3, 2), we also illustrate a technique that yieldsan upper bound for W (k, r) in general.

Figure 3: (a) A representatively 2-colored 5-block has two elements of the same color andpoint to a third element in the same block which would complete a 3-AP if it were not coloreddifferently. (b) Since there are 32 possible 2-colored 5-blocks, having 33 5-blocks forces arepeat. (c) Two 2-term AP’s of different colors point to the same location ‘X’ which wouldextend both AP’s, guaranteeing a 3-term monochromatic AP when there are only 2 colors.

Proposition 18. W (3, 2) ≤ 325.

10

Proof. We initially consider a 2-coloring of [n], for sufficiently large n. It turns out thatn ≥ 325 is large enough, but for now we could assume we have a 2-coloring of N. Anarbitrary contiguous 5-member subsequence, which we refer to as a block, has two colorsappearing at least twice in the first 3 elements. In addition, these two elements “point” toa third element in the block which completes a 3-AP if it is of the same color. There are32 possible blocks, but without loss of generality, we illustrate the proof using the top blockof Figure 3(a), by coloring the first and third position red, and the fifth position blue. Theremaining two elements are colored arbitrarily to avoid a 3-AP. The important part of theconstruction is that we may choose the first two elements of the same color and of distanced from each other, but the third element completing the 3-AP might be a different color.

Coloring 5-blocks with 2 colors yields 32 possible 5-blocks. We want to look for the firsttime a block is repeated; this is equivalent to asking for W (2, 25), which is 33. As shownin Figure 3(b), within 33 5-blocks there are two identical 5-block which are at a distance ofb ≤ 32 5-blocks apart. The proof concludes by considering the block pointed to by theseidentical 5-blocks. As illustrated in Figure 3(c), the first red in the first 5-block and thesecond red in the second 5-block point to X, and the blues in the first and second 5-blockspoint to X, forcing a 3-term AP since X must be colored red or blue. The constructionrequired 2b + 1 5-blocks which contain up to 325 elements; therefore W (3, 2) ≤ 325.

The extravagant construction in the proof of Prop. 18 begs the question as to whetherW (3, 2) is much smaller. An easy lower bound of W (3, 2) > 8 is obtained by the example

R R B B R R B B,

which has no 3-term AP’s. The upper bound is determined by examining a 2-colored finitesequence of appropriate length.

Proposition 19. W (3, 2) = 9.

Proof. By the above example on a sequence of length 8, it suffices to show that any 2-coloredsequence of length 9 has a 3-term AP. Focus initially on the center element, which we assumeis red.

· · · · R · · · ·The colors on either side cannot both be red without giving a 3-term AP. By symmetry,assume that the left color is blue.

· · · B R · · · ·

We separate now into two cases based on the color of the element to the right of the centralred element. If it is blue, we have the following situation:

· X · B R B · Y · ,

and any choice of X and Y forces a 3-term monochromatic AP. If the element to the rightof the central red element is red, we have the following situation:

Y · · B R R X · · ,

11

which forces a blue in position X in order to avoid a 3-term red AP, which in turn forces ared in position Y in order to avoid a 3-term blue AP. Now we have

R Z Y B R R B · X ,

which forces a blue in position X, then another blue in position Y, and then a red in positionZ, in order to avoid 3-term monochromatic AP’s at each stage. We are left with the sequence

R R B B R R B · B ,

with no way to color the last position in order to avoid a 3-term monochromatic AP. ThereforeW (3, 2) = 9.

The other known nontrivial values of W (k, r) are W (4, 2) = 35, W (5, 2) = 178, W (3, 3) =27, and W (3, 4) = 76. The proof of Theorem 17 involves an induction on blocks of integersconstructed in a similar fashion as in Prop. 18, and will be further illustrated by the nontrivialcase of W (3, 3).

Figure 4: (a) A representatively-colored W (2, 3)-block has two elements of the same color,and point to an element in the next block, together forming a block of length 2W (2, 3).(b) There are 32W (2,3) 3-colorings of blocks of length 2W (2, 3), and W

(2, 32W (2,3)

)blocks

guarantee two identically colored blocks, which point to a position in a third block. Thethree blocks together form a super-block. (c) Enough super-blocks guarantees two identicallycolored super-blocks which point to a position in a third which is forced to complete amonochromatic 3-term AP.

Theorem 20. W (3, 3) ≤ 209984 (313124 + 1).

Proof. All colored blocks and elements in this proof are viewed as taken from [n] for somesufficiently large n to be determined. First consider a 3-coloring of [W (2, 3)], which guaran-tees a monochromatic 2-term AP, as illustrated in Figure 3(a). This AP points to a thirdelement within [2W (2, 3)], which is either red, creating a red 3-term AP, or another color.Without loss of generality, assume it is blue, as in Figure 3(a).

Now consider blocks of size [2W (2, 3)], of which there are 32W (2,3) kinds, since each elementmay be 3-colored. Among W

(2, 32W (2,3)

)such blocks there must be a duplicate, as in Figure

3(b). By considering 2W(2, 32W (2,3)

)such blocks, two duplicate blocks point to a position

in a third block in two different ways, and so its color must be yellow, or else it completesa monochromatic 3-term AP. Now this entire structure is a single super-block, as in thebottom of Figure 3(b).

There must be two identical super-blocks in the first

W(2, 32W(2,32W (2,3))

)super-blocks. Extending to twice as many super-blocks ensures a focal position X in a thirdsuper-block which can extend a red, blue, or yellow 2-term AP. Therefore

2W(2, 32W(2,32W (2,3))

)= 2W

(2, 32W(2,38)

)12

= 2W(2, 32(38+1)

)= 2W

(2, 313124

)= 2

(313124 + 1

)super-blocks suffice to force a monochromatic 3-term AP. Each super-block contains 2W

(2, 32W (2,3)

)blocks, which each contain 2W (2, 3) elements, for a total of

2(313124 + 1

)· 2W

(2, 32W (2,3)

)· 2W (2, 3) = 2

(313124 + 1

)· 13124 · 8

= 209984(313124 + 1

)3-colored elements to force a monochromatic 3-term AP.

The construction in the theorem is extravagantly wasteful, coming in much higher thanthe known value of W (3, 3) = 27. The extremely large numbers from this and relatedconstructions for bounding W (k, r) must be expressed in a way that exploits the fact thatthey are primitive recursive. Let W (k) := W (k, 2) and recall the tower notation in Section2. The following is conjectured.

Conjecture 4 ($1000 (verify source, tower size)). W (k) ≤ T2(k).

The following result appears in [28].

Theorem 21 (Shelah, 1988). W (k) ≤ T2(T2(T2(T2(T2(1))))).

This result was improved in [14].

Theorem 22 (Gowers, 2001).

W (k) ≤ 22222k+9

.

One immediate question concerning van der Waerden’s Theorem is which part containsthe k-term AP. This was answered by Szemeredi as follows.

Theorem 23. For any A ⊆ N, define d(A) = |A ∩ {1, . . . ,m}|/m. If

lim supm→∞

d(A) > 0,

then A has a k-term AP of arbitrary length k.

Roth studied the case k = 3 in 1954 [23] (check ref), and Szemeredi the case k = 4 in1969 [29].

One open problem is to minimize the size of a set of integers which when colored forcesan AP. For instance, if we consider 2-coloring 27 of the first 37 integers in the followingpattern:

100100110 11 . . . 11︸ ︷︷ ︸19 1’s

011001001,

a 4-term AP is forced. This is because the numbers on the edges are not in as many AP’s asthe ones in the center. We might want to look for k-AP’s forced when all but one memberis monochromatic, or a k-AP which has only 2 of a possible r colors.

13

Remarks

• (Schur’s Theorem) For m ∈ N large enough, any partition [m] = C1 ∪ · · · ∪Cr intor color classes forces the existence of a monochromatic solution to x + y = z.

• (quoted by Ron as Schur, 1911) It is sufficient to consider that the partition

[der!e] = C1cup · · · ∪ Cr

in order to guarantee a non sum-free monochromatic color class. This was determinedduring a search for a modular version of Fermat’s last theorem.

• We could also consider the equation x + y = 2z as a problem of this “sum-free” type.This is equivalent to the existence of a 3-term AP, provided that the x, y, z are distinct,since z = (x + y)/2.

• In 1933, Rado showed that any partition-regular system of equations must have amonochromatic solution when N is finitely colored. An equation

∑aixi = 0 is partition-

regular provided that it can be solved in 0’s and 1’s but not all 0’s. Examples are, asabove, x + y = z and x + y = 2z.

• (Exercise) Determine if x + y = 3zx is partition-regular, especially in the case of 2colors.

• (Conjecture, $100) Show that x2 + y2 = z2 is partition-regular. In particular, isthere a monochromatic solution when N is 2-colored?

• Rodl showed that x−1 + y−1 = z−1 is partition-regular and a monochromatic solutionis forced with an r-coloring.

Notes

From [15]:

Historical Note. I. Schur, working on the distribution of quadratic residues in Zp,first conjectured the result proved by van der Waerden. Van der Waerden heardof the conjecture through Baudet, a student at Gottingen at the time, and hasreferred to his result as Baudet’s Conjecture in the literature. A brief account ofSchur’s contribution is given by A. Brauer in the preface to I. Schur-GesammelteAbhandlungen (Springer-Verlag, 1973).

The book of Schur mentioned is [25, 26, 27].

To do list:

• reference for Baudet and Schur

14

4 Turan Theory

15

5 More Turan Theory

The treatment of Turan Theory begun in Section 4 is continued here. The Turan numbert(n,G) indicates the minimum number of edges that must be contained in a graph H on nvertices before G is forced to appear as a subgraph. In other words,

t(n, G) = max |{E(H) : |V (H)| = n, G 6⊂ H}| .

Turan Theory and Ramsey Theory are related by considering an edge k-coloring of thecomplete graph in which one color class of edges is forced to have high density.

Fact 1. If kt(n, G) <(

n2

), then any edge k-coloring of Kn contains a monochromatic G.

One color class is forced to have at least(

n2

)/k edges, which forces a graph H with edges

of this color which is dense enough to force the presence of G.

Theorem 24. If

kt(n, G) > 2

(n

2

)log n,

then rk(G) > n.

Instead of showing this theorem, we give the following stronger version, which arisesfrom a random packing k colored copies of H into Kn in order to show that there is an edgek-coloring of Kn which does not force a monochromatic G.

Theorem 25. If (n

2

)(1− t(n,G)(

n2

) )k

< 1, (4)

then rk(G) > n.

Proof. Let H have n vertices and t(n,G) edges, with G 6⊂ H. Consider k copies of H placedrandomly on Kn. This is done for each of k colors by permuting the vertices of H randomlyand assigning the current color to an edge {u, v} ∈ E(Kn) if {u, v} is present in the randomlyarranged H. Without loss of generality, multiply-colored edges in Kn retain the first assignedcolor. Now for each edge {u, v} in Kn and each iteration i in the placement of colored H’s,the probability that {u, v} is assigned the ith color is

|E(H)|(n2

) =t(n,G)(

n2

) .

The probability that {u, v} is left uncolored by all k iterations is(1− t(n,G)(

n2

) )k

.

16

If the expected number of edges left uncolored, which is(n

2

)(1− t(n, G)(

n2

) )k

,

is less than 1, the existence of a coloring of this type in which no edge is left uncolored isascertained. In particular, there is no monochromatic G in this edge k-coloring, becauseG 6⊂ H, and the set of edges of a particular color form a subgraph of H.

Theorem 25 is stronger than Theorem 24, because (4) is satisfied when

1− t(n,G)(n2

) <

(n

2

)−k

t(n, G) >

(n

2

)−(

n

2

)−k

,

which is stronger by a factor of log n when k is a fixed constant. The full implications ofTheorem 25 for bounding Ramsey numbers may not yet be completely explored.

We now compute specific Turan numbers, starting with cycles C3 and C4, and thenconsidering the complete bipartite graph Kr,s.

Exercise 3. t(n, C3) = bn/2cdn/2e.

Proof. The complete bipartite graph Kbn/2c,dn/2e has bn/2cdn/2e edges and no odd cycle,giving t(n,C3) ≥ bn/2cdn/2e. The reverse direction proceeds by showing that when morethan bn/2cdn/2e edges are present, there is either a triangle or we may remove arbitrarilymany odd cycles from the graph (which is impossible).

Suppose H has n vertices and at least bn/2cdn/2e+1 edges. It is impossible for H to bebipartite, because the largest number of edges in a bipartite graph on n vertices is achievedby Kbn/2c,dn/2e with bn/2cdn/2e edges. Suppose to the contrary that H has no C3. Let C bethe smallest odd cycle of H, and label its vertices {1, . . . , 2k + 1} with edges {i, i + 1} for1 ≤ i < 2k + 1 and {1, 2k + 1}. No other edges are present by assumption of C being thesmallest odd cycle. A vertex u not in C may only be incident to at most k vertices in C;otherwise a triangle is formed. Thus there are at most k(n− 2k − 1) edges between C andV (H) \ C. In total, removing C and all edges incident to C from H causes the removal ofat most

k(n− 2k − 1) + 2k + 1 = nk − 2k2 + k + 1

edges. The remaining graph H ′ has n′ := n− 2k − 1 vertices. We now argue that H ′ has atleast bn′/2cdn′/2e+ 1 edges to complete the proof.

If n is even, n′ is odd, and the number of remaining edges is at least

17

n2

4+ 1− (nk − 2k2 + k + 1) =

n2 − 4nk + 8k2 − 4k

4

≥ n2 − 4nk + 4k2 − 2n + 4k + 4

4(5)

=(n− 2k − 2)(n− 2k)

4+ 1

= bn′/2cdn′/2e+ 1

where (5) follows from 2k + 1 ≤ n. If n is odd, n′ is even, and the number of remainingedges is at least

(n− 1)(n + 1)

4+ 1− (nk − 2k2 + k + 1) =

n2 − 4nk + 8k2 − 4k

4

≥ n2 − 4nk − 2n + 4k2 + 4k + 5

4(6)

=(n− 2k − 1)2

4+ 1

= bn′/2cdn′/2e+ 1

where (6) follows from 2k + 1 ≤ n and k ≥ 2. At no matter what stage of removal ofsmallest odd cycles, H is always non-bipartite with a positive number of edges. This is acontradiction, and so H has a triangle and t(n,G) ≤ bn/2cdn/2e.

n even: n2/4n odd: (n− 1)(n + 1)/4

each edge contributes to n− 2 trianglesthere are

(n3

)triangles so having

(n3

)/(n − 2) = n(n − 1)/6 edges is sufficient to force a

triangle. But t(n, C3) = bn/2cdn/2e.

non-bipartite forces a cycle of length 2k +1. If there is no triangle including any verticesfrom this cycle, then k > 1. If k = 2 the cycle is C5, and there may not be any interior edgesof the cycle. Consider the shortest cycle of odd length. There are no interior edges to thiscycle, because they would create a shorter odd cycle. The cycle consists of 2k + 1 edges,which can be completed by having 2 edges to the same vertex outside the cycle. There aren − (2k + 1) such vertices, and so there can be at most (n − (2k + 1))(2k + 1) + (2k + 1)edges incident to the cycle. This is

2kn + n− (2k + 1)2 + 2k + 1 = 2kn + n− 4k2 − 4k − 1 + 2k + 1

= 2kn + n− 4k2 − 2k

= n(2k + 1)− 2k(2k + 1) = (n− 2k)(2k + 1)

edges at most. When n is even, this leaves at least n2/4− (n− 2k)(2k + 1) = (n2 + 8nk +4n − 16k2 − 8k)/4 = (n(n + 8k + 4) − 2k(8k + 4))/4 edges in the resulting graph. For theinduction hypothesis, we require this to be at least (n2 − 4nk + 4k2 − 2n + 4k)/4

18

Removing these 2k +1 vertices from the graph results in a graph with n−2k−1 verticesn even: (n− 2k − 2)(n− 2k)/4n odd: (n− 2k − 1)2/4

19

6 Turan Theory (4/29/02)

We now return to Turan’s original theorem that was proved in 1941. There are several proofs,and in this lecture, two proofs are presented. First we need to define the Turan graph, Tn,r.

Definition 6. The Turan graph Tn,r is the complete r-partite graph on n vertices so thatevery partite set has size dn/re or bn/rc. If n = rm + s, then Tn,r has s parts with m + 1vertices each and r− s parts with m vertices each. Note that Tn,r cannot contain a completegraph on r + 1 vertices.

For example, Tn,2 is the complete biparite graph with part sizes dn/2e and bn/2c, and byMantel’s theorem, it is the unique largest (in terms of number of edges) graph on n verticeswith no triangle. Turan’s theorem generalizes this to say that the unique extremal graphnot containing a Kr+1 is exactly the Turan graph, Tn,r. The proof of uniqueness is left as anexercise.

Exercise 4. Prove that among all r-partite graphs on n vertices, the Turan graph Tn,r isthe unique graph with the maximum number of edges.

There are two ways of proving this — either compute a formula for the number of edgesin an r-partite graph on n vertices with differing part sizes and maximize it, or use a localpertubation argument to start from some r-partite graph and reach Tn,r.

We now state and prove Turan’s theorem.

Theorem 26. (Turan [?]) Among all n vertex graphs with no r +1-clique, Tn,r is the uniquegraph with the maximum number of edges.

Proof. Suppose G is a graph with no (r + 1)-clique. We will prove that there is an r-partitegraph, H with the same number of vertices as G such that the number of edges in H is atleast the number of edges in G. This together with the Exercise establishes the theorem.

The claim is proved by induction on r. For the base case, when r = 1, e(G) = e(H) = 0.Let r > 1, and consider a vertex v in G of maximum degree, ∆. Let G′ denote the graphinduced by the neighborhood of v, N(v). Clearly, G′ contains no r-clique. By the inductionhypothesis, there is an (r−1)-partite graph, H ′ with vertex set N(v) such that e(H ′) ≥ e(G′).

Let A denote the set V (G) − N(v). Form H by making every vertex in A adjacent toevery vertex in H ′. Since A has no edges in it, H is an r-partite graph contaning no Kr+1.Now,

e(G) = #edges inside N(v) + #edges outside N(v)

≤ e(G′) +∑x∈A

dG(x)

≤ e(H ′) + ∆(n−∆)

= e(H).

Notice that for equality to hold, e(G′) = e(H ′) and every vertex in A must have degree∆. If we require H ′ to be Tn,r−1 by the induction hypothesis, we can conclude that e(G) isuniquely maximized by Tn,r. This completes the proof.

20

Roughly, the number of edges in Tn,r is (1− 1r−1

)(

n2

). If n = rm + s, then the number of

edges can be computed exactly.

Exercise 5. Verify that the number of edges in Tn,r where n = rm + s and 0 ≤ s ≤ r− 1 isgiven by the expression (1− 1/r)n2/2− s(r − s)/2r. Need to double-check this, RR

We now give a different proof of Turan’s theorem proved by Li and Li [?] that uses analgebraic approach. We first define a graph polynomial and the notion of independence in agraph.

Definition 7. Given a graph G with vertex set v1, . . . , vn and e edges. The graph polynomialfG of G is defined by fG(x1, . . . , xn) =

∏(vi,vj)∈E(G)(xi − xj).

An independent set in a graph G is a set of vertices that are pairwise non-adjacent. Theindependence number, i(G) is the maximum size of an independent set in G.

G has independence number less than k + 1 if and only if every set of k + 1 verticescontains at least one edge. In other words, the polynomial fG vanishes whenever k + 1variables are set equal.

Definition 8. Given integers k, n with 1 ≤ k ≤ n, let I(k + 1, n) denote the ideal of thering Z[x1, . . . , xn] consisting of the polynomials that vanish whenever k + 1 variables are setequal.

Then clearly, i(G) < k +1 if and only if fG ∈ I(k +1, n). Li and Li showed that the idealI(k + 1, n) is generated by polynomials ∆(P ) of the form

∆(P ) =k∏

m=1

∏i,j∈Pm,i<j

(xi − xj)

, where P = {P1, . . . , Pk} runs through all partitions of the set {1, . . . , n} into k subsets(some of which may be empty). They also showed that the Delta(P )s of the lowest degreecorrespond to partitions P of the set {1, . . . , n} into k subsets of as nearly equal size aspossible. In other words, the ideal I(k, n) is generated by the polynomials ∆(P ) where P isa partition of {1, . . . , n} into k subsets of as nearly equal cardinality as possible.

With this theorem, the criterion for independence can be restated as follows.

Theorem 27. A graph G has independence number i(G) ≤ k + 1 if and only if fG =∑H gH fH , where H is the union of k vertex disjoint cliques and gH ∈ I(k + 1, n).

Independence is a complementary notion to clique size. A graph G has independencenumber k if and only if its complement has clique size k. The complement of the Turangraph Tn,k is a disjoint union of k cliques whose sizes are almost equal, which correspondsto the graph H described in the theorem above.

Fix an integer k for 1 ≤ k ≤ n. If we write n = qk + r for 0 ≤ r ≤ k, then the resutlof Li and Li implies that every non-zero polynomial in I(k + 1, n) must have degree at leastk(

q2

)+ rq, which is the degree of the generators ∆(P ). Considering the complements of the

21

graphs associated with the Delta(P )s, we conclude that a graph with clique number lessthan k + 1 can have at most(

n

2

)− k

(q

2

)− rq =

k − 2

2(k − 1)(n2 = r2) +

(r

2

)edges, which is Turan’s theorem.

This proof shows that the Turan graphs are in some sense “fundamental pieces” sincethey correspond to the generators of the ideal I(k, n).

We now return to Turan problems for graphs other than cliques. It is interesting to lookat the different stages at which different graphs are forced. We have seen that forcing a Kk

requires basically (1 − 1/k)(

n2

)edges, and the figure below shows the thresholds (known or

conjectured) for other graphs.

22

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