Math 262: Topics in Combinatorial Mathematics

Transcription

1 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

2 Contents 1 Introduction to Ramsey Theory (4/1/02) Definitions The party problem Other Ramsey numbers Open problems for Ramsey numbers Super Six, Part 1 (4/3/02) Problem 1: Ramsey s Theorem Problem 2: Schur s Theorem Problem 3: Van der Waerden s Theorem Van der Waerden s Theorem (Ron Graham, 4/8/02) 9 4 Turán Theory 15 5 More Turán Theory 16 6 Turán Theory (4/29/02) 20 i

3 1 Introduction to Ramsey Theory (4/1/02) Ramsey Theory is the study of unavoidable configurations in families of graphs. A typical problem 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 one sense, Ramsey Theory is a generalization of the Pigeonhole Principle; for instance, in 2- coloring the edges of the star S 2n+1 on 2n + 1 vertices, it is impossible to avoid obtaining a monochromatic S n+1. The seminal paper in Ramsey Theory by Erdős 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 integer k, the set ( V k) 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 ( V 2) of edges. Hypergraphs are a generalization of graphs in which edges may contain any number of vertices. Definition 2. A hypergraph H = (V, E) consists of a set V of vertices and a set E 2 V hyperedges. of We may think of 2 V as ( ) ( V 0 V V ), whence the generalization of graphs on edges consisting of 2 vertices. A hypergraph is r-uniform provided that every hyperedge consists of r vertices, whereby E ( ) V r. The complete graph K n is the graph for which E = ( V 2), and the complete r-uniform hypergraph K n (r) is the graph for which E = ( ) V r. The basic Ramsey s Theorem is as follows. Theorem 1 (Ramsey s Theorem). Given l, r, k N, there exists an n 0 (l, r, k) so that whenever K n (r) is edge k-colored, n > n 0 guarantees the existence of a monochromatic subgraph K (r) l. The original proof showed existence of n 0, but the actual values of n 0 for various parameters 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 to as the party problem : given 6 people at a party, it is guaranteed that 3 of them are either mutually 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 any 2-coloring of K n contains either a monochromatic K k or a monochromatic K l. 1

4 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 the upper bound, consider the graph K 6. Each vertex has degree 5, and so by switching red and blue if necessary, we may assume a particular vertex v has 3 red edges adjacent to it (see Figure 1). Now consider the edges of the 3-clique among vertices a, b, and c. If one of these edges is red, then that edge forms a red triangle with edges coming from v. Otherwise, none of the edges are red, and there is a blue triangle with a, b, and c. Therefore R(3, 3) 6. a 4 red v b? 4 red 4? red? c 4 Figure 1: Gives an upper bound for R(3, 3) by proving the existence of a monochromatic triangle. In order to show R(3, 3) > 5, it is sufficient to demonstrate a red-blue coloring of K 5 which has no monochromatic triangles. Consider the coloring in Figure 2, where drawn edges are red and missing edges are blue. 4 red red 4 4 blue interior edges red red 4 4 red Figure 2: Gives a lower bound for R(3, 3) by demonstrating a coloring of K 5 with no monochromatic triangle. It is impossible to find a monochromatic 3-clique using the coloring in Figure 2. Choosing any 3 vertices, two of them must be adjacent using edges on the perimeter, and so there is a red edge. But two of them must not be adjacent using edges on the perimeter, and so there is a blue edge. Thus no triangle is monochromatic, and R(3, 3) > 5. 2

5 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. A dynamic survey of results on Ramsey numbers is maintained at [21]. The generalized Ramsey number R(c 1,..., c k ), where k colors are used, is defined as follows. Definition 4. The Ramsey number R(l 1,..., l k ; r) is the smallest number n such that any hyperedge k-coloring of K n (r) with the colors c i for 1 i k forces a K (r) l i of color c i for some i. When r is omitted from the above definition, the coloring is done on graphs rather than hypergraphs; for instance, R(3, 3; 2) = R(3, 3). Furthermore, the notation R k (l) is often used to denote R(l,, l), with l taken k times in succession. The following is due to Greenwood and Gleason ([16]). Proposition 3. R(3, 3, 3) = 17. Proof. To show R(3, 3, 3) 17, isolate one vertex v 0 in K 17 and consider the 16 edges leaving v 0. At least 6 of these edges must be of the same color; say, blue. The K 6 induced by these 6 vertices adjacent to v 0 via blue edges either has a blue edge, or not. If so, a blue triangle is formed with v 0. If not, any edge 2-coloring of K 6 forces a monochromatic triangle, by Theorem 2. To show R(3, 3, 3) 16, we must exhibit a 3-colored K 16 with no monochromatic triangle. Again fix a vertex v 0, 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 yellow edges defines 5 neighbors of v 0 ; denote the corresponding K 5 s by K5 red, K5 blue, and K yellow 5, respectively. Color K yellow 5 according to Figure 2. Color K5 red and K5 blue similarly, except with yellow replacing red or blue, respectively. Color the remaining edges from K5 x to K y 5, by {red, yellow, blue} \ {x, y}. The result is a 3-coloring of K 16 with no monochromatic triangles. Other results include 51 R 4 (3) 64 (initial work in [6] showed R 4 (3) 50), R(4, 4; 3) = 13, and the following 2-coloring generalization. Proposition 4. Theorem 5. c 1 k log k R(3, k) c k 2 log k. 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. Consider the complete graph K n, fix a vertex v 0, and consider the n 1 edges incident to v 0. Suppose k and l correspond to the colors red and blue, respectively. Any edge {red, blue}-coloring of K n gives either at least R(k 1, l) red edges or at least R(k, l 1) blue edges incident to v 0. In either case, by definition of R(k 1, l) and R(k, l 1), a red K k or a blue K l is 3

6 guaranteed. If there are at least R(k 1, l) red edges incident to v 0, the K R(k 1,l) defined by the vertices adjacent to v 0 along red edges have either a red K k 1 or a blue K l. If they have a red K k 1, throwing in v 0 creates the red K k. 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 + l 3 + k 2 k 1 ( ) k + l 2 =. k 1 A critical graph, usually K n, is one for which the Ramsey number is achieved, or in other words, smallest graph which forces the configuration in question. In fact, the first inequality in Theorem 5 is strict if both values R(k 1, l) and R(k, l 1) are even, because no K n is critical for n odd. The Ramsey number R(k, k) has been studied extensively; the current lower bound is R(k, k) k e 2 2k/2 (1 + o(1)). An easier lower bound which illustrates common techniques is the following. Theorem 6. R(k, k) 2 k/2. Proof. Color the edges of K n in two colors, red and blue. There are 2 (n 2) 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 (n 2) ( k 2)+1 S-bad colorings. The total number of bad colorings over all choices of S is at most ( n k ) 2 (n 2) ( k 2)+1, including double counting. Then there is a good coloring if ( ) 2 (n 2) n > 2 (n 2) ( k 2)+1, (1) k where we wish to find some n = n(k) satisfying (1). In particular, we may choose n to satisfy ( ) n 2 1 (k 2) < 1 k n k 2 k2 /2 < 1 n < 2 k/2. In particular, we have shown that for n < 2 k/2, there is a 2-coloring of K n with no monochromatic K k, and so R(k, k) 2 k/2. 4

7 Using Theorem 5 to bound R(k, k) above, we have R(k, k) ( ) 2k 2 c22k 2, (2) k 1 k 1 by employing the Stirling approximation n! = n n e n 2πn. Combining Theorem 6 and (2) and taking the kth root throughout yields the following. Corollary 7. 2 (R(k, k)) 1/k Open problems for Ramsey numbers Several outstanding questions remain for Ramsey numbers, including the following conjectures of Erdős with money attached. Conjecture 1 (Erdős, $100). The following limit exists: lim R(k, 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]): ( n c 1 log n ) k+1 2 n k 1 < c 2 (1 + o(1)). (3) log k 1 n A special case is possibly strengthened by the following corollary, which would reduce the order of the gap in (3) from n 5/2 to n 3. Conjecture 2 (Erdős, $250). R(4, n) > c 1n 3 log c 2 n. These and other conjectures of Erdős appear in [7]. 5

8 2 Super Six, Part 1 (4/3/02) This lecture begins a survey of six problems in Ramsey Theory which demonstrate circumstances 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 later lecture. 2.1 Problem 1: Ramsey s Theorem For the first theorem, recall Definition 4 of Ramsey numbers from Section 1. Theorem is that Ramsey numbers exist. Ramsey s Theorem 8 (Ramsey s Theorem). Given c, r, and k 1,..., k c from N, there exists an n = n(k 1,..., k c ; r) such that any r-hyperedge c-coloring of K n (r) must contain a monochromatic K (r) k i for some i. Recall from last lecture, Corollary 7 gives us that 2 (R(k, k)) 1/k 4. The bounds on R(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 cyclic coloring, the vertices [n] are arranged consecutively around the circumference of a circle, with n and 1 adjacent. The color assigned to an edge {i, j} is determined by i j, corresponding to distinct chord lengths. The coloring exhibited in Figure 2 is cyclic: {i, j} is colored red when ch(i, j) = 1, and blue otherwise. Specifically, if a cyclic coloring on 44 vertices can be demonstrated which has no monochromatic K 5, Exercise 1 is settled in the affirmative. The following constrained Ramsey problem uses an upper bound cited by the Guinness Book of World Records as the largest number. Consider the k-hypercube Q k on vertices V = {0, 1} k, and build the related graph G k with vertices V and edges E = ( V 2). Let f(4) be the minimum k such that any 2-coloring of G k must contain a monochromatic K 4 in a plane. A restatement of this problem is available in [18, p. 54]. The best and only upper bound for f(4) is known as Graham s number, and is a 64-times repeated tower of powers of Problem 2: Schur s Theorem Schur s Theorem treats partitions of the first n positive integers into t parts, and whether or not a part contains a triple x, y, z such that x + y = z. A part which contains no such triple is called sum-free. Theorem 9 (Schur s Theorem). Given t N, there is a maximum number n = S(t) so that there exists a partition of [n] into t classes, each of which is sum-free. 6

9 S(t) is known as the Schur number. Clearly, if [n] can be partitioned into t sum-free classes, then so can [n ], for n < n. Also, if [n] = S(t) and n > n, there does not exist a partition of [n ] into t sum-free classes. Schur proved Theorem 9 in [24] by showing that S(t) < t!e, and also gave the following lower bound, with the cases for equality shown in [17]. Theorem 10 (Schur number lower bound). with equality for t = 1, 2, 3. S(t) 1 2 ( 3 t 1 ), Example 1. S(2) = 4. Consider the partition [4] = {1, 4} {2, 3}, which is sum-free, showing S(2) 4. Now consider any partition of [5] into to sets A and B. WLOG, 1 A, and so 2 B in order for A to be sum-free. Thus 4 A is forced, and then 3 B is forced. There is nowhere to put 5 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}} demonstrates that S(3) 13, as does Theorem 10. The partition P = {π 1, π 2, π 3 } induces a cyclic coloring of K 14 by coloring {i, j} with color i provided that i j π i. Suppose to the contrary that a monochromatic triangle on vertices a < b < c results. Then b a, c a, and c b are all in π i for some 1 i 3, and since (b a) + (c b) = (c a), π i fails to be sum-free, contradicting the choice of partition. This formulation gives us in general a way to bound S(t) in terms of R t (3). Theorem 11 (Schur s Ramsey Theorem). S(t) R t (3) 2. Proof. Let P = {π 1,..., π t } be a partition of S(t) into t sum-free parts. Induce a cyclic coloring on K S(t)+1 by coloring {i, j} with color c where i, j π c. As described in the discussion, the presence of a monochromatic triangle would correspond to a partition which is not sum-free, and so R t (3) > S(t) + 1. In the theorem, the difference of 2 is due to one vertex being added in the cyclic coloring so that the set of chord lengths is all of [S(t)]. Then R t (3) is the smallest value for which a monochromatic 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 shown in [12]. We may relax our goals and attack variations of Schur s Theorem, such as the following challenging exercise, which is thought to be completely open, yet tractable for small computational examples. Exercise 2 (Research question). What is the smallest subset of Z such that any partition into t parts has some part not sum-free? 7

10 2.3 Problem 3: Van der Waerden s Theorem (coordinate references from mathworld site) Van der Waerden s Theorem concerns partitioning either N or the first n positive integers so as to try to avoid arithmetic progressions of a specified length. Theorem 12 (van der Waerden s Theorem). Any partition of N into r parts π 1,..., π r ensures the existence of arbitrarily long arithmetic progressions in some π i. 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 = W r (k) such that any partition of [n] into r parts guarantees a k-term arithmetic progression in some part. W r (k) is known as van der Waerden s number. The existence of W r (k) implies well-definedness, but the proof of existence and the proof of Theorem 12 will be delayed until Section 3. Defining W (k) = W 2 (k), we have the following: W r (2) = r +1, W (3) = 9, W (4) = 34, and W (5) = 178. Berlekamp (add reference) proved the lower bound k2 k W (k), and Shelah (add reference) claimed a $500 Erdős prize by showing that W (k) is bounded above by a (check usage) primitive-recursive function as follows. Let T 2 (n) be defined for all n N by the recurrence T 2 (n) = 2 T 2(n 1) for n 1 and the initial condition T 2 (0) = 1. Then T 2 (n) can be thought of as a tower of 2 s: T 2 (n) = Shelah s prize-winning bound is W (k) T 2 (T 2 (T 2 (T 2 (T 2 (1))))). This is interpreted as a tower of 2 s, where the number of 2 s in the tower is the tower of 2 s of the next level in. The following improved bound by Gowers in (cite gowers paper) won a $1000 Erdős prize as well as a Fields Medal. Theorem 13 (van der Waerden s number bound). W (k) k+9. R. L. Graham both proposed the following conjecture and placed a bounty on it worth $1000. (check for reference) Conjecture 3 (Graham, $1000). Things to do: W (k) 2 k2. possibly add references to Rödl, 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 where tower notation is defined 8

11 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 a conjecture of Schur. We state the theorem as follows. Theorem 14 (van der Waerden s Theorem). For all r N, all possible partitions N = C 1 C r force a k-term arithmetic progression in some C i for all k. A k-term arithmetic progression (k-ap for short) has the form a, a + d,..., a + (k 1)d for real numbers a and d. Theorem 14 immediately raises the question as to whether the infinite 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 the finite with the following Compactness Principle, which is presented in the context of van der Waerden s Theorem, but can be extended to colorings of hypergraphs. The following proof borrows from [15, Theorem 4]. Theorem 15 (Compactness Principle (König infinity lemma?)). Suppose that for each n N, there exists an r-partition of [n] for which no part contains a k-term AP. Then N contains no k-term AP. Proof. For each n, consider the partition of [n] into r parts as an r-coloring of [n] by the function χ n : [n] [r]. By the assumption of the theorem, for each n let χ n define a partition of [n], no part of which contains a k-term AP. We desire to construct a coloring χ from the χ n 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 infinitely often as χ n (1). Of these choices which are defined on 2, at least one color appears as χ n (2) for infinitely many n s; choose this color as χ (2). For each j, continue refining this set of n s by throwing away j 1 and defining χ (j) as a color which appears as χ n (j) for infinitely many n s. Now consider a k-term AP a 1,..., a k. Then χ agrees with χ n for some n a k on [a k ], and the AP is not monochromatic for χ n, 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-term AP from can be considered a hyperedge of the graph H on countably many vertices. This leads to the generalization that appears as [15, Theorem 4] in terms of infinite hypergraphs where each hyperedge is on finitely many vertices. Another specific problem of this type appears in [13], in which the unavoidable substructure is a sequence which appears as the subsequence of another larger sequence. Theorem 16 (Compactness Principle). Let H = (V, E) be a hypergraph where all X E are finite (but V may be countably infinite). Suppose that, for all W V, W finite, χ(h W ) r. Then χ(h) r. 9

12 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 order to 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)] = C 1 C r forces a k-term AP in some C i. r We may abbreviate the statement of this theorem as [W (k, 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 part is a k-term AP. For all r, W (2, r) = r + 1 by the Pigeonhole Principle, since at least one part contains two elements which comprise a 2-term AP. The first nontrivial we examine is W (3, 2). In showing an upper bound for W (3, 2), we also illustrate a technique that yields an upper bound for W (k, r) in general. Figure 3: (a) A representatively 2-colored 5-block has two elements of the same color and point to a third element in the same block which would complete a 3-AP if it were not colored differently. (b) Since there are 32 possible 2-colored 5-blocks, having 33 5-blocks forces a repeat. (c) Two 2-term AP s of different colors point to the same location X which would extend both AP s, guaranteeing a 3-term monochromatic AP when there are only 2 colors. Proposition 18. W (3, 2)

13 Proof. We initially consider a 2-coloring of [n], for sufficiently large n. It turns out that n 325 is large enough, but for now we could assume we have a 2-coloring of N. An arbitrary contiguous 5-member subsequence, which we refer to as a block, has two colors appearing at least twice in the first 3 elements. In addition, these two elements point to a third element in the block which completes a 3-AP if it is of the same color. There are 32 possible blocks, but without loss of generality, we illustrate the proof using the top block of Figure 3(a), by coloring the first and third position red, and the fifth position blue. The remaining two elements are colored arbitrarily to avoid a 3-AP. The important part of the construction is that we may choose the first two elements of the same color and of distance d 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 first time a block is repeated; this is equivalent to asking for W (2, 2 5 ), which is 33. As shown in Figure 3(b), within 33 5-blocks there are two identical 5-block which are at a distance of b 32 5-blocks apart. The proof concludes by considering the block pointed to by these identical 5-blocks. As illustrated in Figure 3(c), the first red in the first 5-block and the second red in the second 5-block point to X, and the blues in the first and second 5-blocks point to X, forcing a 3-term AP since X must be colored red or blue. The construction required 2b 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 whether W (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 finite sequence 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-colored sequence of length 9 has a 3-term AP. Focus initially on the center element, which we assume is 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 central red 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 right of the central red element is red, we have the following situation: Y B R R X, 11

14 which forces a blue in position X in order to avoid a 3-term red AP, which in turn forces a red 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 position Z, 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. Therefore W (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 integers constructed in a similar fashion as in Prop. 18, and will be further illustrated by the nontrivial case 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 3 2W (2,3) 3-colorings of blocks of length 2W (2, 3), and W ( 2, 3 2W (2,3)) blocks guarantee two identically colored blocks, which point to a position in a third block. The three blocks together form a super-block. (c) Enough super-blocks guarantees two identically colored super-blocks which point to a position in a third which is forced to complete a monochromatic 3-term AP. Theorem 20. W (3, 3) ( ). Proof. All colored blocks and elements in this proof are viewed as taken from [n] for some sufficiently large n to be determined. First consider a 3-coloring of [W (2, 3)], which guarantees a monochromatic 2-term AP, as illustrated in Figure 3(a). This AP points to a third element 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 3 2W (2,3) kinds, since each element may be 3-colored. Among W ( 2, 3 2W (2,3)) such blocks there must be a duplicate, as in Figure 3(b). By considering 2W ( 2, 3 2W (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 completes a monochromatic 3-term AP. Now this entire structure is a single super-block, as in the bottom of Figure 3(b). There must be two identical super-blocks in the first W (2, 3 2W(2,32W (2,3) ) ) super-blocks. Extending to twice as many super-blocks ensures a focal position X in a third super-block which can extend a red, blue, or yellow 2-term AP. Therefore ( 2W 2, 3 2W(2,32W (2,3) ) ) = 2W (2, 3 2W(2,38 ) ) 12

15 ( ) = 2W 2, 3 2(38 +1) = 2W ( 2, ) = 2 ( ) super-blocks suffice to force a monochromatic 3-term AP. Each super-block contains 2W ( 2W 2, 3 (2,3)) blocks, which each contain 2W (2, 3) elements, for a total of 2 ( ) 2W ( 2, 3 2W (2,3)) 2W (2, 3) = 2 ( ) = ( ) 3-colored elements to force a monochromatic 3-term AP. The construction in the theorem is extravagantly wasteful, coming in much higher than the known value of W (3, 3) = 27. The extremely large numbers from this and related constructions for bounding W (k, r) must be expressed in a way that exploits the fact that they are primitive recursive. Let W (k) := W (k, 2) and recall the tower notation in Section 2. The following is conjectured. Conjecture 4 ($1000 (verify source, tower size)). W (k) T 2 (k). The following result appears in [28]. Theorem 21 (Shelah, 1988). W (k) T 2 (T 2 (T 2 (T 2 (T 2 (1))))). This result was improved in [14]. Theorem 22 (Gowers, 2001). W (k) k+9. One immediate question concerning van der Waerden s Theorem is which part contains the k-term AP. This was answered by Szemerédi as follows. Theorem 23. For any A N, define d(a) = A {1,..., m} /m. If lim sup d(a) > 0, m then A has a k-term AP of arbitrary length k. Roth studied the case k = 3 in 1954 [23] (check ref), and Szemerédi the case k = 4 in 1969 [29]. One open problem is to minimize the size of a set of integers which when colored forces an AP. For instance, if we consider 2-coloring 27 of the first 37 integers in the following pattern: } 11.{{.. 11} , 19 1 s a 4-term AP is forced. This is because the numbers on the edges are not in as many AP s as the ones in the center. We might want to look for k-ap s forced when all but one member is monochromatic, or a k-ap which has only 2 of a possible r colors. 13

16 Remarks (Schur s Theorem) For m N large enough, any partition [m] = C 1 C r into r 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 [ er! ] = C 1 cup C r in order to guarantee a non sum-free monochromatic color class. This was determined during 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 a monochromatic solution when N is finitely colored. An equation a i x i = 0 is partitionregular provided that it can be solved in 0 s and 1 s but not all 0 s. Examples are, as above, x + y = z and x + y = 2z. (Exercise) Determine if x + y = 3zx is partition-regular, especially in the case of 2 colors. (Conjecture, $100) Show that x 2 + y 2 = z 2 is partition-regular. In particular, is there a monochromatic solution when N is 2-colored? Rödl showed that x 1 + y 1 = z 1 is partition-regular and a monochromatic solution is forced with an r-coloring. Notes From [15]: Historical Note. I. Schur, working on the distribution of quadratic residues in Z p, first conjectured the result proved by van der Waerden. Van der Waerden heard of the conjecture through Baudet, a student at Göttingen at the time, and has referred to his result as Baudet s Conjecture in the literature. A brief account of Schur s contribution is given by A. Brauer in the preface to I. Schur-Gesammelte Abhandlungen (Springer-Verlag, 1973). The book of Schur mentioned is [25, 26, 27]. To do list: reference for Baudet and Schur 14

17 4 Turán Theory 15

18 5 More Turán Theory The treatment of Turán Theory begun in Section 4 is continued here. The Turán number t(n, G) indicates the minimum number of edges that must be contained in a graph H on n vertices before G is forced to appear as a subgraph. In other words, t(n, G) = max {E(H) : V (H) = n, G H}. Turán Theory and Ramsey Theory are related by considering an edge k-coloring of the complete graph in which one color class of edges is forced to have high density. Fact 1. If kt(n, G) < ( n 2), then any edge k-coloring of Kn contains a monochromatic G. One color class is forced to have at least ( n 2) /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 then r k (G) > n. ( ) n kt(n, G) > 2 log n, 2 Instead of showing this theorem, we give the following stronger version, which arises from a random packing k colored copies of H into K n in order to show that there is an edge k-coloring of K n which does not force a monochromatic G. Theorem 25. If ( ) ( n 1 2 ) k t(n, G) ( n < 1, (4) 2) then r k (G) > n. Proof. Let H have n vertices and t(n, G) edges, with G H. Consider k copies of H placed randomly on K n. This is done for each of k colors by permuting the vertices of H randomly and assigning the current color to an edge {u, v} E(K n ) if {u, v} is present in the randomly arranged H. Without loss of generality, multiply-colored edges in K n retain the first assigned color. Now for each edge {u, v} in K n and each iteration i in the placement of colored H s, the probability that {u, v} is assigned the ith color is E(H) ( n ) = 2 t(n, G) ( n 2). The probability that {u, v} is left uncolored by all k iterations is ( 1 ) k t(n, G) ( n. 2) 16

19 If the expected number of edges left uncolored, which is ( ) ( n 1 2 ) k t(n, G) ( n, 2) is less than 1, the existence of a coloring of this type in which no edge is left uncolored is ascertained. In particular, there is no monochromatic G in this edge k-coloring, because G 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) ( n 2) < t(n, G) > ( ) k n 2 ( ) n 2 ( ) k n, 2 which is stronger by a factor of log n when k is a fixed constant. The full implications of Theorem 25 for bounding Ramsey numbers may not yet be completely explored. We now compute specific Turán numbers, starting with cycles C 3 and C 4, and then considering the complete bipartite graph K r,s. Exercise 3. t(n, C 3 ) = n/2 n/2. Proof. The complete bipartite graph K n/2, n/2 has n/2 n/2 edges and no odd cycle, giving t(n, C 3 ) n/2 n/2. The reverse direction proceeds by showing that when more than n/2 n/2 edges are present, there is either a triangle or we may remove arbitrarily many odd cycles from the graph (which is impossible). Suppose H has n vertices and at least n/2 n/2 + 1 edges. It is impossible for H to be bipartite, because the largest number of edges in a bipartite graph on n vertices is achieved by K n/2, n/2 with n/2 n/2 edges. Suppose to the contrary that H has no C 3. Let C be the smallest odd cycle of H, and label its vertices {1,..., 2k + 1} with edges {i, i + 1} for 1 i < 2k + 1 and {1, 2k + 1}. No other edges are present by assumption of C being the smallest 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 and V (H) \ C. In total, removing C and all edges incident to C from H causes the removal of at most k(n 2k 1) + 2k + 1 = nk 2k 2 + k + 1 edges. The remaining graph H has n := n 2k 1 vertices. We now argue that H has at least n /2 n /2 + 1 edges to complete the proof. If n is even, n is odd, and the number of remaining edges is at least 17

20 n (nk 2k2 + k + 1) = n2 4nk + 8k 2 4k 4 n2 4nk + 4k 2 2n + 4k (n 2k 2)(n 2k) = = n /2 n /2 + 1 where (5) follows from 2k + 1 n. If n is odd, n is even, and the number of remaining edges is at least (n 1)(n + 1) (nk 2k 2 + k + 1) = n2 4nk + 8k 2 4k 4 n2 4nk 2n + 4k 2 + 4k (n 2k 1)2 = = n /2 n /2 + 1 where (6) follows from 2k + 1 n and k 2. At no matter what stage of removal of smallest odd cycles, H is always non-bipartite with a positive number of edges. This is a contradiction, and so H has a triangle and t(n, G) n/2 n/2. n even: n 2 /4 n odd: (n 1)(n + 1)/4 (5) (6) each edge contributes to n 2 triangles there are ( ( n 3) triangles so having n 3) /(n 2) = n(n 1)/6 edges is sufficient to force a triangle. But t(n, C 3 ) = n/2 n/2. non-bipartite forces a cycle of length 2k + 1. If there is no triangle including any vertices from this cycle, then k > 1. If k = 2 the cycle is C 5, and there may not be any interior edges of the cycle. Consider the shortest cycle of odd length. There are no interior edges to this cycle, 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 are n (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 4k 2 4k 1 + 2k + 1 = 2kn + n 4k 2 2k = n(2k + 1) 2k(2k + 1) = (n 2k)(2k + 1) edges at most. When n is even, this leaves at least n 2 /4 (n 2k)(2k + 1) = (n 2 + 8nk + 4n 16k 2 8k)/4 = (n(n + 8k + 4) 2k(8k + 4))/4 edges in the resulting graph. For the induction hypothesis, we require this to be at least (n 2 4nk + 4k 2 2n + 4k)/4 18

21 Removing these 2k + 1 vertices from the graph results in a graph with n 2k 1 vertices n even: (n 2k 2)(n 2k)/4 n odd: (n 2k 1) 2 /4 19

22 6 Turán Theory (4/29/02) We now return to Turán s original theorem that was proved in There are several proofs, and in this lecture, two proofs are presented. First we need to define the Turán graph, T n,r. Definition 6. The Turán graph T n,r is the complete r-partite graph on n vertices so that every partite set has size n/r or n/r. If n = rm + s, then T n,r has s parts with m + 1 vertices each and r s parts with m vertices each. Note that T n,r cannot contain a complete graph on r + 1 vertices. For example, T n,2 is the complete biparite graph with part sizes n/2 and n/2, and by Mantel s theorem, it is the unique largest (in terms of number of edges) graph on n vertices with no triangle. Turán s theorem generalizes this to say that the unique extremal graph not containing a K r+1 is exactly the Turán graph, T n,r. The proof of uniqueness is left as an exercise. Exercise 4. Prove that among all r-partite graphs on n vertices, the Turán graph T n,r is the unique graph with the maximum number of edges. There are two ways of proving this either compute a formula for the number of edges in an r-partite graph on n vertices with differing part sizes and maximize it, or use a local pertubation argument to start from some r-partite graph and reach T n,r. We now state and prove Turán s theorem. Theorem 26. (Turán [?]) Among all n vertex graphs with no r + 1-clique, T n,r is the unique graph 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-partite graph, H with the same number of vertices as G such that the number of edges in H is at least 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 graph induced by the neighborhood of v, N(v). Clearly, G contains no r-clique. By the induction hypothesis, 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 to every vertex in H. Since A has no edges in it, H is an r-partite graph contaning no K r+1. Now, e(g) = #edges inside N(v) + #edges outside N(v) e(g ) + x A d G (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 T n,r 1 by the induction hypothesis, we can conclude that e(g) is uniquely maximized by T n,r. This completes the proof. 20

23 Roughly, the number of edges in T n,r is (1 1 edges can be computed exactly. r 1 )( n 2). If n = rm + s, then the number of Exercise 5. Verify that the number of edges in T n,r where n = rm + s and 0 s r 1 is given by the expression (1 1/r)n 2 /2 s(r s)/2r. Need to double-check this, RR We now give a different proof of Turán s theorem proved by Li and Li [?] that uses an algebraic approach. We first define a graph polynomial and the notion of independence in a graph. Definition 7. Given a graph G with vertex set v 1,..., v n and e edges. The graph polynomial f G of G is defined by f G (x 1,..., x n ) = (v i,v j ) E(G) (x i x j ). An independent set in a graph G is a set of vertices that are pairwise non-adjacent. The independence 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 vertices contains at least one edge. In other words, the polynomial f G vanishes whenever k + 1 variables are set equal. Definition 8. Given integers k, n with 1 k n, let I(k + 1, n) denote the ideal of the ring Z[x 1,..., x n ] consisting of the polynomials that vanish whenever k + 1 variables are set equal. Then clearly, i(g) < k + 1 if and only if f G I(k + 1, n). Li and Li showed that the ideal I(k + 1, n) is generated by polynomials (P ) of the form (P ) = k m=1 i,j P m,i<j (x i x j ), where P = {P 1,..., P k } 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 degree correspond to partitions P of the set {1,..., n} into k subsets of as nearly equal size as possible. In other words, the ideal I(k, n) is generated by the polynomials (P ) where P is a 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 f G = H g Hf H, where H is the union of k vertex disjoint cliques and g H I(k + 1, n). Independence is a complementary notion to clique size. A graph G has independence number k if and only if its complement has clique size k. The complement of the Turán graph T n,k is a disjoint union of k cliques whose sizes are almost equal, which corresponds to 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 resutl of Li and Li implies that every non-zero polynomial in I(k + 1, n) must have degree at least k ( q 2) + rq, which is the degree of the generators (P ). Considering the complements of the 21

24 graphs associated with the Delta(P )s, we conclude that a graph with clique number less than k + 1 can have at most ( ) ( ) n q k rq = k 2 ( ) r 2 2 2(k 1) (n2 = r 2 ) + 2 edges, which is Turán s theorem. This proof shows that the Turán graphs are in some sense fundamental pieces since they correspond to the generators of the ideal I(k, n). We now return to Turán problems for graphs other than cliques. It is interesting to look at the different stages at which different graphs are forced. We have seen that forcing a K k requires basically (1 1/k) ( n 2) edges, and the figure below shows the thresholds (known or conjectured) for other graphs. 22

25 References [1] Leonard D. Baumert and Solomon W. Golomb. Backtrack programming. J. Assoc. Comput. Mach., 12: , [2] Béla Bollobás. Extremal graph theory. Academic Press Inc. [Harcourt Brace Jovanovich Publishers], London, [3] Béla Bollobás. Extremal graph theory with emphasis on probabilistic methods. Published for the Conference Board of the Mathematical Sciences, Washington, DC, [4] Béla Bollobás. Extremal graph theory. In Handbook of combinatorics, Vol. 1, 2, pages Elsevier, Amsterdam, [5] Béla Bollobás. Modern graph theory. Springer-Verlag, New York, [6] F. R. K. Chung. On the Ramsey numbers N(3, 3,..., 3; 2). Discrete Math., 5: , [7] F. R. K. Chung and R. L. Graham. Erdős on graphs. A K Peters Ltd., Wellesley, MA, His legacy of unsolved problems. [8] F. R. K. Chung and R. L. Graham. Forced convex n-gons in the plane. Discrete Comput. Geom., 19(3, Special Issue): , Dedicated to the memory of Paul Erdős. [9] F. R. K. Chung and C. M. Grinstead. A survey of bounds for classical Ramsey numbers. J. Graph Theory, 7(1):25 37, [10] P. Erdős and G. Szekeres. A combinatorial problem in geometry. Compositio Math., 2: , [11] Harold Fredricksen. Schur numbers and the Ramsey numbers N(3, 3,, 3; 2). J. Combin. Theory Ser. A, 27(3): , [12] Harold Fredricksen and Melvin M. Sweet. Symmetric sum-free partitions and lower bounds for Schur numbers. Electron. J. Combin., 7(1):Research Paper 32, 9 pp. (electronic), [13] Harvey M. Friedman. Long finite sequences. J. Combin. Theory Ser. A, 95(1): , [14] W. T. Gowers. Erratum: A new proof of Szemerédi s theorem [Geom. Funct. Anal. 11 (2001), no. 3, ; ]. Geom. Funct. Anal., 11(4):869, [15] Ronald L. Graham, Bruce L. Rothschild, and Joel H. Spencer. Ramsey theory. John Wiley & Sons Inc., New York, second edition, A Wiley-Interscience Publication. [16] R. E. Greenwood and A. M. Gleason. Combinatorial relations and chromatic graphs. Canad. J. Math., 7:1 7,

26 [17] Richard K. Guy. Unsolved problems in number theory. Springer-Verlag, New York, second edition, Unsolved Problems in Intuitive Mathematics, I. [18] Paul Hoffman. The man who loved only numbers. Hyperion Books, New York, The story of Paul Erdős and the search for mathematical truth. [19] D. Kleitman and L. Pachter. Finding convex sets among points in the plane. Discrete Comput. Geom., 19(3, Special Issue): , Dedicated to the memory of Paul Erdős. [20] J. Pach and G. Tóth. A generalization of the Erdős-Szekeres theorem to disjoint convex sets. Discrete Comput. Geom., 19(3, Special Issue): , Dedicated to the memory of Paul Erdős. [21] Stanislaw Radziszowski. Small ramsey numbers. Electron. J. Combin., [22] Wolfram Research. Happy end problem. [23] K. F. Roth. On certain sets of integers. J. London Math. Soc., 28: , [24] I. Schur. Über die Kongruenz xm +y m z m (mod p). Jahresber. Deutsche Math.-Verein., 25: , [25] Issai Schur. Gesammelte Abhandlungen. Band I. Springer-Verlag, Berlin, Herausgegeben von Alfred Brauer und Hans Rohrbach. [26] Issai Schur. Gesammelte Abhandlungen. Band II. Springer-Verlag, Berlin, Herausgegeben von Alfred Brauer und Hans Rohrbach. [27] Issai Schur. Gesammelte Abhandlungen. Band III. Springer-Verlag, Berlin, Herausgegeben von Alfred Brauer und Hans Rohrbach. [28] Saharon Shelah. Primitive recursive bounds for van der Waerden numbers. J. Amer. Math. Soc., 1(3): , [29] E. Szemerédi. On sets of integers containing no four elements in arithmetic progression. Acta Math. Acad. Sci. Hungar., 20:89 104, [30] G. Tóth and P. Valtr. Note on the Erdős-Szekeres theorem. Discrete Comput. Geom., 19(3, Special Issue): , Dedicated to the memory of Paul Erdős. [31] Douglas B. West. Introduction to graph theory. Prentice Hall Inc., Upper Saddle River, NJ,