(n 1)/2. Proof. Choose a random tournament. The chance that it fails to have S k is bounded above by this probability.

Transcription

1 : Preliminaries Asymptotic notations f = O(g) means f cg. f = o(g) means f/g 0. f = Ω(g) means f cg. f = Θ(g) means c 1 g f c g. f g means f/g 1. 1 p e p for all nonnegative p, close for small p. The Stirling approximation: k! πk(k/e) k. Some bounds on ( ) n k : ( n ) ( ) k n nk ( n e ) k k k k!. k On central binomial coefficients: n 1 n ( ) n 4n. n πn Let S n be the sum of n iid uniform {±1} random variables. Then: ( ) n 1 E[ S n ] = n 1 n = ( /π + o(1)) n. (n 1)/ 1: The basic method The Ramsey number R(k, l) is the least n such that for any -colouring of the vertices of the complete graph K n on n vertices, there is always either a blue K k or a red K l. Prop 1.1.1: If ( n k) 1 ( k ) < 1 then R(k, k) > n. Proof. Want to show that there s a chance you fail to get a monochromatic subgraph from a random colouring. Add up the probabilities of getting a monochromatic subgraph at each opportunity, to get ( n k) 1 ( k ). It is possible that none of these occur. A tournament on a set V of n players is an orientation of the complete graph K V. That is, every pair of players has a unique winner. A tournament could have property S k : for every group of k players, there is one that beats them all. Thm 1..1: If ( n k) (1 k ) n k < 1 then there is a tournament with S k of size n. Proof. Choose a random tournament. The chance that it fails to have S k is bounded above by this probability. A dominating set in a graph (V, E) is a set U of vertices to which every vertex v V U is connected by an edge. Thm 1..: A graph on n vertices with minimum degree δ > 1 has a dominating set of at most n[1 + log(δ + 1)]/(δ + 1) vertices. 1

2 Proof. Fix a probability p. Choose vertices of with probability p. The expected size of the set {vertices chosen} {vertices not neighbouring the chosen set} is at most pn + (1 p) δ+1 n, by linearity of expectation (also note: alteration). Thus there is a dominating set of size at most this amount. Using 1 p e p to simplify, and then minimizing, we get the result. A cut in a graph is simply a partition of the edges into two nonempty disjoint subsets. The size of a cut is the number of edges whose endpoints lie in opposite sides. The edge connectivity is the minimum size of a cut. A hypergraph (V, E) is a finite set V of vertices, and a subset E V. It is n-uniform if each edge contains n vertices. It has property B, or is two-colourable, if there is a two-colouring of the vertices so that no edge is monochromatic. Let m(n) be the minimum number of edges of an n-uniform hypergraph which is not two-colourable. Prop 1.3.1: Every n-uniform hypergraph with less that n 1 edges is two-colourable, so that m(n) n 1. Proof. Randomly colour the vertices. The probability of failing to give a valid -colouring is less than n 1 1 n. Thm 1.3.: m(n) < (1 + o(1)) e log 4 n n. Proof. We need to construct a small n-uniform hypergraph which is not -colourable. Let v and m be to be determined. Then choose a set V of size v for the vertices. A random subset S of size n has probability ( ( ( a n) + b ( n) )/ v n) of being monochromatic (in a colouring with a and b of each colour), which is at least p = ( ) ( v/ n / v ) n, by convexity (if v is even). Now choose m random sets of size n independently. The probability that a given colouring is a -colouring is at most (1 p) m, so the probability that there exists a -colouring is at most v (1 p) m. Thus, if this quantity is less than 1, we have m(n) m. Now it s asymptotics. A (k, l)-system is a family F = {(A i, B i )} h i=1 pairs of sets1, such that A i = k, B i = l, and A i B j is empty iff i = j, for 1 i, j h. Thm 1.3.3: If F = {(A i, B i )} h i=1 is a (k, l)-system, then h ( ) k+l k. Proof. Let X = i (A i B i ), and randomly order X. Let X i be the event that the elements of A i precede those of B i. If these events are pairwise disjoint, we have 1 P[X i ] = h/ ( ) k+l k. Suppose then that both X i and X j can occur simultaneously. Then there s an ordering such that A i precedes B i and A j precedes B j. WLOG the last element of A i precedes that of A j, but then A i precedes B j, contradicting that A i B j is nonempty. This is sharp, writing F = {(A, X \ A) : A X, A = k}, where X = k + l. A subset A of an abelian group is sum-free if (A + A) A is empty. Thm 1.4.1: Every set A of n nonzero integers contains a sum-free subset of size at least n/3. Proof. Choose a prime p = 3k + far larger than the largest absolute value of any element of A. Embed the question in the field F = Z/pZ, where it is unchanged. In this field, there is a large sum-free subset C = {k + 1,..., k + 1}, which has at least 1/3 of the elements. Choose a random x F, and consider xa C. The average size of this set is at least A /3, and it is always sum-free! 1 Really, pairs of subsets of an arbitrary set.

3 For F a family of subsets of {1,..., n}, let d(f) be the number of unordered pairs of disjoint elements of F. Thm 1.5.1: Let F be a family of m = (1/+δ)n subsets of {1,..., n}. Then d(f) < m δ /. Proof. Suppose for contradiction that d(f) m δ /. Choose an integer t, to be determined. Choose A 1,..., A t independently and with repetition from F. The point is that it will be likely that U = A i has at least n/ elements, and that there are at least n/ distinct subsets of X disjoint from U, a contradiction. For the first, it s really unlikely that U will be contained in any given subset S of size n/, so: P[ U n/] S =n/ ( ) P[all A i S] n n/ t = n(1 δt). m Note: One has that A i could be any of m options, and at most n/ of these lie in each S. We used a very clumsy approximation n { S = n/}. Now to the other side. We have a function v : F N which maps B to the number of disjoint pairs its a member of. Then v takes average value d(f)/m m 1 δ /. Let Y be the r.v. returning the number of B F disjoint to U. Then we want to see that Y often exceeds n/. Applying convexity: ( ) v(b) E[Y ] = m 1 t t ( ) m 1 t t m 1 δ / = t m 1 tδ / = t+(1/+δ)n(1 tδ /). m We d like to bound p = P[Y n /]. As Y < m, we ve got an inequality pm + (1 p) n/ E[Y ]. This shows that p is pretty big, and in fact, for some good t, bigger than n(1 δt). The Probabilistic Lens: The Erdős-Ko-Rado theorem A family F of sets is intersecting if any A, B F satisfy A B. We are interested in large intersecting families of k-element subsets of {0,..., n 1} (where k n/). One option is the set of k-element subsets containing a given element. Erdős-Ko-Rado: This is best possible any intersecting family of k-element subsets of an n-element set has size at most ( n 1 k 1). Proof. Suppose F is such. View {0,..., n 1} as Z/nZ. Let A 0 = {0,..., k 1} and define translates A s = A 0 + s. Then F contains at most k of the A s. [Suppose A 0 F, the other A t intersecting can be arranged in k 1 pairs which are disjoint.] Now for any k-element subset A of Z/nZ and a random i Z/nZ, the probability that A + i F is at most k/n. Thus, if we randomise A uniformly amoungst the k-sets, the probability that A + i F is still at most k/n. Now A + i is uniformly chosen from amoungst the k-sets, so that the probability a random k-element subset is in F is at most k/n, giving the result. 3

4 : Linearity of expectation A hamiltonian path is a path in a graph which visits each vertex exactly once. A hamiltonian cycle is such which returns to its starting vertex. Thm.1.1: There s a tournament T with n players and at least n! (n 1) Hamiltonian paths. Proof. For this is the expected number of Hamiltonian paths in a random Tournament. A bipartite graph is one in which the vertices can be partitioned into two disjoint subsets, such that the induced subgraph on each has no edges. Thm..1: Any graph with e edges contains a bipartite subgraph (not necessarily complete) with at least e/ edges. Proof. For each vertex, randomly assign it to either A or B, to give a random partition. On average, there should be e/ crossing edges. Thm..: If G has n vertices and e edges, then it contains a bipartite subgraph with at least en/(n 1) edges. [Similarly, get e(n + 1)/(n + 1) edges given n + 1 vertices.] Proof. Choose A uniformly from all n-element subsets of V. Same deal. There s something nasty which I don t get about..3. Let K a,b be the graph (A B, A B ) where A = a and B = b. Thm.3.1: There s a -colouring of K n with at most ( n a) 1 ( a ) monochromatic Ka. Proof. This is the expected number of monochromatics in a random -colouring. Thm.3.: Similarly, there s a -colouring of K n with at most ( m a )( n b) 1 ab monochromatic K a,b. Balancing vectors: Thm.4.1: Suppose that v 1,..., v n R n all have length one. Then there is a signed sum ±ɛ 1 v 1 ± ɛ v ± ± ɛ n v n of length at least (and one with length at most) n. Proof. For one can calculate that the average norm-squared random signed sum is n. Thm.4.: Let v 1,..., v n R n have length at most one. Fix weights p 1,..., p n [0, 1], and let w = p i v i be the weighted sum. Then there exists S {1,..., n} with w i S v i n/. Proof. Let ɛ i be one with probability p i, else zero. Let v = ɛ i v i, and let X = w v. The expectation of X is calculated just like the variance of ɛ i, with a bunch of inner products multiplied in, but all these inner products are less than one in magnitude, so E[X] n/4. Unbalancing lights: Given an n n array of lights, some on and some off, and n switches, one for each row and column, that toggle each light therein. Then it s possible to flick switches in order to ensure that there are ( /π + o(1))n 3/ more lights on than off. Proof. Randomly flick the switches for each column. Let R i be the difference of the number on and off in the i th row. This is has distribution S n the sum of n iid uniform {±1} random variables. Now E[ S n ] = ( /π + o(1)) n. Thus E[ R i ] is the quantity we desire.????? 4

5 The Probabilistic Lens: Brégman s theorem The permanent of a matrix is the same sum as the determinant, taken without the signs. We want to bound the permanent of an n n {0, 1}-matrix A with r i ones in the i th row. Note that one might be interested in doing this as the permanent of the matrix is the number of cycle coverings in the digraph associated with A. Brégman s theorem: per(a) (r i!) 1/ri. Proof. To solve this problem, we use the geometric mean. That is, if Y is a random variable, we define G(Y ) := exp(e(log Y )). Note that G(XY ) = G(X) G(Y ) even when X and Y are not independent. Note that if Y takes values a i with probabilities p i, we have G(Y ) = a pi i. Just as E(X) is the mean of the conditional expectations, G(Y ) is the geometric mean of the conditional geometric means. Define an ordered transversal of A to be a pair (α, β) Σ n Σ n of permutations such that A α(i),β(i) = 1 for all i. That is, a sequence of ones contributing one to the permanent, taken with an order. Define a transversal to be a permutation γ such that A i,γ(i) = 1. Let O be the set of ordered transversals, and let T be the set of transversals. There s a function u : O T, forgetting the order, whose fibres each have order n!. There s a projection π 1 : O Σ n. Choose o = (α, β) randomly from O. For i = 1,..., n, remove the α(i) th row and the β(i) th column (containing the i th one in the ordering), and let R i be the number of ones remaining in the α(i) th row just before it is removed. Define L = R i. The random variable L should overestimate per(a), in the sense that per(a) G(L). We ll come back and prove this. We calculate: G[L u(o) = σ] = G[R i u(o) = σ] = (r i ) 1/ri (r i 1) 1/ri (1) 1/ri (R i is uniform on {1,..., r i }) = (r i!) 1/ri which is independent of σ, so G[L] = (r i!) 1/ri (each σ is equally likely) Now we now should show that per(a) G(L). We show this conditionally on π 1 (o) = τ for any τ Σ n. That is, we fix the order of deletion of the rows, and randomise the order of deletion of the columns. Let r = r τ(1) nonzero entries in the τ(1) th row, in the e 1,..., e r columns. Let t i be the perm-cofactor opposite the i th nonzero entry. The probability that we first delete the (e i ) th column first is P[β(i) = e i π 1 (o) = τ] = t i /per(a). Thus (letting t = t i /r = per(a)/r): G[L π 1 (o) = τ] = r r G[rR R n π 1 (o) = τ, β(1) = e i ] ti/per(a) i=1 r (t i ) ti/per(a) (by induction) i=1 ( r ) 1/per(A) = r (t i ) ti i=1 r (( t t) r) 1/per(A) = rt = per(a) (convexity) 5

6 3: Alterations Recall that R(k, l) > n iff there is a two-colouring of the edges of K n with neither a red K k nor blue K l. Thm 3.1.1: For any integer n, R(k, k) > n ( n k) 1 ( k ). Proof. For choosing a random two-colouring of K n, ( n k) 1 ( k ) is the average number of monochromatic K k. There s some colouring with at most this many; remove a vertex from each. There are a couple of analogous results on off-diagonal Ramsey numbers. The independence number α(g) of a graph G is the maximal size of a collection of vertices of G sharing no edges. Thm 3..1: Let G have n vertices and nd/ edges (d 1). Then α(g) n/(d). Restatement: In G has v vertices and e edges with e n/, then α(g) (n/) /e. Proof. Assign each vertex independently a probability of p on being put in a set S. Then the average number of elements of S is pn, and the average number of edges in the induced subgraph is p e. We thus obtain a randomised method of returning an independent subset of G with an average of pn p e elements. This quantity is maximised at p = n/(e) [0, 1]. Combinatorial geometry: We are interested in finding ways to distribute n points in a unit square such that all of the triangles that can be formed from the n points are large in area. For S I, let T (S) be the area of the smallest triangle with corners in S, and let T (n) be the largest value of T (S) as S ranges over subsets of I of size n. Thm 3.3.1: There is a set of n points is the unit square such that T (S) 1/(10n), so that T (n) = Ω(n ). Proof. For any ɛ > 0, suppose that P, Q, R are chosen uniformly from I. Then it is an easy exercise to bound P[Area(P QR) ɛ] 16πɛ. Now choose at random n points in I. Then expected number of triangles with area less that 1/(10n) is at most ( ) n 3 16π/(10n) < n. There is a configuration with less than n small triangles, and we delete a vertex from each of these. A construction of Erdős: T (n) 1/((n 1) ) when n is prime. Proof. Graph the parabola y = x in (Z/nZ) after embedding (Z/nZ) in [0, n 1] as the lattice points. As n is prime, no three points are collinear, and each triangle has integer coordinates, so has area a multiple of 1/. Note that in fact, T (n) = Ω(log n/n ). Given a bounded measurable subset C R d, the packing number of X is δ(c) := µ(c) lim x f(x)/xd where f(x) is the largest number of copies of C that fit inside [0, x] d. Thm 3.4.1: Let C be convex and centrally symmetric around the origin. Then δ(c) d 1. 6

7 Proof. We d like to choose the centres of the copies of C randomly inside the box B = [0, x] d. Choosing p, q B at random, (C +p) intersects (C +q) iff p q C, by central symmetry. Conditioning on fixed q, this occurs with probability at most µ(c)/x d, and thus the unconditional probability of an intersection is at most (/x) d µ(c). Now choose n points in B at random, and put a copy of C at each. Then average number of intersections is at most (n /)(/x) d µ(c), so removing some, we obtain n (n /)(/x) d µ(c) points. Choose n to maximise this quantity, and note that the fact that some of the copies may hang over the side of the box is insignificant in the limit. A greedy algorithm: By simply adding new copies of C until we can no longer go on, we see that δ(c) d. Proof. Note that if we choose p 1,..., p m a maximal subset of B such that the C +p i are disjoint. Then by maximality, there can be no point q such that q p i / c for all i (else we can add q to the collection). Thus n copies of C cover B, from which we can derive the result. Recolouring: recall that we wrote m(n) > m iff every n-uniform hypergraph with m edges is -colourable, i.e. has property B. Thm 3.5.1: If there is a probability p such that k(1 p) n + k p < 1, then m(n) > n 1 k. Cor 3.5.: m(n) = Ω( n n/ log n). Proof of 3.5.: It s good enough to find the largest p such that k p < 1/ and then plug it into k(1 p) n < 1/, to see that k satisfying the hypotheses of can be chosen fairly large. Proof of 3.5.1: It s enough to fix a hypergraph with n k edges, and prove that its vertices can be -coloured. We give three sources of randomness. For each vertex, we flip a fair red / blue coin and a (p, 1 p) recolour / don t recolour coin. We also randomly order the vertices. All this happens once and for all at the start. The idea is as follows. Colour the vertices according to their red/blue coins. Let D be the set of dangerous vertices those that lie in a monochromatic edge. Go through the vertices in D in their (random) order and recolour the coin in question iff it is still dangerous (it still lies in a monochromatic edge) and its second coin displays recolour. Note here that coins which are not initially considered dangerous are never candidates for recolouration. The point is that we can bound the failure probability by k(1 p) n + k p. We ll bound the probability that there is a red edge at the end of the process. The point is that there is a red edge iff one of the following occur, for e E: A e : e was initially red and remained so forever (P[A e ] = n (1 p) n ). C e : e was not initially red but became so. Now if C e occurs, one can see that e blames f (written B ef ), for some f E. That is: e ends up red; f starts out blue; e f = {v} the two edges share exactly one vertex, v; v is the last vertex of e to switch to red; when v changed colour, f was still blue. Now suppose that e, f are edges with e f = {v}. We wish to bound P[B ef ]. Fix the order σ on the vertices, and calculate conditionally. Write i (resp. j) for the number of vertices of e (resp. j) which precede v in σ. Then [ ] 1 [ P[B ef σ] p n+1 ] [ (1 p) j] [ [ (n i+1)] ( ) ] i p Here the factors are, in order: 7

8 1. v must have blue and recolour ;. all other vertices of f must have blue ; 3. the vertices of f which precede v must have don t recolour ; 4. the vertices of e after v must have red ; 5. the vertices of e before v must either have red or have both blue and recolour. In particular, we have P[B ef ] 1 n p E[(1 + p) i (1 p) j ]. The result will follow if we can see that the expectation is at most one. This is so, by the following trick. Let I be a set of size n 1, and let ϕ : T e {v} and ψ : T f {v} be bijections. Let I i (resp. J i ) be the indicator random variable for the event that ϕ(i) (resp. ψ(i)) precedes v. Then [ ] E[(1 + p) i (1 p) j ] = E (1 + p) I i (1 p) Ji = E [ (1 + p) Ii (1 p) Ji] 1 (independence) As there are less than ( n 1 k) pairs of edges, the probability of any failure of type C e is at most k p/. Similarly, the probability of any A e occurring is at most k(1 p) n /. There s some stuff here about continuous time, which I haven t read. The Probabilistic Lens: High Girth and High Chromatic Number The girth of a graph G is the size of its shortest cycle. We ll write α(g) for the independence number (the size of the largest independent set). The chromatic number χ(g) is the least number of colours needed to colour the vertices so that no edge is monochromatic. Erdős (1959): For all k, l there exists a graph G with girth(g) > l and χ(g) > k. Proof. The idea is as follows. Fix θ < 1/l, let p = n θ 1, and let G G(n, p). This graph, for large n, often has few (< n/) cycles shorter than l, and a small (< (3 log n)/p) independence number. Removing a vertex for each short cycle, we have a graph G of size at least n/ with girth at least l, and small independence number. As chromatic number and independence number are inverse (χ(g ) G /α(g )), we see that the resulting graph also has high chromatic number (for n 0). Then if X is the average number of cycles of length at most l: E[X] = l i=3 n(n 1) (n i + 1) p i i l i=3 n θi i = o(n) so that P[X n/] = o(1). Let x = (3 log n)/p, so that P[α(G) x] ( n )(1 p) (x) [ < ne p(x 1)/] x = o(1) x Then for n 0 some choice of G, we have girth(g ) l, G n/, and α(g ) 3n 1 θ log n. As n grows even larger, we get χ(g ) as large as we like. 8

9 4: The second moment Chebyshev s Inequality: P[ X µ λσ] λ. Proof. σ = E[(X µ) ] λ σ P[ X µ λσ]. If X = X i, then Var[X] = i,j Cov[X i, X j ]. If the X i are indicator random variables, then Var[X i ] E[X i ], so that Var[X] E[X] + i j Cov[X i, X j ]. Let ν(n) be the number of distinct primes dividing n. Then almost all n have ν(n) close to log log n. Thm 4..1: Let ω(n) arbitrarily slowly. Then the number of x in {1,..., n} such that is o(n). ν(x) log log n > ω(n) log log n Proof. Choose such x at random. Let X p be the indicator random variable for p x for p n 1/10 and let X = X p. Now E[X p ] = 1/p + O(1/n) so that E[X] = ( ( )) 1 1 p + O = log log n + O(1). n And similarly we can see that p M Var[X p ] = log log n + O(1) and Cov[X p, X q ] 1 n p M ( 1 p + 1 ). q This shows that p q Cov[X p, X q ] [ o(1), o(1)], so that the covariances are negligible. Now we can apply Chebyshev s inequality: [ P X log log n > ω(n) ] log log n < ω(n) + o(1) 0. Thm 4.. is a refinement of this result. Non-negative integer valued random variables have some handy properties. Let Y and X 1, X,... be non-negative integer valued random variables, whose limit (meaningless) we denote by X. When we make a claim about X, we mean that the claim tends towards being corret as n. P[Y > 0] E[Y ]. If E[X n ] 0 then X = 0 almost always. Thm 4.3.1: P[Y = 0] Var[Y ] E[Y ]. Proof. Chebyshev s inequality gives a stronger result: P[ Y µ µ] σ µ. Cor 4.3. & 4.3.3: If Var[X n ] = o(e[x n ] ) then, almost always: X > 0, and X E[X]. Proof. The first is just a restatement of For the second, note that we could have written, for any ɛ > 0, P[ Y µ ɛµ] σ ɛ µ 0. 9

10 Sums of indicator random variables: Let X = X X m be a sum of indicator random variables for events A i. Write i j iff i j and A i and A j are not independent, in which case, Cov[X i, X j ] P[A i A j ]. Define := i j P[A i A j ], so that Var[X] E[X] +. Cor 4.3.4: If E[X] and = o(e[x] ) then, almost always: X > 0, and X E[X]. We say that the indicators X 1,..., X n are symmetric if for each i j there s a measurepreserving autommorphism of the probability space sending A i to A j. Then = E[X], where := j i P[A j A i ] (independent of i). Cor 4.3.5: If E[X] and = o(e[x]) then, almost always: X > 0, and X E[X]. In particular, this applies when (E[x], and) sufficiently many of the A i and A j are independent (or have low probability), that conditioning on one of the A i doesn t change the expected number of events A j that occur significantly. A function r is a threshold function for a property P of graphs if whenever p(n) r(n) q(n) then G(n, p) almost always fails to have P, and G(n, q) almost always has P. The clique number ω(g) of a graph G is the maximal number of vertices of a clique, where a clique is a subgraph which is a complete graph. Thm 4.4.1: The property ω(g) 4 has threshold function n /3. Proof. For each S a set of 4 vertices, let A S be the event that S is a clique, indicated by X S. Let X = X S, so that E[X] = ( n 4) p 6 n 4 p 6 /4. Suppose first that p(n) n /3. Then E[X] = o(1) so that X = 0 almost surely. Suppose instead that p(n) n /3, in which case E[X]. Then we d like to show that = o(e[x]), so that X E[X] almost surely, completing the proof. But this is easy, as there are very few dependencies, as S S iff S S is or 3, and for fixed S, this happens for less than n and n S respectively. We find [ 1 n p 5 + np 3 = n p ] n 3 p 3 n 4 p 6 = o(n 4 p 6 ) = o(e[x]). The density ρ(h) of a graph H = (V, E) is simply E / V. We say that H is balanced if every subgraph of H has density no greater than that of H. Thm 4.4.: Let H be a balanced graph with v vertices and e edges. Let A(G) be the event that H is a subgraph of G (not necessarily induced). Then p = n v/e is a threshold function for A. Proof. For each S a set of v vertices, let A S be the event that G S contains H as a subgraph, and let X S be the indicator. Let X = X S. To see that this almost never occurs for p n v/e, we note that E[X] = o(1) in this case. 10

11 Suppose instead that p n v/e. Then, like last time, E[X] = Θ(n v p e ), and = v 1 i= T S =i P[A T A S ] Now suppose that T S = i and we know that every edge possible appears in S. Any possible appearance of H with vertices in T (i.e. bijection of the vertices T with those of H, of which there are v! in some sense) has at most ie/v edges wholly contained in S, as H is balanced. Thus there is at most a probability of p e ie/v that this happens. Thus we have P[A T A S ] = O(p e ie/v ). As there are O(n v i ) choices of such T : v 1 v 1 = O(n v i p e ie/v ) = O(n v p e ) 1 i/v = o(n v p e ) i= i= (as n v p e ). This is o(e[x]), so we re done. There s some more stuff here, to which I should come back. Clique number: Fix edge probability 1/ and consider the clique number ω(g). The following is quite subtle. Let f(k) be the expected number of k-cliques, ( n k) ( k ). This drops below one at k log n. Thm 4.5.1: Let k = k(n) satisfy k log n and f(k) [so that the average number of k-cliques is unbounded, but the average number of rk-cliques is less than one for some fixed r]. Then almost always ω(g) k. Proof. For each k-set S, let A S be the event S is a clique, and X S the indicator. Let X = XS. We need to see that almost always X > 0. Well E[X] = f(k), so we consider: = k 1 P[A T A S ] = T S i= ( k i )( n k k i Detailed calculation shows that is o(e[x]), and we re done. ) ((k ) ( i )) Cor 4.5.: There exists k = k(n) so that P[ω(G) = k or k + 1] 1. Proof. Let k(n) be as large as possible so that f(k(n)) n. Observe that for k log n, f(k + 1)/f(k) = n 1+o(1), so that for n 0, the function f is decreasing by ratios of at least, say, n.9. Then f(k(n) + ) < n.4 0. Then, almost surely there is no (k(n) + )-clique, and by there is almost surely a k(n)-clique. Distinct sums: Let f(n) denote the largest k for which there is a set {x 1,..., x k } {1,..., n} with distinct sums. That is, for which the sums i X x i, for S {1,..., k}, are distinct. Thm 4.6.1: f(n) < log n + 1 log log n + O(1). Proof. Fix a set {x 1,..., x k } in {1,..., n} with distict sums, and let X the sum of a random subset. Then: µ := E[X] = x x k and σ : = Var[X] = x x k 4 Noting that f(n) < nf(n), we can show that f(n) < log n + log log n + O(1). n k 4 11

12 Chebyshev s inequality gives the left hand inequality, and as X takes any given value with probability at most k, we have the right hand inequality, in: [ 1 λ P X µ < λn ] ( k/ k λn ) k + 1 ( which implies n k 1 λ ) 1 λ. k This holds for any λ > 0, and choosing any λ > 1, for some fixed γ we have n k k 1/ γ. Now apply log ( ) + 1 log log ( ) to both sides, and note that RHS > k constant. 1