The cover time of sparse random graphs.

The cover time of sparse random graphs Colin Cooper Alan Frieze May 29, 2005 Abstract We study the cover time of a random walk on graphs G G n,p when p = c n, c > 1 We prove that whp the cover time is asymptotic to c c c 1 1 Introduction Let G = V, E be a connected graph, let V = n, and E = m For v V let C v be the expected time taken for a simple random walk W on G starting at v, to visit every vertex of G The cover time C G of G is defined as C G = max v V C v The cover time of connected graphs has been extensively studied It is a classic result of Aleliunas, Karp, Lipton, Lovász and Rackoff [1] that C G 2mn 1 It is also known see Feige [7], [8], that for any connected graph G 1 o1n C G 1 + o1 4 27 n3 In this paper we study the cover time of the random graph, G G n,p It was shown by Jonasson [11] that whp 1 a C G = 1 + o1n if np b If c > 1 is constant and np = c then C G > 1 + αn for some constant α = αc Thus Jonasson has shown that when the expected average degree n 1p grows faster than, a random graph has the same cover time whp as the complete graph K n, whose cover time is determined by the Coupon Collector problem Whereas, when np = Ω this is not the case In this paper we sharpen Jonasson s results for the case np = c where ω = c 1 This condition on ω ensures that whp G n,p is connected, see Erdős and Rényi [6] Department of Mathematical and Computing Sciences, Goldsmiths College, London SW14 6NW, UK Department of Mathematical Sciences, Carnegie Mellon University, Pittsburgh PA15213 Supported in part by NSF grant CCR-9818411 1 A sequence of events E n is said to occur with high probability whp if lim n PrE n = 1 1

Theorem 1 Suppose that np = c = + ω where ω = c 1 and c=o1 If G G n,p, then whp c C G c c 1 It follows from the results of Broder and Karlin [4] that if c 1 = Ω1 then whp C G = On This is because G n,p is then an expander whp Thus it was known that whp C G = Θn for this value of p Our paper improves on previous results by giving asymptotically exact values for the cover time When c 1 = Ω1 we give the exact constant of multiplication in C G = Θn Also, when c 1 = ω where ω = o we establish the novel result that C G n log The challenge is to find a technique which can make an accurate average case analysis of the random walk In the next section we give some properties that hold whp in G n,p In Section 4 we show that a graph with these properties has a cover time described by Theorem 1 2 Outline of proof In Section 3 we describe the properties of G n,p that will be needed for our proof We call graphs with these properties typical and estimate the cover time of typical graphs For a walk starting at vertex u and a vertex v u and a time s, we define A s v to be the event that W u has not visited v by time s We then argue that C u t + PrA s v 1 v V s>t for any t > 0, see equation 14 We then write PrA s v = s PrA i v A i 1 v 2 and spend much of the time estimating PrA i v A i 1 v for i not too small i=1 Heuristically, a random walk of length s under the conditioning A s v should be much like a random walk on H = G[V \ {v}] Now the probability of A s v when we forbid a visit to v in the first s 1 steps is close to w N G v d H w 2 EH 1 d H w Here, d H w is the degree of v in H and d Hw 2 EH is the steady state of a random walk on H N Gv is the set of neighbours of v in G If H is an expander and s much larger than the mixing time then the sum of these probabilities is the probability that W u is at a neighbour of v at time s 1 1 The factor degree H w is then the probability we move to v at the next step The main part of the analysis in Lemma 3 goes to show that PrA i v A i 1 v 1 dv 2m 2

for large i This is then used in 2 to estimate PrA s v from above and below The upper bound follows from 1 after some calculation The lower bound uses the Chebyshev inequality and requires the estimate PrA s v 1 A s v 2 PrA s v 1 PrA s v 2 for suitably chosen pairs v 1, v 2 3 Properties of G n,p Let dv be the degree of vertex v V and let δ, denote the minimum and maximum degree Let δ v 2 be the minimum degree of a neighbour of v, excluding neighbours of degree one Let distu, v denote the distance between the vertices u, v of the graph G Let c = cn > 1 where c > 1 We say that a graph G is typical if it has the structural properties P0 P7 given below The proof of the following lemma is given in the Appendix Lemma 1 Let p = c n where ω = c 1 and c=o1 Then whp G G n,p is typical P0: G is connected P1: G 0 = c + 10 and δg { 1 c 1 + e 500 α c > 1 + e 500 where α = α /2 and α > e 600 satisfies c 1 = α logce/α P2: Call a vertex small if its degree is at most /20 There are at most n 1/3 small vertices and no two small vertices are within distance of each other log 2 P3: For L V, L 4, let H = G[V \ L] be the subgraph of G induced by V \ L For S V \ L let e H S, S be the number of edges of H with one end in S and the other in S = V \ L S For all H G such that δh 1, and for all S V \ L, S n/2, e H S, S d H S 1 6 P4: Let Dk be the number of vertices of degree k in G Let n 1 Dk = n p k 1 p n 1 k k denote the expected size of Dk in G n,p Define K 0 = {k [1, 0 ] : Dk 2 } K 1 = {1 k 15 : 2 Dk log } K 2 = {k [16, 0 ] : 2 Dk 2 } K 3 = [1, 0 ] \ K 0 K 1 K 2 3

P4a: If k K 3 then 1 2Dk Dk 2Dk, and = 0 k K 0 Dk log 2 k K 1 4 k K 2 P4b: If ω 2/3 then K 1 = and min{k K 2 } 1/2 and K 2 = Olog P5: The number of edges m = mg of G satisfies m 1 2 cn n 1/2 P6: Let k = c 1, V = {v : dv = k } and let B = {v V 10 : distv, w for some w V, w v} Then log 2 Let X = {v : δ v α } Then V 1 2 Dk and B 1 10 Dk V X 1 10 Dk P7 Call a cycle short if its length is at most 10 log The minimum distance between two short cycles is at least and the minimum distance between a small vertex and a short cycle is at least log 10 log P8: G has at least one triangle and at least one 5-cycle 4 The cover time of a typical graph In this section G denotes a fixed graph with vertex set [n] which satisfies P0 P8 and u is some arbitrary vertex from which a walk is started For a subgraph H of G let W u,h denote a random walk on H which starts at vertex u and let W u,h t denote the walk generated by the first t steps Let X u,h t be the vertex reached at step t and let P t u,h v = PrX u,ht = v Let π u,h v be the steady state probability of the random walk W u,h Note that properties P1 and P8 guarantee that the Markov chain associated with the walk is irreducible and aperiodic and therefore it has a steady state For an unbiased random walk on a connected graph H with mh edges, π H v = π u,h v = d Hv 2mH where d Hv denotes degree in H Let H = Hv = G[V \ {v}] if v is not a neighbour of a vertex w of degree 1, and let Hv = G[V \ {v, w}] if v has a neighbour w of degree 1 Note that P2 rules out a vertex having two neighbours of degree 1 For a subgraph H let N H v be the neighbourhood of v in H ie N H v = N G v V H When H = G we drop the H from the above notation and often drop the u as well Lemma 2 Let G be typical, then there exists a sufficiently large constant K > 0 such that if τ 0 = K then for all v V, and for all u, x H = Hv, after t τ 0 steps P t u,h x π u,hx = On 10 3 4

Proof The conductance Φ of the walk W u,h is defined by ΦW u,h = min πs1/2 e H S : S d H S It follows from P3 that the conductance Φ of the walk W u,h satisfies Φ 1 6 Now it follows from Jerrum and Sinclair [10] that P t u,h x π u,hx = O t n 1/2 1 Φ2 4 2 For sufficiently large K, the RHS above will be On 10 at τ 0 We remark that there is a technical point here The result of [10] assumes that the walk is lazy, and only makes a move to a neighbour with probability 1/2 at any step This leaves the steady state distribution unchanged, halves the conductance and 4 remains true For us it is sufficient simply to keep the walk lazy for 2τ 0 steps until it is mixed This is negligible compared to the cover time For v u V, let A t v be the event that W u,g t does not visit v Lemma 3 a If t > 2τ 0 and δ v 2 then PrA t v PrA t v 1 1 δv 1 2 t 2τ0 1 dv O PrA 2τ0 δ v 2m v 2 t 2τ0 δv 1 dv + O PrA 2τ0 δ v 1 2m v b Suppose that v, v V \ X see P6 and that distv, v > 10 Then log 2 1 k PrA 2τ0 v A 2τ0 v t 2τ0 1 1 + O m Proof The main idea of the proof is to show that the variation distance between a random walk of length τ 0 on H and a random walk on G, conditioned not to visit v is sufficiently small a Fix w v and y N H v Let W k y denote the set of walks in Hv which start at w, finish at y, are of length 2τ 0 and which leave a vertex in the neighbourhood N H v exactly k times Note that the walk can leave y N H v without necessarily leaving N H v Let W k = y W ky and let W = w 0, w 1,, w 2τ0 W k y Let ρ W = PrX w,gs = w s, s = 0, 1,, 2τ 0 PrX w,h s = w s, s = 0, 1,, 2τ 0 5 Then δv 1 k 1 ρ W δ v 5

This is because PrX w,h s = w s X w,h s 1 = w s 1 PrX w,g s = w s X w,g s 1 = w s 1 = { 1 ws 1 / N G v d G w s 1 d G w s 1 1 If E = {X w,g τ v, 0 τ 2τ 0 } then PrE = PrW w,g 2τ 0 = W k 0 W W k = ρ W PrW w,h 2τ 0 = W k 0 W W k where p k = We will show in Lemma 4, below, that k 0 p k δv 1 δ v k W W k PrW w,h 2τ 0 = W = PrW w,h 2τ 0 W k w s 1 N G v p 0 + p 1 + p 2 1 O 1 6 which immediately implies that PrE p 0 + p 1 1 1 + p 2 1 1 2 1 1 2 O 1 δ v δ v δ v Now fix y and write PrX w,g 2τ 0 = y E = k 0 = k 0 W W k y W W k y PrW w,g 2τ 0 = W PrE 1 ρ W PrW w,h 2τ 0 = W PrE 1 Now if p k,y = PrW w,h W k y PrX w,h 2τ 0 = y then = PrW w,h 2τ 0 leaves a vertex of N H v k times X w,h 2τ 0 = y k 0 We will show in Lemma 4, below, that δv 1 k p k,y PrX w,g2τ 0 = y E PrE 1 δ v PrX w,h 2τ 0 = y p 0,y + p 1,y + p 2,y 1 O 1 7 and so δv 1 δ v 2 1 O PrX w,g 2τ 0 = y E PrX w,h 2τ 0 = y δv δ v 1 2 1 + O 6

Taking w as X u,g t 2τ 0 1, and conditioning on A t 2τ0 1v, we deduce that δv 1 δ v 2 1 O PrX u,g t 1 = y A t 1 v PrX w,h 2τ 0 = y δv δ v 1 2 1 + O Therefore 2 δv 1 PrA t v A t 1 v 1 + O P 2τ 0 δ v 1 w,hv y 1 dy y N H v 2 δv 1 dy 1 1 1 = 1 + O δ v 1 2m 2dv + O n 10 dy y N H v 2 δv 1 1 1 + O dv 1 δ v 1 2m 2dv dy y N H v 2 δv 1 dv = 1 + O δ v 1 2m Here we use P2 to see that y N H v Similarly, 1 dy 40dv PrA t v A t 1 v 1 δv 1 δ v 2 1 dv O 2m and the lemma follows immediately Lemma 4 Equations 6,7 are valid Proof Clearly, we only need to prove 7 and so fix y N H v The main idea is to show that a random walk of length 2τ 0 from w to y is close in distribution to a random walk of length τ 0 from w to x followed by the reversal of a random walk W 3 of length τ 0 from y to x, where x is chosen from the the steady state of a walk Each of W 1, W 3 is basically a random walk of length τ 0 and it easy to estimate the number of returns to N H v for such walks Let Wa, b, t denote the set of walks in H from a to b of length t and for W Wa, b, t let PrW = PrW a,h t = W Then for x V H we have PrX w,h τ 0 = x X w,h 2τ 0 = y = and with W 3 equal to the reversal of W 2, W 1 Ww,x,τ 0 W 2 Wx,y,τ 0 = π 1 x,h W 1 Ww,x,τ 0 W 2 Wx,y,τ 0 PrW 1 PrW 2 PrWw, y, 2τ 0 PrW 1 π x,h PrW 2 PrWw, y, 2τ 0 7

= π 1 x,h π y,h = = W 1 Ww,x,τ 0 W 3 Wy,x,τ 0 PrW 1 PrW 3 PrWw, y, 2τ 0 π 1 x,h π y,h PrWw, y, 2τ 0 PrWw, x, τ 0PrWy, x, τ 0 π 1 x,h π y,h PrWw, y, 2τ 0 π x,h On 10 2 = π x,h On 9 It follows that the variation distance between X w,h τ 0 and a vertex chosen from the steady state distribution π H is On 8 Now given x = X w,h τ 0, W w,h τ 0 is a random walk of length τ 0 from w to x and W 2 = x = X w,h τ 0, X w,h τ 0 + 1,, y = X w,h 2τ 0 is a random walk of length τ 0 from x to y For W Wy, x, τ 0 let QW be the probability that y, X w,h 2τ 0 1,, X w,h τ 0 = W Then we have Thus if W = w 1, w 2,, w τ0 = x then QW = 1 + On 8 π x,hprw reverse PrWx, y, τ 0 = 1 + On 8 π y,h PrW PrWx, y, τ 0 = 1 + On 8 π y,h π x,h PrW π x,h PrWx, y, τ 0 = 1 + On 8 π y,h π x,h PrW π y,h PrWy, x, τ 0 QW X w,h τ 0 = x = 1 + On 8 PrW PrWy, x, τ 0 and so the distribution of W reverse 2 is within variation distance On 8 of that of a random walk of length τ 0 from y to a vertex x chosen with distribution π H Thus the variation distance between the distribution of a random walk of length 2τ 0 from w to y and that of W 1, W3 reversed is On 8 where W 1, W 3 are obtained by i choosing x from the steady state distribution and then ii choosing a random walk W 1 from w to x and a random walk W 3 from y to x Furthermore, the variation distance between the distribution of W 1 and a random walk of length τ 0 from w is On 9 Similarly, the variation distance between distribution of W 3 and a random walk of length τ 0 from y is On 9 Now consider W 1 and let Z t be the distance of X w,h t from v We observe from P2 and P7 that except for at most one value ā J = [1, ] we have 2log 2 and this will enable us to prove PrZ t+1 = a + 1 Z t = a 1 20, a I \ ā PrW 1 or W 3 make a return to N H v = O1/ 8 and this implies 7 Note that a move from N H v to N H v has to be counted as a return here 8

To prove 8, let t 0 be the first time that W 1 visits N H v We have to estimate the probability that W 1 returns to N H v later on and so we can assume wlog that w N H v ie Z 0 = 1 It follows from P2 and P7 that PrZ i = i + 1, i = 1,, 6 Z 0 = 1 1 40 6 9 To check this consider two possiblilites: Let N 7 v denote the set of vertices within distance 7 or less of v in G a N 7 v does not contain a small vertex Since there is at most one edge joining two vertices in N 7 v, we see that PrZ i+1 > Z i = 1 40 for i = 1,, 6 and 9 follows b On the other hand, if there is a small vertex x in N 7 v then with probability 1 20 the first move from w takes us further away from x and 9 follows as before If Z 3 = 4 and there is a return to N H v then there exists τ τ 0 such that Z τ = 4, Z τ+1 = 3 and Z τ+2 3 If there is no small vertex within distance 4 of v then P2 and P7 imply τ0 Pr τ τ 0 : Z τ = 4, Z τ+1 = 3, Z τ+2 3 = O 2 10 If there is a unique small vertex within distance 4 of v and Z 6 = 7 and there is a return to N H v then there exists τ τ 0 such that Z τ = 7, Z τ+1 = 6 and Z τ+2 = 5 no short cycles close to v now We can then argue as in 10 that the probablity of this O τ0 2 This completes the proof of part a of the lemma b We simply run through the proof as in a, replacing v by v, v : H = Hv, v = G {v, v }, N H v, v = N G v N G v The proof of 7 remains valid because v, v are far apart 41 The upper bound on cover time From here on, A 1, A 2, are a sequence of unspecified positive constants Let t 0 = 2m log c c 1 We now prove for typical graphs, that for any vertex u V C u t 0 + om 11 Let T G u be the time taken to visit every vertex of G by the random walk W u Let U t be the number of vertices of G which have not been visited by W u at step t We note the following: PrT G u > t = PrU t > 0 min{1, E U t }, 12 C u = E T G u = t>0 PrT G u > t 13 It follows from 12,13 that for all t C u t + E U s = t + PrA s v 14 s>t v V s>t 9

Now, by Lemma 3, for s > 2τ 0, PrA s v where α is as in P1 Then from P4, δv 1 2 1 A 1 δ v exp sdv 1 A 2 2m s 2τ0 dv PrA 2τ0 2m v, if δ v α E U s 3 i=1 k K i dv=k v / X PrA s v + v X PrA s v 3 Dk exp kdv 1 A 2 + PrA s v 2m i=1 k K i v X T 3 s + T 1 s + T 2 s + T X s 15 where and n 1 T 3 s = 2 n k=1 T i s = n 1 k k K i Dke p k 1 p n 1 k e sk 2m sk 1 A 2 2m 1 A 2,, i = 1, 2 T X s = δv 1 2 1 dv 1 O δ v 2m v X 2 { δv } 1 2 exp A 3 sdv δ v 2m v X s 2τ0 Now for γ > 0, s=t 0 +1 e γs = 1 e γ 1 e γt 0 γ 1 e γt 0 16 10

Let λ = t 0 2m 1 A 2 Applying 16 we get s=t 0 +1 We have used the estimation, n 1 n T 3 s 3m k k=1 n 1 k p k 1 p n k 1 e kλ 6 m n 1 n p eλ p k+1 1 p n k 1 e k+1λ k + 1 k=1 < 7 m c 1 p + pe λ n p c 1 mn 7 c 1 e np+npe λ me 2A 2 8 c 1 = om 17 c 1 npe λ c c 1 + 1 c 1 1 + On 1 c 1 A2 / Note that we have used c 1 to get the second line Continuing we get s=t 0 +1 T 1 s A 4 m log 2 k k K 1 2A 2 1 + c 1 e kλ = om 18 since either i ω 2/3 and K 1 = or ii ω < 2/3 and e λ 1 o1 1/3 s=t 0 +1 T 2 s A 5 m 4 k k K 2 e kλ = om 19 since either i ω 2/3 and min{k K 2 } 1/2 and K 2 = Olog or ii ω < 2/3 and e λ 1 o1 1/3 11

Note now that δ v 2 and if v X see P6 then from P2 dv /20 Thus s=t 0 +1 T X s s=t 0 +1 v X v X v X v X 10m dv exp 200m exp 200m { exp sdv 10m { t 0dv 10m c 1 } } { t 0 200m c } /201 by P2 = om 20 since either i c 1+e 500 and X = or ii c < 1+e 500, in which case we use c 1/c e 500 As C G = max u V C u, the upper bound on C G now follows from 11, 15, 17, 18, 19, 20 and 14 with t = t 0 42 The lower bound on cover time For any vertex u, we can find a set of vertices S such that at time t 1 = t 0 1 ɛ, ɛ 0, the probability the set S is covered by the walk W u tends to zero Hence T G u > t 1 whp which implies that C G 1 o1t 0 We construct S as follows Let k, V, B be as defined in Property P6 Let S = V \ B X and let ɛ = 10 log c 1 log c/c 1 3 = o1 and δ = S Note that Dk = Ω n c 1 lnc/c 1 = Ω a 21 c 1 for any constant a > 0 Then P6 implies that S = Ω a for any constant a > 0 Now for v, w u let A t v, w be the event that W has not visited v or w by step t Let Q S be given by Q = {v S : PrA 2τ0 v < 1 δ, or PrA 2τ0 v, w < 1 δ 2, for some w S } Now in time 2τ 0, W can visit at most 2τ 0 + 1 vertices and so PrA 2τ0 v 2τ 0 + 1 and 2τ0 + 1 PrA 2τ0 v, w 2 v V v,w V Thus Q 2τ 0 + 1 δ + 2τ 02τ 0 + 1 21 1 δ 2 = o S 12

Therefore, if S = S \ Q, S Dk 3 Let St denote the subset of S which has not been visited by W by time t Now E St v S 1 1 + A 6 k t 2τ0 PrA 2τ0 2m v Setting t = t 1 we have n c 1 log c/c 1 E St 1 = Ω exp k c 1 2m t 1 nɛc 1 log c/c 1 = Ω c 1 = Ω 9 22 Let Y v,t be the indicator for the event that W u t has not visited vertex v at time t As v, w S are not adjacent, and have no common neighbours, when we delete v, w the total degree of Hv, w is 2m 2dv 2dw, and dv = dw = k It follows from Lemma 3b that for v, w S It follows therefore that from 22 and 23 E Y v,t1 Y w,t1 PrSt 1 0 E St 1 2 E St 1 2 = 1 k t1 2τ 0 1 1 + O m 1 + o1e Y v,t1 E Y w,t1 23 E S t1 S t1 1 E St 1 2 1 = 1 o1 + E S t1 1 Acknowledgement We thank Johan Jonasson for pointing out a significant error in earlier draft References [1] R Aleliunas, RM Karp, RJ Lipton, L Lovász and C Rackoff, Random Walks, Universal Traversal Sequences, and the Complexity of Maze Problems Proceedings of the 20th Annual IEEE Symposium on Foundations of Computer Science 1979 218-223 [2] BBollobás, Random graphs Second edition, Cambridge University Press 2001 [3] BBollobás, TFenner and AMFrieze, An algorithm for finding Hamilton paths and cycles in random graphs, Combinatorica 7 1987 327-341 [4] AZ Broder and AKarlin, Bounds on the cover time, Journal of Theoretical Probability 2 1989 101-120 [5] C Cooper and AM Frieze, Crawling on web graphs, to appear in Proceedings of STOC 2002 13

[6] P Erdős and A Rényi, On random graphs I, Publ Math Debrecen 6 1959 290-297 [7] U Feige, A tight upper bound for the cover time of random walks on graphs, Random Structures and Algorithms 6 1995 51-54 [8] U Feige, A tight lower bound for the cover time of random walks on graphs, Random Structures and Algorithms 6 1995 433-438 [9] SJanson, T Luczak and ARuciński, Random graphs, Wiley 2000 [10] M Jerrum and A Sinclair The Markov chain Monte Carlo method: an approach to approximate counting and integration In Approximation Algorithms for NP-hard Problems D Hochbaum ed PWS 1996 482-520 [11] J Jonasson, On the cover time of random walks on random graphs, Combinatorics, Probability and Computing, 7 1998, 265-279 5 Appendix: Typical graph properties A proof of P0,P1 can be found in Bollobás [2] or Janson, Luczak and Ruciński [9] A proof of P2 can be found in [3] P3: Case of 1 s = S n/c We first prove that whp e G S, S s log Now Pr S : e G S, S s log n s 2 ne s p s log spe s s log s 2 log 2n log exp s log log e cse s = on 2 s log By property P0 and the definition of H, both G and H contain no isolated vertices and hence ds > 0 We write es, S/dS = 1 2eS, S/dS Partition S into sets S 1 and S 2, where S 1 are the vertices of S of degree at most /10 Let T 1 be the neighbour set of S 1 in S 2 and let T 2 be the neighbour set of S 1 in S By property P1 the set S 1 induces no edges, and the neighbours of vertices of S 1 are distinct Thus 2e H S, S d H S 2 T 1 + S 2 log 2 T 1 + T 2 + S 2 /10 L 2 + log /10 = o1 Now use d H S = e H S, S + 2e H S, S Case of n/c s n/2 14

The expected value of e H S, S is at least µ = sn s 4p Thus from Chernoff bounds, for fixed s, n Pr S : e H S, S µ/2 e c sn s 4 9 n s exp s = on 2 We note that E d H S = 2 s 2 p + sn s L p Thus Pr S : d H S 3 n 2 E d HS s exp c 18 e c 20 s s = on 2 c 20 ne log s ne log s Thus e H S, S d H S 1 2 3 2 2 s 2 sn s 4p 1 p + sn sp 6 P4a: First observe that Pr k K 0 : Dk > 0 Dk = K 0 1 2 = O k K 0 Then Similarly, Pr k K 1 : Dk > log 2 k K 1 Pr k K 2 : Dk > 4 k K 2 Dk log 2 = O 1 log Dk 1 4 = O A simple calculation gives that for our range of values of p E DkDk 1 = Dk 1 2 + O n Thus VarDk = Dk 1 + O Applying the Chebychef inequality we see that Dk n PrDk 1 4VarDk Dk or Dk 2Dk = 4 2 Dk 2 Dk 1 + O So, as K 3 = O, Pr k K 3 : Dk 1 1 2 Dk or Dk 2Dk = O 15 n

P4b: The sequence Dk, k 0 is unimodal and Dk + 1 Dk c k + 1 when k = O 24 Moreover, for k 0 there is a positive constant A = Ak such that ce k Dk An 1 c 1 25 k k1/2 Suppose first that there exists k K 1 K 2 such that k < 1/2 It follows from 25 that c 1 < 1/3, for if c 1 1/3 and k < 1/2 then Dk = o 2 Now suppose that k K 2 implies k 1/2 Observe from 25 that both D c c 1/3 and D c + c 1/3 are much greater than 2 Thus either k c c 1/3 or k c + c 1/3 In either case, we see from iterating 24 that K 2 = Olog P5: This follows immediately from Chernoff bounds P6: From 21 we see that c 1 K 3 for c constant That V 1 2 Dk now follows from P3 Now B {v, w V 2 10 : distv, w d = } Therefore log 2 E B Dk 2 d n k p k+1 = odk, k=1 and the second part of P6 follows from the Markov inequality The third part is a similar first moment calculation P7: A proof of similar results can be found in [3] P8: The expected number of triangles tends to infinity and we can use the Chebychef inequality to show that one exists whp The same argument will work for 5-cycles 16