Traveling in randomly embedded random graphs.

Transcription

1 Traveling in ranomly embee ranom graphs. Alan Frieze, Wesley Pegen Department of Mathematical Sciences, Carnegie Mellon University, Pittsburgh PA 15213, U.S.A. February 13, 2018 Abstract We consier the problem of traveling among ranom points in Eucliean space, when only a ranom fraction of the pairs are joine by traversable connections. In particular, we show a threshol for a pair of points to be connecte by a geoesic of length arbitrarily close to their Eucliean istance, an analyze the minimum length Traveling Salesperson Tour, extening the Bearwoo- Halton-Hammersley theorem to this setting. 1 Introuction The classical Bearwoo-Halton-Hammersley theorem [2] see also Steele [21] an Yukich [22]) concerns the minimum cost Traveling Salesperson Tour through n ranom points in Eucliean space. In particular, it guarantees the existence of an absolute though still unknown) constant β such that if X 1, X 2..., is a ranom sequence of points, uniformly istribute in the -imensional cube [0, 1], the length T X n,1 ) of a minimum tour through X 1,..., X n satisfies T X n,1 ) β n 1 a.s. The present paper is concerne still with the problem of traveling among ranom points in Eucliean space. In our case, however, we suppose that only a ranom) subset of the pairs of points are joine by traversable connections, inepenent of the geometry of the point set. In particular, we stuy ranom embeings of the Erős-Rényi-Gilbert ranom graph G n,p into the -imensional cube [0, 1]. We let X n enote a uniformly ranom set of points X 1, X 2,..., X n [0, 1], an we enote by X n,p the ranom graph whose vertex set is X n an whose pairs of vertices are joine by eges each with inepenent probability p. Eges are weighte by the Eucliean istance between their points, an we are intereste in the total ege-weight require to travel about the graph. Research supporte in part by NSF grant DMS Research supporte in part by NSF grant DMS 1

2 Figure 1: Paths in an instance of X n,p for = 2, n = 2 30, an p = 10 n, 25 n, n, an n, respectively. In each case, the path rawn is the shortest route between the vertices x an y which are closest to the SW an NE corners of the square. See Q. 2, Section 5.) This moel has receive much less attention than the stanar moel of a ranom geometric graph, efine as the intersection graph of unit balls with ranom centers X i, i [n], see Penrose [17]. We are only aware of the papers by Mehrabian [14] an Mehrabian an Wormal [15] who stuie the stretch factor of X n,p. In particular, let x y enote the Eucliean istance between vertices x, y, an istx, y) enote their istance in X n,p. They showe consiering the case = 2) that if n1 p), then the stretch factor istx, y) sup x,y X n,p x y tens to with n. As a counterpoint to this, our first result shows a very ifferent phenomenon when we pay attention to aitive rather than multiplicative errors. In particular, for p log n n, the istance between a typical pair of vertices is arbitrarily close to their Eucliean istance, while for p log n, the istance nlog log n) 2 between a typical pair of vertices in X n is arbitrarily large Figure 1). We write log k x for log x) k.) In particular, this means that when log n n p < 1 ε, the supremum in the stretch factor theorem of Mehrabian an Wormal is ue just to pairs of vertices which are very close together. Theorem 1.1. Let ω = ωn). We have for 2: 1 log a) For p n ω log log n) 2 n an u = X 1, v = X 2, we have istu, v) ω 8e, a.a.s.1 b) For p ω log n n, we have a.a.s. that uniformly for all pairs of vertices u, v X n, istu, v) = u v + o1). Theorem 1.1 means that, even for p quite small, it is not that much more expensive to travel from one vertex of X n,p to another than it is to travel irectly between them in the plane. On the other han, there is a ramatic epenence on p if the goal is to travel among all points. Let T X n,p ) enote the length of a minimum length Traveling Salesperson tour in X n,p, i.e. a minimum length Hamilton cycle. 1 A sequence of events E n occurs asymptotically almost surely a.a.s.) if lim n PrE n) = 1. 2

3 Theorem 1.2. There exists a sufficiently large constant K > 0 such that for all p = pn) such that p K log n n, 2, we have that ) 1) T X n,p ) = Θ n 1 p 1/ a.a.s. Recall that fn) = Θgn)) means that fn) is boune between positive constant multiples of gn) log n+log log n+ωn) for sufficiently large n.) As the threshol for G n,p to be Hamiltonian is at p = n see e.g. Bollobás [3]), this theorem covers nearly the entire range of p for which a TSP tour exists a.a.s. Finally, we exten the asymptotically tight BHH theorem [2] to the case of X n,p for any constant p. To formulate an almost surely statement, we let X N,p enote a ranom graph on a ranom embeing of N = {1, 2,..., } into [0, 1], where each pair {i, j} is inepenently present as an ege with probability p, an consier X n,p as the restriction of X N,p to the first n vertices {1,..., n}. Theorem 1.3. If 2 an p > 0 is constant, then there exists β,p > 0 such that T X n,p ) β,p n 1 a.s. Karp s algorithm [13] for a fining an approximate tour through X n extens to the case X n,p, p constant as well: Theorem 1.4. For fixe 2 an p constant, then there is an algorithm that a.s. fins a tour in X n,p of value 1 + o1))β,p n 1)/ in polynomial time, for all n N. 2 Traveling between pairs In this section, we prove Theorem Proof of Theorem 1.1a) Outline of proof This is straightforwar. We show by the first moment metho that any path between u an v with many eges must contain a significant number of long eges an hence must be as long as claime. We then show that a.a.s. there are no paths between u an v without many eges. Proof proper Let ν enote the volume of a -imensional unit ball; recall that ν is boune ν ν 5 < 6 for all ). Let an ege be long if its length is at least l 1 = there exists a path with k eges, k k 0 = log n 2 log log n ωlog log n)2 4e log n. Let ε = 1 log log n an let A be the event that from u to v that uses at most εk long eges. Then 3

4 2) Pr A) ) ) ) n k ωlog log n) k 1)! p k 2 ) 1 ε)k ν k 1 εk 4e log n k k 0 n k 1 p k e ) ) εk ωlog log n) 2 ) 1 ε)k ν ε 4e log n k k 0 ) 1 ν 1 ε log ε n e ) k ε n 4e ) 1 ε) ε k k 0 ) k 1 6e +o1) = o1), n k k 0 after using 2 an log ε n = e. 4e 2 Explanation of 2): Choose the k 1 interior vertices of the possible path an orer them in one of k 1)! n k 1) ways as u1, u 2,..., u k 1 ). Then p k is the probability that the eges exist in G n,p. Now choose the short eges e i = u i 1, u i ), i I in one of k 1 ε)k) = k εk) ways an boun the probability that these eges are short by ball of raius l 1, center u i 1 for i I. ν ωlog log n) 2 4e log n ) ) 1 ε)k viz. the probability that u i is mappe to the Now a.a.s. the shortest path in G n,p from u to v requires at least k 0 eges: Inee the expecte number of paths of length at most k 0 from u to v can be boune by k 0 ) ) n k 1)! p k 1 k 0 k log n k 1 n ω log log n) 2 = o1). So a.a.s. k=1 istu, v) εk 0 l 1 = k=1 ε log n ωlog log n)2 2 log log n 4e log n = ω 8e. 2.2 Proof of Theorem 1.1b) Outline of proof 1 We first consier two points u, v such that u v γ = log log n. We then consier a set of small isjoint balls with centers on the line joining u, v. We argue that a.a.s. i) all of these balls contain relatively) giant components, ii) there is an ege joining any pair of giant components insie each ball, iii) the iameter of each of these giant components is small an iv) there is an ege between u an one of the g giant components X closest to u an an ege between v an one of the g giant components Y closest to v. This gives a path consisting of an ege from u to the giant component X plus a walk insie X plus an ege to the giant component Y plus an ege to v. Because the balls are small the length of this path is close to u v. We reuce the case where u v γ to the first case. Proof proper We begin by consiering the case of vertices u, v at istance u v γ. Letting δ = 1 log n, then, for 4

5 v F u,v F v,u u Figure 2: Fining a short path. sufficiently large n, we can fin a set B of at least 2C δ, C = γ 8, isjoint balls of raius δ centere on the line from u to v, such that C δ of the balls are closer to u than v, an C δ balls are closer to v than u Figure 2). Denote these two families of C δ balls by F u,v an F v,u. The sets B, F u,v an F v,u are fixe for the rest of the argument.) Given a ball B F {u,v} = F u,v F v,u, the inuce subgraph G B on vertices of X lying in B is a copy of G N,p, where N = NB) is the ranom) number of vertices lying in B. Let [ ] N0 S B be the event that NB) 2 +1, 2N 0 where N 0 = ν δ n. Diviing by 2 +1 accounts for points close to the bounary of [0, 1].) Now NB) is istribute as the binomial Binn, q) where q ν δ [2, 1]. bouns will thus be useful: The following Chernoff 3) 4) PrBinM, p) 1 ε)mp) e ε2 Mp/2 for 0 ε 1. PrBinM, p) 1 + ε)mp) e ε2 Mp/3 for 0 ε 1. The bouns 3) an 4) imply that for B F {u,v}, Pr S B ) e Ωnδ) = e n1 o1). This gives us that a.a.s. S B occurs for all pairs u, v X with u v γ. We now argue that for all B B: A) All subgraphs G B for B F {u,v} have a giant component X B, containing at least N 0 /2 +2 vertices. Inee, the expecte average egree in G B is Np = Ωω) an with probability 1 e n1 o1) we have N = n 1 o1) ) an at this value the giant component is almost all of B a.a.s. In particular, since S B occurs, we have that 5) Pr X B N 0 /2 +2 S B ) e ΩN 0) e Ωδ n) = on 3 ). 5

6 See [3] for the first inequality in 5). This can be inflate by n 2 2C log n) to account for pairs u, v an the choice of B F {u,v}. B) There is an ege between X B an X B for all B, B F {u,v}. Inee, the probability that there is no ege between X B, X B, given A), is at most 1 p) N 2 0 /22+2 e Ωδ2 n 2 p) e n1 o1). This can be inflate by n 2 C log n) 2 to account for all pairs u, v an all pairs B, B. C) For each B F {u,v}, the graph iameter iamx B ) the maximum number of eges in any shortest path in X B ) satisfies Pr iamx B ) > 100 log N ) 0 n 3. logn 0 p) This can be inflate by n 2 2C log n) to account for pairs u, v an the choice of B F {u,v}. Fernholz an Ramachanran [5] an Rioran an Wormal [20] gave tight estimates for the iameter of the giant component, but we nee this cruer estimate with a lower probability of being exceee. We prove this later in Lemma 2.1. It will be convenient for the proof of Lemma 2.1 to assume that N 0 p = Olog N 0 ). There is no loss in generality because Theorem 1.1b) hols a fortiori for larger p. This follows from a stanar coupling argument, involving aing ranom eges to increase the ege probability. Part C) implies that with high probability, for any u, v at istance γ an all B F {u,v} an vertices x, y X B, 6) istx, y) 200δ log N 0 logn 0 p) 200 log n log n log log n + log ν log ω + log ν As the giant components X B B F u,v ) contain in total at least C δ probability that u has no neighbor in these giant components is at most N 0 = C p) Cν nδ 1 /2 +2 e Cν npδ 1 /2 +2 = n ωcν /2 +2. = o1) ν nδ 1 vertices, the In particular, the probability is small after multiplication by n 2, an thus a.a.s., for all pairs x, y X n,p, x has a neighbor in X B for some B F u,v an y has a neighbor in X B for some B F v,u. Now by part B) an equation 6), we can fin a path u, w 0, w 1,..., w s, z t, z t 1,..., z 1, z 0, v from u to v where the w i s are all in some X B for B F u,v an the total Eucliean length of the path w 0,..., w s tens to zero with n, an the z i s are all in some X B for some B F v,u, an the total Eucliean length of the path z 0,..., w t tens to zero with n. Meanwhile, the Eucliean segments corresponing to the three eges u, w 0, w s, z t, an z 0, v lie within δ of isjoint segments of the line segment from u to v, an thus have total length u v + 6δ, giving 7) istu, v) u v + 6δ + o1) = u v + o1). We must also hanle vertices u, v X n,p with u v < γ. Given such a pair, we let B u, B v enote any choice of balls of raius γ such istb u, B v ) γ, istb u, u), istb v, v) γ + 2). These bouns 6

7 are chosen to make such a choice trivially possible, even when u, v are close to a corner.) Observe that we have: where C u, C v enote the giant components of B u, B v, an x y means that {x, y} is an ege of X n,p, Pr u, v X n,p, w C u, z C v such that u w, v z) 1 with n since a.a.s we have that B u an B v contain at least ν nγ /2 +2 points for all u, v X n,p an we have that 1 2n 2 1 p) n ν γ / In particular, we can a.a.s for all pairs u, v X n,p fin w u within istance γ + 4) of u, z v within Eucliean istance γ + 4) of v, such that Now, we can use the previous case 7) to see that γ w z 2 + 8)γ. istu, v) 2 + 9)γ + 6δ + o1) = o1). In particular, istu, v) u v = o1). We complete the proof of Theorem 1.1 by proving Lemma 2.1. Suppose that Np = ω, ω = Olog N) an let C 1 enote the unique giant component of size N on) in G N,p, that q.s. 2 exists. Then for L large, Pr iamc 1 ) L log N ) ON L/10 ). log Np Proof. Let Bk) be the event that there exists a set S of k vertices in G N,p that inuces a connecte subgraph in which more than half of the vertices have less than ω/2 neighbors outsie S. Then for k = on) we have ) N 8) PrBk)) p k 1 k k 2 2 k PrBinN k, p) ω/2) k/2 k 9) ek ω k ) k/2 pk 2 2k e N k)p ω/2)2 /2N k)p) ω, from 3) with ε = 1 2N k)p, ek ω k pk 2 2k e.99ω ω/2)2 /2ω ) k/2 p 1 2eωe ω/20 ) k Ne kω/21. Explanation of 8): N k ) bouns the number of choices for S. We then choose a spanning tree T for S in k k 2 ways. We multiply by p k 1, the probability that T exists. We then choose half the vertices X of S in at most 2 k ways an then multiply by the probability that each x X has at most ω/2 neighbors in [N] \ S. If κ = κl) = L log N log Np then 9) implies that PrBκ)) N 1 L. 2 A sequence of events E n occurs quite surely q.s. if Pr E n) = On K ) for all positive constants K. 7

8 Next let Dk) = D N k) be the event that there exists a set S of size k for which the number of eges es) containe in S satisfies es) 2k. Then, PrDk)) N k Since ω = Olog N) we have that q.s. ) k 2) 2k ) p 2k Ne k keω 2N k [κ1), N 3/4 ] such that Dk) occurs. ) ) 2 k ke 3 ω 2 ) k =. 4N Now let Bk 1, k 2 ) = k 2 k=k 1 Bk) an Dk 1, k 2 ) = k 2 k=k 1 Dk), an suppose that Bk 1, k 2 ) Dk 1, k 2 ) oes not occur, where k 1 = κl/4) an k 2 = N 3/4. Fix a pair of vertices v, w an efine sets S 0, S 1, S 2,... where S i is the set of vertices at istance i from v. If there is no i k 1 with w S i then we must have S k1 an S k1 k 1 where S t = t i=0 S i for t 0. This is because v, w C 1 an C 1 is connecte an so S i+1 S i + 1. We also see that k 1 S t N 3/4 implies that S t+1 ω S t /10. Inee, if S t+1 < ω S t /10 then S t+1 has at most ω + 10) S t /10 vertices an more than ω S t /4 eges, contraiction. Thus if L is large, then we fin that there exists t k 1 + κ3/4) N 3/4 such that S t N 3/4 an so also that S t 1 o1))n 3/4. Now apply the same argument from w to create sets T 0, T 1,..., T s, where either we reach v or fin that T s N 3/4 where s k 1 + κ3/4). At this point the eges between S t an T s are unconitione an the probability there is no S t : T s ege is at most 1 p) N 3/2 = Oe ΩN 1/2) ). 3 Traveling among all vertices Our first aim is to prove Theorem 1.3; this will be accomplishe in Section 3.2, below. In fact, we will prove the following general statement, which will also be useful in the proof of Theorem 1.2: Theorem 3.1. Let Y 1 [0, 1] enote a set of points chosen from any fixe istribution, such that the carinality Y = Y 1 satisfies EY ) = µ > 0 an PrY k) Cρk for all k, for some C > 0, ρ < 1. For t > 0 let Y t enote a ranom set of points in [0, t] obtaine from the union of t inepenent copies Y 1 + x x {0,, t 1} ). If p > 0 is constant, 2, an Yt,p enotes the ranom graph on Yt with inepenent ege probabilities p, then β > 0 epening on p an the process generating Y1 ) such that i) T Y t,p) βt a.a.s., an ii) T Y t,p) βt + ot ) q.s. 3 3 In this context On ω1) ) is replace by Ot ω1) ). 8

9 Note that as a probabilistic statement, Part i) above asserts that there exists a choice for o1) a function of t, say, tening to 0) such that 1 o1))βt T Y t,p) 1 + o1))βt hols a.a.s. Similarly for Part ii), the statement asserts the existence of a suitable fixe choice of ot ) a function of t, whose ratio to t tens to 0). The restriction Pr Y 1 k) Cρ k simply ensures that we have exponential tail bouns on the number of points in a large number of inepenent copies of Y 1 : Observation 3.2. For the total number T n of points in n inepenent copies of Y1, we have for some absolute constant A C,ρ > 0, Pr T n µn > δµn) < e A C,ρδ 2 µ 2n. This is straightforwar to prove, but we o not have a reference an so we give a sketch proof in the appenix. Note that the conitions on the istribution of Yt are satisfie for a Poisson clou of intensity 1, an it is via this case that we will erive Theorem 1.3. Other examples for which these conitions hol inclue the case where Yt is simply a suitable gri of points, or is a ranom subset of a suitable gri of points in [0, t], an we will make use of this latter case of Theorem 3.1 in our proof of Theorem 1.2. Outline of proof of Theorem 3.1 Our proof uses subaitivity, but some of the stanar properties of the classical case e.g., monotonicity) fail in our setting, requiring us to use inuction on to achieve the result. For technical reasons see also Question 4 of Section 5) Theorems 3.1 an 1.3 are given just for 2, an before beginning with the inuction, we must carry out a separate argument to boun the length of the tour in 1 imension. When = 1 all we can prove is an On) boun on the length of the minimum tour. We o this by examining a natural greey algorithm for fining a tour. This is the content of Lemma 3.4. After this we prove a sort of Lipschitz conition for the tour length, see Lemma 3.7. This will substitute for monotonicity. After this we can push ahea using subaitivity. 3.1 Bouning the expecte tour length in 1 imension We begin with the following simple lemma. Lemma 3.3. Let σ be a permutation of [n], an let lσ) = n 1 i=1 σ i+1 σ i. Then lσ) < σ n + 4 invσ), where invσ) is the number of inversions in σ i.e. {i, j) : i < j an σ i > σ j } ). Proof. We prove this by inuction on n. It is trivially true for n = 1 since in this case lσ) = 0. Assume now that n > 1, an given a permutation σ of [n], consier permutation σ of [n 1] obtaine by truncation: { σ i σ i if σ i < σ n = σ i 1 if σ i > σ n 9

10 We have by inuction that 10) lσ ) σ n invσ ). Now observe that lσ) = lσ ) + σ n σ n 1 + {i σ i < σ n < σ i+1 OR σ i > σ n > σ i+1 } lσ ) + σ n σ n 1 + 2invσ) invσ )), since {i σ i < σ n < σ i+1 } an {i σ i+1 < σ n < σ i } are each boune by invσ) invσ ). Recalling that invσ) = invσ 1 ), we have invσ) invσ ) = n σ n. Since σ n 1 σ n 1, 10) gives that lσ) σ n invσ ) + σ n σ n 1 + 2invσ) invσ )) = σ n invσ ) + 2invσ) n + σ n ) + σ n σ n 1 + 2invσ) invσ )) = σ n invσ) 2n + 2σ n + σ n σ n 1 = σ n + 4 invσ) 2n σ n σ n 1 σ n σ n 1 ) σ n + 4 invσ). For the 1-imension case of Theorem 1.3, we have, roughly speaking, a 1-imensional string of points joine by some ranom eges. Lemma 3.3 allows us to prove the following lemma, which begins to approximate this situation. Lemma 3.4. Consier the ranom graph G = G n,p on the vertex set [n] with constant p, where each ege {i, j} EG) is given length i j N. Let Z enote the minimum length of a Hamilton cycle in G starting at vertex 1, assuming one exists. If no such cycle exists let Z = n 2. Then there exists a constant A p such that EZ) A p n an Z A p n, q.s. Proof. We first write G = G 1 G 2 G 3 where the G i are inepenent copies of G n,p1, where 1 p = 1 p 1 ) 3. Note that p 1 p/3. We consier G 1 to be the restriction to [n] of a ranom graph G = G N,p1 on the set of all natural numbers, an will begin by constructing a path in G via the following algorithm: We start with v 1 = 1. Then for j 1 we let φj) = min {k N : k / {v 1, v 2,..., v j } an {v j, k} EG 1 )} an let v j+1 = φj) i.e. we move from v j to the lowest inex k that has not been previously visite. This constructs an infinite path in N, an we efine j 0 by j 0 = max {j N : i j = v i n}. In particular, observe that P 1 = v 1, v 2,..., v j0 is a path in G 1 of length Λ 1 = j 0 1 j=1 v j+1 v j. It is convenient to exten the sequence v 1,..., v j0 to a permutation of [n]; to o this, we let σ i = v i for i j 0, an then let σ j0 +1,..., σ n be [n] \ {v 1,..., v j0 } in increasing orer. Applying Lemma 3.3 to σ gives that the length Λ 1 of the initial part corresponing to the path is at most lσ) < n + 4 invσ). So we woul like to boun invσ). 10

11 Observe first that Prj 0 n k) n1 p 1 ) k. This is because at j 0 we fin that v j0 has no neighbors in the set of unvisite vertices an the existence of such eges is unconitione at this point. So, 11) j 0 n log2 n p 1 q.s. Now let α j = {i > j : σ i < σ j } for all 1 j n, so that invσ) = α 1 + α α n. In fact, by our efinition of σ, we have that α j = 0 for all j > j 0, giving invσ) = α 1 + α α j0, since σ contains no inversions of i < j where i > j 0. Next we efine an approximation a j to α j. We let a j = {t < v j : t / V j } for all j 1, where V j = {v 1, v 2,..., v j }. Observe that α j a j for j j 0. Moreover, 12) Pra j = k) = 1 p 1 ) k p 1 for k 0. To see this, observe that the vertex v j was chosen as the leftmost vertex available to the algorithm, an etermining this vertex involves querying eges which have not yet been conitione by the running of the algorithm. Observe that 12) hols even when conitioning on any previous history of the algorithm. So a 1 = 0 an a 2, a 3,... is a sequence of inepenent copies of Geop 1 ) 1 where Geop 1 ) is the geometric ranom variable with probability of success p 1. We thus have: j 0 j 0 n 13) Einvσ)) E α j E a j E a j n 1 p 1. p 1 j=0 j=0 Moreover, stanar concentration arguments give then that 14) invσ) j 0 j=1 α j j 0 j=1 a j j=0 n a j n q.s. p 1 j=1 It follows from Lemma 3.3, 13), an 14) that ) 4 15) EΛ 1 ) n 3 p 1 ) an Λ 1 n 1 + 4p1 q.s. It remains to show that there is a Hamilton cycle of length not much greater then Λ 1. We show that we can cheaply insert all σ j, j > j 0 into P 1 by replacing eges by paths of length two. We let I = { i j 0 : v i v i 1, v i v i+1 n 2/3} an m = I. It follows from 15) that m n 0 On 1/3 ) q.s. For J = {σ j0 +1,..., σ n }, our aim is to use the eges of G 2 to insert J into the path v 1,..., v j0 = σ 1,..., σ j0, thus extening it to all of G 1. Let v j J be a vertex to be inserte. Without loss of generality, assume v j n/2; the reverse case is essentially the same. We procee by examining v j 1, v j 2, v j 3,... until we fin a l satisfying i) l I 11

12 ii) {v l, v j }, {v l 1, v j } EG 2 ). When examining a vertex v l for which l satisfies i), there is a p 2 1 chance that v l satisfies ii). In particular, after examining log 2 n vertices v = v l which satisfy i), we will q.s. fin a v which satisfies both i) an ii). Thus there is q.s. a vertex v l v j satisfying i) an ii) with v j v l log 2 n+on 1/3 ). Thus, we can replace the ege v l 1, v l ) by a path v l 1, v j, v l to q.s. incorporate v j into our path at a cost of at most O log 2 n + n 1/3 + n 2/3). Finally, 11) implies that we can insert all vertices of J in this manner one-by-one even when we avoi re-using a caniate v l, this only increases the patching istance by at most a log2 n p 1 aitive factor). In the en, we get a Hamilton path x 1, x 2,..., x n in G 1 G 2 q.s. with v 1..., v j0 a subsequence of x 1,..., x n ) an the total ae cost over Λ 1 is q.s. On 2/3 log 2 n). There is only an exponentially small probability that we cannot fin G 3 -eges {x 1, x j+1 }, {x j, x n } which now give us a Hamilton cycle; since the maximum value of of Z is just n 2, this gives EZ) A p n, as esire. We have: Corollary 3.5. Suppose that we replace the length of ege i, j) in Lemma 3.4 by ξ i + + ξ j 1 where ξ 1, ξ 2,..., ξ n are ranom variables with mean boune above by µ an exponential tails. If ξ 1,..., ξ n are inepenent of G n,p then EZ) A p µn. Proof. The boun on the expectation follows irectly from Lemma 3.4 an the linearity of expectation. Let us observe now that we get an upper boun ET Yt,p)) 1 A p t on the length of a tour in 1 imension. We have ET Yt,p)) 1 = E T Yt,p) 1 Y 1 t,p = n ) Pr Yt,p 1 = n). n=0 When conitioning on Yt,p 1 = n, we let P 1 < P 2 < < P n [0, t] be the points in Yt,p. 1 We choose k {0, n 1} uniformly ranomly an let ξ i = P k+i+1 P k+i, where the inices of the P j are evaluate moulo n. We now have Eξ i ) 2t n for all i, an Corollary 3.5 gives that an thus E T Y 1 t,p) Y 1 t,p = n ) A p n 2t n, 16) E T Y 1 t,p) ) 2A p t. 3.2 The asymptotic tour length Our proof of Theorem 3.1 uses recursion, by iviing the [t] cube into smaller parts. However, since our ivisions of the cube must not cross bounaries of the elemental regions Y1, we cannot restrict ourselves to subivisions into perfect cubes in general, the integer t may not have the ivisors we like). 12

13 To this en, if L = T 1 T 2 T where each T i is either [0, t] or [0, t 1], we say L is a - imensional near-cube with sielengths in {t 1, t}. For 0, we efine the canonical example L := [0, t] [0, t 1] for notational convenience, an let Φ, p t) = E T Y t,p L ) ). so that Φ pt) := Φ, p t) = Φ,0 p t + 1). In the unlikely event that Yt,p L is not Hamiltonian, we take T Y t,p L ) = t+1, for technical reasons. Our first goal is an asymptotic formula for Φ: Lemma 3.6. There exists β = β p, > 0 such that Φ, p t) βt. The proof of this is eferre until after the proof of Corollary 3.9 below. The proof is by inuction on 2. We prove the base case = 2 along with the general case. We begin with a technical lemma. Lemma 3.7. For every fixe p, 0, there is a constant F p, > 0 such that 17) Φ, p t) Φ, 1 p t) + F p, t 1 for all t sufficiently large. Here we interpret Φ, p t) = Φ,0 p t) for < 0. In particular, this implies that there is a constant A p, > 0 such that 18) Φ pt + h) Φ pt) + A p, ht 1 for sufficiently large t an 1 h t. Proof. We let S enote the subgraph of Y t,p L inuce by the ifference L \ 1 L. By ignoring the th coorinate of S if > 0 an the th coorinate otherwise, we obtain the 1) imensional set πs), for which inuction on or equation 16) if = 2) implies an expecte tour T S) of length Φ 1, 1 p t) β 1 t 1, an so changing notation, we can write p Φ 1, 1 p t) D p, 1 t 1. We have that ET S)) ET πs)) + 1/2 E πs) ) D p, 1 t 1 + 1/2 t 1. The first inequality stems from the fact that the points in L \ 1 L have a coorinate in [t 1, t]. Now if Y t,p L 1 an S are both Hamiltonian, then we have 19) T Yt,p L ) T Y t,p L 1 ) + T S) + O t) 13

14 which gives us the Lemma, by linearity of expectation. We have 19) because we can patch together the minimum cost Hamilton cycle H in Yt,p L 1 an the minimum cost path P in S as follows: Let u 1, v 1 be the enpoints of P. If there is an ege u, v of H such that u 1, u), v 1, v) is an ege in Yt,p then we can create a cycle H 1 through Yt,p L 1 P at an extra cost of at most 2 1/2 t. The probability there is no such ege is at most 1 p 2 ) t/2, which is negligible given the maximum value of T Yt,p L ). On the other han, because p is a constant, the probability that either of Yt,p L 1 or S is not Hamiltonian is exponentially small in t, see for example [9]), which is again negligible given the maximum value of T Yt,p L ). This completes the proof of 17). To obtain 18) we use 17) to write Φ, p t + h) Φ,0 p t + h) + F p, t + h) 1 = Φ pt + h 1) + F p, t + h) 1 Φ pt) + F p, h t + i) 1. i=0 Our argument is an aaptation of that in Bearwoo, Halton an Hammersley [2] or Steele [21], with moifications to aress ifficulties introuce by the ranom set of available eges. First we introuce the concept of a ecomposition into near-cubes. Allowing near-cube ecompositions is necessary for the en of the proof, beginning with Lemma 3.10). Simplifications relying on Bounary Functionals as in Yukich [22] o not appear to be available ue to missing eges. We say that a partition of L into m near-cubes S α with sielengths in {u, u + 1} inexe by α [m] is a ecomposition if for each 1 b, there is an integer M b such that, letting { au if a < M b f b a) =. a M b )u + 1) + M b u if a M b. we have that S α = [f 1 α 1 1), f 1 α 1 )] [f 2 α 2 1), f 2 α 2 )] [f α 1), f α )]. Observe that so long as u t, L always has a ecomposition into near-cubes with sielengths in {u, u + 1}. Inee, if t = ru s for 0 s < u then we can take M b = s for b an M b = s 1 for b >, unless s = 0, in which case M b = u 1. First we note that tours in not-too-small near-cubes of a ecomposition can be paste together into a large tour at a reasonable cost: Lemma 3.8. Fix δ > 0, an suppose t = mu for u = t γ for δ < γ 1 m, u Z), an suppose S α α [m] ) is a ecomposition of L. We let Y,α t,p := Y t,p S α. We have T Y t,p L ) α [m] T Y,α t,p ) + 4m u with probability at least 1 e Ωu p 2). Proof. Let B, C enote the events B = { α : Y,α t,p is not Hamiltonian } 14

15 { } C = α : Y,α t,p u δu, an let E = B C. Now PrB) m e Ωup) an, by Observation 3.2, PrC) m e Ωu) an so PrE) e Ωup). Assume therefore that E occurs. Each subcube S α will contain a minimum length tour H α. We now orer the subcubes {S α } as T 1,..., T m, such that for S α = T i an S α = T i+1, we always have that the Hamming istance between α an α is 1. Our goal is to inuctively assemble a tour through the subcubes T 1, T 2,..., T j from the smaller tours H α with a small number of aitions an eletions of eges. Assume inuctively that for some 1 j < m we have ae an elete eges an foun a single cycle C j through the points in T 1,..., T j in such a way that i) the ae eges have total length at most 4 ju an ii) we elete one ege from τt 1 ), τt j ) an two eges from each τt i ), 2 i j 1. To a the points of T j+1 to create C j+1 we elete one ege u, v) of τt j ) C j an one ege x, y) of τt j+1 ) such that both eges {u, x}, {v, y} are in the ege set of Y t,p. Such a pair of eges will satisfy i) an ii) an the probability we cannot fin such a pair is at most 1 p 2 ) u /2 1)u /2. Thus with probability at least 1 e Ωu p 2) we buil the cycle C m with a total length of ae eges 4 m u. Linearity of expectation an the upper boun t +1 on T Y t,p) when there is no tour) now gives a short-range recursive boun on Φ pt) when t factors reasonably well: Corollary 3.9. For all large u an 1 m u 10 m, u N ), for some constant B. Φ pmu) m Φ pu) + B p, u) Proof of Lemma 3.6. Note that here we are using a ecomposition of [mu] into m subcubes with sielength u; near-cubes are not require. To get an asymptotic expression for Φ pt) we now let Choose u 0 large an such that β = β p, = lim inf t Φ pu 0 ) β + ε u 0 Φ pt) t. an then efine the sequence u k, k 1 by u 1 = u 0 an u k+1 = u 10 k for k 0. Assume inuctively that for some i 0 that for A p, as in Lemma 3.7 an B p, as in Corollary 3.9, ) Φ pu i ) i 2 A p, 20) u β + ε + + B p, i u j uj 1. This is true for i = 0, an then for i 0 an 0 u u i an m [u i 1, u i+1 ] we have j= 1 21) Φ pmu i + u) mu i + u) Φ pmu i ) + A p, umu i ) 1 mu i ), from Lemma 3.7, 15

16 22) m Φ pu i ) + B p, u i ) + A p, umu i ) 1 mu i ), from Corollary 3.9, ) i 2 A p, β + ε + β + ε + j= 1 i 1 j= 1 u j A p, u j + B p, u 1 j + B p, u 1 j ) + B p, ui 1 + A p, m,. by inuction, Putting m = u i+1 /u i an u = 0 into 21) an 22) completes the inuction. We euce from 20), 21) an 22) that for i 0 we have ) Φ pt) A p, t β + ε + + B p, u j u 1 β + 2ε for t J i = [u i 1 u i, u i u i+1 + 1)] j j= 1 Now i=0 J i = [u 2 0, ] an since ε is arbitrary, we euce that 23) β = lim t Φ pt) t, We can conclue that Φ pt) βt, which, together with Lemma 3.7, completes the proof of Lemma 3.6, once we show that β > 0 in 23). To this en, we let ρ enote Pr Y1 1), so that E Y t ) ρt. We say x {0,..., t 1} is occupie if there is a point in the copy Y1 + x. Observing that a unit cube [0, 1] + x x {0,..., t 1} ) is at istance at least 1 from all but 3 1 other cubes [0, 1] + y, we certainly have that the minimum tour O 3 1 length through Yt is at least, where where O is the number of occupie x. Linearity of expectation now gives that β > ρ/3 1), completing the proof of Lemma 3.6. Before continuing, we prove the following much cruer version of Part ii) of Theorem 3.1: Lemma For any fixe ε > 0, T Y t,p) t +ε q.s. Proof. We let m = t 1 ε/2, u = t/m, an let {Yτ,p,α } be a ecomposition of Yt,p into m nearcubes with sielengths in {u, u + 1}. We have that q.s. each Yτ,p,α has i) u points, an ii) a Hamilton cycle H α. We can therefore q.s. boun all T Yτ,p,α ) by u u, an Lemma 3.8 gives that q.s. T Yt,p) 4ut + 4m u. Proof of Theorem 3.1. We consier a ecomposition {S α } α [m] ) of Y t into m near-cubes of sie-lengths in {u, u + 1}, for γ = 1 ε 2, m = tγ, an u = t/m. Lemma 3.6 gives that Let S γ Y t,p) = E T Y,α t,p ) βu βt 1 γ). α [m] min { T Y,α t,p ), 2t1 γ)+ε)}. 16

17 Note that S γ Y t,p) is the sum of t γ ientically istribute boune ranom variables. Now, since q.s. T Y,α t,p ) 2t1 γ)+ε) for all α by Lemma 3.10, we have that q.s. S γ Yt,p) = α T Y,α t,p ). Applying Theorem 1 of Hoeffing [11] for the sum of inepenent boune ranom variables, we see that for any ξ > 0, we have Pr S γ Yt,p) m ET Yu,p)) 2ξ 2 ) ξ) 2 exp 4m 2 t 21 γ)+ε). Putting ξ = t ε for small ε, we see that 24) S γ Y t,p) = βt + ot ) q.s. Note next that Lemma 3.8 implies that 25) T Y t,p) S γ Y t,p) + δ 2 where δ 2 = ot ) q.s. It follows from 24) an 25) an the fact that Pr Yt = t ) = Ωt /2 ) that 26) T Yt,p) βt + ot ) q.s. which proves part ii) of Theorem 3.1. Of course, we have from Lemma 3.6 that 27) ET Yt,p)) = βt + δ 1 where δ 1 = ot ), an we show next that that this together with 25) implies part i) of Theorem 3.1, that: 28) T = T Yt,p) = βt + ot ) a.a.s. We choose 0 δ 3 = ot )) such that 0 δ 2, δ 1 = oδ 3 ). Let I = [βt δ 3, βt + δ 2 ]. Then we have βt + δ 1 = ET Y t,p) T Y t,p) βt + δ 2 ) PrT Y t,p) βt + δ 2 ) + ET Y t,p) T Y t,p) I) PrT Y t,p) I)+ ET Y t,p) T Y,α t,p ) βt δ 3 ) PrT Y t,p) βt δ 3 ). Now ε 1 = ET Y t,p) T Y t,p) βt + δ 2 ) PrT Y t,p) βt + δ 2 ) = Ot ω1) ) since Y t,p 2 1/2 t an PrT Y t,p) βt + δ 2 ) = Ot ω1) ), from 26). So, if λ = PrT Y t,p) I) then we have or βt + δ 1 ε 1 + βt + δ 2 )λ + βt δ 3 )1 λ) λ δ 1 ε 1 + δ 3 δ 2 + δ 3 an this proves 28) completing the proof of Theorem = 1 o1),

18 Proof of Theorem 1.3. We now let W t,p be the graph on the set of points in [0, t] which is the result of a Poisson process of intensity 1. Our first task is to boun the variance Vt) of T W t,p). Here we follow Steele s argument [21] with only small moifications. We approximate T W 2t,p ) as the sum over 2 half-size cubes of T Wt,p) an use this to show that V2 k t) k=1. This eals with n of the form 2 k t for some value 2 k t) 2 of t an we then have to fill in the gaps. Let E t enote the event that 29) T W2t,p) T W,α t,p ) t. α [2] Observe that Lemma 3.8 with m = 2, u = t/2 implies that 30) Pr E t ) e Ωt p). We efine the ranom variable λt) = T W t,p) + 10t, an let λ i enote inepenent copies of λt). Conitioning on E t, we have from 29) that λ2t) 2 i=1 λ i t) 6t2 2 i=1 λ i t). In particular, 30) implies that letting Υt) = Eλt)) = Ωt ) see 27)) an Ψt) = Eλt) 2 ), we have for sufficiently large t that 2 Ψ2t) E t,p ) t + 21t For we have = 2 i=1 α [2] T W,α Eλ i t) 10t ) 2 ) + 2 i j )t Eλ i t) 10t ) Eλ j t) 10t )+ 2 i=1 Eλ i t) 10t ) )t ) 2 = 2 Eλt) 10t ) 2 ) ) Eλt) 10t ) )t Eλt) 10t ) )t ) 2 = 2 Ψt) )Υt) 2 Ωt Eλt)) + Ot 2 )) 2 Ψt) )Υt) 2. Vt) := VarT W t,p)) = Ψt) Υt) 2, V2t) 2t) 2 1 Vt) 2 t 2 Υt)2 t 2 Υ2t)2 2t) 2. Now with t 1 arbitrary, summing over 2 k t for k = 0,..., M 1 gives M k=1 V2 k t) 2 k t) 2 1 M 1 V2 k t) Υt)2 2 2 k t) 2 t 2 Υ2M t) 2 Υt)2 2 M t) 2 t 2 k=0 18

19 an so, solving for the first sum, we fin 31) M k=1 V2 k t) 2 k t) ) 1 ) Vt) 2 t 2 + Υt)2 t 2 <. Still following Steele, we let Nt) be the Poisson counting process on [0, ). We fix a ranom embeing U of N in [0, 1] as u 1, u 2,... an a ranom graph U p where each ege is inclue with inepenent probability p. We let U n,p enote the restriction of this graph to the first n natural numbers. In particular, note that U Nt ),p is equivalent to W t,p, scale from [0, t] to [0, 1]. Thus, applying Chebychev s inequality to 31) gives, in conjunction with Lemma 3.6, that ) t2 k T U Nt2 Pr k ) ),p) t2 k ) β p, > ε < an so for t > 0 that k=0 T U Nt2 32) lim k ) ),p) k t2 k ) 1 = β p, a.s. Now choosing some large integer l, we have that 32) hols simultaneously for all the finitely many) integers t S P = [2 l, 2 l+1 ); an for 2 l r R, we have that 33) r [2 k t, 2 k t + 1)) for t S l an some k. We simply choose k such that 2 l 2 k r < 2 l+1.) Unlike the classical case p = 1, in our setting, we o not have monotonicity of T U n,p ). Nevertheless, we show a kin of continuity of the tour length through T U n,p ): Lemma For all ε > 0, δ > 0 such that for all 0 k < δn, we have 34) T U n+k,p ) < T U n,p ) + εn 1, q.s. Proof. We consier cases accoring to the size of k. Case 1: k n 1 3. Note that we have T U n+1,p ) < T U n,p ) + q.s., since we can q.s. fin an ege in the minimum tour though U n,p whose enpoints are both ajacent to n + 1). n 1 3 applications of this inequality now give 34). Case 2: k > n 1 3. In this case the restriction R of U n+k,p to {n + 1,..., k} is q.s. with respect to n) Hamiltonian [4]. In particular, by Theorem 3.1, we can q.s. fin a tour T though R of length 2βpk 1. Finally, there are q.s., eges {x, y} an {w, z} on the minimum tours through U n,p an R, respectively, such that x w an y z in U n+k,p, giving a tour of length This gives the lemma with δ = ε/3β) / 1). T U n+k,p ) T U n,p ) + 2β p, k

20 Let ε l, l = 1, 2,... be a sequence of positive reals, tening to aero. We apply Lemma 3.11 with ε = ε l an δ = 1+ 1 t ) 1 = O t ), assuming t is large. Then we have 2k t) r 2 k t) 1+δ) = 2 k t+1)) by 33), an using the fact that 1 2δ)Nr ) < N1 δ)r ) < N1 + δ)r ) < 1 + 2δ)Nr ) q.s. with respect to r), gives that for large enough l, we have q.s. T U Nt+1)2 k ) ),p) ε l r 1 < T U Nr ),p) < T U Nt2 k ) ),p) + ε l r 1, an so iviing by r 1 an using 32) an taking limits we fin that a.s. β p, 2ε l lim inf r T U Nr )) r 1 lim sup r T U Nr )) r 1 β p, + 2ε l. Since l may be arbitrarily large, we fin that T U Nr lim )) r r 1 = β p,. Now the elementary renewal theorem guarantees that N 1 n) n, a.s. So we have a.s. T U n,p ) lim = lim r n 1 T U NN 1 n)),p) r N 1 n)) 1 N 1 n)) 1 n 1 = β p, 1 = β p,. 3.3 The case pn) 0 We will in fact show that 1) hols q.s. for np ω log n, for some ω. That we also get the statement of Theorem 1.2 can be seen by following the proof carefully, but this also follows as a consequence irectly from the appenix in Johannsson, Kahn an Vu [12]. We first show that q.s. 35) T X n,p ) = Ωn 1)/ /p 1/ ). Let Y 1 enote the number of vertices whose closest G n,p -neighbor is at istance at least 1 np) 1/. Observe first that if r = 1/np) 1/ then with probability at least 1 ν r p ) n 1 e ν, there are no points within istance 1/np) 1/ of any fixe v X n,p. Thus EY 1 ) ne ν /2 an one can use the Azuma-Hoeffing inequality to show that Y 1 is concentrate aroun its mean. Thus q.s. T X n,p ) n 1)/ e ν /4p 1/, proving 35). We will for convenience prove the following theorem. After which Theorem 1.2 follows in a couple of lines. 20

21 Theorem Let Y 1 [0, t] enote a set of points chosen via a Poisson process of intensity one in [0, t] where t = n 1/. Then there exists a constant γ p such that T Y t,p) γ p t p 1/ q.s. Proof. We let p 0 = p 1 = p/3 an p i = p 1 /2 i 1, i = 2,..., k = log 2 t an efine p k+1 so that 1 p = k+1 j=1 1 p j). We let G i = Yt,p i, i = 0, 1,..., k + 1, where each G i has the same vertex set Y1, but in which the eges are inepenently chosen. Observe that with this choice, we have that Yt,p ecomposes as Yt,p = k+1 i=0 G i. Figure 3: Applying Theorem 3.1 to prove Theorem 3.12.) The shae S α are those which are heavy with not too few vertices) an typical with a large cycle). Theorem 3.1 implies that we can fin the short) cycle of bol eges through the S α these eges inicate the presence of patchable pairs which can be use to construct a large cycle. We then repeat this process on remaining vertices with a new set of ranom eges, with coarser an coarser ivisions of the square, to cover most vertices with large cycles, which must then be merge. Proof Strategy: We partition [0, t] into small subcubes. Most of these will inuce subgraphs of G 1 that contain relatively large cycles. We then use Theorem 3.1 to fin a tour through these subcubes that can be use to patch together the subcube cycles into one large cycle H 1. We then use this iea an coarser an coarser partitions into subcubes to create cycles H 2, H 3,..., H l where we use G j eges to create H j. We then use the extensions an rotations of Pósa [18] an the eges of G 0 to merge the H i an the relatively few vertices not in any of the H i into a hamilton cycle of the require cost. We begin by constructing a large cycle, using only the eges of G 1. We choose ε small an then choose an absolute constant K sufficiently large for subsequent claims. In preparation for an inuctive argument we let t 1 = t, T 1 = t 1, m 1 = T 1 p 1 /K) 1/ an consier the partition 1 = {S α } α [m 1 ] ) of [0, t] into m 1 subcubes of sie length u = t/m 1. Note that t will not change throughout the inuction). 21

22 Now each S α contains K/p 1 vertices, in expectation an so it has at least 1 ε)k/p 1 vertices with probability 1 e ΩK/p1) = 1 o1). Let α be heavy if S α has at least this many vertices, an light otherwise. Let Γ α be the subgraph of G 1 inuce by S α. If α is heavy then for any ε > 0 we can if K is sufficiently large fin with probability at least 1 e ΩK/p1) = 1 o1), a cycle C α in Γ α containing at least 1 ε) 2 K/p 1 vertices. This is because when α is heavy, Γ α has expecte average egree at least 1 ε)k see [8] Chapter 6.3 for an explanation). We say that a heavy α is typical if it Γ α contains a cycle with 1 ε) S α X eges; otherwise it is atypical. We boun the length of aa C α by c α = S α X t 1 1/2 /m 1, where the secon factor is the Eucliean iameter of S α. We now let N enote the set of vertices in C α, where the union is taken over all typical heavy α. Our aim is to use Theorem 3.1ii) to prove that we can q.s. merge the vertices N into a single cycle C 1, without too much extra cost, an using only the eges of G 1. Letting q α = P rs α is typical) 1 ε, we make each typical heavy α available for this roun with inepenent probability 1 ε 1 q α, so that the probability that any given α is available is exactly 1 ε. This is of course rejection sampling.) Now we can let Y = Y1 in Theorem 3.1 be a process which places a single point at the center of [0, 1] with probability 1 ε, or prouces an empty set with probability ε. Let now Y α α m 1 ) be the inepenent copies of Y which give Ym 1. Given two cycles C 1, C 2 in a graph G we say that eges u i = x i, y i ) C i, i = 1, 2 are a patchable pair if f x = x 1, x 2 ) an f y = y 1, y 2 ) are also eges of G. Given x Y α, y Y β, we let x y whenever there exist two isjoint patchable pairs σ α,β between C α, C β. Observe that an ege between two vertices of Y1 is then present with probability q α,β PrBin1 ε) 4 K 2 /4p 2 1, p 2 1) 2) 1 ε. In particular, this graph contains a copy of Ym 1,1 ε), for which Theorem 3.1ii) gives that q.s. we have a tour of length B 1 m 1 for some constant B 1; in particular, there is a path P = α 1, α 2,..., α M ) through the typical heavy α with at most this length. Using P, we now merge its cycles C αi, i = 1, 2,..., M into a single cycle. Suppose now that we have merge C α1, C α2,..., C αj into a single cycle C j an have use one choice from σ αj 1,α j to patch C αj into C j 1. We initially ha two choices for patching C αj+1 into C αj, one may be lost, but one at least will be available. Thus we can q.s. use G 1 to create a cycle H 1 from C α1, C α2, by aing only patchable pairs of eges, giving a total length of at most ) 36) 2T 1 t 1 1/2 + m 1 B 1 + t 1 1/2 LT 1 1/2 m 1 m 1 p 1/, 1 where L = 4K 1/. The first term in 36) is a boun on the total length of the cycles C α where α is available, assuming that α c α Y t,p 2t. The secon smaller term is the q.s. cost of patching these cycles to create H 1. This consists of the length of P plus a term bouning the cost of joining the center of a subcube C α to the ens of an ege of C α. This latter value, bouns the cost of aing C α to H 1. Having constructe H 1, we will consier coarser an coarser subivisions D i of [0, t] into m i subcubes, an argue inuctively that we can q.s. construct, for each 1 i l for suitable l, vertex isjoint cycles H 1, H 2,..., H l satisfying: P1 T i 3εT i 1 for i 2, where T j = t j 1 i=1 H i, 22

23 P2 the set of points in the αth subcube in the ecomposition D i occupie by vertices which fail to participate in H i is given by a process which occurs inepenently in each subcube in D i, an P3 the total length of each H i is at most LT i 1/2. p 1/ i Note that H 1, above, satisfies these conitions for l = 1. Assume inuctively that we have constructe such a sequence H 1, H 2,..., H j 1 j 2). We will now use the G j eges to construct another cycle H j. Suppose now that the set T j of points that are not in j 1 i=1 H i satisfies T j = T j t 1 / log t. We let m j = T j p j /K) 1/ an t j = T 1/ j. The expecte number of points in a subcube will be K/p j but we have not exercise any control over its istribution. For i 2, we let α [m i ] be heavy if S α contains at least εk/p j points. Now we want K to be large enough so that εk is large an that a heavy subcube has a cycle of size 1 ε) T j S α with probability at least 1 ε, in which case, again, it is typical. We efine Γ j as the set of typical heavy pairs {α, β} for which there are at least two isjoint patchable pairs between the corresponing large cycles. Applying the argument above with T j, t j, m j, Γ j replacing T 1, t 1, m 1, Γ 1 note that P2, above, ensures that Theorem 3.1 applies) we can q.s. fin a cycle H j with at least 1 3ε)T j vertices an length at most LT j 1/2, giving inuction hypothesis part P3. Part P1 is satisfie since the light subcubes only p 1/ j contribute ε fraction of points to T j, an we q.s. take a 1 ε) fraction of the heavy subcubes. Finally, Part P2 is satisfie since participation in H j is etermine exclusively by the set of ajacency relations in G j T j, which is inepenent of the positions of the vertices. Thus we are guarantee a sequence H 1, H 2,..., H l as above, such that T l+1 < t 1 / log t. The total length of H 1, H 2,..., H l is at most l LT i 1/2 i=1 p 1/ i L31/ t p 1/ i=1 ) t 3 i 2 i/ ε i 1 = O p 1/. We can now use G 0 to finish the proof. It will be convenient to write G 0 = 2 i=0 A i where A i, i = 1, 2, 3 are inepenent copies of Yt,q where 1 p 0 = 1 q) 3. Also, let R = {x 1, x 2,..., x r } = Yt,p \ l i=1 H i. We first create a Hamilton path containing all vertices, only using the eges of A 1 A 2 an the extensionrotation algorithm introuce by Pósa [18]. We begin by eleting an arbitrary ege from H 1 to create a path P 1. Suppose inuctively that we have foun a path P j through Y j = H 1 H ρj X j, where X j R, at an ae cost of Ojt). We let V j enote the vertices of P j an promise that V l+r = Y t,p. We also note that V j V 1 = Ωt ) for j 1. At each stage of our process to create P j+1 we will construct a collection Q = {Q 1, Q 2,..., Q r } of paths through V j. Let Z Q enote the set of enpoints of the paths in Q. Roun j of the process starts with P j an is finishe when we have constructe P j+1. If at any point in roun j we fin a path Q in Q with an enpoint x that is an A 2 -neighbor of a vertex in y / V j then we will make a simple extension an procee to the next roun. If x H i then we elete one of the eges in H i incient with y to create a path Q an then use the ege x, y) to concatenate Q, Q to make P j+1. If y R then P j+1 = Q + y. If Q = v 1, v 2,..., v s ) Q an v s, v 1 ) A 1 then we can take any y / V j an with probability at least 1 1 q) s = 1 Ot ω1) ) fin an ege y, v i ) A 2. If there is a cycle H i with H i V j = then 23

24 we choose y H i an elete one ege of H i incient with y to create a path Q an then we can take P j+1 = Q, v i, v i 1,..., v i+1 ) an procee to the next roun. Failing this, we choose any y R \ V j an let P j+1 = y, v i, v i 1,..., v i+1 ) an procee to the next roun. Note that this is the first time we will have examine the A 2 eges incient with y. We call this a cycle extension. Suppose now that Q = v 1, v 2,..., v s ) Q an v s, v i ) A 1 where 1 < i < s 1. The path Q = v 1,..., v i, v s, v s 1,..., v i+1 ) is sai to be obtaine by a rotation. v 1 is the fixe enpoint. We partition Q = Q 0 Q 1 Q k0, k 0 = log t where Q 0 = {P j } an Q i is the set of paths that are obtainable from P j by exactly i rotations with fixe enpoint v 1. We let N i enote the set of enpoints of the paths in Q i, other than v 1, an let ν i = N i an let N Q = i N i. We will prove that q.s. 37) ν i 1 100q implies that ν i+1 ν i t q 300. It follows from this that q.s. we either en the roun through a simple or cycle extension or arrive at a point where the paths in Q have Ωt ) istinct enpoints. We can take an arbitrary y / V j an fin an A 2 neighbor of y among N Q. The probability we cannot fin a neighbor is at most 1 q) Ωt) = Ot ω1) ). Once we prove 37) we will have shown that we can create a Hamilton path through Yt,p from H 1, H 2,..., H l, R at an extra cost of O 1/2 t log t l + t 1 / log t) = Ot ). Explanation: Each ege ae because of a rotation or through extening the current path costs at most 1/2 t. The log t factor comes from the fact that each path is obtaine by at most k 0 rotations. We exten into another cycle at most l 1 times an into R at most t 1 / log t times. We will not have use any A 3 eges to o this. Proof of 37): We first prove that in the graph inuce by A 1 we q.s. have 38) S 1 100q implies that N A 1 S) S t q 100. Here N A1 S) is the set of vertices not in S that have at least one A 1 -neighbor in S. Inee, if s 0 = 1 100q = on = t ) then ) Pr S : N A1 S) < S t q 100 s 0 s=1 s 0 s=1 s 0 s=1 t s t ) ) Pr Bint s, 1 1 q) s ) st q ) Pr Bin t s, sq ) ) st q s t e s q) e Ωt = Ot ω1) ). Now 37) hols for i = 0 because q.s. each vertex in Y t,p is incient with at least t q/2 A 1 eges. Given 38) for i = 0, 1,..., i 1 we see that ν ν i 1 = oν i ). In which case 38) implies that ) s 100 ν i+1 N A 1 N i ) ν ν i 1 ) 2 t qν i o1) 24