Percolation of averages in the stochastic mean field model: the near-supercritical regime

For a complete graph of size $n$, assign each edge an i.i.d.\ exponential variable with mean $n$. For $\lambda>0$, consider the length of the longest path whose average weight is at most $\lambda$. It was shown by Aldous (1998) that the length is of order $\log n$ for $\lambda<1/\mathrm{e}$ and of order $n$ for $\lambda>1/\mathrm{e}$. In this paper, we study the near-supercritical regime where $\lambda = \mathrm{e}^{-1} +\eta$ with $\eta>0$ a small fixed number. We show that there exist two absolute constants $c^*, C^*>0$ such that with high probability the length is in between $n \mathrm{e}^{-C^*/\sqrt{\eta}}$ and $n \mathrm{e}^{-c^*/\sqrt{\eta}}$. Our result corrects a non-rigorous prediction of Aldous (2005).


Introduction
Let W n be a complete undirected graph of n vertices where each edge is assigned an independent exponential weight with mean n; this is referred to as the stochastic mean-field model.For a (self-avoiding) path path π = (v 0 , v 1 , . . ., v m ), define its length len(π) and average weight A(π) by len(π) = m , and where W (u,v) is the weight of the edge (u, v).For λ > 0, let L(n, λ) be the length of the longest path with average weight below λ, i.e., L(n, λ) = max{len(π) : A(π) ≤ λ, π is a path in SMF n model}.
Theorem 1.1.Let λ = 1/e + η where η > 0. Then there exist absolute constants c * , C * , η * > 0 such that for all η ≤ η * , lim n→∞ P ne −c * / √ η ≤ L(n, λ) ≤ ne −C * / √ η = 1 . (1.1) The study of the object L(n, λ) was initiated by Aldous [1] where a phase transition was discovered at the threshold e −1 .It was shown that with high probability L(n, λ) is of order log n for λ < e −1 and L(n, λ) is of order n when λ > e −1 .The critical behavior was established in [4], where it was proved that with high probability L(n, λ) is of order (log n) 3 when λ is around e −1 within a window of order (log n) −2 .Our Theorem 1.1 describes the behavior in the near-supercritical regime, and in particular states that L(n, λ)/n is a stretched exponential in η with η = λ − e −1 ↓ 0.
A highly related question is the length for the cycle of minimal mean weight, which was studied by Mathieu and Wilson [9].An interesting phase transition was found in [9] with critical threshold e −1 on the mean weight.Further results on this problem will be proved in a forthcoming paper [5].
Another related question is the classical traveling salesman problem (TSP), where one minimizes the weight of the path subject to passing through every single vertex in the graph.For the TSP in the mean-field set up, Wästlund [12] established the sharp asymptotics for more general distributions on the edge weight, confirming the Krauth-Mézard-Parisi conjecture [10,11,8].Indeed, it is an interesting challenge to give a sharp estimate on L(n, λ) for e −1 < λ < λ * (here λ * is the asymptotic value for TSP), interpolating the critical behavior and the extremal case of TSP.
Another question of similar flavor is on the tree with average weight below a certain threshold, where a phase transition is proved in [1].The extremal case of the question on the tree with minimal average weight is the well-known minimal spanning tree problem, where a ζ(3) limit is established by Frieze [7].Main ideas of our proofs.A straightforward first moment computation as done in [1] implies that the critical threshold for the average weight is 1/e, and that lim n→∞ P L(n, λ) = O(log n) = 1 when λ < 1/e.For λ > 1/e, a sprinkling method was employed in [1] to show that with high probability L(n, λ) = Θ(n), where the author first proved with high probability there exist a large number of paths with average weight slightly above 1/e and then used a certain greedy algorithm to connect these paths into a single long path with average weight slightly above 1/e.However, the method in [1] was not able to describe the behavior at criticality.In [4] (see also [9] for the cycle with minimal average weight), a second moment computation was carried out restricted to paths of average weight below 1/e and with the maximal deviation (see (3.2) below for definition of the maximal deviation) at most O(log n), thereby yielding that with high probability L(n, 1/e) = Θ((log n) 3 ).A crucial fact responsible for the success of the second moment computation is that the length of the target path is Θ((log n) 3 ) √ n.As such, a straightforward adaption of this method would not be able to succeed in the regime considered by this paper.For TSP, where one studies paths (cycles) that visit every single vertex, is in a sense analogous to the question of finding the minimal value λ for which L(n, λ) = n with high probability.Wästlund [12] showed that the minimum average cost of TSP converges in probability to a positive constant by relaxing it to a certain linear optimization problem.But it seems difficult to extend his method to incomplete TSP i.e. when the target object is the minimum cost cycle having at least pn many edges for some p ∈ (0, 1).Since our problem is in a sense dual to incomplete TSP in the regime we are interested in, the method of [12] does not seem to be suitable for our purpose either.In the current work, our method is inspired by the (first and) second moment method from [4,9] as well as the sprinkling method employed in [1].
In order to prove the upper bound, our main intuition is that we would have a larger number of short and light paths (a light path refers to a path with small average weight -slightly above 1/e) than we would typically expect if L(λ, n) were greater than e −C * / √ η n.Formally, let = c 1 η where c 1 is a small positive constant, and consider the number of paths (denoted by N ) with length and total weight no more than λ − c 2 √ for a positive constant c 2 .We call such a path a downcrossing.A straightforward computation should give EN = O(1)n e −c 3 / √ η for a positive constant c 3 depending on c 1 and c 2 .Now we consider the number of paths (denoted by N δ ) of length δ(λ)n and average weight at most λ.Such paths have two possibilities: (1) The path contains substantially more than EN many downcrossings, which is unlikely by Markov's inequality.(2) The path does not have substantially more than EN many down crossings.This is also unlikely for the following reasons: (a) A straightforward first moment computation gives that EN δ = O(n)e c 4 δnη for a constant c 4 > 0; (b) The number of down crossings in a typical path of this kind should dominate a Binomial random variable Bin(δn/ , c 5 ) where c 5 > 0 is an absolute constant (since in the random walk bridge, every sub-path of size has a positive chance to have such a downcrossing).If we choose δ suitably large as in Theorem 1.1, we are suffering a probability cost for the constraint on the number of downcrossings (probability for a binomial much smaller than its mean) and this probability cost is of magnitude e −c 6 δn/ for a constant c 6 > 0 depending in c 1 .If we choose c 1 small enough this probability cost kills the growth of e c 4 δnη in EN δ .Therefore, paths of this kind do not exist either.The detailed are carried out in Section 2.
For the lower bound, our proof consists of two steps.In light of the preceding discussion, we cannot hope to directly apply a second moment method from [4,9] to show the existence of a light path that is of length linear in n.As such, in the first step of our proof we prove that with high probability there exists a linear (in n) number of disjoint paths, each of which has weight slightly below λ and is of length e c 7 / √ η for an absolute constant c 7 > 0. This is achieved by two second moment computations, which are expected to succeed as the length of the path under consideration is √ n (indeed remains bounded as n → ∞).In the second step, we propose an algorithm which, with probability going to 1, strings together a suitable collection of these short light paths to form a light path of length e −c 8 / √ η n for an absolute constant c 8 > 0. Our algorithm is similar to the greedy algorithm (or in a different name exploration process) employed in [1].But in order to ensure that the additional weight introduced by these connecting bridges only increases the average weight of the final path by an amount of O(η), we have to use a more delicate algorithm.The details are carried out in Section 3. Notation convention.For a graph G, denote by V (G) and E(G) the set of vertices and edges of G, respectively.For a path π, we use the notation π to denote the sequence of vertices in the path as well as the underlying graph (which is a path).This should be clear from the context.The weight of an edge e in W n is W e and we denote the total weight of a path π by W (π). Π is the collection of all paths π of length ∈ [n].We let λ = 1/e + η where η is a positive number.A path is called λ-light if its average weight is at most λ, and a path is called (λ, C)-light if its total weight is at most λ − C √ where is length of the path.For any nonnegative integer or real valued variable v, the phrases "for large v" or "when v is large" would mean that there exists an absolute positive constant v 0 such that the corresponding statement holds whenever v ≥ v 0 .In cases where some other variable v (or a list of several variables) is involved and the statement holds for any fixed value of v and v ≥ v 0 where v 0 is a function of v, we will add "(given v )" after the above phrases.In a similar vein we use "for small v" or "when v is small" if the corresponding statement holds for 0 < v ≤ v 0 .Throughout this paper the order notations O(.), Θ(.) and o(.) are assumed to be with respect to n → ∞ while keeping all the other involved parameters (such as , η etc.) fixed.We will use C 1 , C 2 , . . . to denote for constants, and each C i will denote the same number throughout of the rest of the paper.Acknowledgements.We thank David Aldous for very useful discussions.

Proof of the upper bound
Let η be a constant multiple of η whose precise value is to be selected, and let C be a constant to be selected later.Set = 1/η and let N η ,C be the number of "(λ, C)-light" paths of length .As outlined in the introduction, we shall first control N η ,C .
It is clear that the distribution of the total weight of a path of length k follows a Gamma distribution Γ(k, 1/n), where the density f θ,k (z) of Gamma(k, θ) is given by f θ,k (z) = θ k z k−1 e −θz /(k − 1)! for all z ≥ 0, θ > 0 and k ∈ N.
(2.1) By (2.1) and the Stirling's formula, we carry out a straightforward computation and get that where C 0 (η) → 1 as η → 0, and α C is a positive constant depending on C. Furthermore both of 1 + o(1), C 0 (η) are less than or equal to 1.
We also need a bound on the second moment of N η ,C to control its concentration around E(N η ,C ).For γ ∈ Π , define F γ to be the event that γ is (λ, C)-light.Then clearly we have N η ,C = γ∈Π 1 Fγ .In order to compute EN 2 η ,C , we need to estimate P(F γ ∩ F γ ) for γ, γ ∈ Π .In the case E(γ) ∩ E(γ ) = ∅, we have F γ and F γ independent of each other and thus P(F γ |F γ ) = P(F γ ).In the case |E(γ) ∩ E(γ )| = j > 0, we have (2.3) Altogether, we obtain that (2.4) Combined with (2.2), it yields that As a consequence of Markov's inequality (applied to |N η ,C − EN η ,C | 2 ), we get that Next, we show that any long λ-light path should have a large number of sub-paths which are (λ, C)-light.Let π be a path of length δn for some δ > 0. Denote its successive edge weights by X 1 , X 2 , . . .X δn and let S k = k i=1 X i for 1 ≤ k ≤ δn.Throughout this section probabilities of events involving edge weights of π will be assumed to be conditioned on "{A(π) ≤ λ}" although we will omit that phrase except in formal expressions and lemmas.Now divide π into edge disjoint sub-paths of length (with the last subpath of length possibility less than in the case does not divide δn) and call each such sub-path a C-downcrossing if it is (λ, C)-light.The following well-known result about exponential random variables (see, e.g., [3,Theorem 6.6]) will be very useful.
Lemma 2.1.Let W 1 , W 2 , . . ., W N be i.i.d.exponential random variables with mean 1/θ, and let follows Gamma(N ; θ) distribution, and they are independent of each other.Here 1 N is the N -dimensional vector whose all entries are 1.
We will also require the following simple lemma which we prove for completeness.Lemma 2.2.Let Z 1 , Z 2 , . . ., Z N be i.i.d.exponential random variables with mean 1 and let S N = N i=1 S N .Then , for all α > 0 . (2.8) Proof.By Markov's inequality, we get that for any α > 0 and 0 < θ < 1,  1)) respectively for all µ > 0. So it is plausible to expect that the probability of the event )} to be bounded away from 0 uniformly for small η and large n (given η) regardless of the weights of first (k − 1) edges of π.The formal statement is in the next lemma.
Proof.Notice that it suffices to prove that there exists an absolute constant c > 0 such that uniformly for µ > 0 and large L (given ) To this end, we see that for L > where the last equality follows from Lemma 2.1.Since distribution of S S L does not depend on the mean of the underlying X j 's, we can in fact assume that X j 's are i.i.d.exponential variables with mean 1 for purpose of computing (2.9).By (2.8), we have So for − 6 √ > 0, we get (2.10) By central limit theorem there exist absolute numbers 0 , c > 0 such that P(S ≤ − 6.5 √ ) ≥ c for ≥ 0 .Hence from (2.10) it follows that for any ≥ 0 there exists L 0 = L 0 ( ) such that the right hand side of (2.9) is at least c for L ≥ L 0 .Lemma 2.3 motivates us to use a binomial approximation for the number of downcrossings that occur, say, up to first half of the path.One scenario that might undermine our estimate is when some Λ k is significantly above λ.But that would imply a substantial drop in S k for some 1 ≤ k ≤ δn/2, which occurs with small probability.Lemma 2.4.Denote by E n,η the event that Λ k is more than λ + √ η for some 1 ≤ k ≤ δn 2 .Then for any 0 < η < 1/4 and 0 < δ 0 < 1 there exists a positive integer n 0 = n 0 (δ 0 , η ) such that, P(E n,η |A(π) ≤ λ) ≤ 2ne −δnη /16 for all δ ≥ δ 0 and n ≥ n 0 . (2.11) where the last inequality holds since we are conditioning on S δn ≤ λδn and λ ≤ 1 when η < 1/4.Therefore, we get Now we evaluate the right hand side of (2.12).Analogous to (2.9) in the proof of Lemma 2.3, we can assume without loss of generality that X 1 , X 2 , . . .X L are i.i.d.exponential variables with mean 1.It is routine to check that Thus, for all 1 ≤ k ≤ δn/2 we get ≤ e −δnη /16 + e −δnη /16 , where the second inequality follows from (2.8) and (2.7) respectively.Combined with (2.12), it gives that An application of a union bound over k completes the proof of the lemma.
Proof of Theorem 1.1: upper bound.Let Ñ π η ,δ,n denote the number of 1-downcrossings in the path π.Let η 0 be given in the statement of Lemma 2.3 and let n 0 be the maximum value of the n 0 's as stated in Lemmas 2.3 and 2.4 for some fixed δ 0 in (0, 1).Assume that η < 1/4 ∧ η 0 .Roughly, the argument goes as follows: By Lemma 2.4 we see that with large probability Λ k ≤ λ + √ η for all k between 1 and δn/2 .On E c n,k,η , it holds that the event √ for small η .Thus, we can use Lemma 2.3 to obtain a binomial approximation for Ñ π η ,δ,n on E c n,k,η .Formally, where α 1 is an absolute constant (see (2.2)) and the inequality follows from Lemma 2.4.Therefore, by Lemma 2.3, we get that Next let us define a new event Ξ η,δ 0 ,n by Then, Combined with (2.14), this completes the proof of the upper bound.
3 Proof of the lower bound

Existence of a large number of vertex-disjoint light paths
For η, ζ 1 ∈ (0, 1), let W * n ⊆ W n be a complete graph of (1−ζ 1 η)n vertices.This subsection is devoted to prove that there exists a large number of vertex-disjoint λ-light paths with high probability.For ∈ N, denote by Π * the set of all paths of length in W * n .For π ∈ Π * and some ζ 2 > 0, define where M (π) is the maximum deviation of π away from the linear interpolation between the starting and ending points, formally given by A similar class of events were considered in [4,9] in order for second moment computation.As the authors mentioned in these papers, the factor W (π)/λ provides some technical ease in view of the following property which is a consequence of Lemma 2.1: We will call a path π ∈ Π * good if G π occurs.Denote by N * the total number of good paths, i.e., N * = π∈Π * 1 Gπ .In order to carry out second moment analysis we need to control the correlation between 1 Gπ and 1 G π where π, π are from Π * .It is plausible that such correlation depends on the number of common edges for π and π .But the number of common edges on its own is not sufficient to characterize the correlation and we need another measurement to supplement it.This is discussed in detail in [4,9] and some of their results will be used.For a subset S of E(π), call the connected components of the underlying graph G(S) of S as S-components.For two paths π and π , define a functional θ(π, π ) to be the number of S-components where S = E(π) ∩ E(π ).This new quantity, together with |E(π) ∩ E(π )|, turns out to be sufficient for bounding the correlation between 1 Gπ and 1 G π from above.The following simple but useful result ([4, Lemma 2.9]) follows from the fact that G(S) is acyclic with θ(π, π ) many components: Since θ(π, π ), |E(π) ∩ E(π )| controls the correlation between 1 Gπ and 1 G π , it makes sense to partition Π * based on the value of this pair.More formally for integers i ≤ j, define the set A i,j as We need a number of lemmas from [4].
Then there exist absolute constants c * , C * > 0 such that for all r ≥ 1 and n ≥ r 2 , and M n be defined as in the previous lemma.Then for all z j such that where C * is the constant from Lemma 3.2 and C 3 > 0 is an absolute constant.
Remark 3.4.(1) Notice that the bounds in Lemma 3.2 and 3.3 do not depend on the particular mean of Z i 's due to Lemma 2.1.(2) Lemma 3.3 is same as Lemma 3.2. in [4] except that in the latter q is restricted to be at most n−10r.But we can easily extend this to all q ≤ n−1.To see this assume n−1 ≥ q ≥ n−10r.Then the right hand side in (3.6) becomes at least C 3 p n e C * n/r 2 e −10C * /r .Now from Lemma 3.2 we get p n e C * n/r 2 ≥ 1.So the right hand side in (3.6) is bigger than C 3 e −10C * whenever n − 1 ≥ q ≥ n − 10r.Increasing C 3 if necessary we can make this number bigger than 1 and thus Lemma 3.3 follows.
By second moment computation, we can hope to show that N ∼ EN with high probability.Then the main challenge is to prove that a large fraction of the good paths are mutually vertexdisjoint with high probability.To this end, we consider a graph G n where each vertex corresponds to a good path and an edge is present when the corresponding good paths intersect at one vertex at least.In this context, it is equivalent to the existence of a large independent subset (i.e., a subset that has no edge among them) in the graph.The following simple lemma is sometimes referred to as Turán's theorem, and can be proved simply by employing a greedy algorithm (see, e.g., [6]).Lemma 3.5.Let G = (V, E) be a finite, simple graph with V = ∅.Then G contains an independent subset of size at least |V | 2 /(2|E| + |V |).Notice 2|E| is the total degree of vertices in G.
In light of the preceding lemma, we wish to show that with high probability the total degree of vertices in G n is not big.For this purpose, it is desirable to show that the typical number of good paths that intersects with a fixed good path π ∈ Π * is not big.Thus, we need to estimate ,π is the collection of all paths π sharing at least one vertex with π.In light of discussions preceding to (3.4), we will first estimate P(G π |G π ) for a specific value of the pair (θ(π, π ), |E(π) ∩ E(π )|).Lemma 3.6.Let π ∈ Π * and π ∈ A i,j with 1 ≤ i ≤ j ≤ .Then there exist absolute constants η 1 , C 4 > 0 such that for 0 < η < η 1 , ζ 2 > 1 ∨ 2C * /e and ≥ ζ 2 2 /η we have Proof.Denote by S and S the sets E(π) ∩ E(π ) and E(π ) \ E(π) respectively.By standard calculus, there exists 0 < η 1 ≤ 1 such that 1 + eη ≥ e (1+e/2)η for all 0 < η < η 1 .Note that where Since the maximum deviation of a good path from its linear interpolation between starting and ending edges is at most Consequently when j ≤ − 1, where C 4 > 0 is an absolute constant and the last inequality used (2.1).For the second term in the right hand side of (3.8), we can apply (3.3) and Lemma 3.3 to obtain where C 4 > 0 is an absolute constant the last inequality follows from (2.1).Plugging the preceding inequality into (3.10) and using the fact !≤ e √ ( /e) (Stirling's approximation) Combined with (3.9), it yields that Since ζ 2 ≥ 2C * /e and η < η 1 we have The case j = can also be easily accommodated.To this end let us first compute P(G π ).It follows from (2.1) and Lemma 3.2 that Applying Stirling's formula again, we get that for ζ 2 ≥ 2C * /e and η < η 1 , Proof.By Lemmas 3.6 and 3.1, we get that for 1 ≤ i ≤ j ≤ , where ξ(η, , i, j) is a number depending only on (η, , i, j) (so in particular, ξ(η, , i, j) does not depend on n).It is also clear that Combined with (3.13), it yields (3.12).It remains to prove (3.11).To this end, we note that the major contribution to the term π ∈Π * ,π P(G π |G π ) comes from those paths π with θ(π, π ) = 1 or |V (π ) ∩ V (π)| = 1 .Thus, we revisit (3.13) for the case of i = 1.By Lemmas 3.6 and 3.1 again, we get that 1≤j≤ where the last two inequalities follow from the facts that ζ 1 < 1/4 and e −η/2 ≤ 1 − η/4 whenever 0 < η < 1.We still need to consider paths that share vertices with π but no edges.For 1 ≤ i ≤ , define B i to be the collection of paths which shares i vertices with π but no edges, i.e., We need an upper bound on the size of B i .To this end notice that there are +1 i many choices for V (π ) ∩ V (π) as cardinality of the latter is i and these vertices can be placed along π in at most +1 i i! many different ways.Also the number of ways we can choose the remaining + 1 − i vertices is at most N +1−i .Multiplying these numbers we get Since the edge sets are disjoint, P(G π |G π ) = P(G π ) for all π ∈ B i and 1 ≤ i ≤ .So we have Combined with (3.14), it completes the proof of (3.11).
We will now proceed with our plan of finding a large independent subset of G n .For any two paths π and π in Π * , define an event As an immediate consequence of Lemma 3.7, we can compute an upper bound of E(N ) as follows: In view of Lemma 3.5, we now wish to control the deviation of N and N .For this purpose, it suffices to show that E(N ).The latter has already been addressed by (3.12).For the former we need to estimate contributions from terms like P(H π 1 ,π 2 ∩ H π 3 ,π 4 ) in the second moment calculation for N .Our next lemma will be useful for this purpose.
Proof.We orient each path from left to right in the order of appearance of vertices in its defining sequence.Denote by H the graph (π 1 ∪ π 2 ) ∩ (π 3 ∪ π 4 ) and by S the graph π 3 ∩ π 4 .Suppose S has exactly k + 1 components and let v 1 , v 2 , . . ., v k+1 be their right endpoints along π 4 .Also let v k+1 be the rightmost vertex of π 4 that belongs to S. It follows from these definitions that for all 1 ≤ i ≤ k (when k > 0) the edge e i having v i as its left endpoint along π 4 lies in E(π 4 ) \ E(π 3 ).We also have |V (S)| = j + k + 1 since E(S) = j and S is acyclic with k + 1 components.Now define a new graph H with V ( H) = V (H) and E( H) = E(H) \ 1≤i≤k {e i }.Due to acyclicty of π 3 and π 4 , any cycle C in H must contain edges from both E(π 4 )/E(π 3 ) and E(π 3 )/E(π 4 ).Thus C must also contain an e i for some 1 ≤ i ≤ k.Consequently H is acyclic with at least j − k many edges.Thus |V (H)| ≥ j − k + 1. Adding this to |V (S)| = j + k + 1, we get (3.17).
We are now well-equipped to prove the main lemma of this subsection.For convenience of notation, write f ( , η) = e −1000ζ Then there exists an absolute constant C 6 > 0 such that, Proof.Let h( , η) = C 5 e 1000ζ 2 / √ η 7 /η 3 .By Lemma 3.9 and (3.16), we assume without loss that where g ,η (n) is defined as in Lemma 3.9.
, by Lemma 3.5 we get that the graph G n has an independent subset of size at least Therefore, with high probability |S η, | ≥ n/2h( , η) which leads to (3.19) for C 6 = 1/2C 5 .

Connecting short light paths into a long one
By Lemma 3.19, throughout this subsection we assume without loss that there exist C 6 f ( , η)n many disjoint good paths in Π * .The remaining part of our scheme is to connect a fraction of these disjoint good paths in a suitable way to form a light and long path γ.In order to describe our algorithm for the construction of γ, we need a few more notations.Denote a collection of disjoint good paths from Π * with maximum cardinality by Π * G, , and denote the vertex sets V (W * n ) and . . in some arbitrary way.Our aim is to build up the path γ in step-by-step fashion starting from π 1 .In each step we will connect γ to π j by a path of length 2 whose middle vertex is in V 2 .These paths will be referred to as bridges.To leverage additional flexibility we also demarcate two segments of length /4 one on each end of the paths π j 's which we call end segments.These end segments will allow us to "choose" endpoints of π j 's while connecting them (as such, it is possible that we only keep half of the vertices of π j in γ).A vertex v will be said to be adjacent to a path or an edge if it is an endpoint of that path or edge.If an edge e has exactly one endpoint in S, we denote by v e,S that endpoint.The following algorithm, referred to as BRIDGE(U, , δ), will construct a long path γ.
Intialization. γ = π 1 , P 1 is the set of all vertices which are in end segments of π j 's for j ≥ 2, P 2 = V 2 , P 3 = ∅ and designate an end segment of γ as the open end γ O and let v be the endpoint of γ not in γ O .Vertices in P 2 will be the middle vertices of bridges.Now repeat the following sequence of steps δn/ − 1 times: Step 1. Find the lightest edge e between γ O and P 2 , remove v e,P 2 from P 2 and include it in P 3 .Repeat this step U times.These edges will be called predecessor edges (so at the end of this step, |P 3 | = U ).
Step 2. Find the lightest edge between P 3 and P 1 .Call it e .Then v e ,P 1 comes from an end segment of some path in Π * G, say π.Step 3. The edge e and the unique pedecessor edge adjacent to v e ,P 3 defines a path b of length 2 (so b connects a vertex in γ O to a vertex in π).Let w be the endpoint of π not in the end segment that v e ,P 1 came from.Then there is a unique path γ in the tree γ ∪ b ∪ π between v and w.Set γ = γ and γ O = the end segment of π containing w.
Step 4. Remove the vertices on the end segments of π from P 1 and reset P 3 at ∅. Proof.A crucial observation is that, at the beginning of each iteration the edges between P 2 and γ O are still unexplored.The same is true for the edges between P 3 and P 1 at the end of Step 1. Consequently their weights are i.i.d.Exp(1/n) regardless of the outcomes from the previous iterations.Therefore, all the bridge weights are independent of each other.Now suppose the mean and variance of each bridge weight is bounded above by 2 η and σ 2 respectively and we emphasize that the latter does not depend on n.By Markov's inequality it then follows that with probability tending to 1 as n → ∞ the total contribution to the weight of γ by the bridges does not exceed 3 η × δn/ .We are now ready to bound the average weight A(γ) of γ.Let i be the length of the segment selected by the algorithm in the i-th iteration.We see that its weight can be no more than λ i + 2ζ 2 / √ η, since the segment is chosen from a good path of average weight at most .We can assume this restriction on since Lemma 3.10 remains valid as long as ≥ ζ 2 2 /η.Indeed, later we will specify the value of and it will satisfy the condition ≥ ζ 2 /η 3/2 .It remains to bound the mean and variance of each bridge weight by 2 η and a certain σ 2 = σ 2 (η, , U ) > 0 (which is independent of n), respectively.Let us consider the bridge obtained from the m-th iteration where 1 ≤ m ≤ δn/ − 1.Let e be the lightest edge between P 3 , P 1 and e be the predecessor edge adjacent to e (for this iteration).By discussions on independence at the beginning of the proof, it follows that W e and W e are independent of each other and also of the weights of bridges already chosen.Since these weights are minima of some collections of i.i.d.exponentials, they will be of small magnitude provided that we are minimizing over a large collection of exponentials, i.
where the last inequality holds for ≥ 20 and large n (given η, U ).By the same line of arguments, we get that the its variance is bounded by a number that depends only on η, and U (so in particular independent of n).By straightforward calculation, we see that the right hand side of (3.

Figure 1 :
Figure 1: Illustrating an iteration of BRIDGE for U = 2 and = 4
2 / √ η η 3 / 5 and U δn/ ≤ ζ 1 ηn 2 .(3.21) Assume for now that (3.21) holds.Since W e is minimum of U × |P 1 | many independent Exp(1/n) random variables, it is distributed as Exp(U |P 1 |/n).As for W e , it is bounded by the maximum weight of the U predecessor edges.From properties of exponential distributions and description of the algorithm it is not difficult to see that this maximum weight is distributed as E 1 + E 2 + . . .E U , where E i+1 is exponential with rate (|P 2 | − i) × 1/n × /4 .By (3.20), we can then bound the expected weight of the bridge from above by