Successive shortest paths in complete graphs with random edge weights

Consider a complete graph $K_n$ with edge weights drawn independently from a uniform distribution $U(0,1)$. The weight of the shortest (minimum-weight) path $P_1$ between two given vertices is known to be $\ln n / n$, asymptotically. We define a second-shortest path $P_2$ to be the shortest path edge-disjoint from $P_1$, and consider more generally the shortest path $P_k$ edge-disjoint from all earlier paths. We show that the cost $X_k$ of $P_k$ is asymptotically $(2k+\ln n) / n$. Specifically, $X_k / (2k/n+\ln n/n) \stackrel{\mathrm{p}}{\to} 1$ uniformly for all $k \leq n-1$. We also show an analogous result when the edge weights are drawn from an exponential distribution.


Introduction
It is a standard problem to find the shortest s-t path in a graph, i.e., the cheapest path P 1 between specified vertices s and t, and its cost X 1 , where the cost of a path is the sum of the costs of its edges. We will use the terms "cost" and "weight" interchangeably, and reserve "length" for the number of edges in a path.
Consider the complete graph G = K n with each edge {u, v} having weight w(u, v), where the w(u, v) are i.i.d. random variables with exponential distribution Exp(1) or uniform distribution U (0, 1) (we consider both versions). In this random setting, a well-known result of Janson [Jan] is that as n → ∞, X 1 ln n/n p −→ 1. (1.1) We define the second cheapest path P 2 , with cost X 2 , to be the cheapest s-t path edge-disjoint from P 1 , and in general define P k , with cost X k , to be the cheapest s-t path edge-disjoint from P 1 ∪ · · · ∪ P k−1 , provided such a path exists. We also think of this as finding path P k after the preceding paths' edges have been destroyed. Our question is how the costs X k behave in the limit as n → ∞ (this limit implicit throughout). Our main result is the following.
Theorem 1.1. In the complete graph K n with i.i.d. uniform U (0, 1) edge weights, with X k the cost of the kth cheapest path, X k 2k/n + ln n/n p −→ 1 (1.2) uniformly for all k ≤ n − 1. That is, for any ε > 0, asymptotically almost surely, for every k = 1, . . . , n − 1, 1 − ε ≤ X k 2k/n + ln n/n ≤ 1 + ε. (1.3) As discussed shortly, in contrast to many cases, the result for the uniform distribution does not extend immediately to all distributions with positive density at 0. However, we have a corresponding result for exponentially distributed edge weights. Let W (k) be the distribution of weight of the kth cheapest edge out of a vertex.
Theorem 1.2. In the complete graph K n with i.i.d. exponential edge weights with mean 1, X k 2 E W (k) + ln n/n Note that E W (k) = k i=1 1 n−i in the exponential case (see e.g. Lemma 4.2). In the uniform case, E W (k) = k/n, so (1.2) in Theorem 1.1 can also be written as (1.4).
1.1. Intuition. The intuitive picture is that path P k should use the kth cheapest edges out of s and t, whose costs are denoted W s (k) and W t (k) respectively. Then, if we ignore previous paths' use of other edges in G \ {s, t}, by (1.1), the opposite endpoints of these two edges should be connected by a path of cost about ln n/n. This suggests that X k ≤ W s (k) + W t (k) + ln n/n, and this is our guiding intuition. Obviously, the path P k does not have to use the kth cheapest edge, its middle section may cost more or less than ln n/n, and as earlier paths use up edges, the costs of these middle sections may rise. It is true, though, that (summing only to k − 1 on the right-hand side to avoid doubly counting edge {s, t}), and we use this in proving the lower bounds on X k (in Section 7 for uniform and Section 8.7 for exponential) and, more surprisingly, in proving the upper bounds on X k for large k (in Section 5 generically, the details treated in Sections 6 and 8).
Our upper bounds are obtained by reasoning as follows. Janson [Jan] analyses the shortest s-t path, and shortest-path tree (SP tree or SPT) on s, in the randomly edge-weighted graph G = K n , showing that the cost of P 1 is asymptotically almost surely, almost exactly ln n/n. When the path P 1 is deleted, this prunes away a root-level branch of the SP tree. The SP tree is a uniform random tree, and using known properties of such trees (see for example [SFH06]) it is not hard to show that what remains of the SP tree is likely to be large; capitalising on this we can find an almost equally cheap path P 2 . This line of argument also shows that there remains a cheap path after deleting P 2 , but we need to know what happens when we delete the true second-shortest path P 2 , and at this point the argument fails because it gives no characterisation of P 2 , only of P 2 . We do know, however, that P 2 is cheap (no more expensive than P 2 ), and of course uses just one edge incident to each of s and t, and we will show that deleting any edge set with these properties (including P 2 as a possibility) must still leave a cheap path P 3 , and so forth. This "adversarial" deletion argument is developed in Section 3.2 to prove Theorem 1.1.

1.2.
Context. The question fits with a broad research theme on optimisation (and satisfiability) problems on random structures. The novel element here is the "robustness" aspect of finding cheap structures even after the cheapest has been removed, and in this we were motivated by a recent study by Janson and Sorkin [JS19] of the same question for successive minimum spanning trees (MSTs), again for K n with uniform or exponential random edge weights. The results for shortest paths and MSTs are dramatically different. For MSTs, it is a celebrated result of Frieze [Fri85] that as n → ∞ the cost of the MST T 1 satisfies w(T 1 ) p −→ ζ(3) def = ∞ k=1 1/k 3 , and [JS19] shows that each subsequent tree's cost has w(T 1 ) p −→ γ k with the γ k strictly increasing (and 2k − 2 √ k < γ k < 2k + 2 √ k). That is very different from the case here, for paths, where for k = o(ln n) we have X k asymptotically equal to X 1 .
1.3. Edge weight distributions. As remarked earlier, in many contexts (including for the length X 1 of a shortest path) the result for any distribution with positive density at 0 follows immediately from that for the uniform distribution U (0, 1), but that is not the case for the successive paths considered here. Remark 1.3. Janson proves the X 1 case in the exponential model but provides standard "blackbox" reasoning that it holds also for the uniform distribution, for any distribution with density 1 at 0 (i.e., with cumulative distribution function (CDF) P(X ≤ x) = x + o(x) for x 0), and, after simple rescaling, for any distribution with positive density at 0. Simply, if there is a path of cost o(1) in some such model, each edge w must also cost o(1), and, coupling with the uniform distribution by replacing w with w = F (w), with F the CDF, w ≤ (1 + o(1))w, and thus the same path is similarly cheap in the uniform model. By the same token, if a path is cheap in any model, the same path has asymptotically the same cost in any other model, and thus the cheapest paths have asymptotically the same cost.
Remark 1.4. In our setting this argument does not apply: to find path P k we must know the nature of the k − 1 previous paths; their costs are not enough. For k = o(n), however, the standard argument applies within our proofs, since the proofs rely only on edges of cost o(1). However, for larger k there are genuine difficulties. Our argument for the exponential case, in Section 8, largely parallels that for uniform but requires new calculations for the upper bound, and one new idea for the lower bound (in Section 8.7). It is not clear for what other edge-weight distributions (even those with positive density near 0) (1.4) will hold.
2. Open problems 2.1. Expectations and the Poisson multigraph model. We have not studied the expectation E X k , but we observe that it is undefined for k large.
Remark 2.1. With n even, E X k is defined for all k ≤ n/2 and undefined for all k > n/2.
Proof. Consider the n − 2 length-2 s-t paths. Any path P k can destroy (share an edge with) at most two such paths (since P k uses just one edge incident to each of s and t). Also, the single-edge path {s, t} is destroyed only by the path P k consisting of just this edge. So, for P 1 , . . . , P k to destroy all length-1 and length-2 paths requires k ≥ (n − 2)/2 + 1 = n/2, so for k ≤ n/2, certainly path P k exists.
Conversely, a construction described in 1892 by Lucas [Luc92,, which he attributes to Walecki, shows that a complete graph K 2r can be decomposed into r edge-disjoint Hamilton paths whose 2r terminals are all distinct. Decompose G = K n \ {s, t} in this way, then link s to one "start" terminal of each such path and t to the other "end" terminal, giving (n − 2)/2 edge-disjoint s-t paths. The edge {s, t} gives another path, for n/2 paths in all. The only edges not used by these paths are a star from s to the Hamilton paths' ends, and another star from t to their starts, and as there are no other unused edges to connect these two stars, there is no further s-t path. With nonzero probability, the edge weights are such that P 1 , . . . , P n/2 are these n/2 paths, so that P n/2+1 does not exist.
The issue of possible non-existence of paths P k for k > n/2 is obviated if, as in [JS19], we work in a Poisson multigraph model. Here, each pair of vertices {u, v} of K n is joined by infinitely many edges, whose weights are drawn from a Poisson process of rate 1 (so that the cheapest {u, v} edge has exponentially distributed cost of mean 1). By construction, in this model every s-t path is always available (just at higher cost).
Conjecture 2.2. In the Poisson multigraph model, X k 2k/n+ln n/n p −→ 1 uniformly for all k ≤ n−1, and E X k 2k/n+ln n/n → 1 for all k ≤ n − 1. Actually, in this model there is no need to stop at k = n − 1, but it is not clear how far out we can go (especially preserving uniform convergence).
2.2. Related questions. Some related questions may also be interesting. One is to characterise successive shortest paths that are vertex-disjoint rather than edge-disjoint. Another is, rather than finding the k successive shortest paths, to find the k edge-disjoint (or vertex-disjoint) paths that cumulatively have minimum cost; the analogous problem for minimum spanning trees is considered in [FJ18].

Upper bound for small k
In this section we prove the upper bound of Theorem 1.1 for all k = o( √ n); larger values are treated in the next section.
As discussed in the introduction, we can characterise the cheapest path P 1 and subsequent paths that are cheap but not necessarily cheapest, putting us at a loss to characterise what remains on deletion of a subsequent cheapest path. We address this in this section. We show a construction of a subgraph R of G designed to give s-t paths, show that its s-t paths are all cheap, and show that no deletion of edges from R subject to certain constraints can destroy all these paths. We show that the union of the k shortest paths satisfies these constraints, and use this to prove Theorem 1.1. t s Figure 1. Cartoon of a robust subgraph R of G, showing the vertices s and t, their respective structures R s and R t including shortest-path trees represented by triangles (some "failed" and thus not shown), and the cheap edges connecting R s and R t .
Specifically, we will define a structure R, sketched in Figure 1, that has many cheap and spread-out paths between s and t, within which we will always find a cheap path. A crucial point is that each step of the construction occurs in a complete induced subgraph of G of size n − o(n) with all edges unconditioned. e We will show that, inductively assuming that for all i ≤ k, the same holds for i = k + 1. We will do so by showing that after deleting k paths, each of cost ≤ (1 + ε)(2k/n + ln n/n) from G, some or all of whose edges may lie in R, there remains a path in R satisfying the same cost bound, and so this must also be true of P k+1 . Before going into detail let us sketch the construction of R. We first build up a tree R s on s, starting from s at level 0, the opposite endpoints of edges out of level i forming level i + 1. We will always choose "cheap" edges, but not always the cheapest ones, as explained later. From s we will choose k + r 0 cheap edges; from each of these k + r 0 level-1 vertices we choose r 1 cheap edges; from each of the (k + r 0 )r 1 level-2 vertices we choose r 2 cheap edges; and on each of the (k + r 0 )r 1 r 2 level-3 vertices we construct a shortest-path tree comprising d vertices. We do a similar construction on t to form R t . Finally, we link R s and R t using cheap edges between their shortest-path trees. The values of the parameters r 0 , r 1 , r 2 and d are given in (3.8), (3.10), (3.14) and (3.16), and it is confirmed in Section 3.9 that |V (R)| = O((k + r 0 )r 2 r 2 d) = o(n), a fact we rely on in the construction.
We will repeatedly use the following Chernoff bound, which in fact holds under more general conditions; see for example [Jan02, Theorem 1, eq. (4)].
3.1. Cheap paths are short. We show that, w.h.p., every cheap path in G is also short. The following lemma asserts the contrapositive. The result is used in (3.6) to restrict the number of edges the adversary can delete.
Lemma 3.2. In both the uniform and exponential models, with probability 1 − O n −1.19 , simultaneously for all l with ln n ≤ l < n, every s-t path of length l has cost ≥ l/(9n).
Proof. We start with the uniform distribution. Here, with X = l i=1 X i , X i ∼ U (0, 1) i.i.d., P(X ≤ a) = P(X ≥ l −a) by 0-1 symmetry of U (0, 1) variables. By Markov's inequality, for any t > 0, P(X > b) ≤ E(e tX )/e tb , and the sum of l uniform random variables has MGF (moment generating function) E(e tX ) = ((e t − 1)/t) l , so substituting t = 9n. Then, over the ≤ n l choices for such a path, the number M l of "short paths" satisfies Summing over l ≥ ln n, the probability that there is a "light" path of any such length is O (e/9) ln n = O n −1.19 .
Since an Exp(1) random weight X can be obtained from a U (0, 1) r.v. X by setting X = − ln(1 − X) > X, the exponential weight stochastically dominates the uniform, so the result for uniform immediately implies that for exponential.
3.2. Adversarial edge deletions. As noted in the introduction, we introduce an edge-deleting adversary whose powers allow it to delete the paths P 1 , . . . , P k , but which is more easily characterised than those paths are. We now specify what the adversary is permitted to do. Let s := 2k + ln n.
(3.2) (From context it should be easy to distinguish this use of s from that as the source of an s-t path.) Let w 0 be the "target cost" of a path, namely Define a "heavy" edge to be one of cost Assuming that there are no "failures" (that the high-probability assertions in 1.1 and elsewhere hold), the edge set P 1 ∪ · · · ∪ P k has total weight ≤ k · (1 + ε)w 0 . The number of heavy edges in these paths is therefore at most Also, as each of the k paths deleted is to have cost ≤ (1 + ε)w 0 , modulo the one-time failure probability O n −1.19 from Lemma 3.2, by that lemma each path has length ≤ (1 + ε)w 0 · 9n < 10s. Thus, the length of all k paths taken together is at most 10ks < 10s 2 . (3.6) And of course the k paths include exactly k edges incident on each of s and t. (3.7) Subject to the absence of failures, the edge set P 1 ∪· · ·∪P k satisfies all three of the constraints (3.5), (3.6), and (3.7) on heavy edges, all edges, and "incident" edges. An adversary who can delete any edge set subject to these constraints is able to delete P 1 ∪ · · · ∪ P k , which is all we require. However, for convenience we will give the adversary more power. At the root of R we will allow it to delete edges subject only to (3.7); at level 1, additional edges subject only to (3.5); and at levels 2 and 3 and for "middle" edges, additional edges subject only to (3.6).
We will show how to choose the parameters of R so that every s-t path in R has cost ≤ (1 + ε)w 0 , and so that R is "robust": after the adversarial deletions, at least one path remains. Specifically, we will arrange that there remains a path in which the "root" edge incident to s costs ≤ k n + 1 9 εw 0 , the edge out of level 1 is heavy but has cost ≤ 1 9 εw 0 , the edge out of level 2 may be light or heavy and also has cost ≤ 1 9 εw 0 , the path through the SP tree has total cost ≤ 1 2 ln n n + 1 9 εw 0 , the central edge joining this to the opposite SP tree adds cost ≤ 1 9 εw 0 , and the continuation of this path to t has the symmetrical properties. It is immediate that such a path has total cost ≤ (2k + ln n)/n + 9 · 1 9 εw 0 = (1 + ε)w 0 . (But see (3.25) for confirmation, after the construction is detailed.) 3.3. Level 0, cheapest edges. On s, add to R the k + r 0 edges of lowest cost, excluding {s, t} from consideration, with r 0 = 1 10 εs = Θ(s). (3.8) Consider this step a failure if any selected edge has cost greater than k n + 1 9 εw 0 . There are n = n − 2 = (1 − o(1))n edges under consideration, with weights i.i.d. U (0, 1), and failure occurs iff the number X of edges with weights in the interval [0, k n + 1 9 εw 0 ] is smaller than k + r 0 . Note that X ∼ Bi(n , k n + 1 9 εw 0 ), thus E X = (1 − o(1)) (k + 1 9 εs), and failure means that X < k + r 0 , i.e., that X E X < (1 + o(1)) k + r 0 k + 1 9 εs = (1 + o(1)) k + 1 10 εs k + 1 9 εs , which by s > 2k is By Lemma 3.1, then, the probability of failure is where we have used that ε is constant and subsumed its value into the Θ expressions. So, modulo the given failure probability, every selected edge incident on s has cost ≤ k n + 1 9 εw 0 , and after the adversarial deletion of k of these edges, r 0 remain. The selection of edges conditions the costs of the other edges incident on s, but none will play any role in the analysis.
The purpose of the next two levels is to expand the number of edges to the point where the adversary cannot delete all of them, because of the heavy-edge budget (out of level 1) and the total-edge budget (out of level 2 and beyond). At the same time, we try to minimise the number of vertices introduced into the construction so that it will remain o(n) for as large a k as possible.
3.4. Level 1, cheapest heavy edges. From each neighbour v of s along the edges just added, add to R the r 1 := 10, 000/ε 2 = Θ(1) (3.10) cheapest heavy edges from v to any of the n vertices not yet added, as before also excluding vertex t. Consider this step a failure if any added edge has cost greater than 1 9 εw 0 . For each neighbour v there are n = (1−o(1))n edges under consideration, with weights i.i.d. U (0, 1), and failure occurs iff the number X of edges with weights in the interval [ 1 11 εw 0 , 1 9 εw 0 ] is smaller than r 1 . Note that X ∼ Bi(n , ( 1 9 − 1 11 )εw 0 ), thus E X = (1 − o(1)) ( 1 9 − 1 11 )εs = Θ(s). Failure means that X < r 1 < E X/2, so by Lemma 3.1 the probability of failure for a given v is ≤ exp(−Θ(s)). This edge selection conditions the costs of the other edges incident on each v, but none will play any role in the analysis. The adversary must leave r 0 edges out of the root, expanding to r 0 r 1 ≥ 1 10 εs · 10, 000/ε 2 = 1, 000s/ε (heavy) edges out of level 1, of which (by (3.5)) he can delete at most 12s/ε, leaving (very generously calculated) at least ρ 1 := 120s/ε = Θ(r 0 r 1 ) (3.12) edges out of level 1. The vertices at the opposite endpoints of these edges constitute level 2.
3.5. Level 2, cheapest edges. From each level 2 vertex v in turn, add to R the cheapest r 2 edges to any of the n vertices not yet added, again also excluding vertex t from consideration.
Here choose r 2 so as to make ρ 2 := ρ 1 r 2 = 12s 2 , (3.13) namely taking r 2 = 12s 2 ρ 1 = 12s 2 120s/ε = 1 10 εs = Θ(s). (3.14) Consider this step a failure if any added edge has cost greater than 1 9 εw 0 . For each neighbour v there are n = (1−o(1))n edges under consideration, with weights i.i.d. U (0, 1), and failure occurs iff the number X of edges with weights in the interval [0, 1 9 εw 0 ] is smaller than r 2 . Note that X ∼ Bi(n , 1 9 εw 0 ), thus E X = (1 − o(1)) 1 9 εs = Θ(s). Failure means that X < r 2 < 0.99 E X, so by Lemma 3.1 the probability of failure for a given v is ≤ exp(−Θ(s)). The number of level-2 vertices v is (k + r 0 )r 1 = O(s) so by the union bound the probability of any failure is This edge selection conditions the costs of the other edges incident on each v, but none will play any role in the analysis. The adversary had to leave at least ρ 1 edges out of level 1, expanding to ρ 1 r 2 = ρ 2 = 12s 2 edges out of level 2, of which by (3.6) he can delete at most 10s 2 , leaving at least 2s 2 edges out of level 2. The vertices at the opposite endpoints of these edges constitute level 3.
3.6. Level 3, shortest-path trees. We now grow each level-3 vertex v to a tree T v with d vertices, including v, choosing Controlling the lengths of the paths in T v would allow a choice of d as large as √ n, but we make it smaller to keep the number of vertices in R as small as possible (and thus keep it to o(n) for s as large as possible).
Here it will be convenient to work with exponentially rather than uniformly distributed edge weights. There are various easy ways to arrange this. We do so by temporarily replacing each uniform weight w with a weight w = − ln(1 − w); it is standard that these transformed weights are exponentially distributed, and that w ≥ w. We construct a shortest-path tree (SPT) of order d using the transformed weights; it will not be an SPT for the original weights, but its paths will be short under the original weights, which is all that we care about.
Define the radius rad(T v ) of an SPT T v to be the distance from v to the furthest vertex. The radius is described by the following claim, which we phrase in a generic setting with n vertices and a root vertex s. Claim 3.3. In a complete graph K n with i.i.d. exponential edge weights with mean 1, the radius X = rad(T s ) of a shortest-path tree T s of order d is where the X i are independent random variables with X i ∼ Exp(i(n − i)).
Proof. Following [Jan], think of the process of finding shortest paths from s to other vertices as first-passage percolation or "infection spreading" starting from s. Let dist(v) be the distance (shortest path length) from v to s. Let R := R(r) be the set of vertices within radius r of s; we think of gradually increasing r, starting with r = 0 where R = {s}. It is well known that each edge (v, u) ∈ R(r) × (V \ R(r)) has exponentially distributed weight W conditioned by W + dist(v) ≥ r, and that these random lengths are independent. This can be seen by imagining that infection has spread to radius r from s, including to the vertex v and additionally a length r − d(v) further along the edge (v, u), and appealing to the memoryless property of the exponential distribution; it can also be verified by analyzing Dijkstra's algorithm in this randomised setting. It follows that the distance X 1 to the vertex nearest s is distributed as X 1 ∼ Exp(n − 1); the additional distance to the next vertex is X 2 with X 2 ∼ Exp(2(n − 2)) and independent of X 1 (for total distance X 1 + X 2 ); and when there are i vertices in the tree, the additional distance to the next is X i ∼ Exp(i(n − i)), with all the X i independent, for total distance as claimed.
Since (3.17) is monotone increasing in d, the larger the d, the greater the probability of failure, so in the next paragraphs we will pessimistically take d to be √ n (ignoring integrality since √ n is large). In this case, applying Claim 3.3 to T v , constructed in a complete graph of order n = n(1 − o(1)), the expectation of X is (3.18) Thus, failure of T v implies that To bound the probability of this event we require one more lemma (also used later in proving Lemma 4.2). (3.20) for any λ = 1 − ε ≤ 1, and for any ε > 0,

19) (and thus that of T v failing) is at most
using that a * = n = (1 − o(1))n, µ is given by (3.18), and ε = Θ(1). The total number of trees built is N = (k + r 0 )r 1 r 2 , which, with reference to (3.2), (3.8), (3.10), (3.14), is Θ(s 2 ). By (3.23), each tree independently fails with at most some probability p = o(1). Thus, the number of trees surviving dominates Bi(N, (1)). Failure at level 3 means that at least 0.01s 2 = Θ(N ) trees fail, thus the number surviving is at most some λ(1 − Θ(1)), which by Lemma 3.1 has probability exp(−Ω(s 2 )). (3.24) Remark 3.5. When construction of a tree T v rooted at a level-3 vertex v is finished, the edge between any vertex a of T v and any vertex b in V \ V (T v ) has weight w(a, b) that -still in the uniform model with edge weights temporarily transformed to be exponentially distributed - are independent, over all choices of a and b.
Call R s the now-complete construction on s. Note that there is no conditioning on edges between the remaining vertices; in particular, the SPT infection process never looked at edges between uninfected vertices. 3.7. Symmetric construction on vertex t. Just as we have constructed R s , we now make a similar construction R t for vertex t, with the same branching factors out of levels 0, 1, and 2 and similar SPTs on level-3 vertices. Since the number n of vertices available after constructing R s still satisfies n = (1 − o(1))n, and because the construction on s did not look at nor condition any edge between these vertices, the construction on t enjoys the same properties as that on s.

3.8.
Edges between the trees on s and t. It remains only to complete paths between s and t, which we do by adding cheap edges (where present) between the SPTs in R s and those in R t .
Let T u be an SPT rooted at a level-3 vertex u of R s , and T v one rooted at a level-3 vertex v of R t . Let a and b be any vertices in T u and T v respectively. By Remark 3.5, edge {a, b} gives a u-to-b path with cost rad(T u ) + X a,b , the collection of all the excesses X a,b being i.i.d. each with distribution X a,b ∼ Exp(1). Thus, {a, b} gives a u-to-v path with cost ≤ rad(T u )+X a,b +rad(T v ).
Select, and add to the full construction R, any such "middle edge" {a, b} having X a,b ≤ 1 9 εw 0 . This completes the construction of R.
We now confirm that, assuming that the construction was successful, any s-t path in R through successful SPTs has cost ≤ (1 + ε)w 0 . (Remember that there may be some unsuccessful SPTs.) By assumption of success, any level-0 edge on s or t has cost ≤ k n + 1 9 εw 0 , any level-1 edge has cost ≤ 1 9 εw 0 , and any level-2 edge also has cost ≤ 1 9 εw 0 . Each successful level-3 tree T in R s or R t has radius rad(T ) ≤ (1 + 2 9 ε) 1 2 ln n/n ≤ 1 2 ln n/n + 1 9 εw 0 , and each selected middle edge {a, b} connects the roots of two trees at an excess cost (above the sum of the two radii) of X a,b ≤ 1 9 εw 0 . The total of the 9 upper bounds in question is 2 · k n + 2 · 1 2 ln n/n + 9 · 1 9 εw 0 = (1 + ε)w 0 . (3.25) 3.10. Robustness of R. We now show that, after the deletion of the k cheapest paths in G, there remains at least one path in R (that uses successful SPTs). Recall from Section 3.2 that deletion of the k cheapest paths in G is conservatively modeled as an adversarial deletion subject to: (3.7), the deletion of exactly k edges incident on each of s and t; (3.5), the number of heavy edges deleted at level 1; and (3.6), the total number of edges deleted elsewhere in R (at levels 2 and 3, and joining R s and R t ).
Without loss of generality we may assume that the adversary does not delete an edge within an SPT T , nor a middle edge from such a tree to a facing one, since deleting the level-2 edge into the level-3 root of T destroys more paths in R at the same budgetary cost. By the assumption of success, there are at most 0.01s 2 failed SPTs on each of s and t, and for simplicity we will deal with them by imagining all trees to be successful but allowing the adversary his choice of this many SPTs to delete; by the argument above we can model this as deletion of edges into the roots of these trees, and simply add 0.02s 2 to this budget.
Let us now allow the adversary to delete k edges from each of s and t, 12s/ε edges out of level 1 for each (double-counting the heavy-edge budget), and 10.02s 2 edges out of level 2 for each (again double-counting). Can he destroy all s-t paths? We have not yet made any high-probability structural assertion about the middle edges, so this is a probabilistic question: what is the probability, over the randomness still present in the middle edges, that there is an adversarial deletion destroying all paths?
Of the k + r 0 = O(s) edges on s, the adversary chooses k to delete; there are at most 2 O(s) ways to do so. Any choice leaves Θ(r 0 r 1 ) = Θ(s) edges out of level 1, of which the adversary is able to delete a positive fraction, again in at most 2 O(s) ways. Any choice leaves Θ(s 2 ) edges out of level 2, of which the adversary is able to delete a positive fraction, in at most 2 O(s 2 ) ways. The adversary makes a similar set of choices on t, but still this comes to just 2 O(s 2 ) possible outcomes in all.
A given deletion choice destroys all paths precisely if it leaves no middle edge of excess ≤ 1 9 εw 0 . (Remember that, w.l.o.g., we have excluded deletions in and between the SPTs at level 3.) By construction, any deletion choice leaves Θ(s 2 ) edges out of level 2 and thus, by (3.16), Θ(s 2 d) = Ω( √ ns ln n) vertices in SPTs in each of R s and R t , for Ω(ns ln n) potential middle edges. A middle edge is selected if its excess cost (in the exponential model) is w = − ln(1 − w) ≤ 1 9 εw 0 , i.e., if 1−w ≥ exp(− 1 9 εw 0 ), thus is rejected with probability exp(− 1 9 εw 0 ). There is no path only if every potential edge is rejected, which happens w.p. ≤ exp(− 1 9 εw 0 · ns ln n) = exp(−Ω(s 2 ln n)). Taking the union bound over all adversarial choices, the probability than any choice leaves no paths is 2 O(s 2 ) exp(−Ω(s 2 ln n)) = exp(−Ω(s 2 ln n)).
(3.26) This is dominated by the failure probabilities exp(−Θ(s)) for other steps.
3.11. Success for every k, and the inductive hypothesis. We have shown that, for any k = o( √ n), w.h.p. we generate a robust structure R in which, even after adversarial deletions This confirms that the inductive hypothesis holds, w.h.p., simultaneously for all k to which the argument applies, namely all k = o( √ n).
3.12. Limitation to small k. We have established Theorem 1.1 up to any k = o( √ n), and the construction of R k was tailored to such values. For levels using heavy edges, fanout is limited to O(s). On the other hand, the meet-in-the-middle argument requires that each side grow large, to Ω( n/s). Thus, for small k, a more-than-constant number of levels is needed. Summing heavy edges over this many levels would exceed the target weight (1 + ε)w 0 , so light edges are needed. The adversary may delete Θ(s 2 ) light edges, so the construction must contain at least this many. The construction explicitly required each light edge to lead to a new vertex, and we do not readily see how to do otherwise as long as we are using shortest-path trees, thus intrinsically limiting s (thus k) to O( √ n). For larger k, however, we can obtain sufficient heavy-edge fanout in constant depth, permitting a simpler construction described in Section 5.

Edge order statistics
In this section we establish results on order statistics needed in later sections. Let {W (k) } n−1 k=1 be the order statistics of n − 1 i.i.d. random variables, variously uniform U (0, 1) or exponential Exp(1). We choose n − 1 rather than n as the parameter both because many expressions are more natural in this parametrisation, and because this way W (k) is the cost of the kth cheapest edge incident to a fixed vertex v ∈ K n . The following lemma is used in Section 6.3. where the sign in the Ω is taken as positive. The probability that any interval I i contains less than 0.999 points is, by the union bound, at most, 1/L ·exp(−Ω(Ln)) = exp(−Ω(n 0.01 )) = o(1) as desired.
The following lemma is used in (6.10) and (7.5).
simultaneously for all k in the range a ≤ k ≤ n − 1.
Proof. Without loss of generality, we may assume that a ≤ n/10. Exponential case. It is standard that, where Z i ∼ Exp(i) are independent exponential r.v.s, we may generate the W (k) as Using a superscripted E to highlight the exponential model, W (k) has mean the change by 1 in the logarithms' arguments avoids ln(0) when k = n − 1 and remains asymptotically correct. By (3.22), By the union bound, it suffices to show that the sum over k from a to n − 1 of the RHS of (4.3) is o(1). We treat the sum in two ranges. For k ≤ n 2 , (n − k)µ k ≥ n 2 · k n = k 2 . Thus, since a = ω(1). For k > n 2 , for brevity letk = n − k. Then µ k ∼ ln n − ln(k) by (4.2) and where the explicit inequality factors out thek = 1 term, from which, sincek/n ≤ 1/2, the later terms decrease geometrically. This concludes the exponential case.
Uniform case: Let U i ∼ U (0, 1) be i.i.d. uniform random variables and W i ∼ Exp(1) i.i.d. exponential random variables. Because the exponential distribution has CDF F (x) = 1 − exp(−x), we may couple the two sets of variables as ). Now using superscript U to distinguish the uniform model, the mean is well known to be We want to show that with high probability, for all k in the range a ≤ k ≤ n − 1, or equivalently, From the exponential case already proved, taking error bound ε/2, we know that w.h.p., for all k, k , so it suffices to show that, for all k (deterministically), k . This is so. Using (4.6), (4.2), and convexity of f ,

5.
Upper bound for large k, sketch 5.1. Introduction. To address larger values of k we use a different construction, generating s-t paths of length 4. A straightforward extension of the previous argument to this construction would let us get up to k = n − f (n) for an arbitrarily slowly growing function f , but not to k = n − 1 because it requires k + 1/ε 2 edges incident on each of s and t (thus requires that k + 1/ε 2 ≤ n − 1). Getting all the way to k = n − 1 requires a couple of additional ideas. First, we observe that most of the adversary's cost budget must be spent on edges incident to s and t, leaving less to delete other edges, thus allowing a smaller structure R to be sufficiently robust. In particular, the k cheapest paths from s to t must use edges incident on s of total weight at least is the cost of the ith cheapest edge incident to v. (We may omit the superscript when it is either generic or clear from context.) One technical detail is that, where R includes the k + r 0 cheapest edges incident to s, we will control W (k+r 0 ) − W (k) directly, using results on order statistics from Section 4, rather than through a high-probability upper bound on W (k+r 0 ) and a high-probability lower bound on W (k) . Finally, it is no longer adequate to allow path costs to exceed their nominal values by an ε = Θ(1) factor, as such large excesses would swell the adversary's budget too quickly, so we more tightly control the excess cost of each path P k as a function of k (and n, implicitly).
The details later will be clearer if we sketch the argument now, with most details but without the calculations. We will argue for k from n 4/10 to n − 1. (We must start with some k = o(n 1/2 ) since that is as far as the "small k" argument extended, and we need k = ω(n 1/3 ) since below this the new construction's path costs would exceed the 2k/n target.) 5.2. Structure R. Figure 2 illustrates the robust structure R = R k after adversarial deletion of root edges, as discussed in item (5.6) below. The construction is based on parameters r 0 = r 0 (k) and ε k to be defined later. Start with R consisting of just the vertices s and t. Add to R the k + r 0 edges incident on s of lowest cost, and let V s be the set of opposite endpoints of these edges. Do the same for t, generating vertex set V t . Take M := V (G) \ {s, t} as a collection of "middle vertices".
Note that V s , V t , and M may well have vertices in common, but our analysis will use a subgraph of R where the relevant subsets of these three sets are disjoint, and it is easier to understand the construction imagining them to be disjoint. Add to R each edge e in M × V s and M × V t that is "heavy but not too heavy", with cost W (e) ∈ (ε k , 2ε k ). This concludes the construction of the structure R.

Path weights. It is immediate that every s-t path in R has cost at most
We will show (in (6.11) for uniform and (8.15) for exponential) that, subject to the nonoccurrence of certain unlikely failure events, (5.2) is at most We will show in Section 5.6 that, after deletion of the first k paths, there remains an s-t path in R (again subject to non-occurrence of unlikely failure events), whereupon it follows that (5.4) 5.4. Adversary. We define an adversary who is "sufficiently strong" to delete the first k paths. Assume an absence of failure events so that (5.4) holds, and so that for i ≤ n 4/10 we have the upper bounds on X i from (1.3). Then, hiding a sum of the ln n/n terms of (1.3) in the o( ) Thus, the first k paths' edges have total weight at most U k . Furthermore, the first k paths' edges incident on s and t are all distinct except, possibly, for the edge {s, t}. Therefore, not counting edge s-t at all, the cost of these "incident" edges is at least (5.6) (In proving Claim 8.2 we will use a slightly different lower bound I k on the weight of the incident edges.) It follows that the first k paths' "middle edges" (edges other than the incident edges) cost at most U k − I k . We will explicitly define a budget B k satisfying (5.7) We will allow the adversary to delete any k edges in G incident on each of s and t, possibly including the edge s-t (enough to let it delete the incident edges of the first k paths), and to delete any other edges in G of total cost at most B k (enough to let it delete the middle edges of the first k paths). Thus, the adversary is sufficiently strong to delete the first k paths.
The adversary's allowable deletions in G mean that also in R it deletes at most k edges incident on each of s and t, and middle edges of total cost at most B k . 5.5. Budgets B k . The budgets B k will be defined explicitly in the details. For the model with uniformly distributed edge weights we will do so in two ranges of k, corresponding to Claims 6.1 and 6.2, and likewise in the model with exponentially distributed edge weights, corresponding to Claims 8.1 and 8.2. For Claims 6.2 and 8.2 we will establish (5.7) directly.
For Claims 6.1 and 8.1 we will establish (5.7) by the following reasoning; we will only need to check (5.8), (5.9), and (5.10) below. We will show that the budgets satisfy (5.8) (Roughly speaking, given B k we will set ε k as small as possible while keeping R k robust to the adversary with budget B k . Then, we will set B k+1 as small as possible, namely by taking equality in (5.8). Behind the scenes, we derive B k by solving the differential-equation equivalent of (5.8) satisfied with equality.) We will show that (5.7) is satisfied in the base case, by showing that Then, (5.7) is established for all k by induction on k: which by (5.5), (5.6), and the inductive hypothesis (5.7) is (5.10) ≤ B k+1 (by (5.8)). (5.11) To justify (5.10) it suffices to show that W (k+1) − W (k) is at most 0.1ε k , and we do so in (6.12) for the uniform case and in (8.17) for the exponential case. In both cases, r 0 = ω(1), and W (k+r 0 ) − W (k+1) = O(ε k ) (as used in going from (5.2) to (5.3)), making this conclusion unsurprising. 1 5.6. Robustness of R. We wish to make R robust against the adversary, so that after the deletions just described, R should retain an s-t path w.h.p., so that (5.4) holds and X k+1 is small. It will suffice to show that, to delete all s-t paths in R, after deletion of k edges incident on each of s and t, an adversary would still have to delete middle edges of total cost more than B k , (5.12) and thus it is powerless to do so. Obtaining this robustness requires choosing ε k sufficiently large in the construction. With reference to Figure 2, on deletion of any k edges on each of s and t, the level-1 sets are in effect pruned to V s and V t , each of cardinality r 0 . Should V s and V t have vertices in common, or if t ∈ V s or s ∈ V t , then there is an s-t path. So, assume that V s and V t are disjoint and do not contain s nor t. Consider only middle vertices M ⊆ M not appearing in V s nor V t , i.e., Consider a choice of the k deletions on s and t to be fixed in advance. (We will eventually take a union bound over all such choices.) The weights of edges in M ×V s and M ×V t have not even been observed yet, so each has (unconditioned) U (0, 1) distribution, all are independent (by distinctness of the edges), and thus each such edge is included in R with probability ε k , independently.
A vertex v ∈ M is connected to V s by Z s v ∼ Bi(r 0 , ε k ) (5.13) edges, with mean (5.14) Define Z t v symmetrically, and note that Z s v and Z t v are i.i.d. Intuitively, if λ is small, Z s v is usually 0, is 1 with probability about λ, and rarely any larger value. So, the probability that v is connected to both V s and V t is about λ 2 , in which case to destroy s-t paths through v the adversary must delete an edge of cost at least ε k . So, to delete all s-t paths, over the nearly n vertices in M the adversary would have to delete edges of expected total weight at least ε k n λ 2 . (5.15) We will choose ε k so that ε k n λ 2 > B k , (5.16) which hopefully will ensure (see Remark 5.1) that a path must remain (i.e., that R is robust). Let us give an exemplary back-of-the-envelope calculation. In the uniform case we expect W (k) to be about k/n, so letting r 0 = ε k n means that W (k+r 0 ) − W (k) will be about ε k , justifying (5.3). Then (5.14) gives λ = ε 2 k n, so (5.16) indicates that we need to take ε 5 k n 3 > B k . As noted after (5.8), roughly speaking, we obtain B k and ε k by solving this and (5.8) with equality as a system of differential equations.
Remark 5.1. This intuitive argument proves to be essentially sound, but to make it rigorous will take some work. Chiefly, P(Z s v > 0) is of course not exactly E Z s v = λ even when λ is small, and we will also have to consider the case when λ is large. Also, where the intuition is based on expectations, we must calculate the probability of the "failure" event that all paths can be deleted at a cost less than B k . Finally, we must take the union bound of this failure event over all choices of root edges at s and t (but, as in the small-k case, this turns out to change nothing).

Upper bound for large k, uniform model
In this section we fill in the details of the steps from Section 5 and show that they conclude the proof of the upper bound in Theorem 1.1. Specifically, to control the path weights (these emphasised keywords match section titles) we must show that (5.2) is at most (5.3). For the adversary we need only show (5.7); as noted earlier, for large k (Claim 6.2) we will do this directly, while for medium k (Claim 6.1) we will argue that the budgets B k satisfy (5.9) and (5.10). And for robustness we will prove that the probability of failure is small (i.e., it is unlikely that the adversary can destroy all s-t paths in R k ).
6.1. Claims, and implications for Theorem 1.1. We first state the two precise claims we make for large k, in two ranges. We use symbolic constants C B , C ε , C B , and C ε in the claims and the proofs, as it makes the calculations clearer. Whenever we encounter an inequality that the constants must satisfy, we will highlight with a parenthetical "check" that they do so. and ε k = C ε n −3/5 B k 1/5 , with C B = 32 and C ε = 5. Then, asymptotically almost surely, simultaneously for all k in this range, Remark: In proving Claim 6.1 we will set r 0 := ε k n. (6.2) From the definitions of B k and ε k in Claim 6.1, both are increasing in k, and we will make frequent use of the following inequalities. For n sufficiently large, B k = Θ k 5/4 n −3/4 + n −1/5 (6.3) B k ≥ B n 4/10 ≥ 2n −1/5 (6.5) ε k = Θ k 1/4 n −3/4 + n −16/25 (6.6) ε k ≥ ε n 4/10 ≥ 1.14C ε n −16/25 . (6.8) Claim 6.2. For k ∈ (n − 14 √ n, n − 2], let B k = C B √ n and ε k = C ε n −1/6 (6.9) with C B = 78 and C ε = 5. Then, asymptotically almost surely, simultaneously for all k in this range, X k+1 ≤ W s (k+1) + W t (k+1) + 8ε k . Remark: In proving Claim 6.2 we will set r 0 := 1. Note that here B k and ε k are constants independent of k, but we retain the subscript for consistency with the notation of Section 5.1.
We will prove the two claims shortly.
We now prove the two claims, by filling in the details for Sections 5.2 to 5.6. 6.2. Structure R. With reference to Section 5.2, all that we need to confirm is that k + r 0 ≤ n − 1. For Claim 6.1, by hypothesis k ≤ n − 14 √ n, and provided that 1.01C ε C 1/4 B ≤ 13 (check), by (6.7) ε k ≤ 13n −1/2 , whereupon r 0 = ε k n ≤ 13 √ n. For Claim 6.2, with r 0 = 1, k + r 0 ≤ n − 1 is immediate. 6.3. Path weights. With reference to Section 5.3, we establish that the bound (5.3) holds w.h.p. simultaneously for all k ≥ n 4/10 . With W (k) representing the cost of the kth cheapest edge incident on some fixed vertex (which we will take to be s and then t in turn), it suffices to show that holds with high probability for all k ≥ n 4/10 . For Claim 6.2, with r 0 = 1, (6.11) is immediate. For Claim 6.1, with r 0 = ε k n, generate the variables W (k) by placing n − 1 points uniformly at random on the unit interval I, associating W (k) with the kth smallest point. It suffices to show that, w.h.p., each interval (W (k+1) , W (k+1) + 1.1ε k ) contains at least r 0 points. For all k ∈ [n 4/10 , n − 14 √ n − 1 ], Claim 6.1 has ε k ≥ n −0.99 by (6.8), so Lemma 4.1 shows that w.h.p. every interval of length ≥ 1.1ε k in [0, 1] contains at least r 0 def = ε k n points, and in particular this holds for all the intervals (W (k+1) , W (k+1) + 1.1ε k ). We assume henceforth that the graph G is "good" in the sense that (6.11) holds for all k ≥ n 4/10 for vertices s and t, and that for all k ≤ n 4/10 we have the upper bounds on X k from (1.3), as proved to hold w.h.p. in Section 3. 6.4. Adversary. With reference to Section 5.4, we need only verify (5.7), and this will be done in the next subsection. 6.5. Budgets B k . With reference to Section 5.5, we first establish (5.10). This follows from The reasoning for this is the same as for (6.11): each interval of length 0.1ε k contains at least one point. The parameters are trivial to check.
6.6. Minimum of two binomial variables. Before addressing robustness of the structure R, we require a lemma (Lemma 6.4) on the minimum Z of two i.i.d. binomial Bi(n, p) random variables. There is a genuine difference in the cases when the common mean λ = np is large or small: if λ is large then Z is likely to be close to λ, making E Z = Θ(λ); if λ is small then Z will most often be 0, occasionally 1 (with probability about λ 2 ), and rarely anything larger, making E Z = Θ(λ 2 ). The lemma relies on the following property of the median of a binomial random variable. (A weaker form of (6.14) and thus of Lemma 6.4 can be obtained from Lemma 3.1 in lieu of using the median.) In this discrete setting Med(X) is not unique: it can be any value m for which P(X ≤ m) ≥ 1/2 and P(X ≥ m) ≥ 1/2. [Ham95] defines it uniquely as the smallest integer m such that P(X ≤ m) > 1/2; as desired, this gives P(X ≥ Med(X)) = 1−P(X ≤ Med(X)−1) ≥ 1−1/2 = 1/2. (For other results on the binomial median see Kaas and Buhrman [KB80], in particular, Corollary 1. Stronger results for the Poisson distribution are given by Choi [Cho94], proving a conjecture of Chen and Rubin, and by Adell and Jodrá [AJ05].) Lemma 6.4. Let Z 1 , Z 2 be i.i.d. Bi(n, p) random variables, Z := min(Z 1 , Z 2 ) and λ := E Z 1 = np.
In the second case we again use independence, and here The last inequality comes from minimising 1−exp(−x) x over 0 ≤ x ≤ 2; the function is decreasing so the minimum is at x = 2. 6.7. Robustness in Claim 6.1. With reference to Section 5.6, let us complete the robustness argument for Claim 6.1, showing that (5.12) holds with high probability. Here we have taken r 0 = ε k n, so that the number of edges from a middle vertex to V S (see (5.13)) is Z s v ∼ Bi(ε k n, ε k ), with mean λ = r 0 ε k = ε 2 k n (see (5.14)). Recall that if λ is small we expect (see (5.15)) that to destroy all paths the adversary will have to delete edges of total weight at least ε k n λ 2 = ε 5 k n 3 , which will exceed B k . And, if λ is large, then each Z v will have expectation close to λ = ε 2 k n, for a total cost ε k n times larger, namely ε 3 k n 2 , and again this exceeds B k . We now replace these rough calculations with detailed probabilistic ones, applying Lemma 6.4 to Z v in the two cases of λ small and large.
For the adversary to delete all s-t paths via v, he must delete at least edges, and to destroy all paths he must delete at least N := v∈M Z v edges. As described in Section 5.6, we imagine a fixed deletion of k edges on each of s and t giving neighbour sets V s and V t and a set M of middle vertices; we will eventually take a union bound over all such choices.
6.8. Robustness in Claim 6.2. Again, our aim is to establish robustness of R by showing that (5.12) holds with high probability, and the argument is similar to but simpler than that of Section 6.7. Since r 0 = 1, both V s and V t have size 1. For a vertex v ∈ M , let Z v be the number of paths from V s to V t via v. There is only one such possible path, hence To destroy all s-t paths the adversary must delete at least edges. N stochastically dominates Bi(0.99n, ε 2 k ), which has expectation 0.99ε 2 k n. We declare the event N ≤ 0.98ε 2 k n a failure. Assuming success, destroying all s-t paths would cost at least ε k N ≥ 0.98ε 3 k n. This exceeds B k , since 0.98ε 3 and 0.98C ε 3 > C B (check). The probability of failure is P N ≤ 0.98ε 2 k n = exp(−Ω(ε 2 k n)) = exp(−Ω(n 2/3 )). (6.20) Over all rounds and adversary choices, using that k+r 0 r 0 = k+1 1 ≤ n, the total failure probability is at most k k + r 0 r 0 2 · P(N ≤ 0.98ε 2 n) ≤ (14 √ n) n 2 exp(−Ω(n 2/3 )) (by (6.20)) = o(1).

Lower bound
In this section, we establish the lower bound in (1.3) of Theorem 1.1. Section 7.1 establishes the lower bound on X k directly for k ≤ √ ln n. Values k ≥ √ ln n are treated in the subsequent parts. In Section 7.2, Lemma 7.1 establishes a lower bound on the running totals S k , (7.1) In Section 7.3, Lemma 7.2 obtains a lower bound on X k using Lemma 7.1's lower bound on S k , the previously established upper bound on X k from Theorem 1.1, and the monotonicity of X k .
7.1. Lower bound for small k. We begin with k ≤ √ ln n. For any fixed ε > 0, we know from [Jan] that w.h.p.
Assuming that (7.2) holds, it follows immediately, and deterministically, that for all k ≤ √ ln n, 3) The first inequality holds because the sequence X k is monotone increasing, the next by assumption on X 1 , the next by k = o(ln n).

7.2.
Lower bound on the running totals.
Lemma 7.1. For any ε > 0, w.h.p., simultaneously for every k ≤ n − 1, 2i + ln n n . (7.4) Proof. Write W s (i) and W t (i) for the order statistics of edge weights out of s and t, respectively. By Lemma 4.2, w.h.p., and we will assume throughout the proof that (7.5) holds. We prove the assertion in two ranges of k.
For ln 11/10 n ≤ k ≤ n − 1, the k paths must use at least k − 1 edges on each of s and t, all distinct (k edges each, ignoring the edge {s, t} if it is used). Then, using (7.5), we get that w.h.p., for all k in the range, 2i + ln n n (see below) (7.7) 2i + ln n n . (7.8) To justify (7.7) it suffices to show that the first sum in (7.6) is of strictly larger order than the other terms. The first sum is at least k i=k/2 2i/n = Ω(k 2 /n), which since k ≥ ln 11/10 n is also Ω(k ln 11/10 n/n) and Ω(ln 22/10 n/n); we will use all three formulations. The second term is of order O(k ln n/n), negligible compared with the middle formulation. The third term is O(ln 2/3 n/n), negligible compared with the last formulation. And the fourth term, of order O(k/n), is negligible compared with the first formulation.
For 1 ≤ k ≤ ln 11/10 n, let δ = ε/3 and let G = G − s − t. Let N s and N t be the endpoints of the cheapest ln 3 n edges out of s and t respectively. Note that these sets are independent of the edge weights of G .
If any path P i , i ≤ k, uses a root edge (edge incident on s or t) not amongst the ln 3 n cheapest edges of s or t, then by (7.5) this edge costs at least (1 − ε) ln 3 n/n, thus S k ≥ (1 − ε) ln 3 n/n. Then (7.4) follows because this is larger than the RHS of (7.4), namely Θ((k 2 + k ln n)/n) = O(ln 11/5 n/n) for this range of k. Thus we may assume that for all i ≤ k, each path P i goes via some s ∈ N s , t ∈ N t .
For s ∈ N s , t ∈ N t , define A(s , t ) to be the event that t is one of the (n − 2) 1−δ nearest vertices of s , by cost, in G . Clearly, for each pair s , t , P(A(s , t )) = (n − 2) −δ . Let A be the union of these events, i.e., the event that any such pair has this property. By the union bound, We assume henceforth that A does not hold: the ln 3 n cheapest root edges at s and t do not happen to sample any "nearest" pairs in G .
Since (1 + o(1))(1 − δ) 2 ≥ (1 − 3 4 ε), by the union bound this implies By standard arguments, e.g. Remark 1.3, this also implies the same for the uniform model in which we are working. Thus w.h.p., for all s ∈ N s , t ∈ N t , we have d G (s , t ) ≥ (1 − 3 4 ε) ln n; assume this holds. We already assumed that each path P i , i ≤ k, goes via some s ∈ N s , t ∈ N t , so its non-root edges contribute at least d G (s , t ) ≥ (1 − 3 4 ε) ln n/n to S k . Then, for all k in this range, n (by (7.5)) (7.9) 2i + ln n n .
and 6, the "middle edges" are of cost o(1) for all k, but the "incident edges" have larger cost for k large. In particular, for large k, (6.11) will no longer hold in the exponential case until we adjust r 0 and ε k appropriately.) For the lower bound too, the argument in Section 7 carries over for all k = o(n). The lower bounds L k on the prefix sums S k derived in Sections 7.1 and 7.2 carry over to the exponential case because the edge costs are equal to within 1 + o(1) factors in the two models. The upper bounds U k on the prefix sums are simply the sums of the individual upper bounds on X k , and we have just argued that these change only by a 1 + o(1) factor. Section 7.3 only uses L k and U k to derive lower bounds on X k , so with these both changed only by 1 + o(1) factors, its results carry over verbatim.
Our task, then, is to prove the upper and lower bounds in Theorem 1.2 for larger k. For the upper bound, arguing for k > n 0.4 (there is no advantage to a larger starting value), we use the same approach as for the uniform model in Section 5.
For the lower bound, we argue for k ≥ n 9/10 . Unfortunately, the method used in Section 7 for the uniform distribution does not extend; let us explain why. The lower bound there came from (7.11), tX k ≥ L k − U k−t , valid for any functions L and U with L k ≤ S k ≤ U k . Here, we would take L k as the sum I k of incident edges as in (5.6) and U k as the sum of the X k upper bounds as in (5.4). Recall that we defined B k so that B k ≥ U k − L k , as in (5.7). Then we can rewrite the previous lower bound approach as For large k, W (k) and therefore X k are Θ(ln n). Since the B k grow to size Θ(n 1/2 ) (in the exponential case as well as the uniform case), we are thus limited by the second term to t = Ω(n 1/2+o(1) ). However, from (4.2), such a large value of t would mean that the average given by the first term is significantly different from W (k) .
The desired lower bound would be immediate if we could claim that P k necessarily used the kth cheapest edge on s (of cost W s (k) ) or a later one, and likewise for t. We will prove something close to this. We argue in Section 8.7 that every pair of vertices (excluding both s and t) is joined by a path of cost at most δ (for some small δ to be specified) that is edge-disjoint from all P i , i = 1, . . . , n − 1. We will show that this implies that path P k uses an edge on s that is at most δ cheaper than W s (k) , and likewise for t, yielding a sufficient lower bound.
8.1. Claims, and implications for Theorem 1.2. In order to establish upper bounds on X k in the exponential model, we use the same structure R k as described in 5.2. Then (5.4) follows as before, and we can continue to define U k as in (5.5). For convenience definē k = n − k. (8.1) As before we will treat k in two ranges, and we start now with the smaller range. with C B = 44 and C ε = 4. Then, asymptotically almost surely, Remark: In proving Claim 8.1 we will set because it roughly equates W (k+r 0 ) − W (k) and ε k ; see (4.2). In this regime integrality is not an issue: r 0 is large, per (8.9). It is clear that both B k and ε k in (8.2) are increasing in k, even over the larger range k ∈ [0, n]. We will make use of the following bounds, holding for n sufficiently large. Here, (8.5) uses that at k = n − Θ( √ n),k 3/5 dominates 2n 1/25 , while (8.6) takes k = 0.
Proof of the upper bounds in Theorem 1.2. Analogous to the argument in Section 6.1, it is sufficient to check that ε k = o(E W (k) ). Since E W (k) ∼ ln n n−k ≥ k n (see (4.2)), it is enough to show that ε k = o(k/n).
8.4. Robustness in Claim 8.1. With reference to Section 5.6, we complete the robustness argument for Claim 8.1, showing that 5.12 holds with high probability. Here we have taken r 0 = ε kk , so the number of edges from a middle vertex to V S (see (5.13)) is Z s v ∼ Bi(ε kk , ε k ), with mean λ = r 0 ε k = ε 2 kk (8.18) (see (5.14)). Recall that if λ is small we expect (see (5.15)) that to destroy all paths the adversary will have to delete edges of total weight at least ε k n λ 2 = ε 5 k nk 2 , which will exceed B k . And, if λ is large, then each Z v will have expectation close to λ = ε 2 kk , for a total cost ε k n times larger, namely ε 3 k nk, and again this exceeds B k . We now show the details of these rough calculations, including the probabilistic details, applying Lemma 6.4 to Z v in the two cases of λ small and large.
For the adversary to delete all s-t paths via v, he must delete at least Z v := min(Z s v , Z t v ) edges, and to destroy all paths he must delete at least N := v∈M Z v edges. As described in Section 5.6, we imagine a fixed deletion of k edges on each of s and t, giving neighbour sets V s and V t and a set M of middle vertices, eventually taking a union bound over all such choices.
8.5. Budgets in Claim 8.2. We now establish (5.7) for the parameters of Claim 8.2. Section 8.3 showed that (5.7) holds for k up to k := n − √ n , the point where Claim 8.1 ends and just before Claim 8.2 begins, so in particular B k ≥ U k − I k . For the regime of Claim 8.2, we redefine I k from (5.6). Recall that I k is a lower bound on the edges incident to s and t used by the first k paths. Previously, the sum defining I k in (5.6) went to k − 1 to avoid double counting the {s, t} edge. In this regime, however, we need the sum to go k, as the W (i) increase rapidly. The weight of the {s, t} edge is distributed as Exp(1), thus w.h.p. it costs at most n 0.01 . For k > k , define so that w.h.p. I k is a lower bound on the incident edges: the n 0.01 term resolves the potential double-counting of {s, t}. We are now ready to check that (5.7) holds. Following the derivation of (6.13), for k from k + 1 to n − 2, 7ε k − (W s (k ) + W t (k ) − n 0.01 ) (see (5.5), (5.6), and (8.23)) ≤ 114 √ n + √ n · 7C ε n −1/6 + n 0.01 (see (8.5) and (8.10)) ≤ 115 √ n ≤ B k (see (8.10)), (8.24) using that C B ≥ 115 (check).
8.6. Robustness in Claim 8.2. Again, our aim is to establish robustness of R by showing that (5.12) holds with high probability, and the argument is similar to but simpler than that for robustness in Claim 8.1. Since r 0 = 1, both V s and V t have size 1. For a vertex v ∈ M , let Z v be the number of paths from V s to V t via v. There is only one such possible path, hence To destroy all s-t paths the adversary must delete at least N := v∈M Z v edges. N stochastically dominates Bi(0.99n, ε 2 k ), with expectation at least 0.99ε 2 k . We declare the event N ≤ 0.98ε 2 k n a failure. Assuming success, destroying all s-t paths would cost at least ε k N ≥ 0.98ε 3 k n. This exceeds B k , since by (8.10) and 0.98C ε 3 > C B (check), The probability of failure is P N ≤ 0.98ε 2 k n = exp(−Ω(ε 2 k n)) = exp(−Ω(n 2/3 )). 8.7. Lower bound. As argued in the introduction of this section, for any k = o(n), the lower bound follows from the uniform case. Thus it is sufficient if we show the lower bound for k ≥ n 9/10 , which we do now.
Remark 8.3. With high probability, for every pair of vertices u and v in G = G − s − t, there is a u-v path in G of cost at most δ = 20n −1/6 that is edge-disjoint from P 1 , . . . , P n−1 .
Proof. The proof of Claim 8.2 showed that w.h.p., for all k in the claim's range (up to k = n−2), there is a cheap s-t path (of cost given by (8.11)) disjoint from P 1 , . . . , P k , because for a given pair of neighbours u, v of s and t, there is a u-v path in G that is edge-disjoint from these k paths and has cost at most 4ε k = 20n −1/6 def = δ (see (8.10)). The existence of a k + 1st s-t path limits k to n − 2 since after that there are no new neighbours u and v of s and t, but the rest of the argument extends to k = n − 1.
In particular, extending the definition (8.10) of B k and ε k to k = n − 1, the derivation of (8.24) extends without change and shows that the budget B n−1 covers the middle edges of all paths P 1 , . . . , P n−1 , and the robustness argument also extends and shows (8.25) to hold for k = n − 1. Since the failure probability in (8.25) is exponentially small, and there are fewer than n 2 pairs {u, v} in G , w.h.p. there is a cheap path (of cost ≤ δ) for every pair.
For the remainder of this section we assume that the high-probability conclusion of Remark 8.3 holds.
Let H s k be the weight of the heaviest edge incident to s used by the first k paths, and let L s k be the weight of the lightest edge incident to s not used by the first k paths. Define H t k and L t k likewise. We claim that for all k from 1 to n − 1, with δ = 20n −1/6 as in Remark 8.3, We argue by contradiction. Given k, let P i , i ≤ k, be the path using the edge of weight H s k . By Remark 8.3, we can construct an s-t path Q whose s-incident edge is the one of weight L s k , whose t-incident edge is the same as that of P i , and whose middle edges cost at most δ and are not used in P 1 , . . . , P n−1 . This path Q is cheaper than P i : its s-incident edge is cheaper by H s k − L s k > δ, its t-incident edge has the same cost, and its middle edges (costing at most δ) cost at most δ more than those of P i . Also, Q is edge-disjoint from the first i − 1 paths: its s-incident edge L s k is not used even by the first k paths, the middle edges are disjoint from those of all n − 1 paths, and its t-incident edge is that used by P i (so not used by a previous path). Thus, Q should have been chosen in preference to P i , a contradiction, establishing (8.26).
Trivially, H s k ≥ W s (k) . Thus, from (8.26), For k ≤ n−2, the edge of P k+1 incident to s costs at least L s k and the edge incident to t at least L t k . If P k+1 is not the single-edge path {s, t} these two edges are distinct, so that X k+1 ≥ L s k +L t k . If P k+1 is the single-edge path {s, t} then P k is not, and X k+1 ≥ X k ≥ L s k−1 + L t k−1 . Either way, by (8.27), Recall that we are concerned here with k ≥ n 9/10 . By Lemma 4.2, for all such k, and for any γ > 0, w.h.p. W (k) ≥ (1 − γ)µ k where µ k def = E W (k) . Since the exponential random variable is stochastically greater than the uniform, µ k > k/n = Ω(n −1/10 ), while δ = 20n −1/6 = o(µ k ). From (4.2) it is clear that µ k−1 ∼ µ k+1 (for any k = ω(1)), and we subsume the asymptotic error into the constant γ. Thus, from (8.28), for any γ > 0, w.h.p., for all k ≥ n 9/10 , completing the proof of the lower bound in Theorem 1.2.