A Constant-Factor Approximation for Directed Latency in Quasi-Polynomial Time

12/12/2019 ∙ by Zachary Friggstad, et al. ∙ University of Waterloo University of Alberta 0

We give the first constant-factor approximation for the Directed Latency problem in quasi-polynomial time. Here, the goal is to visit all nodes in an asymmetric metric with a single vehicle starting at a depot r to minimize the average time a node waits to be visited by the vehicle. The approximation guarantee is an improvement over the polynomial-time O(log n)-approximation [Friggstad, Salavatipour, Svitkina, 2013] and no better quasi-polynomial time approximation algorithm was known. To obtain this, we must extend a recent result showing the integrality gap of the Asymmetric TSP-Path LP relaxation is bounded by a constant [Köhne, Traub, and Vygen, 2019], which itself builds on the breakthrough result that the integrality gap for standard Asymmetric TSP is also a constant [Svensson, Tarnawsi, and Vegh, 2018]. We show the standard Asymmetric TSP-Path integrality gap is bounded by a constant even if the cut requirements of the LP relaxation are relaxed from x(δ^in(S)) ≥ 1 to x(δ^in(S)) ≥ρ for some constant 1/2 < ρ≤ 1. We also give a better approximation guarantee in the special case of Directed Latency in regret metrics where the goal is to find a path P minimize the average time a node v waits in excess of c_rv, i.e. 1/|V|·∑_v ∈ V (c_v(P)-c_rv).



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1 Introduction

We investigate the Directed Latency problem (), a vehicle routing problem where we are to route a single vehicle to serve a set of clients/nodes. Unlike the standard Traveling Salesman problem () where the goal is to minimize the length of the route, in the goal is to minimize the average time a node waits to be served. Formally, in we are given an asymemtric metric space where is a set of node nodes, is the depot node, and gives asymmetric metric distances over . That is, for any two nodes , for any node , and for any three nodes . Our goal is to find a Hamiltonian path starting at the depot to minimize where denotes the total cost of all edges on the subpath of . This sometimes called the Traveling Repairman problem.

Our main contribution is the first constant-factor approximation for in quasi-polynomial (i.e. time. A key technical contribution towards this is generalizing recent work by Köhne, Traub, and Vygen [11] to give constant-factor integrality gap bounds for a slight weakening of the standard LP relaxation for Asymmetric TSP-Path (). We also get explicit constants for the special case of in so-called regret metrics where the goal is to minimize when is a symmetric metric (i.e. ). That is, we want to minimize the average time each node waits in excess of their shortest-path distance from . Such an instance can be cast as special case of by using regret distances , which form an asymmetric metric.

The algorithm we present is based on a time-indexed linear programming (LP) relaxation, much like the approach taken in

[14] for the Minimum Latency problem in symmetric metrics. Roughly speaking, our approach uses variables for pairs where is a node to be visited and is the time they should be visited. Other variables indicate transitions between nodes at different times.

1.1 Related Work

Nagarajan and Ravi first studied and obtained an approximation guarantee of in time for any constant [13], which extends easily to an -approximation in quasi-polynomial time where (roughly speaking) is an upper bound on the integrality gap of the natural Held-Karp LP relaxation for / They also also showed is bounded by . Friggstad, Salavatipour, and Svitkina improved the approximation guarantee for and the upper bound on the integrality gap for to [7]. This is currently the best polynomial-time approximation for and no better quasi-polynomial time approximation was known before our work. If the metric is symmetric, constant-factor approximations are know. The first was given by Blum et al. [3], the best guarantee so far is a 3.59-approximation by Chaudhuri et al. [6].

Post and Swamy studied LP relaxations for the undirected minimum latency problem [14]. Using time-indexed LP relaxations, they obtain improved approximations for the multi-depot variant and also recover the 3.59-approximation for the single-vehicle version using an LP relaxation. We build off one of their LP relaxations in this work.

The integrality gap for has seen some improvements since [7]. In [9], it is shown the integrality gap is in fact . Recently, [11] shows the integrality gap is in fact . Specifically, they show the gap is at most where is the integrality gap for the Held-Karp relaxation for standard ATSP. Prior to this, Svensson, Tarnawski, and Vegh showed is bounded by a constant [15], though we remark that the current best bound on is very large. Currently, the best lower bound on is 2 [5].

1.2 Results and Techniques

Our main result is the following. Throughout, we let denote .

Theorem 1.1.

For some constant , there is a -approximation for running in time time.

To discuss this, we first introduce some notation. For a directed graph and some , we let , and . If the graph is clear from the context, we may omit the subscript . We often identify an asymmetric metric with the complete directed graph over nodes having edge costs for distinct . For a path and a node on , let be the cost of the subpath of .

We first scale the distances in the metric be polynomially-bounded integers. Standard scaling techniques allow us to do this.

Theorem 1.2.

For any constant , if there is an -approximation for instances of where each is a positive integer bounded by a polynomial in and where for nodes , then there is an -approximation for general instances of .

So we may assume all distances are integers bounded as such. Let and notice that is bounded by a polynomial in . Any Hamiltonian path in the metric has length at most , so all nodes in the optimum solution are visited by time . For brevity, let .

We begin with essentially the same time-indexed LP relaxation that was used in [14] for the undirected minimum latency problem, specifically (LP3) in their work. The variables in the time-indexed relaxation are the following: for and let indicate that we visit at time exactly , let indicate we finished traversing edge at time exactly . This is slightly different than [14] where they let indicate was traversed by time . Note, we omit Constraints (14) from [14], one can easily show they are implied by our slightly different approach.


It is easy to check that an optimal solution naturally corresponds to an integral solution to (LP-Latency) with the same cost as the latency of . The constraints admit an efficient separation oracle simply by checking for each and if the minimum cut has capacity at least when using capacities for the edges.

Our proof of Theorem 1.1 proceeds by bucketing clients based on their fractional latencies, finding low-cost paths for these buckets, and stitching these paths together to form our final path. Our advantage over [7] comes from the fact that we guess the nodes appearing at distances roughly along the optimum path , plus their exact visiting times along . We add constraints to (LP-Latency) to reflect these guesses. For each , consider the nodes that are at least, say, -visited before is visited: call this the bucket for . With a bit of modification, the restriction of (LP-Latency) to the times before is visited induces an LP solution with cost for the natural LP relaxation that covers all to an extent of at least 2/3. That is, we get a solution to the following LP relaxation for for .


The integrality gap of the case was proven to be constant in [11]. At this point, we need a stronger integrality gap bound.

Theorem 1.3.

For some absolute constant that is independent of , the integrality gap of () is at most .

In [7], it was shown that if then the integrality gap of () is unbounded even if we strengthen it to have an in-flow of 1 for each (but still have the relaxed cut constraints). As a side note, we also show the dependence on is asymptotically correct as approaches 1/2.

Theorem 1.4.

There is an instance of where the integrality gap of () on that instance is for every even if we strengthen the LP with constraints for each .

Returning to the idea behind the proof of Theorem 1.1, once we have these paths we must bound the cost of stitching the last node of to the first node after on . This is where guessing plays the most prominent role, we show that strengthening the LP with our guess ultimately implies this new edge used to stitch to has cost , as required.

Our final result is an improved approximation in the case that the metric is the regret metric of an undirected metric, which we simply call regret metrics.

Theorem 1.5.

The integrality gap of () in regret metrics is at most and we can find a path whose cost is at most times the value of an optimum LP solution.

We then work out an explicit constant for approximating in regret metrics.

Theorem 1.6.

There is a quasi-polynomial time 778-approximation for in regret metrics.

While this constant is large, it it considerably better than what we would obtain if we simply used Theorem 1.3 and the current-best bound on , which would lead to an approximation guarantee in the tens of thousands.

Outline of the Paper
Section 2 proves Theorem 1.1 and discusses how Theorem 1.6 would follow from Theorem 1.5. The scaling result itself (Theorem 1.2) is fairly standard, it’s proof is found in Appendix A. Section 3 proves Theorem 1.3. Finally, Theorem 1.6 is proven in Section 5.

2 An -Approximation in Quasi-Polynomial Time

Recall, by Theorem 1.2, we may assume distances are integers bounded by a polynomial in and that for distinct nodes . We also let , which is an upper bound on the cost of any Hamiltonian path. Our algorithm starts by guessing the last node visited by an optimal solution at some time in the interval111One can show the geometric factor of 2 is optimal for our analysis, so we fix it now. (if any) and its exact distance for each . Let if no such node exists for this interval. For any , we then know that no node is visited at any time in if and, if , we also know no node is visited at a time in the interval so we mark these times as forbidden. Let be admissible buckets corresponding to intervals where the optimum visits at least one node. Let be a parameter we optimize later.

Algorithm 1 (Directed Latency: -approximation in time).


Input: asymmetric metric with integer distances at most .

Output: an -rooted path

  1. [label=D0., ref=D0, topsep=0ex, itemsep=0.5ex, leftmargin=*]

  2. For every choice (guess) of for each and for each such where , perform the following steps. Let be the forbidden times for this guess and the admissible buckets.

    1. [label*=0., ref=1.0, topsep=0ex, itemsep=0ex, leftmargin=*]

    2. Get an optimal extreme point solution to the (LP-Latency) strengthened with the following additional constraints: 1) for each and 2) for each and . If the LP is infeasible, abort this guess of .

    3. For each , let be the minimum time such that . For , let .

    4. For each , use the LP-based approximation from Theorem 1.3 to get an path spanning .

    5. Let be the path obtained by concatenating the paths in increasing order of , and shortcutting past repeat occurrences of .

  3. Return the best path found for all guesses where the strengthening of (LP-Latency) for that guess was feasible.


Let be an optimum solution and consider the iteration where is consistent with . Let be an optimum LP solution for the strengthening of (LP-Latency) by the constraints in step (1a). Clearly this strengthened LP is feasible and the value of the solution is at most , the latency of .

Note for each that is well-defined by Constraints (1). Ultimately, we will show the path visits each by time . We begin by showing this suffices to get a constant-factor approximation.

Lemma 2.1.

Let be a path and be such that for each . Then the latency of is at most .


Fix some . By our choice if , which yields

So, . ∎

2.1 Bounding the Latency of

In the remainder of the proof it is convenient to view a “time-expanded” graph . The nodes are pairs with and and an edge connects to if . Observe is acyclic. We can then view as assigning values to edges of : the edge has value and cost .

The constraints of (LP-Latency) mean constitutes one unit of -preflow in . Let be the greatest index in . Considering the constraints from (1a), we see and for all . Thus, must be a flow with value 1 in ending at . Since the support of the flow is acyclic in and since one unit of flow passes through every node in for each , no flow skips past node . That is, no edge in supports any -flow if for some , nor does any edge support any -flow if yet or yet for some .

We start by showing we can compute low-cost paths covering each bucket. Before doing so, we recall a famous splitting-off theorem by Mader. The following is a slight specialization of one such result.

Theorem 2.2 (Mader [12]).

Let be an Eulerian, directed graph with, perhaps, parallel edges such that the connectivity for every is at least . Then for every there is some such that in the graph , the connectivity for every remains at least .

For brevity, let denote the integrality gap of ().

Lemma 2.3.

For each , we can compute a Hamiltonian path in with cost in polynomial time.


We first show there is a feasible LP solution for in with cost . Let

be a vector over edges of the metric given by

for . As discussed above, the truncation of to times constitutes one unit of flow from to in , so is then one unit of flow in the metric. Further, since the cost of an edge is in , the cost of this flow is, in fact, exactly which is at most .

Next we verify for each with . Consider some . Constraint (2), the fact that , and the fact that for shows . Since is an flow and , then flow conservation shows .

Much like in [1] for the Prize-Collecting TSP-Path problem, one can use a standard splitting off result from [12] to shortcut past nodes not in to get solution for () for in the graph (with start node and end node ), also with cost at most . That is, we may assume is rational as is a rational vector (being part of an extreme point of an LP with rational coefficients). Let be an integer such that is integral. Consider the graph with nodes where is a new node. The edges of consist of copies of edge for each , and edges from to and also from to (each having cost 0). Note the connectivity for each is at least .

For each , we iteratively perform the splitting off procedure from Theorem 2.2 for . The total cost of the edges does not increase by the triangle inequality (note the edges that are removed and added all lie in the metric over ), and the connectivity remains at least for each . After doing this for each , we are left with a multigraph of total edge cost cost no more than the total cost of all edges in . Further, if we remove all and edges, we still get the connectivity to any other is at least . If denotes the number of copies of edges in this new graph, setting for each yields a feasible LP solution for () in the metric graph over (with start node and end node ) with cost at most . Note that we do not actually need to perform this step in our algorithm, this analysis is simply proving there is a low-cost solution to .

So, the optimal solution to () in (starting at and ending at ) has value at most . By Theorem (1.3), we can then efficiently find a Hamiltonian path in with cost at most . ∎

Lemma 2.4.

Let and be two paths constructed in Step (1c) for consecutive indices . Let be the first node on after and recall is the last node of . Then .


Note that means . So by definition of . All units of -flow in the acyclic graph pass through and also through . So the restriction of to edges in with constitutes one unit of flow that supports . Therefore, a path decomposition of this restriction of includes on some path. Any such path has cost exactly . By the triangle inequality, . ∎

Next, we bound the latency of each along the final obtained by concatenating the paths for increasing indices and shortcutting past all but the first occurrence of .

Lemma 2.5.

for any .


Consider any and say it lies on . To reach along , we traverse paths for plus the “stitching” edges for consecutive indices , . By Lemma (2.3) and Lemma (2.4), the latency of along can be bounded by . ∎

Set and note Theorem 1.3 implies is bounded by a constant. So by The proof of Theorem 1.1 then follows readily from Lemmas (2.1) and (2.5) and the fact that is bounded by a polynomial in .

The integrality gap bound in [15] for ATSP is very large. Using our approach (even with a better ) still produces an approximation ratio in the tens of thousands using our framework. We give an improved bound for regret metrics below.

Proof of Theorem (1.5).

Choosing and using the integrality gap bound from Theorem 1.5 yields in this regret metrics. Then using Lemma 2.5 and choosing sufficiently small in Theorem 1.2 yields a 778-approximation. ∎

3 Bounding the Integrality Gap of ()

Consider nodes with two distinguised and asymmetric metric distances between points of . We consider () for the Asymmetric TSP Path problem where the goal is to find the cheapest Hamiltonian path. As mentioned earlier, the integrality gap is unbounded if [7], so we focus on the case . As in [11], we start withthe dual of ().


Naturally, our proof borrows many steps from Köhne, Traub, and Vygen [11] but there are a number of significant differences.

For a vector over the edges of the directed metric (when viewed as a complete, directed graph), let . Similarly, for a vector over cuts of the metric let . From now on, we focus on the graph . The proofs of Propositions 3.1, 3.2, and 3.4 are very similar to proofs in [11] and are omitted or just sketched in this paper.

Proposition 3.1.

Given any optimal dual solution , one can find an optimal dual solution with being laminar in polynomial time.

In other words, we can modify to be laminar without changing using efficient uncrossing techniques. The proof is exactly the same as the proof in [11] essentially because the set of feasible solutions to () does not change if we select different .

The next proposition is almost identical to one in [11], but we omit the case in the statement. In fact, the result may not be true for this case , we handle that separately below.

Proposition 3.2.

Let be an optimum primal solution and let and . For any with , any topological ordering of the strongly connected components of satisfies:

  • ,

  • , and

  • for any .

We sketch the proof of Proposition 3.2 so the reader is assured it holds, though the proof is essentially the same.

Proof sketch.

Because is a tight set, . Further, . All edges in entering must lie in because is the first node in the topological ordering. Thus, , so equality must hold throughout and as we are working in the support of . A similar statement shows .

For we note simply because the are topologically ordered. Inductively, we have and each edge in is already proven to lie in for some . So we see and, thus,

So, again, equality must hold throughout. ∎

We use a different observation to address the case that was omitted from Proposition 3.2.

Proposition 3.3.

In any topological ordering of the strongly connected components of , for each there is some edge with .


This is easy for and . For example, we have and all edges from lie in . Thus, so the cheapest edge in has cost at most . We finish by observing as . A similar argument works for , so we now assume .

We quickly introduce notation. For an index let and . Let denote for . With this notation, let , and . We have as is the disjoint union of the sets defining ). On the other hand, and . Therefore, so .

So the cheapest edge has

Proposition 3.4.

Let be the support graph of an optimum solution to () and an optimum dual with laminar. For any and any with being reachable from in , there is a path in that crosses each set at most twice for .

Again, the proof is the same as that in [11] which only relies on Proposition 3.2 for (i.e. not on the case that we omitted from the proposition in our setting). We sketch the argument briefly to ensure the reader this still holds with the omission of in the statement of Proposition 3.2.


Consider any path contained in . Suppose is maximal among all such sets where re-enters after it exits . Let be the first node of in and the last node of in (it could be or ). Inductively, replace the portion of with an path in that enters and leaves every set at most once for . Repeat for all such maximal . ∎

3.1 Constructing the Path

Let denote the optimum solution value to (). Recall we let denote an upper bound on the integrality gap of the standard Held-Karp relaxation for ATSP. We will prove the following lemma later.

Lemma 3.5.

An optimal dual solution with being laminar and can be computed in polynomial time.

Using this, we now turn to the main result of this section. Note, we are choosing simplicity in presentation over optimizing the constants in the guarantee.

Proof of Theorem 1.3.

Complementary slackness ensures every satisfies . Consider the edge support graph . Modify to get an ATSP instance by adding a new node and edges with cost and