A Randomized Rounding Algorithm for the Asymmetric Traveling Salesman Problem

Let V be a set of n vertices and let c : V × V → R + be a cost function. We assume the triangle inequality: c i,j ≤ c i,k +c k,j for all vertices i, j, k. The asymmetric traveling salesman problem (ATSP) is to solve min π v∈V c v,π(v) over all cyclic permutations π on V . A subgraph { (v, π(v)) : v ∈ V } is called a tour; we seek a minimum cost tour.

We will use the following standard notation. For U ⊆ V , define

For any vector x ∈ R V ×V + and F ⊆ V × V , we use the notation x(F ) = e∈F x e .

The Held-Karp linear programming relaxation of ATSP is as follows.

By standard shortcutting arguments, we may assume that x (v,w) ≤ 1 for all v, w, and that x(δ + ({v})) = 1 for all v.

Several polynomial-time algorithms [3] [9, pp. 125] [4] [2] are known for computing a tour whose cost is at most a factor O(log n) larger than the optimum. In addition, several proofs [10,8] are known showing that the integrality gap of the Held-Karp relaxation is O(log n). This note provides another such algorithm and another such proof of the integrality gap.

Let x be any feasible solution to this linear program. Since x is balanced at every vertex, this implies that x is Eulerian, i.e., x(δ + (U )) = x(δ -(U )) for all U ⊆ V . So x is a fractional solution for which all cuts are perfectly balanced. We now use x to construct an integral solution z for which all cuts are nearly-balanced, i.e., for each cut U V , z(δ

The first observation is that x has equivalent cut values to an undirected graph. Formally, for

Proof. Define y {v,w} = (x v,w + x w,v )/2 for all v, w. Then

as required.

We now apply a random sampling result of Karger [5]. For convenience, we reprove it here in our notation. For any undirected graph with minimum cut value c, Karger [5] shows that the number of cuts of value at most αc is less than n 2α . This result applies to the graph induced by y and hence, by Claim 1, also to x:

To round x, we must first scale it so that its minimum cut value is large. Let G be the directed, weighted, multigraph obtained from x by taking K := 100 ln n parallel copies of each edge, each of the same weight as in x. Let c i be the value of the i th cut, ordered such that K ≤ c 1 ≤ c 2 ≤ • • • . We will construct a directed multigraph H by taking each edge of G with probability proportional to its weight. The expected number of edges chosen by H in the i th cut is c i . Let p i be the probability that the actual number of edges chosen in the i th cut diverges from its expectation by more than an ǫ fraction. By a Chernoff bound, p i ≤ 2e -ǫ 2 c i /3 . We will ensure that no cut diverges significantly from its expectation by choosing ǫ appropriately and applying a union bound. Define ǫ = 1/10. Since c i ≥ K = 100 ln n, we have p i ≤ n -3 for all i. For the small cuts, we use the bound

For the large cuts, we use a different bound. Eq. ( 1) implies that c n 2α ≥ αK. Letting i = n 2α , we have c i ≥ K ln i/(2 ln n), and hence p i ≤ i -3/2 . Thus

Combining Eq. ( 2) and Eq. ( 3) shows that with probability 1 -O(1/n), no cut in H diverges from its expectation by more than an ǫ fraction. The expected cost of H is K • c T x, so a Chernoff bound again implies that the cost of H is O(log n) • c T x with high probability.

Let z ∈ Z V ×V + be the vector giving the total weight of the edges in the multigraph H. Assuming that no cut in H diverges significantly from its expectation, we have

The last inequality follows because ǫ < 1/3. Thus z is a nearly-balanced graph with high probability.

We now make z Eulerian by “patching it up” with another graph w. That is, we seek another vector w ∈ Z V ×V + such that z + w is connected and Eulerian -an integral, feasible solution to the Held-Karp relaxation of ATSP. Indeed, we show that a subgraph of z can be used to patch up z.

Consider the transshipment problem on V where each vertex v has demand b(v) := z(δ + (v))z(δ -(v)). Hoffman’s circulation theorem [7,Corollary 11.2f] implies that there exists a subgraph of z giving a feasible, integral transshipment for these demands iff the capacity of each cut is at least its demand:

This inequality is implied by Eq. ( 4), so the desired transshipment w exists, and its cost is at most c T z. Thus z + w gives a connected, Eulerian graph of cost at most 2 c T z, which is O(log n) • c T x, as argued above. By shortcutting, we obtain a tour of no worse cost. If x is an optimum solution of the linear program then the resulting tour is at most a factor O(log n) larger than the optimum tour. Consequently, the integrality gap of this linear program is at most O(log n).

We now show that the analysis of the algorithm given above is tight to within constant factorswe give an example where we must choose K to be Ω(log n) in the first step of the algorithm. This condition is necessary not only to ensure the nearly-balanced condition but also to ensure that H is connected.

Consider any extreme point x such that x a < 2 3 for every arc a ∈ A and let E be the support

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