A primal-dual approximation algorithm for the Asymmetric Prize-Collecting TSP

Abstract

We present a primal-dual ⌈log(n)⌉-approximation algorithm for the version of the asymmetric prize collecting traveling salesman problem, where the objective is to find a directed tour that visits a subset of vertices such that the length of the tour plus the sum of penalties associated with vertices not in the tour is as small as possible. The previous algorithm for the problem (V.H. Nguyen and T.T Nguyen in Int. J. Math. Oper. Res. 4(3):294–301, 2012) which is not combinatorial, is based on the Held-Karp relaxation and heuristic methods such as the Frieze et al.’s heuristic (Frieze et al. in Networks 12:23–39, 1982) or the recent Asadpour et al.’s heuristic for the ATSP (Asadpour et al. in 21st ACM-SIAM symposium on discrete algorithms, 2010). Depending on which of the two heuristics is used, it gives respectively 1+⌈log(n)⌉ and $3+ 8\frac{\log(n)}{\log(\log(n))}$ as an approximation ratio. Our algorithm achieves an approximation ratio of ⌈log(n)⌉ which is weaker than $3+ 8\frac{\log(n)}{\log(\log(n))}$ but represents the first combinatorial approximation algorithm for the Asymmetric Prize-Collecting TSP.

Appendix

In this section, we give arguments showing that we can always obtain a nonnegative optimal dual solution for the minimum cost assignment problem provided that the vector cost is nonnegative. We give a copy here of the assignment problem in Sect. 3.4.1.

(19)

(20)

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The cost $\bar{c}_{e}$ is supposed to be nonnegative for all e∈A _B .

Suppose that we solve this problem by the Hungarian algorithm and the tableau on which the Hungarian algorithm is executed has the vertices in B ⁺ as its rows and those in B ⁻ as it’s columns. Suppose that the u variables correspond to the rows of the tableau (hence associated with the vertices in B ⁺) and the v variables correspond to its columns (hence associated with the vertices in B ⁻). Each iteration of the Hungarian algorithm consists of the two following steps:

Step 1.:

Augment dual variables such that the reduced cost of every arc is decreased by a some positive value γ.

Step 2.:

Find a minimum cover of the tableau which is a subsets of rows and columns. Decrease the dual variables corresponding to the rows and columns in the cover by γ to maintain the non-negativity of the reduced costs.

Note that, we can do the job in Step 1 by augment all components of u by γ. In Step 2, the value of some components of u may be decreased but only by γ. Hence, we can assume that u is always nonnegative. At the end of the algorithm, we obtain a dual optimal solution (u ^∗,v ^∗) where u ^∗≥0 and some components of v ^∗ may be negative. We reiterate the Algorithm in Table 1 while there exists negative components of v ^∗.

Table 1 Algorithm making u ^∗ and v ^∗ nonnegative

Lemma 4

Y′⊆Y.

Proof

As X is the set of the rows containing all 0s at the columns in Y′, the rows assigned to the columns in Y′ in the optimal solution should be among the rows in X, i.e. the columns assigned to the rows in X should include those of Y′. Hence, Y′⊆Y. □

Lemma 5

The dual objective does not change after each iteration.

Proof

It suffice to observe that |X|=|Y|. □

Lemma 6

At the beginning of every iteration, the components of u ^∗ in X are always strictly positive. And at the end, none of them becomes negative.

Proof

Let $c^{r}_{ij}$ denote the reduced cost of the element in the row i and the column j for all i∈B ⁺ and for all j∈B ⁻. Hence $c^{r}_{ij}=\bar{c}_{ij}-u^{*}_{i}-v^{*}_{j}$. Suppose that at the beginning of some iteration, the dual variable $u^{*}_{i}$ of some row i∈X is non-positive. By hypothesis, there exists a column k∈Y′ such that the reduced cost $c^{r}_{ik}=0$. As $c^{r}_{ik}=\bar{c}_{ik}-u^{*}_{i}-v^{*}_{k}=0$, we have $\bar{c}_{ik}=u^{*}_{i}+v^{*}_{k}$. And as $v^{*}_{k}$ is strictly negative and the original cost $\bar{c}_{ik}$ is nonnegative, that leads to a contradiction. Hence, at the beginning of every iteration, the components of u ^∗ in X are always strictly positive.

Now suppose at the end of some iteration, the dual variable $u^{*}_{i}$ of some row i∈X becomes strictly negative. Let k∈Y′ be any column in Y′ such that $c^{r}_{ik}=0$. Hence $\bar{c}_{ik}=u^{*}_{i}+v^{*}_{k}$. But by the Algorithm in Table 1, the value of $v^{*}_{k}$ is at most 0 at the end of the iteration, this implies that $\bar{c}_{ik}=u^{*}_{i}+v^{*}_{k}<0$. Contradiction. □

Theorem 2

The Algorithm in Table 1 returns a nonnegative dual optimal solution for (A) in O(n ²) time.

Proof

In each iteration of the algorithm, the computation of α and the modification of u ^∗ and v ^∗ cost at most O(n). By the algorithm, after each iteration, at least one more component of v ^∗ becomes nonnegative, thus there are at most n iterations. In addition, by Lemma 6, u ^∗ is always nonnegative and by Lemma 5, the objective value remains the same. Hence, at the end of the algorithm, u ^∗ and v ^∗ are nonnegative and dual optimal. □