A Fast $(2 + 2/7)$-Approximation Algorithm for Capacitated Cycle Covering

Abstract

We consider the capacitated cycle covering problem: given an undirected, complete graph G with metric edge lengths and demands on the vertices, we want to cover the vertices with vertex-disjoint cycles, each serving a demand of at most one. The objective is to minimize a linear combination of the total length and the number of cycles. This problem is closely related to the capacitated vehicle routing problem (CVRP) and other cycle cover problems such as min-max cycle cover and bounded cycle cover. We show that a greedy algorithm followed by a post-processing step yields a $(2 + 2/7)$-approximation for this problem by comparing the solution to a polymatroid relaxation. We also show that the analysis of our algorithm is tight and provide a $2 + \epsilon$ lower bound for the relaxation.

Appendices

A Sketch of the Proof of Lemma 1

Pick an arbitrary root r for T. Then we perform the following splitting-off procedure (similar to Algorithm A in [15]).

As long as the vertex set V(T) of the tree T remains large, we iterate the following. Let v be maximally far away from r with the property that $V(T_v)$ is large, where $T_v$ is the subtree rooted at v. Let $w_1, \ldots , w_l$ be the children of v. Since $b(V(T_v)) = b(v) + \sum _{i = 1}^{l}{b(V(T_{w_l}))}$, we must have that $b(v) \ge 1/2$ or there exists a set $N \subseteq \{1,\ldots ,l\}$ with $\sum _{i \in N}{b(V(T_{w_i}))} \in [1/2, 1]$. In the first case we split off a singleton tree $(\{v\},\emptyset )$ covering the vertex v and replace v in T by a Steiner vertex, i.e. we set its demand to zero. In the second case we split off a tree covering all vertices contained in the subtrees $T_{w_i}$ for $i \in N$; the Steiner tree for this set of terminals contains v as a Steiner vertex and for $i \in N$ contains the edge $\{v,w_i\}$ and the subtree $T_{w_i}$. Thus we then remove these subtrees from T.

Let $T_1, \ldots , T_{k - 1}$ be the Steiner trees split off during this algorithm and let $T_k$ be the remaining tree. Moreover, let $R_1, \ldots , R_k$ be the respective terminal sets of these Steiner trees. Then we know that $b(R_i) \ge 1/2$ for all $i \le k - 2$ and $b(R_{k - 1}) + b(R_k) \ge 1$. Thus $2 b(V) = 2 \sum _{i = 1}^{k}{b(R_i)} \ge k.$

B Sketch of the Proof of Lemma 4

The key part is showing that r is indeed submodular. For this let $F' \subseteq F \subseteq E$ be arbitrary and $e \in E \setminus F$. We need to show that

$$ r(F' \cup \{e\}) - r(F') \ge r(F \cup \{e\}) - r(F). $$

Let $A_1, A_2 \in \mathcal {C}(F)$ be the two components of F joined by e. Moreover, let $A'_1, A'_2 \in \mathcal {C}(F')$ be the same for $F'$. We can assume that $A'_1 \subseteq A_1$ and $A'_2 \subseteq A_2$ since $F' \subseteq F$. Then one can show that

$$\begin{aligned} r(F \cup \{e\}) - r(F)&= \max \{1, b(A_1)\} + \max \{1, b(A_2)\} - \max \{1, b(A_1 \cup A_2)\} \end{aligned}$$

and

$$\begin{aligned} r(F' \cup \{e\}) - r(F')&= \max \{1, b(A'_1)\} + \max \{1, b(A'_2)\} - \max \{1, b(A'_1 \cup A'_2)\}. \end{aligned}$$

So the submodularity of r reduces to observation that the expression

$$ \max \{1, x\} + \max \{1, y\} - \max \{1, x + y\} $$

is non-increasing in x and y for $x,y\ge 0$.

C Sketch of the Proof of Lemma 8

Note that the partition $\mathcal {C}$ in iteration i of Algorithm 2 is the same as in iteration i of Algorithm 1; here we assume wlog. that the edges are sorted in the same order in both algorithms. Hence, we apply lines 7–10 of Algorithm 2 precisely for those edges $e_i$ for which we set $x_{e_i}$ in line 7 of Algorithm 1. We define $E'$ as in Sect. 4 and also define $C^u_e\subseteq V$ for an edge $e\in E'$ and a vertex $u\in e$ as before.

Let x be the output of Algorithm 1. It is easy to verify that the choice of which edges we include in the forest F in lines 7–10 of Algorithm 2 is such that we minimize

$$\begin{aligned} \sum _{e\in F} 2\cdot \ell (e) + \sum _{e\in E\setminus F} \sum _{u\in e} \gamma \cdot \max \{1 - 2 b(C^{u}_e), 0\} \end{aligned}$$

(8)

among all set F with $\{ e\in E : x_e =1\} \subseteq F \subseteq \{e\in E : x_e > 0\}$. By Lemma 7 there exists such an edge set F where (8) is at most $(2+2/7)\cdot \ell (x) + 2/7 \cdot \gamma \cdot (|V| - x(E))$. Hence, also the edge set F computed by Algorithm 2 fulfills this bound. By Lemma 6 this implies the claimed bound (7).

D Proof of Theorem 3

For $n \in \mathbb {N}$ with $n\ge 4$, let $G = (V,E)$ be the complete graph on the vertices $v_1, \ldots , v_n$ with the metric $\ell$ on V given by $\ell (v_i, v_j) := \frac{1}{4} |i - j|,$ i.e. $(G, \ell )$ is the metric closure of a path. Assign uniform demands of $b(v) := 1/4$ to every vertex v and let $\gamma := 1$. Then we observe that $\mathrm {LP}(G, \ell , b, \gamma ) = \frac{7}{16} n$. See Fig. 2.

But now consider what Algorithm 2 does on this instance. Assume that the edges are sorted such that $e_i = \{v_i, v_{i + 1}\}$ for all $i \in \{1, \ldots , n - 1\}$. The algorithm will then buy the edges $e_1$ to $e_3$. But it will not buy any other edge as

$$\textstyle \gamma \max \{1 - 2 b(v_{i + 1}), 0\} = \frac{1}{2} = 2 \ell (\{v_i, v_{i + 1}\}) $$

for all $i \in \{1, \ldots , n - 1\}$. So the condition in line 9 is never satisfied except for the first three iterations of the loop. Hence, any CCCP solution which is “contained” in the connected components of F (i.e. it does not contain a cycle $C_i$ where $V(C_i)$ is not connected in (V, F)), must contain at least $n - 4$ singleton cycles.

Fig. 2.

An optimum solution to the tree cover LP (3) for instance from the proof of Theorem 3 for $n=12$. For every solid edge e we have $x_e = 1$ and for every dotted edge e we have $x_e = 3 / 4$.

Finally, we conclude that any such CCCP solution has a cost of at least

$$\begin{aligned} n - 4\ =\ \frac{n - 4}{\frac{7}{16} n} \mathrm {LP}\ \ge \ \Bigl (\frac{16}{7} - \epsilon \Bigr ) \mathrm {LP}\ =\ \Bigl (2 + \frac{2}{7} - \epsilon \Bigr ) \mathrm {LP}\end{aligned}$$

for n large enough.