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Approximation Algorithms for Solving the Constrained Arc Routing Problem
in Mixed Graphs
Honglin Ding, Jianping Li, Ko-Wei Lih
PII: S0377-2217(14)00381-6
DOI: http://dx.doi.org/10.1016/j.ejor.2014.04.039
Reference: EOR 12290
To appear in: European Journal of Operational Research
Received Date: 22 May 2013
Accepted Date: 26 April 2014
Please cite this article as: Ding, H., Li, J., Lih, K-W., Approximation Algorithms for Solving the Constrained Arc
Routing Problem in Mixed Graphs, European Journal of Operational Research (2014), doi: http://dx.doi.org/
10.1016/j.ejor.2014.04.039
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Approximation Algorithms for Solving the
Constrained Arc Routing Problem
in Mixed Graphs
Honglin Ding
Department of Mathematics
Yunnan University
Kunming 650091, P.R. China
Email:[email protected]
Jianping Li∗
Department of Mathematics
Yunnan University
Kunming 650091, P.R. China
Email:[email protected]
Ko-Wei Lih
Institute of Mathematics
Academia Sinica
Taipei 10617, Taiwan
Email:[email protected]
May 1, 2014
Abstract
Given a mixed graph G with vertex set V , let E and A denote the sets of edges
and arcs, respectively. We use Q+ and Z+ to denote the sets of positive rational
numbers and positive integers, respectively. For any connected mixed graph G =
(V,E ∪A; w; l, u) with a length function w : E ∪A → Q+ and two integer functions
∗Corresponding author, Tel: (0086-871) 65033701, Fax: (0086-871) 65033700
1
l, u : E ∪ A → Z+ satisfying l(e) ≤ u(e) for each e ∈ E ∪ A, we are asked to
determine a minimum length tour T traversing each e ∈ E ∪ A at least l(e) and at
most u(e) times. This new constrained arc routing problem generalizes the mixed
Chinese postman problem. Let n = |V | and m = |E ∪ A| denote the number of
vertices and edges (including arcs), respectively. Using network flow techniques,
we design a (1 + 1/l0)-approximation algorithm in time O(n2m3 log n) to solve this
constrained arc routing problem such that l(e) < u(e) holds for each edge e ∈ E,
l(e) ≤ u(e) holds for each arc e ∈ A and l0 = min{l(e) | e ∈ E}. In addition, we
present two optimal combinatorial algorithms in times O(n3) and O(nm2 log n) to
solve this problem for the cases A = ∅ and E = ∅, respectively.
Keywords: Combinatorial optimization; arc routing; lower/upper demand bound;
approximation algorithm; combinatorial algorithm.
1 Introduction
In most arc routing problems, the objective is to determine a least-cost traversal of a
set of specified edges or arcs of a mixed graph. Such problems occur in a variety of
practical contexts and have attracted attention from both mathematicians and operations
researchers. The earliest arc routing problem was the Konigsberg seven bridges problem,
which is to determine whether there exists a closed tour traversing exactly once each
of the seven bridges on the Pregel river through Konigsberg. After Euler solved the
Konigsberg seven bridges problem in the negative, he was concerned almost exclusively
with the existence of a tour. The question of determining such a tour was addressed and
solved by Hierholzer and Wiener [22]. The reader is referred to Fleischner [10] for an
interesting account of this problem.
Another closely related well-known problem is the so-called Chinese postman problem
(CPP) posed by Guan [21]. In contrast to the Konigsberg seven bridges problem, Guan
addressed a natural question that is to determine a minimum length tour traversing
each edge of a weighted graph at least once. The CPP is a version of the arc routing
problem specialized to graphs. Using Edmonds’ combinatorial algorithm for solving the
maximum weighted matching problem [6], Edmonds and Johnson [7] designed an optimal
2
combinatorial algorithm to solve the CPP.
Closely related to the CPP is the directed Chinese postman problem (DCPP), which
is a version of the arc routing problem specialized to directed graphs, or simply digraphs.
Using a combinatorial algorithm in [12] for solving the transportation problem, Edmonds
and Johnson [7] also designed another optimal combinatorial algorithm to solve the DCPP.
The mixed Chinese postman problem (MCPP), listed in Garey and Johnson [14] as
the problem ND25, is a common generalization of the CPP and the DCPP. In the MCPP,
an input graph may contain both undirected and directed edges. A directed edge is also
referred to as an arc. The objective is to determine a tour that traverses at least once
every edge in either direction and every arc in the direction of that arc. Even though
the CPP and the DCPP are both polynomially solvable [7], Papadimitriou [29] showed
that the MCPP becomes NP-complete. Because a number of practical applications, such
as mail delivery, snow removal, and trash pick-up, can be modeled as instances of the
MCPP, it is important to design good approximation algorithms for this problem.
Quite a few papers about the MCPP have appeared in the literature over the past
four decades. Edmonds and Johnson [7] presented the first approximation algorithm for
the MCPP. Christofides et al. [3] designed an exact algorithm for the MCPP based on
a branch-and-bound algorithm using Lagrangian bounds. Afterwards, Frederickson [13]
showed that the approximation algorithm due to Edmonds and Johnson [7] is indeed a
2-approximation algorithm to solve the MCPP. Frederickson provided a mixed strategy
algorithm for the MCPP that is a 5/3-approximation algorithm. As far as we know,
the best approximation algorithm to solve the MCPP is a 3/2-approximation algorithm
due to Raghavachari and Veerasamy [31]. Comprehensive surveys, latest advances and
new challenges on the MCPP are available in [5, 19, 34]. Integer and linear programming
formulations of the postman problems have generated great interest in recent years [26, 32].
Ralphs [32] showed that a linear relaxation of the MCPP has optimal solutions that are
half-integral. One can use this fact to derive a 2-approximation algorithm for the MCPP.
It is interesting to note that the implementation of Nobert and Picard [26] has been
used for scheduling snow removal in Montreal. Corberan et al. [4] presented a heuristic
algorithm for the MCPP, using greedy randomized adaptive search procedure techniques.
Goel and Gruhn [16] recently considered a general vehicle routing problem (GVRP) that
3
generalizes the MCPP, where the GVRP is a rich vehicle routing problem incorporating
various complexities found in real-life applications. Other interesting materials related to
the GVRP are referred to [15, 23].
In the literature, one may either traverse some required edges or arcs exactly once in a
minimum length feasible tour of a mixed graph or show the non-existence of such a tour.
This becomes a variant of the MCPP. In order to widen the scope of possible applications
for arc routing models, the times that an edge or arc is traversed should be given a
reasonable range rather limited to at least once. We should consider a generalization of
the MCPP for which some required edges or arcs would be traversed at least t1 and at
most t2 times (1 ≤ t1 ≤ t2) in a tour. This new model generalizes the above-mentioned
variant of the MCPP. In this paper, we consider the constrained arc routing problem in
connected mixed graphs (CARM) defined as follows. Let G = (V, E ∪ A; w; l, u) be a
connected mixed graph defined on vertex set V such that E is the set of all edges and A is
the set of all arcs. Let n = |V | and m = |E ∪A| denote the number of vertices and edges
(including arcs), respectively. Let Q+ and Z+ denote the set of positive rational numbers
and the set of positive integers, respectively. For a length function w : E ∪ A → Q+ and
two integer functions l, u : E ∪ A → Z+ satisfying l(e) ≤ u(e) for each e ∈ E ∪ A, we
are asked to determine a tour T traversing each e ∈ E ∪ A at least l(e) and at most u(e)
times such that the objective is to minimize the total length w(T ) =∑
e∈E∪A t(e)w(e),
where t(e) denotes the number of times that the edge or arc e is traversed by T .
There are two special cases to note. (1) If l(e) = u(e) = 1 for each edge e of the
required subset of E ∪ A, then the CARM problem for this version becomes the above-
mentioned variant; (2) If l(e) = 1 and u(e) = +∞ for each edge e ∈ E∪A, then the CARM
problem for this version specializes to the MCPP. It follows that the CARM problem is
also NP-complete.
A capacitated Chinese postman problem (CCPP) was posed by Golden and Wong [20]
as follows. Consider a mixed graph G = (V, E ∪ A; q; c). Vertex v1 represents a depot
where identical vehicles of capacity L are based. Each e ∈ E ∪ A has a positive demand
q(e) and a cost c(e) and can be traversed any number of times by any number of vehicles.
A vehicle route is said to be feasible if it starts and ends at the depot v1 and the total
demand of the edges and arcs traversed by the vehicle does not exceed L. The CCPP looks
4
for a set of feasible vehicle routes of least total cost such that each edge or arc appears in
at least one route and is traversed by exactly one vehicle. Many authors [2, 19, 25, 35]
studied the capacitated arc routing problem (CARP) that is an extension of the CCPP.
Recently, Lysgaard and Wohlk [24] also considered a variation of the CARP where the
objective is to minimize the sum of arrival times at the customers, for a fixed starting
time of each route.
It is easy to see that the CCPP restricted to the case of one vehicle of capacity
L ≥ ∑e∈E∪A q(e) becomes the MCPP. Christofides et al. [3] proposed a heuristic procedure
for obtaining a reasonable solution to the CCPP. Golden and Wong [20] showed that even
finding a 3/2-approximation algorithm to solve the CCPP remains NP-complete.
We know that the CARM problem and the CCPP generalize the MCPP, but each of
them does not follow from the other. Since the CCPP can be formulated as a set covering
problem [18], Theorem 4.4 in Feige [9] implies that the CCPP would not be approximated
within (1 − ε) log n unless NP ⊆ TIME(nO(log log n)), where ε ∈ (0, 1). However, the
CARM problem can be approximated within a constant performance ratio as shown in
the following text. We will show that two special versions of the CARM problem can be
optimally solved in polynomial times.
The remainder of this paper is organized as follows. In Section 2, we present ter-
minology for the description of our algorithms, and then establish some key lemmas to
ensure the correctness of our algorithms. In Section 3, we study the CARM problem
that can be approximated within performance ratio 1 + 1/l0 and in time O(n2m3 log n),
where l(e) < u(e) holds for each edge e ∈ E, l(e) ≤ u(e) holds for each arc e ∈ A
and l0 = min{l(e) | e ∈ E}. In Section 4, we study the CARM problem for the case
A = ∅ that can be solved optimally by our combinatorial algorithm in time O(n3). In
Section 5, we study the CARM problem for the case E = ∅ that can be solved optimally
by our combinatorial algorithm in time O(nm2 log n). In Section 6, we summarize our
conclusions.
5
2 Terminology and Key Lemmas
A mixed graph G = (V, E ∪A) contains both edges (unordered pairs of vertices in E) and
arcs (ordered pairs of vertices in A). A mixed graph G is called a multigraph if G contains
parallel edges or arcs, but no self-loops. Parallel edges (or arcs) are called multi-edges (or
multi-arcs). In this case, the set E (or A) is a multiset of edges (or arcs). A mixed graph
G is called a simple mixed graph if G contains neither self-loops nor parallel edges or arcs,
but G may have an edge and an arc with the same end-vertices. The term “edge” is also
used to refer both edge and arc of a mixed graph. Thus, an arc is regarded as a directed
edge. A mixed graph G = (V,E ∪ A) is said to be an undirected graph (or simply, graph)
if A = ∅ and a directed graph (or simply, digraph) if E = ∅, respectively.
In the sequel, G always denotes a mixed graph. A walk in G is an alternating sequence
W = x1e1x2e2x3 · · ·xk−1ek−1xk of vertices xi and edges or arcs ej from G such that either
the two end-vertices of the edge ei are xi and xi+1 or the tail of the arc ei is xi and the
head of the arc ei is xi+1 for every 1 ≤ i ≤ k − 1. This means that this walk W traverses
every edge ei in either direction and every arc ei in the direction of that arc. We say that
W is a walk from x1 to xk or an (x1, xk)-walk. A walk W is closed if x1 = xk, and we
always refer such a closed walk as a tour or circuit.
If the vertices of a walk W are distinct, then W is called as a path. If the vertices
x1, x2, . . . , xk−1 are distinct and x1 = xk, where k ≥ 3, W is called as a cycle. Two paths
P1 and P2 are said to be vertex-disjoint (or edge-disjoint) if their vertex sets (or edge sets)
are disjoint, respectively. Similarly, we can define two directed paths being arc-disjoint.
A mixed graph G = (V, E ∪A) is connected if, for any two vertices x and y in G, there
exists a path Px,y from x to y, where each edge e = uv on Px,y may be traversed in either
direction of this edge, but each arc e = (u, v) on Px,y may only be traversed from u to v.
In addition, when G is a (mixed) weighted graph and Px,y is a path from x to y in G, we
denote w(Px,y) to be the sum of weights of all edges and arcs in the path Px,y, and we
refer dG(x, y) as the weight of a shortest path from x to y, i.e., dG(x, y) = min{w(Px,y)
| Px,y is a path from x to y in G}. On the other hand, a digraph D = (V, A) is strongly
connected if, for any two vertices x and y in D, there exists a directed path Px,y from x
to y and at the same time a directed path P ′y,x from y to x.
6
For any vertex v in a mixed graph G = (V, E ∪A), the degree of the vertex v, denoted
by dG(v) or d(v), is the number of edges and arcs that are incident to v. For any vertex
v in a mixed graph G = (V,E ∪ A), the vertex v is even if its degree dG(v) is even,
otherwise the vertex v is odd. For any vertex v in a digraph D = (V,A), denote δ+D(v)
the set of arcs leaving v and δ−D(v) the set of arcs entering v, respectively. The out-degree
of a vertex v, denoted by d+D(v), is the number of arcs that have v as their common tail,
and the in-degree of a vertex v, denoted by d−D(v), is the number of arcs that have v as
their common head. Equivalently, d+D(v) = |δ+
D(v)| and d−D(v) = |δ−D(v)|, then the number
d+D(v) + d−
D(v) is the degree of a vertex v.
In a mixed graph G = (V,E∪A), an Eulerian tour is a closed walk that starts at some
vertex v0, traverses each edge or arc exactly once, and returns to the same vertex v0. A
mixed graph is Eulerian if it contains such an Eulerian tour. In addition, if a digraph
D = (V, A) contains a directed Eulerian tour, then this digraph D = (V, A) is called a
directed Eulerian graph, or simply an Eulerian digraph.
A matching in a graph G = (V, E) is a set M of pairwise disjoint edges (i.e., the
end-vertices of edges are all distinct), and each edge in M is called a matching edge. In
addition, M is called a perfect matching if all vertices in V are covered by M , i.e., the
matching M satisfies |M | = |V |/2. Moreover, for a weighted graph G = (V, E; w), M is
called a maximum matching in G if the weight w(M) =∑
e∈M w(e) is maximized among
all matchings in G. Similarly, for a weighted graph G = (V,E; w) and a matching M of
G, M is called a minimum cost perfect matching if M is a perfect matching such that the
weight w(M) =∑
e∈M w(e) is minimized among all perfect matchings in G.
For any network, a synonym of digraph, N = (V, A; u; c; s, t) with a capacity function
u : A → R+, a cost function c : A → R+, a source s and a sink t, a flow from s to t
(briefly, an (s, t)-flow) is a function f from A to the set R+0 of non-negative real numbers,
i.e., f : A → R+0 , satisfying the following conditions
(1) (the capacity constraint) 0 ≤ f(e) ≤ u(e) for each arc e ∈ A;
(2) (the flow conservation)∑
e∈δ+D(v) f(e) =
∑e∈δ−D(v) f(e) for each vertex v ∈ V −{s, t};
(3) v(f) =∑
e∈δ+D(s) f(e) − ∑
e∈δ−D(s) f(e) ≥ 0 for the source s.
7
We call v(f) the value of an (s, t)-flow f , and we define its cost c(f) =∑
e∈A c(e)f(e).
In addition, if the value f(e) is integral for each arc e ∈ A, we call an (s, t)-flow f as an
integral (s, t)-flow.
For a network N = (V,A; l, u; s, t) with a demand function l : A → R+0 and a capacity
function u : A → R+, a function f : A → R+0 is called a circulation in N if the function f
satisfies l(e) ≤ f(e) ≤ u(e) for each arc e ∈ A and the flow conservation rule (2) holds at
every vertex v in N . When f is a circulation in N , the cost of f is defined to be the value
c(f) =∑
e∈A c(e)f(e). Similarly, we call a circulation f to be integral if f(e) is integral
for each arc e ∈ A.
For any real number r, let �r� (or r�) denote the greatest (or smallest) integer less
(or greater) than or equal to r.
The following key lemmas are needed to design our algorithms later on.
Lemma 1 [22] Suppose that G = (V,E) is a connected graph. Then G is an Eulerian
graph if and only if every vertex v of G has even degree.
If a graph G = (V, E) is Eulerian, we use an algorithm due to Edmonds and John-
son [7] in time O(m) to determine an Eulerian tour on G. We call such an algorithm the
Eulerian-Cycle algorithm.
Lemma 2 [10] Suppose that D = (V, A) is a strongly connected digraph. Then D is an
Eulerian digraph if and only if every vertex v of D has equal in- and out-degrees.
If a digraph D = (V, A) is Eulerian, we use another algorithm due to Edmonds and
Johnson [7] in time O(nm) to determine a directed Eulerian circuit on D. We call such
an algorithm the Eulerian-Circuit algorithm.
Lemma 3 (Goldberg and Tarjan [17]) Let N = (V,A; l, u; c) be a network with a demand
function l : A → R+0 , a capacity function u : A → R+ and a cost function c : A → R+. A
minimum cost circulation f in N can be determined in time O(n2m3 log n). In addition, if
the demand function l(·) and the capacity function u(·) are both integral, then a minimum
cost circulation f in N is integral, and it can be determined with the same running time.
We refer a combinatorial algorithm due to Goldberg and Tarjan [17] to solve the
minimum cost circulation problem as the Cycle-Canceling algorithm.
Lemma 4 (Ford and Fulkerson [12]) Suppose that N = (V, A; u, c; s, t) is an instance of
8
the minimum cost flow problem. Let f1 be a minimum cost (s, t)-flow of value k0 in N
and f2 an (s, t)-flow of value 1 along an augmentation (s, t)-path P in N ′(f1) of least cost,
where N ′(f1) is the incremental weighted flow network of the flow f1. Then f1 + f2 is a
minimum cost (s, t)-flow of value k0 + 1 in N . In addition, if the capacity function u(·)is integral and f1 is a minimum cost integral (s, t)-flow of value k0 in N , then f1 + f2 is a
minimum cost integral (s, t)-flow of value k0 + 1 in N .
By Lemma 4 and the Belleman-Ford algorithm [1] in time O(nm) to determine an
augmentation (s, t)-path P in N ′(f1) of least cost, we can design an algorithm in time
O(nmk1) to determine a minimum cost (s, t)-flow of integral value k1 in N . We call such
an algorithm the Successive-Shortest-Path algorithm.
The Successive-Shortest-Path algorithm is not a polynomial-time algorithm to
solve the minimum cost flow problem. The capacity scaling algorithm in [27] is a variant
of the Successive-Shortest-Path algorithm, and it solves the minimum cost flow
problem in time O(nm2 log k1). In this paper, we need a more efficient algorithm in [28] to
solve the integral minimum cost (s, t)-flow problem, that is a variant of the capacity scaling
algorithm. We call such an algorithm the Enhanced-Capacity-Scaling algorithm.
Lemma 5 [28] The Enhanced-Capacity-Scaling algorithm solves the capacitated
minimum cost flow problem in time O(m log nS(n,m)), where S(n,m) is the time required
to solve a shortest path problem with n vertices and m arcs. In addition, if the capacity
function u(·) is integral, then a minimum cost integral (s, t)-flow can be determined with
the same running time.
3 The Constrained Arc Routing Problem in Mixed
Graphs
The constrained arc routing problem in mixed graphs (CARM) was defined in the intro-
duction section. We first give an integer programming formulation of the CARM problem
as follows. Each mixed graph G = (V,A∪E) is associated with a digraph←→G = (V,A∪B)
such that B = E+∪E− and each edge e ∈ E is replaced by two arcs e+ ∈ E+ and e− ∈ E−
of opposite direction with the same weight w(e+) = w(e−) = w(e). Since G is a connected
9
mixed graph, we can obtain a necessary and sufficient condition for any instance of the
CARM problem as follows. A nonnegative integral circulation x in←→G corresponds to an
arc routing of G (i.e., a feasible solution to the instance G of the CARM problem) if and
only if (1) l(e) ≤ x(e) ≤ u(e) for each arc e ∈ A, and (2) l(e) ≤ x(e+) + x(e−) ≤ u(e) for
each edge e ∈ E, where x(e) is the number of times that the arc e ∈ A is traversed in the
arc routing of G and x(e+) + x(e−) is the number of times that edge e ∈ E is traversed
in the arc routing of G for the CARM problem, respectively.
Thus, we can establish an integer program, denoted by CARM(G), which is equivalent
to the CARM problem for the instance G = (V,A ∪ E; w; l, u).
CARM(G) min∑
e∈A∪B w(e)x(e)
subject to∑
e∈δ−←→G
(v) x(e) − ∑e∈δ+←→
G(v) x(e) = 0 for each v ∈ V (1)
l(e) ≤ x(e) ≤ u(e) for each e ∈ A (2)
l(e) ≤ x(e+) + x(e−) ≤ u(e) for each e ∈ E (3)
x(e) ≥ 0 for each e ∈ A ∪ B (4)
x(e) integer for each e ∈ A ∪ B (5)
Now, we construct a new weighted digraph G = (V , A) from the mixed graph G =
(V, E ∪ A; w; l, u) as follows. At the beginning, let V := V and A := A. We replace each
edge e = ab ∈ E, by a directed subgraph De consisting of four vertices a, b, ye, ze and five
arcs (a, ye), (b, ye), (ye, ze), (ze, a), (ze, b) (see Fig.1), where we add the two new vertices
ye and ze into V and the five new arcs into A, and we define lower and upper bounds
and length weights on these five arcs as follows: l(a, ye) = l(b, ye) = l(ze, a) = l(ze, b) = 0,
l(ye, ze) = l(e), u(a, ye) = u(b, ye) = u(ye, ze) = u(ze, a) = u(ze, b) = u(e), and w(a, ye) =
w(b, ye) = w(ze, a) = w(ze, b) = 0, w(ye, ze) = w(e). Thus, each arc in A of the new
weighted digraph G belongs to A or to a directed subgraph De for some e ∈ E, but not
both.
� � ����
�����������
���������
�
����������
���������
�
a b a b
ye
zeFig.1 The construction of De
10
For each new arc e ∈ A of the network G = (V , A; w; l, u), the values l(e), u(e) and
w(e) may be regarded as demand, capacity and unit cost of e, respectively.
Lemma 6. Let G = (V, E ∪ A; w; l, u) be an instance of the CARM problem such that
l(e) < u(e) holds for each edge e ∈ E and l(e) ≤ u(e) holds for each arc e ∈ A. There
exists a feasible tour T in G if and only if there exists an integral circulation f in the
network G = (V , A; w; l, u).
Proof. A feasible tour T to the instance G of the CARM problem can be transformed into
a feasible solution x of the corresponding integer program CARM(G), and vice versa. We
only need to prove that there exists a feasible solution x of the integer program CARM(G)
if and only if there exists an integral circulation f in the network G = (V , A; w; l, u).
Necessity. Suppose that x is a feasible solution of the integer program CARM(G).
We construct an integral circulation f in the network G = (V , A; w; l, u) as follows. For
each arc e ∈ A, let f(e) = x(e). And for each edge e = ab ∈ E, without loss of
generality, we may assume e+ = (a, b) and e− = (b, a), then let f(a, ye) = f(ze, b) = x(e+),
f(b, ye) = f(ze, a) = x(e−) and f(ye, ze) = x(e+) + x(e−). It is easy to verify that f is an
integral circulation in G.
Sufficiency. Suppose that f is an integral circulation in G = (V , A; w; l, u). First, we
may construct a new integral circulation f from f in G that satisfies further properties.
For each arc e ∈ A, let f(e) = f(e). For each edge e = ab ∈ E, we examine either the
pair f(a, ye) and f(ze, b) or the pair f(b, ye) and f(ze, a). If f(a, ye) = f(ze, b) (implying
f(b, ye) = f(ze, a)), let f(e′) = f(e′) for each arc e′ ∈ {(a, ye), (ze, b), (b, ye), (ze, a), (ye, ze)}.If f(a, ye) �= f(ze, b), we may assume f(a, ye) > f(ze, b), then we denote θ1 =
⌊f(a,ye)−f(ze,b)2
⌋
and θ2 =⌈f(a,ye)−f(ze,b)
2
⌉. We consider two possibilities.
(1) If f(ye, ze) = l(e), let f(a, ye) = f(a, ye) − θ1, f(ze, b) = f(ze, b) + θ2, f(b, ye) =
f(b, ye) + θ2, f(ze, a) = f(ze, a) − θ1 and f(ye, ze) = f(ye, ze) − θ1 + θ2. It follows that
l(e) = f(ye, ze) ≤ f(ye, ze) ≤ f(ye, ze) + 1 = l(e) + 1 ≤ u(e)
(2) If f(ye, ze) > l(e), let f(a, ye) = f(a, ye) − θ2, f(ze, b) = f(ze, b) + θ1, f(b, ye) =
f(b, ye) + θ1, f(ze, a) = f(ze, a) − θ2 and f(ye, ze) = f(ye, ze) − θ2 + θ1. It follows that
l(e) ≤ f(ye, ze) − 1 ≤ f(ye, ze) ≤ f(ye, ze) ≤ u(e)
11
It is easy to verify that f is also an integral circulation in G and f(a, ye) = f(ze, b) for
each edge e = ab ∈ E.
Now, we can construct a feasible solution x to the integer program CARM(G) as
follows. For each arc e ∈ A, let x(e) = f(e). For each edge e = ab ∈ E, without loss
of generality, we may assume e+ = (a, b) and e− = (b, a), then we let x(e+) = f(a, ye)
and x(e−) = f(b, ye). Since f is an integral circulation in G, it is easy to see that the
constraint condition (1) in the integer program CARM(G) holds. The constraint condition
(2) holds obviously. For each edge e ∈ E, the constraint condition (3) also holds since
l(e) ≤ x(e+) + x(e−) = f(ye, ze) ≤ u(e). The constraint conditions (4) and (5) clearly
hold. Thus, x is a feasible solution of the integer program CARM(G).
Edmonds and Karp [8] designed an algorithm to solve the minimum cost circulation
problem, but their algorithm is based on a capacity scaling method and it can be imple-
mented to run in time O(m(m+n log n) log U), where U = max{u(e) | e ∈ A}. Tardos [33]
was the first to design a strongly polynomial-time algorithm to solve the minimum cost
circulation problem. In this paper, we invoke a simpler and more efficient combinatorial
algorithm due to Goldberg and Tarjan [17] to solve the minimum cost circulation problem.
It is stated in Lemma 3 and referred to as the Cycle-Canceling algorithm.
Using Lemmas 3 and 6, we can design an approximation algorithm to solve the CARM
problem as follows.
Algorithm 1. CARM
Input: A connected mixed graph G = (V,E ∪ A; w; l, u) with a length function w :
E ∪ A → Q+ and two integer functions l, u : E ∪ A → Z+ satisfying l(e) ≤ u(e) for
each e ∈ E ∪ A.
Output: A constrained arc routing T and its total length.
Begin
Step 1 Construct a new weighted digraph G = (V , A) from the mixed graph G =
(V,E ∪ A; w; l, u) as above.
12
Step 2 Using the Cycle-Cancelling algorithm in [17], determine a minimum cost
integral circulation f in G. If no such circulation exists, then output “no feasible
solution to this instance of the CARM problem”, and stop.
Step 3 Construct a new integral circulation f in G associated with the integral circulation
f as follows
(1) For each arc e ∈ A, let f(e) = f(e).
(2) For each edge e = ab ∈ E, let f(ye, ze) = f(ye, ze) − θ1 + θ2, where θ1 =⌊ |f(a,ye)−f(ye,b)|
2
⌋and θ2 =
⌈ |f(a,ye)−f(ye,b)|2
⌉; and
if (f(a, ye) > f(ye, b)) then
denote f(a, ye) = f(a, ye)−θ1, f(ze, b) = f(ze, b)+θ2, f(b, ye) = f(b, ye)+θ2
and f(ze, a) = f(ze, a) − θ1.
else
denote f(a, ye) = f(a, ye)+θ2, f(ze, b) = f(ze, b)−θ1, f(b, ye) = f(b, ye)−θ1
and f(ze, a) = f(ze, a) + θ2.
Step 4 Construct a feasible solution x to the integer program CARM(G) as follows
(1) For each arc e ∈ A, let x(e) = f(e).
(2) For each edge e = ab ∈ E, do
if (e+ = (a, b) and e− = (b, a)) then
denote x(e+) = f(a, ye) and x(e−) = f(b, ye).
else
denote x(e−) = f(a, ye) and x(e+) = f(b, ye).
Step 5 Construct a digraph GD = (V, A) so that there are x(e) multi-arcs from the vertex
a to the vertex b in GD for each arc e = (a, b) in←→G .
Step 6 Using the Eulerian-Circuit algorithm in [7], find a directed Eulerian tour TD
on GD.
Step 7 For each arc e in TD, remove its direction only if e /∈ A, and then construct a
(mixed) tour T in G.
13
Step 8 Output the tour T and the value w(T ) =∑
e∈A∪B w(e)x(e).
End
For an instance G = (V, E ∪ A; w; l, u) of the CARM problem, let OPT denote the
optimal value and OUT the output value of the algorithm CARM, respectively.
Theorem 1 For any instance G = (V,E∪A; w; l, u) of the CARM problem, if l(e) < u(e)
holds for each edge e ∈ E and l(e) ≤ u(e) holds for each arc e ∈ A, then the algorithm
CARM can produce a feasible arc routing to satisfy OUT ≤ (1 + 1/l0) · OPT , where
l0 = min{l(e) | e ∈ E}, and the running time of the algorithm CARM is O(n2m3 log n).
Proof. We first obtain a minimum cost integral circulation f in G by executing the
Cycle-Canceling algorithm [17] on the weighted digraph G. From the proof of Lemma 6,
we know that f constructed at Step 3 is an integral circulation in G and x constructed
at Step 4 is a feasible solution to the integer program CARM(G). By the definitions of x
and the digraph GD, we obtain w(T ) =∑
e∈A∪B w(e)x(e) = w(f).
Since an optimal solution for the instance G of the CARM problem must correspond
to an integral circulation in G, then Step 2 works correctly. Because f is a minimum cost
integral circulation in the weighted digraph G, we have
w(f) =∑e∈A
w(e)f(e) ≤ OPT.
On the other hand, by Step 3 of the algorithm CARM, we have the facts f(e) = f(e) for
each arc e ∈ A and the inequalities f(ye, ze) ≤ f(ye, ze) = f(ye, ze)−θ1 +θ2 ≤ f(ye, ze)+1
for each edge e = uv ∈ E. Thus, we can obtain
14
w(T ) = w(f) =∑e∈A
w(e)f(e) +∑e∈E
∑e′∈De
w(e′)f(e′)
=∑e∈A
w(e)f(e) +∑e∈E
w(ye, ze)f(ye, ze)
≤∑e∈A
w(e)f(e) +∑e∈E
w(ye, ze)(f(ye, ze) + 1)
=∑e∈A
w(e)f(e) +∑e∈E
w(ye, ze)f(ye, ze) +∑e∈E
w(ye, ze)
=∑e∈A
w(e)f(e) +∑e∈E
∑e′∈De
w(e′)f(e′) +∑e∈E
w(e)
= w(f) +∑e∈E
w(e)
≤ w(f) +1
l0
∑e∈E
w(e)l(e)
≤ OPT +1
l0· OPT
= (1 +1
l0) · OPT.
where l0 = min{l(e) | e ∈ E}.It is obvious to see that the running time of the algorithm CARM heavily depends on
the running time of the Cycle-Canceling algorithm [17]. According to Lemma 3, we
know that the running time of the algorithm CARM is O(n2m3 log n).
This establishes the theorem.
The following special version of the CARM problem can be derived from the proof of
Theorem 1.
Corollary 1 Given a digraph D = (V,A; w; l, u) with two integer functions l, u : A → Z+
satisfying l(e) ≤ u(e) for each arc e ∈ A and a length function w : A → Q+, the algorithm
CARM produces an optimal arc routing for D and its running time is O(n2m3 log n).
It is well known that Raghavachari and Veerasamy [31] designed a 3/2-approximation
algorithm to solve the mixed Chinese postman problem. We propose a conjecture for the
CARM problem to conclude this section.
Conjecture 1 It would admit a 3/2-approximation algorithm to solve the CARM prob-
lem.
15
4 The Constrained Edge Routing Problem in Graphs
In this section, we consider the constrained edge routing problem in graphs (CERG), i.e.,
a special version of the constrained arc routing problem in mixed graphs (CARM) for
the case A = ∅. The CERG problem is a generalization of the Chinese postman problem
(CPP).
It is known that the CPP is solvable in polynomial time by the following strategy.
(1) When a graph G is Eulerian, the Eulerian-Cycle algorithm due to Edmonds
and Johnson in [7] produces an Eulerian tour as an optimal solution.
(2) When a graph G is not Eulerian, we must add some edges from G to create a new
Eulerian graph G′. The edges added to create G′ must be of least total length, and an
Eulerian tour in the augmented graph G′ is an optimal solution. As shown in [7], these
edges of least total length can be determined by solving a minimum cost perfect matching
problem on the complete graph H whose vertices are the odd-degree vertices of G.
It is known that each edge in a graph G is used at most twice in any optimal solution
for the CPP. A similar argument can establish the following.
Lemma 7 For any graph G = (V,E; w; l, u) as an instance of the CERG problem, each
edge e in G is used at most l(e) + 1 times in any optimal solution.
For convenience, we may assume u(e) = l(e) + 1 when l(e) < u(e) for each edge e ∈ E
of any instance G = (V, E; w; l, u) for the CERG problem.
In order to describe our combinatorial algorithm to optimally solve the CERG problem,
we need two auxiliary constructions. We construct a multigraph Gl = (V, El), where the
multiset El contains each edge e of E exactly l(e) times. Let Eu = {e ∈ E | l(e) < u(e)}.Then we construct a weighted graph Gu = (V, Eu; wu), where the length function wu :
Eu → Q+ satisfies wu(e) = w(e) for each edge e ∈ Eu.
By Lemma 7, we can design a combinatorial algorithm to solve the CERG problem
by the following strategy.
(1) If needed, find a subset E ′u of Eu such that the augmented multigraph G′ =
(V, El ∪E ′u) is an Eulerian graph, having the minimum total length w(E ′
u) =∑
e∈E′uw(e).
(2) Construct an Eulerian tour on G′. We may call the set E ′u as in [21] a priority set
to the instance G = (V,E; w; l, u) of the CERG problem, or simply a priority set.
16
In our algorithm to solve the CERG problem, we need an algorithm due to Floyd [11]
in running time O(n3) to find all shortest paths among all odd-degree vertices and an-
other algorithm due to Edmonds [6] in running time O(n3) to find a minimum cost
perfect matching in an appropriate graph. We refer these two algorithms the Floyd
algorithm [11] and the Edmonds algorithm [6], respectively.
An algorithm to solve the CERG problem is designed as follows.
Algorithm 2. CERG
Input: A connected graph G = (V, E; w; l, u) with a length function w : E → Q+ and
two integer functions l, u : E → Z+ satisfying l(e) ≤ u(e) for each e ∈ E.
Output: A minimum length edge routing T on G to satisfy the constraint.
Begin
Step 1 Define El to be a multiset consisting of each edge e of E exactly l(e) times. And
construct the multigraph Gl = (V, El).
Step 2 If (the connected multigraph Gl is Eularian) then
Determine an Eulerian tour T on Gl by means of the Eulerian-Cycle algo-
rithm [7], and stop. The tour T corresponds to an optimal solution on G.
Step 3 Construct a graph Gu = (V, Eu; wu), where Eu = {e ∈ E | l(e) < u(e)} and
wu : Eu → Q+ satisfies wu(e) = w(e) for each edge e ∈ Eu.
Step 4 Determine the set Vo of odd-degree vertices in Gl. Construct a graph H =
(Vo, EH ; wH) with vertex set Vo and edge set EH , where an edge vivj ∈ EH if and
only if there exists a path in Gu to connect the two odd-degree vertices vi and vj,
and then define the cost wH(vi, vj) of edge vivj in H to be the length of a shortest
path Pvi,vjbetween vi and vj in Gu.
Step 5 If (there exists no perfect matching of cardinality |Vo|/2 in H) then
Output “no feasible solution on G”, and stop.
17
Step 6 Use the Edmonds algorithm [6] to find a minimum cost perfect matching M of
cardinality |Vo|/2 in H.
Step 7 Set G′ := Gl. For each edge vivj in the minimum cost perfect matching M , add
to G′ all edges that lie on the shortest path Pvi,vjbetween vi and vj in Gu.
Step 8 Determine an Eulerian tour T on G′ by means of the Eulerian-Cycle algo-
rithm [7], and stop. The tour T corresponds to an optimal solution on G.
End
Using the algorithm CERG, we obtain the following result for the CERG problem.
Theorem 2 The algorithm CERG works correctly to solve the CERG problem and its
running time is O(n3).
Proof. We first prove that the algorithm CERG produces a feasible solution to the
instance G = (V,E; w; l, u) of the CERG problem. Any two shortest paths produced
at Step 4 corresponding to two distinct edges in the minimum cost perfect matching M
must be edge-disjoint, for otherwise another perfect matching M ′ in H can be constructed
satisfying wH(M ′) < wH(M), a contradiction (the proof in details is similar to that in [6],
we omit it here). Then we add to the augmented graph G′ all edges that lie on the shortest
paths Pvi,vjbetween vi and vj in Gu, where vivj corresponds to a matching edge in M .
Thus, G′ becomes an Eulerian graph and the Eulerian tour T on G′ produced at Step 8
is a feasible solution to the instance G = (V, E; w; l, u).
Next, we prove that the Eulerian tour T produced at Step 8 is a minimum total
length solution to the instance G = (V, E; w; l, u) of the CERG problem. We may assume
that T ∗ = (V,E∗) is an optimal tour to G. We shall prove w(T ) ≤ w(T ∗), equivalently,∑
vivj∈M
∑e∈Pvi,vj
w(e) = wH(M) ≤ w(T ∗ − El).
We may assume that the multigraph Gl = (V, El) contains exactly 2k vertices of odd
degree, namely vi1 , vi2 , . . ., vi2k−1, vi2k
, and the vertex set Vo of the graph H at Step 4
consists of precisely these 2k vertices, hence 2k = |Vo|. We consider the reduced graph
H ′ = T ∗ − El, where each vertex of odd degree in H ′ is also a vertex in Vo.
Since El ⊆ E∗ ⊆ (El ∪ Eu) and⋃
vivj∈M E(Pvi,vj) ⊆ Eu, the graph H ′ contains no
cycle. To the contrary, we may suppose that if C were a cycle in H ′, then the reduced
18
graph T ∗ − E(C), satisfying w(T ∗ − E(C)) < w(T ∗), would be another feasible solution
to the instance G of the CERG problem, which contradicts the optimality of w(T ∗).
Since the graph H ′ contains no cycle, we obtain the fact that H ′ is a forest. Then
the edges in H ′ can be decomposed into k edge-disjoint simple paths by modifying graph
searching algorithm [30] in the following recursive steps (note that H ′ contains at least
two vertices of degree 1).
(1) For r = 1, choose any vertex of degree 1 in H ′, namely vj2r−1 , and by using the
similar graph searching algorithm in [30] to find the next vertex of odd degree along sole
path from vj2r−1 in H ′, namely vj2r , we can construct a simple path P ′vj2r−1
,vj2rto connect
vj2r−1 and vj2r in H ′.
(2) Let H ′ := H ′ − E(P ′vj2r−1
,vj2r). Using the similar argument in (1), we can choose
two new vertices of odd degree in the new graph H ′, namely vj2r+1 and vj2r+2 , and then
construct a simple path P ′vj2r+1
,vj2r+2to connect vj2r+1 and vj2r+2 in H ′. Let r := r + 1 and
execute (2) repeatedly until r = k.
Now, without loss of generality, we assume that the edges in H ′ are decomposed into k
edge-disjoint simple paths P ′vj2r−1
,vj2r, r = 1, 2, . . . , k, such that the edge set {vj1vj2 , vj3vj4 ,
. . ., vj2k−1vj2k
} is a perfect matching in H. It follows that dGu(vj2r−1 , vj2r) ≤ w(P ′vj2r−1
,vj2r),
r = 1, 2, . . . , k, and w(H ′) =∑k
r=1 w(P ′vj2r−1
,vj2r).
Since M is a minimum cost perfect matching in H = (Vo, EH ; wH), we obtain
wH(M) =∑
vivj∈M
∑e∈Pvi,vj
w(e) ≤k∑
r=1
dGu(vj2r−1 , vj2r) ≤k∑
r=1
w(P ′vj2r−1
,vj2r) = w(T ∗ − El).
It implies w(T ) = w(El) + wH(M) ≤ w(El) + w(T ∗ − El) = w(T ∗). This shows that
the Eulerian tour T produced at Step 8 has a minimum total length to the instance
G = (V, E; w; l, u) of the CERG problem.
Finally, the running time of the algorithm CERG can be determined as follows.
(1) Since each of the l(e) multi-edges of G can be constructed within a constant time,
the multigraph Gl = (V,El) at Step 1 needs time O(m) to be constructed.
(2) Since each of the l(e) multi-edges of G can be traversed within a constant time,
an Eulerian tour T on G at Step 2 needs at most time O(m) to be determined.
(3) The graph Gu = (V, Eu) at Step 3 needs time O(m) to be constructed.
19
(4) The graph H at Step 4 needs time O(n3) to be constructed, by using the Floyd
algorithm [11].
(5) A minimum cost perfect matching M in H at Steps 5 and 6 needs time O(n3) to
be found.
(6) The augmented graph G′ at Step 7 needs at most time O(n2) to be constructed.
(7) An Eulerian tour T on G′ at Step 8 needs at most time O(n2) to be determined.
Hence, the total running time of the algorithm CERG is O(n3). This establishes the
theorem.
Using the algorithm CERG, we can obtain the following immediately.
Corollary 2 For any connected graph G = (V,E; w; u), where a length function w :
E → Q+ and an integer function u : E → Z+, the algorithm CERG can either produce
a minimum length tour traversing each edge e at least once and at most u(e) times or
determine that there is none.
5 The Constrained Arc Routing Problem in Digraphs
In this section, we consider the constrained arc routing problem in digraphs (CARD), that
is the constrained arc routing problem in mixed graphs specialized to the case E = ∅.This problem is a generalization of the directed Chinese postman problem (DCPP).
We know that the DCPP is solvable in polynomial time. A solution exists if and
only if the digraph D is strongly connected. The techniques used to solve the DCPP are
similar to those used for the undirected case. If D is symmetric, i.e., each vertex in D
has equal in- and out-degrees, then the directed Chinese postman tour constructed by the
Eulerian-Circuit algorithm [7] corresponds to a directed Eulerian circuit in D.
If D is not symmetric, then copies of some arcs of D must be added to D so that the
augmented digraph D′ becomes symmetric. A directed Eulerian circuit in D′ corresponds
to a directed Chinese postman tour in D. The digraph D′ can be obtained from D by
solving a transportation problem described in [12].
In order to describe our algorithm to optimally solve the CARD problem, we need
two auxiliary constructions. First, we construct a multidigraph Dl = (V,Al), where
20
the multiset Al contains each arc e of A exactly l(e) times. Let Au denote the subset
{e ∈ A | l(e) < u(e)}. Next, we construct a weighted digraph Du = (V,Au; wu), where
the length function wu : Au → Q+ satisfies wu(e) = w(e) for each arc e ∈ Au. For
any feasible solution for the CARD problem, find some multi-arcs from Au to create the
multiset A′u such that the augmented multidigraph D′ = (V, Al ∪ A′
u) is Eulerian.
Modifying the steps to solve the DCPP, we can solve the CARD problem by the
following strategy.
(1) If needed, find a multiset A′u of Au such that the augmented multidigraph D′ =
(V, Al ∪ A′u) is Eulerian, having the minimum total length w(A′
u) =∑
e∈A′uw(e).
(2) Construct a directed Eulerian circuit in D′.
In our algorithm to solve the CARD problem, we needed a combinatorial algorithm
in [28] to solve the minimum cost integral (s, t)-flow problem, that was referred the
Enhanced-Capacity-Scaling algorithm in Lemma 5.
We design an algorithm to solve the CARD problem as follows.
Algorithm 3. CARD
Input: A strongly connected digraph D = (V,A; w; l, u) with a length function w : A →Q+ and two integer functions l, u : A → Z+ satisfying l(e) ≤ u(e) for all arcs e ∈ A.
Output: A minimum length arc routing T on D to satisfy the constraint.
Begin
Step 1 Let Al denote the multiset consisting of each arc e of A exactly l(e) times. And
construct the multidigraph Dl = (V, Al).
Step 2 For each vertex v ∈ V , compute σDl(v) = d−
Dl(v) − d+
Dl(v).
Step 3 If (Dl is strongly connected and σDl(v) = 0 for each vertex v ∈ V ) then
Determine a directed Eulerian circuit T on Dl by using the Eulerian-Circuit
algorithm in [7], and stop. The tour T corresponds to an optimal solution on D.
Step 4 Construct the digraph Du = (V,Au; wu), where Au = {e ∈ A | l(e) < u(e)} and
wu : Au → Q+ satisfies wu(e) = w(e) for each arc e ∈ Au.
21
Step 5 Let I = {vi ∈ V | σDl(vi) > 0} and J = {vj ∈ V | σDl
(vj) < 0}.
Step 6 Construct a network N = (V ′, A′; u′; c; s, t) from the digraph Du = (V, Au; wu)
satisfying the following.
(1) Let V ′ = V ∪ {s, t} and A′ = Au ∪ AI ∪ AJ . The source s and the sink t are
two vertices not belonging to V so that AI = {(s, vi) | vi ∈ I} and AJ = {(vj, t) |vj ∈ J}.
(2) Define the capacity function u′ : A′ → Z+ as follows
u′(e) =
⎧⎪⎪⎪⎨⎪⎪⎪⎩
u(e) − l(e) if the arc e ∈ Au,
σDl(vi) if the arc e = (s, vi) ∈ AI ,
−σDl(vj) if the arc e = (vj, t) ∈ AJ .
(3) Define the cost function c : A′ → Q+ as follows
c(e) =
⎧⎪⎪⎪⎨⎪⎪⎪⎩
w(e) if the arc e ∈ Au,
0 if the arc e = (s, vi) ∈ AI ,
0 if the arc e = (vj, t) ∈ AJ .
(4) Let σ0 =∑
vi∈I σDl(vi).
Step 7 If (there exists no (s, t)-flow of value σ0) then
Output “no feasible solution in D”, and stop.
Step 8 Using the Enhanced-Capacity-Scaling algorithm in [28], determine a min-
imum cost integral (s, t)-flow f of value σ0 in N .
Step 9 Construct the multiset of arcs Af = {(x, y)1, (x, y)2, . . . , (x, y)f(x,y) | (x, y) ∈Au and f(x, y) > 0}.
Step 10 Construct the multidigraph Df := (V,Al ∪ Af ).
Step 11 Determine a directed Eulerian circuit T on Df by the Eulerian-Circuit
algorithm in [7], and stop. The tour T corresponds to an optimal solution on D.
End
22
Using the algorithm CARD, we obtain the following result for the CARD problem.
Theorem 3 The algorithm CARD works correctly to solve the CARD problem, and its
running time is O(nm2 log n).
Proof. First, we prove the output tour T produced by the algorithm CARD is a feasible
directed circuit to the digraph D = (V, A; w; l, u) of the CARD problem. If T is produced
at Step 3, it is easy to verify that T is an optimal solution to the digraph D = (V, A; w; l, u).
Now, we assume that output T is produced at Step 11. Since f is a minimum cost integral
(s, t)-flow of value σ0, where σ0 =∑
vi∈I σDl(vi), we obtain the following three facts.
(1)∑
x:(vi,x)∈Au
f(vi, x) − ∑x:(x,vi)∈Au
f(x, vi) = f(s, vi) = σDl(vi) for each vertex vi of I;
(2)∑
y:(y,vj)∈Au
f(y, vj) −∑
y:(vj ,y)∈Au
f(vj, y) = f(vj, t) = −σDl(vj) for each vertex vj of J ;
(3)∑
x:(x,z)∈Au
f(x, z) =∑
y:(z,y)∈Au
f(z, y) for each vertex z of V − (I ∪ J).
For each vertex vi ∈ I, we have
d+Df
(vi) = d+Dl
(vi) +∑
x:(vi,x)∈Au
f(vi, x)
= d+Dl
(vi) +∑
x:(x,vi)∈Au
f(x, vi) + σDl(vi)
= d+Dl
(vi) +∑
x:(x,vi)∈Au
f(x, vi) + (d−Dl
(vi) − d+Dl
(vi))
= d−Dl
(vi) +∑
x:(x,vi)∈Au
f(x, vi)
= d−Df
(vi).
For each vertex vj ∈ J , we have
d−Df
(vj) = d−Dl
(vj) +∑
y:(y,vj)∈Au
f(y, vj)
= d−Dl
(vj) +∑
y:(vj ,y)∈Au
f(vj, y) − σDl(vj)
= d−Dl
(vj) +∑
y:(vj ,y)∈Au
f(vj, y) − (d−Dl
(vj) − d+Dl
(vj))
= d+Dl
(vj) +∑
y:(vj ,y)∈Au
f(vj, y)
= d+Df
(vj).
23
For each vertex z of V − (I ∪ J), we have
d+Df
(z) = d+Dl
(z) +∑
y:(z,y)∈Au
f(z, y)
= d−Dl
(u) +∑
x:(x,z)∈Au
f(x, z)
= d−Df
(z).
This shows that d+Df
(z) = d−Df
(z) for each vertex z in Df . Since D is strongly connected,
so is Df . Thus, Df is an Eulerian digraph. Obviously, the output tour T on Df produced
at Step 11 is a feasible solution to the instance D = (V,A; w; l, u) of the CARD problem.
Next, we prove that the Eulerian tour T produced at Step 11 is a minimum total
length solution to the instance D = (V, A; w; l, u). We may assume that T ∗ = (V, A∗)
is an optimal directed tour on D = (V, A; w; l, u) such that each arc e in A appears in
T ∗ exactly lT ∗(e) times, where l(e) ≤ lT ∗(e) ≤ u(e), and that w(T ∗) =∑
e∈T ∗ w(e) is
minimized.
Let the directed subgraph H ′ = D[T ∗ − Al] be obtained from the multidigraph H =
T ∗ − Al by replacing multi-arcs joining the same pair of end-vertices by one arc. Since
T ∗ is an optimal directed tour in D, by the fact that Al is a multiset of arcs of A, the
multidigraph H contains no directed circuit, otherwise the optimal value w(T ∗) would be
reduced. It follows that the reduced digraph H ′ is an acyclic subdigraph of the digraph Du.
Now, let I ′ = {vi ∈ V | d+H(vi) > d−
H(vi)} and J ′ = {vj ∈ V | d−H(vj) > d+
H(vj)}.Since T ∗ is an optimal directed tour in D and it traverses all arcs in Al, and for every
vertex v ∈ V , we have d+T ∗(v) = d+
Dl(v) + d+
H(v), d−T ∗(v) = d−
Dl(v) + d−
H(v) and d+T ∗(v) =
d−T ∗(v). It follows that d+
H(v) − d−H(v) = d−
Dl(v) − d+
Dl(v) = σDl
(v). In addition, we have
known I ′ = I and J ′ = J such that∑
vi∈I(d+H(vi) − d−
H(vi)) =∑
vi∈I σDl(vi) = σ0 and
∑vj∈J(d+
H(vj) − d−H(vj)) =
∑vj∈J σDl
(vj) = −σ0.
We define a network N∗ = (V (H ′)∪{s, t}, A′′∪AI∪AJ ; u∗; c; s, t) from H ′ = D[T ∗−Al]
so that A′′ is the arc set of H ′ (clearly, A′′ ⊆ Au), AI = {(s, vi) | vi ∈ I} and AJ = {(vj, t) |vj ∈ J}. The capacity function u∗ : A′′ ∪ AI ∪ AJ → Z+ is defined as follows
u∗(e) =
⎧⎪⎪⎪⎨⎪⎪⎪⎩
lT ∗(e) − l(e) if the arc e ∈ A′′,
σDl(vi) if the arc e = (s, vi) ∈ AI ,
−σDl(vj) if the arc e = (vj, t) ∈ AJ .
24
and the cost function c : A′′ ∪ AI ∪ AJ → Q+ is defined as follows
c(e) =
⎧⎪⎪⎪⎨⎪⎪⎪⎩
w(e) if the arc e ∈ A′′,
0 if the arc e = (s, vi) ∈ AI ,
0 if the arc e = (vj, t) ∈ AJ .
Now, we can construct an integral (s, t)-flow of value σ0 in N∗ as follows. For each
arc e ∈ A′′, we define f ∗(e) = lT ∗(e) − l(e), f ∗(s, vi) = σDl(vi) for (s, vi) ∈ AI and
f ∗(vj, t) = −σDl(vj) for (vj, t) ∈ AJ . Since T ∗ is an optimal directed tour in D, it is easy
to verify that f∗ is an integral (s, t)-flow of value σ0 in N∗ (and in N) and its cost is
c(f ∗) =∑
e∈A′′ w(e) · (lT ∗(e) − l(e)). It follows that c(f ∗) +∑
e∈Alw(e) =
∑e∈T ∗ w(e).
Since f is a minimum cost integral (s, t)-flow of value σ0 at Step 8 of the algorithm
CARD, we have w(Af ) = c(f) ≤ c(f∗). It follows that w(T ) =∑
e∈Auw(e)f(e) +
∑e∈Al
w(e) = c(f)+∑
e∈Alw(e) ≤ c(f ∗)+
∑e∈Al
w(e) =∑
e∈T ∗ w(e). Hence, the directed
Eulerian tour T produced at Step 11 is a minimum total length solution to the instance
D = (V, A; w; l, u) of the CARD problem.
Finally, the running time of algorithm CARD can be determined as follows.
(1) For each arc e in D, the l(e) multi-arcs at Step 1 need a constant time to be
constructed. Hence, the multidigraph Dl = (V,Al) needs time O(m) to be constructed.
(2) All σDl(v) at Step 2 need at most time O(m) to compute.
(3) For each arc e in D, the l(e) multi-arcs at Step 3 need a constant time to traverse.
A directed Eulerian tour T on D needs at most time O(nm) to be determined.
(4) The digraph Du = (V, Au) at Step 4 needs time O(m) to be constructed.
(5) The network N at Steps 5 and 6 can be constructed in time O(m).
(6) A minimum cost integral (s, t)-flow f in N at Steps 7 and 8 needs time O(nm2 log n)
to be found. The Belleman-Ford algorithm [1] used to solve the shortest path problem
needs time O(nm).
(7) The augmented multidigraph Df at Steps 9 and 10 needs at most time O(m) to
be constructed.
(8) For each arc e in D, the l(e) multi-arcs need a constant time to traverse. A directed
Eulerian tour T on Df at Step 11 needs at most time O(nm) to be determined.
Hence, the running time of algorithm CARD is O(nm2 log n). This establishes the
theorem.
25
Using the algorithm CARD, we can obtain the following immediately.
Corollary 3 For any strongly connected digraph D = (V, A; w; u), where a length function
w : A → Q+ and an integer function u : A → Z+, the algorithm CARD can either produce
a minimum length directed tour traversing each arc e at least once and at most u(e) times
or determine that there is none.
6 Conclusions
In this paper, we considered the constrained arc routing problem in mixed graphs (CARM)
and obtained the following three results.
(1) For any instance G = (V,E ∪ A; w; l, u) of the CARM problem, where l(e) < u(e)
holds for each edge e ∈ E, l(e) ≤ u(e) holds for each arc e ∈ A and l0 = min{l(e) | e ∈ E},we have designed a (1+1/l0)-approximation algorithm in time O(n2m3 log n) to solve the
CARM problem.
(2) We presented a combinatorial algorithm in time O(n3) to optimally solve the
CARM problem for the case A = ∅.(3) We provided a combinatorial algorithm in time O(nm2 log n) to optimally solve
the CARM problem for the case E = ∅.
Acknowledgments
We are very grateful to the two anonymous reviewers for giving us good suggestions and
comments so that we are able to improve our manuscript substantially.
H.L. Ding and J.P. Li are supported by the National Natural Science Foundation of
China [No.61063011], the Project of First 100 High-level Overseas Talents of Yunnan
Province and IRTSTYN. K.W. Lih is supported by NSC [No.102-2115-M-001-010] of
Taiwan.
26
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29
Title: Approximation Algorithms for Solving the Constrained Arc Routing Problem in Mixed Graphs
Research Highlights:
We consider the constrained arc routing problem that generalizes the mixed Chinese postman problem. he objective is to determine a minimum length tour T traversing each edge e at least l(e) and at most u(e) times. We design an (1+1/l_0)-approximation algorithm in time O(n^2m^3\log n) to solve this problem, where l(e)<u(e) holds for each edge e\in E, l(e)<u(e) holds for each arc e\in A and l_0=\min\{l(e) \mid e\in E\}, l_0=\min\{l(e) \mid e\in E\} We also present two optimal combinatorial algorithms in times O(n^3) and O(nm^2\log n) to solve this problem for the cases A=\emptyset and E=\emptyset, respectively.
