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Mixed Postman Problem


1. Introduction

The mixed postman problem (MPP) is a generalization of the Chinese Postman Problem (CPP). In the MPP, the input graph may contain both undirected edges and arcs (directed edges). The objective is to find a tour that traverses every edge at least once and traversed directed edges only in the direction of the arc.

Even though both undirected and directed versions of the CPP are polynomial time solvable, Papadimitriou showed MPP is NP-hard.


2. Definitions

Mixed graph: a graph may contain both undirected edges and arcs (directed edges).G(V,E,A,C): V-vertices E-edges A-arcs C-costs on E and A.


3. Properties of Eulerian graphs

A graph is Eulerian if there is a tour that traverses each edge of the graph exactly once.

If a mixed graph satisfies both the conditions, there exists a tour using each edges and arcs just once. And we can find the directed or undirected Eulerian tour in polynomial time. So here we are interested in is to find a set of additional edges and arcs of minimum total cost that can be added to mixed graph to make it Eulerian and identify the tour over the resulting graph.

Basically, the output is a Eulerian graph H that contains the input graph G as a subgraph. So each edge of H can be classified either as an original edge or as a duplicated edge. Also, each arc of H is either an original arc, a duplicated arc an oriented edge, or a duplicated and oriented edge.

Frederickson presented an approximation algorithm for MPP called Mixed algorithm. The algorithm comprises two heuristics called Mixed1 and Mixed2. Both of them are 2-approximation algorithm for MPP.


4. Mixed1: 2 approximation algorithm ( G.N.Frederickson 1979 )

Mixed1 (example see Fig1)

  1. Modify the graph to make the degree of each vertex even by duplicating edges and arcs with minimum cost.(call EVENDEGREE)
  2. Make the indegree of each vertex equal to the outdegree by orienting some edges and duplicated arcs and oriented edges with minimum cost.(call INOUTDEGREE)
  3. Because the second step may not necessarily maintain even degree of each vertex, adjust the graph to maintain the even degree and the indegree equals to outdegree by not increasing the cost.(call EVENPARITY)

Algorithm MIXED1

Input: Mixed graph G = (V, E, A) with cost function C; Output: A postman tour

  1. Call EVENDEGREE
  2. Call INOUTDEGREE
  3. Call EVENPARITY
  4. Find an Eulerian tour over the new graph

Algorithm EVENDEGREE
Input: Mixed graph G = (V, E, A) with cost function C;
Output: A mixed graph G = (V, E', A'), E ⊆ E' A ⊆ A' , and the degree of each vertex, ignoring arc direction, is even.
1. Identify the vertices of odd degree
2. Find all shortest paths between the odd vertices, ignoring arc direction
3. Perform a minimum-cost matching of the odd vertices using the shortest path distance
4. Duplicate the arcs and edges in the shortest paths used in the matching.
5. A'= A ∪ {duplicated arcs};E'=E ∪ {duplicated edges}
The numer of the odd vertices is even, so we could do the minimum-cost matching among them. And duplicating the arcs and edges in the shortest paths used in the matching doesn't affect the degree of even degree, because the shortest pathes either not pass those vertices, or increase the degree of them by even number.

Algorithm INOUTGREE
Input: Mixed graph G(V, E,A) with cost function C;
Output: M: every arc of G at least once and oriented edges; U ⊆ E:edges that have not been oriented and inserted into M; Every vertex has equal indegree and outdegree.
1. E' ← A ∪ E1 ∪ E2
E1: for every edge (u,v) ∈ E, add arcs <u,v> and <v,u> to E1
E2: same as E1
2. bv ← Indegree(v)-Outdegree(v)   ∀ v
3. Formulating as a folw problem:
min z = s ∈ A CsXs + s ∈ E1 CsXs
s.t.
{Xs| s ∈ E' and s is directed away from v} - {Xs| s ∈ E' and s is directed towardsfrom v} = bv   ∀ v
Xs ≤ 1   ∀ s ∈ E2
Xs ≥ 0   ∀ s ∈ E'
4. Initialize M equal to A. ( M includes thoes originally directed arcs )
For each arc s in A, insert Xs addicional copies of s into M; ( Duplicate some arcs )
For each oriented edge s in E1, insert Xs copies of s into M; ( Oriente and duplicate some edges )
For each pair of corresponding oriented edges s1 = <v,w> and s2 = <w,v> in E2, if Xs1 = 1 then insert Xs1 copies of s1 to M; if Xs2 = 1 then insert Xs2 copies of s1 to M ( Oriente some edges )
5. Initialize U to be empty. For each pair of corresponding oriented edges s1 = <v,w> and s2 = <w,v> in E2, if Xs1 + Xs1 = 0 , then insert (v,w) to U. ( This indicates (v,w) remain undirected .)

The arcs of E2 are used to represent the fact that the edges of E can be oriented for free by the flow problem. Therefor, E2 is not included in the objective function, but the capacity of its edges is limited to 1; The arcs of E1 represent the edges of E that are duplicated and oriented by the flow problem, hence the variables of E1 appear in the objective function .

The optimal solution of flow problem are always integral, so optimal solution of this problem is integer too .

Without loss of generality, we assume for each pair of corresponding oriented edges s1 = <v,w> and s2 = <w,v> in E2, Xs1 and Xs2 are not equal to 1 at same time. Because if a solution uses both arcs corresponding to an edge, then deleting both of these arcs retains feasibility of the solution without increasing the cost.

Algorithm EVENPARITY
Input: Mixed graph G = (V, E, A) and multiset M and U;
Output: Multiset M' and U' satisfying the same criteria as sets M and U from INOUTDEGREE, such that vertices in (V,U',M') are of even degree.
1. Identify the set of odd-degree vertices V' w.r.t (V, U, M). M' ← M\ and U' ← U). :: 2. Let \( M'' ⊆ M be the set of additional arcs and oriented edges created by INOUTDEGREE.
3. Call ADJUSTCYCLES (ADJUSTCYCLES identifies cycles consisting of alternating paths in M'' and U, with each path anchored at each end by an odd vertex and without considering the directions of arcs and oriented edges. Then the arcs and oriented edges on the cycles will be either duplicated or deleted, and edges on the path will be oriented. This process change the parity of odd vertices while maintaing indegree equal to outdegree and without increasing the cost)
While V' is not empty
Selet v from V' and set vs ← v
While v is in V'
Remove v from V'
Repeat
Remove <v,w> from M''
If <v,w> is directed toward w then
insert a duplicate copy of <v,w> to M'
else
delete a copy of <w,v> from M' and set v ← w
Until v is in V'
Remove v from V'
Repeat
Remove (v,w) from U'; Orient (v,w) from v to w and insert it into M'; Set v ← w
Until v is in V' or v = vs

Each odd vertex v has an odd number of elements from M( M''v), because #edgesv + #arcsv + #M''v is odd and #edgesv + #arcsv is even because this is the result got from EVENDEGREE. Here #edgesv represents the number of edges incident on v and arcsv represents the number of arcs incident on v, before INOUTDEGREE. Using the similar reasoning method, we know that every odd vertex has odd number of elements from M and odd number of elements from U; every even vertex has even number of elements from M'' and even number of elements from U

EVENPARITY will change the parity of odd vertices only because it applies to an even number of the arcs and oriented edges incident on an even vertex, and to an odd number of the arcs and edges incident on an odd vertex

M' and U' will not cost more than M and U. Suppose the cost of additional copies of edges and arcs in M' is greater than that in M. Then reverse the direction of each edge taken from U, insert two copies of each arc or oriented edge that was deleted, and delete two copies of each edge or arc that was inserted, which is a feasible solution of INOUTDEGREE and has less cost than M and U. This is impossible because M and U were created by a minimum-cost network flow algorithm

upload:zfu_mixed1.gif

The approximation ration of Mixed1 is 2 and it is tight

  1. C*: optimal solution (minimum cost)
  2. C1: the solution of Mixed1 on G
  3. C2: the solution of Mixed1 on G2, which results from duplicating every edge and arc of G

Let G2 be the graph results from duplicating every edge and arc of G. Every vertex of G2 has even degree, we only need run INOUTDEGREE and EVENPARITY for G2, which gets the total cost C2. C2 is optimal solution for G2 and doubling the optimal solution is a feasible solution of INOUTDEGREE, so C2 ≤ 2C*
Let G1 be the graph results from EVENDEGREE of G. Since EVENDEGREE employs a minimum-cost matching, no edge will be duplicated more than once. Thus G1 is a subgraph of G2 and the cost C1 got from applying Mixed1 on G1 is no more than C2
Therefore, C1 ≤ C2 ≤ 2C*

And the 2-approximation ratio is approachable, see the following example.The cost got from Mixed1 is
4+12ε (see b); and the optimal cost is 2+10ε (see c). When ε is small enough, the ration is 2.

upload:zfu_mixed1-worst.gif


5. Mixed2: 2 approximation algorithm ( G.N.Frederickson 1979 )

Mixed2 (example see Fig3)

  1. Make the indegree of each vertex equal to the outdegree by orienting some edges and duplicated arcs and oriented edges with minimum cost.(call INOUTDEGREE)
  2. The degree of each vertex is make even , while preserving the property that for each vertex the indegree equals the outdegree.(call LARGECYCLES)

Algorithm MIXED2

Input: Mixed graph G = (V, E, A) with cost function C; Output: A postman tour

  1. Call INOUTDEGREE
  2. Call LARGECYCLES

Algorithm LARGECYCLES
Input: Output of INOUTDEGREE
Output: A tour covering all edges and arcs of G
1. Identify the vertices with odd degree in the graph G'=(V,U) (U is the set of undirected edges got from INOUTDEGREE ).
2. Find all shortest paths between the odd vertices over the graph G''=(V,E) (V is the set of undirected edges in original graph).
3. Perform a minimum-cost matching of the odd vertices using the shortest path distances. Then insert the edges used in the matching into U.
4. Find a traversal of M and U.

The number of odd vertices is always even because 2(#edges + #arcs)=i is odd vertices #edges and arcs incident on i + j iseven vertices #edges and arcs incident on j . Since the latter part is a even number, to make the equation hold, the number of odd vertices has to be even, which make the matching always possible.

upload:zfu_mixed2.gif

The approximation ration of Mixed2 is 2 and it is tight

  1. C*: optimal solution (minimum cost)
  2. C: the solution of Mixed1 on G
  3. C0: the cost of all the edges and arcs in original graph
  4. C': the cost of the graph G' after INOUTDEGREE
  5. C'': the cost of additional edges added in LARGECYCLES

The graph resulting from assigning the direction to all the edges of optimal solution has the same cost as optimal cost C* and it is a feasible solution of INOUTDEGREE since it has equal indegree and outdegree for every vertex and it includes every edges( may be oriented) and arcs of original graph at least once. C' is the optimal solution of INOUTDEGREE, so C ≤ C*'.
In LARGECYCLES step , we perform the minimum-cost matching, which duplicate every edges at most once. Hence C'' ≤ C0.
Therefore C=C'+C''≤ C*+C ≤ 2C*.

And the 2-approximation ratio is approachable, see the following example.The cost got from Mixed2 is
2+3ε (see b); and the optimal cost is 4+2ε (see c). When ε is small enough, the ration is 2.

upload:zfu_mixed2-worst.gif


5. 3/5 approximation algorithm ( G.N.Frederickson 1979 )

If we examine the worst-cases of MIXED1 and MIXED2, we discover the following facts: Algorithm MIXED2 produces an optimum tour for the worst-case of MIEXED1 in Fig2; Algorithm MIXED1 produces an optimum tour for worst-case of MIXED2 in Fig4. This suggests that we could combine two approach to get better performance.

Algorithm GENERALMIXED

Input: Mixed graph G = (V, E, A) with cost function C; Output: A postman tour

  1. Call MIXED1
  2. Call MIXED2
  3. Select the tour of smaller cost

Now lets examine the approaximation ratio of GENERALMIXED.

C*: cost of optimal tour
CM: cost of edges and arcs in set M produced by INOUTDEGREE on the original graph

LEMMA: The cost C of the tour generated by Algorithm MIXED2 is at most 2C*-CM
Proof: When we apply MIXED2, the cost of edges not in M after INOUTDEGREE is at most C*-CM. And since in the worst case each edge in U will be duplicate in LARGECYCLES. So the total cost of undirected edges after MIXED2 is at most 2(C*-CM). And total cost is at most CM + 2(C*-CM)=2C*-CM.

LEMMA: The cost C of the tour generated by Algorithm MIXED1 is at most C*+2CM
Proof:Let G1 be the graph results from applying EVENDEGREE to G and E' be the set of undirected edges in G1. The total cost CE of E' is no larger than the cost of all arcs and edges in G1, which is no larger than C*.
Let M1 be the multiset of arcs and oriented edges such that for every element in M there are exactly two copies in M1. Hence, M1 achieves equal indegree and outdegree at every vertex in G1. Therefore, the cost of all arcs and edges in the graph resulting from INOUTDEGREE is bounded by CE+2CM, which is at most c*+2CM.

If the cost CM is large relative to C*, then MIXED2 is an appropriate approach, otherwise MIXED1 is an appropriate approach.

THEOREM: Algorithm GENERALMIXED produces a tour such that C/C* ≤ 5/3
PROOF:

if CM > C*/3 then consider the Algorithm MIXED2:
C/C* ≤ (2C*-CM)/C* ≤ (2C*-C*/3)/C* = 5/3
if CM ≤ C*/3 then consider the Algorithm MIXED1:
C/C* ≤ (2CM+C*)/C* ≤ (2C*/3 + C*)/C* = 5/3

We cannot find a tigh example of approximation ratio 5/3. Acturally the tight ratio is 3/2 (see reference 2).

Reference:

[1] Greg N. Frederickson, Approximation Algorithms for Some Postman Problems , J.ACM, 26(1979), pp.538-554
[2] B. Raghavachari and J. Veerasamy, A 3/2-Approximation Algorithm For the Mixed Postman Problem , SIAM J. Discrete Math. Vol. 12, No. 4, pp. 425-433


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Last edited March 19, 2004 6:29 pm by sx260-12.ipam.ucla.edu (diff)
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