Covering Trains by Stationsor The power of Data Reduction

Karsten Weihe, ALEX98, 1998

Presented by Yantao Song

Overview Problem description Data Reduction Computational study and

experiment results

Problem Given a set of trains, select a set of

stations such that every train meets at least one of these stations and the number of selected trains is minimum.

Formal Problem Description Given an undirected graph G=(V, E), paths p1,

p2……pn in G, and a partition V=V1∪ V2∪ …∪ Vm of V into m disjoint vertex classes.

A PCV (path-cover by vertices) is a subset such that every path pi meets at least one vertex in V ‘.

The problem is to find a PCV V’ of minimum size |V’|. More specifically, among all PCVs of minimum size,

V’ should maximize the vector ( | V’ ∩ V1|, | V’ ∩ V2|, …, | V’ ∩ Vm|) lexicographically.

This problem is NP-Hard.

VV '

Path pl is an ordered sequence (v1l, …, vnl

l ) of vertices such that {vi

l, vi+1l} ∈ E for i = 1, …, nl –

1. Vertices and edges may occur more than once in

the same path. If an edge occurs more than once, it may occur

several times with the same direction, or opposite direction.

It’s possible that two paths are exactly equal, or exact reverse of another path.

Without losing generality, we can assume that every edge belongs to some paths.

Paper’s background The data in this paper comes from the central

German train railroad company. Paths are the trains in the time schedules. V is the union of all stations met by at least one of

the trains. We have one edge {v,w} ∈ E iff v, w are directly

connected vertices by at least one train path. Purpose: find a minimum number of stations. It may be desirable to prefer some stations over

other stations. So we have to “maximize the vector ( | V’ ∩ V1|, | V’ ∩ V2|, …, | V’ ∩ Vm|) lexicographically” as described above.

Data Reduction For a vertex v∈ V , P(v) denotes

the set of all paths pi meeting v. For a path pi, V(pi) denotes the

ordinary set of vertices met by pi, which is unordered and don’t allow repetitions of vertices.

Vertex’s dominance and equivalence

Dominance: Let i, j∈{1,…,m}, v∈Vi w∈Vj , if i<j and P(v)=P(w) or i>=j and P(w) P(v), then we say that v dominate w.

Equivalence: if P(v)=P(w) and i=j, v and w is equivalent.

Path’s dominance and equivalence

Dominance: Let i, j∈{1,…,k}, pi pj , if V(pi) V(pj), then we say that pi dominate pj.

Equivalence: if V(pi) = V(pj), pi , pj

is equivalent.

Procedure of reducing vertex Remove v from V, and all edges

incident to v from E. If u, w∈V are incident to v, there is

a path pi which contains u-v-w or w-v-u as a subpath, then an edge {u, w} should be added into E.

All occurrences of v in paths are removed.

Procedures of reducing a path Remove pi from path set. Every edge e∈E which doesn’t

belong to any path afterward is removed.

Every vertex v∈V whose P(v) is empty afterwards is removed.

If the vertex/path is dominated by or equivalent to some other vertices/paths. Then it’s feasible to be reduced.

At the early stage of reduction, use non-exhaustive vertex reduction; at the end of reduction, use exhaustive reduction.

After reducing, we can get an irreducible core. An optimal solution to an irreducible core is also

an optimal solution to the original instance. Then we use the brute-force approach to solve

the problem.

Computational Study Experiments based on real world

data of Europe train network.

Train classes Class 0: high-speed trains; Class 1: other international or long-

distance trains; Class 2: regional trains; Class 3: local trains; Class 4: other trains;

Before Reduction

After reduction

For some instances which consist of trivial connected components, even we can get solution for problem only by data reduction.

For the other cases, the number of non-trivial connected components and size of these components are essential to complexity of the problem.

Conclusion and Discussion In this case, the size of problem can be

reduced to 10% of original size.The reduction algorithm is very efficient for this case.

This is an extreme case, can’t be extended to all cases with so high efficiency. But it give us an case that even the problem is NP-Hard, but we still can solve it in affordable time for some real world cases.