Lecture Notes Discrete Optimization - Applied Mathematics

4. Shortest Paths
4.1 Introduction
We next consider shortest path problems. These problems are usually defined for directed
networks. Let G =(V,E) be a directed graph with cost function c : E → R. Consider a
(directed) path P=〈v 1 ,...,v k 〉 from s=v 1 to t = v k . The length of path P is defined as
c(P)=∑ k−1
i=1 c(v i,v i+1 ). We can then ask for the computation of an s,t-path whose length
is shortest among all directed paths from s to t. There are different variants of shortest
path problems:
1. Single source single target shortest path problem: Given two nodes s and t, determine
a shortest path from s to t.
2. Single source shortest path problem: Given a node s, determine all shortest paths
from s to every other node in V.
3. All-pairs shortest path problem: For every pair(s,t)∈ V ×V of nodes, compute a
shortest path from s to t.
The first problem is a special case of the second one. However, every known algorithm for
the first problem implicitly also solves the second one (at least partially). We therefore
focus here on the single source shortest path problem and the all-pairs shortest path
problem.
4.2 Single Source Shortest Path Problem
We consider the following problem:
Single Source Shortest Path Problem (SSSP):
Given: A directed graph G=(V,E) with cost function c : E →R and a source
node s∈ V.
Goal: Compute a shortest path from s to every other node v∈ V.
Note that a shortest path from s to a node v might not necessarily exist because of the
following two reasons: First, v might not be reachable from s because there is no directed
path from s to v in G. Second, there might be arbitrarily short paths from s to v because
of the existence of an s,v-path that contains a cycle of negative length (which can be
traversed arbitrarily often). We call a cycle of negative total length also a negative cycle.
The following lemma shows that these are the only two cases in which no shortest path
exists.
Lemma 4.1. Let v be a node that is reachable from s. Further assume that there is no
path from s to v that contains a negative cycle. Then there exists a shortest path from s to
v which is a simple path.
Proof. Let P be a path from s to v. We can repeatedly remove cycles from P until we
obtain a simple path P ′ . By assumption, all these cycles have non-negative lengths and
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