By Goldberg A.V.
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Additional resources for Combinatorial Optimization Lecture Notes
At all times a pre ow f and a distance labeling d are maintained and push and relabel operations are applied until there are no active nodes. Lemma 6. If f is a pre ow and d is a distance labeling, t is not reachable from s in the residual graph Gf . Proof. Suppose there is an augmenting path s = v0 v1 ::: vl = t. We may assume the path is simple (otherwise remove any loops in the path) so that l < n. Since d is a distance labeling we know that for any arc (vi vi+1) on the path, d(vi) d(vi+1) + 1.
2. Problem De nition Given a directed and strongly connected graph G = (V E ) and weight function f : E ! R, let the weight of a path P be X w(P ) = f (e) and the mean weight of P be e2P m(P ) = X f (e) e2P jP j where jP j is the number of arcs of the path P . 42 2. PROBLEM DEFINITION 43 The problem is to nd a cycle C in G such that for any cycle C in G, m(C ) m(C ). Note that if the graph is not strongly connected, we can determine the minimum mean cycle for each strongly connected component, and then take the least of these.
23. Example graph. Example 4. 23 as an example. There are two cycles in G, zuyt and tuy . m(zuyt) = (2 = 1 = 1 ; 2)=4 = 1=2 m(tuy) = (1 + 1 ; 2)=3 = 0=3 = 0: Clearly, cycle tuy has the minimum mean weight. Let Fk (v ) be de ned as the total weight of a minimum path from an arbitrarily xed \reference vertex" s to vertex v , having exactly k arcs. If no such path exists with exactly k arcs, then Fk (v ) is de ned to be 1. The following theorem relates a minimum mean cycle with those minimum paths from a reference node.