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By Tero Harju

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This problem is akin to Euler’s problem and to the shortest path problem. Let G be a graph with a weight function α : EG → R + . The Chinese postman problem is to find a minimum weighted tour in G (starting from a given vertex, the post office). If G is eulerian, then any Euler tour will do as a solution, because such a tour traverses each edge exactly once and this is the best one can do. In this case the weight of the optimal tour is the total weight of the graph G, and there is a good algorithm for finding such a tour: Fleury’s algorithm: • Let v0 ∈ G be a chosen vertex, and let W0 be the trivial path on v0 .

Sm } be a family of finite nonempty subsets of a set S. ) A transversal (or a system of distinct representatives) of S is a subset T ⊆ S of m distinct elements one from each Si . As an example, let S = [1, 6], and let S1 = S2 = {1, 2}, S3 = {2, 3} and S4 = {1, 4, 5, 6}. For S = {S1 , S2 , S3 , S4 }, the set T = {1, 2, 3, 4} is a transversal. If we add the set S5 = {2, 3} to S , then it is impossible to find a transversal for this new family. The connection of transversals to the Marriage Theorem is as follows.

Otherwise, since now dG (v) is even for all v, G has a vertex v1 with dG (v1 ) ≥ 4. Let e1 e2 . . et be an Euler tour of G, where ei = vi vi+1 (and vt+1 = v1 ). Define 1, if i is odd , α( ei ) = 2, if i is even . Hence the ends of the edges ei for i ∈ [2, t − 1] are incident with both colours. All vertices are among these ends. The condition dG (v1 ) ≥ 4 guarantees this for v1 . Hence the claim holds in the eulerian case. (2) Suppose then that G is not eulerian. We define a new graph G0 by adding a vertex v0 to G and connecting v0 to each v ∈ G of odd degree.

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