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By Knopp M.I. (ed.)

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3. If G is eulerian, then any trail of G constructed by Fleury’s algorithm is an Euler tour of G. Proof. Exercise. ⊓ ⊔ If G is not eulerian, the poor postman has to walk at least one street twice. , if one of the streets is a dead end, and in general if there is a street corner of an odd number of streets. We can attack this case by reducing it to the eulerian case as follows. An edge e = uv will be duplicated, if it is added to G parallel to an existing edge e′ = uv with the same weight, α(e′ ) = α(e).

2 Hamiltonian graphs 31 4 4 3 3 2 1 4 3 3 2 2 2 1 2 2 2 3 3 1 3 2 Above we have duplicated two edges. The rightmost multigraph is eulerian. There is a good algorithm by EDMONDS AND JOHNSON (1973) for the construction of an optimal eulerian supergraph by duplications. Unfortunately, this algorithm is somewhat complicated, and we shall skip it. 2 Hamiltonian graphs In the connector problem we reduced the cost of a spanning graph to its minimum. There are different problems, where the cost is measured by an active user of the graph.

Notice that if W = e1 e2 . . en is an Euler tour (and so EG = {e1 , e2 , . . , en }), also ei ei+1 . . en e1 . . ei−1 is an Euler tour for all i ∈ [1, n]. A complete proof of the following Euler’s Theorem was first given by HIERHOLZER in 1873. 1 (EULER (1736), HIERHOLZER (1873)). A connected graph G is eulerian if and only if every vertex has an even degree. ⋆ Proof. (⇒) Suppose W : u − → u is an Euler tour. Let v (= u) be a vertex that occurs k times in W . Every time an edge arrives at v, another edge departs from v, and therefore dG (v) = 2k.

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