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File: EulerCycle ********> width="48%" !! Informally ... Given a network, or graph, an Euler Cycle is a tour from vertex to vertex, finishing where it started, and traversing every edge exactly once. In contrast: * An Euler Path traverses every edge but need not finish where it started; * A Hamilton Cycle visits every vertex exactly once and finishes where it started * ... so the start/end vertex is, in some sense, visited twice * ... except that starting at a vertex is not regarded as "a visit"; and * A Hamilton Path includes every vertex exactly once, * ... and need not finish where it started. ******** width="4%" ******** width="48%" !! More formally ... Given a graph $G=(V,E)$ where $|E|=m,$ an Euler Cycle is a list $L=[v_0,v_1,...,v_m]$ of vertices such that: * $v_0=v_m$ * That is, the cycle ends where it started * For every $i\in{1,2,...,m},$ $(v_{i-1},v_i)\in{E}$ * That is, going from vertex to vertex is along an edge * For all $e=(u,v)\in{E},$ there exists exactly one $i$ such that either $u=v_i$ and $v=v_{i+1}$ or /vice/versa./ * That is, every edge is traversed exactly once. A graph with an Euler Cycle is said to be "Eulerian." ********< ---- !! Observations ********> width="48%" This definition does not require the graph to be connected, but in the case of a graph with more than one component, all but at most one component must contain exactly one vertex and no edges. The "degree" of a vertex is the number of edges incident to it, and it is a well-known theorem that a connected graph is Eulerian if and only if every vertex has even degree. A graph has an Eulerian Path that is not a cycle if and only if every vertex has even degree except for two. ******** width="4%" ******** width="48%" The well-known Bridges of Koenigsberg problem asks if the road network of a particular town is Eulerian. Some versions require an Euler Cycle, but many ask only for an Euler Path. Neither is possible, as proven by LeonhardEuler in 1736. * https://en.wikipedia.org/wiki/Seven_Bridges_of_Königsberg TheDoodleTheorem says that a planar Eulerian graph can be two-coloured, and hence is bi-partite. ********<