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[1] [2] Harris graphs were introduced in 2013 when, at the University of Michigan, Harris Spungen conjectured that a tough, Eulerian graph would be sufficient to be Hamiltonian. [3] However, Douglas Shaw disproved this conjecture, discovering a counterexample with an order of 9 and a size of 14. [1] Currently, there are 241,375 known Harris ...
This tour corresponds to a Hamiltonian cycle in the line graph L(G), so the line graph of every Eulerian graph is Hamiltonian. Line graphs may have other Hamiltonian cycles that do not correspond to Euler tours, and in particular the line graph L(G) of every Hamiltonian graph G is itself Hamiltonian, regardless of whether the graph G is ...
A connected graph has an Euler cycle if and only if every vertex has even degree. The term Eulerian graph has two common meanings in graph theory. One meaning is a graph with an Eulerian circuit, and the other is a graph with every vertex of even degree. These definitions coincide for connected graphs. [2]
The proof is an easy consequence of Euler's formula. [1] [2] As a corollary of this theorem, if an embedded planar graph has only one face whose number of sides is not 2 mod 3, and the remaining faces all have numbers of sides that are 2 mod 3, then the graph is not Hamiltonian.
In graph theory, an n-dimensional De Bruijn graph of m symbols is a directed graph representing overlaps between sequences of symbols. It has m n vertices, consisting of all possible length-n sequences of the given symbols; the same symbol may appear multiple times in a sequence. For a set of m symbols S = {s 1, …, s m}, the set of vertices is:
In one direction, the Hamiltonian path problem for graph G can be related to the Hamiltonian cycle problem in a graph H obtained from G by adding a new universal vertex x, connecting x to all vertices of G. Thus, finding a Hamiltonian path cannot be significantly slower (in the worst case, as a function of the number of vertices) than finding a ...
The cycle space, also, has an algebraic structure, but a more restrictive one. The union or intersection of two Eulerian subgraphs may fail to be Eulerian. However, the symmetric difference of two Eulerian subgraphs (the graph consisting of the edges that belong to exactly one of the two given graphs) is again Eulerian. [1]
Illustration for the proof of Ore's theorem. In a graph with the Hamiltonian path v 1...v n but no Hamiltonian cycle, at most one of the two edges v 1 v i and v i − 1 v n (shown as blue dashed curves) can exist. For, if they both exist, then adding them to the path and removing the (red) edge v i − 1 v i would produce a Hamiltonian cycle.