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A Hamiltonian cycle (or Hamiltonian circuit) is a cycle that visits each vertex exactly once. A Hamiltonian path that starts and ends at adjacent vertices can be completed by adding one more edge to form a Hamiltonian cycle, and removing any edge from a Hamiltonian cycle produces a Hamiltonian path.
A verifier algorithm for Hamiltonian path will take as input a graph G, starting vertex s, and ending vertex t. Additionally, verifiers require a potential solution known as a certificate, c. For the Hamiltonian Path problem, c would consist of a string of vertices where the first vertex is the start of the proposed path and the last is the end ...
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.
(Hamilton originally thought in terms of moves between the faces of an icosahedron, which is equivalent by duality. This is the origin of the name "icosian". [3]) Hamilton's work in this area resulted indirectly in the terms Hamiltonian circuit and Hamiltonian path in graph theory. [4]
The Hamiltonian paths are in one-to-one correspondence with the minimal feedback arc sets of the tournament. [5] Rédei's theorem is the special case for complete graphs of the Gallai–Hasse–Roy–Vitaver theorem, relating the lengths of paths in orientations of graphs to the chromatic number of these graphs. [6]
Another version of Lovász conjecture states that . Every finite connected vertex-transitive graph contains a Hamiltonian cycle except the five known counterexamples.. There are 5 known examples of vertex-transitive graphs with no Hamiltonian cycles (but with Hamiltonian paths): the complete graph, the Petersen graph, the Coxeter graph and two graphs derived from the Petersen and Coxeter ...
Furthermore, the longest path problem is solvable in polynomial time on any class of graphs with bounded treewidth or bounded clique-width, such as the distance-hereditary graphs. Finally, it is clearly NP-hard on all graph classes on which the Hamiltonian path problem is NP-hard, such as on split graphs, circle graphs, and planar graphs.
The "compulsory" edges of the fragments, that must be part of any Hamiltonian path through the fragment, are connected at the central vertex; because any cycle can use only two of these three edges, there can be no Hamiltonian cycle. The resulting Tutte graph is 3-connected and planar, so by Steinitz' theorem it is the graph of a polyhedron. In ...