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The problems of finding a Hamiltonian path and a Hamiltonian cycle can be related as follows: 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 ...
A Hamiltonian cycle, Hamiltonian circuit, vertex tour or graph cycle is a cycle that visits each vertex exactly once. A graph that contains a Hamiltonian cycle is called a Hamiltonian graph . Similar notions may be defined for directed graphs , where each edge (arc) of a path or cycle can only be traced in a single direction (i.e., the vertices ...
This is more general than the Hamiltonian path problem, which only asks if a Hamiltonian path (or cycle) exists in a non-complete unweighted graph. The requirement of returning to the starting city does not change the computational complexity of the problem; see Hamiltonian path problem.
Alternatively, the problem can be solved by performing a binary search or sequential search for the smallest x such that the subgraph of edges of weight at most x has a Hamiltonian cycle. This method leads to solutions whose running time is only a logarithmic factor larger than the time to find a Hamiltonian cycle. [1]
A Hamiltonian cycle in a directed graph is a cycle that passes through each vertex of the graph exactly once. The following Lparse program can be used to find a Hamiltonian cycle in a given directed graph if it exists; we assume that 0 is one of the vertices.
The knight's tour problem is an instance of the more general Hamiltonian path problem in graph theory. The problem of finding a closed knight's tour is similarly an instance of the Hamiltonian cycle problem. Unlike the general Hamiltonian path problem, the knight's tour problem can be solved in linear time. [4]
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.
This argument also gives an algorithm for finding the Hamiltonian path. More efficient algorithms, that require examining only O ( n log n ) {\displaystyle O(n\log n)} of the edges, are known. The Hamiltonian paths are in one-to-one correspondence with the minimal feedback arc sets of the tournament. [ 5 ]