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Hamiltonian path problem : ウィキペディア英語版
Hamiltonian path problem

In the mathematical field of graph theory the Hamiltonian path problem and the Hamiltonian cycle problem are problems of determining whether a Hamiltonian path (a path in an undirected or directed graph that visits each vertex exactly once) or a Hamiltonian cycle exists in a given graph (whether directed or undirected). Both problems are NP-complete.〔 A1.3: GT37–39, pp. 199–200.〕
There is a simple relation between the problems of finding a Hamiltonian path and a Hamiltonian cycle. In one direction, the Hamiltonian path problem for graph G is equivalent to the Hamiltonian cycle problem in a graph H obtained from G by adding a new vertex and connecting it 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 Hamiltonian cycle.
In the other direction, the Hamiltonian cycle problem for a graph G is equivalent to the Hamiltonian path problem in the graph H obtained by copying one vertex v of G, v', that is, letting v' have the same neighbourhood as v, and by adding two dummy vertices of degree one, and connecting them with v and v', respectively.〔http://math.stackexchange.com/questions/7130/reduction-from-hamiltonian-cycle-to-hamiltonian-path/1290804#1290804〕
The Hamiltonian cycle problem is also a special case of the travelling salesman problem, obtained by setting the distance between two cities to one if they are adjacent and two otherwise, and verifying that the total distance travelled is equal to n (if so, the route is a Hamiltonian circuit; if there is no Hamiltonian circuit then the shortest route will be longer).
==Algorithms==
There are ''n''! different sequences of vertices that ''might'' be Hamiltonian paths in a given ''n''-vertex graph (and are, in a complete graph), so a brute force search algorithm that tests all possible sequences would be very slow. There are several faster approaches. A search procedure by Frank Rubin〔.〕 divides the edges of the graph into three classes: those that must be in the path, those that cannot be in the path, and undecided. As the search proceeds, a set of decision rules classifies the undecided edges, and determines whether to halt or continue the search. The algorithm divides the graph into components that can be solved separately. Also, a dynamic programming algorithm of Bellman, Held, and Karp can be used to solve the problem in time O(''n''2 2''n''). In this method, one determines, for each set ''S'' of vertices and each vertex ''v'' in ''S'', whether there is a path that covers exactly the vertices in ''S'' and ends at ''v''. For each choice of ''S'' and ''v'', a path exists for (''S'',''v'') if and only if ''v'' has a neighbor ''w'' such that a path exists for (''S'' − ''v'',''w''), which can be looked up from already-computed information in the dynamic program.〔.〕〔.〕
Andreas Björklund provided an alternative approach using the inclusion–exclusion principle to reduce the problem of counting the number of Hamiltonian cycles to a simpler counting problem, of counting cycle covers, which can be solved by computing certain matrix determinants. Using this method, he showed how to solve the Hamiltonian cycle problem in arbitrary ''n''-vertex graphs by a Monte Carlo algorithm in time O(1.657''n''); for bipartite graphs this algorithm can be further improved to time O(1.414''n'').〔.〕
For graphs of maximum degree three, a careful backtracking search can find a Hamiltonian cycle (if one exists) in time O(1.251''n'').〔.〕
Because of the difficulty of solving the Hamiltonian path and cycle problems on conventional computers, they have also been studied in unconventional models of computing. For instance,
Leonard Adleman showed that the Hamiltonian path problem may be solved using a DNA computer. Exploiting the parallelism inherent in chemical reactions, the problem may be solved using a number of chemical reaction steps linear in the number of vertices of the graph; however, it requires a factorial number of DNA molecules to participate in the reaction.〔.〕

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