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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 (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.
Ullmann (1976) describes a recursive backtracking procedure for solving the subgraph isomorphism problem. Although its running time is, in general, exponential, it takes polynomial time for any fixed choice of H (with a polynomial that depends on the choice of H).
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 ...
The requirement of returning to the starting city does not change the computational complexity of the problem; see Hamiltonian path problem. Another related problem is the bottleneck travelling salesman problem: Find a Hamiltonian cycle in a weighted graph with the minimal weight of the weightiest edge.
The classic textbook example of the use of backtracking is the eight queens puzzle, that asks for all arrangements of eight chess queens on a standard chessboard so that no queen attacks any other. In the common backtracking approach, the partial candidates are arrangements of k queens in the first k rows of the board, all in different rows and ...
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.
Thus, each two consecutive permutations in the sequence generated by the Steinhaus–Johnson–Trotter algorithm correspond in this way to two vertices that form the endpoints of an edge in the permutohedron, and the whole sequence of permutations describes a Hamiltonian path in the permutohedron, a path that passes through each vertex exactly ...