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The Hamiltonian cycle problem is similar to the Hamiltonian path problem, except it asks if a given graph contains a Hamiltonian cycle. This problem may also specify the start of the cycle. The Hamiltonian cycle problem is a special case of the travelling salesman problem, obtained by setting the distance between two cities to one if they are ...
In an undirected graph, the edge to the parent of a node should not be counted as a back edge, but finding any other already visited vertex will indicate a back edge. In the case of undirected graphs, only O(n) time is required to find a cycle in an n-vertex graph, since at most n − 1 edges can be tree edges.
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 ...
Cycle detection is the problem of finding i and j, given f and x 0. Several algorithms are known for finding cycles quickly and with little memory. Robert W. Floyd 's tortoise and hare algorithm moves two pointers at different speeds through the sequence of values until they both point to equal values.
If the graph is connected, it finds a minimum spanning tree. It is a greedy algorithm that in each step adds to the forest the lowest-weight edge that will not form a cycle. [2] The key steps of the algorithm are sorting and the use of a disjoint-set data structure to detect cycles. Its running time is dominated by the time to sort all of the ...
An Eulerian cycle, [note 1] also called an Eulerian circuit or Euler tour, in an undirected graph is a cycle that uses each edge exactly once. If such a cycle exists, the graph is called Eulerian or unicursal. [4] The term "Eulerian graph" is also sometimes used in a weaker sense to denote a graph where every vertex has even degree.
The windy postman problem is a variant of the route inspection problem in which the input is an undirected graph, but where each edge may have a different cost for traversing it in one direction than for traversing it in the other direction. In contrast to the solutions for directed and undirected graphs, it is NP-complete. [11] [12]
The union of the tree and the matching is a cycle, with no possible shortcuts, and with weight approximately 3n/2. However, the optimal solution uses the edges of weight 1 + ε together with two weight-1 edges incident to the endpoints of the path, and has total weight (1 + ε)(n − 2) + 2, close to n for small values of ε. Hence we obtain an ...