Search results
Results From The WOW.Com Content Network
An Eulerian orientation of an undirected graph G is an assignment of a direction to each edge of G such that, at each vertex v, the indegree of v equals the outdegree of v. Such an orientation exists for any undirected graph in which every vertex has even degree, and may be found by constructing an Euler tour in each connected component of G ...
Since the graph corresponding to historical Königsberg has four nodes of odd degree, it cannot have an Eulerian path. An alternative form of the problem asks for a path that traverses all bridges and also has the same starting and ending point. Such a walk is called an Eulerian circuit or an Euler tour. Such a circuit exists if, and only if ...
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.
A Hamiltonian cycle around a network of six vertices Examples of Hamiltonian cycles on a square grid graph 8x8. In the mathematical field of graph theory, a Hamiltonian path (or traceable path) is a path in an undirected or directed graph that visits each vertex exactly once.
A k-barnacle is a path of length k between two nodes where every node on the path has degree 2. Flowering is the process of adding a 2-barnacle between two nodes on the shortest path between two odd-degree nodes. Flowering a tough, non-Hamiltonian graph that has an even number of nodes with odd degrees produces a Harris graph. [2]
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 ...
Euler stated the fundamental results for this problem in terms of the number of odd vertices in the graph, which the handshaking lemma restricts to be an even number. If this number is zero, an Euler tour exists, and if it is two, an Euler path exists. Otherwise, the problem cannot be solved.
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.