Search results
Results from the WOW.Com Content Network
For graphs that are allowed to contain loops connecting a vertex to itself, a loop should be counted as contributing two units to the degree of its endpoint for the purposes of the handshaking lemma. [2] Then, the handshaking lemma states that, in every finite graph, there must be an even number of vertices for which is an odd number. [1]
The degree sum formula states that, given a graph = (,), = | |. The formula implies that in any undirected graph, the number of vertices with odd degree is even. This statement (as well as the degree sum formula) is known as the handshaking lemma. The latter name comes from a popular mathematical problem, which is to prove that in any group ...
From the handshaking lemma, a k-regular graph with odd k has an even number of vertices. A theorem by Nash-Williams says that every k ‑regular graph on 2k + 1 vertices has a Hamiltonian cycle. Let A be the adjacency matrix of a graph. Then the graph is regular if and only if = (, …,) is an eigenvector of A. [2]
Language links are at the top of the page across from the title.
A directed graph is weakly connected (or just connected [9]) if the undirected underlying graph obtained by replacing all directed edges of the graph with undirected edges is a connected graph. A directed graph is strongly connected or strong if it contains a directed path from x to y (and from y to x ) for every pair of vertices ( x , y ) .
Download as PDF; Printable version; In other projects ... Help. Pages in category "Lemmas in graph theory" The following 5 pages are in this category, out of 5 total ...
The total degree is the sum of the degrees of all vertices; by the handshaking lemma it is an even number. The degree sequence is the collection of degrees of all vertices, in sorted order from largest to smallest. In a directed graph, one may distinguish the in-degree (number of incoming edges) and out-degree (number of outgoing edges). [2] 2.
To decide if a graph has a Hamiltonian path, one would have to check each possible path in the input graph G. 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.