In this graph, an even number of vertices (the four vertices numbered 2, 4, 5, and 6) have odd degrees. The sum of degrees of all six vertices is 2 + 3 + 2 + 3 + 3 + 1 = 14, twice the number of edges.
In graph theory, a branch of mathematics, the handshaking lemma is the statement that, in every finite undirected graph, the number of vertices that touch an odd number of edges is even. For example, if there is a party of people who shake hands, the number of people who shake an odd number of other people's hands is even.[1] The handshaking lemma is a consequence of the degree sum formula, also sometimes called the handshaking lemma,[2] according to which the sum of the degrees (the numbers of times each vertex is touched) equals twice the number of edges in the graph. Both results were proven by Leonhard Euler (1736) in his famous paper on the Seven Bridges of Königsberg that began the study of graph theory.[3]
Beyond the Seven Bridges of Königsberg Problem, which subsequently formalized Eulerian Tours, other applications of the degree sum formula include proofs of certain combinatorial structures. For example, in the proofs of Sperner's lemma and the mountain climbing problem the geometric properties of the formula commonly arise. The complexity class PPA encapsulates the difficulty of finding a second odd vertex, given one such vertex in a large implicitly-defined graph.
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In graph theory, a branch of mathematics, the handshakinglemma is the statement that, in every finite undirected graph, the number of vertices that touch...
equilateral triangle. There are infinitely many deltahedra. By the handshakinglemma, each deltahedron has an even number of faces. Only eight deltahedra...
considering appropriate parameters for circulant graphs. From the handshakinglemma, a k-regular graph with odd k has an even number of vertices. A theorem...
T of G. Let O be the set of vertices with odd degree in T. By the handshakinglemma, O has an even number of vertices. Find a minimum-weight perfect matching...
The total degree is the sum of the degrees of all vertices; by the handshakinglemma it is an even number. The degree sequence is the collection of degrees...
the whole graph is connected (otherwise no tour exists), and by the handshakinglemma it has an even number of odd vertices, so a T-join always exists....
in which all vertices have odd degree, an argument related to the handshakinglemma shows that the number of Hamiltonian cycles through any fixed edge...
between this lower bound and the n/6 upper bound. It follows from the handshakinglemma, proven by Leonhard Euler in 1736 as part of the first paper on graph...
the graph G {\displaystyle G} is not connected.) According to the handshakinglemma, every connected component of an undirected graph has an even number...
matroid, an abstract generalization of Eulerian graphs Five room puzzle Handshakinglemma, proven by Euler in his original paper, showing that any undirected...
Combinatorial Theory, 9 (1): 54–59, doi:10.1016/S0021-9800(70)80054-0 By the handshakinglemma, a face-regular polyhedron with an odd number of faces must have faces...
the sum of the numbers of neighbors of all vertices, which (by the handshakinglemma) is proportional to the number of input edges. van Rooij & Wilf (1965)...
Argument") is the class of problems whose solution is guaranteed by the handshakinglemma: any undirected graph with an odd degree vertex must have another...
vertices (such as the odd complete graphs); for such graphs, by the handshakinglemma, k must itself be even. However, the inequality χ′ ≥ m/β does not...
is self-dual, is one example. Another given by Harary involves the handshakinglemma, according to which the sum of the degrees of the vertices of any...
{\displaystyle n!} vertices. Its degree is n − 1, hence, according to the handshakinglemma, it has 1 2 n ! ( n − 1 ) {\displaystyle {\dfrac {1}{2}}~n!\left(n-1\right)}...
in the graph, and must find a second vertex of odd degree. By the handshakinglemma, such a vertex exists; finding one is a problem in NP, but the problems...
thrackle in which every vertex has at most two neighbors, by the handshakinglemma the number of edges is at most the number of vertices. Based on Erdős'...
edges on the Hamiltonian cycle (which must have even length by the handshakinglemma) and a third color for all remaining edges. Therefore, all snarks...
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