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The sum of degrees of all six vertices is 2 + 3 + 2 + 3 + 3 + 1 = 14, twice the number of edges. In graph theory , 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 two convex polygons P and Q in the plane with m and n vertices, their Minkowski sum is a convex polygon with at most m + n vertices and may be computed in time O(m + n) by a very simple procedure, which may be informally described as follows. Assume that the edges of a polygon are given and the direction, say, counterclockwise, along the ...
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).
Three points property: Three points either lie on a line or lie on a circle. Pythagoras' theorem: In a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides. [1] An easy formula for these properties is that in any three points in any shape, there is a triangle formed.
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 ...
For a polyhedron, the defect at a vertex equals 2π minus the sum of all the angles at the vertex (all the faces at the vertex are included). If a polyhedron is convex, then the defect of each vertex is always positive. If the sum of the angles exceeds a full turn, as occurs in some vertices of many non-convex polyhedra, then the defect is ...
The maximum independent set problem is the special case in which all weights are one. In the maximal independent set listing problem, the input is an undirected graph, and the output is a list of all its maximal independent sets. The maximum independent set problem may be solved using as a subroutine an algorithm for the maximal independent set ...
Each vertex of the independent set is adjacent to vertices of the matching, and each vertex of the matching is adjacent to vertices of the independent set. [2] Because of this decomposition, and because odd graphs are not bipartite, they have chromatic number three: the vertices of the maximum independent set can be assigned a single color, and ...