What is the maximum possible number of edges of a graph with n vertices and k components?
Table of Contents
- 1 What is the maximum possible number of edges of a graph with n vertices and k components?
- 2 How do you find the number of edges with vertices?
- 3 How do you find the vertices and edges of a graph?
- 4 How do you know how many components a graph has?
- 5 How do you find the number of edges on a planar graph?
What is the maximum possible number of edges of a graph with n vertices and k components?
ni −k). Theorem 4. A simple graph with n vertices and k components can have at most have (n−k)(n− k+1)/2 edges. As a direct application of Theorem 4, we have the following result.
How do you find the number of edges with vertices?
The sum of the vertex degree values is twice the number of edges, because each of the edges has been counted from both ends. In your case 6 vertices of degree 4 mean there are (6×4)/2=12 edges.
What is the minimum number of edges necessary in a simple planar graph with 15 regions?
In a simple planar graph, degree of each region is >= 3. So, we have 3 x |R| <= 2 x |E|. Thus, Minimum number of edges required in G = 23. Get more notes and other study material of Graph Theory.
How do you calculate the number of edges?
In a complete graph, every pair of vertices is connected by an edge. So the number of edges is just the number of pairs of vertices.
How do you find the vertices and edges of a graph?
Graph Theory Definition − A graph (denoted as G = (V, E)) consists of a non-empty set of vertices or nodes V and a set of edges E. A vertex a represents an endpoint of an edge. An edge joins two vertices a, b and is represented by set of vertices it connects.
How do you know how many components a graph has?
A graph can be partitioned into pieces each of which is connected. Each piece is called a component. For example, the graph above has two components— a, b, c, d is one and e is the other. has three components: a, b is one, c, d is a second, and e is a third.
How can you prove that if a graph has exactly 2 vertices of an odd degree there must be a path joining these 2 vertices?
Since G contains exactly two vertices of odd degree, we must have v∈J. This implies there is a path from u to v. Let G be a graph with exactly 2 vertices of odd degree u and v. If G has a cycle (Cn for n≥3), then delete it (this preserves the parity of the vertex degrees).
How do you find the number of edges with vertices and faces?
The edges of a polyhedron are the edges where the faces meet each other. The vertices are the corners of the polyhedron. Euler’s Formula tells us that if we add the number of faces and vertices together and then subtract the number of edges, we will get 2 as our answer. The formula is written as F + V – E = 2.
How do you find the number of edges on a planar graph?
Put the vertices in a unit circle equally spaced. That is, on the n complex roots of the equation zn−1=0. Each vertex can visit exactly n−1 other vertices, for a total of n(n−1). But each edge was counted twice (from vi to vj and from vj to vi) so the exact maximum is n(n−1)/2.