In a connected component of a graph with Mi vertices, the maximum number of edges is
MiC2 or Mi(Mi-1)/2. So if we have k components and each component has Mi vertices
then the maximum number of edges for the graph is M1C2+M2C2+...+MKC2.
Of course the sum of Mi as i goes from 1 to k must be n since the sum of the vertices in each component is the sum of all the vertices in the graph which you gave as n.
Where MC2 means choose 2 from M and there are M(M-1)/2 ways to do that.
Let G be a complete graph with n vertices. Consider the case where n=2. With only 2 vertices it is clear that there will only be one edge. Now add one more vertex to get n = 3. We must now add edges between the two old vertices and the new one for a total of 3 vertices. We see that adding a vertex to a graph with n vertices gives us n more edges. We get the following sequence Edges on a graph with n vertices: 0+1+2+3+4+5+...+n-1. Adding this to itself and dividing by two yields the following formula for the number of edges on a complete graph with n vertices: n(n-1)/2.
A star graph, call it S_k is a complete bipartite graph with one vertex in the center and k vertices around the leaves. To be a tree a graph on n vertices must be connected and have n-1 edges. We could also say it is connected and has no cycles. Now a star graph, say S_4 has 3 edges and 4 vertices and is clearly connected. It is a tree. This would be true for any S_k since they all have k vertices and k-1 edges. And Now think of K_1,k as a complete bipartite graph. We have one internal vertex and k vertices around the leaves. This gives us k+1 vertices and k edges total so it is a tree. So one way is clear. Now we would need to show that any bipartite graph other than S_1,k cannot be a tree. If we look at K_2,k which is a bipartite graph with 2 vertices on one side and k on the other,can this be a tree?
You count the edges
57
the center of the cult. Not the edges or outside. Not necessarily inside, but rather absolutely central, at the center most point, equally positioned in diameter from the edges of the cult.
5 vertices and 8 edges.5 vertices and 8 edges.5 vertices and 8 edges.5 vertices and 8 edges.
Pretty simple really: Vertices are corners and edges are boundaries so, a hexagon has six of each.
No. Edges join vertices; or, put another way, edges meet at vertices.
A heptahedron has 7 faces. It can have 6 vertices and 11 edges, or7 vertices and 12 edges, or8 vertices and 13 edges, or9 vertices and 14 edges, or10 vertices and 15 edges.
A point would have neither edges or vertices
Edges are the lines that connect the vertices. The vertices are the actual points where the edges meet.
Vertices are points (corners) and edges are lines that connect vertices
n-k-1
12 edges, 8 vertices
# of vertices: 6 # of edges: 5
A triangle has 3 edges and 3 vertices. A triangular prism has 9 edges and 6 vertices.
heck if I know * * * * * 8 edges and 5 vertices (not vertices's).