Question

Make a general conjecture about the minimum number of edges in a graph with n vertices and r components, where n, r >= 1. Then prove this conjecture.

Answer #1

How many vertices and edges does the complete tripartite graph
K_m,n,p have? Prove your conjecture.

Prove that your conjecture will hold for any connected
graph.
If a graph is connected graph, then the number of
vertices plus the number of regions minus two equals the number of
edges

Let G be an undirected graph with n vertices and m edges. Use a
contradiction argument to prove that if m<n−1, then G is not
connected

A simple undirected graph consists of n vertices in a single
component.
What is the maximum possible number of edges it could have?
What is the minimum possible number of edges it could have?
Prove that your answers are correct

Prove that a bipartite simple graph with n vertices must have at
most n2/4 edges. (Here’s a hint. A bipartite graph would have to be
contained in Kx,n−x, for some x.)

please solve it step by step. thanks
Prove that every connected graph with n vertices has at least
n-1 edges. (HINT: use induction on the number of vertices
n)

Call a graph on n vertices dendroid if it has n edges and is
connected. Characterize degree sequences of dendroids.

Consider the complete bipartite graph Kn,n with 2n vertices. Let
kn be the number of edges in Kn,n. Draw K1,1, K2,2 and K3,3 and
determine k1, k2, k3. Give a recurrence relation for kn and solve
it using an initial value.

Let G be a connected simple graph with n vertices and m edges.
Prove that G contains at least m−n+ 1 different subgraphs
which are polygons (=circuits). Note: Different polygons
can have edges in common. For instance, a square with a diagonal
edge has three different polygons (the square and two different
triangles) even though every pair of polygons have at least one
edge in common.

In lecture, we proved that any tree with n vertices must have n
− 1 edges. Here, you will prove the converse of this statement.
Prove that if G = (V, E) is a connected graph such that |E| =
|V| − 1, then G is a tree.

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