- Among all simple graphs (no loops, no parallel edges) with π = 100 vertices, determine (with justification) the minimum possible and the maximum possible values for π, the number of edges such a graph could have.
- LetπΊbethesimplegraphwithvertexset
π = {00,01,02,10,11,12,20,21,22}; |π| = 9and where vertices ππ and ππ are joined by an edge when exactly one of the following conditions holds (so there are no loops):
π = π or π = π.
A. Sketch πΊ; you are allowed to do this by hand and then copy your sketch electronically into your PDF submission.
B. Determine π, the number of edges of πΊ.
3. The βgrid graphβ π π,π
π . For instance, π
π 3,4
is the cartesian product of the two paths π and π
is drawn below:
A. In terms of π and π , find a formula for the number of vertices of the grid graph π .
π,π
B. In terms of π and π , find a formula for the number of edges of the
grid graph π . π,π
4. Recall that a graph πΊ is βcubicβ if and only if it is 3-regular.
- Show that there exists no cubic graph with an odd number of
vertices.
- For every integer π β₯ 3, show that there exists a simple cubic
graph (no loops, no parallel edges) with 2π vertices. One way to do this is to produce a construction, i.e., give a set of 2π vertices and a recipe for when vertices are joined by edges for constructing such graphs.
