CS 453 Assignment 2 solved

Description

1. Among all simple graphs (no loops, no parallel edges) with 𝑛 = 100
vertices, determine (with justification) the minimum possible and the
maximum possible values for 𝑚, the number of edges such a graph could
have.
2. Let 𝐺 be the simple graph with vertex set
𝑉 = {00, 01, 02, 10, 11, 12, 20, 21, 22}; |𝑉| = 9
and where vertices 𝑎𝑏 and 𝑐𝑑 are joined by an edge when exactly one of
the following conditions holds (so there are no loops):
𝑎 = 𝑐 or 𝑏 = 𝑑.
A. Sketch 𝐺; you are allowed to do this by hand and then copy your sketch
electronically into your PDF submission.
B. Determine 𝑚, the number of edges of 𝐺.
3. The “grid graph” 𝑃𝑟,𝑠
is the cartesian product of the two paths 𝑃𝑟 and
𝑃𝑠
. For instance, 𝑃3,4
is drawn below:
A. In terms of 𝑟 and 𝑠, find a formula for the number of vertices of the
grid graph 𝑃𝑟,𝑠
.
B. In terms of 𝑟 and 𝑠, find a formula for the number of edges of the
grid graph 𝑃𝑟,𝑠
.
4. Recall that a graph 𝐺 is “cubic” if and only if it is 3-regular.
A. Show that there exists no cubic graph with an odd number of
vertices.
B. For every integer 𝑛 ≥ 3, show that there exists a simple cubic
graph (no loops, no parallel edges) with 2𝑛 vertices. One way to do
this is to produce a construction, i.e., give a set of 2𝑛 vertices and a
recipe for when vertices are joined by edges for constructing such
graphs.