Homework 5: Solutions
1. We will do an induction on the number of edges E.
Base case: E = 0. If there are no edges then the graph is bipartite as we can colour every
vertex either black or white.
Induction step: Assume every vert
Homework 6: Solutions
1. In order to get from graph G1 to G2 we repeatedly either add in a vertex to an edge
increasing both the number of vertices and number of edges by 1, or delete a vertex of degree
2 decreasing both the numb
Homework 9: Solutions
1. Drawing a graph with a vertex for each lecture, and en edge between them if they must
not coincide we get the following.
Notice there is a K4 subgraph so we need at least 4 colours. If we co
Homework 1: Solutions
1. Well prove this by weak induction.
Base case: 28 = 256 256 = 44 .
Induction step: Now assume the result is true for n = k, so 22k k 4 . For n = k + 1
(k + 1)4 = k 4 + 4k 3 + 6k 2 + (4k + 1) k 4 + k 4 + k
Homework 4: Solutions
(a) Hamiltonian: For the graph above trace the red path shown.
Not Eulerian: Since there exists vertices of odd degree the graph is not Eulerian by
(b) Eulerian: Since the degree of eve
Homework 1 - due January 5th
1. Prove that n! 2n for all n 4.
2. Prove that 1 + x + + xn1 +
for all n 1.
3. Prove that xn 1 = (x 1)(xn1 + + x + 1) for all n 1.
4. Prove that 5 divides 34n 1 for all n 1.
Homework 8: Solutions
1. We will do a strong induction on the number of edges E.
Base case: E = 1. Let G be a simple graph with n vertices where n 2 and one edge e.
Then by deletion-contraction and Proposition 6 we have
PG (k) =
Homework 3: Solutions
1. Each edge has 2 ends, and each end contributes 1 to the total sum of the degrees. Hence
if there are e edges in the graph the total sum of the degrees is 2e, which is even.
Let G be a graph with total sum
Homework 2: Solutions
1. If we cross each bridge only once as we go in and out of each landmass, we use up
two bridges so the number of bridges emerging from each landmass must be even to find
a route through the town crossing ea
Homework 7: Solutions
1. The chromatic polynomial for the graph on the left is k(k 1)(k 2)(k 2 3k + 3) and
on the right is k(k 1)3 (k 2).
2. We have that (Kn e) = n 1. This is because since Kn e contains Kn1 as a
Homework 10: Solutions
1. The total number of spanning trees (G) = number of spanning trees that dont include
e + number of spanning trees that include e.
The former is equal to the number of spanning trees that would exist if e
Homework 12: Solutions
1. For the lower bound remove F and get a minimal spanning tree below of weight 7. Hence
a lower bound is 7 + 4 + 5 = 16.
For the upper bound a minimal spanning tree is as below. Starting at C we get CDBFEA
Homework 11: Solutions
(a) The number of pairs (T, e) is the number of spanning trees of Kn multiplied by the
number of edges in a spanning tree. By Cayleys formula and Q51 this is nn2 (n 1).
(b) By Proposition 1, the number o