049059 - Homework 3
Assignment 1
Brook’s Theorem implies that the cliques and the odd cycles are the only (𝑘 − 1)-regular 𝑘critical graphs.
Answer
Suppose that there is a graph G, which is K-1 regular and K-critical, and is neither a clique nor
an odd cycle.
From Brook's theorem, we know that𝝌(𝑮) ≤ ∆(𝑮), but𝝌(𝑮) = 𝐾, and∆(𝑮) = 𝑲 − 𝟏, in
contrast to Brook's theorem.
So the graph has to be either a clique or an odd cycle.
Assignment 2
For 𝑛-vertex graph𝐺:
1.
2.
3.
4.
5.
the degree of 𝑃𝐺 (𝑘) is 𝑛,
the coefficient of 𝑘 𝑛 is 1
and that of 𝑘 𝑛−1 is |𝐸(𝐺)|,
the sign of the coefficients is alternating,
And the free coefficient is zero.
Answer
1. From the theorem we've seen in class, each graph's chromatic function is a sum of
chromatic functions of cliques. The largest clique in it is of size n, and it has the
chromatic function: P  cliquen   k   k  1  ...   k  n  1 , which is a polynomial
that has a degree of n. The other cliques has a lower degree, so the sum of all the
polynomials has k n in it.
2. There is only one n-sized clique in the sum, so it will be the only element with a
degree of n. We saw that the function of this clique has an element of degree n, and
it's kn.
3. For each graph, the chromatic function is the sum of two chromatic functions: one is
the chromatic function of a graph of the same size n, and another graph of size n-1.
Since we can identify graphs until we get a clique, we get n(n-1)/2 - |E| graphs of
size n-1. Those graphs are also identified until becoming cliques, yielding more
graphs of size n-2. In graphs of size n-2 and below, the coefficient of 𝑘 𝑛−1 is 0. In
graphs of size n-1, the coefficient is 1 (proved in (2)). In the clique of size n, the
coefficient is 0 - 1 - 2 -… - (n - 1) = n(n-1) / 2. Overall, the coefficient of 𝑘 𝑛−1 is:
 n  n  1

coeff n 1  PG   coeff n 1  clique n   
 E   coeff n 1  graph of size n -1  ...
2


...   number of smaller graphs   coeff n 1  smaller graphs  

n  n  1  n  n  1


 E  1  0 | E |
2
2


4. Proof by induction on the number of edges.
For a single edge the function is P(k) = k. The claim holds.
For a two edge graph, the function is P(k) = k2 – k. The claim also holds.
Assume the claim is correct for every graph with n edges or less. For some graph G
with n+1 edges, we know that the subgraph G' created by removing some edge exy
from x to y from G has a function which is a polynomial with alternating signs. We
add the edge we previously removed. We have 𝑘 − |𝑁(𝑥) ∪ 𝑁(𝑦)| options to color
this edge, and overall we have (𝑘 − |𝑁(𝑥) ∪ 𝑁(𝑦)|) ∙ 𝑃𝐺 ′ (𝑘) options to color the
graph. This is the chromatic function of G, and it's easy to see that its coefficients'
sign is alternating: 𝑃𝐺 (𝑘) = (𝑘 − 𝑡) ∙ (𝑘 𝑛 + 𝑎𝑛−1 𝑘 𝑛−1 + ⋯ ), where t is some
natural number (number of neighbors of both x and y) and the coefficients are:
𝑏𝑖 = 𝑎𝑖 − 𝑡 ∙ 𝑎𝑖−1
We know that ai and ai-1 alternate in signs, so sign(bi) = sign(ai), and we know that ai
have alternating signs. The claim is correct for n+1 edges as well, and
consequentially for any graph.
5. Since for every graph the chromatic function can be represented as the sum of
chromatic functions of cliques, and the claim is correct for every clique (i.e. the
polynomial is P  cliquen   k   k  1  ...   k  n  1 , which has zero as its free
coefficient), the sum of all zeros is zero.
Assignment 3
Show that a 𝑑-regular graph 𝐺(𝑉, 𝐸) is 𝑑-edge colorable iff 𝐸 can be partitioned into perfect
matchings.
Answer
First direction:
If a d-regular graph G is d-edge colorable, E can be partitioned into perfect matchings.
Since every vertex of G has d neighbors, all colors are used in every vertex. Let's take the
group of edges colored in color #1. This group has edges on all vertices: if it doesn't, there is
a vertex without an edge colored with #1, and that contradicts G being d-regular since there
aren't enough colors without #1 for all its edges. The group also inspires a perfect matching,
since none of the edges touch the same vertex, due to the definition of edge-coloring.
Second direction:
If a d-regular graph G has a perfect matching, it's d-edge colorable.
We prove this by induction over d:
Base: d=1. Trivial, since every vertex has one neighbor the edges themselves are the
perfect matching.
Assumption: assume that the claim is correct for some natural finite d.
Step: for a d+1-regular graph G: we know the graph has a perfect matching M. If we remove
M from G's edges, we remove a single edge from every vertex, and we have a graph in which
all vertices has d neighbors. From the assumption there is a d-edge coloring for this graph.
We apply this coloring over G, and color M in color #d+1.
This is a valid edge coloring: if two edges with the same color touch the same vertex, they
could have colors #1-#d, which contradicts the assumption since the coloring for G' isn't
valid. The other option is two edges colored with #d+1, but that would contradict the perfect
matching. So the claim is correct for d+1 as well.
Assignment 4
Show by an appropriate edge coloring that 𝜒 ′ (𝐾𝑚,𝑛 ) = max{𝑚, 𝑛}.
Answer
We mark the vertices in X as v1…vm, and the vertices in Y as u1…un. Assume without the loss
of generality that m≥n. We create the following coloring:
Edges colored in color #1: ui vi |1  i  n
Edges colored in color #2: ui vi 1 |1  i  n
x  1, m: Edges colored in color #x: ui vi  x 1 |1  i  n
We see that for each edge coming out of every vertex in Y, there is a different color. We can
also see that every two edges with the same color goes to a different vertex in X since i
changes. All edges are colored since we colored m*n edges, which connects all vertices in X
to all vertices in Y. We used m colors since x  1, m and 1, m   m .
Assignment 5
Given graph 𝐺(𝑉, 𝐸), |𝑉| = 𝑛, |𝐸| = 𝑚, show that 𝜒 ′ (𝐺) ≥ ⌈2𝑚⁄𝑛⌉.
Answer
We calculate the average number of edges per vertex. The graph has m edges, and each
touches two vertices. The average number of edges per vertex is 2m / n, since there are less
than 2m possibilities of edges in G. The max number is ΔG since the graph is given, and the
average has to be lower than that. So ⌈2𝑚⁄𝑛⌉ ≤ ΔG ≤ 𝜒 ′ (𝐺) from what we saw in class.
Assignment 6
Eight schoolgirls go for a walk in pairs every day. Can they arrange their outgoing so that
every girl has a different companion at every day of the week?
Answer
Let G be a complete graph where |V|=girls group. We color the edges. If G has an edge
coloring for 7 colors they can have a different companion each day. We mark each vertex for
the girl it represents: v1 for girl #1, and so on.
Since every vertex in the graph has 7 edges out of it and into others, we can color it the
following way:
v1v2, v3v4, v5v6, v7v8 – color 1
v1v3, v2v4, v5v7, v6v8 – color 2
v1v4, v2v3, v5v8, v6v7 – color 3
v1v5, v2v6, v3v7, v4v8 – color 4
v1v6, v2v5, v3v8, v4v7 – color 5
v1v7, v2v8, v3v6, v4v5 – color 6
v1v8, v2v7, v3v5, v4v6 – color 7.
This is a valid edge coloring, so this is a valid arrangement for them to switch partners along
the week.