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Homework 11
Math 490 - Spring 2015
Solutions
#1. Do Problem 10.8cd (i.e., skip parts (a) and (b)) on page 278 of the textbook.
Solution:
a) Consider the graph K1,4 . This is a complete bipartite graph where one of the partite sets
contains one vertex of degree 4, and the other partite set contains 4 vertices of degree 1.
Since K1,4 has edges, we know that χ(K1,4 ) ≥ 2, and since K1,4 is bipartite, we know that
χ(K1,4 ) ≤ 2. Therefore we have χ(K1,4 ) = 2. Moreover, we have ∆(K1,4 ) = 4. Therefore,
we have ∆(K1,4 ) = 2χ(K1,4 ).
d) Consider the complete graph K2 . This graph has χ(K2 ) = 2 and ∆(K2 ) = 1. Therefore
we have χ(K2 ) = 2∆(K2 ).
#2. Do Problem 10.2ac (i.e., skip part (b)) on page 278 of the textbook, EXCEPT find
the chromatic INDEX of each graph instead of the chromatic number.
Solution:
a) We will show that χ0 (P ) = 4, where P is the Peterson graph. We know from Vizing’s
Theorem that χ0 (P ) ≤ 4, so we need to prove that χ0 (P ) 6= 3. That is, we need to show that
it is impossible to color P using only 3 colors: red, blue, and green. Consider the Peterson
graph shown below.
First, we will color the outer ring of the graph. We have already seen that this requires
three colors. Since it is impossible to use all three colors twice, there must be a color which
is used only one time on the outer ring. Suppose without loss of generality that this color is
red. Also, without loss of generality, suppose that the edge on the outer ring clockwise from
red is green. Then the outer ring must look as shown below. (Remember that red can only
be used one time in this ring.)
Now we can color the “spokes” connecting the outer ring to the inner star. Because we
only have three colors, and each of the spokes is incident with edges of two different colors,
these edges are forced to be colored as in the picture below.
Now, the horizontal edge of the star must be colored green, since it is incident with red
and blue edges. Therefore the graph looks like this.
However, now consider the edge which runs from the bottom left to the middle right of the
star. This edge is incident with green, blue, and red edges. Therefore there is no way in which
the edge can be colored with any of our three colors. Since all of our choices were forced by
the fact that the outer ring contains only one red edge, there cannot be any way to ever color
the edges of the Petersen graph with only 3 colors. Therefore we must have χ0 (P ) ≥ 4. This,
along with the χ0 (P ) ≤ 4 bound we obtained from Vizing’s Theorem, implies that χ0 (P ) = 4.
b) For this problem, we have the same issue as last week that the definition of Wn in the
problem is slightly different than the definition I gave in class. Like last week, I’ll use the
definition from class in these solutions, but you can still get full credit for using the definition
in the book.
We will show that χ0 (Wn ) = ∆(Wn ) = n − 1. Here, we’re using the definition that Wn
is constructed by taking the graph Cn−1 and adding one extra vertex which is adjacent to
all the others. For this solution, let the vertices of Cn−1 be labeled as v1 , v2 , v3 , etc. in
clockwise order, and let u be the extra vertex. For the edges incident with u, we will color
uv1 with color 1, uv2 with color 2, and so on until uvn−1 has color n − 1. To color the
outside cycle, we will color edge v1 v2 with color 3, v2 v3 with color 4, and so on until edge
vn−2 vn−1 receives color 1 and edge vn−1 v1 receives color 2. This is a valid (n − 1)-coloring,
and therefore Vizing’s Theorem says that χ0 (Wn ) = n − 1.
If it’s hard to see what the coloring in the above paragraph is, think of the wheel graph
Wn as having n − 1 triangles around a central point. After coloring the spokes all different
colors, we do the following. Visit the triangles moving clockwise, and for each triangle, color
the outside edge the same color as the most counter-clockwise edge of the previous triangle.
For example, we would color the wheel graph W6 as below.
#3. Suppose that a graph G has 5 vertices of degree 6 and 8 vertices of degree 7. There
are no other vertices in the graph. Prove that χ0 (G) = 8.
Solution:
The graph G has n = 13 vertices, and has ∆(G) = 7. These numbers immediately give us
(n − 1)∆(G)/2 = 42. Using the First Theorem of graph theory, we can calculate that G has
m = 43 edges. Since m > (n−1)∆(G)/2, Theorem 10.13 tells us that χ0 (G) = 1+∆(G) = 8.
#4. a) The only two trees of order 4 are P4 and K1,3 . Find the chromatic polynomial for
each of these trees.
b) All of the possible trees of order 6 are shown in Figure 4.3 on page 88 of the textbook.
Find the chromatic polynomial for each of these trees.
c) Based on your answers to parts (a) and (b), can you make a conjecture about the chromatic polynomials of trees in general? You DO NOT need to prove your conjecture.
Note: You can find all the chromatic polynomials in this problem by pure reasoning. You
do not need to use the formula involving deleting and contracting edges. You may use that
formula if you like, but it’s not necessary.
Solution:
a) I won’t go through all the details here, but you should find that PP4 (k) = PK1,3 (k) =
k(k − 1)3 .
b) Again, I won’t go through the details, but if T is any of the trees in the figure, then its
chromatic polynomial is PT (k) = k(k − 1)5 .
c) In general, if T is a tree with n vertices, then PT (k) = k(k − 1)n−1 .
#5. Find a formula for PG (k) for the graph G = K2,5 .
Note: The comment for Problem 4 applies here too!
Solution:
The graph K2,5 is shown below, where two vertices have been labeled.
u
v
Suppose first that u and v have different colors. Then there are k possible colors for vertex
u and k − 1 possible colors for vertex v. The remaining 5 vertices can each be colored with
any of the other k − 2 colors. So there are k(k − 1)(k − 2)5 colorings in which u and v have
different colors.
If u and v have the same color, then there are k possible colors for u and 1 possible color
for v. Each of the 5 remaining vertices can be colored with any of the k − 1 remaining colors.
So there are k(k − 1)5 colorings in this situation. If we add these amounts together, the total
number of ways to color K2,5 with k colors is PK2,5 = k(k − 1)(k − 2)5 + k(k − 1)5 .