Mathematics
for Computer Scientists
Solutions to Selected Exercises from
Volume II: Number Theory, Modular Arithmetic and
Graph Theory
Twelfth Edition,
and 2004, 2005, 2006 Quizzes, Midterm Exams, and Final Exams with Solutions
S. A. Rankin & I. J. W. Robinson
c
Copyright 2007
by Kinson Publishing, Ltd.
All rights reserved.
1
CHAPTER II
Graph Theory
Solutions for the Exercises of Section II.10
1. Exactly which of the graphs shown below are not planar?
•
•
G1 :
•
•
G2 :
•
•
•
•
•
•
G4 :
•
•
•
•
•
•
G3 :
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
G5 :
G6 :
•
•
G7 :
•
•
•
•
•
•
•
A: G3 , G6 , G7
D: G2 , G3 , G6 , G7
•
•
B: G2 , G3 , G7
E: None of A, B, C, D
•
C: G2 , G3 , G6
Solution. G1 is a plane graph, hence planar. G2 is the Petersen graph, which we
know to be nonplanar. G3 is K3,3 , so by Kuratowski’s theorem, G3 is not planar.
Since G6 is isomorphic to G3 , it follows that G6 is not planar either. G4 and G5 are
plane graphs, hence planar, and finally, since G7 is isormorphic to G5 , it follows
that G7 is planar as well. Thus G2 , G3 and G6 are not planar, while all the rest
are. The answer is C.
2. If graphs G and H are subdivision-equivalent, exactly which of the following statements are true?
(i) If G is a tree, then H is a tree.
(ii) If G has a vertex of degree 2, then H has a vertex of degree 2.
(iii) If G is planar, then H is planar.
A: (i), (iii)
B: (i), (ii)
C: (i)
D: (ii), (iii)
E: None of A, B, C, D
2
Planar Graphs
II.10
Solution.
(i) is true. For the subdivision operations do not create cycles or
components, and so if G and H are subdivision equivalent and G has is without
cycles, then H is without cycles.
(ii) is false, since the 2-chain and the 1-chain are subdivision equivalent.
(iii) is true by Kuratowski’s theorem. Alternatively, one could observe that if it
is possible to embed G, then the subdivision operations can be performed on the
edge curves of this plane graph to obtain an embedding of H.
The answer is A.
3. Let G be a simple, connected 4-regular planar graph. If G has 12 edges, how many
faces are there in a plane representation of G?
A: 8
B: 12
C: 4
D: 5
E: None of A, B, C, D
Solution. Since G is connected, Euler’s formula tells us that νG − εG + fG = 2.
From the handshake lemma, we know that 4νG = 2(12) = 24, so νG = 6. Thus
6 − 12 + fG = 2, or fG = 8. The answer is A.
7. Let G be a connected simple 4-regular plane graph with 8 faces. How many vertices
does G have?
A: 18
B: 12
C: 10
D: 6
E: None of A, B, C, D
Solution. Let n = νG and m = εG . Then 4n = 2m, so 2n = m, and n−m+8 = 2,
so m = n + 6. Thus 2n = n + 6, so n = 6. The answer is D.
8. For positive integers m and n, exactly which of the conditions (i)-(iv) below are
individually sufficient to ensure that the complete bipartite graph Km,n is planar?
(i)
(ii)
(iii)
(iv)
A: (i), (ii), (iii)
B: (i), (iii)
min{ m, n } ≤ 2;
mn < 6;
max{ m, n } < 3;
m 6= 3 and n 6= 3.
C: (iv)
D: (ii), (iii)
E: None of A, B, C, D
Solution.
Since Km,n is a subgraph of Kr,s if m ≤ r and n ≤ s, we see by
Kuratowski’s theorem that Km,n is not planar if m ≥ 3 and n ≥ 3. Consider the
case when at least one is less than 3, say m ≤ 3. If m = 1, then the graph is a
star graph, which is a tree and hence planar, while if m = 2, then we may apply n
elementary subdivision operations to remove the n vertices of degree two, so that
we are left with a graph on two vertices, with n edges joining the two vertices.
II.10
Graph Theory
3
This is a planar graph as well, so Km,n is planar when m = 2. Thus Km,n is planar
if and only if at least one of m and n is less than 3. This is the case in (i), (ii),
(iii) but not (iv), so the answer is A.
15. Let G be a connected simple plane graph with 12 faces, all of degree 3.
a) How many edges does G have?
Solution. Since the sum of the degrees of all faces is equal to twice the number
of edges in G, we have (3)(12) = 2εG , so εG = 18.
b) How many vertices does G have?
Solution. By Euler’s formula, νG −εG +fG = 2, and we are given that fG = 12,
while we have determined above that εG = 18. Thus νG = 2 + 18 − 12 = 8.
c) Prove that G cannot be a regular graph.
Solution. Suppose that G is a k-regular graph. By the handshake lemma, we
would have 8k = 2(18) = (4)(9), so 2k = 9. Since this is impossible, G cannot
be regular.
16. Show that if G is a plane graph with at most 11 faces and for which every vertex
has degree at least 3, then G has a face f with degG (f ) ≤ 4.
Solution. Suppose that every face of G has degree at least 5. Then 2 εG ≥ 5 fG ,
or εG ≥ 5 fG /2. As well, we have 2 εG ≥ 3 νG , and so νG ≤ 2 εG /3. From Euler’s
formula, we obtain νG − εG + fG = κG + 1 ≥ 2, so
fG ≥ 2 + ε G − ν G ≥ 2 + ε G −
2 εG
ε
5 fG
=2+ G ≥2+
.
3
3
(2)(3)
5f
f
Thus fG − G ≥ 2, or G ≥ 2, which implies that fG ≥ 12. But fG ≤ 11, and
6
6
so it follows that not every face can have degree greater than or equal to five.
19. Draw a simple planar graph with 9 vertices which is regular of degree 4 and which
has exactly three faces of degree 4, with all other faces of degree 3.
Solution.
•
•
• •
•
• •
•
•
21. Draw three connected simple 3-regular plane graphs with 8 vertices, no two of
which are isomorphic. Each edge should be drawn as a straight line segment.
4
Planar Graphs
II.10
Solution.
•
• •
• •
•
•d
•g
•
e• f
•
•
a•
•d
c
•h •
•h
a•
•c
•f
b
≡
≡
g
•d
•
•g
h•
•c
e•
b•
d•
•f
•c
a•
f•
•e
a•
•b
≡
≡
a
d
•
•
b
•
c• •h
•
e
d•
•g
•
•b
h•
e•
a•
•c
h
•
e•
•g
g
•
•f
•
b
•
f
≡
f
•
a
•
g•
h•
•c
•b
d•
•e
In each column, all graphs in the column are isomorphic to each other. We claim
that the graph in the first column is not isomorphic to any graph in either the
second column or the third column. To see this, observe that the graph in the first
column has no 3-cycle, while the graphs in the second and third columns do have
a 3-cycle.
Furthermore, no graph in the second column is isomorphic to any graph in the
third column. This follows from the observation that the graphs in the second
column have a 5-cycle, while the graphs in the third column do not have a 5-cycle,
the observation that the graphs in the second column each contain exactly two
3-cycles, while the graphs in the third column each contain exactly four 3-cycles.
25. Let G be a connected 3-regular planar graph such that every face of G has degree
3. How many edges does G have?
II.10
Graph Theory
5
Solution. By the handshake lemma, 2εG = 3νG . By the analogous result for faces
of a plane graph, 2εG = 3fG . Finally, by Euler’s formula, 2 = νG − εG + fG , so
2 = 2/3εG − εG + 2/3εG = 1/3εG , so εG = 6.
Note that since εG = 6, we have νG = (2)(6)/3 = 4, so G is a graph on 4 vertices
with 6 edges. G can’t have loops or multiple edges, since otherwise G would have
a face of degree 1 or 2. Thus G is simple. The only simple graph on 4 vertices
with 6 edges is K4 .
30. Let G be any graph formed from K7 by deleting a 7-cycle. Prove that G is not
planar.
Solution. Since all 7! permutations of { 1, 2, 3, 4, 5, 6, 7 } give automorphisms of
K7 (that is, Aut(K7 is isomorphic to S7 ), any two graphs formed in this way will
be isomorphic. Thus the problem is independent of the choice of 7-cycle. Let us
draw K7 by equally spacing seven points around the circumference of a circle, with
the resulting seven arcs of the circle denoting edges in K7 , then proceed to join
each pair of non-adjacent (on the circumference of the circle) points by a straight
line segment. Delete the arcs on the circumference of the circle (which form a
7-cycle) to form G, shown in (a) below. We shall try to find a subgraph of G
that is subdivision-equivalent to either K5 or K3,3 . If we are successful, then by
Kuratowski’s theorem, G is not planar. Note that G is 4-regular with 7 vertices.
If we hope to find a subgraph of G that is subdivision-equivalent to K5 , then we
need to find five vertices of degree at least 4, and either delete the remaining two
vertices or delete edges incident to the remaining two vertices so that they end up
with degree 2 in the subgraph. But we would have to delete two edges incident to
each of the two remaining vertices, so we must delete at least one edge incident to
one that is not incident to the other, which means that in the subgraph that we are
constructing, one of the remaining five vertices now has degree less than 4. Thus
we will not find any subgraph of G that is subdivision-equivalent to K5 . We shall
therefore concentrate on finding a subgraph of G that is subdivision-equivalent to
K3,3 , which means that we want six vertices of degree at least 3, where we will
either delete the seventh or remove edges incident to it to bring its degree down to
2. The result of deleting one vertex (it doesn’t matter which one) is shown in (b)
below. In (b), we see that four of the six vertices have
degree three, so they must
7
not have any incident edges removed. We have 2 − 7 − 4 = 21 − 11 = 10 edges
still, so we must delete one. The only candidate is the edge joining the two vertices
of degree 4, and the subgraph of G that we get by further deleting this edge is
shown in (c). This is a graph with six vertices and 9 edges, so if it subdivisionequivalent to K3,3 , it must be equal to K3,3 . However, the graph shown in (c) has
3-cycles, so it is not bipartite. Thus we may not delete a vertex in order to form
our subgraph of G that is subdivision-equivalent to K3,3 . It follows that we must
choose one vertex and remove edges incident to the chosen vertex to bring it down
to degree 2, in the hopes that the other six vertices have degree at least degree 3
in the resulting graph and any that have degree greater than 3 can have incident
6
Vertex Colourings
II.10
edges removed to bring the degree down to 3, with the hope that in the end, we
have a copy of a subgraph that is identical to K3,3 except that one edge of K3,3 is
represented by two edges joined by a vertex of degree 2 in the subgraph. We show
such a subgraph in (d) below, with the vertices in one cell labelled r and in the
other cell labelled b.
r
•
•
•
r• • •r
•
•
•
•
•
•
•
•
(a)
•
•
•
•
•
•
•
(b)
•
•
•
b
•
(c)
b
•
b
(d)
Solutions for the Exercises of Section II.11
1. What is the chromatic number of the graph G shown below?
•
•
•
•
•
•
•
A: 1
B: 2
C: 3
•
D: 4
E: None of A, B, C, D
Solution. Since G is non-null bipartite, its chromatic number is 2. The answer is
B.
2. What is the chromatic number for the graph G
shown at the right?
A: 4
B: 3
C: 2
D: 1
•
•
•
•
•
•
E: None of A, B, C, D
Solution.
Colour the top left 3-cycle with three colours, then colour the uncoloured vertex in the second 3-cycle with the colour of the vertex at top left, then
copy the colouring of the two vertices on the left side to the two on the right side,
as in
g
•
b
•
•
•
•
g
•
r
r
g
Since G contains a 3-cycle, χ(G) ≥ 3, and since we are able to 3-vertex-colour G,
χ(G) ≤ 3. Thus χ(G) = 3.
The answer is B.
II.11
Graph Theory
7
3. Let G be a simple graph. Prove that χ(G) ≥ 3 if and only if G has a cycle of odd
length.
Solution. Since G is simple, it is loop-free and so any cycle of odd length has length
at least 3 (a priori, there may not be any cycles of odd length). Now χ(G) ≥ 3
if and only if χ(G) 6= 1 and χ(G) 6= 2. Since G is simple, χ(G) = 1 if and only
if G is a null graph (that is, G has no edges), and by Theorem 12.3, a loop-free
graph G has χ(G) = 2 if and only if G is non-null bipartite. Thus χ(G) ≥ 3 if and
only if G is neither null nor bipartite. By Theorem 7.6, a loop-free graph is not
bipartite if and only if it has a cycle of odd length. Thus χ(G) ≥ 3 if and only if G
is non-null and has a cycle of odd length. Finally, if G has a cycle of odd length,
then G is non-null. Thus we obtain that for a simple graph G, χ(G) ≥ 3 if and
only if G is non-null and has a cycle of odd length, if and only if G has a cycle of
odd length.
4. a) Prove that the Grötzsch graph (Figure 10.10) has chromatic number 4.
The Grötzsch graph G is shown to the left. We note first of
all that since G has an odd cycle (5-cycle), χ(G) ≥ 3. If G
r
•
has a 3-vertex-colouring, say with colours r, g and b, then
r
g
•
b•
g • one of the colours, say r, appears exactly once on the outer
b•
• •
5-cycle. By symmetry, we may assume that the top vertex of
• •
the outer 5-cycle is coloured r. Since r can not appear on the
•b
g•
outer 5-cycle again, we observe that as we work our way in
the clockwise direction around the outer 5-cycle, the vertices must be coloured
in an alternating fashion with g and b. By symmetry again, we may assume
that the pattern is g, b, g, b (as shown in the illustration at left). But then
the colouring of the top 3 vertices of the inner 5-star is completely determined,
with all three colours appearing on these 3 vertices. This establishes that the
central vertex can not possibly be coloured without using a fourth colour, so
we have a contradiction. Thus there is no 3-vertex-colouring of the Grötzsch
graph, which proves that χ(G) ≥ 4. A 4-vertex-colouring of G using r, g, b and
y is shown below, which establishes that χ(G) ≤ 4. It follows that χ(G) = 4,
as required.
Solution.
r
•
b•
r
• g •g
b•
•y •
r• •b
•
•b
g
b) Prove that if e is any edge in the Grötzsch graph G, then G − e has chromatic
number 3 (which implies that G is a critical 4-chromatic graph).
Solution.
8
Vertex Colourings
r
•
b
r
• b• • •g •g
•r
g• •b
•b
g•
r
•
e
(i)
g
b•
• e •g
b•
b
•r •
g• •b
•
•r
g
(ii)
II.11
r
•
r
b• g • g •g
• •er •
g• •b
•
•b
g
(iii)
In (i) above, we delete an edge e on the outer 5-cycle. The resulting graph still
has odd (5-) cycles, so its chromatic number is at least 3. In (i), a 3-vertexcolouring is shown, so the chromatic number of this graph is 3. By symmetry,
the same result will be obtained if any of the edges on the outer 5-cycle are
deleted.
In (ii) above, we delete an edge e between a vertex on the outer 5-cycle and
a vertex on the inner 5-star. The resulting graph still has odd (5-) cycles, so
its chromatic number is at least 3. In (ii), a 3-vertex-colouring is shown, so
the chromatic number of this graph is 3. By symmetry, the same result will
be obtained if any edge that connects a vertex on the outer 5-cycle to a vertex
on the inner 5-star are deleted.
Finally, in (iii) above, we delete an edge e which joins a vertex on the inner
5-star to the star’s central vertex. The resulting graph still has odd (5-) cycles,
so its chromatic number is at least 3. In (iii), a 3-vertex-colouring is shown,
so the chromatic number of this graph is 3. By symmetry, the same result will
be obtained if any of the edges of the inner 5-star are deleted.
Since this covers all possible cases, we have established that for any edge e of
the Grötzsch graph G, χ(G − e) = 3, whence G is a 4-critical graph.
8. The wheel graph Wn (see Definition 7.2) is defined to be the join of the n-cycle Cn
and the null graph N1 . The vertex of N1 is called the hub of the wheel. Thus Wn
has n + 1 vertices and 2n edges. Prove that for n ≥ 2, Wn is 3-chromatic if n is
even, and 4-chromatic if n is odd.
Solution. Let n ≥ 2 be an integer and let k = χ(Wn ). Consider any k-vertexcolouring of Wn . Let r denote the colour used for the vertex which is the hub of the
wheel. Since the hub vertex is adjacent to each vertex on the rim (and the rim is
an n-cycle), we see that r can’t be used to colour any vertex on the rim, whence we
have a (k−1)-vertex-colouring of Cn . It follows that χ(Wn )−1 = k−1 ≥ χ(Cn ), or
χ(Wn ) ≥ 1 + χ(Cn ). Similarly, if we have any vertex-colouring of Cn , then we may
use a new colour for the hub vertex, thereby establishing that 1 + χ(Cn ) ≥ χ(Wn ),
whence χ(Wn ) = 1 + χ(Cn ). The result now follows from Proposition 12.4.
II.11
Graph Theory
9
c
•
9. What is χ(G) if G is the following graph:
e
•
a
•d
•
•
b
Solution. Since G contains K3 , χ(G) ≥ χ(K3 ). Here is a 3-vertex-colouring of G,
whence χ(G) ≤ 3 and so χ(G) = 3.
r
•
r
•
b
•
•b
•
g
10. a) Construct a connected simple cubic graph G for which
(i) χ(G) = 4;
Solution. K4 is connected, simple, 3-regular, and since χ(Kn ) = n for
any positive integer n, we have χ(K4 ) = 4.
(ii) χ(G) = 3;
Solution. r•
g•
b•
•b Since the graph has a 3-cycle, its chromatic number
•r
is at least 3, and we have shown a 3-vertex-colouring,
•g
so its chromatic number is in fact equal to 3.
(iii) χ(G) = 2;
Solution.
K3,3 , which has chromatic number equal to 2 since it is a
non-null bipartite graph. It is connected, simple and 3-regular.
b) Is K4 the only connected simple cubic graph with chromatic number 4?
Solution. Suppose that G is a connected simple 3-regular graph. Then G is
not a cycle. Since Kn is regular of degree n − 1, if G is complete, then G is
(isomorphic to) K4 . Suppose that G is not K4 , so that G is not a complete
graph. By Brook’s theorem, χ(G) ≤ 3, so χ(G) 6= 4. Thus if G is a connected
simple cubic graph with chromatic number equal to 4, then G is K4 .
11. What is the chromatic number of the Petersen graph? Establish your answer.
Solution. We provide a 3-vertex-colouring of the Petersen graph G, which establishes that χ(G) ≤ 3.
10
Vertex Colourings
II.11
g
b•
•
•r
b
•g •
b• •g
•r
g•
r•
Since the Petersen graph has odd (5-)cycles, its chromatic number is at least 3,
whence the Petersen graph has chromatic number 3.
17. Determine all simple graphs G with the property that G has chromatic number 3,
but for any edge e of G, G − e has chromatic number 2.
Solution. Let G be such a graph. Then G is not bipartite, but if any edge is
removed, the result is bipartite. Since any forest is bipartite, G is not a forest,
whence G has cycles. If every cycle is of even length, then G is bipartite, so G has
a cycle of odd length. Let C be a cycle of odd length in G. If there is an edge e
of G that is not on C, then C is a subgraph of G − e, whence G − e has a cycle of
odd length and is therefore not bipartite, which is not possible. Thus every edge
of G is on C, whence G has one connected component which is an odd cycle, with
any other compenents being trivial.
22. a) Prove that for any positive integer n, αKn = 1.
Solution. Any vertex by itself comprises an independent set, and since any
two vertices are adjacent, no independent set of Kn can contain more than one
vertex. Thus αKn = 1.
b) For any positive integers m, n with m ≥ n, prove that αKm,n = m.
Solution. No independent subset of a complete bipartite graph can contain
vertices from each cell of the bipartite partition, since each vertex of one cell
is adjacent to every vertex of the other cell. On the other hand, any subset
of either cell is independent, so each cell is independent, and thus the larger
cell (both if m = n) will be the largest independent set in Km,n . Note that if
m > n, then Km,n has a unique largest independent subset.
26. Let C = { cn | n ∈ N } be a set of colours. Use induction to prove that for
each positive integer n, there exists a tree Tn for which there exists a vertex vn
and a sequencing v1 , v2 , . . . , vkn = vn , where kn = νTn , such that when the greedy
algorithm is run on this sequence, vn receives colour cn and no vertex of Tn receives
colour ci for i > n (which proves that for every positive integer n, there exists a
tree T and a sequencing of the vertices of T such that when the greedy algorithm
is applied to T using this sequence, n colours get used–even though it is always
possible to colour any non-trivial tree with exactly two colours).
Solution. For n = 1, we may take the null graph on a single vertex v1 . Suppose
now that n ≥ 1 is an integer for which the hypothesis holds for all integers k,
II.12
Graph Theory
11
1 ≤ k ≤ n. Let Tk be a tree with vertex sequence vk,1 , vk,2 , . . . , vk,nk = vk such
that when the greedy algorithm is applied to Tk with this sequence, vk receives
colour ck and no vertex receives colour ci for any i > k (so that the set of colours
that get used is { c1 , c2 , . . . , ck }. Let vn+1 denote a new vertex, and join vn+1 to
each vertex vk , k = 1, . . . , n. Let the resulting tree be denoted by Tn+1 . Sequence
the vertices of Tn+1 by concatenating the sequences for each Tk , 1 ≤ k ≤ n in
order of index, so that the sequence for T1 comes first, then the sequence of T2 ,
and so on, and then adjoin vn+1 at the end. When the greedy algorithm is applied
to Tn+1 with this sequence, vk will receive colour ck for k = 1, 2, . . . , n, but no
colours with index greater than n will have been used, so when it is time to colour
vn+1 , every colour c1 through cn appears on a vertex adjacent to vn+1 and thus
the next available colour, cn+1 , will be assigned to vn+1 . The result follows now
by induction.
27. Prove that for each positive integer k, there exists a tree T such that if the greedy
algorithm is applied with the vertices arranged in a degree sequence, then k colours
will be used. Hint: Work with your solution to Exercise 26.
Solution. Let n be a positive integer and let T be a tree with a vertex sequence
v1 , v2 , . . . , vk as described in Exercise 26. Adjoin sufficiently many vertices of
degree 1 to vk−1 to ensure that degT (vk−1 ) > degT (kn ). Then adjoin sufficiently
many vertices of degree 1 to vk−2 to ensure that degT (vk−2 ) > degT (vk−1 ), and so
on.
When the vertices of the resulting tree are listed in order of non-increasing degrees,
the order will still be v1 , v2 , . . . , vk , followed by all the new vertices of degree 1.
The greedy algorithm will still use n colours when applied.
Solutions for the Exercises of Section II.12
1. Let G be a plane graph. If H is a natural plane embedding of the geometric dual
G∗ of G, then exactly which of the following statements are true?
(i)
(ii)
(iii)
(iv)
(v)
H is connected.
If G is connected, then H is connected.
If H is connected, then G is connected.
|FG | = |VH |.
|VG | = |FH |.
A: (i), (ii), (iv)
B: (i), (iii), (v)
D: (i), (ii), (v)
E: None of A, B, C, D
Solution.
(i) is true. This is a theorem in the text.
C: (ii), (iv), (v)
12
Face colouring plane graphs
II.12
(ii) is true since H is always connected.
(iii) is false whenever G is not connected.
(iv) is true since by definition, FG = VH .
(v) is not true. Let G be a plane null graph on two vertices, so that H is null on
one vertex. Then H has one face.
Thus (i), (ii), and (iv) are true, while (iii) and (v) are false. The answer is A.
2.
•
Consider the plane graphs G1 and G2 shown in
Figure 1.
•
•
a) Prove that G1 and G2 are isomorphic.
•
e
•d
•
•d
a•
•c
•b
•c
a•
•b
•
•
•
•
•
(i) G1
(ii) G2
Figure 1.
Solution. Observe that if the vertices in each
graph are labelled as follows:
e
•
•
then the following table is the adjacency table for each graph, whence the two
graphs are isomorphic.
Vertex
a
b
c
d
e
Adjacent to
b, d, e
a, d, c
b, d
a, c, e
a, d
b) Prove that G∗1 and G∗2 are not isomorphic.
Solution. G∗1 and G∗2 are shown below.
•
◦
•
•◦
•
◦
◦
◦
(i) G∗1
•
◦
◦
•
•
•◦
◦
◦
(ii) G∗2
The degree sequence of G∗1 is 5, 3, 3, 3, while the degree sequence for G∗2 is
4, 4, 3, 3, whence G∗1 and G∗2 are not isomorphic.
3. Either construct a plane graph with 5 faces such that any two faces have an edge
in common, or prove that this is impossible.
Solution. If such a graph existed, then its dual would be a planar graph on 5
vertices with the property that between any two vertices, there is at least one edge.
II.13
Graph Theory
13
Such a graph would have K5 as a subgraph, and since K5 is not planar, this is not
possible.
6. Prove that for each positive integer n, Wn is isomorphic to Wn∗ (see Definition 7.2
for the definition of Wn ).
Solution.
•65431
•
•
•
•
•2
•
7. Determine χ∗ (G), the face-chromatic number of the plane graph G shown below.
•
•
•
•
•
•
Solution. Since G is 2-face-colourable if and only if G is eulerian, and G is not
eulerian (since it has two vertices of odd degree), we see that χ∗ (G) ≥ 3. Colour
the face of degree 4 with colour r, then colour the adjacent face of degree 3 with
colour b and the other face of degree 3 with colour r. Now colour the exterior face
with colour g. This is a 3-face-colouring of G, so χ∗ (G) ≤ 3. Thus χ∗ (G) = 3.
9. a) Prove that for any positive integer n, K2n,2 is both 2-vertex-colourable and
2-face-colourable.
Solution. By Theorem 3, K2n,2 is 2-vertex-colourable. As well, every vertex
of K2n,2 has degree either 2n or 2, whence by Theorem 10.4, K2n,2 is eulerian.
But then, since K2n,2 is a map, it follows from Corollary 13.11 that K2n,2 is
2-face-colourable.
b) Give an example of a simple connected non-null plane graph G which is both
2-vertex-colourable and 2-face-colourable, but which is not isomorphic to K2n,2
for any positive integer n.
Solution. We know that G must be bipartite and eulerian, so every vertex
must have positive even degree. Consider C6 , which is bipartite and eulerian.
Since C6 has 6 edges, if C6 were to be isomorphic to K2n,2 for some positive
integer n, we must have 4n = 6. But there is no integer n with this property,
so C6 is not isomorphic to K2n,2 for any positive integer n.
Solutions for the Exercises of Section II.13
1. Exactly which of the following are true for every plane graph G and its dual G∗ ?
14
Edge Colourings
II.13
(i) If G∗ is simple, then G has no cut-edges.
(ii) If G∗ is simple, then G is simple.
(iii) If G is non-null eulerian, then G∗ is bipartite.
A: (i), (ii)
B: (ii), (iii)
C: (i), (iii)
D: (ii)
E: None of A, B, C, D
Solution. (i) is true, since if e is a cut-edge of G, then e∗ is a loop in G∗ , and G∗
has no loops.
(ii) is false, since if e in G is a loop, then G∗ is a cut-edge of G∗ , so for example,
if G consists of a single vertex, with a loop at that vertex, then G is not simple,
but G∗ is K2 , hence simple.
(iii) is true, because an eulerian graph is connected, and a non-null connected
graph is eulerian if and only if its dual is bipartite.
Thus (i) and (iii) are true. The answer is C.
2. What is the chromatic index of the graph G shown below?
•
•
•
•
•
•
•
A: 1
B: 2
C: 3
•
D: 4
E: None of A, B, C, D
Solution. By Vizing’s theorem, the chromatic index of G4 is either 3 or 4. Let
us see if we can edge-colour G4 with 3 colours. Since the outer cycle is even we
can edge-colour it with 2 colours, say red and green. Colour the inner 4-cycle also
with red and green. Now colour the two arcs that connect the inner 4-cycle to the
outer 4-cycle with yellow. The result is a 3-edge-colouring of G4 , so the chromatic
index of G4 is 3. The answer is C.
3. Prove that χ0 (Wn ) = n if n ≥ 3 (see Exercise 11.8).
Solution. By Vizing’s theorem, we have n ≤ χ0 (Wn ) ≤ n + 1, so it suffices to
prove that Wn can be n-edge-coloured if n ≥ 3. Let the hub be denoted by v and
for each i = 1..n, colour the edge from v to i with colour ci . It remains to colour
the edges of the rim. Use the colours ci , i = 1..n, in order to colour the edges
sequentially, starting with the edge from 2 to 3. Thus the edge from i to i + 1
modulo n is coloured with colour ci−1 for i = 1..n (again, all indices are calculated
modulo n). At vertex i, the three incident edges are coloured ci , ci−1 and ci−2
modulo n, so no two adjacent edges are coloured with the same colour. Thus we
have an n-edge-colouring of Wn , whence χ0 (Wn ) = n.
II.13
Graph Theory
15
4. What is the chromatic index of the Petersen graph?
Solution. Since the Petersen graph G is a 3-regular graph, the corollary to Vizing’s
theorem tells us that 3 ≤ χ0 (G) ≤ 4. We shall rule out the possibility that the
chromatic index is 3. Suppose to the contrary that we have a 3-edge-colouring of G.
Since the outer cycle (see the diagram below) is an odd cycle, by Corollary 14.8 we
see that all 3 colours must appear on the outer 5-cycle. Suppose the colours are r,
g and b. Since it is not possible to have three edges coloured the same (for if three
edges were coloured the same, then at least two will be adjacent), there will be
one colour that appears just once on the outer 5-cycle, while the other two colours
will each appear twice. Without loss of generality (since any permutation of the
colours will transform a 3-edge-colouring into another 3-edge-colouring), suppose
that r is the colour that appears just once on the outer 5-cycle. By the symmetry
of the Petersen graph, we may assume that the edge on the outer 5-cycle that is
coloured r is the one in the clockwise direction from the vertex on the top of the
5-cycle. We may further assume (interchange g and b if not) that the next edge in
the outer 5-cycle in the clockwise direction is coloured g. Then, in the clockwise
direction from the edge coloured r, the full colouring of the outer 5-cycle must be
r, g, b, g, b. This completely determines the colours of the edges that join the
inner 5-cycle (drawn like a star in the diagram below) to the outer 5-cycle.
•
g
b
•
•
r
•
g
r
•
•
r
g b
•
g
•
b
•
g
r
•
Once these are filled in, we see that the colour of several of the edges of the
inner 5-cycle is determined. In particular, the two edges of the inner 5-cycle that
are incident to the top right vertex of the star must both be g, since they are
each incident to an edge coloured b and an edge coloured r. But then we have
two adjacent edges coloured g, whence this is not a 3-edge-colouring of G. This
contradiction follows from the assumption that there was a 3-edge-colouring of G,
so we conclude that there is no 3-edge-colouring of G. Thus χ0 (G) = 4.
5. Prove that a hamiltonian cubic graph has chromatic index 3.
Solution. Let G be a hamiltonian cubic graph. By Vizing’s theorem, we know
that 3 ≤ χ0 (G) ≤ 4. We shall prove that G can be 3-edge-coloured, whence
χ0 (G) = 3, as required. Now, since G is 3-regular, the handshake lemma tells us
that 2εG = 3νG . It follows that there is a positive integer k such that νG = 2k and
εG = 3k. Let C be a hamilton cycle for G. Then C has length 2k. By Proposition
14.9, we know that C can be 2-edge-coloured. Let C be 2-edge-coloured with
colours r and g. At each vertex, there are 3 incident edges, two of which are on C
16
Edge Colourings
II.13
and already coloured, one r and one b. We may colour the third edge b. The other
end point of this third edge is also on C, and is adjacent to to other edges at that
endpoint, both on C and therefore already coloured, one r and one b. This process
will colour all k remaining edges, so that at each vertex, we have 3 incident edges,
one of each colour. Thus we have 3-edge-coloured G.
6. For positive integers m ≥ n, what is χ0 (Km,n )? Justify your answer.
Solution. Since the graph is bipartite, we have χ0 (Km,n ) = ∆Km,n . The vertices in
the bipartite cell of size m each have degree n, while the vertices in the bipartite
cell of size n each have degree m. Since m ≥ n, we have ∆Km,n = m. Thus
χ0 (Km,n ) = m if m ≥ n.
11. a) Let G be a 3-regular connected graph with a cut-edge e and let K be a component of G − e. Prove that K has an odd number of vertices.
Solution. Let r = νK . Then K has r − 1 vertices of degree 3 and one vertex
of degree 2 (the one that e was incident to in G), so by the handshake lemma,
3(r − 1) + 2 = 2εK . Thus 3(r − 1) is even, so r − 1 is even. But then r is odd,
as required.
b) Let G be a simple 3-regular graph with a cut-edge. Prove that χ0 (G) = 4.
Solution. Without loss of generality, we may assume that G is connected
(otherwise, consider a component of G that contains a cut-edge). Since G is
simple, it follows from Vizing’s theorem that 3 ≤ χ0 (G) ≤ 4, so it suffices to
prove that G has no 3-edge-colouring. Suppose to the contrary that G does
have a 3-edge-colouring. Let e be a cut-edge of G, and let K be a component
of G − e that contains one of the endpoints of e. In the 3-edge-colouring of
G, choose one of the two colours that is different from the colour assigned to
e, and call this colour i. In G, each vertex of the subgraph K has exactly
one incident edge of colour i, and since i was not the colour assigned to e,
it follows that in K, each vertex has exactly one incident edge with colour i.
This provides a partition of the set of vertices of K into cells of size 2, where
two vertices of K are put in the same cell if and only if they are endpoints
of an edge of colour i. This implies that K has an even number of vertices.
But by the first part of this question, K has an odd number of vertices. This
contradiction followed from the assumption that G had a 3-edge-colouring. We
conclude therefore that there is no 3-edge-colouring for G.
c) Give an example of a 3-regular connected simple plane graph that has a cutedge.
Solution. Take any 3-regular simple plane graph, select an edge in the boundary of the exterior face, and perform the elementary-subdivision operation of
creating a vertex of degree 2 on the selected edge. Draw a second copy of the
II.14
Graph Theory
17
resulting graph in the exterior face of the first copy, and join the two vertices
of degree 2. The result is a simple plane 3-regular graph for which the newly
created edge is a cut-edge. For example, let us perform this construction on
K4 .
•
•
•
•
•
•
•
•
•
•
13. Use Grinberg’s theorem (Theorem 14.13) to prove that the Herschel graph G drawn
below is not hamiltonian (see ? )..
•
•
•
•
•
•
•
•
•
•
•
Solution. All nine faces of the Herschel graph have face degree 4, so Grindberg’s
theorem states that if C is a hamiltonian cycle for the Herschel graph, then 2(2φ4 −
f4 ) = 0, where φ4 is the number of faces inside C, and f4 = 9, the number of faces
in the graph. But this implies that 2φ4 = 9, which is not possible. Thus the
Herschel graph does not have a hamiltonian cycle.
Solutions for the Exercises of Section II.14
1. What is the chromatic polynomial for the graph
G shown at the right?
•
•
•
•
•
•
A: k 2 (k − 1)2 (k 2 + 3k − 4)
B: k(k − 1)(k − 2)2 (k 2 − 3k + 3)
D: (k − 1)2 (k − 2)2 (k − 3)2
E: None of A, B, C, D
C: k(k − 1)2 (k − 2)3
Solution. We have
PG(k) =
=
•
•
•
•
•
•
=
•
•
•
2 •
[• • ]
•
2
•
•
[k 2 (k − 1)2 (k − 2)2 ] [k(k − 1)3 − k(k − 1)(k − 2)]
= k(k − 1)(k − 2)2 (k 2 − 3k + 3)
k 2 (k − 1)2
The answer is B.
2. What is the chromatic polynomial of the graph shown to the right?
•
•
•
•
•
•
18
Chromatic Polynomials
II.14
A: k 2 (k − 1)2 (k − 2)2
B: k(k − 1)2 (k − 4)(k 2 − 3k + 3)
C: (k − 1)2 (k − 2)(k 4 + k + 1)
D: k(k − 1)(k 2 − 3k + 3)2
E: None of A, B, C, D
Solution.
PG(k) =
•
•
•
=
•
•
• 2
•
[• • ]
[k(k − 1)3 − k(k − 1)(k − 2)]2
= k(k − 1)[k 2 − 3k + 3]2
=
k(k − 1)
•
•
•
so the answer is D.
3. If G is a simple graph with chromatic polynomial PG (k) = k(k − 1)(k 2 − 3k + 3),
then the number of edges in G is:
A: 3
B: 4
C: 5
D: 6
E: Can’t be determined
Solution. This will be the absolute value of the coefficient of k 3 , which is −3−1 =
−4. Thus such a graph will have 4 edges (and 4 vertices).
In fact, PC (k) = k(k − 1)(k 2 − 3k + 3). Suppose now that G is any graph with
4
chromatic polynomial k(k − 1)(k 2 − 3k + 3). Then G has 4 vertices and 4 edges.
Furthermore, if G were a tree, then PG(k) = k(k − 1)3 , which is not the case. Thus
G has a cycle. Since G is simple, there are only two possiblities: G is a 3-cycle with
a fourth vertex joined by an edge to one of the vertices of the 3-cycle, or else G is a
4-cycle. The first graph would have chromatic polynomial PC (k)PI (k)/PK (k) =
1
3
2
k(k − 1)(k − 2)k(k − 1)/k = k(k − 1)2 (k − 2) 6= k(k − 1)(k 2 − 3k + 3), so G must
be C4 .
The answer is B.
4. Exactly which of the following polynomials are not chromatic polynomials?
(i) k 3 − 2k 2 + 2k − 1;
(ii) k 3 − 3k 2 + 2k;
(iii) k 3 − 2k 2 + 3k.
A: (i)
B: (iii)
C: (ii), (iii)
D: (i), (iii)
E: None of A, B, C, D
Solution. (i) can’t be the chromatic polynomial of any graph since chromatic
polynomials have no constant term.
II.14
Graph Theory
19
(ii) k 3 − 3k 2 + 2k = k(k 2 − 3k + 2) = k(k − 1)(k − 2), which is the chromatic
polynomial of K3 .
(iii) Since 1 is not a root of k 3 − 2k 2 + 3k (that is, the sum of the coefficients is
not zero), any simple graph for which this is the chromatic polynomial would be
1-vertex-colourable, which can only happen if the graph is a null graph. But such
a graph would also have to have 3 vertices and 2 edges, so it could not be a null
graph. Therefore this polynomial can’t be the chromatic polynomial of any graph.
Thus neither (i) nor (iii) is the chromatic polynomial of any graph, while (ii) is
the chromatic polynomial of K3 . The answer is D.
5. Which of the following polynomials are chromatic polynomials?
(i)
(ii)
(iii)
(iv)
A: (iii)
B: (iv)
k 4 − 5k 3 + 4k
k 4 − 2k 3 + 3k 2
k 4 − 7k 3 + 6k 2
k4 − k3
C: (i)
D: (ii)
E: None of A, B, C, D
Solution. (i) is not, since it has nonzero coefficient of k 3 and k, but zero coefficient
for k 2 .
(ii) is not, for if it were, then any graph that had (ii) as its chromatic polynomial
would have chromatic number 1, and the only graphs with chromatic number 1
are the null graphs, the graphs with no edges. But since the coefficient of the next
to highest power of x is −2, any graph that had (ii) as its chromatic polynomial
would have 2 edges and thus not be a null graph.
(iii) is not, for any graph that had (iii) as its chromatic polynomial would have 4
vertices and 7 edges, but the largest possible number of edges in a 4 vertex simple
graph is 42 = 6.
(iv) is the chromatic polynomial of the 4 vertex graph which consists of two single
vertex components and one component equal to K2 .
The answer is B.
6. If G is a simple graph with chromatic polynomial PG(k) = k(k − 1)(k 2 − 3k + 3),
then the number of edges in G is:
A: 2
B: 3
C: 4
D: 5
ν −1
E: 6
Solution. (−1)εG is the coefficient of k G in the chromatic polynomial of G,
where νG is the degree of the chromatic polynomial. Thus νG = 4, so we want the
20
Chromatic Polynomials
II.14
coefficient of k 3 . The terms which involve k 3 are −3k 3 − k 3 = −4k 3 , so εG = 4.
The answer is C.
8. Find the chromatic polynomial of K5,2 .
Solution.
[
 • 
• 
•
•
K5,2 ] =  • •e •  =  • • • 
•
•
•
h •• i5
[ • • ]5
• •
+
=
[ • • ]4
[ • ]4
5
5
5


•
•
+ • •
by Theorem 15.6
•
•
by Theorem 15.11
k 5 (k − 1)5
k (k − 1) (k − 2)
+
by Proposition 15.8
k 4 (k − 1)4
k4
= k(k − 1)(k − 2)5 + k(k − 1)5 = k(k − 1)[(k − 2)5 + (k − 1)4 ]
=
= k(k − 1)(k 5 − 9k 4 + 37k 3 − 74k 2 + 76k − 31)
= k 7 − 10k 6 + 46k 5 − 111k 4 + 150k 3 − 107k 2 + 31k.
10. Let G be a simple graph whose chromatic polynomial is k 3 (k − 1)2 .
a) How many vertices does G have?
Solution. The number of vertices is equal to the degree of the chromatic
polynomial, and since the degree of k 3 (k − 1)2 is 5, G has 5 vertices.
b) How many edges does G have?
Solution. The number of edges of G is equal to the magnitude of the coefficient
ν −1
of k G
in the chromatic polynomial of G. Since PG(k) = k 3 (k − 1)2 =
k 3 (k 2 − 2k + 1) = k 5 − 2k 4 + k 3 , we see that G must have 2 edges.
c) Draw an example of a simple graph whose chromatic polynomial is k 3 (k − 1)2 .
Solution.
Since such a graph has 5 vertices and only 2 edges, it is not
connected. If we take the 2-chain, which has 2 edges and 3 vertices, and throw
in a couple of null components, we end up with a graph with 5 vertices and 2
edges, as shown below.
•
•
•
•
•
The chromatic polynomial of a disconnected graph is just the product of the
chromatic polynomials of its components, so we have PG(k) = k(k−1)2 (k)(k) =
k 3 (k − 1)2 , as required.
14. Prove that the graphs G and H shown in Figure 1 are non-isomorphic, but they
have the same chromatic polynomial.
II.14
Graph Theory
21
•
•
• •
• •
• •
• •
•
(ii) H
•
(i) G
Figure 1.
Solution. The degree sequence for G is 4, 4, 3, 3, 3, 3, while the degree sequence for
H is 4, 4, 4, 3, 3, 2. Since isomorphic graphs must have identical degree sequences,
we see that G and H are not isomorphic.
For the chromatic polynomials, we have
h
i h
i •
•
• •
• •
• • = • •
• • +
•
•
•
h
• i2
• 2
• •
•
[• •]
2
• •
by Theorem 15.6
• •
by Theorem 15.11
[•]
k (k − 1)2 (k − 2)2 (k − 3)2 k 2 (k − 1)2 (k − 2)2
=
+
k
k(k − 1)
2
2
2
= k(k − 1)(k − 2) (k − 3) + k(k − 1) (k − 2)2
=
+
= k(k − 1)(k − 2)2 [(k − 3)2 + (k − 1)]
= k(k − 1)(k − 2)2 (k 2 − 5k + 8)
= k 6 − 10k 5 + 41k 4 − 84k 3 + 84k 2 − 32k,
and
h
•
• •
• •
•
i
=
=
h
h
•
• •
• •
•
•
•
•
ih
i
• •
• •
•
•
i
[• •]
h • i h•
•
•
•
• •
•
•
•
= (k − 2)
+[
•
• •
2
]
[• •]
+
• •
• •
•
•
[•]
i
+ (k − 1)
•
• •
2
[• •]
(k − 2)k 2 (k − 1) (k − 2)2 (k − 3)2 (k − 1)k 2 (k − 1)2 (k − 2)2
=
+
k(k − 1)
k(k − 1)(k − 2)
= k(k − 1)(k − 2)2 (k − 3)2 + k(k − 1)2 (k − 2)2
= k(k − 1)(k − 2)2 [(k − 3)2 + (k − 1)]
= k(k − 1)(k − 2)2 (k 2 − 5k + 8),
which is the same as PG(k), as required.
17. Prove that if G is a graph on n vertices for which PG(k) = k(k − 1)n−1 , then G is
22
Chromatic Polynomials
II.14
a tree.
Solution. First, we note that G is connected. For if G has components G1 , G2 , . . . , Gt ,
then by Proposition 15.5, PG(k) = PG (k)PG (k) · · · PG (k), while by Theorem
t
1
2
15.9, PG (k) has zero constant term for each i = 1, 2, . . . , t, so every term of dei
gree less than t in PG(k) has zero coefficient. Since the coefficient of k in PG(k)
is equal to (−1)n−1 , it follows that t = 1 and so G is connected. Now, from the
binomial theorem, we see that the coefficient of k n−2 in (k − 1)n−1 is (−1)(n − 1),
and so the coefficient of k n−1 in PG(k) is (−1)(n − 1). By Theorem 15.9, the
absolute value of the coefficient of k n−1 in PG is the number of edges in G, so
εG = n − 1 = νG − 1. By Theorem 5.4 (iii), G is a tree.
19. Prove that C4 is the only simple graph with chromatic polynomial k(k − 1)(k 2 −
3k + 3).
Solution. First, we note that in fact, PC (k) = k(k − 1)(k 2 − 3k + 3). Suppose
4
now that G is any graph with chromatic polynomial k(k − 1)(k 2 − 3k + 3). Then G
has 4 vertices and 4 edges. Furthermore, if G were a tree, then PG(k) = k(k − 1)3 ,
which is not the case. Thus G has a cycle. Since G is simple, there are only two
possiblities: G is a 3-cycle with a fourth vertex joined by an edge to one of the
vertices of the 3-cycle, or else G is a 4-cycle. The first graph would have chromatic
polynomial PC (k)PI (k)/PK (k) = k(k − 1)(k − 2)k(k − 1)/k = k(k − 1)2 (k − 2) 6=
1
3
1
k(k − 1)(k 2 − 3k + 3), so G must be C4 .
23. Let G be a simple graph. Prove that for any k < 0, PG(k) 6= 0. More precisely,
prove that for k < 0, PG(k) > 0 if n is even, while PG(k) < 0 if n is odd.
Solution. By Theorem 15.9, there exist positive integers aκ , . . . , an such that
G
P
PG(k) = ni=κ (−1)n−i ai k i . Thus for k < 0, if we set k = −t, then t > 0 and
G
P
P
we have PG(k) = PG(−t) = ni=κ (−1)n−i ai (−t)k i = ni=κ (−1)n−i (−1)i ai ti =
G
G
P
P
(−1)n ni=κ ai ti . Since ni=κ ai ti > 0, it follows that for k < 0, PG(k) 6= 0 and
G
G
the sign of PG(k) is that of (−1)n .
Exam and Quiz Solutions
CHAPTER III
Exam and Quiz Solutions
23