Introduction to Graph Theory Second Edition Solution Manual (by Douglas B. West)

8/3/2019 Introduction to Graph Theory Second Edition Solution Manual (by Douglas B. West)

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INTRODUCTION TO GRAPH THEORY

SECOND EDITION (2001)

SOLUTION MANUAL

SUMMER 2005 VERSION

c DOUGLAS B. WEST

MATHEMATICS DEPARTMENT

UNIVERSITY OF ILLINOIS

All rights reserved. No part of this work may be reproduced

or transmitted in any form without permission.

NOTICE

This is the Summer 2005 version of the Instructor’s Solution Manual for

Introduction to Graph Theory, by Douglas B. West. A few solutions have

been added or clarified since last year’s version.

Also present is a (slightly edited) annotated syllabus for the one-

semester course taught from this book at the University of Illinois.This version of the Solution Manual contains solutions for 99.4% of

the problems in Chapters 1–7 and 93% of the problems in Chapter 8. The

author believes that only Problems 4.2.10, 7.1.36, 7.1.37, 7.2.39, 7.2.47,

and 7.3.31 in Chapters 1–7 are lacking solutions here. There problems are

too long or difficult for this text or use concepts not covered in the text; they

will be deleted in the third edition.

The positions of solutions that have not yet been written into the files

are occupied by the statements of the corresponding problems. These prob-

lems retain the (−), (!), (+), (∗) indicators. Also (•) is added to introducethe statements of problems without other indicators. Thus every problem

whose solution is not included is marked by one of the indicators, for ease

of identification.

The author hopes that the solutions contained herein will be useful to

instructors. The level of detail in solutions varies. Instructors should feel

free to write up solutions with more or less detail according to the needs of

the class. Please do not leave solutions posted on the web.

Due to time limitations, the solutions have not been proofread or edited

as carefully as the text, especially in Chapter 8. Please send corrections [email protected]. The author thanks Fred Galvin in particular for con-

tributing improvements or alternative solutions for many of the problems

in the earlier chapters.

This will be the last version of the Solution Manual for the second

edition of the text. The third edition will have many new problems, such

as those posted at http://www.math.uiuc.edu/ west/igt/newprob.html . The

effort to include all solutions will resume for the third edition. Possibly

other pedagogical features may also be added later.

Inquiries may be sent to [email protected]. Meanwhile, the authorapologizes for any inconvenience caused by the absence of some solutions.

Douglas B. West









1 Chapter 1: Fundamental Concepts Section 1.1: What Is a Graph? 2

1.FUNDAMENTAL CONCEPTS

1.1. WHAT IS A GRAPH?

1.1.1. Complete bipartite graphs and complete graphs. The complete bipar-

tite graph K m,n is a completegraphif and onlyif m = n = 1 or {m, n} = {1, 0}.1.1.2. Adjacency matrices and incidence matrices for a 3-vertex path.

0 1 1

1 0 0

1 0 0

0 1 0

1 0 1

0 1 0

0 0 1

0 0 1

1 1 0

1 1

1 0

0 1

1 1

0 1

1 0

1 0

1 1

0 1

0 1

1 1

1 0

1 0

0 1

1 1

0 1

1 0

1 1

Adjacency matrices for a path and a cycle with six vertices.

0 1 0 0 0 0

1 0 1 0 0 0

0 1 0 1 0 0

0 0 1 0 1 0

0 0 0 1 0 1

0 0 0 0 1 0

0 1 0 0 0 1

1 0 1 0 0 0

0 1 0 1 0 0

0 0 1 0 1 0

0 0 0 1 0 1

1 0 0 0 1 0

1.1.3. Adjacency matrix for K m,n.

m n

m 0 1

n 1 0

1.1.4. G ∼= H if and only if G ∼= H . If f is an isomorphism from G to H ,

then f is a vertex bijection preserving adjacency and nonadjacency, and

hence f preserves non-adjacency and adjacency in G and is an isomor-phism from G to H . The same argument applies for the converse, since the

complement of G is G.

1.1.5. If every vertex of a graph G has degree 2, then G is a cycle—FALSE.

Such a graph can be a disconnected graph with each component a cycle. (If

infinite graphs are allowed, then the graph can be an infinite path.)

1.1.6. The graph below decomposes into copies of P4.

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1.1.7. A graph with more than six vertices of odd degree cannot be decom-

posed into three paths. Every vertex of odd degree must be the endpoint

of some path in a decomposition into paths. Three paths have only six

endpoints.

1.1.8. Decompositions of a graph. The graph below decomposes into copies

of K 1,3 with centers at the marked vertices. It decomposes into bold and

solid copies of P4 as shown.

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•

1.1.9. A graph and its complement. With vertices labeled as shown, two

vertices are adjacent in the graph on the right if and only if they are not

adjacent in the graph on the left.

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a

b c

d

e

f g

h

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•

a

f d

g

c

h b

e

1.1.10. The complement of a simple disconnected graph must be connected—

TRUE. A disconnected graph G has vertices x, y that do not belong to apath. Hence x and y are adjacent in G. Furthermore, x and y have no com-

mon neighbor in G, since that would yield a path connecting them. Hence



































































































































































































































































































293 Chapter 5: Coloring of Graphs Section 5.3: Enumerative Aspects 294

currence for the chromatic polynomial, we begin with the compatible pairs

for G − e and consider the effect of adding e. If ( D, f ) is a compatible pair

for G − e such that f (u) = f (v), say f (u) < f (v), then e must be oriented

from u to v to obtain an orientation of G compatible with f . The result

is indeed acyclic, else it has a directed v, u-path along which the value f

must step downward at some point. Conversely, we can delete e from a

compatible pair for G with f (u)

=f (v) to obtain a compatible pair for

G − e. Hence the compatible pairs with differing labels for u and v are in1-1 correspondence in G and G − e.

Now consider pairs with f (u) = f (v). It suffices to show that each

such pair for G − e becomes a compatible pair for G by adding e oriented

in at least one way, and that for η(G · e, k ) of these, both orientations of e

yield compatible pairs for G. For the first statement, consider an arbitrary

compatible pair ( D, f ) with f (u) = f (v) for G − e, and suppose neither

orientation for e yields a compatible pair for G This requires D to have



orientation for e yields a compatible pair for G. This requires D to have

both a u, v-path and a v, u-path, which cannot happen since D is acyclic.

For the second statement, suppose that ( D, f ) is a compatible pair for Gwith f (u) = f (v) and that the orientation obtained by reversing e is also

compatible with f . Then D − e has neither a u, v-path nor a v, u-path,

and contracting e yields a compatible pair for G · e. Conversely, given a

compatible pair for G · e, we can split the contracted vertex to obtain a

compatible pair for G − e with f (u) = f (v) so that orienting e in either

direction yields a compatible pair for G.

















311 Chapter 6: Planar Graphs Section 6.2: Characterization of Planar Graphs 312

6.2.3. A convex embedding. This is the graph of the icosahedron. It is

3-connected and has a convex embedding in the plane.

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•

••

•

• •

•

• •

6.2.4. Planarity of graphs.

•••

••

••

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••

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•

6.2.6. Fary’s Theorem.

a) Every simple polygon with at most 5 vertices contains a point that

sees every point in the polygon. In a convex polygon, by definition, the

segment joining any pair of points lies entirely in the polygon. Hence every

point in a convex polygon sees the entire polygon.

Proof 1. If a 4-gon is not convex, then the vertex opposite the interior



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••••

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The first graph is planar; a straight-line embedding with every face

convex appears below. The second graph is nonplanar. It has many sub-

graphs that are subdivisions of K 3,3; one is shown below.

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•

6.2.5. The minimum number of edges that must be deleted from the Petersen graph to obtain a planar subgraph is 2. The drawing on the left below

illustrates a subdivision of K 3,3 in the Petersen graph. Since this does not

use every edge of the Petersen graph, the graph obtained by deleting one

edge from the Petersen graph is still nonplanar (all edges are equivalent,

by the disjointness description of the Petersen graph).

Deleting two edges from the Petersen graph yields a planar subgraph

as shown on the right below.

reflex angle (exceeding 180 degrees) sees the entire polygon. A non-convex

pentagon has one or two reflex angles, and if two they may be consecutive

or not. The cases are illustrated below.

• • •

Proof 2. Triangulate the polygon by adding chords between corners

that can see each other. This can be done by adding one chord to a 4-gonand by adding two to a 5-gon, with cases as illustrated above. The resulting

triangles have one common vertex. Since a corner of a triangle sees the

entire triangle, the common corner sees the entire region.

b) Every planar graph has a straight-line embedding. By induction on

n(G), we prove the stronger statement that the edges of any plane graphG can be “straightened” to yield a straight-line embedding of G without

changing the order of incident edges at any vertex. The statement is true

by inspection for n(G) ≤ 3.

For n(G) ≥ 4, we may assume that G is a triangulation, since anyplane graph can be augmented to a maximal plane graph, and deleting

extra edges in a straight-line embedding of the maximal planar supergraph

yields a straight-line embedding of the original graph. Every planar graph

has a vertex of degree at most 5; let x be such a vertex in G.

Since G is a triangulation, the neighborhood of G is a cycle C , and

G − x has C as a face boundary. By the induction hypothesis, G − x has

a straight-line embedding with C as a polygonal face boundary. By part



























339 Chapter 7: Edges and Cycles Section 7.1: Line Graphs and Edge›Coloring 340

7.EDGES AND CYCLES

7.1. LINE GRAPHS & EDGE›COLORING

7.1.1. Edge-chromatic number and line graph for the two graphs below.

The labelings are proper edge colorings, the number of colors used is the

maximum degree, so the colorings are optimal.

•

• •

2

7.1.4. For every graph G , χ (G) ≥ e(G)/α(G). In a proper edge-coloring,

each color class has at most α(G) edges. The lower bound follows because

all e(G) edges must be colored.

7.1.5. The Petersen graph is the complement of L(K 5). The vertices of L(K 5)

are the edges in K 5, which can be named as the 2-element subsets of [5].

Two such pairs are adjacent in the Petersen graph if and only if they are

disjoint, which is the condition for them being nonadjacent in L(K 5).

7.1.6. The line graph of the Petersen graph has 10 triangles. For a simple

graph G, there is a triangle in L(G) for every set of three edges in G that

share one common endpoint and for every set of three edges that form

a triangle in G. The Petersen graph has no triangles, so the latter type

does not arise. However, the Petersen graph has 10 triples of edges with a

common endpoint, one for each of its vertices.

7.1.7. P5 is a line graph. The complement of P5 is a 5-cycle with a chord.

It is the line graph of a 4-cycle with a pendant edge.



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2

3

0

3

0 1

0

2•

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0

1

2

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1

2

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1

2

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7.1.2. χ (Qk ) = (Qk ) , by explicit coloring. In the cube Qk , the edges

between vertices differing in coordinate j form a complete matching. Over

the k choices of j , these partition the edges.

7.1.3. χ (C n K 2) = 3. The lower bound is given by the maximum degree.

For the upper bound, when n is even colors 0 and 1 can alternate alongthe two cycles, with color 2 appearing on the edges between the two copies

of the factor C n. When n is odd, colors 0 and 1 can alternate in this way

except for the use of one 2. Color 2 appears on all cross edges except those

incident to edges on the cycles with color 2, as shown below.

• • • • • •

• • • • • •2 2 0 1 2 2

0 1 2 0 1

0 1 2 0 1

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•

••

•

•

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•

H P5 = L( H )

7.1.8. The line graph of K m,n is the cartesian product of K m and K n . For

each edge xi y j in K m,n , we have a vertex (i, j) in L(K m,n); these are also

the vertices of K m K n . Pairs (i, j) and (k , l) in V (K m K n) are adjacent inK m K n if and only if i = k or j = l. This is the same as the condition for

adjacency in L(K m,n), because xi y j and xk yl share an endpoint in K m,n if and

only if i = k or j = l.

7.1.9. A set of vertices in the line graph of a simple graph G form a clique

if and only if the corresponding edges in G have one common endpoint or form a triangle. Let S be the corresponding set of edges in G, and choosee ∈ S. If all other elements of S intersect e at the same endpoint of e, we

have one common endpoint. Otherwise, we have edges f and g such that

f shares endpoint x with e and g shares endpoint y with e. Since f and

g must share an endpoint, they share their other endpoint z and complete

a triangle. Since no single vertex lies in all of e, f , g, no additional edge of

the simple graph G can share a vertex with all of these.











































































































































































511 Chapter 8: Additional Topics Section 8.6: Eigenvalues of Graphs 512

Let S be the set of switches flipped an odd number of times; the flips

of other vertices have no effect. Since also si changes li and li begins off,li is on at the end if and only if its closed neighborhood in G has an odd

number of vertices in S. Thus, the problem is equivalent to finding an

odd dominating set S. This form has a combinatorial proof by Gallai (see

Lovasz [1979], Exercise 5.17).

The proof using linear algebra is much shorter. Let B be the augmented

adjacency matrix A(G) + I (add 1 to each diagonal entry). If x is the inci-

dence vector of S in a binary vector space, then Bx is the incidence vector of

the set of lights on after flipping the switches at S (since arithmetic is mod-

ulo 2). The problem is to show that 1 ∈ T , where T = { Bx ∈ Zn2}. We prove

the more general statement that if B is a symmetric binary matrix, with

vector u along the diagonal, then u ∈ T . (see Problem 798, Nieuw Archief

voor Wiskunde (4) 9 (1991), 117-118) The solution given in the Monthly

(1993, p. 806) is as follows.

We show that every vector orthogonal to T is also orthogonal to u.

This implies that u is in the orthogonal complement of the orthogonal

complement of T and hence is in T itself. Thus we want to show that

if n

i=1x

i B

i, j =0 for all j , then n

i=1x

iu

i =0, where all computation is

modulo 2.

Multiplying the vector 0 by x yieldsn

j=1n

i=1 xi Bi, j x j = 0. By the

symmetry of B, the off-diagonal entries contribute nothing, and we obtainni=1 x2i Bi,i = 0. Since x2i = xi in binary, this reduces to 0 = n

i=1 xi Bi,i =ni=1 xiui , which completes the proof.



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Introduction to Graph Theory Second Edition Solution Manual (by Douglas B. West)

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