16.5. COLORINGS
211
CHAPTER 1
1.2. V(G) = {1,2,3,4,5,6) and E(G) = {12,15,23,24,46,45).
1.3. (a) and (c).
1.4. E(G) = {AC,AD,AE, BC, B E , C D , DE).
1.5. (a) 3; (b) 6; (c) 10; (d)
)(;
= n(n
- 1)/2.
1.8. ms.
1.11. (a) degvl = 2 ; degvz=3; degvs=3; degv4=3; degv5 = 4 ; d e g v 6 = 1 ; degv7=2;
degu8 = 2.
(b) 4,3,3,3,2,2,2,1.
(c) The sum of the degrees is 20 and the number of edges is 10. Hence, the sum of
the degrees is exactly twice the number of edges.
1.12. Yes; the number of degrees listed is the order of G. The sum of the degrees is twice
the size of G.
1.13. Suppose there are n 2 2 people in the group. Represent this situation by a graph
where the vertices correspond to the people and an edge joins two vertices if the
corresponding people are acquainted. We now consider the degrees of the vertices in
this graph, which we call the "acquaintance graph." The degree of each vertex must
belong to (0, 1, . . . ,n - 1). If some vertex has degree n - 1, then it is adjacent to every
other vertex, and so no vertex has degree 0. Conversely, if some vertex has degree 0,
then it is adjacent to no other vertex, and so no vertex has degree n - 1. Thus the
vertex degrees all belong to either the set {0,1,. . . ,n - 2) or the set {1,2,. . . ,n - 1)
but not to both sets. There are only n - 1 distinct elements in each of these sets
and, since there are n vertices, at least two of them must have the same degree. The
corresponding people have the same number of acquaintances.
1.14. Represent this situation by a graph where the vertices correspond to the people and
each edge represents a handshake. We now consider the degrees of the vertices of this
graph. Each person has shaken hands with at most six people. So the seven people who
have answered you are represented by vertices of degree 0,1,2,3,4,5 and 6. Number
each of these vertices according to its degree and denote the vertex representing you
by 7. Then (vertex) 6 is adjacent to all other vertices except 0, and 1 is adjacent
only to 6. Note that 0 and 6 represent a couple. The vertex 5 is adjacent to all other
vertices except 0 and 1, and so 1 and 5 represent a couple. Now vertex 4, which
cannot be adjacent to any of the vertices O , 1 or 2, must be adjacent to vertices 7,6,5
and 3 (and so 4 and 2 represent a couple). Note that vertex 3 is adjacent to 6,5 and
4 and therefore not to 7. So your partner is represented by vertex 3 and has shaken
three hands. Since vertex 7 is adjacent to vertices 6,5 and 4, it has degree 3, and so
you have shaken three hands.
CHAPTER 16. PROBABILISTIC METHODS
212
1.15. Let G be an r-regular graph, and so degv = r for all v E V(G). We now count the
edges of G. Observe that the number q(G) of edges in G is the number of edges that
are incident with vertices in Vl. Hence, q(G) is the sum of the degrees of the vertices
in Vl , namely rlVl 1. However we can also count the number of edges of G by counting
the edges incident with vertices in fi, and so q(G) is the sum of the degrees of the
61= Ihl.
vertices in h,namely r l h ( . Thus, rlVlI = rlfil, or, equivalently, 1
1.16. (a) This sequence contains an odd number of odd terms, and is therefore not graphical.
(b) Applying Algorithm 1.4 to the sequence
we get
Since the sequence s2 contains a negative integer, the sequence sz, and therefore
the sequence s, is not graphical.
(c) Applying Algorithm 1.4 to the sequence
s : 3,3,2,2,2,2,1,1 weget
s'l :
sl :
2,1,1,2,2,1,1
2,2,2,1,1,1,1 (reordering)
sg :
s3 :
O, l, l, l l
1,1,1,1,0 (reordering)
Since the graph Gg = 2K2UK1 has degree sequence sg, sg is a graphical sequence.
By the Havel-Hakimi Theorem, each of the sequences s2,sl and s is in turn
graphical. To construct a graph with degree sequence s2, we proceed in reverse
from S; to s2, observing that a vertex should be added to Gg SO that it is adjacent
to one vertex of degree 0 in Gg. We thus obtain a graph Gq with degree sequence
s2 (or s;). Note that Gz S 3K2. To construct a graph with degree sequence s l ,
we proceed in reverse from s; to sl, observing that a vertex should be added to
G2 so that it is adjacent to two vertices of degree 1 in Gz. We thus obtain a
graph G1 with degree sequence sl (or si). (One possibility for G1 is Kg U 2K2.)
Proceeding from s i to s, we add a new vertex joining it to two vertices of degree 1
and one vertex of degree 2 in GI. This gives a graph G with degree sequence s.
16.5. COLORINGS
(d) Applying Algorithm 1.4 to the sequence
s : 7,4,3,3,2,2,2,1,1,1 weget
s::
sl :
3,2,2,1,1,1,0,1,1
3,2,2,1,1,1,1,1,0 (reordering)
s; :
1,1,O , l , 131, 170
1,1,1,1,1,1,0,0 (reordering)
s2 :
Since the graph G2 = 3Kz U 2K1 has degree sequence s2, s 2 is a graphical
sequence. By the Havel-Hakimi Theorem, each of the sequences sl and s is in
turn graphical. To construct a graph with degree sequence sl, we proceed in
reverse from sb to sl, observing that a vertex should be added to G2 SO that it
is adjacent to two vertices of degree 1 and one vertex of degree 0 in Gz. We thus
obtain a graph G1 with degree sequence sl (or si). Proceeding from si to s, we
add a new vertex joining it to one vertex of degree 3, two vertices of degree 2,
three vertices of degree 1, and one vertex of degree 0 in GI. This gives a graph
G with degree sequence s.
Suppose that the sequence s: dl, d2,. . . ,d, is graphical. Then there exists a graph G
with degree sequence s. Thus, the vertices of G can be labelled vl, 712,. . . ,v, such
that degc vi = d, for all i. We observe, therefore, that vi is adjacent with 4 vertices
of G and not adjacent with n - 4 - 1 vertices of G. Thus in the complement of G
the vertex vi is not adjacent with di vertices of G and adjacent with n - d, - 1. Hence,
degcvi=n-4-1
(1 5 i I n ) . ~ h u s , ~ h a s d e g r e e s e q u e n c e s l : n - d l - l , n - d 2 1,. . ,n - d, - 1. This shows that s l is graphical. Similarly, if sl is graphical, then
so too is s.
.
1.18 ( a ) There are four nonisomorphic graphs of order 3. (Of all graphs on three vertices,
there is one graph with no edges, one with one edge, one with two edges, and
one with three edges.)
( b ) There are eleven nonisomorphic graphs of order 4. (Of all graphs on four vertices,
there is one graph with no edges, one with one edge, two with two edges (either
the two edges are adjacent or not), three with three edges, two with four edges,
one with five edges, and one with six edges.)
1.19. The mapping $ : V(Fl)
+ V(F2)
defined by
is an isomorphism.
1.20. There are several isomorphisms. For example, the mapping $ : V(G) 4 V(H) defined
by
.
i ( a ) = 1, 4(b) = 4, $ ( c ) = 7, 4(d) = 3, $(el = 6, 4 ( f ) = 2, 4(9) = 5,
is an isomorphism.
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CHAPTER 16. PROBABILISTIC METHODS
1.21. No. To see this, look at the four vertices of degree 2. In the first graph they are joined
in pairs, whereas in the second graph none of them is joined to any other.
1.22 (a) True (b) False (see Exercise 1.19) (c) True, (d) False (see Exercise 1.19).