205
Case 2. {x, y} = {u, v}, say x = u and y = v. We show that u and v are connected
in G. Since the order of G is at least 3, there is a vertex w in G such that w ̸= u, v.
Since G − v is connected, G − v contains a u − w path P ′ . Furthermore, since G − u is
connected, G − u contains a w − v path P ′′ . Therefore, P ′ followed by P ′′ produces a
u − v walk. Thus G contains a u − v path and so u and v are connected in G.
For the converse, let u and v be two vertices of G such that d(u, v) = diam(G). We
claim that G − u and G − v are both connected. Suppose that this is not the case. Then
at least one of G − u and G − v is disconnected, say G − v is disconnected. Therefore,
G − v contains two vertices x and y that are not connected in G − v. However, since G
is connected, the vertices u and x are connected in G, as are u and y.
Let P ′ be an x − u geodesic in G and let P ′′ be a u − y geodesic in G. Since dG (u, v) =
diam(G), the vertex v cannot lie on either P ′ or on P ′′ , so P ′ and P ′′ are paths in G − v.
The path P ′ followed by P ′′ produces an x − y walk W in G − v. Thus G − v contains
an x − y path and so x and y are connected in G − v. This is a contradiction.
(b) The statement is false. Consider G = P5 .
38. Proof. Assume, to the contrary, that G contains a vertex x such that deg x ≡ 0 (mod 3).
Since G is connected, G contains a w − x path, say P = (w = w0 , w1 , . . . , wk = x). Let t
be the smallest positive integer such that deg wt ≡ 0 (mod 3). Then deg wt−1 ̸≡ 0 (mod 3).
However, then, deg wt−1 + deg wt ̸≡ 0 (mod 3), a contradiction.
Section 12.3. Eulerian Graphs
1. (a) No; otherwise, the stroll would only contain two of the three bridges incident with B.
(b) No; otherwise, there would have to be at least four bridges incident with B.
(c) This provides a somewhat different solution of the Königsberg Bridge Problem.
2. Yes. The graph modeling this situation is an Eulerian graph.
3. Proof. Assume first that G contains an Eulerian trail T . Thus T is a u − v trail for some
distinct vertices u and v. We now construct a new connected graph H from G by adding a
new vertex x of degree 2 and joining it to u and v. Then C = (T, x, u) is an Eulerian circuit
in H. Thus, every vertex of H is even and so only u and v have odd degrees in G = H − x.
For the converse, we proceed in a similar manner. Let G be a connected graph containing
exactly two vertices u and v of odd degree. We show that G contains an Eulerian trail T ,
where T is either a u − v trail or a v − u trail. Add a new vertex x of degree 2 to G and join
it to u and v, calling the resulting graph H. Therefore, H is a connected graph all of whose
vertices are even. Thus H is an Eulerian graph containing an Eulerian circuit C. Since it is
irrelevant which vertex of C is the initial (and terminal) vertex, we assume that C is an x − x
circuit. Since x is incident only with the edges ux and vx, one of these is the first edge of C
and the other is the final edge of C. Deleting x from C results in an Eulerian trail T of G that
begins either at u or v and ends at the other.
4. The graphs G2 and G3 have Eulerian circuits, G1 has an Eulerian trail while G4 and G5 have
neither.
5. See Figure 74.
6. (a) True. This follows from Theorem 12.44(1).
(b) False. The resulting multigraph may be disconnected.
(c) True. This follows from Theorem 12.42.
206
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Figure 74: The graphs in Exercises 5
7. Proof. If n = 3, then C 3 = K 3 , which is disconnected and therefore is not Eulerian. Thus,
we may assume that n ≥ 4. Observe that Cn is 2-regular of order n and so C n is a connected
(n − 3)-regular graph. Thus C n is Eulerian if and only if n − 3 is even; while n − 3 is even if
and only if n is odd.
8. Proof. In the (connected) graph Ks,t , every vertex has degree s or degree t. Thus Ks,t is
Eulerian if and only if both s and t are even.
9. Since Qn is connected and n-regular, Qn is Eulerian if and only if n is even.
10. Only K8 has this property. Suppose that some other graph G of order 8 has this property.
Necessarily, G must be connected. Then G contains a vertex u that is not adjacent to all other
vertices of G. Then u is adjacent only to odd vertices and not adjacent only to even vertices.
Let w be an even vertex of G and let x be a vertex adjacent to w. Then G − x contains odd
vertices and so is not Eulerian. This is a contradiction.
11. Proof. Suppose that there is an Eulerian graph of order n ≥ 3, where at most two vertices
have the same degree. We consider two cases.
Case 1. n is even, say n = 2k for some k ≥ 2. Then each vertex of G has one of the
k − 1 degrees 2, 4, . . . , 2k − 2. Then the order of G is at most 2(k − 1) = 2k − 2 < n, which is
impossible.
Case 2. n is odd, say n = 2k + 1 for some k ≥ 1. Then each vertex of G has one of the k
degrees 2, 4, . . . , 2k. Then the order of G is at most 2k < n, which is impossible.
12. (a) Suppose that n is even. Then n = 2ℓ for some ℓ ≥ 2. Then every vertex of G has one
of the ℓ − 1 degrees 2, 4, . . . , 2ℓ − 2. If three vertices have one of these degrees and at
most two vertices have any other degree, then n ≤ 3 + 2(ℓ − 2) = 2ℓ − 1 < n, which is
impossible.
(b) Since n is odd by (a), it follows that n = 2ℓ + 1 for some positive integer ℓ. We consider
two cases.
Case 1. ℓ is even, say ℓ = 2k. Then n = 2ℓ + 1 = 2(2k) + 1 = 4k + 1.
Case 2. ℓ is odd, say ℓ = 2k − 1. Then n = 2ℓ + 1 = 2(2k − 1) + 1 = 4k − 1.
(c) Suppose that n = 4k + 1. Let V (G) = {v1 , v2 , . . . , v4k+1 }. At least two vertices have
degree 4k, say v1 and v2 and at least two vertices have degree 2, say v4k and v4k+1 . If
another vertex has degree 4k, then no vertices have degree 2. Therefore, exactly two
vertices have degree 4k and exactly two vertices have degree 2. At least two vertices have
degree 4k −2 and at least two vertices have degree 4. Let v3 and v4 have degree 4k −2 and
let v4k−2 and v4k−1 have degree 4. If another vertex has degree 4k − 2, then no vertices
have degree 4. Continuing in this manner, we see that exactly two vertices have degree
4k − 2s and exactly two vertices have degree 2s + 2 for s = 0, 1, . . . , k − 2. Finally, exactly
two vertices have degree 2k + 2 and at least two vertices have degree at least 2k. Since
the number of vertices of each degree 2, 4, 2k − 2, 2k + 2, . . . , 4k is two, there are exactly
three vertices of degree 2k. (The argument is similar when n = 4k − 1.)
207
13. The statement is true. Proof. Since G is neither an Eulerian circuit nor an Eulerian trail, G
contains 2k odd vertices, where k ≥ 2. Observe that degH v = 2k and all other vertices of H
are also even. Thus H is Eulerian.
14. Each of the graphs K 3,4 , W 5 and P4 + P5 is a disconnected graph with exactly two components.
The components of K 3,4 are K3 and K4 . The components of W 5 are K1 and C5 . The
components of P4 + P5 are P4 and P5 .
15. See Figure 75.
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Figure 75: The graphs in Exercise 15
16. Note that Q4 = Q3 × K2 (see Figure 76).
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Figure 76: The graphs in Exercise 16
17. Let m be the size of Qn . Since Qn is an n-regular graph of order 2n , it follows that n · 2n = 2m
and so m = n · 2n−1 .
18. (a) The graphs R2 and R3 are disconnected graphs with exactly two components. The
components of R2 are both K2 ; while the components of R3 are both K4 .
!"
(b) The degree of each vertex of R4 is 42 = 6 and so the size of R4 is 48.
19. Proof. Let Gi be an ri -regular graph of order ni (i = 1, 2, 3). Since G1 is Eulerian, r1 is even.
Since G1 is Eulerian, n1 − r1 − 1 is even. Thus n1 is odd. Since G2 is not Eulerian, r2 is odd
and so n2 is even. Similarly, r3 is odd and n3 is even. Observe that
(1) every vertex of G1 in G has degree r1 + n2 + n3 , which is even,
(2) every vertex of G2 in G has degree r2 + n1 + n3 , which is even,
(3) every vertex of G3 in G has degree r3 + n1 + n2 , which is even.
Since G is connected and every vertex has even degree, G is Eulerian.
208
20. (b) is true. Since n is odd, r is even and n − 1 − r is even. In H, every vertex of G has degree
r + 2 and every vertex of G has degree (n − 1 − r) + 2, both of which are even. Both u and v
have degree 2n + 1. Thus H is a connected graph having exactly two vertices of odd degree.
21. The statement is false.
Claim: If G and H are not Eulerian, then G + H is not Eulerian.
Proof. Observe that if v ∈ V (G), then degG+H v = degG v + |V (H)|; while if v ∈ V (H),
then degG+H v = degH v + |V (G)|. Assume, to the contrary, that G + H is Eulerian. Thus
degG+H v is even for each vertex v of G+ H. Since G is not Eulerian, it follows that G contains
an odd vertex and so the order of H is odd. However, this implies that every vertex of G is
odd and so the order of G is even. Since H is not Eulerian, H contains an odd vertex and so
the order of G is even, which is impossible.
22. This situation can be modeled by a multigraph in which only the vertices A and F have odd
degree (see Figure 77). Thus such a stroll is possible, beginning at A or F and ending at the
other.
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Figure 77: The multigraph in Exercise 22
23. Floyd is lying as the multigraph modeling this situation has 4 odd vertices.
Section 12.4. Hamiltonian Graphs
1. The graph in Figure 78 models the art exhibit consisting of seven rooms labeled A, B, C, D,
R, S, T. A visitor enters the art exhibit by going from the outside into room A.
D
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Figure 78: Strolling through an art exhibit
(a) It is not possible to walk through the exhibit, entering each room exactly once, returning
to room A and then exiting since the graph in Figure 78 is not Hamiltonian.
(b) Once a visitor has entered room A, it is not possible to stroll through the art exhibit,
passing through each doorway exactly once, returning to room A and then exiting since
the graph in Figure 78 is not Eulerian.
2. The graphs G1 and G4 are Hamiltonian, while G2 and G3 are not. The graph G3 contains a
cut-vertex; while for S = {t, u, v} in G2 , k(G2 − S) = 4 > |S| = 3.