k-Ordered Hamiltonian
Graphs
Lenhard Ng*
DEPARTMENT OF MATHEMATICS
HARVARD UNIVERSITY
CAMBRIDGE, MASSACHUSETTS
Michelle Schultz†
DEPARTMENT OF MATHEMATICS AND STATISTICS
WESTERN MICHIGAN UNIVERSITY
KALAMAZOO, MICHIGAN
ABSTRACT
A hamiltonian graph G of order n is k-ordered, 2 ≤ k ≤ n, if for every sequence
v1 , v2 , . . . , vk of k distinct vertices of G, there exists a hamiltonian cycle that encounters
v1 , v2 , . . . , vk in this order. Theorems by Dirac and Ore, presenting sufficient conditions
for a graph to be hamiltonian, are generalized to k-ordered hamiltonian graphs. The
existence of k-ordered graphs with small maximum degree is investigated; in particular, a family of 4-regular 4-ordered graphs is described. A graph G of order n ≥ 3 is
k-hamiltonian-connected, 2 ≤ k ≤ n, if for every sequence v1 , v2 , . . . , vk of k distinct vertices, G contains a v1 -vk hamiltonian path that encounters v1 , v2 , . . . , vk in this order. It
is shown that for k ≥ 3, every (k + 1)-hamiltonian-connected graph is k-ordered and a result of Ore on hamiltonian-connected graphs is generalized to k-hamiltonian-connected
c 1997 John Wiley & Sons, Inc.
graphs. * Present Address: Department of Mathematics, Massachusetts Institute of Technology, Cambridge, MA 02139-4307.
† Present Address: Department of Mathematical Sciences, University of Nevada, Las Vegas, Las
Vegas, NV 89154-4020.
Journal of Graph Theory Vol. 24, No. 1, 45 57 (1997)
c 1997 John Wiley & Sons, Inc.
CCC 0364-9024/97/010045-13
46 JOURNAL OF GRAPH THEORY
1. INTRODUCTION
One of the best known and most studied areas in graph theory is hamiltonian graph theory. A
hamiltonian cycle of a graph G is a cycle containing every vertex of G, and a hamiltonian graph
is a graph possessing a hamiltonian cycle. It is obvious that for every two vertices u and v of a
hamiltonian graph G and every hamiltonian cycle C, we can traverse C so that u is encountered
prior to v. Indeed, for each sequence S of three distinct vertices of G, there exists a hamiltonian
cycle C of G (indeed, for any hamiltonian cycle C), we can traverse C in one of its two directions
so that the vertices of S are encountered in the order specified. These observations lead us to the
main concept of this paper.
Let G be a hamiltonian graph of order n. For a positive integer k with k ≤ n, we say that G is
k-ordered if for every sequence S: v1 , v2 , . . . , vk of k distinct vertices, there exists a hamiltonian
cycle C of G such that the vertices of S are encountered on C in the specified order. As we have
observed, every hamiltonian graph is both 2-ordered and 3-ordered. Furthermore, a hamiltonian
graph G of order n is n-ordered if and only if G = Kn . Also, if G is k-ordered, then G is
l-ordered for every positive integer l < k.
Our first result gives a necessary condition for a graph to be k-ordered.
Proposition 1. Let G be a hamiltonian graph of order n ≥ 3. If G is k-ordered, 3 ≤ k ≤ n,
then G is (k − 1)-connected.
Proof. Suppose, to the contrary, that G is not (k − 1)-connected. Then there exists a cutset
S = {v1 , v2 , . . . , vl } of G, where l ≤ k − 2. Let vl+1 and vl+2 be vertices belonging to distinct
components of G − S. Consequently, every vl+1 − vl+2 path in G contains vertices of S, so G
contains no hamiltonian cycle containing the vertices of the sequence v1 , v2 , . . . , vl , vl+1 , vl+2 in
this order. Hence G is not (l + 2)-ordered and so is not k-ordered.
An immediate corollary now follows.
Corollary 2. If G is a k-ordered hamiltonian graph, then δ(G) ≥ k − 1.
2. SUFFICIENT CONDITIONS FOR k -ORDERED GRAPHS
Many sufficient conditions have been given for hamiltonian graphs. Two of the best known are
due to Dirac [1] and Ore [2].
Theorem A (Dirac). Let G be a graph of order n ≥ 3. If deg v ≥ n/2 for every vertex v of G,
then G is hamiltonian.
Theorem B (Ore). Let G be a graph of order n ≥ 3. If for every pair u, v of nonadjacent
vertices of G, deg u + deg v ≥ n, then G is hamiltonian.
In this section we present sufficient conditions for a hamiltonian graph to be k-ordered. Before
presenting a lemma that will aid us, we first establish some notation. We write v1 , (x1 ), v2 ,
(x2 ), . . . , vk−1 , (xk−1 ), vk to indicate a v1 −vk path that contains the k vertices v1 , v2 , . . . , vk and
possibly up to k − 1 additional vertices, namely, x1 , x2 , . . . , xk−1 . Thus for 1 ≤ j ≤ k − 1, (xi )
indicates that some vertex, which we denote by xi , may or may not be present on the path. So for
each i(1 ≤ i ≤ k − 1), the vertices vi and vi+1 are either adjacent or vi , xi , vi+1 is a path. For
example, u, v, (x), y indicates the path u, v, y or the path u, v, x, y.
k-ORDERED HAMILTONIAN GRAPHS 47
Lemma 3. Let G be a graph of order n ≥ 3 and let S: v1 , v2 , . . . , vk be a sequence of k distinct
vertices of G, where 3 ≤ k ≤ n. If
deg u + deg v ≥ n + 2k − 7
for every pair u, v of nonadjacent vertices of G, then G contains a v1 − vk path of the type
v1 , (x1 ), v2 , (x2 ), . . . , (xk−1 ), vk or a vk − vk−1 path of the type vk , v1 , (x1 ), v2 , (x2 ), . . . ,
(xk−2 ), vk−1 .
Proof. We first consider the vertices v1 and v2 . Of course, if v1 and v2 are adjacent, then G
contains the path v1 , v2 ; otherwise, deg v1 + deg v2 ≥ n + 2k − 7 ≥ n − 1, which implies the
existence of a vertex x1 mutually adjacent to v1 and v2 , and so v1 , x1 , v2 is a path in G. In any
case, G contains a path v1 , (x1 ), v2 .
Now, we consider the vertex v3 . If v2 v3 ∈ E(G), then G contains the path v1 , (x1 ), v2 , v3 .
Suppose then that v2 v3 6∈ E(G). Thus deg v2 + deg v3 ≥ n + 2k − 7 by hypothesis. Since G has
order n, the vertices v2 and v3 are mutually adjacent to at least 2k − 5 vertices. If k ≥ 4, then
G contains a vertex x2 distinct from v1 and x1 (if it exists) such that x2 is mutually adjacent to
v2 and v3 . Hence G contains the path v1 , (x1 ), v2 , (x2 ), v3 . Suppose then that k = 3 and that G
contains no vertex distinct from v1 and x1 that is mutually adjacent to v2 and v3 . If v3 is adjacent
to v1 , then v3 , v1 , (x1 ), v2 is a path of G. If v3 is not adjacent to v1 , then v2 and v3 are mutually
adjacent to x1 and to no other vertex, while v2 is adjacent to v1 . However, then, v1 , v2 , x1 , v3 is
a path in G.
Proceeding inductively, we assume then that we have constructed a path v1 , (x1 ), v2 , (x2 ), . . . ,
(xj−2 ), vj−1 in G. Suppose first that j ≤ k − 1. We show that G contains a path v1 , (x1 ), v2 ,
(x2 ), . . . , vj−1 , (xj−1 ), vj . Of course, if vj−1 is adjacent to vj , then G contains such a path.
Suppose then that vj−1 vj 6∈ E(G). If G contains a vertex xj−1 distinct from the vertices on
the path v1 , (x1 ), v2 , (x2 ), . . . , (xj−2 ) such that xj−1 is mutually adjacent to vj−1 and vj , then
G contains a path of the desired type; otherwise vj−1 and vj are mutually adjacent to at most
2j − 4 vertices and G contains at least n + 2k − 2j − 1 vertices. Since k ≥ j + 1, this produces
a contradiction, and the desired claim is verified.
Finally, suppose that j = k, that is, assume that we have constructed a path v1 , (x1 ),
v2 , (x2 ), . . . , (xk−2 ), vk−1 in G. We show that G contains either a path v1 , (x1 ), v2 , (x2 ), . . . ,
vk−1 , (xk−1 ), vk or a path vk , v1 , (x1 ), v2 , (x2 ), . . . , (xk−2 ), vk−1 . If either vk vk−1 ∈ E(G) or
vk v1 ∈ E(G), then we have the desired result; thus we assume that vk is adjacent to neither vk−1
nor v1 . Also, if vk−1 and vk are mutually adjacent to some vertex other than the (at most) 2k − 5
vertices on the path (x1 ), v2 , (x2 ), . . . , vk−2 , (xk−2 ), the proof is complete; so we may assume
that this is not the case. Hence vk−1 and vk are mutually adjacent to at most 2k −5 vertices. Since
G has order n, it follows that vk−1 and vk are mutually adjacent to exactly 2k − 5 vertices (so all
of the vertices x1 , x2 , . . . , xk−2 exist) and every other vertex of G is adjacent to exactly one of
vk−1 and vk . In particular, vk−1 and vk are adjacent to both vk−2 and xk−2 , and G contains the
path v1 , x1 , v2 , x2 , . . . , vk−2 , vk−1 , xk−2 , vk , producing the desired result.
If G is a graph of order n ≥ 3 and k is an integer with 3 ≤ k ≤ n such that deg u + deg v
≥ n + 2k − 6 for every pair u, v of nonadjacent vertices of G, then it follows readily that G
is (2k − 4)-connected and, further, that every pair of nonadjacent vertices share at least 2k − 4
neighbors. In fact, we show that this same condition implies that G is k-ordered hamiltonian.
Theorem 4. Let G be a graph of order n ≥ 3 and let k be an integer with 3 ≤ k ≤ n. If
deg u + deg v ≥ n + 2k − 6
48 JOURNAL OF GRAPH THEORY
for every pair u, v of nonadjacent vertices of G, then G is a k-ordered hamiltonian graph.
Proof. First, notice that when k = 3, the hypothesis implies that G is hamiltonian, by Ore's
Theorem. So G is 3-ordered, and thus, we may assume that k ≥ 4. Now let v1 , v2 , . . . , vk be
a sequence of k distinct vertices of G. We show that there exists a hamiltonian cycle of G that
encounters these k vertices in the given order.
By Lemma 3, there exists either a v1 − vk path of the type v1 , (x1 ), v2 , (x2 ), . . . , (xk−1 ), vk or
a vk − vk−1 path of the type vk , v1 , (x1 ), v2 , (x2 ), . . . , (xk−2 ), vk−1 . We refer to these as paths of
type I or II. In either case, G contains a v1 − vk−1 path of the type v1 , (x1 ), v2 , (x2 ), . . . , (xk−2 ),
vk−1 . We claim that G has a hamiltonian path containing a subpath of type I or II. Suppose,
to the contrary, that this is not the case. Then there is a path P : u1 , u2 , . . . , ul of maximum
order containing a subpath of type I or II but l < n, with ua = v1 and ub = vk−1 , where
k − 2 ≤ b − a ≤ 2k − 4.
By assumption, there exists a vertex v of G that is not on P and v 6= vi for i = 1, 2, . . . , k.
The defining property of P guarantees that every vertex adjacent to ul belongs to P . Observe
that neither u1 nor ul is adjacent to v, for otherwise v can be added immediately before u1 or
after ul to produce a path having more vertices than P and containing a subpath of type I or II,
contrary to hypothesis.
Suppose that u is a vertex that is adjacent to ul . Then u = ui , where 1 ≤ i ≤ l − 1.
If ui is adjacent to ul for 1 ≤ i ≤ a − 1, then ui+1 is not adjacent to v; for otherwise,
v, ui+1 , ui+2 , . . . , ul , ui , ui−1 , . . . , u1 is a path having more vertices than P and containing a
subpath of type I or II. If ui is adjacent to ul for b ≤ i ≤ l − 2, then ui+1 is not adjacent to
v; for otherwise, u1 , u2 , . . . , ui , ul , ul−1 , . . . , ui+1 , v is a path having more vertices than P and
containing a subpath of type I or II.
The vertex ul is adjacent to ul−1 and possibly the vertices ua , ua+1 , . . . , ub−1 . So this is a
total of 2k − 3 vertices that ul is or may be adjacent to. However, for each additional vertex to
which ul is adjacent, a vertex is ruled out to which v is adjacent. The vertex v is not adjacent
to u1 , ul , or itself, and since deg ul ≥ 2k − 3, the maximum number of vertices to which v is
adjacent is reduced by deg ul − (2k − 3). Therefore
deg v ≤ n − 3 − (deg ul − (2k − 3)),
which implies that deg ul + deg v ≤ n + 2k − 6. However, since ul and v are not adjacent,
deg ul + deg v ≥ n + 2k − 6. Thus deg ul + deg v = n + 2k − 6, which further implies that
(1) b − a = 2k − 4, (2) the path ua , ua+1 , . . . , ub is v1 , x1 , v2 , x2 , . . . , vk−2 , xk−2 , vk−1 , (3) the
vertices ua , ua+1 , . . . , ub−1 are adjacent to ul , and (4) the vertices ua+1 , ua+2 , . . . , ub are adjacent
to v. In particular, since k ≥ 4, it follows that v2 and v3 are adjacent to ul and x2 is adjacent to
v. Now the path u1 , u2 , . . . , ua−1 , v1 , x1 , v2 , v, v3 , x3 , . . . , xk−2 , vk−1 , ub+1 , ub+2 , . . . , ul , x2
is a path having more vertices than P and containing a subpath of type I or II, contradicting
the defining property of P . Hence, as claimed, G has a hamiltonian path Q: w1 , w2 , . . . , wn
containing a subpath of type I or II. Whether Q contains a subpath of type I or II, the path Q
contains the subpath v1 , (x1 ), v2 , (x2 ), . . . , (xk−2 ), vk−1 , which does not contain vk .
Now, suppose, to the contrary, that G does not have a hamiltonian cycle containing the vertices
v1 , v2 , . . . , vk in this order. Certainly, then, w1 wn 6∈ E(G). We now consider two cases according
to whether vk appears before v1 or after vk−1 in Q.
Case 1. The vertex vk appears before v1 in Q. Let v1 = wa and vk−2 = wb in Q, so
k − 3 ≤ b − a ≤ 2k − 6. Since deg w1 + deg wn ≥ n + 2k − 6 and deg w1 ≤ n − 1, it follows
that deg wn ≥ 2k − 5.
k-ORDERED HAMILTONIAN GRAPHS 49
If wn is adjacent to wi for 2 ≤ i ≤ a − 1, then w1 is not adjacent to wi+1 ; for otherwise G has the hamiltonian cycle w1 , wi+1 , wi+2 , . . . , wn , wi , wi−1 , . . . , w1 that contains the
vertices v1 , v2 , . . . , vk in this order, contrary to assumption. Also, if wn is adjacent to wi for
b ≤ i ≤ n − 2, then w1 is not adjacent to wi+1 ; for otherwise G has the hamiltonian cycle
w1 , w2 , . . . , wi , wn , wn−1 , . . . , wi+1 , w1 that contains the vertices v1 , v2 , . . . , vk in this order,
which again is impossible.
The vertex w1 is not adjacent to wn (and, of course, is not adjacent to itself). The vertex wn
may be adjacent to the b − a + 1 vertices wa , wa+1 , . . . , wb−1 , but for every vertex other than
these to which wn is adjacent, the number of vertices to which w1 may be adjacent is reduced by
1; that is,
deg w1 ≤ n − 2 − (deg wn − (b − a + 1))
or
deg w1 + deg wn ≤ n − 1 + b − a ≤ n + 2k − 7,
which produces a contradiction.
Case 2. The vertex vk appears after vk−1 in Q. In this case we let wa = v2 and wb = vk−1 .
The proof is now similar to that given in Case 1 and is consequently omitted.
Theorem 4 can be considered as an analogue of Ore's Theorem B to k-ordered hamiltonian
graphs. As a consequence of Theorem 4, we have an analogue of Dirac's Theorem A.
Corollary 5. Let G be a graph of order n ≥ 3 and let k be an integer such that 3 ≤ k ≤ n. If
deg v ≥
n
+k−3
2
for every vertex v of G, then G is a k-ordered hamiltonian graph.
In the case where k = 4, the first nontrivial case, we have the following.
Corollary 6. Let G be a graph of order n ≥ 4. If deg v ≥ (n + 2)/2 for every vertex v of G,
then G is a 4-ordered hamiltonian graph.
Although Theorem 4 is not sharp in general, we do have a lower bound for the desired degree
condition.
Theorem 7. For each integer k ≥ 4, there exists a graph Hk of order n ≥ 2k − 1 having deg u
+ deg v ≥ n + 3bk2 c − 5 for every pair u, v of nonadjacent vertices and such that Hk is not
k-ordered.
Proof. First, suppose that k is even. We construct Hk as follows. Let V (Hk ) = X ∪ Y ∪ Z,
where |X| = k, |Y | = n − 2k + 1, and |Z| = k − 1. Let X = {v1 , v2 , . . . , vk } and let C
denote the k-cycle v1 , v2 , . . . , vk , v1 . Further let hXi ∼
= Kk − E(C), hY i ∼
= Kn−2k+1 , and
∼
hZi = Kk−1 . Finally, let Hk be the graph obtained from hXi, hY i, and hZi by joining each
vertex of Y to each vertex vi of X with i even and by joining each vertex of Z to each vertex
of X and to each vertex of Y . For n = 14, the graph H6 is shown in Figure 1, where the lines
between the sets X and Z and between Y and Z represent that all possible edges exist between
vertices of the two sets.
The graph Hk is hamiltonian. A pair u, v of nonadjacent vertices in Hk consists either of
two vertices of X that are consecutive on the cycle C or of a vertex of Y and a vertex vi of
X with i odd. In the first case, deg u + deg v = n + 2k − 7 and in the second, deg u +
deg v = n + (3k/2) − 5 so that sincek ≥ 4 implies n + 2k − 7 ≥ n + (3k/2) − 5, it follows
that deg u + deg v ≥ n + (3k/2) − 5 for every pair u, v of nonadjacent vertices in Hk . Finally,
50 JOURNAL OF GRAPH THEORY
FIGURE 1.
we show that Hk is not k-ordered. Suppose, to the contrary, that Hk is k-ordered. Then there
is a hamiltonian cycle C 0 in Hk that encounters the vertices of X in the order v1 , v2 , . . . , vk .
Further, C 0 can be decomposed into k paths P1 , P2 , . . . , Pk , where Pi is a vi -vi+1 path for
i = 1, 2, . . . , k − 1, and Pk is a vk -v1 path. For each odd integer j(1 ≤ j ≤ k), the vertex vj is
only adjacent to other vertices of X and to vertices of Z. Hence each path Pi (1 ≤ i ≤ k) must
contain at least one vertex of Z and since Z contains only k−1 vertices, we have our contradiction.
− 5 = n + 3bk2 c − 5
Now if k ≥ 5 is odd, then Hk−1 is a graph with deg u + deg v ≥ n + 3(k−1)
2
for every pair of nonadjacent vertices and since Hk−1 is not (k − 1)-ordered, it is not k-ordered.
This collection of graphs shows that the best possible degree condition of Theorem 4 lies
between n + 3bk2 c − 4 and n + 2k − 6. Actually, for n sufficiently large, we have an improvement
on this upper bound; however, we omit the proof since it is similar to that of Theorem 4.
Theorem 8. Let G be a graph of order n ≥ 2k, where k ≥ 5 is an integer. If
deg u + deg v ≥ n + 2k − 7
for every pair u, v of nonadjacent vertices of G, then G is k-ordered.
3. EXISTENCE OF 4-ORDERED GRAPHS OF SMALL SIZE
By Corollary 6 if G is a graph of order n ≥ 4 each of whose vertices has degree at least (n + 2)/2,
then G is 4-ordered. By Corollary 2, however, it is conceivable that there are 3-regular graphs that
are 4-ordered. Indeed, this is the case since K4 and K3,3 are 3-regular, 4-ordered graphs. Whether
there is an infinite class of 3-regular, 4-ordered graphs is not known. With the exception of K4 ,
every 3-regular, 4-ordered graph is triangle-free, for suppose that G is a 3-regular, hamiltonian
graph of order at least 6 that contains a triangle. Then G must contain a triangle with vertices
u, v, and w, such that ux, vy ∈ E(G) and x 6= y. However, then, there is no hamiltonian
cycle of G that encounters the vertices v, x, u, y in this order. On the other hand, the 3-cube is a
hamiltonian bipartite 3-regular graph that is not 4-ordered either. We next show that an infinite
class of 4-regular, 4-ordered graphs exists.
Theorem 9. For every integer n ≥ 5, there exists a 4-regular, 4-ordered graph of order n.
Proof. Certainly K5 has the desired property and by Corollary 6, the graph of the octahedron
(the unique 4-regular graph of order 6) is 4-ordered. So let n ≥ 7. Let Cn denote the n-cycle
k-ORDERED HAMILTONIAN GRAPHS 51
FIGURE 2.
u1 , u2 , . . . , un , u1 . We now define Gn to consist of Cn together all edges ui uj such that the
distance dCn (ui , uj ) = 3. The graphs G7 and G8 are shown in Figure 2. All graphs Gn (n ≥ 7)
are 4-regular and hamiltonian. It remains only to show that each graph Gn is 4-ordered. To
simplify the argument, we denote each vertex ui by i(1 ≤ i ≤ n).
Let S = {v1 , v2 , v3 , v4 } be a set of four vertices of G, where we assume, without loss of
generality, that 1 = v1 < v2 < v3 < v4 ≤ n. For each ordering of the vertices of S, we show
that there exists a hamiltonian cycle encountering them in the prescribed ordering. There are six
(cyclic) orderings of the vertices of S. From the construction of Gn , it is obvious that Gn contains
a hamiltonian cycle that encounters the vertices of S in the order v1 , v2 , v3 , v4 and by traversing
this cycle in reverse direction, there is a hamiltonian cycle encountering the vertices of S in the
order v1 , v4 , v3 , v2 .
Next we show that Gn contains a hamiltonian cycle that encounters the vertices of S in the
order v1 , v2 , v4 , v3 . We consider four cases.
Case 1. The number v4 − v3 ≡ 0 (mod 3). In this case
1 = v1 , 2, 3, . . . , v2 , . . . , v3 − 1, v3 + 2, v3 + 5, . . . , v4 − 1, v4 ,
v4 − 3, v4 − 6, . . . , v3 , v3 + 1, v3 + 4, . . . , v4 + 1, v4 + 2, . . . , n, 1
is a hamiltonian cycle with the desired property.
Case 2. The number v4 − v3 ≡ 2 (mod 3). The hamiltonian cycle
1 = v1 , 2, 3, . . . , v2 , . . . , v3 − 1, v3 + 2, v3 + 5, . . . , v4 , v4 − 1,
v4 − 4, . . . , v3 + 1, v3 , v3 + 3, v3 + 6, . . . , v4 + 1, v4 + 2, . . . , n, 1
encounters the vertices of S in the required order.
Case 3. The number v4 − v3 ≡ 1 (mod 3) and v4 6= n. Here we see that
1 = v1 , 2, 3, . . . , v2 , . . . , v3 − 1, v3 + 2, v3 + 5, . . . , v4 − 2, v4 + 1, v4 ,
v4 − 3, . . . , v3 + 1, v3 , v3 + 3, v3 + 6, . . . , v4 − 1, v4 + 2, v4 + 3, . . . , n, 1
is a hamiltonian cycle of Gn encountering the vertices of S in the order v1 , v2 , v4 , v3 . Observe
that if v4 = n − 1, then this cycle concludes with v4 − 1, 1.
Since Gn has a hamiltonian cycle C encountering the vertices of S in the order v1 , v2 , v4 , v3
if and only if Gn has a cycle C 0 containing the vertices of S in the order v1 , v3 , v4 , v2 , it follows
from Cases 1, 2, and 3 that the proof will be complete if (1) v2 − v1 ≡ 0 (mod 3), (2) v2 − v1 ≡ 2
(mod 3), or (3) v2 − v1 ≡ 1 (mod 3) and v3 6= v2 + 1. Hence, only one case remains.
Case 4. The numbers v1 , v2 , v3 , and v4 satisfy the conditions v4 −v3 ≡ 1 (mod 3), v2 −v1 ≡ 1
(mod 3), v4 = n, and v3 = v2 + 1. Since n ≥ 7, either v4 − v3 ≥ 4 or v2 − v1 ≥ 4. We assume
52 JOURNAL OF GRAPH THEORY
the former since the argument in the latter case is similar (and symmetric). The cycle
1 = v1 , 4, 7, . . . , v2 − 1, v2 , v2 − 3, v2 − 6, . . . , 2, n − 1,
n = v4 , 3, 6, . . . , v3 , v3 + 1, v3 + 2, . . . , n − 2, 1
is hamiltonian and encounters the vertices of S in the desired order.
The existence of a hamiltonian cycle in Gn encountering the vertices of S in the order
v1 , v2 , v4 , v3 implies the existence of a hamiltonian cycle in Gn encountering the vertices of
S in the order v1 , v3 , v4 , v2 . The proof in four cases we have just given also shows that a hamiltonian cycle of Gn exists that encounters the vertices of S in the order v4 , v1 , v3 , v2 or, equivalently,
in the order v1 , v3 , v2 , v4 . Consequently, there is a hamiltonian cycle of Gn that encounters the
vertices of S in the order v1 , v4 , v2 , v3 as well.
4. k -ORDERED HAMILTONIAN-CONNECTED GRAPHS
One of the best known families of hamiltonian graphs is the class of hamiltonian-connected
graphs. A graph G is hamiltonian-connected if for every pair u, v of distinct vertices, G contains
a u-v hamiltonian path. Every hamiltonian-connected graph of order at least 3 is hamiltonian.
Ore [3] showed that if G is a graph of order n and deg u + deg v ≥ n + 1 for every pair u, v of
nonadjacent vertices of G, then G is hamiltonian-connected.
We define a graph G of order n ≥ 3 to be k-ordered hamiltonian-connected, or more simply
k-hamiltonian-connected, 2 ≤ k ≤ n, if for every sequence v1 , v2 , . . . , vk of k distinct vertices,
G contains a v1 -vk hamiltonian path that encounters the vertices v1 , v2 , . . . , vk in this order. It is
clear that every hamiltonian-connected graph of order at least 3 is 3-hamiltonian-connected and
also that a k-hamiltonian-connected graph is l-hamiltonian-connected for 2 ≤ l ≤ k − 1. We
first describe a relationship between such classes of graphs and k-ordered hamiltonian graphs.
Theorem 10. Every (k + 1)-hamiltonian-connected graph, where k ≥ 3, is a k-ordered hamiltonian graph.
Proof. Let k ≥ 3 be an integer and let G be a (k + 1)-hamiltonian-connected graph. Furthermore, let v1 , v2 , . . . , vk be a sequence of k distinct vertices of G. Since G is connected and
has order at least k + 1, there exists a vertex u of G different from v1 , v2 , . . . , vk such that u is
adjacent to vi for some i with 1 ≤ i ≤ k. Since G is (k + 1)-hamiltonian-connected, G contains
a vi -u hamiltonian path P that encounters the vertices vi , vi+1 , . . . , vk , v1 , v2 , . . . , vi−1 , u in this
order, where the sequence is v1 , v2 , . . . , vk , u when i = 1. Hence P together with the edge vi u
is a hamiltonian cycle of G that encounters the vertices v1 , v2 , . . . , vk in this order. So G is a
k-ordered hamiltonian graph.
Recall that every hamiltonian-connected graph of order at least 4 is 3-connected. Hence 4hamiltonian-connected graphs are 3-connected. The next result follows from Proposition 1 and
Theorem 10.
Corollary 11. Every k-hamiltonian-connected graph, k ≥ 5, is (k − 2)-connected.
The converse of Theorem 10 is not true in general. Indeed, there is no positive integer k such
that every k-ordered hamiltonian graph is even hamiltonian-connected. Let k ≥ 3 be given.
Then the graph Kk−1,k−1 is k-ordered is k-ordered (although not (k + 1)-ordered) but is not
hamiltonian-connected. To see that Kk−1,k−1 is k-ordered, let S be a sequence of k distinct
vertices of Kk−1,k−1 whose partite sets are denoted by V1 and V2 . For example, for k = 8 with
the vertices labeled as V1 = {ui |1 ≤ i ≤ 7} and V2 = {wi |1 ≤ i ≤ 7}, we may choose the
k-ORDERED HAMILTONIAN GRAPHS 53
sequence S: u6 , w2 , w3 , u3 , u7 , w5 , w1 , w4 . Then this sequence can be represented as a sequence
whose terms are either 1 or 2 such that there is at least one 1 and at least one 2, where a vertex
of S in V1 is simply replaced by a 1 and a vertex of S in V2 is replaced by a 2. For our example,
we obtain 1, 2, 2, 1, 1, 2, 2, 2. Then for each 1 in the sequence not followed by 2, we insert
a 20 immediately after this term, and similarly, for each 2 or 20 not followed by 1, we insert 10 .
We now have a sequence S 0 that alternates between 1 (or 10 ) and 2 (or 20 ), and this sequence
contains at most 2k − 2 terms. Again for our example, the sequence S 0 is 1, 2, 10 , 2, 1, 20 , 1, 2,
10 , 2, 10 , 2. We now construct an appropriate hamiltonian cycle by beginning at the first vertex
of S and proceed along S 0 , where we take the next term of S whenever 1 or 2 is encountered, we
take a vertex of V1 not in S when 10 is encountered, and we take a vertex of V2 not in S when
20 is encountered. A hamiltonian cycle can now be completed. To complete our example, the
hamiltonian cycle constructed is u6 , w2 , u1 , w3 , u3 , w6 , u7 , w5 , u2 , w1 , u4 , w4 , u5 , w7 , u6 .
Theorem 10 is best possible in the sense that for k ≥ 4, a k-hamiltonian-connected graph need
not be k-ordered hamiltonian.
Theorem 12. For each integer k ≥ 4, there exists a k-hamiltonian-connected graph that is not
k-ordered.
Proof. Consider G = C̄k + Kk−1 , which has order 2k − 1. We show first that G is khamiltonian-connected. For k = 4, it is routine to show that G is 4-hamiltonian-connected; so
we may assume that k ≥ 5. Let V1 denote the vertices of C̄k and V2 the vertices of Kk−1 . Let S:
v1 , v2 , . . . , vk be a sequence of k distinct vertices of G. We consider three cases.
Case 1. The vertices v1 , vk ∈ V1 . In this case, we follow the procedure presented earlier that
shows Kk−1,k−1 is k-ordered. Replace the sequence S by a sequence whose terms are either 1
or 2, where vi is replaced by j (j = 1 or j = 2) if vi ∈ Vj . In this case, the new sequence starts
and ends with 1. For each 1 in the sequence that is not followed by 2, with the exception of the
last, we insert a 20 . Similarly, for each 2 or 20 not followed by 1, we insert a 10 . We now have a
sequence that alternates between 1 (or 10 ) and 2 (or 20 ) necessarily of odd length. If the length of
this sequence is less than 2k − 1, then insert alternating 10 and 20 prior to the last 1 until the length
of the sequence is 2k − 1. The new sequence is S 0 , which describes a v1 -vk hamiltonian path P
in G, namely, start with v1 and then continue with the next vertex of S if the corresponding term
of S 0 is 2, or continue with a vertex of V2 not in S if the next term of S 0 is 20 . In general, if we
have arrived at a particular vertex of P , which corresponds to a term of S 0 , then the next vertex
of P is the next vertex of S if the corresponding term of S 0 is 1 or 2; while the next vertex of P
is a vertex of V1 (or V2 ) not on S and not previously on P if the corresponding term of S 0 is 10
(or 20 ).
Case 2. The vertices v1 , vk ∈ V2 . In this case, we proceed as in Case 1 to construct a sequence
that alternates between 1 (or 10 ) and 2 (or 20 ) that begins and ends with 2. Further the sequence
has odd length at most 2k − 3. If the length is less than 2k − 3, insert alternating 20 and 10 prior to
the last 2. This sequence S 0 describes a v1 -vk path of order 2k − 3 in G. Two vertices of V1 are
not on this path and neither of these is a vertex of S. Since k ≥ 5, each such vertex is adjacent to
a vertex of V1 and thus can now be appropriately inserted into the path. Hence we have a v1 -vk
hamiltonian path in G that encounters the vertices of S in the appropriate order.
Case 3. The vertex v1 ∈ V1 and vk ∈ V2 or vk ∈ V1 and v1 ∈ V2 . The proof in this case is
similar to that of Case 2, the difference being that there will be one vertex (instead of two) that
needs to be inserted to obtain the desired v1 -vk hamiltonian path.
That G is not k-ordered follows since G is a special case of the graphs Hk that were defined
in the proof of Theorem 7.
54 JOURNAL OF GRAPH THEORY
Next, we present a result that generalizes the theorem stating that a graph G of order n ≥ 3 such
that deg u + deg v ≥ n + 1 for every pair u, v of nonadjacent vertices is hamiltonian-connected.
Since 3-hamiltonian-connected is equivalent to hamiltonian-connected, we consider k ≥ 4. First,
using a proof technique analogous to the proof of Lemma 3, we obtain the following.
Lemma 13. Let G be a graph of order n ≥ 3 and let v1 , v2 , . . . , vk be a sequence of k distinct
vertices of G, where 3 ≤ k ≤ n. If
deg u + deg v ≥ n + 2k − 6
for every pair u, v of nonadjacent vertices, then G contains a v1 -vk path of the type v1 , (x1 ), v2 ,
(x2 ), . . . , (xk−1 ), vk .
With this, we can verify our degree condition for k-hamiltonian connected graphs.
Theorem 14. Let G be a graph of order n ≥ 4 and let k with 4 ≤ k ≤ n be an integer. If
deg u+ deg v ≥ n + 2k − 6 for every pair u, v of nonadjacent vertices of G, then G is khamiltonian-connected.
Proof. Let v1 , v2 , . . . , vk be a sequence of k distinct vertices of G. By Lemma 13, there exists
a v1 -vk path of the type v1 , (x1 ), v2 , (x2 ), . . . , (xk−1 ), vk . We claim that G contains a v1 -vk hamiltonian path that has a subpath of the type v1 , (x1 ), v2 , (x2 ), . . . , (xk−3 ), vk−2 . Suppose, to the contrary, that this is not the case. Then there is a v1 -vk path P : v1 = u1 , u2 , . . . , ul = vk of maximum
length, where l < n − 1, that contains a subpath of the type v1 , (x1 ), v2 , (x2 ), . . . , (xk−3 ), vk−2 .
Let ub = vk−2 . Then b ≤ 2k − 5. We consider two cases.
Case 1. The vertex vk−2 = ub is adjacent to a vertex v that is not on P . If ub+1 is also adjacent
to v, then
u1 , u2 , . . . , ub , v, ub+1 , . . . , ul
is a path contradicting the defining property of P . Hence ub+1 is not adjacent to v and, by
hypothesis, deg ub+1 + deg v ≥ n + 2k − 6. If there exists a vertex w not belonging to P such
that w is mutually adjacent to v and ub+1 , then
u1 , u2 , . . . , ub , v, w, ub+1 , . . . , ul
is a path that contradicts the defining property of P . Hence no such vertex w exists and so v and
ub+1 are mutually adjacent only to vertices on P . If for some i with b + 2 ≤ i ≤ l − 1, the vertex
ui is adjacent to v, then ui+1 is not adjacent to ub+1 ; for otherwise, the path
u1 , u2 , . . . , ub , v, ui , ui−1 , . . . , ub+1 , ui+1 , . . . , ul
is a path that contradicts the defining property of P .
Thus v is adjacent to ub = vk−2 and possibly to the vertices u1 , u2 , . . . , ub−1 , ul . But for each
vertex x other than u1 , u2 , . . . , ub , ul to which v is adjacent, a vertex is ruled out to which ub+1
may be adjacent. Since deg v ≥ b + 1 and since ub+1 is not adjacent to v or itself, we have
deg ub+1 ≤ n − 2 − (deg v − (b + 1)),
so
deg ub+1 + deg v ≤ n + b − 1.
k-ORDERED HAMILTONIAN GRAPHS 55
Since b ≤ 2k − 5, we have
deg ub+1 + deg v ≤ n + 2k − 6
and hence deg ub+1 + deg v = n + 2k − 6. This implies that (1) b = 2k − 5, (2) the path
u1 , u2 , . . . , ub is v1 , x1 , v2 , x2 , . . . , xk−3 , vk−2 , (3) the vertices u1 , u2 , . . . , ub−1 , and ul are adjacent to v, and (4) the vertices u1 , u2 , . . . , ub−1 , and ul are adjacent to ub+1 .
In particular, v is adjacent to vk−3 , and ub+1 is adjacent to xk−3 = ub−1 . Thus
u1 , u2 , . . . , vk−3 , v, ub , xk−3 , ub+1 , ub+2 , . . . , ul
is a path that contradicts the defining property of P .
Case 2. The vertex vk−2 = ub is adjacent only to vertices on P . Let v be a vertex not on P .
First, we show that v is adjacent to none of the vertices ub+1 , ub+2 , . . . , ul . Assume, instead, that
v is adjacent to at least one of ub+1 , ub+2 , . . . , ul . Then since v is not adjacent to ub , there exists
a minimum integer j (b ≤ j ≤ l − 1) such that v is adjacent to uj+1 and not adjacent to uj . (In
fact, v is not adjacent to ub , ub+1 , . . . , uj .) Hence, by hypothesis, deg uj + deg v ≥ n + 2k − 6.
If v is adjacent to some vertex w not on P , then uj is not adjacent to w; for otherwise the path
u1 , u2 , . . . , uj , w, v, uj+1 , uj+2 , . . . , ul
contradicts the defining property of P . If v is adjacent to ui+1 , where j + 2 ≤ i ≤ l − 1, then uj
is not adjacent to ui ; for otherwise, the path
u1 , u2 , . . . , uj , ui , ui−1 , . . . , uj+1 , v, ui+1 , ui+2 , . . . , ul
contradicts the choice of P .
Hence v is adjacent to uj+1 and possibly to the vertices u1 , u2 , . . . , ub−1 ; but for each additional
vertex to which v is adjacent, a vertex is ruled out to which uj may be adjacent. Since uj is not
adjacent to v or itself, and since deg v ≥ b, we have
deg uj ≤ n − 2 − (deg v − b),
so
deg uj + deg v ≤ n + 2k − 7,
producing a contradiction. Thus v is adjacent to none of the vertices ub+1 , ub+2 , . . . , ul .
So deg v ≤ b − 1 + n − l − 1 = n − l + b − 2. Also by assumption deg ub ≤ l − 1, and so
deg v + deg ub ≤ n + b − 3 ≤ n + 2k − 8,
which is a contradiction since v and ub are not adjacent.
An immediate consequence of Theorem 14 is presented next.
Corollary 15. Let G be a graph of order n ≥ 4 and let k (4 ≤ k ≤ n) be an integer. If
deg v ≥ n/2 + k − 3 for every vertex v of G, then G is k-hamiltonian-connected.
Thus in the first nontrivial case, when k = 4, if the degree of each vertex in a graph G is
at least (n + 2)/2, then G is 4-hamiltonian-connected. It is not known whether the condition
deg v ≥ (n + 1)/2 for every vertex v of G implies that G is 4-hamiltonian-connected or, indeed,
whether deg v ≥ n/2 for every vertex v of G and G hamiltonian-connected implies G is 4hamiltonian-connected.
56 JOURNAL OF GRAPH THEORY
As was the case for the Ore-type condition for k-ordered hamiltonian graphs, Theorem 14 is
not sharp in general. However, we present a lower bound for the desired degree condition.
Theorem 16. For each integer k ≥ 4, there exists a graph Gk of order n ≥ 2k − 1 having
deg u + deg v ≥ n + k − 4 for every pair u, v of nonadjacent vertices and such that Gk is not
k-hamiltonian-connected.
Proof. Let k ≥ 4 be an integer and let n ≥ 2k − 1 be an integer having the same parity
as k. Define Gk = 2K(n−k+2)/2 + Kk−2 . Let X and Y represent the two sets of vertices
with hXi = hY i = K(n−k+2)/2 and let Z denote the remaining vertices of Gk . If u and v
are nonadjacent vertices of Gk , then one of u, v is in X and the other is in Y and so deg u +
deg v ≥ n + k − 4. To see that Gk is not k-hamiltonian-connected, consider the sequence
v1 , v2 , . . . , vk of distinct vertices, where vi ∈ X if i is odd and vi ∈ Y if i is even. Since
n ≥ 2k − 1, it follows that (n − k + 2)/2 ≥ dk/2e. Suppose, to the contrary, that Gk is
k-hamiltonian-connected. Then there exists a v1 -vk hamiltonian path P and P contains k − 1
internally disjoint paths P1 , P2 , . . . , Pk−1 , where Pi is a vi − vi+1 path for i = 1, 2, . . . , k − 1.
Each such path must contain a different vertex of Z, and this contradicts that |Z| = k − 2.
From Theorems 14 and 16, the best possible Ore-type degree condition for k-hamiltonianconnectedness lies between n + k − 3 and n + 2k − 6. Hence in the first nontrivial case when
k = 4, it is not known whether the best possible Ore-type condition is n + 1 or n + 2.
5. OPEN PROBLEMS
In Section 2, it is determined that the best possible degree condition for an analogue of Ore's
Theorem to k-ordered graphs of order n lies between n + 3bk/2c − 4 and n + 2k − 6.
Problem 1. Determine the best possible degree condition for Theorem 4.
In Section 3, an infinite class of 4-regular 4-ordered graphs is presented, but it is not known
whether there exists an infinite class of 3-regular 4-ordered graphs.
Problem 2. Determine whether there is an infinite class of 3-regular 4-ordered graphs.
Section 4 introduces k-hamiltonian-connected graphs.
Problem 3. Determine the best possible degree condition for Theorem 14.
Problem 4. Study the existence of small degree k-hamiltonian-connected graphs.
ACKNOWLEDGMENTS
The authors are grateful to Gary Chartrand for creating the concept of k-ordered hamiltonian
graphs and for suggesting this problem to us. We also thank Kiran Kedlaya for many fruitful
discussions. Furthermore, the research of the first author was conducted while he was a participant
k-ORDERED HAMILTONIAN GRAPHS 57
in the 1994 Research Experience for Undergraduates at the University of Minnesota-Duluth,
directed by Joseph Gallian and sponsored by the National Science Foundation (DMS-9225045)
and the National Security Agency (MDA 904-91-H-0036). The research of the second author
was supported in part by Office of Naval Research Grant N00014-91-J-1060.
References
[1]
G. A. Dirac, Some theorems on abstract graph, Proc. London Math. Soc. 2 (1952), 69–81.
[2]
O. Ore, Note on Hamiltonian circuits, Amer. Math. Monthly 67 (1960), 55.
[3]
O. Ore, Hamilton connected graphs, J. Math. Pures. Appl. 42 (1963), 21–27.
Received July 25, 1995