Noname manuscript No.
(will be inserted by the editor)
Cycle Lengths in Hamiltonian Graphs with a Pair of
Vertices Having Large Degree Sum
Michael Ferrara · Michael S. Jacobson ·
Angela Harris
the date of receipt and acceptance should be inserted later
Abstract A graph of order n is said to be pancyclic if it contains cycles of
all lengths from three to n. Let G be a hamiltonian graph and let x and y be
vertices of G that are consecutive on some hamiltonian cycle in G. Hakimi and
Schmeichel showed [2] that if d(x) + d(y) ≥ n then either G is pancyclic, G has
cycles of all lengths except n − 1 or G is isomorphic to a complete bipartite
graph. In this paper, we study the existence of cycles of various lengths in a
hamiltonian graph G given the existence of a pair of vertices that have a high
degree sum but are not adjacent on any hamiltonian cycle in G.
Keywords hamiltonian cycle · pancyclic
Mathematics Subject Classification (2000) 05C38 · 05C45
1 Introduction
In this paper we consider only graphs without loops or multiple edges. Let
|V (G)| denote the order of G and |E(G)| the size of G. For a vertex v ∈ V (G),
let N (v) denote the set of neighbors, or neighborhood, of v and let d(v) denote
the degree of v. If H is a subgraph of G, we will let dH (v) denote the degree of
v in H. Furthermore, let δ(G) and ∆(G) denote the minimum and maximum
degree of G, respectively. If x and y are vertices of G the distance between x
and y in G, denoted distG (x, y), is the length of the shortest x − y path in G.
M. Ferrara and M.S. Jacobson
University of Colorado Denver, Denver, CO 80217
A. Harris
University of Wisconsin, Whitewater, Whitewater, WI 53190
2
In this paper, we will consider cycles to have an implicit clockwise orientation. With this in mind, given a cycle C and a vertex x on C, we let x+ denote
the successor of x under this orientation and let x− denote the predecessor.
For any other vertex y on C, we let xC + y denote the path from x to y on C
in the clockwise direction and xC − y denote the path from x to y on C in the
counterclockwise direction. We will let distC (x, y) denote the distance from x
to y on C without consideration for the implicit orientation.
A spanning cycle in a graph G is called a hamiltonian cycle, and if such a
cycle exists, we say that G is hamiltonian. If a graph G of order n contains
cycles of all lengths ℓ where 3 ≤ ℓ ≤ n we say that G is pancyclic. Let σ2 (G)
denote the minimum degree sum over all pairs of nonadjacent vertices in G.
Ore’s Theorem [3], one of the classic results pertaining to hamiltonian graphs,
states the following.
Theorem 1 (Ore 1960) If G is a graph of order n ≥ 3 with σ2 (G) ≥ n then
G is hamiltonian.
Frequently, the condition σ2 (G) ≥ n is referred to as the Ore condition.
This theorem has lead to an extensive use of degree sum conditions to study
cycle properties in graphs. In [1], Bondy showed that the Ore condition was
sufficient to assure much more than a hamiltonian cycle in G
Theorem 2 If G is a graph of order n ≥ 3 with σ2 (G) ≥ n then either G is
pancyclic or G is isomorphic to K n2 , n2 . Consequently, if σ2 (G) ≥ n + 1, G is
pancyclic.
In the same paper, Bondy stated the following ”metaconjecture”:
Conjecture 1 (Bondy’s Metaconjecture) Almost any nontrivial condition
on a graph which implies that the graph is hamiltonian also implies that the
graph is pancyclic. There may be a simple family of exceptional graphs.
In light of Conjecture 1, it has been of interest to examine the relationship
between hamiltonicity and pancyclicity in graphs.
Ore’s Theorem and Theorem 2 both consider graphs where σ2 is high
enough to assure a considerable amount of global density. In this paper, we
wish to relax the Ore condition and examine the cycle structure of hamiltonian
graphs that may be globally sparse yet have a small number of vertices with
higher degree. Hakimi and Schmeichel proved the following theorems in [2],
which serve to motivate our investigation.
Theorem 3 Let G be a graph of order n with V (G) = {v0 , . . . , vn−1 } and
hamiltonian cycle v0 . . . vn−1 v0 . If d(v0 ) + d(vn−1 ) ≥ n then G is either pancyclic, bipartite or missing only an (n − 1)-cycle.
3
Theorem 4 Let G be a graph of order n with V (G) = {v0 , . . . , vn−1 } and
hamiltonian cycle v0 . . . vn−1 v0 . If d(v0 ) + d(vn−1 ) ≥ n + 1 then G is pancyclic.
These results require only a single pair of vertices having high degree sum,
however they must be consecutive on some hamiltonian cycle in G. In this
paper, our goal is to extend Theorems 3 and 4 by examining hamiltonian
graphs containing a pair of vertices that have a high degree sum but may lie
far apart on any hamiltonian cycle.
2 Absolute Bounds
We begin by determining the degree sum of two vertices on a hamiltonian cycle,
regardless of their placement, required to assure that the graph is pancyclic.
The necessary degree sum turns out to be quite large. We begin by stating
two useful lemmas.
Lemma 1 Let G be a hamiltonian graph of order n ≥ 5 with hamiltonian
cycle C. If there exists a vertex x in V (G) of degree n − 2, then G is pancyclic.
Proof Let C = xv1 . . . vn−1 x and let vi , for some 2 ≤ i ≤ n − 2, be the vertex
nonadjacent to x. If xvℓ is an edge of G, then the cycle C ′ = xvℓ C − x is a
cycle of length ℓ + 1 in G, so our assumption implies that G contains cycles
of all lengths except possibly a cycle of length (i + 1). However, since n ≥ 5,
there exist vertices vj and vj+i−1 in G that are both adjacent to x, and hence
xvj C + vj+i−1 x is an (i + 1)-cycle. Consequently, G is pancyclic.
Lemma 2 Let G be a hamiltonian graph of order n ≥ 8 with hamiltonian
cycle C. If there exists a vertex x in V (G) of degree n − 3, then either G is
pancyclic or G contains a cycle of each length except (n − 1).
Proof Let C = xv1 . . . vn−1 x and let vi , vj , 2 ≤ i < j ≤ n − 2, be the vertices
of G that are not adjacent to x. As above, we may conclude that G contains
cycles of all lengths except possibly cycles of length (i + 1) and (j + 1). If
i = 2 and j = n − 2 then G contains cycles of each length except possibly
n − 1. For all other i, j, since n ≥ 8, there exist distinct vertices xk and xℓ
such that xk , xk+i−1 , xℓ , xℓ+j−1 are all adjacent to x. Then xxk C + xk+i−1 x
is an (i + 1)-cycle in G and xxℓ C + xℓ+j−1 x is a (j + 1)-cycle in G, implying
that G is pancyclic.
Theorem 5 Let G be a hamiltonian graph of order n ≥ 9. If there exist
x, y ∈ V (G) such that d(x) + d(y) ≥ 2n − 7, then G is pancyclic. This result
is sharp.
4
Proof Assume that G is not pancyclic, and let C = v0 v1 . . . vn−1 v0 , where
x = v0 , be a hamiltonian cycle in G. Since d(x) + d(y) ≥ 2n − 7, we may
assume without loss of generality that d(x) is at least n − 3. Lemmas 1 and
2 therefore imply that x has degree exactly n − 3 and furthermore, since by
assumption G is not pancyclic, that G contains cycles of each length except
n − 1. As such, we can see that v2 and vn−2 must be the vertices of G that
are not adjacent to x. We now consider the vertex y, which has degree at least
n − 4.
It is useful to note that if y was either vn−1 or v1 , this would place x and
y consecutive on a hamiltonian cycle, implying that G would be pancyclic by
Theorem 4. Thus, we assume y is neither vn−1 nor v1 . We also note that y has
at most three nonadjacencies, and two of these must be y −− and y ++ since G
does not contain a cycle of length n − 1. Then at least one of yvn−1 and yv1
is an edge of G. If y ̸= v2 , v3 , v4 , vn−4 , vn−3 , vn−2 , this implies that one of the
following is an (n − 1)-cycle in G:
y ++ C + xy −− C − v1 yy + y ++ ,
or
y −− C − xy ++ C + vn−1 yy − y −− .
Suppose then that y = v2 , v3 or v4 . Then either yvn−1 or yvn−2 is an edge
of G. If yvn−1 is an edge of G, then yvn−1 C − y ++ xC + y is an (n − 1)-cycle in
G, and if yvn−2 is an edge of G, then yvn−2 C − y + xC + y is an (n − 1)-cycle in
G.
Similarly, if y = vn−4 , vn−3 or vn−2 then either yv1 or yv2 is an edge of
G. If yv1 is an edge of G, then yv1 C + y −− xC − y is an (n − 1)-cycle in G and
if yv2 is an edge of G, then yv2 C + y − xC − y is an (n − 1)-cycle in G. Having
exhausted all cases, we may conclude that G is pancyclic.
To see that this result is sharp, consider the graphs G1 and G2 constructed
as follows:
Let V (G1 ) = {v0 , . . . vn−1 } and construct the hamiltonian cycle C =
v0 v1 . . . nn−1 v0 in G1 . Selecting x = v0 and y = vn−3 we let xvi be an edge for
each i ̸= 0, 2, n − 5, n − 2 and yvj be an edge for each j ̸= 2, n − 5, n − 3, n − 1.
Then d(x) + d(y) = 2n − 8, but G1 contains no (n − 1)-cycle.
Let V (G2 ) = {v0 , . . . vn−1 } and construct the hamiltonian cycle C =
v0 v1 . . . nn−1 v0 in G1 . Selecting x = v0 and y = vn−4 we let xvi be an edge for
each i ̸= 0, 2, n − 6, n − 2 and yvj be an edge for each j ̸= 2, n − 6, n − 4, n − 2.
Then d(x) + d(y) = 2n − 8, but G2 contains no (n − 1)-cycle.
In both G1 and G2 , the distance between x and y on the hamiltonian
cycle C was at most four. If we require that this distance be at least five, the
necessary degree sum for pancyclicity can be reduced slightly.
5
Theorem 6 Let G be a graph of order n ≥ 9 with hamiltonian cycle C. If
there exist x, y ∈ V (G) such that distC (x, y) ≥ 5 and d(x) + d(y) ≥ 2n − 9,
then G is pancyclic. This result is best possible.
Proof Assume that G is not pancyclic, and let C = xv1 . . . vn−1 x. Since d(x) +
d(y) ≥ 2n − 9, we may assume without loss of generality that x has degree at
least n − 4. By Lemmas 1 and 2 and the assumption that G is not pancyclic,
we may assume that x has degree at most n − 3.
Suppose that x degree exactly n−3 and therefore, by Lemma 2, G contains
cycles of each length except n − 1. Then we can see that v2 and vn−2 are the
vertices of G that are not adjacent to x. We now consider the vertex y, which
has degree at least n − 6.
Since distC (x, y) ≥ 5 we know that y ∈
/ {vn−4 , vn−3 , vn−2 , vn−1 , v0 , v1 , v2 ,
v3 , v4 }. If at least one of yy −− or yy ++ is an edge of G, then G contains an (n−
1)-cycle and is pancyclic, so we assume that neither y −− nor y ++ are adjacent
to y. If yv1 ∈ E(G) or yvn−1 ∈ E(G), then xy − C − v1 yC + x or xC + yvn−1 y ++ x
is an (n − 1)-cycle, so we may also assume that yv1 , yvn−1 ∈
/ E(G). Since y
can have at most five nonadjacencies, one of yvn−2 and yv2 is an edge of G,
and G contains the (n − 1)-cycle xy + C + vn−2 yC − x or xC − yv2 C + y − x. Thus,
if x has degree exactly n − 3, G is pancyclic.
Suppose, then, that x has degree n − 4 and let vi , vj and vℓ be the nonneighbors of x in G. If i ̸= n − j, n − ℓ, j ̸= n − ℓ, and i, j, ℓ ̸= n2 , then the
cycle C ′ = xvn−m C + x is a cycle of length m + 1, where m = i, j, ℓ, and G is
pancyclic.
Without loss of generality, suppose then that i = n2 . If j, ℓ ̸= 2, n+2
2 , the
cycle xvi+1 C − v2 x has length i + 1. If j or ℓ = 2 or if j or ℓ = n+2
2 , then
G contains either the (i + 1)-cycle xvi−1 C + vn−2 x or xvi−2 C + vn−3 x. Now we
must find the cycles of lengths j+1 and ℓ+1. If j ̸= n−ℓ, G contains the (j+1)and (ℓ + 1)-cycles C ′ = xvn−m C + x as above, where m = j, ℓ. If j = n − ℓ ̸= 2,
then xvm+1 C − v2 x or xvn−m−1 C + vn−2 x is a cycle of length m + 1, where
m = j, ℓ. If j = n − ℓ = 2, we will find an (n − 1)-cycle by using the vertex
y. As above, we know that y ∈
/ {vn−4 , vn−3 , vn−2 , vn−1 , v0 , v1 , v2 , v3 , v4 }, and
that yy −− , yy ++ , yv1 , and yvn−1 are not edges of G. Now, y can have at most
four nonadjacencies, so G contains the (n − 1)-cycle xy + C + vn−2 yC − x.
Thus, we may assume without loss of generality that i = n − j (and hence
i, j ̸= n − ℓ). Observe that if ℓ ̸= n2 the (ℓ + 1)-cycle C ′ = xvn−ℓ C + x is
contained in G . If ℓ = n2 an (ℓ + 1)-cycle can be found in G as described
above. We proceed by considering three cases.
−
Case 1: Assume i ̸= 2, n−1
2 . If ℓ ̸= 2, i + 1, j + 1, then xvi+1 C v2 x is an
−
(i + 1)-cycle and xvj+1 C v2 x is a (j + 1)-cycle. If ℓ = 2, i + 1, or j + 1, then
xvi+2 C − v3 x is an (i + 1)-cycle and xvj+2 C − v3 x is a (j + 1)-cycle.
6
n+1
−
Case 2: Assume i = n−1
2 and j = 2 . If ℓ ̸= 2, 3, j + 1, then xvi+2 C v3 x
−
is a cycle of length i + 1 and xvj+1 C v2 x is a cycle of length j + 1. If ℓ = 2, 3,
or j + 1, then xvi−1 C + vn−3 x is a cycle of length i + 1 and xvj−2 C + vn−2 x is
a cycle of length j + 1.
Case 3: Assume i = 2 and j = n − 2. As noted above, if at least one of
yy −− or yy ++ is an edge of G, then G contains an (n − 1)-cycle, so we will
assume that these edges are not in G.
If ℓ = y − , then if yv1 , or yvn−2 ∈ E(G), the (n − 1)-cycle xy −− C − v1 yC + x
or xC + yvn−2 C − y + x is contained in G. If yv1 , yvn−2 ∈
/ E(G), then yvn−1 ∈
E(G), and G contains the (n − 1)-cycle xC + yvn−1 C − y ++ x. If ℓ = y + , the
argument is similar, so we assume that ℓ ̸= y − , y + .
If yv1 ∈ E(G) or yvn−1 ∈ E(G), then xy − C − v1 yC + x or xC + yvn−1 y ++ x
is an (n − 1)-cycle, so we may also assume that yv1 , yvn−1 ∈
/ E(G). Since y can
have at most four nonadjacencies, yv2 must be an edge of G, and G contains
the (n−1)-cycle xC − yv2 C + y − x. Thus, if x has exactly three nonadjacencies, G
is pancyclic. Having exhausted all cases, we may conclude that G is pancyclic.
To see that this result is sharp, consider the graph G∗ defined as follows:
Let V (G∗ ) = {v0 , . . . vn−1 } and construct the hamiltonian cycle C =
v0 v1 . . . nn−1 v0 in G∗ . Selecting x = v0 and y = vn−k−1 for some k satisfying
4 ≤ k ≤ n−2
2 , we let xvi be an edge for each i ̸= 0, 2, n−k−3, n−k+1, n−2 and
yvj be an edge for each j ̸= 2, n−k−3, n−k−1, n−2. Then d(x)+d(y) = 2n−10,
but G∗ contains no (n − 1)-cycle.
3 Distance-dependent Bounds
Theorems 5 and 6 provide sharp results, and are important for completeness.
The fact that the required degree sum is so large, while in some ways surprising,
also makes them somewhat unsatisfying. We return to the notion introduced
in [2] and examine the cycle structure of G when x and y are allowed to lie
farther apart on a hamiltonian cycle. We begin by considering the case where
the distance between x and y is two.
Theorem 7 Let G be a graph of order n with hamiltonian cycle C. If there
exist x, y ∈ V (G) such that distC (x, y) = 2 and d(x) + d(y) ≥ n + 1, then G
is pancyclic.
Proof Let C = v0 v1 . . . vn−1 v0 , where x = v0 and y = vn−2 . We separate the
proof into two cases based on whether or not xy is an edge of G.
Suppose that xy is an edge of G and consider the graph G′ = G − vn−1 .
This graph contains a hamiltonian cycle C ′ = x, v1 , . . . , vn−3 , y, x and dG′ (x)+
7
dG′ (y) ≥ n − 1. Consequently, by Theorem 3 G′ is either pancyclic, bipartite,
or missing only an (n − 2)-cycle. If G′ is pancyclic, then G is also pancyclic
and we are done.
Suppose then that G′ is bipartite, and note that that x and y must lie in
opposite partite sets of G′ . Since G′ is hamiltonian, each partite set of G′ must
have order n−1
2 . Furthermore, if we let U and W denote these partite sets and
assume, without loss of generality, that x ∈ U and y ∈ W then the fact that
dG′ (x) + dG′ (y) = n − 1 implies that x is adjacent to each vertex in W and y is
adjacent to each vertex in U . Since x and y are consecutive on the hamiltonian
′
cycle C ′ , we see that for each 2 ≤ t ≤ n−1
2 there is a cycle of length 2t in G
(and hence G) that contains the edge xy. To obtain cycles of odd length in G
we simply replace xy with the path xvn−1 y to obtain a cycle of length 2t + 1
for each t. These cycles, along with C and the triangle xvn−1 yx, imply that G
is pancyclic.
Now suppose that both G′ and G do not contain an (n − 2)-cycle. We
then consider the following pairs of possible edges in G′ : (v0 v1 , vn−2 vn−4 ),
(v0 v2 , vn−2 vn−3 ) and (v0 vi , vn−2 vi−2 ), where 3 ≤ i ≤ n − 3. For convenience,
we will refer to this collection of pairs as A0 . We claim that if both elements
of any of the pairs in A0 are in E(G′ ), then G′ would contain a cycle of length
n − 2.
Indeed, were both elements of either (v0 v1 , vn−2 vn−4 ) or (v0 v2 , vn−2 vn−3 )
edges in G′ , it is not difficult to see that G′ must contain a cycle of length
n − 2. If both elements of (v0 vi , vn−2 vi−2 ), where 3 ≤ i ≤ n − 3, were edges in
G, then v0 vi C + vn−2 vi−2 C − v0 is an (n − 2)-cycle in G′ .
Consequently at most one element of each pair in A0 is an edge of G′ , and
hence of G. Since there are n − 3 pairs and v0 vn−2 , v0 vn−1 and vn−2 vn−1 are
all edges of G, dG (x) + dG (y) ≤ n − 3 + 4 = n + 1. Therefore equality must
hold and exactly one element from each pair in A0 must be an edge of G.
Similarly, examine the pairs (v0 v1 , vn−2 vn−5 ), (v0 v2 , vn−2 vn−4 ),
(v0 v3 , vn−2 vn−3 ) and (v0 vi , vn−2 vi−2 ), where 4 ≤ i ≤ n − 3. For convenience,
we will refer to this collection of pairs as A1 . As is the case with A0 , at most
one element from each pair in A1 is an edge of G′ . Using an argument similar
to that given above, this implies that exactly one element from each pair in
A1 must be an edge of G.
By assumption, vn−2 vn−3 is an edge of G and therefore, as (vn−2 vn−3 , v0 v3 )
is in A1 , v0 v3 cannot be an edge of G. As (v0 v3 , vn−2 v1 ) is in A0 , vn−2 v1 must
be an edge of G. However, it then follows that v1 v2 . . . vn−2 v1 is an (n−2)-cycle
in G, contradicting our supposition. Thus if xy is an edge of G, G is pancyclic.
Suppose that xy is not an edge of G. Note first that since dG (x) + dG (y) ≥
n+1, one of x or y must have degree at least n+1
2 , and hence must be adjacent
to two consecutive vertices on C. This implies that G contains a triangle.
8
Assume then that G does not contain a cycle of some fixed length ℓ ≥ 4.
Then at most one element of each the pairs (v0 vi , vn−2 vℓ−4+i ), where 1 ≤ i ≤
n − ℓ + 1, and each of the pairs (v0 vn−ℓ+j , vn−2 vj ), where 1 ≤ j ≤ ℓ − 3, is
an edge of G. Indeed, were it the case that for some i, both v0 vi , vn−2 vi+ℓ−4
were edges in G, then v0 vi C + vi+ℓ−4 vn−2 C + v0 would be a cycle of length ℓ in
G. We handle the situation where both v0 vn−ℓ+j and vn−2 vj are edges in G
in a similar manner.
Since there are n − 2 of these pairs and both v0 vn−1 and vn−2 vn−1 are
edges of G, d(x) + d(y) ≤ n − 2 + 2 = n, which contradicts the hypothesis that
d(x) + d(y) ≥ n + 1. Hence, if xy is not an edge of G, G is pancyclic.
The following corollary is an immediate consequence of Theorem 4 and
Theorem 7, since the conditions imply that there must be two vertices of
degree sum at least n + 1 at distance one or two apart on a hamiltonian cycle
of G.
Corollary 1 If G is hamiltonian with more than
n+1
2 , then G is pancyclic.
n
3
vertices of degree at least
As the vertices x and y are moved farther apart on the hamiltonian cycle, it
becomes possible for cycle lengths to be omitted from the graph. Those cycle
lengths that may be missing from G are determined both by the degree sum
of x and y along with the distance between x and y on the hamiltonian cycle.
Theorem 8 Let G be a graph of order n with hamiltonian cycle C and let
k be an integer satisfying 1 ≤ k ≤ n−2
2 . If there exist x, y ∈ V (G) such that
distC (x, y) = k + 1 and d(x) + d(y) ≥ n + k, then G contains cycles of all
lengths ℓ, where 3 ≤ ℓ ≤ n − k. This result is sharp.
Proof Let C = v0 v1 . . . vn−1 v0 , where x = v0 and y = vn−k−1 . If there is a
vertex z in the path P = vn−k vn−k+1 . . . vn−1 that is adjacent to both x and
y, then the cycle C ′ = xC + yzx is a hamiltonian cycle of the graph G′ induced
by V (G) − V (P ) + z, which has order n − k + 1. In G′ , distC ′ (x, y) = 2 and
dG′ (x) + dG′ (y) ≥ n + k − 2(k − 1) = (n − k + 1) + 1, so by Theorem 7 G′
is pancyclic. Consequently, G contains cycles of all lengths ℓ, where 3 ≤ ℓ ≤
n − k + 1.
Suppose that there is no vertex z ∈ V (P ) that is adjacent to both x and
y. Since d(x) + d(y) ≥ n + k, x and y have at least k common neighbors, all of
which must be in V (G) − {V (P ), x, y}. We separate the proof into two cases
based on whether or not xy is an edge of G. For both cases we will consider
the graph G′′ induced by V (G) − V (P ).
Suppose that xy is an edge of G. The cycle C ′′ = xC + yx is a hamiltonian
cycle of G′′ with dC ′′ (x, y) = 1 and dG′′ (x) + dG′′ (y) ≥ n + k − k = n. Since
9
|V (G′′ )| = n − k and k ≥ 1, G′′ is pancyclic by Theorem 4 and so G contains
cycles of all lengths ℓ, where 3 ≤ ℓ ≤ n − k.
Therefore, suppose that xy is not an edge of G. An observation similar to
that made in the proof of Theorem 7 yields that G contains a cycle of length 3.
As such, we will suppose that G contains no cycle of length ℓ, for some fixed ℓ
between 4 and n − k. Recall that x and y are nonadjacent, share no neighbors
in P and have a degree sum of at least n + k in G. As a result, these two
vertices have at least k + 2 common neighbors in G′′ and dG′′ (x) + dG′′ (y) ≥ n.
We consider pairs of possible edges in G, as above. If 4 ≤ ℓ ≤ k + 2, let A0
denote the collection of pairs (v0 vi , vn−k−1 vℓ+i−4 ), where 1 ≤ i ≤ n − k − ℓ + 2,
and let A1 denote the pairs (v0 v(n−k−1)−(ℓ−2)+j , vn−k−1 v1+j ), where 1 ≤ j ≤
ℓ − 4. The reader should note that vn−k−1 vℓ−3 appears in pairs from both A1
and A2 , while vn−k−1 v1 appears in neither. Additionally, it is useful to note
that each v0 vi , for 2 ≤ i ≤ n − k − 2 appears exactly once in A0 ∪ A1 as does
each vn−k−1 vj , save the ones already mentioned, for 2 ≤ j ≤ n − k − 2.
We claim that if more than one element from any of these pairs is an edge,
then G would contain a cycle of length ℓ. Indeed, consider (v0 vi , vn−k−1 vℓ+i−4 )
in A0 , where 1 ≤ i ≤ n − k − ℓ + 2, and assume that both elements of this pair
were edges in G. Note that since x and y have at least k +2 common neighbors,
there must be at least one such common neighbor v ′ that does not lie in the
set {vi , . . . , vℓ+i−4 }. Then v0 v ′ vn−k−1 vℓ+i−4 C − vi v0 is a cycle of length ℓ in G.
Next, consider (v0 v(n−k−1)−(ℓ−2)+j , vn−k−1 v1+j ), where 1 ≤ j ≤ ℓ − 4, and
assume that both elements of this pair were edges in G. Then
v0 C + vj+1 xn−k−1 C − v(n−k−1)−(ℓ−2)+j v0 is an ℓ-cycle in G.
Hence exactly one element from each of the n − k − 2 pairs in A0 ∪ A1 can
be an edge of G. Taking into account that v1 vn−k−1 may be an edge of G, and
recalling that either x or y, but not both, may be adjacent to each vertex in
P , we have that d(x) + d(y) ≤ n − k − 2 + k + 1 = n − 1. This contradicts the
hypothesis that d(x) + d(y) ≥ n + k.
If k + 3 ≤ ℓ ≤ n − k, then arguments similar to those given previously yield
that for each of the pairs (v0 vi , vn−k−1 vℓ−k+i−3 ), where 1 ≤ i ≤ n − ℓ + 1, and
(v0 vn−ℓ+1−j , vn−k−1 vk+2−j ), where n − ℓ + 2 ≤ j ≤ n − 2k − 1, at most one
element is an edge of G. There are n − 2k − 1 of these pairs and none of the 2k
possible edges v0 vn−k−i and vn−k−1 vi , where 1 ≤ i ≤ k, lie in any pair. This
implies that d(x) + d(y) ≤ n − 2k − 1 + 2k + k = n + k − 1, which contradicts
the hypothesis that d(x) + d(y) ≥ n + k.
Therefore G contains cycles of all lengths ℓ for 3 ≤ ℓ ≤ n − k.
This result is sharp in the sense that, for 2 ≤ k ≤ n+4
7 , the graph in Figure
1 (for readability the v’s are suppressed in Figures 1 and 2) has two vertices
x and y satisfying d(x) + d(y) ≥ n + k, yet there are no cycles longer than
n − k. We also note that if we lower the degree sum of x and y, we may miss
10
Fig. 1 d(x) + d(y) ≥ n + k, but G has no cycle of length longer than n − k
Fig. 2 d(x) + d(y) = n + k − 1, but G contains no cycle of length
n−k+3
2
some cycles of length ℓ for 3 ≤ ℓ ≤ n − k. As an example see Figure 2 in which
d(x) + d(y) = n + k − 1, but there is no cycle of length n−k+3
, provided that
2
2 ≤ k ≤ n5 .
4 Conclusion
In this paper, we give several conditions that assure a large variety of cycle
lengths in hamiltonian graphs containing a pair of vertices with high degree
sum. Along these lines, it may be of interest to investigate the effect of larger
sets of vertices with high degree sum on the cycle spectrum of a hamiltonian
graph. The addition of other structural properties in addition to hamiltonicity
and a pair of vertices with a high degree sum may also result in some interesting
lines of inquiry.
Acknowledgements M.S. Jacobson and A. Harris were partially supprted by NSF grant
DGE-0742434.
11
References
1. J. A. Bondy, Pancyclic Graphs, J. Combin. Theory Ser. B, 11 (1971), 80-84.
2. S. L. Hakimi, E. F. Schmeichel, A Cycle Structure Theorem for Hamiltonian Graphs,
J. Combin. Theory Ser. B, 45 (1988), 99-107.
3. O. Ore, A Note on Hamilton Circuits, Amer. Math. Monthly, 67 (1960), 55.