AbstractFor n∈N and D⊆N, the distance graph PnD has vertex set {0,1,…,n−1} and edge set {ij∣0≤i,j≤n−1,|j−i|∈D}. Note that the important and very well-studied circulant graphs coincide with the regular distance graphs.A fundamental result concerning circulant graphs is that for these graphs, a simple greatest common divisor condition, their connectivity, and the existence of a Hamiltonian cycle are all equivalent. Our main result suitably extends this equivalence to distance graphs. We prove that for a finite set D of order at least 2, there is a constant cD such that the greatest common divisor of the integers in D is 1 if and only if for every n, PnD has a component of order at least n−cD if and only if for every n≥cD+3, PnD has a cycle of order at least n−cD. Furthermore, we discuss some consequences and variants of this result

Penso, Lucia Draque

Rautenbach, Dieter

Szwarcfiter, Jayme Luiz

For $n\in \mathbb{N}$ and $D\subseteq \mathbb{N}$, the distance graph $P_n^D$ has vertex set $\{ 0,1,\ldots,n-1\}$ and edge set $\{ ij\mid 0\leq i,j\leq n-1, |j-i|\in D\}$. Note that the important and very well-studied circulant graphs coincide with the regular distance graphs. A fundamental result concerning circulant graphs is that for these graphs, a simple greatest common divisor condition, their connectivity, and the existence of a Hamiltonian cycle are all equivalent. Our main result suitably extends this equivalence to distance graphs. We prove that for a set $D$, there is a constant $c_D$ such that the greatest common divisor of the integers in $D$ is $1$ if and only if for every $n$, $P_n^D$ has a component of order at least $n-c_D$ if and only if for every $n$, $P_n^D$ has a cycle of order at least $n-c_D$. Furthermore, we discuss some consequences and variants of this result

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Long cycles and paths in distance graphs

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Discrete Mathematics 310 (2010) 3417–3420
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Discrete Mathematics
journal homepage: www.elsevier.com/locate/disc
Note
Long cycles and paths in distance graphs
Lucia Draque Penso a, Dieter Rautenbach a,∗, Jayme Luiz Szwarcfiter b
a Institut für Optimierung und Operations Research, Universität Ulm, D-89069 Ulm, Germany
b Universidade Federal do Rio de Janeiro, Instituto de Matemática, NCE and COPPE, Caixa Postal 2324, 20001-970 Rio de Janeiro, RJ, Brazil
a r t i c l e i n f o
Article history:
Received 23 November 2009
Received in revised form 19 July 2010
Accepted 20 July 2010
Available online 11 August 2010
Keywords:
Circulant graph
Distance graph
Connectivity
Hamiltonian cycle
Hamiltonian path
a b s t r a c t
For n ∈ N and D ⊆ N, the distance graph PDn has vertex set {0, 1, . . . , n − 1} and edge set{ij | 0 ≤ i, j ≤ n− 1, |j− i| ∈ D}. Note that the important and very well-studied circulant
graphs coincide with the regular distance graphs.
A fundamental result concerning circulant graphs is that for these graphs, a simple
greatest common divisor condition, their connectivity, and the existence of a Hamiltonian
cycle are all equivalent. Our main result suitably extends this equivalence to distance
graphs. We prove that for a finite set D of order at least 2, there is a constant cD such that
the greatest common divisor of the integers in D is 1 if and only if for every n, PDn has a
component of order at least n− cD if and only if for every n ≥ cD+3, PDn has a cycle of order
at least n− cD. Furthermore, we discuss some consequences and variants of this result.
© 2010 Elsevier B.V. All rights reserved.
1. Introduction
Circulant graphs form an important and very well-studied class of graphs [2,17,18,22]. They are Cayley graphs of cyclic
groups and have been proposed for numerous network applications such as local area computer networks, large area
communication networks, parallel processing architectures, distributed computing, and VLSI design. Their connectivity and
diameter [5,2,17,18], cycle and path structure [1,3,4,6], and further graph-theoretical properties have been studied in great
detail. Polynomial time algorithms for isomorphism testing and recognition of circulant graphs have been long-standing
open problems which were completely solved only recently [15,21].
For n ∈ N and D ⊆ N, the circulant graph CDn has vertex set [0, n− 1] = {0, 1, . . . , n− 1} and the neighbourhood NCDn (i)
of a vertex i ∈ [0, n− 1] in CDn is given by
NCDn (i) = {(i+ d) mod n | d ∈ D} ∪ {(i− d) mod n | d ∈ D}.
Clearly, we may assume max(D) ≤ n2 for every circulant graph CDn .
Our goal here is to extend a fundamental result concerning circulant graphs to the similarly defined yet more general
class of distance graphs: For n ∈ N and D ⊆ N, the distance graph PDn has vertex set [0, n− 1] and
NPDn (i) = {i+ d | d ∈ D and (i+ d) ∈ [0, n− 1]} ∪ {i− d | d ∈ D and (i− d) ∈ [0, n− 1]}
for all i ∈ [0, n− 1]. Clearly, we may assume max(D) ≤ n− 1 for every distance graph PDn .
Every distance graph PDn is an induced subgraph of the circulant graph C
D
n+max(D). More specifically, distance graphs are
the subgraphs of sufficiently large circulant graphs induced by sets of consecutive vertices. Conversely, our following simple
observation from [11] shows that every circulant graph is in fact a distance graph.
Proposition 1 ([11]). A graph is a circulant graph if and only if it is a regular distance graph.
∗ Corresponding author.
E-mail addresses: lucia.penso@uni-ulm.de (L.D. Penso), dieter.rautenbach@uni-ulm.de, dieter.rautenbach@tu-ilmenau.de (D. Rautenbach),
jayme@nce.ufrj.br (J.L. Szwarcfiter).
0012-365X/$ – see front matter© 2010 Elsevier B.V. All rights reserved.
doi:10.1016/j.disc.2010.07.020
3418 L.D. Penso et al. / Discrete Mathematics 310 (2010) 3417–3420
Proof. Clearly, every circulant graph CDn is regular and isomorphic to the distance graph P
D′
n for D
′ = D ∪ {n− d | d ∈ D}.
Now let PDn be a regular distance graph. Let D = {d1, d2, . . . , dk} with d1 < d2 < · · · < dk ≤ n − 1. Since the vertex 0
has exactly k neighbours D, PDn is k-regular.
Let i ∈ [1, k]. The vertex di−1 has exactly i−1 neighbours jwith j < di−1. Hence, di−1 has exactly k+1− i neighbours
jwith j > di − 1 which implies (di − 1)+ dk+1−i ≤ n− 1. The vertex di has exactly i neighbours jwith j < di. Hence, di has
exactly k− i neighbours jwith j > di which implies di + dk+1−i > n− 1.
We obtain di + dk+1−i = n for every i ∈ [1, k] which immediately implies that PDn is isomorphic to the circulant graph
CD
′
n for D
′ = {d ∈ D | d ≤ n2 }. 
Distance graphs lack the symmetry and algebraic interpretation of circulant graphs and the algebraicmethods used in [15,21]
will not apply to them. In view of Proposition 1, the recognition of distance graphs will be at least as difficult as the
recognition of circulant graphs.
Originally, coloring problems for infinite distance graphs were studied by Eggleton et al. [13,14]. In fact, most research
on distance graphs focused on colorings [8,10,12,19,23].
One of the most fundamental results for circulant graphs is the following beautiful equivalence.
Theorem 2 ([5,7,16]). For n ∈ N and a finite set D ⊆ N, the following statements are equivalent.
(i) CDn is connected.
(ii) The greatest common divisor gcd({n} ∪ D) of the integers in {n} ∪ D equals 1.
(iii) CDn has a Hamiltonian cycle.
In 1970 Lovász [9,20,24] asked whether every connected vertex-transitive graph has a Hamiltonian path. Since circulant
graphs are clearly vertex transitive, Theorem 2 is a positive example for this well-studied problem [9,24].
In the present paper, we suitably extend Theorem2 to distance graphs.While connectivity and hamiltonicity of circulants
are equivalent to a simple necessary gcd-condition, we prove that a similar condition for distance graphs is only equivalent
to the existence of a large component and a long cycle. We also discuss consequences and variants of our result.
2. Cycles and paths in distance graphs
We immediately proceed to our main result. The residue of an integer n ∈ Zmodulo d ∈ Nwill be denoted by n mod d.
Theorem 3. For a finite set D ⊆ N with |D| ≥ 2, the following statements are equivalent.
(i) There is a constant c1(D) such that for every n ∈ N, the distance graph PDn has a component of order at least n− c1(D).
(ii) gcd(D) = 1.
(iii) There is a constant c2(D) such that for every n ∈ N with n ≥ c2(D) + 3, the distance graph PDn has a cycle of order at least
n− c2(D).
Proof. (i)⇒ (ii): Let n be such that n is even and n > 2c1(D). By (i), more than half the vertices are in the same component
of PDn . By the pigeonhole principle, there is some i ∈ [0, n−2] such that the two vertices i and i+1 are in the same component
of PDn . This implies that there is a path in P
D
n from i to i+ 1. Hence 1 is an integral linear combination of the elements in D. It
is a well-known consequence of the Euclidean algorithm that this is equivalent to (ii).
(ii) ⇒ (iii): Let d = max(D). Let C⃗D\{d}d denote the directed circulant graph with vertex set [0, d − 1] where (i, j) is an arc
for i, j ∈ [0, d− 1] if and only if (j− i) mod d ∈ D \ {d}.
Claim. C⃗D\{d}d has a Hamiltonian path.
Proof of Claim. We prove the statement by induction on |D|. For |D| = 2, let D = {d, d′}. Since gcd(d, d′) = 1, we have
{id′ mod d | 0 ≤ i ≤ d− 1} = [0, d− 1], i.e.
0(d′ mod d)(2d′ mod d)(3d′ mod d) . . . ((d− 1)d′ mod d)
is a Hamiltonian path. Now let |D| > 2. Let D′ = D \ {d′} for some d′ ∈ D \ {d}. Let gcd(D′) = g . By induction, C⃗D′\{d}d has
a path P which visits exactly all vertices in [0, d − 1] which are multiples of g . Clearly, we may assume that g ≥ 2. Since
gcd(g, d′) = 1, we have {id′ mod g | 0 ≤ i ≤ g − 1} = [0, g − 1] and concatenating translates of P and arcs of the form
(i, i+ d′) yields a Hamiltonian path of C⃗D\{d}d as desired. 
By the claim, there are a1, a2, . . . , ad−1 ∈ D \ {d} such that
0(a1 mod d)((a1 + a2) mod d)((a1 + a2 + a3) mod d) · · · ((a1 + a2 + · · · + ad−1) mod d)
is a Hamiltonian path of C⃗D\{d}d , i.e.
∑k
i=1 ai

mod d | 0 ≤ k ≤ d− 1

= [0, d−1]. Let A = a1+a2+· · ·+ad−1 ≤ d(d−1).
Clearly, we may assume that n is sufficiently large in terms of D. Ifm ∈ N is such that n− 2d ≤ (2m− 1)d+ A < n, then
ld+
k−
i=1
ai | l ∈ [0, 2m− 1], k ∈ [0, d− 1]

L.D. Penso et al. / Discrete Mathematics 310 (2010) 3417–3420 3419
is a set of 2md distinct vertices which induce a subgraph of PDn that is isomorphic to P2m × Pd. Since P2m × Pd obviously has
a Hamiltonian cycle, PDn has a cycle of order at least 2md ≥ n− d− A ≥ n− d2, i.e. (iii) holds.
(iii)⇒ (i): Since this implication is trivial, the proof is complete. 
We add some comments concerning Theorem 3.
It is easy to see that a distance graph PDn with gcd(D) = 1 and n ≥ 2max(D) + 1 is actually connected. Hence (i) in
Theorem 3 could be replaced by
(i)′ There is a constant c3(D) such that for every n ∈ N with n ≥ c3(D), the distance graph PDn is connected.
For (iii) in Theorem 3, a similar change is not possible, because no lower bound on the order n would imply that PDn has a
Hamltonian cycle. If n as well as all elements of D are odd for instance, then PDn is bipartite and every cycle misses at least
one vertex. In this sense, Theorem 3 is best possible. It follows easily from our proof that c2(D) = O(max(D)2). For the case
that gcd(D) is different from 1, Theorem 3 implies the following corollary.
Corollary 4. For a finite set D ⊆ N with |D| ≥ 2 and g ∈ N, the following statements are equivalent.
(i) There is a constant c4(D) such that for every n ∈ N, the distance graph PDn has a component of order at least ng − c4(D).
(ii) gcd(D) ≤ g.
(iii) There is a constant c5(D) such that for every n ∈ N with n ≥ g(c2(D) + 3), the distance graph PDn has a cycle of order at
least ng − c5(D).
Theorem 3 trivially implies yet another condition which is equivalent to (i), (ii), and (iii) in Theorem 3.
(iv) There is a constant c6(D) such that for every n ∈ N, the distance graph PDn has a path of order at least n− c6(D).
Clearly, such a path can be obtained from the cycle in (iii) by deleting one edge. In fact, if v(l,k) = ld +∑ki=1 ai denotes the
vertices of the induced subgraph P2m × Pd of PDn identified in the proof of ‘‘(ii)⇒ (iii)’’ of Theorem 3, then
v(0,0)v(0,1) · · · v(0,d−1)v(1,d−1)v(1,d−2) · · · v(1,0)v(2,0)v(2,1) · · · v(2,d−1)v(3,d−1)v(3,d−2) · · · v(3,0) · · · v(2m−1,d−1)
specifies a Hamiltonian path of P2m × Pd which proves the following.
Corollary 5. For a finite set D ⊆ N with |D| ≥ 2, the following statements are equivalent.
(i) gcd(D) = 1.
(ii) There are two constants c7(D) and c8(D) such that for every n ∈ N, the distance graph PDn has a path u0u1 . . . ul of order at
least n− c7(D) such that uj > ui for all 0 ≤ i, j ≤ l with j− i ≥ c8(D).
We believe that Corollary 5 can be strengthened as follows.
Conjecture 6. For a finite set D ⊆ N with |D| ≥ 2 and gcd(D) = 1, there is some n ∈ N such that n ≥ 2 and PDn has a
Hamiltonian path with endvertices 0 and n− 1.
As our last result, we verify Conjecture 6 for |D| = 2.
Proposition 7. If d1, d2 ∈ N are such that d1 > d2 and gcd({d1, d2}) = 1, then P {d1,d2}d1+d2+1 has a Hamiltonian path with
endvertices 0 and d1 + d2.
Proof. Consider the sequence i0, i1, . . . , id1+d2 produced by Algorithm 1.
i0 := 0;
n := 0;
n1 := 0;
n2 := 0;
while n < d1 + d2 do
if in ≥ d2 and n2 < d1 − 1 then
in+1 := in − d2;
n2 := n2 + 1;
else
in+1 := in + d1;
n1 := n1 + 1;
end
n := n+ 1;
end
Algorithm 1
Clearly, ij ≥ 0 for j ∈ [0, d1 + d2].
If n1 > d2 + 1 after the termination of the algorithm, then n2 = n− n1 < d1 − 1 and hence
id1+d2 = n1d1 − n2d2 ≥ (d2 + 2)d1 − (d1 − 2)d2 = 2d1 + 2d2.
3420 L.D. Penso et al. / Discrete Mathematics 310 (2010) 3417–3420
Let j ∈ [1, d1+d2−1] bemaximum such that ij+1 = ij+d1. Clearly, ij+1 ≥ d2 and the algorithmwould have set ij+1 = ij−d2
instead, which is a contradiction. Hence, n1 ≤ d2 + 1. Since n2 ≤ d1 − 1, we obtain n1 = d2 + 1 and n2 = d1 − 1. This
implies
id1+d2 = (d2 + 1)d1 − (d1 − 1)d2 = d1 + d2.
If ij > d1 + d2 for some j ∈ [1, d1 + d2 − 1], then let j be largest with this property. Clearly, ij−1 ≥ d2 and at this moment
of the execution of the algorithm n2 < d1 − 1. Therefore, the algorithm would have set ij = ij−1 − d2 instead, which is a
contradiction. Hence, ij ≤ d1 + d2 for all j ∈ [0, d1 + d2].
If ir = is for some r, s ∈ [0, d1 + d2] with s > r , then is − ir = a1d1 − a2d2 = 0 for some a2 ∈ [1, d1 − 1]. This
implies a1d1 = a2d2. Since gcd({d1, d2}) = 1, we obtain that a2 must be a multiple of d1, which is a contradiction. Hence, all
d1 + d2 + 1 integers i0, i1, . . . , id1+d2 are distinct and define the desired Hamiltonian path of P {d1,d2}d1+d2+1. This completes the
proof. 
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Long cycles and paths in distance graphs 

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Long cycles and paths in distance graphs

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