A sufficient condition for the hamiltonian property of digraphs with large semi-degrees

Abstract

Let $D$ be a digraph on $p\geq 5$ vertices with minimum degree at least $p-1$
and with minimum semi-degree at least $p/2-1$. For $D$ (unless some extremal
cases) we present a detailed proof of the following results [12]: (i) $D$
contains cycles of length 3, 4 and $p-1$; (ii) if $p=2n$, then $D$ is
hamiltonian.

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arXiv:1111.1843v1 [math.CO] 8 Nov 2011
A sufficient condition for the hamiltonian property of
digraphs with large semi-degrees
S. Kh. Darbinyan
Institute for Informatics and Automation Problems
Armenian National Academy of Sciences, P. Sevak 1, Yerevan 0014, Armenia
e-mail: samdarbin@ipia.sci.am
Abstract
Let D be a digraph on p ≥ 5 vertices with minimum degree at least p − 1 and with minimum semi-
degree at least p/2 − 1. For D (unless some extremal cases) we present a detailed proof of the following
results [12]: (i) D contains cycles of length 3, 4 and p − 1; (ii) if p = 2n, then D is hamiltonian.
Keywords: Digraphs; semi-degrees; cycles; Hamiltonian cycles
1. Introduction and Terminology
Ghouila-Houri [18] proved that every strong digraph on p vertices and with minimum degree at least
p is hamiltonian. There are many extentions of this theorem for digraphs and orgraphs. In particular,
in many papers, various degree conditions have been obtained for digraphs (orgraphs) to be hamiltonian
or pancyclic or vertex pancyclic (see e.g. [2]-[33]). C. Thomassen [31] proved that any digraph on
p = 2n + 1 vertices with minimum semi-degree at least n is hamiltonian unless some extremal cases,
which are characterized. In [9], we proved that if a digraph D on 2n+1 vertices satisfies the conditions of
this Tomassen’s theorem, then D also is pancyclic (the extremal cases are characterized). For additional
information on hamiltonian and pancyclic digraphs, see the book [1] by J. Bang-Jenssen and G. Gutin.
In this paper we present a detailed proof of the following results.
Every digraph D (unless some extremal cases) on p ≥ 5 vertices with minimum degree at least p − 1
and with minimum semi-degree at least p/2 − 1: (i) D has cycles of length 3, 4 and p − 1; (ii) if p = 2n,
then D is hamiltonian (in [12], we gave only a short outline of the proofs of this results).
In this paper we will consider finite digraphs without loops and multiple arcs. We denote the vertex
set of digraph D by V (D) and its arc set by A(D). We will often use D instead of A(D) and V (D). The
arc from a vertex x to a vertex y will be denoted by xy. If xy is an arc, then we say that x dominates y
(or y is dominated by x). For A, B ⊂ V (D), we define A(A → B) as the set {xy ∈ A(D)/x ∈ A,y ∈ B}
and A(A,B) = A(A → B) ∪ A(B → A). If x ∈ V (D) and A = {x}, we often write x instead of
{x}. For disjoint subsets A and B of V (D), A → B means that every vertex of A dominates every
vertex of B. If C ⊂ V (D), A → B and B → C, then we write A → B → C. The outset of vertex
x is the set O(x) = {y ∈ V (D)/xy ∈ A(D)} and I(x) = {y ∈ V (D)/yx ∈ A(D)} is the inset of x.
Similarly, if A ⊆ V (D) then O(x,A) = {y ∈ A/xy ∈ A(D)} and I(x,A) = {y ∈ A/yx ∈ A(D)}. The
out-degree of x is od(x) = |O(x)| and id(x) = |I(x)| is the in-degree of x. Similarly, od(x,A) = |O(x,A)|
and id(x,A) = |I(x,A)|. The degree of the vertex x in D is defined as d(x) = id(x) + od(x). The
subdigraph of D induced by a subset A of V (D) is denoted by ?A?. All paths and cycles we consider
in this paper are directed and simple. The path ( respectively, the cycle ) consisting of distinct vertices
x1,x2,...,xn( n ≥ 2) and arcs xixi+1, i ∈ [1,n − 1] ( respectively, xixi+1, i ∈ [1,n − 1], and xnx1), is
denoted by x1x2...xn(respectively, x1x2...xnx1). The cycle on k vertices is denoted Ck. For a cycle
Ck= x1x2...xkx1, we take the indices modulo k, i.e., xs= xifor every s and i such that i ≡ smodk.
Two distinct vertices x and y are adjacent if xy ∈ A(D) or yx ∈ A(D) (or both) (i.e, x is adjacent
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with y and y is adjacent with x). The notation A(x,y) ?= ∅ (respectively, A(x,y) = ∅) means that the
vertices x and y are adjacent (respectively, are not adjacent).
The converse digraph← −
D of a digraph D is the digraph obtained from D by reversing all arcs of D.
For an undirected graph G, we denote by G∗symmetric digraph obtained from G by replacing every
edge xy with the pair xy, yx of arcs. Further, C∗(5) is a symmetric digraph obtained from undirected cycle
of length 5. Kn(respectively, Kn,m) denotes the complete undirected graph on n vertices (respectively,
undirected complete bipartite graph, with partite sets of cardinalities n and m), and Kn denotes the
complement of Kn. If G1and G2are undirected graphs, then G1∪ G2is the disjoint union of G1and
G2. The join of G1and G2, denoted by G1+ G2, is the union of G1∪ G2and of all the edges between
G1and G2.
For integers a and b, let [a,b] denote the set of all integers which are not less than a and are not greater
than b. We refer the reader to J.Bang-Jensens and G.Gutin’s book [1] for notations and terminology not
defined here.
2. Preliminaries and Additional notations
Let us recall some well-known lemmas used in this paper.
Lemma 1 ([21]). Let D be a digraph on p ≥ 3 vertices containing a cycle Cm, m ∈ [2,p−1]. Let x be a
vertex not contained in this cycle. If d(x,Cm) ≥ m+1, then D contains a cycle Ckfor all k ∈ [2,m+1].
The following Lemma will be used extensively in the proofs our results.
Lemma 2 ([6]). Let D be a digraph on p ≥ 3 vertices containing a path P := x1x2...xm, m ∈ [2,p−1]
and let x be a vertex not contained in this path. If one of the following holds:
(i) d(x,P) ≥ m + 2;
(ii) d(x,P) ≥ m + 1 and xx1 / ∈ D or xmx1 / ∈ D;
(iii) d(x,P) ≥ m, xx1 / ∈ D and xmx / ∈ D;
then there is an i ∈ [1,m−1] such that xix,xxi+1∈ D, i.e., D contains a path x1x2...xixxi+1...xmof
length m (we say that x can be inserted into P or the path x1x2...xixxi+1...xmis extended from P
with x).
As an immediate consequence of Lemma 2, we get the following:
Lemma 3. Let D be a digraph on p ≥ 4 vertices and let P := x1x2...xm, m ∈ [2,p − 2], be a
path of maximal length from x1 to xm in D. If the induced subdigraph ?V (D) \ V (P)? is strong and
d(x,V (P)) = m + 1 for every vertex x ∈ V (D) \ V (P), then there is an integer l ∈ [1,m] such that
O(x,V (P)) = {x1,x2,...,xl} and I(x,V (P)) = {xl,xl+1,...,xm}.
Now we introduce the following notations.
Notation. For any positive integer n, let H(n,n) denote the set of digraphs D on 2n vertices such that
V (D) = A ∪ B, ?A? ≡ ?B? ≡ K∗
n,A(B → A) = ∅ and for every vertex x ∈ A (respectively, y ∈ B)
A(x → B) ?= ∅ (respectively, A(A → y) ?= ∅).
Notation. For any integer n ≥ 2, let H(n,n − 1,1) denote the set of digraphs D on 2n vertices such
that V (D) = A∪ B ∪{a}, |A| = |B|+1 = n, A(?A?) = ∅, ?B ∪{a}? ⊆ K∗
n, yz, zy ∈ D for each pair of
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vertices y ∈ A, z ∈ B and either I(a) = B and a → A or O(a) = B and A → a.
Notation. For any integer n ≥ 2 define the digraph H(2n) as follows: V (H(2n)) = A ∪ B ∪ {x,y},
?A? ≡ ?B? ≡ K∗
n−1, A(A,B) = ∅, O(x) = {y} ∪ A, I(x) = O(y) = A ∪ B and I(y) = {x} ∪ B.
H′(2n) is a digraph obtained from H(2n) by adding the arc yx.
Notation. Let D6be a digraph with vertex set {x1,x2,...,x5,x} and arc set
{xixi+1/1 ≤ i ≤ 4} ∪ {xxi/1 ≤ i ≤ 3} ∪ {x1x5,x2x5,x5x1,x5x4,x3x2,x3x,x4x1,x4x}.
By D′
6we denote a digraph obtained from D6by adding the arc x2x4.
Note that the digraphs D6and D′
of length 5.
6both are not hamiltonian and each of D6and D′
6contains a cycle
Lemma 4. Let D be a digraph on p ≥ 3 vertices with minimum degree at least p−1 and with minimum
semi-degree at least p/2 − 1. Then
(i) either D is strong or D ∈ H(n,n);
(ii) if B ⊂ V (D), |B| ≥ (p + 1)/2 and x ∈ V (D) \ B, then A(x → B) ?= ∅ and A(B → x) ?= ∅.
3. A sufficient condition for the existence of cycles of length |V (D)| − 1 in digraph D
Theorem 1. Let D be a digraph on p ≥ 5 vertices with minimum degree at least p−1 and with minimum
semi-degree at least p/2 − 1. Then D has a cycle of length p − 1 unless
D ∈ H(n,n) ∪ {[(Kn∪ Kn) + K1]∗,H(2n),H′(2n),C∗(5)}
or elsep = 2n andD ⊆ K∗
n,n.
Proof. By Lemma 4(i), the result is easily verified if D is not strong. Assume that D is strong. Suppose,
on the contrary, that the theorem is not true. In particular, D contains no cycle of length p − 1. Let
C := Cm:= x1x2...xmx1be an arbitrary non-hamiltonian cycle of maximum length in D. It is easy to
see that m ∈ [3,p − 2].
From Lemma 1 and the maximality of m it follows that for each vertex y ∈ B := V (D) \ V (C) and
for each i ∈ [1,m],
d(y,C) ≤ m,d(y,B) ≥ p − m − 1and ifxiy ∈ D,then yxi+1 / ∈ D. (1)
Using d(y,B) ≥ p − m − 1 it is not difficult to show the following claim:
Claim 1. For any two distinct vertices x,y ∈ B if in subdigraph ?B? there is no path from x to y, then
in ?B? there is a path from y to x of length at most 2.
We first prove the following Claims 2 and 3.
Claim 2. The induced subdigraph ?B? is strongly connected.
Proof. Suppose, on the contrary, that ?B? is not strong. Let D1, D2, ..., Ds (s ≥ 2) be the strong
components of ?B? labeled in such a way that no vertex of Didominates a vertex of Djwhenever i > j.
From Claim 1 it follows that for each pair of vertices y ∈ V (D1) and z ∈ V (Ds) in ?B? there is a path
from y to z of length 1 or 2. We choose the vertices y ∈ V (D1) and z ∈ V (Ds) such that the path
y1y2...yk, where y1 := y and yk := z, will have minimum length among all paths in ?B? with origin
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vertex in D1and terminus vertex in Ds. By Claim 1, k = 2 or k = 3. We consider the following tree cases.
Case 1. k < |B| = p − m.
It follows from the maximality of C that if xiy1 ∈ D, where i ∈ [1,m], then A(yk → {xi+1,xi+2,
...,xi+k}) = ∅. Since D is strong, we see that C ?⊆ I(y1). Therefore the vertex yk dose not dom-
inate at least id(y1,C) + 1 vertices of C. On the other hand, we have A(yk → V (D1)) = ∅ and
I(y1) ⊂ C ∪ V (D1). Hence the vertex yk dose not dominate at least id(y1) + 3 vertices. From this
we obtain od(yk) ≤ p − id(y1) − 3 ≤ p/2 − 2, which is a contradiction.
Case 2. k = |B| = 2.
It is easy to see that s = 2, m = p − 2, V (D1) = {y1}, V (D2) = {y2}, I(y1) ⊂ C and
|A(xi→ y1)| + |A(y2→ xi+2)| ≤ 1
for all i ∈ [1,m]. Hence the vertex y2dose not dominate at least id(y1) + 2 vertices. Therefore od(y2) ≤
p − id(y1) − 2 ≤ p/2 − 1. It follows that p = 2n, id(y1) = od(y2) = n − 1 and
y2xi∈ D if and only ifxi−2y1 / ∈ D. (2)
By Lemma 1, it is easy to see that d(y1) = d(y2) = 2n − 1 and od(y1) = id(y2) = n, m ≥ 4. Now we
divide this case into two subcases.
Subcase 2.1. y1→ {xi,xi+1} for some i ∈ [1,m].
Note that, by Lemma 2, without loss of generality, we may assume that xmy1∈ D, y1→ {x2,x3} and
A(x1,y1) = ∅. From this, (1) and (2) it follows that x2y1/ ∈ D, y2x3, y2x4∈ D and A(x2,y2) = ∅. There-
fore, by Lemma 2 we have x1y2∈ D since d(y2,C) = 2n−2 and the vertex y2cannot be inserted into the
path x3x4...xmx1. If x2x1∈ D, then C2n−1= xmy1x2x1y2x4...xm. This contradicts our supposition
that D contains no cycle of length p − 1. Hence, x2x1 / ∈ D. From this and A(x2,y2) = A(x2→ y1) = ∅
it follows that d(x2,{x3,x4,...,xm}) ≥ 2n − 3. Therefore by Lemma 2, xmx2∈ D since the vertex x2
cannot be inserted into the path x3x4...xm. Now it is easy to see that |A(xi→ y1)|+|A(y2→ xi+1)| ≤ 1
for all i ∈ [2,m − 1]. Therefore x3y1 / ∈ D since y2x4∈ D. From this and (2) it follows that y2x5∈ D
and x4y1 / ∈ D. Continuing in this manner, we obtain that A({x5,x6,...,xm−1} → y1) = ∅. Therefore
A({x1,x2,...,xm−1} → y1) = ∅, which is a contradiction.
Subcase 2.2. |A(y1→ {xi,xi+1)| ≤ 1 for all i ∈ [1,m].
Since od(y1) = n, we can assume that O(y1) = {x1,x3,...,x2n−3,y2}. Using this and od(y2) =
id(y1) = n − 1, we obtain I(y1) = {x1,x3,...,x2n−3}. Therefore by (2),
O(y2) = {x2,x4,...,x2n−2}
andI(y2) = {y1,x2,x4,...,x2n−2}.
If xixj∈ D for distinct vertices xi,xj∈ {x1,x3,...,x2n−3}, then C2n−1= y1xixjxj+1...xi−1y2xi+1...
xj−2y1, when |{xi+1,xi+2,...,xj−1}| ≥ 2 and C2n−1= xixjy1y2xj+1xj+2...xi−1xi, when |{xi+1,xi+2,
..., xj−1}| = 1. This contradicts that Cp−1?⊂ D. Thus we have
A(?{x1,x3,...,x2n−3,y2}?) = ∅.
Considering the digraph← −
D, by the same arguments we obtain
A(?{x2,x4,...,x2n−2,y1}?) = ∅.
Therefore D ⊆ K∗
n,n, which contradicts our supposition that the theorem is not true.
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Case 3. k =| B |= 3.
From the minimality of k it follows that y1y3 / ∈ D, s = 3, A({y2,y3} → y1) = ∅ and V (D1) = {y1}.
Hence I(y1) ⊂ C. On the other hand, from the maximality of the cycle C it follows that for each i ∈ [1,m]
ifxiy1∈ D, thenA(y2→ {xi+1,xi+2}) = ∅.
Therefore y2dose not dominate at least id(y1) + 3 vertices, a contradiction. Claim 2 is proved.
Claim 3. At least two distinct vertices of C are adjacent with some vertices of B.
Proof. Assume that Claim 3 is not true. Then exactly one vertex, say x, of C is adjacent with some
vertices of B. Hence for each vertex xi∈ C \ {x} and for each vertex y ∈ B we have
d(xi) = d(xi,C) ≤ 2m − 2andd(y) = d(y,B) + d(y,x) ≤ 2p − 2m.
Since d(xi)+d(y) ≥ 2p−2, we conclude that the inequalities above are equalities. This implies that the
subdigraphs ?C? and ?B ∪{x}? are complete. From d(xi) = 2m−2 ≥ p−1 and d(y) = 2p−2m ≥ p−1,
we obtain that p = 2m − 1. Therefore G ≡ [(Km−1∪ Km−1) + K1]∗, which contradicts our supposition.
This proves Claim 3.
Since D is strong, then A(C → B) ?= ∅ and A(B → C) ?= ∅. Together with Claim 3 this implies that
there are vertices xa?= xb, xa,xb∈ C and x,y ∈ B such that xax, yxb∈ D and
A({xa+1,xa+2,...,xb−1},B) = ∅, ifxb?= xa+1.(3)
To be definite, assume that xb:= x1and xa:= xm−h(0 ≤ h ≤ m−2). We consider the following two cases.
Case 1. xm−h+1?= x1(i.e., h ≥ 1).
Consider the paths P0, P1, ..., Pk (0 ≤ k ≤ h and k is as maximum as possible), where P :=
P0 := x1x2...xm−h and the path Pi, i ∈ [1,k], is extended from the path Pi−1 with a vertex zi ∈
{xm−h+1,xm−h+2,...,xm} \ {z1,z2,...,zi−1}. Note that the path Pi, i ∈ [0,k], contains m − h + i
vertices. It follows that some vertices y1,y2,...,yd∈ {xm−h+1,xm−h+2,...,xm}, where 1 ≤ d ≤ h, dose
not containing the extended path Pk. Therefore, using (3) and Lemma 2, for each z ∈ B and for each yi
we obtain
d(z) = d(z,B) + d(z,C) ≤ 2p − 2m − 2 + m − h + 1 = 2p − m − h − 1
and
d(yi) = d(yi,C) ≤ m + d − 1.
Hence it is clear that
2p − 2 ≤ d(z) + d(yi) ≤ 2p + d − h − 2.
It is not difficult to see that h = d, d(z,C) = m − h + 1, d(yi,C) = m + h − 1 and the subdi-
graphs ?B? and ?{xm−h+1,xm−h+2,...,xm}? are complete. By Lemma 2(ii), we also have xm−h →
B ∪ {xm−h+1,xm−h+2,...,xm} → x1. It is easy to see that h = |B| = p − m ≥ 2 and the path
P = x1x2...xm−h has maximum length among all paths from x1 to xm−h in subdigraph ?C? and in
subdigraph ?B ∪ {x1,x2,...,xm−h}?. Therefore by Lemma 3, there are integers l ∈ [1,m − h] and
r ∈ [1,m − h] such that
O(u,P) = {x1,x2,...,xl},I(u,P) = {xl,xl+1,...,xm−h},
O(z,P) = {x1,x2,...,xr},I(z,P) = {xr,xr+1,...,xm−h}.(4)
for all u ∈ B and for all z ∈ {xm−h+1,xm−h+2,...,xm}.
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