spanning 2-trails from degree sum conditions
TRANSCRIPT
Spanning 2-trails from degree sum
conditions
M. N. Ellingham∗
Department of Mathematics, 1326 Stevenson CenterVanderbilt University, Nashville, TN 37240, U. S. A.
Xiaoya Zha†
Department of Mathematical SciencesMiddle Tennessee State University, Murfreesboro, TN 37132, U.S.A.
Yi ZhangTillinghast–Towers Perrin
175 Powder Forest Drive, Weatogue, CT 06089, U. S. [email protected]
17th December 2001; minor revisions 3rd September 2003
This is a preprint of an article accepted for publication in the Journal of Graph Theory
c© 2004(?) (copyright owner as specified in the Journal)
Abstract
Suppose G is a simple connected n-vertex graph. Let σ3(G) denotethe minimum degree sum of three independent vertices in G (which is∞ ifG has no set of three independent vertices). A 2-trail is a trail thatuses every vertex at most twice. Spanning 2-trails generalize hamiltonpaths and cycles. We prove three main results. First, if σ3(G) ≥ n−1,then G has a spanning 2-trail, unless G ∼= K1,3. Second, if σ3(G) ≥ n,then G has either a hamilton path or a closed spanning 2-trail. Third,if G is 2-edge-connected and σ3(G) ≥ n, then G has a closed spanning2-trail, unless G ∼= K2,3 or K∗2,3 (the 6-vertex graph obtained fromK2,3 by subdividing one edge). All three results are sharp. Theseresults are related to the study of connected and 2-edge-connectedfactors, spanning k-walks, even factors, and supereulerian graphs. Inparticular, a closed spanning 2-trail may be regarded as a connected(and 2-edge-connected) even [2, 4]-factor.
∗Supported by NSF Grant DMS-0070613†Supported by NSF Grant DMS-0070430
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1 Introduction
We use Bondy and Murty’s book [2] for terminology and notation not definedhere, and consider finite simple graphs only.
A walk in a graph will be called a k-walk if each vertex is used no morethan k times (if the walk is closed, so that the same vertex occurs as both
first and last vertex, we count this as just one use of that vertex). A k-trailis a k-walk with no repeated edges. For the purposes of the current paper, a
k-walk or k-trail need not be closed, and need not be spanning. Note that aclosed spanning 1-walk or 1-trail is a hamilton cycle, and an open spanning
1-walk or 1-trail is a hamilton path. A graph with a closed spanning trailhas a spanning Eulerian subgraph, and is sometimes called supereulerian.
The degree of a vertex v in the graph G will be denoted degG(v), or just
deg(v). The set of vertices adjacent to v in G will be denoted NG(v) or justN(v). The degree sum of independent sets of vertices of given cardinality
has often been used to give sufficient conditions for the existence of hamiltonpaths or cycles, or other structures such as spanning trails. The following
notation will be useful:
σk(G) = min{Σki=1deg(vi) | {v1, v2, . . . , vk} is independent in G}.
If there are no independent sets of cardinality k, σk(G) is taken to be ∞.
Our goal in this paper is to give conditions involving σ3 for spanning 2-trails.Spanning k-trails are related to many other types of spanning subgraphs.
Some relationships are shown in Figure 1: a closed spanning k-trail is ageneralization of a hamilton cycle, and a special case of several other types
of spanning subgraphs that have been examined previously. In particular,the existence of a closed spanning k-trail is equivalent to the existence ofa connected (or 2-edge-connected) even [2, 2k]-factor. It is also equivalent
to the existence of a connected covering of the vertices of the graph byedge-disjoint cycles with each vertex appearing in at most k of the cycles.
Spanning k-trails therefore seem to be a natural subject for investigation.Degree sum results (involving σk, for some k) are common as sufficient
conditions for the existence of spanning subgraphs obeying degree and con-nectivity restrictions. The first degree sum result was a condition for hamil-
ton cycles by Ore, generalizing a well known result of Dirac based on theminimum degree.
Theorem A (Ore [17]) If G is an n-vertex graph with n ≥ 3 and σ2(G) ≥n, then G has a hamilton cycle.
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hamilton cycle
closed spanningk-trail
closed spanningk-walk
spanning trail(supereulerian)
even[2, b]-factor
2-edge-conn.[2, b]-factor
connected[2, b]-factor
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Figure 1: Hierarchy of spanning subgraphs
Corollary B If G is an n-vertex graph with σ2(G) ≥ n − 1, then G has ahamilton path.
A result on k-walks that generalizes Theorem A was found by Jackson
and Wormald.
Theorem C (Jackson and Wormald [12]) If G is a connected n-vertex
graph with n ≥ 3 and σk+1(G) ≥ n, where k ≥ 1, then G has a closedspanning k-walk.
For trails, which are the subject of this paper, there are several results
involving σk that guarantee the existence of a spanning trail. However, noneof these give a bound on how often each vertex is used. We begin with two
results for σ2.
Theorem D (Lesniak-Foster and Williamson [15]) If G is an n-vertex
graph with n ≥ 3, δ(G) ≥ 2, and σ2(G) ≥ n − 1, then G has a closedspanning trail.
Since a graph with σ2 ≥ n − 1 has δ ≥ 2 if and only if it is 2-edge-connected, the following is a strengthening of Theorem D.
Theorem E (Benhocine et al. [1]) If G is a 2-edge-connected n-vertex
graph with n ≥ 3 and σ2(G) ≥ (2n + 3)/3, then G has a closed spanningtrail.
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These results have been improved further by Catlin [3] and Chen [5],
provided n is taken to be sufficiently large.The above results deal with closed trails. If the trails may be open, they
are easier to find. Lesniak-Foster and Williamson [15] mention that if G is aconnected n-vertex graph with n ≥ 5 and σ2(G) ≥ n− 2, then G contains a
(possibly open) spanning trail. The following result involving σ3 strengthensthis.
Theorem F (Veldman [19]) If G is a connected n-vertex graph with n ≥ 5
and σ3(G) ≥ n− 1, then G has a (possibly open) spanning trail.
This has been further extended, as follows.
Theorem G (Z. Q. Chen and Y. F. Xue, cited in [6]) If G is a connected
n-vertex graph with σ3(G) ≥ n − 2, then either G has a spanning trail(possibly open), or G is isomorphic to one of K1,4, C++
4 (the 6-vertex graph
obtained by attaching pendant edges to opposite vertices of a 4-cycle), or Anwith n ≥ 4 (see the end of Section 2).
Catlin [4] showed that some stronger conclusions could be reached ifσ3 ≥ n + 1.
Our results in this paper strengthen Theorems E and F, and we obtain
results for 2-trails that parallel Theorem A and Corollary B. Our maintheorems, Theorems 2.6, 3.3 and 4.1, are all sharp. With the exception of
results by Gao and Wormald [11] for triangulations of the plane, projectiveplane, torus and Klein bottle, these are the first results that we are aware
of that deal with spanning k-trails for k ≥ 2.As mentioned earlier, spanning k-trails are related to connected factors
with degree conditions, and also to coverings of the vertices of a graphby cycles. Recent conditions based on degree sums for the existence of
connected factors with degree restrictions include conditions involving σ2 byKouider and Maheo [14] for 2-edge-connected [2, b]-factors, and by Nam [16]for connected [a, b]-factors. Conditions based on degree sums for coverings
of the vertices of a graph by cycles, edges and vertices include work byEnomoto et al. [10] which was extended by Kouider and Lonc [13] and
Saito [18]. Our results on closed 2-trails, Theorems 3.3 and 4.1, can beinterpreted as results on the existence of connected even [2, 4]-factors, or as
results on the existence of connected coverings of the vertices of a graph byedge-disjoint cycles with each vertex in at most 2 of the cycles.
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2 Open 2-trails
We begin by examining the existence of open spanning 2-trails in graphs.The following assumption will be used heavily in this section, and also in
Section 3.
Assumption 2.1 Suppose G is a connected n-vertex graph, and let T =u1u2 . . . up be a 2-trail in G that has the minimum number of edges among
all 2-trails covering the maximum number of vertices.
Lemma 2.2 Suppose Assumption 2.1 holds. Then T is open, i.e., u1 6= up,and its ends u1 and up are used only once.
Proof. Suppose that u1 = uj for some j ≥ 2 (possibly j = p). Then
u2u3 . . . up is a 2-trail covering the same vertices as T but with fewer edges,contradicting Assumption 2.1. Thus, u1 6= up, and u1 is used only once.
Similarly, up is used only once.
Lemma 2.3 Suppose Assumption 2.1 holds. If u ∈ V (T ) is the initial
vertex of an open 2-trail that covers V (T ), then N(u) ⊆ V (T ). In particular,N(u1) ⊆ V (T ) and N(up) ⊆ V (T ).
Proof. Suppose there exists w ∈ N(u) \ V (T ). Let (u = v1)v2v3 . . . vq be
the open 2-trail starting at u and covering V (T ). Since T has the maximumnumber of vertices, v1, v2, . . . , vq ∈ V (T ). Then wv1v2 . . . vq is an open 2-
trail covering more vertices than T , contradicting Assumption 2.1.
Define
N−(u1) ={ui−1 | u1ui ∈ E(G)},N+(up) ={ui+1 | upui ∈ E(G)}.
These are both well-defined since neither u1 nor up is adjacent to itself. Forany vertex v and i = 1 or 2, let Ni(v) be the set of neighbours of v that are
used i times on T .
Lemma 2.4 Suppose Assumption 2.1 holds. If uk ∈ N−(u1) or N+(up)then uk is used only once on T , and N(uk) ⊆ V (T ).
Proof. If uk ∈ N−(u1), then T ′ = ukuk−1 . . .u1uk+1 . . . up is a 2-trail from
uk to up with the same number of vertices and edges as T . Therefore, T ′
satisfies Assumption 2.1. By Lemma 2.2, uk is used only once on T ′ and
hence only once on T , and by Lemma 2.3, N(uk) ⊆ V (T ′) = V (T ). Theargument when uk ∈ N+(up) is similar.
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Lemma 2.5 Suppose Assumption 2.1 holds. Then |N−(u1)| = |N(u1)| +|N2(u1)|, and |N+(up)| = |N(up)|+ |N2(up)|.
Proof. |N(u1)|+ |N2(u1)| is the number of indices i for which ui ∈ N(u1).For each such i, ui−1 ∈ N−(u1), and by Lemma 2.4 this counts every vertex
in N−(u1) exactly once. The proof for |N+(up)| is similar.
The main result of this section, which strengthens Theorem F by addingthe restriction that the spanning trail uses every vertex at most twice, is as
follows.
Theorem 2.6 Let G be a connected n-vertex graph with σ3(G) ≥ n − 1.Then either G has a (possibly open) spanning 2-trail, or G ∼= K1,3.
Proof. Suppose G does not have a spanning 2-trail. Let T satisfy Assump-tion 2.1. Since G is connected, there exists a vertex w ∈ V (G) \ V (T )
adjacent to a vertex of T .
Claim 2.6A There is no closed 2-trail T ′ with V (T ′) = V (T ). In particular,u1 and up are nonadjacent.
Proof. Write T ′ = v1v2 . . . vq, with vq = v1. Since V (T ′) = V (T ), w is
adjacent to some vi, i ≤ q − 1. Then vi+1vi+2 . . . vq−1(vq = v1)v2 . . .viw is a2-trail covering more vertices than T ′ and T , contradicting Assumption 2.1.
By Lemmas 2.2 and 2.3 and Claim 2.6A, u1, up and w are distinctnonadjacent vertices, so deg(w) + deg(u1) + deg(up) ≥ n − 1. Let C =
N−(u1)∪N+(up) ∪N(w).
Claim 2.6B If ui ∈ N−(u1), uj ∈ N+(up), i < j, and either i 6= 1 or j 6= p,then there exists k with i < k < j for which at least one of the followingholds:
(i) uk is used only once on T , and uk ∈ V (T ) \ C, or(ii) uk ∈ N2(u1) and uk is the second occurrence of this vertex on T , or
(iii) uk ∈ N2(up) and uk is the first occurrence of this vertex on T .
Proof. We use induction on j − i = d.Suppose d = 1. If i = 1 or j = p, then u1up ∈ E(G). If i > 1 and j < p,
then u1ujuj+1 . . .upuiui−1 . . . u1 is a closed 2-trail. In either case, Claim2.6A is contradicted. So d cannot be 1, and the result holds vacuously in
this case.Now assume the result holds when j− i < d, and consider the case where
j−i = d ≥ 2. Since i 6= 1 or j 6= p, we may assume without loss of generalitythat i 6= 1.
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Suppose ui+1 is used once on T . If ui+1 ∈ N−(u1) or N+(up) then we
may use induction. If ui+1 ∈ N(w) then since i+ 1 6= 2, wui+1u1u2 . . .up isa 2-trail covering more vertices than T , contradicting Assumption 2.1. So,
ui+1 /∈ C, and (i) holds.Suppose ui+1 is used twice on T . Then ui+1 ∈ N2(u1). If d = 2, then
ui+1 = uj−1 ∈ N(up), and either (ii) or (iii) holds depending on whether ui+1
is the second or first occurrence of this vertex, respectively. So, assume that
d ≥ 3. If ui+1 is being used for the second time, then (ii) holds, so supposeui+1 = ul where i+1 < l. Then T ′ = ui+2ui+3 . . . (ul = ui+1)u1u2 . . . (ui+1 =
ul)ul+1 . . .up is a 2-trail covering V (T ) with the same number of edges as T ,so Assumption 2.1 holds for T ′. By Lemma 2.2, ui+2 is used only once onT ′, and hence only once on T . By Lemma 2.3, N(ui+2) ⊆ V (T ′) = V (T ),
so ui+2 /∈ N(w), and if ui+2 ∈ N−(u1) or N+(up) we may use induction.Therefore, we may assume that ui+2 /∈ C, and (i) holds.
Claim 2.6C If either deg(u1) ≥ 2 or deg(up) ≥ 2, then |N2(u1)|+|N2(up)|+|V (T ) \ C| ≥ |N−(u1)∩N+(up)|+ 1.
Proof. Write N−(u1) ∩N+(up) = {ui1 , ui2, . . . , uim}, where 1 < i1 < i2 <. . . < im < p and m = |N−(u1) ∩ N+(up)|. If m ≥ 1, then applying Claim
2.6B to consecutive pairs of vertices in the sequence u1, ui1, ui2 , . . . , uim, upwe find m + 1 distinct indices k satisfying (i), (ii) or (iii) of Claim 2.6B, of
which there are at most |V (T ) \ C|, |N2(u1)|, and |N2(up)|, respectively.If m = 0 and deg(u1) ≥ 2, we apply Claim 2.6B to the last vertex ui of
N−(u1) on T and uj = up, and we use a similar argument if deg(up) ≥ 2.
Now we may complete the proof of Theorem 2.6. Suppose deg(u1) = 1
and deg(up) = 1. Then deg(w) ≥ n − 3, but w is not adjacent to u1, up oritself, so deg(w) = n− 3, and w is adjacent to all other vertices of G. Sincew has a neighbor on T , p ≥ 3. If p ≥ 4, then w is adjacent to u2 and u3,
and u1u2wu3u4 . . . up is a 2-trail with more vertices than T , a contradiction.Therefore p = 3 and N(w)∩ V (T ) = {u2}. If w has a neighbor x not on T ,
then u1u2wx is a 2-trail with more vertices than T , so N(w) = {u2}. Hence,1 = deg(w) = n− 3, n = 4, and G ∼= K1,3.
Now suppose that deg(u1) ≥ 2 or deg(up) ≥ 2. Then
|C| = |N(w)|+ |N−(u1) ∪N+(up)| by Lemma 2.4
= |N(w)|+ |N−(u1)|+ |N+(up)| − |N−(u1)∩N+(up)|= |N(w)|+ |N(u1)|+ |N2(u1)|+ |N(up)|+ |N2(up)|
−|N−(u1)∩N+(up)| by Lemma 2.5
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≥ n − 1 + |N2(u1)|+ |N2(up)| − |N−(u1) ∩N+(up)|since σ3 ≥ n− 1
≥ n − |V (T ) \ C| by Claim 2.6C
and so |C| + |V (T ) \ C| ≥ n, which is impossible because C and V (T ) \ Care disjoint and w belongs to neither.
This concludes the proof of Theorem 2.6.
The condition σ3(G) ≥ n − 1 in Theorem 2.6 is best possible. Forn ≥ 4, let An be the graph obtained from Kn−3 by choosing a vertex v,
then adding new vertices w, x1, x2 and edges vw, wx1, wx2. These graphsappear in Theorem G, above. For n ≥ 5, An has σ3 = n−2 and no spanningtrail of any kind. Moreover, K1,3, the exceptional graph in Theorem 2.6, is
just A4. As another family of examples, for m ≥ 1 the graph Km,2m+2 hasσ3 = 3m = n − 2 but no spanning 2-trail. Note also that the hypothesis
that G is connected is necessary; it is easy to construct disconnected graphswith σ3 ≥ n− 1.
We also have the following straightforward corollary, which is sharp, atleast when n ≡ 2 (mod 3), as shown by Km,2m+2. As usual, δ(G) denotes
the minimum degree of G, and we use the facts that σ3(G) ≥ 3δ(G) andσ3(G) ≥ 3σ2(G)/2.
Corollary 2.7 Let G be a connected n-vertex graph. If δ(G) ≥ (n − 1)/3
or σ2(G) ≥ 2(n− 1)/3 then G has a (possibly open) spanning 2-trail.
3 Closed 2-trails and hamilton paths
Now we consider the situation where we have a spanning 2-trail, and try to
find a closed spanning 2-trail. The following fact will be used frequently.
Observation 3.1 If u1u2 . . . up is a trail and uk = ul with k < l, then
l ≥ k + 3.
Lemma 3.2 Suppose Assumption 2.1 holds, and uk = ul with k < l.
(i) Suppose k ≤ i ≤ l, and u1ui or upui is in E(G) \ E(T ). If i ≥ k + 1then ui−1 is used only once by T , and if i ≤ l − 1 then ui+1 is used
only once by T .(ii) None of u1uk+1, upuk+1, u1ul−1 or upul−1 is in E(G) \E(T ).
(iii) uk = ul, uk−1, uk+1, ul−1, and ul+1 all exist and are distinct vertices.
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Proof. (i) Suppose k + 1 ≤ i ≤ l, and upui ∈ E(G) \ E(T ). Then there is
a spanning 2-trail T ′ = ui−1ui−2 . . .uk+1(uk = ul)ul+1 . . . upuiui+1 . . .ul−1
(ul = uk)uk−1 . . .u1 with the same number of edges as T . Hence, by Lemma
2.2, T ′, and hence T , uses ui−1 only once. The other cases are similar.
(ii) If one of the given edges is in E(G) \ E(T ), then by (i) uk = ul would
be used only once by T , a contradiction.
(iii) Note that k ≥ 2 and l ≤ p− 1 by Lemma 2.2, so all these vertices exist.All of uk−1, uk+1, ul−1, ul+1 are joined to uk = ul by distinct edges of T , soall five vertices are distinct.
Given a trail T = v1v2 . . . vk, define ρmin(T ) to be the position of the
first repeated vertex, i.e., the smallest i such that vi = vj for some j 6= i, or∞ if T has no repeated vertices. Similarly, define ρmax(T ) to be the largest
i such that vi = vj for some j 6= i, or −∞ if T has no repeated vertices.The main result of this section is the following. Note that the proof
requires that we take care to distinguish between indices of vertices along Tand the vertices themselves, and between elements of E(G) \ E(T ) and ofE(T ).
Theorem 3.3 Let G be a connected n-vertex graph with σ3(G) ≥ n. Thenat least one of the following holds.
(i) G has a hamilton path.(ii) G has a closed spanning 2-trail.
Proof. Suppose G has neither a hamilton path nor a closed spanning 2-trail.Let T = u1u2 . . . up be a spanning 2-trail with (1) fewest edges; (2) subject
to (1), smallest ρmin(T ); and (3) subject to (1) and (2), largest ρmax(T ).Assumption 2.1 holds for T . Since G has no hamilton path, q = ρmin(T ) is
finite, and uq = ut for some t > q.If u1 and up are adjacent, then there is a closed spanning 2-trail, so they
are nonadjacent. We begin by finding a vertex ur+1 which will form thethird vertex of an independent set with u1 and up.
Claim 3.3A We may assume (possibly after a reordering of the verticesand edges of T that does not change ρmin(T ) or ρmax(T )) that there exists
r, 2 ≤ r ≤ p− 4, such that(a) ur = us for some s, r+ 3 ≤ s ≤ p− 1.
(b) r = q or q + 1, so that each of u1, . . . , ur−2 is used only once on T .(c) Either ur−1 is used only once on T , or u1ur and upur are not in E(G)\
E(T ), or both.(d) ur+1 6= u2 and ur+1 6= up−1.
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(e) There is no i ≥ r + 1 such that ur+1ui+1 ∈ E(T ) and at least one of
u1ui or upui is in E(G) \E(T ).
Proof. Take r = q (and s = t). This satisfies (a)–(c) above, since u1, . . . , uq−1
are used only once on T . If uq+1 = u2 then u2 is used twice, so q = 2 giving
uq+1 = u3 = u2, which is impossible. Thus, uq+1 6= u2. If uq+1 = up−1
then, by Lemma 3.2(iii), ut−1 6= uq+1 = up−1. Therefore, by reversing the
segment uquq+1 . . . ut−1ut of T , we may assume that uq+1 6= up−1, and (d)also holds. We shall show that either (e) holds, or we can make another
choice of r.In our arguments we shall suppose that (e) does not hold. In doing this,
we focus on the vertex ui+1 as described in (e). We say ‘Suppose x = ui+1 as
in (e)’ to mean ‘Suppose there is i ≥ r+1 so that x = ui+1, ur+1ui+1 ∈ E(T )and at least one of u1ui or upui is in E(G) \E(T ).’
Suppose uq = ut = ui+1 as in (e). Since i ≥ r + 1 = q + 1, i = t − 1,which contradicts Lemma 3.2(ii) for uq . . . ut.
Suppose uq+2 = ui+1 as in (e). If i = q + 1 then Lemma 3.2(ii) foruq . . .ut is contradicted. So uq+2 = ui+1 for some i 6= q+ 1. By Observation
3.1, either i+ 1 ≤ q − 1 or i + 1 ≥ q + 5. However, the former contradictsq = ρmin(T ), so i + 1 ≥ q + 4 and in particular i+ 1 > q + 2. Now u1ui or
upui contradicts Lemma 3.2(ii) for uq+2 . . .ui+1.If uq+1 is used only once by T , we are finished. So, assume that uq+1 is
used twice, with uq+1 = ul, where l ≥ q + 4 since q = ρmin(T ).
Suppose ul+1 = ui+1 as in (e). If i = l then u1ul = u1uq+1 or upul =upuq+1 contradicts Lemma 3.2(ii) for uq . . . ut. Therefore, ul+1 = ui+1 where
i 6= l. If i > l then u1ui or upui contradicts Lemma 3.2(ii) for ul+1 . . . ui+1,so i < l. Since q = ρmin(T ), i+ 1 ≥ q. By Lemma 3.2(iii), ul = uq+1, uq and
ul+1 = ui+1 are distinct. Therefore, i + 1 6= q or q + 1, so i+ 1 ≥ q + 2, ori ≥ q + 1. Now applying Lemma 3.2(i) to ui in uq+1 . . . ul, we see that ui+1
is used only once by T , a contradiction.Suppose ul−1 = ui+1 as in (e). If i = l − 2 then we choose a new r
(see below). So, suppose i 6= l − 2. Then ul−1 = ui+1 where l − 1 6= i + 1.By an argument similar to the one for ul+1 = ui+1, above, we obtain acontradiction to Lemma 3.2(i).
So, we satisfy conditions (a)–(e) with r = q, unless uq+1 = ul whereu1ul−2 or upul−2 is in E(G) \ E(T ). In that case, reorder T as T ′ =
u1u2 . . . uq(uq+1 = ul)ul−1 . . . uq+2(uq+1 = ul)ul+1 . . .up = v1 . . .vp, andtake r = q + 1. Then vr = uq+1 = ul, and neither v1vr = u1uq+1 nor
vpvr = upuq+1 are in E(G) \ E(T ) by Lemma 3.2(ii) on uq . . . ut. Thus,(a)–(c) are satisfied. Note that by Lemma 3.2(i) for ul−2 on uq+1 . . . ul we
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have that T uses ul−1 = vr+1 only once, so T ′ also uses vr+1 only once. For
(d), since 4 ≤ r + 1 ≤ p− 2 and vr+1 is used only once, vr+1 6= v2 or vp−1.For (e), since vr+1 = ul−1 is used only once, we need only show that neither
vr = vl = ul = uq+1 nor vr+2 = ul−2 can be vi+1 as in (e).Suppose vr = vl = vi+1 as in (e) (applied to T ′). Since i ≥ r+1, i = l−1,
which contradicts Lemma 3.2(ii) for vr . . . vl.Suppose vr+2 = vi+1 as in (e). If i = r+1 then Lemma 3.2(ii) for vr . . . vl
is contradicted. So vr+2 = vi+1 for some i 6= r+ 1. Either i+ 1 ≤ r− 1 = qor i + 1 ≥ r + 5. Since q = ρmin(T ), i + 1 ≥ q. If i+ 1 = q = r − 1, then
by Lemma 3.2(i) applied to vr−1 = vr+2 on vr−1 . . . vr+2, vr = uq+1 is usedonly once on T ′, and hence on T , a contradiction. Thus, i+ 1 ≥ r + 5 andin particular i+ 1 > r + 2. Now v1vi or vpvi contradicts Lemma 3.2(ii) for
vr+2 . . .vi+1.The two reorderings of T mentioned above do not change ρmin(T ) or
ρmax(T ), so this completes the proof of the claim.
Now ur+1 is not adjacent to u1 or up by an element of E(T ), by Claim3.3A(d), or by an element of E(G) \ E(T ), by Lemma 3.2(ii) for ur . . . us.
Therefore, {u1, ur+1, up} is an independent set.For an arbitrary v, define the following sets of integers:
I(v) = {i | i ≤ r and vui ∈ E(G) \E(T )},J(v) = {i | i ≥ r + 1 and vui ∈ E(G) \E(T )},
Observe that
I(u1) ⊆ {3, 4, . . . , r}, J(u1) ⊆ {r+ 1, r+ 2, . . . , p− 1},I(up) ⊆ {2, 3, . . . , r}, J(up) ⊆ {r+ 1, r+ 2, . . . , p− 2}.
Therefore, for v = u1 or up, it makes sense to define
A−(v) = {ui−1 | i ∈ I(v)},B+(v) = {ui+1 | i ∈ J(v)}.
Then, let
C = N(ur+1)∪ A−(u1) ∪A−(up) ∪B+(u1) ∪B+(up).
Claim 3.3B N(ur+1), A−(u1)∪A−(up), and B+(u1)∪B+(up) are pairwise
disjoint.
Proof. Suppose first that v is in both A−(u1)∪A−(up) and B+(u1)∪B+(up).Then v = ui−1 = uj+1 where i ∈ I(u1) ∪ I(up), j ∈ J(u1) ∪ J(up), so that
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i− 1 <= r < j + 1. Then the edge u1ui or upui contradicts Lemma 3.2(ii)
on ui−1 . . . uj+1.Now suppose that v is in both B+(u1) ∪ B+(up) and N(ur+1). Write
v = ui+1 where i ∈ J(u1) ∪ J(up), so that i ≥ r + 1. By Claim 3.3A(e),ur+1ui+1 /∈ E(T ). Since ur+1us ∈ E(T ), i 6= s − 1. We show that in all
cases a spanning 2-trail T ′ with fewer edges than T can be found, giving acontradiction. If r + 1 ≤ i ≤ s− 2 and u1ui ∈ E(G) \E(T ), let
T ′ = upup−1 . . . us+1(us = ur)ur−1 . . . u1uiui−1 . . .ur+1ui+1ui+2 . . . us−1.
A similar trail T ′ can be found if r+ 1 ≤ i ≤ s− 2 and upui ∈ E(G)\E(T ).If i ≥ s and u1ui ∈ E(G) \E(T ), let
T ′ = upup−1 . . . ui+1ur+1ur+2 . . . us . . .uiu1u2 . . . ur−1.
If i ≥ s and upui ∈ E(G) \E(T ), let
T ′ = u1u2 . . . ur−1(ur = us)us+1 . . . uiupup−1 . . . ui+1ur+1ur+2 . . . us−1.
Now suppose that v is in both A−(u1) ∪ A−(up) and N(ur+1). Writev = ui−1, where i ∈ I(u1) ∪ I(up), so that i ≤ r. If ur+1ui−1 ∈ E(T ), then
ui−1 must be used twice on T , so by Claim 3.3A(b), i = r. However, sinceui−1 = ur−1 is used twice, by Claim 3.3A(c) i = r /∈ I(u1) ∪ I(up), which is
a contradiction. Thus, ur+1ui−1 ∈ E(G) \ E(T ), and the result follows byarguments similar to those of the previous paragraph for the case i ≥ s.
This completes the proof of the claim.
Claim 3.3C (a) The map i 7→ ui−1 provides bijections I(u1) → A−(u1),I(up)→ A−(up), and I(u1)∩ I(up)→ A−(u1) ∩A−(up).
(b) The map i 7→ ui+1 provides bijections J(u1) → B+(u1), J(up) →B+(up), and J(u1) ∩ J(up)→ B+(u1) ∩B+(up).
Proof. The arguments for I(u1) → A−(u1), I(up) → A−(up), J(u1) →B+(u1) and J(up)→ B+(up) are similar; we prove the third. By definitionof B+(u1), i 7→ ui+1 maps J(u1) onto B+(u1). We prove it is also one-to-one. If not, then there are i, j ∈ J(u1) with i < j such that ui+1 = uj+1.
But now u1uj ∈ E(G) \E(T ) contradicts Lemma 3.2(ii) for ui+1 . . .uj+1.The arguments for I(u1)∩I(up)→ A−(u1)∩A−(up) and J(u1)∩J(up)→
B+(u1) ∩B+(up) are similar; we prove the latter. The map i 7→ ui+1 mapsJ(u1)∩J(up) into B+(u1)∩B+(up), but we must prove that this map is one-
to-one and onto. It is one-to-one because it is a restriction of the one-to-onemap J(u1)→ B+(u1). To show it is onto, consider any v ∈ B+(u1)∩B+(up).
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Since v ∈ B+(u1), v = ui+1 where i ∈ J(u1), and since v ∈ B+(up),
v = uj+1 where j ∈ J(up). If i < j then upuj ∈ E(G) \ E(T ) contradictsLemma 3.2(ii) for ui+1 . . . uj+1, and if i > j we get a similar contradiction.
Therefore i = j ∈ J(u1) ∩ J(up) maps to v = ui+1.
We now modify our definitions of N(v) and Ni(v) to get N∗(v) and
N∗i (v), where N∗(v) = {w | vw ∈ E(G) \ E(T )}, and N∗i (v) contains thevertices of N∗(v) that are used i times on T , i = 1 or 2. Thus, N∗(u1) =
N(u1) \ {u2}, and N∗(up) = N(up) \ {up−1}.Claim 3.3D If v = u1 or up then |I(v)|+ |J(v)| = |N(v)|+ |N∗2 (v)| − 1,
Proof. In both cases |N∗(v)| = |N(v)| − 1, so
|I(v)|+ |J(v)| = |{i | vui ∈ E(G) \E(T )}|= |N∗1 (v)|+ 2|N∗2 (v)| = |N∗(v)|+ |N∗2 (v)|= |N(v)|+ |N∗2 (v)| − 1.
Claim 3.3E (a) If i ∈ (I(u1)∩ I(up))∪ (J(u1)∩J(up)) then ui ∈ N∗2 (u1)∩N∗2 (up).
(b) |I(u1)∩ I(up)|+ |J(u1) ∩ J(up)| = 2|N∗2 (u1) ∩N∗2 (up)|.Proof. (a) By definition, ui ∈ N∗(u1)∩N∗(up). If ui is not used twice, then
u1u2 . . . upuiu1 is a closed spanning 2-trail, which is a contradiction.
(b) Each element of N∗2 (u1)∩N∗2(up) has the form ui for exactly two numbersi ∈ (I(u1) ∩ I(up)) ∪ (J(u1) ∩ J(up)). By (a), these numbers i include all
elements of (I(u1) ∩ I(up))∪ (J(u1) ∩ J(up)).
Claim 3.3F ur+1 /∈ C.
Proof. Clearly ur+1 /∈ N(ur+1). If ur+1 ∈ A−(u1) ∪ A−(up), then ur+1 =ui−1 where i ≤ r. By Claim 3.3A(b), we must have i − 1 = r − 1, which
contradicts Observation 3.1. If ur+1 ∈ B+(u1) ∪ B+(up), then ur+1 = ui+1
where i ≥ r + 1. Then the edge u1ui or upui in E(G) \ E(T ) contradicts
Lemma 3.2(ii) for ur+1 . . . ui+1.
Claim 3.3G We have
|V (G) \ {ur+1} \ C|+ |N∗2(up) \N∗2 (u1)|+ |N∗2 (u1) \N∗2 (up)| ≤ 1.
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Proof. We have
|C| = |N(ur+1)|+ |A−(u1) ∪A−(up)|+ |B+(u1) ∪B+(up)|by Claim 3.3B
= |N(ur+1)|+ |A−(u1)|+ |A−(up)| − |A−(u1) ∩ A−(up)|+|B+(u1)|+ |B+(up)| − |B+(u1) ∩B+(up)|
= |N(ur+1)|+ |I(u1)|+ |I(up)| − |I(u1) ∩ I(up)|+|J(u1)|+ |J(up)| − |J(u1) ∩ J(up)| by Claim 3.3C
= |N(ur+1)|+ |N(u1)|+ |N(up)|+ |N∗2(u1)|+ |N∗2(up)| − 2
−|I(u1)∩ I(up)| − |J(u1) ∩ J(up)| by Claim 3.3D
≥ n+ |N∗2 (u1)|+ |N∗2 (up)|+ |I(u1)∩ I(up)| − |J(u1) ∩ J(up)| − 2
since σ3 ≥ n= n+ |N∗2 (u1)|+ |N∗2 (up)| − 2|N∗2 (u1) ∩N∗2 (up)| − 2
by Claim 3.3E
= n+ |N∗2 (up) \N∗2 (u1)|+ |N∗2 (u1) \N∗2 (up)| − 2
and the result follows since |V (G)\{ur+1}\C| = n−|C|−1, by Claim 3.3F.
Now we complete the proof of Theorem 3.3. Recall that q = ρmin(T ) ≥ 2,
and let a = ρmax(T ) ≤ p − 1, so that ua+1, ua+2, . . . , up are used onlyonce. Let α1 = |{u1, u2, . . . , uq−1} \ C|, α2 = |{ua+1, ua+2, . . . , up} \ C|,β1 = |N∗2 (up) ∩ {u2}|, and β2 = |N∗2 (u1) ∩ {up−1}|. Because u2 /∈ N∗(u1)and up−1 /∈ N∗(up), Claim 3.3G implies that α1 + α2 + β1 + β2 ≤ 1.
Suppose α1 + β1 = 0. Then α1 = 0, so u1, u2, . . . , uq−1 ∈ C. Sinceu1 ∈ C but u1 /∈ N(ur+1), A−(u1), B+(u1) or B+(up), we get u1 ∈ A−(up),
and hence u2 ∈ N∗(up). Since β1 = 0, u2 must be used only once on T .Hence q ≥ 3. We shall prove that u1, u2, . . . , uq−2 ∈ A−(up). If q = 3we are finished. Suppose that 2 ≤ j ≤ q − 2, and we have proved that
u1, . . . , uj−1 ∈ A−(up). Since uj−1 ∈ A−(up), uj ∈ N∗(up). We know uj isin C, but uj /∈ B+(u1) ∩ B+(up). If uj ∈ A−(u1) then uj+1 ∈ N∗(u1) and
hence
T ′ = u1u2 . . . ujupup−1 . . . uj+1u1
is a closed 2-trail, a contradiction. If uj ∈ N(ur+1) then
T ′ = u1u2 . . .ur−1(ur = us)us+1 . . .upujur+1ur+2 . . . us−2
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has ρmin(T ′) = j < q, contradicting our choice of T . Therefore uj ∈ A−(up).
Repeating this argument, we get uq−2 ∈ A−(up) which means that uq−1 ∈N∗(up). But then, since uq = ut with t > q,
T ′ = u1u2 . . . uq−1upup−1 . . . utut−1 . . . uq+1
has fewer edges than T , a contradiction.Suppose now that α1 + β1 = 1. Then α2 + β2 = 0. By reasoning
symmetric to the above we can show that up ∈ B+(u1), so that up−1 ∈N∗(u1), and up−1 is used only once, so that a = ρmax(T ) ≤ p − 2. Ifa = p− 2, we can use an argument symmetric to the one above, so suppose
that a ≤ p − 3. We know that up−1 ∈ C. By an argument symmetric tothe one above, up−1 /∈ B+(up). Suppose that up−1 ∈ N(ur+1). Consider the
trail
T ′ = ur−1ur−2 . . . u1up−1ur+1ur+2 . . . us . . .up−1up.
Considering (T ′)−1, the reverse of T ′, we have ρmin((T ′)−1) = 2 and thereforeq = ρmin(T ) = 2 by choice of T . Also, by Lemma 2.2, ur−1 is used only
once on T ′ and hence only once on T , so by Claim 3.3A(b), r = q = 2.Now we see that ρmin(T ′) = 2 = ρmin(T ) and ρmax(T ′) = p− 1 > ρmax(T ).
This contradicts our choice of T . Therefore up−1 ∈ B+(u1), so that up−2 ∈N∗(u1). But then the trail
T ′ = u2u3 . . . up−2u1up−1up
has ρmin(T ′) = q − 1 < ρmin(T ), a contradiction.This completes the proof of Theorem 3.3.
The condition σ3(G) ≥ n in Theorem 3.3 is best possible. For alln1, n2, n3 ≥ 1 the graph K1 + (Kn1 ∪ Kn2 ∪ Kn3) (‘+’ denotes join) hasσ3 = n1 +n2 +n3 = n−1 but neither a hamilton path nor a closed spanning
2-trail. These graphs also appear as sharpness examples in [10]. Also, forevery m ≥ 1 the graph Km,2m+1 has σ3 = 3m = n−1 but neither a hamilton
path nor a closed spanning 2-trail. Note also that the hypothesis that Gis connected is necessary; it is easy to construct disconnected graphs with
σ3 ≥ n.We also have the following straightforward corollary, which is sharp, at
least when n ≡ 1 (mod 3), as shown by Km,2m+1.
Corollary 3.4 Let G be a connected n-vertex graph. If δ(G) ≥ n/3 or
σ2(G) ≥ 2n/3 then G has either a hamilton path or a closed spanning 2-trail.
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It is also interesting to note the following, which follows from Theorem
3.3 because a graph with a cutedge cannot have a closed spanning trail.
Corollary 3.5 If G is an n-vertex graph with a cutedge and σ3(G) ≥ n then
G has a hamilton path.
4 Closed 2-trails in 2-edge-connected graphs
In this section we prove our third main result, as follows.
Theorem 4.1 Let G be a 2-edge-connected n-vertex graph with σ3(G) ≥ n.
Then either(i) G is isomorphic to K2,3, or the 6-vertex graph K∗2,3 obtained by subdi-
viding an edge of K2,3, or(ii) G has a closed spanning 2-trail.
Proof. The result holds for n = 1 (whether or not one considers K1 tobe 2-edge-connected), so suppose that n ≥ 2. Assume that G does not
have a closed spanning 2-trail. By Theorem 3.3, G has a hamilton pathT = u1u2 . . .up, where now p = n, and u1up /∈ E(G). Since G is 2-edge-
connected, deg(u1), deg(up) ≥ 2, so that N∗(u1), N∗(up) 6= ∅, and n ≥ 4.Define
η(T ) = min{j − i | ui ∈ N∗(u1), uj ∈ N∗(up)},
which may be positive, zero or negative.
Claim 4.1A There exists a hamilton path T with η(T ) ≤ 0.
Proof. Suppose not. Let T be a hamilton path with η(T ) as small aspossible. Then there exist q, t with t− q = η(T ) > 0 such that uq ∈ N∗(u1),
q ≥ 3, N(u1) ⊆ {u2, u3, . . . , uq}, ut ∈ N∗(up), t ≤ p − 2, and N(up) ⊆{ut, ut+1, . . .up−1}.
Suppose first that there exists some r ≤ q− 1 such that urus ∈ E(G) for
some s ≥ q+ 1. Then r ≥ 2. Choose r as large as possible. If r = q− 1 then
T ′ = us−1us−2 . . .uqu1u2 . . . uq−1usus+1 . . . up
has η(T ′) ≤ t− s < η(T ), a contradiction. So r ≤ q − 2. Define
A−(u1) = {ui−1 | i ≤ r, u1ui ∈ E(G)},B+(u1) = {ui+1 | i ≥ r + 1, u1ui ∈ E(G)}.
Clearly |A−(u1)|+ |B+(u1)| = |N(u1)|.
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Now we claim that A−(u1), B+(u1), N(ur+1) and N+(up) are pairwise
disjoint. Since A−(u1) ⊆ {u1, . . .ur−1}, B+(u1) ⊆ {ur+2, . . .uq+1}, andN+(up) ⊆ {ut+1, . . . , up}, these three sets are disjoint; furthermore, ur+1
belongs to none of them. Therefore we need only consider intersections ofN(ur+1) with the other three sets. If there is ui−1 ∈ A−(u1)∩N(ur+1), then
T ′ = us−1us−2 . . . ur+1ui−1ui−2 . . . u1uiui+1 . . . urusus+1 . . . up
has η(T ′) ≤ t− s < η(T ), a contradiction (this works even if i = 2). If thereis ui+1 ∈ B+(u1) ∩N(ur+1), then because r + 1 ≤ q − 1 and we chose r as
large as possible, i ≤ q − 1, and
T ′ = us−1us−2 . . .ui+1ur+1ur+2 . . . uiu1u2 . . . urusus+1 . . . up
has η(T ′) ≤ t − s < η(T ), a contradiction. If there is ui+1 ∈ N+(up) ∩N(ur+1) then because r + 1 ≤ q − 1, we contradict our choice of r as largeas possible.
Since u1 ∈ A−(u1) and up ∈ N+(up), the above shows us that {u1, ur+1,up} is independent. We therefore have
n ≤ |N(u1)|+ |N(up)|+ |N(ur+1)|= |A−(u1)|+ |B+(u1)|+ |N+(up)|+ |N(ur+1)|
which is impossible, since these four sets are disjoint and ur+1 belongs to
none of them.Therefore, there is no number r as described above, and uq is a cutvertex.
Making the same argument with T reversed, which does not change η(T ),we find that ut is also a cutvertex. If t = q+ 1 then uqut is a cutedge, which
is a contradiction. Thus, t ≥ q + 2. Now {u1, uq+1, up} is independent,and it is easy to see that deg(u1) + deg(uq+1) + deg(up) ≤ n − 1, again a
contradiction.Hence, there must exist T with η(T ) ≤ 0.
Now for each hamilton path T define
θ(T ) = min{i− j | uj ∈ N∗(up), ui ∈ N∗(u1), j ≤ i},
which is finite when η(T ) ≤ 0, and ∞ (the minimum of the empty set) whenη(T ) > 0. Let T be a hamilton path with θ(T ) as small as possible. By Claim
4.1A, θ(T ) is finite, so there exist r, swith s−r = θ(T ), ur ∈ N∗(up) and us ∈N∗(u1). If θ(T ) = 0, we have a closed spanning 2-trail u1u2 . . . upur=su1, and
if θ(T ) = 1, then we have a hamilton cycle u1u2 . . .urupup−1 . . .ur+1=su1.
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So we may assume that θ(T ) ≥ 2, i.e., s ≥ r+2. None of ur+1, ur+2, . . .us−1
are adjacent to u1 or up, by definition of θ(T ). For v = u1 or up, define
A−(v) = {ui−1 | i ≤ r, vui ∈ E(G)},B+(v) = {ui+1 | i ≥ s, vui ∈ E(G)}.
so that |A−(v)|+ |B+(v)| = |N(v)|. Also, let
C = N(ur+1)∪ A−(u1) ∪A−(up) ∪B+(u1) ∪B+(up).
Claim 4.1B A−(u1), A−(up), B+(u1), B+(up) and N(ur+1) are pairwise
disjoint, except that possibly u1 ∈ A−(u1) ∩ A−(up), and possibly up ∈B+(u1) ∩B+(up).
Proof. Since A−(u1), A−(up) ⊆ {u1, . . .ur−1} and B+(u1), B+(up) ⊆ {us+1,. . . up}, the former two are disjoint from the latter two. If there is a vertex
other than u1 in A−(u1) ∩ A−(up), or other than up in B+(u1) ∩ B+(up),then there is uj ∈ N∗(u1) ∩N∗(up) and
T ′ = u1u2 . . . upuju1
is a closed 2-trail, a contradiction.
Now consider the intersections of N(ur+1) with the other sets. If thereis ui−1 ∈ A−(u1)∩N(ur+1) then
T ′ = u1 . . . ui−1ur+1ur+2 . . .upurur−1 . . .uiu1
is a hamilton cycle, a contradiction (this works even when i = 2). If thereis ui−1 ∈ A−(up) ∩N(ur+1) then
T ′ = u1u2 . . . ui−1ur+1ur+2 . . . us . . . upuiui+1 . . . ur
has θ(T ′) ≤ s − r − 1 < θ(T ) because of the edges u1us, urur+1, which is a
contradiction. If there is ui+1 ∈ B+(u1) ∩N(ur+1), then
T ′ = u1u2 . . . urupup−1 . . . ui+1ur+1ur+2 . . . uiu1
is a hamilton cycle, a contradiction. If there is ui+1 ∈ B+(up) ∩ N(ur+1)then
T ′ = u1u2 . . . uiupup−1 . . . ui+1
has θ(T ′) ≤ s− r − 1 < θ(T ) because of the edges u1us, ui+1ur+1, which isa contradiction (this works even when i = p− 1). Thus, N(ur+1) is disjoint
from the other four sets.This ends the proof of the claim.
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Claim 4.1C |V (G) \C| ≤ |A−(u1) ∩A−(up)|+ |B+(u1)∩ B+(up)|.Proof. By Claim 4.1B, N(ur+1), A−(u1) ∪ A−(up) and B+(u1) ∪ B+(up)
are pairwise disjoint, so
|C| = |N(ur+1)|+ |A−(u1) ∪ A−(up)|+ |B+(u1) ∪B+(up)|= |N(ur+1)|+ |A−(u1)|+ |A−(up)| − |A−(u1) ∩ A−(up)|
+|B+(u1)|+ |B+(up)| − |B+(u1) ∩B+(up)|= |N(ur+1)|+ |N(u1)|+ |N(up)| − |A−(u1) ∩A−(up)|
−|B+(u1) ∩ B+(up)|≥ n− |A−(u1) ∩ A−(up)| − |B+(u1) ∩B+(up)| since σ3 ≥ n
and the result follows.
Let γ = |V (G) \C|, α = |A−(u1)∩A−(up)| and β = |B+(u1)∩B+(up)|.Claim 4.1C says that γ ≤ α + β. Since ur+1 ∈ V (G) \ C, we know that
γ ≥ 1. By Claim 4.1B, α = 1 if u2up ∈ E(G)\E(T ), and 0 otherwise, whileβ = 1 if u1up−1 ∈ E(G) \E(T ), and 0 otherwise.
Claim 4.1D θ(T ) = 2, i.e., s = r + 2.
Proof. Suppose that θ(T ) ≥ 3, i.e., s ≥ r + 3. If us ∈ N(ur+1), then
T ′ = u1u2 . . . urupup−1 . . . ur+1usu1
is a closed 2-trail, a contradiction. Thus, us /∈ N(ur+1) and hence us ∈V (G) \ C and γ ≥ 2. Therefore α = β = 1, so that u2up, u1up−1 ∈ E(G) \E(T ). But now
T ′ = u3u4 . . . upu2u1
is a hamilton path with θ(T ′) ≤ 2 < θ(T ) because of the edges u3u2, u1up−1,which is a contradiction.
Claim 4.1E Suppose 1 ≤ k ≤ r− 1, uk ∈ A−(up), and uk+1, . . . , ur−1 ∈ C.
Then for k ≤ j ≤ r− 1, we have uj ∈ A−(up), and for k+ 1 ≤ j ≤ r− 1 wehave uj /∈ N(ur+1) ∪A−(u1).
Proof. We show that uj ∈ A−(up) for k ≤ j ≤ r − 1 by induction on j,
proving the rest along the way. We are given uk ∈ A−(up). Suppose nowthat k + 1 ≤ j ≤ r − 1. By induction, uj−1 ∈ A−(up), so uj ∈ N∗(up). If
uj ∈ N(ur+1), then
T ′ = u1u2 . . .ur+1ujupup−1 . . . ur+2=su1
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is a closed spanning 2-trail, a contradiction. If uj ∈ A−(u1) then
T ′ = u1u2 . . . ujupup−1 . . . uj+1u1
is a hamilton cycle, a contradiction. Therefore uj /∈ N(ur+1)∪A−(u1), but
since uj ∈ C, we must have uj ∈ A−(up).
Claim 4.1F If α = 1 and u2, . . . , ur−1 ∈ C, then(a) u1, u2, . . . , ur−1 ∈ A−(up), and consequently u2, u3, . . . , ur ∈ N∗(up);(b) N∗(u1) ⊆ {ur+2, ur+4, ur+5, . . . , up−1};(c) N∗(ur+1) ⊆ {ur+4, . . . , up−1}.
Proof. Since α = 1, we apply Claim 4.1E with k = 1.
(a) This follows immediately.
(b) Clearly u1, u2 /∈ N∗(u1). By Claim 4.1E, u2, u3, . . . , ur−1 /∈ A−(u1), sou3, . . . , ur /∈ N∗(u1). By definition of θ(T ), ur+1 /∈ N∗(u1). If ur+3 ∈ N∗(u1)
then
T ′ = u2u3 . . . ur+2u1ur+3ur+4 . . . upu21
is a hamilton cycle, a contradiction. We know that up /∈ N∗(u1).
(c) We know that u1, up /∈ N∗(ur+1). By Claim 4.1E, u2, u3, . . . , ur−1 /∈N∗(ur+1). Clearly ur, ur+1, ur+2 /∈ N∗(ur+1). If ur+3 ∈ N∗(ur+1), then
T ′ = u1u2 . . . urupup−1 . . . ur+3ur+1ur+2u1
is a hamilton cycle, a contradiction.
Now we complete the proof of Theorem 4.1. Since θ(T ) = 2, i.e., s =r + 2, we have r + 1 = s − 1. This means that symmetric arguments may
be applied from the two ends of our path T . We examine two cases.
(1) α = 0 and β = 1, or α = 1 and β = 0. By symmetry we may supposethat α = 1 and β = 0. Then γ = 1, so ur+1 is the only vertex that doesnot belong to C. By Claim 4.1F(b) and the fact that β = 0, N∗(u1) ⊆{ur+2, ur+3, . . . , up−2}, and since N∗(u1) 6= ∅, r ≤ p− 4.
Suppose that r ≤ p− 5.
First, we show that ur+4, . . . , up /∈ B+(u1). We know up /∈ B+(u1) sinceβ = 0. If there is uk ∈ B+(u1) with r + 4 ≤ k ≤ p − 1, then by Claim
4.1E (using symmetry) we get ur+3, . . . , uk ∈ B+(u1). In particular, sincek ≥ r + 4, uk−1 ∈ B+(u1), and
T ′ = u2u3 . . . uk−2u1uk−1uk . . . upu2
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is a hamilton cycle, a contradiction. From Claim 4.1F(b), we now know that
N∗(u1) = {ur+2}.Second, we show that if r ≤ p − 6, then ur+4, . . . , up−2 /∈ N(ur+1).
Suppose there is k, r+ 4 ≤ k ≤ p− 2, with uk ∈ N(ur+1). If uk+1 ∈ B+(up)then
T ′ = u1u2 . . . ur+1ukupup−1 . . .uk . . . ur+1u1
is a closed 2-trail, a contradiction. From above, uk+1 /∈ B+(u1). Sinceuk+1 ∈ C, we must have uk+1 ∈ N(ur+1). Now
T ′ = u1u2 . . . urupup−1 . . . uk+1ur+1ukuk−1 . . . ur+2u1
is a hamilton cycle, a contradiction. From Claim 4.1F(c), we now know thatN∗(ur+1) ⊆ {up−1}.
Third, we show that up−1 /∈ N(ur+1). Suppose up−1 ∈ N(ur+1). Ifr ≥ 3, then, by Claim 4.1F(a), u2, u3 ∈ N∗(up), and
T ′ = u1u2upu3 . . . ur+1up−1up−2 . . . ur+2u1
is a hamilton cycle, a contradiction. Therefore, r = 2. If r ≤ p − 6, thenfrom above ur+4 must be in B+(up) since it is not in B+(u1)∪N(ur+1), and
T ′ = u1u2 . . . urupur+3ur+4 . . . up−1ur+1ur+2u1
is a hamilton cycle, a contradiction. Therefore, r = p− 5. Thus, p = n = 7.Besides T , G contains the edges u1u4, u2u7 and u3u6. There are no further
edges incident with u1 or ur+1 = u3. If this is all of G, then {u1, u5, u7}is an independent set with degree sum 6 < n = 7, a contradiction. There-
fore, there is an additional edge incident with at least one of u5 or u7,either to increase the degree sum or destroy the independence of this set.
The only possible extra edges are u5u2, u5u7 and u7u4. The edge u5u2
gives the closed 2-trail u2u3u4u1u2u5u6u7u2, the edge u5u7 gives a hamil-
ton cycle u1u2u3u6u7u5u4u1, and the edge u7u4 gives the closed 2-trailu1u2 . . . u7u4u1, all of which provide contradictions.
So, now we know that ur+4, . . . , up−1 ∈ B+(up) and N∗(ur+1) = ∅. Thus,{u1, ur+1, up−1} is an independent set. Since deg(u1) = deg(ur+1) = 2,deg(up−1 ≥ n − 4, so up−1 is nonadjacent to at most three vertices, and
hence adjacent to at least one of ur or ur+2. If up−1ur ∈ E(G), then
T ′ = urur+1ur+2u1u2 . . . urupur+3ur+4 . . . up−1ur
is a closed 2-trail, a contradiction. If up−1ur+2 ∈ E(G), then
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T ′ = u1u2 . . . ur+2 . . . up−2upup−1ur+2u1
is a closed 2-trail, a contradiction. Thus, we do not have r ≤ p− 5.
Therefore, r = p − 4. Then N(u1) = {u2, ur+2}. By Claim 4.1F(c),N∗(ur+1) = ∅ and so N(ur+1) = {ur, ur+2}. Since β = 0, u1 /∈ N∗(up−1).
Suppose uj ∈ N∗(up−1) where 2 ≤ j ≤ r. By Claim 4.1F(a), uj ∈ N∗(up),so
T ′ = u1u2 . . .ujup−1upujuj+1 . . .ur+2=p−2u1
is a closed 2-trail, a contradiction. We know ur+1 /∈ N∗(up−1). Therefore,N(up−1) = {ur+2=p−2, up}. Now {u1, ur+1, up−1} is an independent set with
degree sum 6, so p = n ≤ 6. Since r = p− 4 ≥ 2, we have p = 6 and r = 2.Besides T , G contains the edges u1u4 and u2u6. If this is all of G, then G ∼=K∗2,3. Otherwise, since {u1, ur+1, up} = {u1, u3, u6} and {u1, ur+1, up−1} ={u1, u3, u5} are independent, G contains the edge u2u4, and has a closedspanning 2-trail u1u2u3u4u5u6u2u4u1, a contradiction.
(2) α = β = 1. Since there is at most one vertex of V (G) \ C other than
ur+1, at least one of D1 = {u1, u2, . . . , ur−1} or D2 = {us+1, us+2, . . . , up} isa subset of C. By symmetry we may suppose that D1 ⊆ C.
If r ≥ 3, then by Claim 4.1F(a) we have u2, u3 ∈ N∗(up), and
T ′ = u1u2upu3u4 . . . up−1u1
is a hamilton cycle, a contradiction.
Therefore, r = 2. If u3up−1 /∈ E(G), then
T ′ = u3u2u1u4u5 . . .up
is a hamilton path with 2 = θ(T ) ≤ θ(T ′) ≤ 2 that falls under case (1), so
we are finished. If u3up−1 ∈ E(G) \E(T ) then
T ′ = u1up−1u3u4 . . . up−1upu2u1
is a closed 2-trail, a contradiction. Therefore, u3up−1 ∈ E(T ) and p = 5.
Besides T , G contains the edges u1u4 and u2u5. If this is all of G, thenG ∼= K2,3. Otherwise, since {u1, u3, u5} is independent, G contains the edge
u2u4 and has a closed spanning 2-trail u1u2u3u4u5u2u4u1, a contradiction.This concludes the proof of Theorem 4.1.
The condition σ3(G) ≥ n in Theorem 4.1 is best possible. Supposen ≥ 7. Take Kn−4, choose (possibly identical) vertices v1, v2, then add new
vertices w1, w2, x1, x2 and edges v1w1, v2w2, w1x1, w1x2, w2x1, w2x2. We get
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two graphs, depending on whether v1 is equal to v2 or not. Both are 2-edge-
connected (the one with v1 6= v2 is actually 2-connected), have σ3 = n − 1,and do not have a closed spanning 2-trail. Also, for every m ≥ 2 the graph
Km,2m+1 is 2-edge-connected (actually, m-connected), has σ3 = 3m = n−1,and has no closed spanning 2-trail. Note also that the hypothesis that G is
2-edge-connected is clearly necessary.We also have the following straightforward corollary, which is sharp,
at least when n ≡ 1 (mod 3), as shown by Km,2m+1. This shows thatTheorem 4.1 strengthens Theorem E.
Corollary 4.2 Let G be a 2-edge-connected n-vertex graph. If δ(G) ≥ n/3or σ2(G) ≥ 2n/3 then G has a closed spanning 2-trail unless G ∼= K2,3 or
K∗2,3.
Results for graphs embedded on surfaces can also be obtained from min-
imum degree conditions. An argument similar to that of Duke [8], usingEuler’s formula to show that δ ≥ n/3, gives the following. We omit the
details. Similar results can be obtained from Corollaries 2.7 and 3.4.
Corollary 4.3 Suppose G is a 2-edge-connected graph embeddable on a sur-
face (compact, without boundary, orientable or nonorientable) of Euler char-acteristic χ < 0, such that δ(G) ≥ 3 +
√9− 2χ. Then G has a closed
spanning 2-trail.
Perhaps Theorem 4.1 can be extended to spanning k-trails for k ≥ 3.
The natural conjecture would be that σk+1 ≥ n, together with sufficientedge-connectivity, guarantees the existence of a closed spanning k-trail. An-other interesting question is whether a toughness condition, together with
sufficient edge-connectivity, guarantees the existence of a closed spanningk-trail. Similar results are known for closed spanning k-walks for k ≥ 2
[9, 12], and a result for k-trails might shed some light on Chvatal’s wellknown conjecture [7] that sufficiently tough graphs have a hamilton cycle.
Acknowledgement
The authors thank Guantao Chen for pointing us to references [10, 13, 18].
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