Post on 13-May-2023
Detour-Saturated Graphs
Lowell W. Beineke,1 J. E. Dunbar,2 and M. Frick3
1INDIANA UNIVERSITY PURDUE UNIVERSITY
FORT WAYNE, INDIANA 46805
E-mail: beineke@ipfw.edu
2CONVERSE COLLEGE
SPARTANBURG, SOUTH CAROLINA 29302
E-mail: jdunbar@carol.net
3UNIVERSITY OF SOUTH AFRICA
PO. BOX 392
UNISA 0003 SOUTH AFRICA
E-mail: frickm@unisa.ac.za
Received April 1, 2002; Revised August 18, 2004
Published online in Wiley InterScience(www.interscience.wiley.com).
DOI 10.1002/jgt.20069
Abstract: A graph is said to be detour-saturated if the addition of any edgeresults in an increased greatest path length. In this paper, we add to therelativelysmall amount that isknownaboutdetour-saturatedgraphs.Ourmainresult is a determinationof all connecteddetour-saturatedgraphswith exactlyone cycle. (The family of detour-saturated trees was found by Kaszonyi andTuza[7].)Wealsoshowthatthesmallestdetour-saturatedgraphofgirth5isthegraph obtainable from the Petersen graph by splitting one of its vertices intothree, each of degree 1. � 2005 Wiley Periodicals, Inc. J Graph Theory 49: 116–134, 2005
Keywords: longest path; detour; detour-saturated; unicyclic
1. INTRODUCTION
This paper concerns graphs G with the property that if any edge is added to G,
there is a path longer than any path in G. Following a precedent set by Kapoor,
Kronk, and Lick [8], we call a longest path in a graph G a detour of G. The
� 2005 Wiley Periodicals, Inc.
116
number of vertices in such a path is called the detour order of G and denoted
�ðGÞ. (By a path in a graph G, we mean any subgraph of G that is a path—it need
not be an induced path.)
A graph G is k-detour-saturated if �ðGÞ � k and �ðGþ xyÞ > k for every pair
of nonadjacent vertices x and y in G. If the specific value of k is not important, we
shall simply use the term detour-saturated. Thus, a graph is detour-saturated if
�ðGþ xyÞ > �ðGÞ for every pair of nonadjacent vertices x and y in G:Kaszonyi and Tuza [7] characterized k-detour-saturated trees. Later, Zelinka
[10] constructed k-detour-saturated graphs of order k þ 1 (also called maximal
non-traceable graphs) and characterized those that are block graphs. However, in
general, very little has been done by way of investigating detour-saturated graphs.
These graphs do play an important role in certain partition problems. For
example, in [4] one family of uniquely partitionable graphs is characterized in
terms of joins of detour-saturated graphs. Results on detour-saturated graphs have
also proved useful in investigating the following intriguing conjecture, known as
the Path Partition Conjecture (PPC). Here Xh i denotes the subgraph induced by X.
Conjecture 1. If G is any graph and ða; bÞ any pair of positive integers such
that �ðGÞ ¼ aþ b, then there exists a partition ðA;BÞ of VðGÞ such that
�ð Ah iÞ � a and �ð Bh iÞ � b:
A graph satifying the requirements of Conjecture 1 is called a �-partitionablegraph. Thus the PPC can be stated as : Every graph is �-partitionable.
Every graph is obviously a spanning subgraph of a detour-saturated graph with
the same detour order. Thus, to prove Conjecture 1, it is sufficient to prove that
every detour-saturated graph is �-partitionable. In [3], results on detour-saturated
graphs were used to prove that every graph that is a join of two graphs is �-
partitionable.
In order to settle the PPC, it seems essential to know more about the cycle
structure of detour-saturated graphs.
The circumference cðGÞ of G is the order of a largest cycle of G: A vertex of
cycle C that has a neighbor in G� VðCÞ is called a clasp vertex of C.
A graph G is semi-pancyclic if either (i) G has a cycle of every order from
d�ðGÞ=2e up to cðGÞ, or (ii) d�ðGÞ=2e > cðGÞ or (iii) G is acyclic. It is proved in
[6] that every semi-pancyclic graph is �-partitionable. It is also shown that if G
has a circumference cycle C with at most d�ðGÞ=2e clasp vertices, then it is �-
partitionable. Thus, the PPC is true if the following conjecture is true.
Conjecture 2. If G is a detour-saturated graph in which every circumference
cycle of G has more than �ðGÞ=2d e clasp vertices, then G is semi-pancyclic.
Non-semi-pancyclic detour-saturated graphs exist (for example, the graph in
Fig. 3), but the ones that we know of all have circumference cycles with few clasp
vertices and, moreover, they are separable (i.e., not 2-connected). We have shown
recently that if every 2-connected graph is �-partitionable, then every graph is �-
partitionable. Thus, the PPC is true if the following conjecture is true.
DETOUR-SATURATED GRAPHS 117
Conjecture 3. Every detour-saturated 2-connected graph is semi-pancyclic.
A bipartite graph G with cðGÞ > d�ðGÞ=2e cannot be semi-pancyclic (since it
has no odd cycles). However, it seems unlikely that a bipartite counterexample to
Conjecture 2 exists. In fact, we suspect the following is true.
Conjecture 4. Every bipartite detour-saturated graph is acyclic.
Of course, bipartite graphs are �-partitionable, but investigating Conjecture 4
could cast some light on Conjectures 2 and 3.
From results in [2] concerning the Ryjacek closure for claw-free graphs, it
follows that every claw-free detour-saturated graph is the line graph of a triangle-
free graph. This provides sufficient information about the cycle structure of claw-
free detour-saturated graphs for proving that Conjecture 2 (and hence the PPC)
holds for claw-free graphs (see [6]).
However, very little is yet known about the cycle structure of detour-saturated
graphs that contain claws and cycles. In this paper, we provide some general
results on detour-saturated graphs and we characterize detour-saturated unicyclic
graphs.
2. ELEMENTARY RESULTS
In this brief section, we provide results that begin to shed light on the structure of
detour-saturated graphs.
If a graph has a hamiltonian path, it is called traceable. Since every component
of a detour-saturated graph is also detour-saturated and the only traceable detour-
saturated graphs are the complete graphs, we shall restrict our attention to con-
nected, non-traceable detour-saturated graphs. The following easy observations
have proved useful.
Proposition 2.1. Let G be a detour-saturated graph. Let x and y be two
nonadjacent vertices in G, and let P be a path of order �ðGÞ þ 1 in Gþ xy. Then
P contains at least one internal vertex of every x-y path in G.
Proof. Suppose Q is an x-y path in G. If P and Q are internally disjoint, then
the path obtained from P by replacing xy with the path Q is a path of order greater
than �ðGÞ in G. &
Corollary 2.2. Let x and y be two nonadjacent vertices in a detour-saturated
graph G and suppose there are 3 internally disjoint x-y paths in G. Then there is a
path in G� fx; yg containing at least one vertex from at least 2 of these paths.
Corollary 2.3. Let G be a detour-saturated graph. If v is a vertex of degree 2 in
G, then its two neighbours are adjacent.
Proof. Let x and y be the two neighbors of v. Suppose xy =2 EðGÞ. Then
Gþ xy contains a path P of order �ðGÞ þ 1 that contains the edge xy: By
118 JOURNAL OF GRAPH THEORY
Proposition 2.1, v is on P. Since xy is also an edge in P, exactly one of the edges
vx and vy, say vx, is an edge of P. So v is an end-vertex of P: But then the path
obtained from P by replacing vxy with xvy is a path of order �ðGÞ þ 1 in G,
which is impossible. &
Lemma 2.4. Suppose G is a triangle-free detour-saturated graph and C is a
cycle in G such that every vertex on C is a clasp vertex. Then at least one com-
ponent of G� VðCÞ has two distinct neighbors on C.
Proof. Suppose, to the contrary, that every component of G� VðCÞ has at
most one neighbor on C. Let C be the cycle y1y2 � � � yny1 and let Li be a longest
path in G� VðCÞ that has an end-vertex adjacent to yi: Let us assume that
jVðL1Þj ¼ mini¼1;...;n
jV Lið Þj:Since G has no triangles, yny2 =2 EðGÞ, and therefore, since G is detour-
saturated, there exists a path P of order � Gð Þ þ 1 in Gþ yny2, which contains the
edge yny2: It follows from Proposition 2.1 that y1 lies on P. We may, therefore,
assume that P consists of L1, followed by the path y1yny2 � � � yr, followed by Lr,
for some r with 2 � r � n� 1: But then the path consisting of Ln, followed by
yny1y2 � � � yr, followed by Lr is a path of order greater than � Gð Þ in G, which is a
contradiction. &
3. SMALLEST DETOUR-SATURATED GRAPHS INCERTAIN CLASSES
By a ‘‘smallest’’ graph in a given class C we mean a graph of minimum order in
C, having the minimum number of edges among the graphs of that order in C.
Let S be the class of connected non-traceable detour-saturated graphs. It is
easy to see that the smallest graph in S is the claw K1;3, the smallest claw-free
graph in S is the net N, and the smallest 2-connected graph in S is the 4-page
book B4: These graphs are shown in Figure 1.
It is proved in [5] that the smallest 2-connected claw-free non-traceable graph
is the graph in Figure 2(a). Thus it follows from Corollary 2.3 that the graph in
Figure 2(b) is the smallest 2-connected claw-free graph in S.
We shall prove that the graph in Figure 3 (denoted by Pr) is the smallest
detour-saturated graph with girth 5.
FIGURE 1.
DETOUR-SATURATED GRAPHS 119
First, we need some definitions and results.
A graph G is hypohamiltonian if G is not hamiltonian and G� v is hamiltonian
for every v 2 VðGÞ. If, in addition, Gþ xy is hamiltonian for every pair x and y of
nonadjacent vertices in G, then G is maximal hypohamiltonian. Souselier (see [1])
proved that PG (the Petersen graph) is the smallest hypohamiltonian graph. It is
easy to check that PG is also maximal hypohamiltonian.
Given a graph G, let x be a vertex with neighbors y1; y2; . . . ; yd. The graph G
split x, denoted by Gs½x�, is the graph obtained from G by replacing x with d
independent vertices x1; x2; . . . ; xd and joining xi to yi; i ¼ 1; 2; . . . ; d. We note
that Pr can be obtained as PGs½x� for any vertex x in PG.
The following result enables us to construct a detour-saturated graph from any
maximal hypohamiltonian graph with minimum degree 3. (Every hypohamilto-
nian graph obviously has minimum degree at least 3.)
Theorem 3.1. Let G be a maximal hypohamiltonian graph and x a vertex of
degree 3 in G. Then Gs½x� is a detour-saturated graph, with detour order
k ¼ jVðGÞj.Proof. Let C be a hamiltonian cycle of G� x and let P be a detour in Gs½x�.
Then P contains every vertex on C and, since G is non-hamiltonian, P contains
at most one of the vertices x1; x2; x3: Therefore, �ðGs½x�Þ ¼ jVðCÞj þ 1 ¼jVðGÞj ¼ k:
Let u and v be any two nonadjacent vertices of Gs½x�. We consider four cases:
FIGURE 2.
FIGURE 3.
120 JOURNAL OF GRAPH THEORY
Case 1. fu; vg � fx1; x2; x3g: Then �ðGs½x� þ uvÞ ¼ jVðCÞj þ 2 ¼ k þ 1:
Case 2. u ¼ xi; v 2 VðCÞ � fy1; y2; y3g: Since xv =2 EðGÞ; Gþ xv has a hamil-
tonian cycle, D. There are two subcases:
2.1. D contains the edge xyi: Then Gs½x� þ uv has a cycle of order jVðDÞj ¼ k.
Hence, since Gs½x� is connected, �ðGs½x� þ uvÞ � k þ 1:2.2. D contains the edge xyj; j 6¼ i: Then Gs½x� þ uv has a path containing all
the vertices of C as well as xi and xj. Thus, �ðGs½x� þ uvÞ� jVðCÞj þ 2 ¼k þ 1.
Case 3. u ¼ xi; v ¼ yj; j 6¼ i: Since G is hypohamiltonian, G� yj is a hamil-
tonian graph. Therefore in Gs½x�, there is a path Q starting at xi containing all the
vertices of VðCÞ � fyjg, and ending at xl; l 6¼ i; j. The path xjyjQ is a path in
Gs½x� þ uv: Thus, �ðGs½x� þ uvÞ� jVðCÞj þ 3 > k þ 1.
Case 4. fu; vg � VðCÞ � fx1; x2; x3g: Gþ uv has a hamiltonian cycle contain-
ing edges yix and xyj; j 6¼ i; hence Gs½x� þ uv has a path starting at xi, containing
all the vertices of the cycle and ending at xj. So again we have �ðGs½x� þ uvÞ�jVðCÞj þ 2 ¼ k þ 1. &
Using the fact that PG is a maximal hypohamiltonian graph, we now prove the
following.
Theorem 3.2. Pr is the smallest detour-saturated graph with girth 5.
Proof. Since PG is regular of degree 3 and Pr ¼ PGs½x� for any x 2 VðPGÞ,it follows from Theorem 3.1 that Pr is detour-saturated and �ðPrÞ ¼ 10. We note
that Pr has 15 edges, order 12 and girth 5.
Now suppose G is a detour-saturated graph with girth gðGÞ ¼ 5 and order at
most 12. It follows from Corollary 2.3 that G has no vertices of degree 2. Let
cðGÞ ¼ c and let C be the circumference cycle v1v2 � � � vcv1 of G: Put R ¼G� VðCÞ. Since G is not traceable, jVðRÞj � 2: We may assume also that G is
connected.
Case 1. c � 7: In this case, since gðGÞ ¼ 5; C has no chords, and hence every
vertex on C is a clasp vertex of C. Therefore, by Lemma 2.4, we may assume that
two vertices, v and w, on C have neighbors in the same component of R: Since
gðGÞ ¼ 5, this is not possible if c ¼ 5, so we assume c ¼ 6 or 7: In either case,
one of the v-w paths on C has only 2 internal vertices, and hence no v-w path has
more than 2 vertices in R; otherwise G has a cycle of order greater than c. Now
we have three internally disjoint v-w paths in G, two on C and one with internal
vertices in R. By Corollary 2.2, there is a path in G� fv;wg that contains vertices
on at least two of those paths. Such a path must have order at least 3 since
otherwise we have a cycle of order less than 5; inspection shows that we then
have a cycle of order greater than c.
DETOUR-SATURATED GRAPHS 121
Case 2. c ¼ 8: Let C be the cycle v1v2 � � � v8v1. Since gðGÞ ¼ 5; the only
chords that C can have are diameters (chords of the form xixiþ4), and C cannot
have diameters from two consecutive vertices. Thus, at most 4 vertices on C have
more than 2 neighbors on C. Since G has no vertices of degree 2, this implies that
C has at least 4 clasp vertices.
We consider two subcases:
2.1. Every vertex of C is a clasp vertex: If three clasp vertices have a common
neighbor, we contradict that gðGÞ ¼ 5. So we must have jVðRÞj ¼ 4, and
every vertex of R is adjacent to two vertices on C. Since gðGÞ ¼ 5;cðGÞ ¼ 8, and jVðGÞj � 12, this is not possible.
2.2. At least one vertex of C is not a clasp vertex: Say v8 is not a clasp vertex.
Then v4v8 is a chord and hence v1v5; v3v7 are not chords. So v1; v3; v5; v7
are clasp vertices. Suppose two of v1; v3; v5; v7 have a common neighbor
x in R. Since gðGÞ ¼ 5, we may assume they are v1 and v5. Then
v4v3v2v1xv5v6v7v8v4 is a cycle of order 9 in G. Therefore, no two of
these clasp vertices have a common neighbor. Hence jVðRÞj � 4. Again,
since jVðGÞj � 12; jVðRÞj ¼ 4. Thus each vertex of R is adjacent to
exactly one of the vertices v1; v3; v5; v7. For i ¼ 1; 3; 5; 7, let xi be the
neighbor of vi in R. Since cðGÞ ¼ 8 and v4v8 is a chord, R is an inde-
pendent set. Since no vertex in R is adjacent to 2 vertices on C, it follows
that every vertex of R has degree 1. Hence no path in Gþ v2v8 contains
more than two vertices of R, so �ðGþ v2v8Þ ¼ 10. But �ðGÞ ¼ 10, since
x1v1v2v3v4v8v7v6v5x5 is a path of order 10 in G, so �ðGÞ ¼ �ðGþ v2v8Þ,a contradiction.
Case 3. c ¼ 9: In this case, jVðRÞj � 3: Since gðGÞ ¼ 5, at least 3 vertices of
C are not incident with chords in C. Hence, at least 3 vertices of C are clasp
vertices.
3.1. �ðGÞ ¼ 11: In this case, jVðRÞj ¼ 3, since otherwise G is traceable. Thus,
every vertex of R has degree 1, otherwise G has more edges than Pr. Now,
if v and w are nonadjacent vertices on C, then every path in Gþ vwcontains at most two vertices of R; therefore �ðGþ vwÞ ¼ 11, and hence
G is not detour-saturated.
3.2. �ðGÞ ¼ 10: In this case, every component of R is a single vertex and C
does not have two consecutive clasp vertices. If some vertex of R is
adjacent to more than one vertex of C, then it must be adjacent to three
vertices of C: But then G has more edges than Pr. Thus, each vertex in R
has degree 1, jVðRÞj ¼ 3 and C has exactly 3 clasp vertices. Suppose none
of v1; v2; v3 is a clasp vertex of C. Since G has no vertices of degree 2 and
no cycles of order less than 5, either v2v6 or v2v7 is a chord of C. In the
first case, any chord incident with v1; and in the second case, any chord
122 JOURNAL OF GRAPH THEORY
incident with v3 will create a cycle of order less than 5. This proves that C
cannot have three consecutive nonclasp vertices, and hence there must be
2 nonclasp vertices on C between any two clasp vertices of C. Now let G0
be the graph obtained from G by replacing R with a single vertex adjacent
to every vertex on C that had a neighbor in R: Then G0 is a non-
hamiltonian graph with girth 5, order 10, and minimum degree at least 3.
It follows that G0 is the Petersen graph and G is Pr. &
Thomassen constructed a maximal hypohamiltonian graph with girth 4, mini-
mum degree 3, and order 32 ([9], Fig. 3). Hence, it follows from Theorem 3.1 that
there exists a detour-saturated graph with girth 4 and order 34. We, therefore,
pose the following questions.
Questions.
(1) What is the smallest order of a detour-saturated graph of girth 4?
(2) Is Pr the smallest triangle-free detour-saturated graph?
(3) Does there exist a detour-saturated graph with finite girth bigger than 5?
4. DETOUR-SATURATED TREES
In their paper, Kaszonyi and Tuza [7] described a family of trees that charac-
terizes k-detour-saturated trees, but they did not provide a complete proof of the
result. Because an understanding of a proof of their result is useful for our
theorem on unicyclic graphs, we provide such a proof here for k � 3. (Note that
for k < 3, the only k-detour-saturated trees are K1 and K2.) The family T k is
obtained by starting with one tree in Tk (called a skeletal tree) and then adding
branches and leaves, while maintaining membership in T k.
First, we define the trees Tk inductively as follows: T3 is the claw and T4, the
double claw. For k � 5; Tk is obtained from Tk�2 by attaching two new leaves at
each of the old leaves (see Fig. 4 for Tk up to k ¼ 8). Note that the center ZðTkÞconsists of one vertex when k is odd and two adjacent vertices, when k is even.
We define the level of a vertex v in a tree Tk to be the order of a shortest path
from it to a central vertex. A leaf is a vertex of degree 1 and an internal vertex is a
vertex of a tree that is not a leaf. If v is an internal vertex, then any component of
Tk � v that does not contain a central vertex is called a branch at v. (Note: this
is more restricted than some definitions of branch.) We observe that when k is
odd and v is the central vertex of Tk, then Tk � v consists of three isomorphic
branches; otherwise, there are just two branches at v and they are isomorphic.
Duplicating a branch B at v consists of adding a branch isomorphic to B at v.
We verify that the tree Tk is k-detour-saturated in the following way. Suppose
xi and xj are nonadjacent vertices in Tk. Let Pi be a detour containing xi but not xj.
Note that Pi has a subpath P0i starting at a leaf and ending at xi. Define Pj and P0
j
similarly. The path obtained from Pi in Tk þ xixj by replacing P0i by P0
jxi has more
vertices than Pi. Thus, the following two propositions are immediate.
DETOUR-SATURATED GRAPHS 123
Proposition 4.1. For each k � 3, the following hold:
(1) Tk is k-detour-saturated.
(2) Every vertex in Tk has degree 1 or 3.
(3) Every vertex and every edge in Tk lies on a detour of Tk:
Proposition 4.2. Let y be an internal vertex of Tk, and let u and v be any pair of
nonadjacent vertices different from y.
(1) Then Tk þ uv has a path of order k þ 1 that avoids one branch at each
child of y, unless u and v lie on different branches of the same child of y.
(2) If k is odd and y is the center of Tk, then Tk þ uv has a path of order k þ 1
that avoids one branch at each neighbor of y, unless u and v lie on
different branches of the same neighbor of y:
The following notation and next three results will be used in proving the
converse portion of Theorem 4.6.
Let T be a k-detour-saturated tree with k � 3, and let z be an internal vertex of
T . Then, by Corollary 2.3, we know z is a vertex satisfying degðzÞ� 3, with
neighbors fx1; x2; . . . ; xdg. Further, let li be the order of a longest path in T
beginning at xi and not containing z. Assume l1 � l2 � � � � �ld.
Lemma 4.3. l1 þ l2 þ 1 ¼ k and l2 ¼ l3.
Proof. Clearly, G contains a path of order l1 þ l2 þ 1; hence l1 þ l2 þ 1 � k:Since T is detour-saturated, there is a path of order k þ 1 in T þ x1x2 that contains
FIGURE 4.
124 JOURNAL OF GRAPH THEORY
the edge x1x2: Any such path has at most l1 þ 2 þ l3 vertices; hence l1 þ l3 þ2 � k þ 1: Since we are assuming that l2 � l3, the result follows. &
Corollary 4.4. Tk � T .
Proof. Since the equalities in Lemma 4.3 hold for any internal vertex of T , an
isomorphic copy of Tk is found by beginning with two opposite leaves u and v on
a path P of order k. We see that since l2 ¼ l3, the neighbor of u or v must be
adjacent to another leaf. Continuing in this way examining all internal vertices of
P until a center vertex of T is reached, we find a copy of Tk. &
A proof of Corollary 4.4 appears in [7].
Corollary 4.5. If l2 � 2, then z has at most one leaf neighbor.
Proof. Suppose to the contrary that z has two leaves u and v. Then, T þ uvmust have a path of order k þ 1, which contains uv. Such a path will have order at
most l1 þ 3 � l1 þ l2 þ 1 ¼ k, a contradiction. &
Finally, let T k be the set of all trees that can be obtained from Tk by
successively duplicating branches and adding at most one leaf to any internal
vertex that is not already adjacent to a leaf. We observe that every internal vertex
of a tree T in T k lies on a subtree isomorphic to Tk.
Kaszonyi and Tuza [7] gave the following characterization of detour-saturated
trees.
Theorem 4.6. B is a k-detour-saturated tree if and only if B 2 T k.
Proof. Let B 2 T k: Clearly, �ðBÞ ¼ k. Now let u and v be any pair of
nonadjacent vertices in B.
Case 1. Both u and v lie in some skeletal tree Tk in B: In this case, it follows
from Proposition 4.1(1) that �ðGþ uvÞ � k þ 1:
Case 2. Exactly one of u; v lies in a skeletal tree Tk in B: Assume v 2 Tk and
u =2 Tk. Then u is a leaf in B and we may assume that its parent, x, lies in Tk: If
x is adjacent to v then, by Proposition 4.1(3), there is a path P of order k in
Tk containing the edge xv. The path obtained from P by replacing the edge xvwith the path xuv is a path in Bþ uv of order k þ 1: If x is not adjacent to v,
then there is a path Q of order k þ 1 in Tk þ xv containing the edge xv. By
replacing the edge xv with the subpath xuv, we obtain a path of order k þ 2 in
Bþ uv.
Case 3. Neither u nor v lies in any skeletal tree Tk in B: Then both u and v are
leaves and, by Corollary 4.5, they do not have the same parent. Let x be the parent
of u, and z the parent of v. If z is adjacent to x, then the edge zx also lies in an
underlying tree Tk of B. By Proposition 4.1(3), there is a path of order k in Tkcontaining the edge xz. If we replace the edge xz with the subpath xuvz, then we
DETOUR-SATURATED GRAPHS 125
create a path in Bþ uv of order k þ 2. If z is not adjacent to x in B, then there is a
path of order k þ 1 in Bþ xz containing the edge xz. Again replacing the edge xz
with the subpath xuvz yields a path in Bþ uv of order k þ 3.
Thus, in any case, Bþ uv contains a path of order greater than k, so B is k-
detour-saturated.
Conversely, assume T is a k-detour-saturated tree for k � 3. We show that T
can be obtained from Tk by successively duplicating branches and attaching a leaf
at any internal vertex that is not yet adjacent to a leaf.
Toward that end, we begin pruning T in the following way: For every vertex
z 2 VðTÞ for which l2 � 2 (in the notation preceding Lemma 4.3), delete any leaf
adjacent to z. We denote the resulting graph by T 0.
Claim. T 0 is k-detour-saturated.
Clearly �ðT 0Þ ¼ �ðTÞ ¼ k. Let u and v be a pair of nonadjacent vertices in T 0
(and hence in T), and let P be a longest path in T þ uv. Suppose that P is not in
T 0 þ uv. Then P must begin (or end) with a deleted leaf of T; call it x. If z is the
neighbor of x, then there must be a path of order at least 3 that has only z in
common with P. Replacing xz with this path gives a longer path than P, which
contradicts the choice of P. Hence, P must be in T 0 þ uv, and since �ðT þ uvÞ �k þ 1; T 0 must be detour-saturated.
Finally, we prune duplicated branches from T 0 as follows: Suppose z 2 VðT 0Þand degðzÞ ¼ d > 3. Let x1; x2; . . . ; xd and l1; . . . ; ld be as in the paragraph
preceding Lemma 4.3. We denote by Di the component of T � z, which contains
xi: Now let T 00 be the tree obtained by removing from T 0 all Di, for i� 4.
Claim. T 00 is k-detour-saturated.
Clearly the detour order of T 00 is k. Let u and v be nonadjacent vertices in
VðT 00Þ. Then there exists a path P of order k þ 1 through uv in T 0 þ uv. Since u
and v are vertices in T 00; P contains vertices from at most three of the Di; of
which at most one has index i � 4. Thus either P is in T 00 or the vertices of P that
are not in T 00 can be replaced by vertices in some Di with i � 3 to yield a path of
order at least k þ 1 in T 00 þ uv.
After repeatedly carrying out this pruning procedure at each vertex of degree
bigger than 3, we obtain a k-detour-saturated tree in which every internal vertex
has degree 3. By Corollary 4.4, this tree is Tk.
Thus T can be obtained from Tk by successively duplicating branches and
attaching a leaf at any internal vertex that is not yet adjacent to a leaf. Hence,
T 2 T k. &
We denote the distance between vertices u and v in a graph G by dGðu; vÞ.From the proof of Theorem 4.6, we have the following result.
Proposition 4.7. Let B 2 T k and let u; v 2 VðBÞ with dBðu; vÞ > 2. Then
�ðBþ uvÞ� k þ 2.
126 JOURNAL OF GRAPH THEORY
5. DETOUR-SATURATED UNICYCLIC GRAPHS
In this section, we provide a characterization of detour-saturated unicyclic graphs.
The family of detour-saturated unicyclic graphs is not easy to describe com-
pletely, and is done in stages. We will show one can always get such a graph from
a detour-saturated tree by replacing a claw with a 3-cycle as shown in Figure 5.
Then it is a matter of appropriately pruning and duplicating branches in such
graphs to get the entire family.
If C is the cycle in a unicyclic graph G, then G� VðCÞ is obviously a forest
and each tree in this forest is attached to exactly one of the vertices of C. If T is a
tree in G� V Cð Þ attached to the vertex v 2 V Cð Þ, we call the maximum of the
distances between v and the vertices in T the elevation of T .
We now determine the order of the cycle in a detour-saturated unicyclic graph.
Lemma 5.1. If G is a detour-saturated unicyclic graph, then its cycle is a
triangle.
Proof. Let C be the cycle in G and suppose C is not a triangle. Then G is
triangle-free, and, since C has no chords, it follows from Corollary 2.3 that every
vertex of C is a clasp vertex. Hence, by Lemma 2.4, G� VðCÞ has a component
with two distinct neighbors on C. But then G is not unicyclic. &
A. Sets of Detour-Saturated Graphs
Let Dk denote the set of connected detour-saturated unicyclic graphs with detour
order k. We now proceed to construct three families of graphs whose union is
contained in Dk. We fix k � 3 and let h ¼ bk�12c (so k ¼ 2hþ 1 or 2ðhþ 1Þ
according as it is odd or even).
For any skeletal tree Tk (defined in the previous section), we apply a ‘claw/
cycle’ operation as follows. For 1 � l � h, let y be a vertex of Tk at level l and let
y1; y2; y3 be its neighbors, with y1 nearest to the center of Tk (if not all three
distances are equal). Delete y, from Tk, and add the three edges between y1; y2;and y3 (see Fig. 5). Call the result TkðlÞ.
Now let Bk be any k-detour-saturated tree, and let BkðlÞ be any unicyclic graph
that results from applying the ‘claw/cycle’ operation using a vertex of degree 3
at level l in some subtree of Bk isomorphic to Tk. Let X k consists of all such
unicyclic graphs.
FIGURE 5. The claw/cycle operation.
DETOUR-SATURATED GRAPHS 127
Lemma 5.2. X k � Dk.
Proof. Let X ¼ BkðlÞ be a graph in X k. We use the notation of Bk; Tk; and
the vertices y; y1; y2; y3 that led to the construction of Bk: We call the cycle C
and refer to y1; y2; y3 as C-vertices and the edges yiyj as C-edges. The result
follows from two claims.
Claim 1. �ðXÞ ¼ k: By Proposition 4.1, the vertex y lies on a detour P of order k
in Tk, and clearly P contains exactly two C-vertices, say y1 and y2: Replacing y by
y3 (with the corresponding edges) in P gives a path of order k in X, so �ðXÞ � k.
Now let R be a path of order r in X. We consider three cases.
Case 1. R contains no C-edges: In this case, R is also a path in Bk.
Case 2. R contains just one C-edge: Suppose the C-edge is yiyj. Replacing it by
the path yiyyj gives a path in Bk.
Case 3. R contains two C-edges: Let the C-edges be yiyj and yjyk: Replacing the
subpath yiyjyk with yiyyk gives a path in Bk.
In each case, it follows that r � k, and so �ðXÞ ¼ k.
Claim 2. X is detour-saturated: Let u and v be nonadjacent vertices in X. Since
not both can be in C, we have two cases.
Case 1. dXðu; vÞ ¼ 2: In this case, u and v must lie in the same component of
Bk � y, and hence every path in Bk þ uv contains at most two neighbors of y.
Since Bk is detour-saturated, there exists a path Q of order k þ 1 in Bk þ uv: If y is
not on Q, then Q is also a path of X þ uv. If y is on Q, then a path of order k þ 1
in X þ uv can be obtained from Q by replacing y with a C-vertex that is not in Q.
Case 2. dXðu; vÞ > 2: Then dBkðu; vÞ > 2, and hence, by Proposition 4.7, there
is a path P in Bk þ uv of order at least k þ 2. If P does not contain y, then P is also
a path in X þ uv. If P contains y, then we may assume that it also contains at least
two neighbors of y, say y1 and y2. Then a path of order k þ 1 in X þ uv can be
obtained from P by replacing the subpath y1yy2 with y1y2:
Thus X is a detour-saturated graph. &
By construction, no C-vertex in a graph in X k has degree 3, but detour-
saturated unicyclic graphs may have such vertices. One example is the net N in
Figure 1. We now define a family Yk of such graphs. Given a graph BkðlÞ with
l < bk�12c, delete all but one of the trees of maximum elevation in Fi (the forest at
yi) for these values of i:
� i ¼ 2 or 3 (or both);
� for k odd and l ¼ 1, any subset of f1; 2; 3g.
The following lemma is a consequence of Proposition 4.2 and Lemma 5.2.
Lemma 5.3. Yk � Dk.
128 JOURNAL OF GRAPH THEORY
To complete our set, we still need more graphs for k � 5. These are obtained
from the graphs Bkðbðk � 1Þ=2cÞ by removing the leaves of a tree of elevation 2
in F1 (there always is such a tree). We let Zk denote this set.
Lemma 5.4. Zk � Dk.
Proof. Let Z 2 Zk: Since Z is derived from the graph Bkðbðk � 1Þ=2cÞ, we
know e2 ¼ 0 and e1 þ 3 ¼ k. Clearly the detour order of Z is k.
Suppose u and v are nonadjacent vertices in Z. We need only consider the case
when u is the tree of order 1 attached to y1: If v is a neighbor of y1 in Z, there is a
path in Z þ uv containing u as well as all three vertices on the cycle and e1
vertices of a tree attached to y1; hence �ðZ þ uvÞ � 1 þ 3 þ e1 ¼ k þ 1:Assume next that v is not a neighbor of y1. Both u and v may be regarded as
vertices of the underlying tree Tk; and d u; vð Þ > 2 in Tk: By Proposition 4.7, there
is a path Q of order at least k þ 2 in Tk þ uv. Next, we consider the triangulated
graph Bkðbðk � 1Þ=2cÞ: If Q contains the vertex y, which was deleted to create
Bkðbðk � 1Þ=2cÞ, we obtain a path P in Bkðbðk � 1Þ=2cÞ by replacing the subpath
yiyyj with the subpath yiykyj; otherwise we let P ¼ Q. Note that in either case the
order of P is at least k þ 2 and P can be regarded as being a path in Z þ uv: If P
contains one of the leaves deleted to create Z, we omit that leaf to yield a path of
order at least k þ 1 in Z þ uv. Thus, Z is detour-saturated, which proves the
lemma. &
Now put Uk ¼ X k [ Yk [ Zk. In Figures 6 and 7, the minimal members of U7
and U8 are depicted.
FIGURE 6.
DETOUR-SATURATED GRAPHS 129
The following result follows from Lemmas 5.2, 5.3, and 5.4.
Theorem 5.5. Uk � Dk.
B. The Converse
We shall now prove the converse of Theorem 5.5. From here on, we assume that
U 2 Dk, i.e., U is a connected detour-saturated unicyclic graph with �ðUÞ ¼ k
and cycle C ¼ y1y2y3y1. For i ¼ 1; 2; 3, let Fi be the forest consisting of all the
trees in G� V Cð Þ attached to yi, and let ei denote the maximum elevation among
all trees in Fi. We assume that e1 � e2 � e3:
Lemma 5.6. e1 þ e2 þ 3 ¼ k and e2 ¼ e3.
Proof. Let Qi be a path of order ei in Fi; starting at a neighbor xi of yi; i ¼1; 2; 3. Then the path Q1y1y2y3Q3 is a path in U, so e1 þ 3 þ e3 � �ðUÞ ¼ k. If
e2 ¼ 0, we are done. Otherwise, since U is detour-saturated, there must be a path
P of order k þ 1 in U þ x2y3: Clearly, jVðPÞj � e1 þ 3 þ 1 þ e3; hence
e1 þ 3 þ e3 � k and so e1 þ 3 þ e3 ¼ k. Since U also has a path of order e1 þ3 þ e2 and e2 � e3, it follows that e1 þ 3 þ e2 ¼ k, and hence e2 ¼ e3: &
It is easy to see that the only graph in D3 is the 3-cycle C3, and the only graphs
in D4 are the stars K1;r with one added edge for r � 4, and these are in X3 and
X4, respectively. Therefore, we assume k � 5.
Our goal at this stage is to show that U is a graph in Uk ¼ X k [ Yk [ Zk.
To that end, we define another graph U� by modifying U under certain
circumstances:
FIGURE 7.
130 JOURNAL OF GRAPH THEORY
Case 1. e1 ¼ e2: At each yi; at which there is only one tree of elevation ei,
duplicate that tree.
Case 2. e1 > e2 > 0: For i ¼ 2; 3, if there is only one tree of elevation ei at yi,
duplicate it.
Case 3. e1 > e2 ¼ 0: If there are trees of exactly two elevations at y1 and one of
them is a leaf, x, add two leaves to x.
In each case, if the stated conditions do not apply, we let U� ¼ U. We note that
if U is a set in Yk, then Case 1 or Case 2 applies, and if U is a set in Zk, then Case
3 applies. Our next objective is to show that in every case, U� 2 X k. First, we
show that U� is a detour-saturated unicyclic graph.
Proposition 5.7. U� 2 Dk:
Proof. If U� ¼ U, we are done, so we may assume that U� is a modification
of U.
Cases 1 and 2. e1 � e2 > 0: Each vertex of U� lies in some induced subgraph
of U� isomorphic to U. Hence �ðU� � k. Suppose U� contains a duplicate of a
tree T attached to yi; then the elevation of T is at most ðk � 1Þ=2, so any path
in U� that traverses T as well as a duplicate of it has order no more than
ððk � 1Þ=2Þ þ 1 þ ððk � 1Þ=2Þ ¼ k: It follows that �ðU�Þ ¼ k:Now let u and v be two nonadjacent vertices of U�. Since U is detour-
saturated, it is sufficient to consider the case where u lies in a tree T attached to yifor some i 2 f1; 2; 3g, and v lies in the duplicate, D, of T : We may assume that
dðyi; uÞ � dðyi; vÞ. By Proposition 4.1, there is a path Q of order k in U con-
taining u. This path obviously has a subpath R from an end-vertex of T to u. Let
Q� be the path in U� þ uv obtained from Q by replacing the path R with a path R�
that goes from a leaf of D to v and then to u. Then Q� has order at least k þ 1;which proves that U� is detour-saturated.
Case 3. e1 > e2 ¼ 0: Since e2 ¼ e3 ¼ 0, we know that k ¼ e1 þ 3. Thus
adding leaves to the tree of order 1 attached to y1 in U does not create a path of
order greater than k, so �ðU�Þ ¼ k.
Let x be the leaf adjacent to y1 in U. Suppose that u and v are nonadjacent
vertices in U�. We only need to consider the case where u is one of the new leaves
attached to x. If v is adjacent to x, it is easy to see that U� þ uv has a path of order
e1 þ 4. On the other hand, if v is not adjacent to x, there is a path P of order k þ 1
in U þ xv, and a path Q of order k þ 2 can be constructed in U� þ uv by
replacing the subpath vx in P by vux. Hence �ðU� þ uvÞ � k þ 1. &
Note that if e1 ¼ e2 in U�, there are two trees of elevation e1 at y1. And if
e2 > 0, there are two trees of order ei at yi for i ¼ 2; 3.
Lemma 5.8. If e1 > e2 > 0, then there are two trees of elevation greater than 1
at y1:
DETOUR-SATURATED GRAPHS 131
Proof. First, suppose there is only one tree, say T at y1. Let x be the neighbor
of y1 in T : Then, clearly, �ðU þ xy2Þ � l1 þ 3 þ l3 ¼ k, a contradiction. Now
suppose that there are two trees with elevations e1 and e01 at y1. Let r1 be the
neighbor of y1 in the tree of elevation e1. Then U þ r1y3 has a path P of
order at least k þ 1. But this means that e01 þ 3 þ e1 � e1 þ e2 þ 4. So e01 � e2 þ1� 2. &
Lemma 5.9. If there are not two trees of elevation e1 at y1 in U�, then there is a
tree of elevation e01 < e1 at y1, and e01 ¼ e2 þ 2.
Proof. If there are not two trees of elevation e1 at y1, then e1 > e2. If e2 ¼ 0,
then there is a tree of order 2 at y1; by the construction of U� (Case 3). If e1 >e2 > 0, then there is a tree of elevation e01 at y1, with 1 < e01 < e1, by Lemma 5.8.
Let u and v be two leaves of the tree with elevation e01. Then U� þ uv has a
path P of order at least k þ 1. Thus 1 þ e01 þ 1 þ e1 � e1 þ e2 þ 4. So e01 � e2 þ 2.
Further there is a path in U� through the tree of elevation e1 and the vertex y1
and the tree with elevation e01. So we must have e1 þ 1 þ e01 � k ¼ e1 þ e2 þ 3.
Hence e01 � e2 þ 2. &
Corollary 5.10. In U� there are always two trees with elevation greater than 1
at y1, unless e1 ¼ 1:
Finally, we let B be the tree obtained from U� by decycling the cycle C, i.e., by
adding a new vertex y to U�, together with the three edges yyi; i ¼ 1; 2; 3, and
deleting the edges of C. The following lemma establishes the fact that U� 2 X k.
Lemma 5.11. B 2 T k.
Proof. We first show that �ðBÞ ¼ k. Clearly �ðBÞ � �ðU�Þ ¼ k. Let P be any
path in B. If P contains y, then jVðPÞj � e1 þ e2 þ 3 ¼ k, while if not, it is in U�.It follows that �ðBÞ � k.
Next, we show that B is detour-saturated, and to this end, let u and v be
nonadjacent vertices in B. There are several cases (and subcases) to consider.
Case 1. y 6¼ u or v:
1.1. u and v are both C-vertices: Then a path P of order k þ 1 can be found in
Bþ uv containing disjoint paths of orders e1 and e2 and the four vertices
y; y1; y2, and y3.
1.2. At most, one of u and v is a C-vertex: Then the edge uv is not in EðU�Þ, and
so U� þ uv contains a path Q of order k þ 1. If Q uses only one C-edge,
say yiyj, then a path longer than Q can be constructed in Bþ uv by
replacing yiyj with yiyyj. On the other hand, if Q contains two C-edges, say
yiyj and yjyk, then replacing yj by y gives a path of order k þ 1 in Bþ uv.
Case 2. y ¼ u (say):
132 JOURNAL OF GRAPH THEORY
2.1. v is in F2 or F3; say F2: Then e2 > 0, so by our construction, there are two
trees of elevation e2 attached to y2. In one that does not contain v, take a
path of order e2 to y2, then the path from y2 to v, then the edges vy and yy1
followed by a path of order e1 in F1. The result is a path of order
e2 þ 4 þ e1 ¼ k þ 1.
2.2. v is in F1: By the construction of U� (Case 1) and Lemma 5.8, we know
that there are at least two trees in F1:2.2.1. All the trees in F1 have elevation e1: Then, as in Case 2.1, we can find
a path of order at least k þ 1 in Bþ uv.
2.2.2. Some tree in F1 has elevation e01 < e1: By Lemma 5.9, we know that
e1 ¼ e2 þ 2. If v is adjacent to y1, we can construct a path in Bþ uvof order at least e1 þ 2 þ e01 ¼ e1 þ e2 þ 4 ¼ k þ 1.
So we assume that v is not adjacent to y1. It follows that U� þ vy1
contains a path Q of order k þ 1. Let a be the number of vertices of Q
that lie in the tree attached to y1 that contains v. If Q contains neither
y2 nor y3, we can construct a path of order k þ 2 in Bþ uv by re-
placing vy1 in Q by vyy1.
On the other hand, if Q contains y2 or y3 (or both), the a vertices of Q
in F1 all lie in one of the trees attached to y1. Let R be a path in B
from y1 to a leaf in the tree in F1, which is avoided by Q. Then R has
at least e01 vertices. Further, since Q contains at least two cycle
vertices, we know that aþ 3 þ e2 � jVðQÞj ¼ k þ 1. We construct a
path P in Bþ uv by using the a vertices on the subpath of Q, which
ends at v, followed by y; y1; R. Thus jvðPÞj � aþ 2 þ e01 ¼ aþ e2 þ4 > k þ 1.
Hence, in any case, B is detour-saturated. &
Corollary 5.12. Dk � Uk.
Combining this with Theorem 5.5, we obtain our characterization.
Theorem 5.13. A connected, unicyclic graph with detour order k is detour-
saturated if and only if it is in Uk.
ACKNOWLEDGMENT
The authors thank the University of South Africa for financial support, which
facilitated reciprocal visits, during which this paper was written.
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