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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