On removable cycles through every edge

On Removable CyclesThrough Every Edge

Manoel Lemos1 and James Oxley2*1DEPARTAMENTO DE MATEMATICA

UNIVERSIDADE FEDERAL DE PERNAMBUCO, RECIFE

PERNAMBUCO 50740-540, BRAZIL

E-mail: [email protected]

2DEPARTMENT OF MATHEMATICS

LOUISIANA STATE UNIVERSITY

BATON ROUGE, LOUISIANA 70803-4918

E-mail: [email protected]

Received March 6, 2001; Revised August 21, 2002

DOI 10.1002/jgt.10082

Abstract: Mader and Jackson independently proved that every 2-connectedsimple graph G with minimum degree at least four has a removable cycle,that is, a cycle C such that G\E(C) is 2-connected. This paper considers theproblem of determining when every edge of a 2-connected graph G, simpleor not, can be guaranteed to lie in some removable cycle. The main resultestablishes that if every deletion of two edges from G remains 2-connected,then, not only is every edge in a removable cycle but, for every two edges,there are edge-disjoint removable cycles such that each contains one of thedistinguished edges. � 2002 Wiley Periodicals, Inc. J Graph Theory 42: 155–164, 2003

——————————————————

Contract grant sponsor: CNPq; Contract grant number: 522910/96-3; Contractgrant sponsor: ProNEx/CNPq; Contract grant number: 664107/97-4; Contractgrant sponsor: CAPES; Contract grant number: AEX1307/00-2; Contract grantsponsor: National Security Agency; Contract grant numbers: MDA904-99-1-0030and MDA904-01-1-0026.*Correspondence to: James Oxley, Department of Mathematics, Louisiana StateUniversity, Baton Rouge, Louisiana 70803-4918. E-mail: [email protected]

� 2002 Wiley Periodicals, Inc.

Keywords: removable cycle; 2-connected graph; matroid

1. INTRODUCTION

If G is a 2-connected simple graph in which the minimum degree � (G) is at least

four, then, as a special case of a result for k-connected graphs, Mader [7] showed

that G has a cycle C such that GnEðCÞ, the graph obtained by deleting the edges

of C, is 2-connected. Independently, Jackson [5] proved that C can be chosen

to avoid a nominated edge e and to have at least � (G)�1 edges. We call a cycle D

in a 2-connected graph H removable if HnEðDÞ is 2-connected. Lemos and

Oxley [6] extended Jackson’s theorem and obtained, as a corollary, that every

simple 2-connected graph with � ðGÞ � 5 has two removable edge-disjoint cycles

each with at least � ðGÞ þ 1 edges.

Although every simple 2-connected graph G with minimum degree at least four

has a removable cycle, the example shown in Figure 1(a), which was obtained

independently by Robertson (in [5]) and Jackson [5], shows that the requirement

that G be simple cannot be dropped. However, Fleischner and Jackson [2] proved

that this requirement could be replaced by the condition that G is planar. Their

result was extended by Goddyn, van den Heuvel, and McGuinness [3] who

proved the following strengthening of a conjecture of Jackson [5].

Theorem 1.1. Let G be a 2-connected graph with minimum degree at least four.

If G has no minor isomorphic to the Petersen graph, then G has two edge-disjoint

removable cycles.

In this paper, we consider when we can guarantee that every edge of a 2-

connected graph G is in some removable cycle. When G is simple, the re-

quirement that � ðGÞ � 4 is not sufficient to ensure the existence of such a family

of removable cycles for if G has an edge cut of size less than four, none of the

FIGURE 1

156 JOURNAL OF GRAPH THEORY

edges in this cut will be in a removable cycle. But even if we strengthen the

minimum-degree requirement to insist that every edge-cut has size at least four,

we shall not be guaranteed that every edge is in a removable cycle. For example,

every edge cut in the graph G in Figure 1(b) has size at least four, but the only

edges that are in removable cycles are those edges that are in 2-cycles. Although

G is not simple, a straightforward modification produces a simple graph G0 that

has no removable cycle meeting either the outer pentagon or the inner pentagram.

Specifically, we replace each pair fx; yg of parallel edges of G by a copy of K5

whose vertex set meets VðGÞ in the endpoints of x.

In the following theorem, the main result of the paper, the hypothesis is strong

enough to ensure not only that every edge is in a removable cycle but also that,

for every two edges, there is a pair of edge-disjoint removable cycles one con-

taining each of the distinguished edges. We observe here that, for the graph G

in Figure 2, every edge is in a removable cycle, but the only removable cycle

containing e also contains f . This example can be modified as above to produce a

simple graph satisfying the same assertions.

Theorem 1.2. Let G be a 2-connected graph. Suppose that GnX is a 2-

connected graph for every 2-subset X of EðGÞ, and let x and y be distinct edges

of G. Then G has edge-disjoint cycles Cx and Cy containing x and y, respectively,

such that both GnEðCxÞ and GnEðCyÞ are 2-connected.

For comparison with Theorem 1.1, we note the following immediate con-

sequence of Theorem 1.2.

Corollary 1.1. Let G be a 2-connected graph such that GnX is 2-connected for

every 2-subset X of EðGÞ. Then G has two edge-disjoint removable cycles.

The proof of Theorem 1.2 will be given in the next section. Section 3 describes

a counterexample to a natural extension of Theorem 1.2 to cographic matroids.

The graph and matroid terminology that is used here will follow Bondy and

Murty [1] and Oxley [8], respectively. A block is a connected graph with no cut-

vertices. A graph with at least three vertices is a block if and only if it is 2-

connected and loopless. For a path � that is not a cycle, if u and v are vertices of

�, then � ½u; v� denotes the subpath of � having u and v as its ends.

FIGURE 2

REMOVABLE CYCLES 157

2. PROOF OF THE MAIN RESULT

The following result is the main tool in the proof of Theorem 1.2. Once it is

proved, it will not be difficult to complete the proof of the theorem.

Proposition 2.1. Suppose that G and H are blocks such that

(i) H is a subgraph of G; and

(ii) GnX is a block for every X � EðGÞ � EðHÞ such that jXj � 2.

Then, for each e in EðGÞ � EðHÞ, either(iii) there is no cycle of G that contains e and is edge-disjoint from H; or

(iv) there is a cycle C of G that contains e such that C is edge-disjoint from

H and GnEðCÞ is a block.

Proof. Suppose that the proposition fails and choose a counterexample

ðG;H; eÞ for which ðjEðGÞj;�jEðHÞjÞ is lexicographically minimal. Then there is

a cycle C of G that contains e and is edge-disjoint from H. Moreover, GnEðCÞ is

not a block. Let B be the block of GnEðCÞ that contains EðHÞ. By the choice of

ðG;H; eÞ, we have that B ¼ H. Choose C such that the connected component of

GnEðCÞ that contains H as a block has the maximum number of edges.

In the rest of the proof, when � is a subpath of C, we define its complement �c

to be the other subpath of C having the same terminal vertices as �. In particular,

�ci and �0c will denote the complements of �i and �0, respectively.

Lemma 2.1. GnEðCÞ is connected.

Proof. Let H0;H1;H2; . . . ;Hn be the connected components of GnEðCÞ and

let H be a block of H0. We want to show that n ¼ 0. Assume that n � 1. For each

i in f1; 2; . . . ; ng, let �i be the shortest subpath of C containing e such that its

terminal vertices are in VðHiÞ. If �i¼C, then �i and Hi have a unique common

vertex, say vi. Since G is a block, vi is not a cut-vertex of G so vi is an isolated

vertex of GnEðCÞ. But then GnX is not a block when X consists of the two edges

of G incident with vi. This contradiction to (ii) implies that �i 6¼ C, so �i has

different terminal vertices. Let �0i be a path in Hi joining the terminal vertices of

�i. Note that �i [ �0i is a cycle C0 of G that contains e and is edge-disjoint from H.

Therefore,

Vð�ci Þ \ VðH0Þ ¼ ;; ð1Þ

otherwise the connected component of GnEðC0Þ that contains H contains EðH0Þ [Eð�c

i Þ, which is contrary to the choice of C.

Now, we shall define an auxiliary graph K such that VðKÞ ¼ f1; 2; . . . ; ng and

ij is an edge of K if and only if i 6¼ j and E(�ci Þ \ E(�c

j Þ 6¼ ;. Let K 0 be a

connected component of K. Let � be the union of all the paths �ci for which i in

VðK 0Þ. Note that � is a subpath of C that does not contain e. We set

V 0 ¼[

i2VðK 0ÞVðHiÞ:


Note that V 0 6¼ VðGÞ because V 0 \ VðH0Þ ¼ ; by the definition of K 0. Let X be

the set of edges of G that join a vertex of V 0 to a vertex of VðGÞ � V 0. By our

construction, X � EðCÞ and, since G is a block, jXj � 2. We shall prove that X is

the set of terminal edges of �c.

First we show that X \ Eð�Þ ¼ ;. Suppose that f 2 X \ Eð�Þ. By definition

of �, the edge f 2 Eð�ci Þ for some i 2 VðK 0Þ. Let f ¼ xy where x 2 V 0 and

y 2 VðGÞ � V 0. By (1), y 62 VðH0Þ. Thus y 2 VðHjÞ for some j in f1; 2; . . . ; ng.

Hence j 62 VðK 0Þ. Let g be the edge of C � ffg incident to y. By definition of �j,

we have that Eð�cj Þ \ ff ; gg 6¼ ;. The definition of K now implies that Eð�c

i Þ \Eð�c

j Þ ¼ ;. Hence g 2 Eð�cj Þ � Eð�c

i Þ and f 2 Eð�ci Þ � Eð�c

j Þ. Thus f is a termi-

nal edge and y a terminal vertex of �ci and hence of �i. Therefore y 2 VðHiÞ.

But y 2 VðHjÞ so j ¼ i. This contradiction implies that X \ Eð�Þ ¼ ;.

We now know that X � Eð�cÞ. Let f 2 X. Then f is incident with some vertex

x of V 0 and x 2 VðKiÞ, say. Since f 2 Eð�cÞ, we deduce that f 2 Eð�iÞ. Now

no internal vertex of �i is in Hi, so x is a terminal vertex of �i. Thus the edge g of

C � f that is incident with x is in Eð�ci Þ and so is in Eð�Þ. Hence f must be a

terminal edge of �c. Thus jXj � 2. But jXj � 2, so jXj ¼ 2 and we have a con-

tradiction since GnX is disconnected. &

Let w1;w2; . . . ;wn be the cut vertices of GnEðCÞ belonging to H. For each k in

f1; 2; . . . ; ng, let Gk be the connected component of GnðEðHÞ [ EðCÞÞ that

contains wk. We set Vk ¼ VðGkÞ.Lemma 2.2. Suppose that fu; vg � Vk \ VðCÞ and that � is a uv-path of Gk.

Let K be the connected component of GknEð�Þ such that wk 2 VðKÞ. If � is the

path in C that joins u and v and does not contain e, then Vð�Þ � Vk or

Vð�Þ \ VðKÞ ¼ ;.Proof. Observe that the union D of �c and � is an Eulerian graph since �c and

� are edge-disjoint paths that have the same terminal vertices. Thus D is a union

of cycles of G. One of these cycles, say C00, contains e. When Vð�Þ \ VðKÞ ¼ ;,

the lemma holds. Therefore, we may suppose that there is a vertex x in Vð�Þ \VðKÞ. Let � be a wkx-path in K. We choose x so that � has minimum length.

Hence Vð�Þ \ Vð�Þ ¼ fxg. In particular, the union of � with either of the paths

�½u; x� or �½x; v� is also a path.

Now suppose that Vð�Þ 6� Vk and choose a vertex w of Vð�Þ � Vk such that its

distance to x in the path � is a minimum. Then, w is a vertex of �½u; x� or �½x; v�.Without loss of generality, we may assume the latter. It follows, by the choice of

w, that Vð�½x;w�Þ � Vk ¼ fwg. Consider the wkw-path � that is the union of � and

�½x;w�.Suppose w 2 VðHÞ. As � is edge-disjoint from H and C00, it follows that

EðHÞ [ Eð�Þ is contained in some block H0 of GnEðC00Þ. This contradicts the

choice of ðG;H; eÞ since H0 has more edges than H. Therefore, w 62 VðHÞ.We may now suppose that w 2 Vl for some l 6¼ k. Let �l be a wwl-path in Gl.

Observe that Vð�lÞ \ Vð�Þ ¼ fwg because Vð�Þ � VðKÞ � Vk and Vð�½x;w�Þ�


Vk ¼ fwg. Thus the union of � and �l is a wkwl-path " that is edge-disjoint from

H and C00. Therefore, EðHÞ [ Eð"Þ is contained in a block H00 of GnEðC00Þ. This

contradicts the choice of ðG;H; eÞ since H00 has more edges than H. We deduce

that w does not exist, so Vð�Þ � Vk. &

Let �k be the longest path in C that has both ends in Vk and does not contain e.

Next we show that

2.3. Vð�kÞ � Vk 6¼ ;:Let X be the set of edges of G that join a vertex of Vk � fwkg to a vertex of

VðGÞ � Vk. Suppose that Vð�kÞ � Vk. Then X � Eð�ckÞ. But the choice of �k

means that only the terminal vertices of �ck are in Vk. Hence each edge of X is

terminal in �ck. Thus jXj � 2. Since GnX has wk as a cut-vertex, it is not difficult

to obtain a contradiction. We conclude that (2.3) holds.

Let ak and bk be the ends of �k. Traverse �k beginning at ak and let vk be the

first vertex that is not in Vk. Let �0k be the shortest subpath of �k that contains vk

and has both ends in Vk. Suppose that �0k joins the vertices a0k and b0k and that ak;

a0k; vk; b0k; bk appear in this order in �k. Note that a0k; vk, and b0k are distinct, but ak

may equal a0k, and bk may equal b0k. We define

Ak ¼ Vð�k½ak; a0k�Þ \ Vk and Bk ¼ Vð�k½b0k; bk�Þ \ Vk:

By the choice of vk, we have that Ak ¼ Vð�k½ak; a0k�Þ.We now define a simple auxiliary graph M with vertex set Ak. If a and a0 are

distinct members of Ak, then aa0 2 EðMÞ if and only if there is no b in Bk such

that Gk has two edge-disjoint paths joining fa; a0g and fb;wkg.

When f is an isthmus of Gk, we denote by Gkð f Þ the connected component of

Gkn f whose vertex set does not contain wk.

Lemma 2.3. Let fa; a0g be a subset of Ak such that if a 6¼ a0, then aa0 2 EðMÞ.Then

(i) Gk has an isthmus that separates fa; a0g from wk.

(ii) If f is an isthmus of Gk that separates fa; a0g from wk such that jVðGkð f ÞÞjis a minimum, then

(a) VðGkð f ÞÞ \ Bk ¼ ;; and

(b) if N is the connected component of M such that fa; a0g � VðNÞ, thenVðGkð f ÞÞ \ Ak � VðNÞ.

Proof. We shall prove (i) and (ii)(a) simultaneously. To do this, we take J ¼Gk and w ¼ wk when there is no isthmus separating fa; a0g from wk; and, when

the hypothesis of (ii)(a) holds, we take J to be Gkð f Þ and w to be the unique

vertex in VðGkð f ÞÞ \ Vðf fgÞ.Next we observe that there is no isthmus in J that separates fa; a0g from w.

This is true by definition when J ¼ Gk. Moreover, if J ¼ Gkð f Þ and g is such an


isthmus, then g is also an isthmus in Gk that separates fa; a0g from wk. Since

GkðgÞ is clearly a subgraph of Gkð f Þ � w, the choice of f is contradicted. Hence

g does not exist.

By the result of the last paragraph, Menger’s Theorem implies that J has an

aw-path � and an a0w-path �0 such that Eð�Þ \ Eð�0Þ ¼ ;.

Next we show that

2.3.1. VðJÞ \ Bk ¼ ;:

If b 2 VðJÞ \ Bk, then choose a path � of minimum length joining b to some

vertex of � or �0, say �0. Let b0 be the common vertex of �0 and � and let � be the

union of �0½a0; b0� and �. Then � and � are edge-disjoint paths of J joining a to w

and a0 to b, respectively. The path � can be completed to an awk-path " of Gk that

is edge-disjoint from � by taking its union with a wwk-path in GknEðJÞ. Hence

aa0 62 EðMÞ and so a ¼ a0. Now apply Lemma 2.2 taking ðu; v; �Þ to be ða0; b; �Þ.The path � implies that a0 2 Vð�k½a0; b�Þ \ VðKÞ, where K is the connected com-

ponent of GknEð�Þ that contains wk. Thus Vð�k½a0; b�Þ \ VðKÞ 6¼ ;. Moreover,

the definitions of Ak and Bk imply that Vð�k½a0; b�Þ 6� Vk. Thus Lemma 2.2 is

contradicted. Therefore b does not exist and so (2.3.1) holds.

As ; 6¼ Bk � VðGkÞ, it follows that J 6¼ Gk. Thus (i) holds. Moreover, J ¼Gkð f Þ and (ii)(a) holds by (2.3.1). Observe that (ii)(b) follows because any path of

Gk joining a vertex of VðGkð f ÞÞ to a vertex of fwkg [ Bk must contain f as an

edge by (ii)(a). It follows by the definition of M that any two distinct elements of

Ak \ VðGkð f ÞÞ are adjacent in M. &

Lemma 2.4. If N is a connected component of M, then there is an isthmus f of

Gk such that

VðGkð f ÞÞ \ Ak ¼ VðNÞ and VðGkð f ÞÞ \ Bk ¼ ;:

Proof. We shall prove the following assertion by induction on jV j.

2.4.1. If V � VðNÞ and N½V� is connected, then there is an isthmus fV of Gk such

V � VðGkð fVÞÞ \ Ak � VðNÞ and VðGkð fVÞÞ \ Bk ¼ ;:

The lemma follows by taking V ¼ VðNÞ in (2.4.1).

Suppose that jVj ¼ 1, say V ¼ fag. Now taking a0 ¼ a in Lemma 2.3(i), we get

that there is an isthmus f of Gk that separates fa; a0g from wk. Choose f so that

jVðGkð f ÞÞj is a minimum. Then (2.4.1) follows from Lemma 2.3(ii) by taking

ffag ¼ f . Now assume that (2.4.1) holds for jV j < n and let jVj ¼ n � 2. Since

N½V� is connected, there is an a in V such that N½V � fag� is connected. By in-

duction, the isthmus fV�fag exists. If a 2 VðGkð fV�fagÞÞ, then we take fV ¼ fV�fagand the result follows. We may now suppose that a 62 VðGkð fV�fagÞÞ. Choose a0

in V � fag such that aa0 2 EðNÞ. By Lemma 2.3(i), there is an isthmus f of Gk


that separates fa; a0g from wk because aa0 2 EðMÞ. Choose such an f for which

jVðGkð f ÞÞj is a minimum.

Note that, since a0 2 V � fag � VðGkð fV�fagÞÞ, any a0wk-path in Gk contains

fV�fag and f as edges. Moreover, since fV�fag separates a from a0 in Gk, it follows

that, in each a0wk-path in Gk, the edge fV�fag is closer to a0 than f is. Thus Gkð f Þhas Gkð fV�fagÞ as a subgraph. Hence V � VðGkð f ÞÞ. The remaining parts of

(2.4.1) follow from Lemma 2.3(ii) by taking fV ¼ f . &

Now let N be a connected component of M. By the previous lemma, there is an

isthmus f of Gk such that

VðGkð f ÞÞ \ Ak ¼ VðNÞ and VðGkð f ÞÞ \ Bk ¼ ;: ð2Þ

Let fx; yg be the subset of VðGkð f ÞÞ \ Ak such that ak; x; y; a0k appear in this

order in �k and all the other vertices of VðGkð f ÞÞ \ Ak lie between x and y on �k.

Next we prove the following:

2.4.2. Every vertex of �k½x; y� is in VðGkð f ÞÞ \ Ak.

Suppose that there is a vertex z of G such that z 2 Vð�k½x; y�Þ but

z 62 VðGkð f ÞÞ. Since Vð�k½ak; a0k�Þ ¼ Ak and fx; yg � Vð�k½ak; a0k�Þ, it follows

that z 2 Ak. By the definition of M, the vertices x and z are in different

components of M. Thus there are edge-disjoint paths, � and �, joining fx; zg to

fwk; bg for some b 2 Bk. First suppose that � and � join x to b and z to wk,

respectively. We know that vk is in the path in C that joins x and b and does not

contain e. Since vk 62 Vk, Lemma 2.2 implies that z and wk are in different

connected components of GknEð�Þ. This contradiction to the fact that � joins z to

wk implies that we may assume that � joins z to b and � joins x to wk.

Let VðffgÞ \ VðGkð f ÞÞ ¼ fwg. Since x 2 VðGkð f ÞÞ, the isthmus f of Gk

separates wk from x. Thus f is an edge of �. Hence w is a vertex of �, and f is not

an edge of z. As y 2 VðGkð f ÞÞ, there is a yw-path g of Gkð f Þ. The union " of �and �½w;wk� is a ywk-path of Gk. Moreover, e is edge-disjoint from z because zand � are edge-disjoint, and Vð�Þ � VðGkð f ÞÞ while Vð�Þ avoids VðGkð f ÞÞ since

f is not an edge of z and z 62 VðGkð f ÞÞ. Now apply Lemma 2.2, letting

ðu; v; �Þ ¼ ðz; b; �Þ. Since �k½z; b� contains vk, it is clear that Vð�k½z; b�Þ 6� Vk.

Therefore, Vð�k½z; b�Þ \ VðKÞ ¼ ;, where K is the connected component of

GknEð�Þ that contains wk. But y 2 Vð�k½z; b�Þ, and the path " joins y to wk so

y 2 VðKÞ. Therefore y 2 Vð�k½z; b�Þ \ VðKÞ. This contradiction implies that

(2.4.2) holds.

We now complete the proof of Proposition 2.1. From (2.4.2), (2), and the

choice of x and y, we have that VðGkð f ÞÞ \ VðCÞ ¼ Vð�k½x; y�Þ. Thus there

are just two edges of C that join a vertex of Gkð f Þ to a vertex outside Gkð f Þ.Delete these edges. The resulting graph has f as an isthmus and has at least three

vertices. Therefore, it is not a block. This contradiction completes the proof of

Proposition 2.1. &


Proof of Theorem 1.2. The graph obtained from G by deleting all the loops

is a block. If we can prove the theorem in the case that G is a block, then it will

follow in general because each loop is itself a removable cycle. Thus assume that

G is a block. First, observe that Gne is a block for every edge e of G. Let x and

y be distinct edges of G. Since Gny is a block, G has a cycle C that contains x

but not y. Now applying Proposition 2.1 taking H to be the block with edge set

fyg, we obtain a cycle Cx of G such that GnEðCxÞ is a block, x 2 EðCxÞ, and

y 62 EðCxÞ. Since GnEðCxÞ is a block, there is a cycle C0 of this graph such that

y 2 EðC0Þ. Applying Proposition 2.1 again, this time taking H to be the cycle Cx,

we obtain a cycle Cy containing y such that EðCxÞ \ EðCyÞ ¼ ; and GnEðCyÞ is

a block. &

3. MATROID (NON)-EXTENSIONS

Mader’s theorem [7], with which we began this paper, implies that every 2-con-

nected simple graph in which every bond has at least four edges has a removable

cycle. In particular:

3.1. Every 2-connected simple graphic matroid M in which every cocircuit has

at least four elements has a removable circuit, that is, a circuit C such that M \C is

2-connected.

The matroid U3;7 shows that one cannot delete the word ‘‘graphic’’ from the

last result. Oxley [8, Problem 14.4.8] asked whether one could replace ‘‘graphic’’

with ‘‘binary’’ in (3.1) and, specializing this, Goddyn, van den Heuvel, and

McGuinness (in an earlier version of [3]) conjectured that (3.1) holds when we

replace ‘‘graphic’’ with ‘‘cographic’’. Lemos and Oxley [6] gave a counterexample

to this conjecture.

We describe this example next for it provides a counterexample to a natural

extension of Theorem 1.2 to cographic matroids. We begin with a copy of K5;5

having vertex classes V1 and V2. Then, for every 3-element set Z that is a subset

of V1 or of V2, adjoin two new degree-3 vertices vZ and wZ making each of them

adjacent to all the members of Z. The resulting graph G has 50 vertices and

145 edges. Evidently G is bipartite and has girth four. Moreover, G is 3-connected

and 3-edge-connected. It is shown in [6, Proposition 2.1] that G has no bond C�

for which the contraction G=C� is 2-connected. Thus M�ðGÞ has no remo-

vable circuit. However, M�ðGÞnX is 2-connected for all 2-subsets X of EðGÞbecause every single-edge contraction of G is 3-connected. Thus, by contrast with

the graphic case dealt with in Theorem 1.2, in the cographic case, the condition

that MnX is 2-connected for all 2-subsets X of EðMÞ is not strong enough to

guarantee the existence of even one removable circuit.

Although Mader’s theorem does not extend to binary or even cographic

matroids, by strengthening the hypothesis slightly, Goddyn and Jackson [4] were


able to extend the theorem to a class of binary matroids that includes cographic

matroids. In particular, they proved that if e is an element of a connected binary

matroid M such that M does not have both the Fano matroid and its dual as

minors and jXj � 5 for all cocircuits X of M not containing e, then M has a

removable circuit C that avoids e such that rðMnCÞ ¼ rðMÞ.

ACKNOWLEDGMENTS

Manoel Lemos was partially supported by CNPq (Grant No. 522910/96-3),

ProNEx/CNPq (Grant No. 664107/97-4), and CAPES (Grant No. AEX1307/00-2).

James Oxley was partially supported by the National Security Agency (Grant

Nos. MDA904-99-1-0030 and MDA904-01-1-0026).

REFERENCES

[1] J. A. Bondy and U. S. R. Murty, Graph Theory with Applications, North-

Holland, New York, 1976.

[2] H. Fleischner and B. Jackson, Removable cycles in planar graphs, J London

Math Soc (2) 31 (1985), 193–199.

[3] L. A. Goddyn, J. van den Heuvel, and S. McGuinness, Removable circuits in

multigraphs, J Combin Theory Ser B 71 (1997), 130–143.

[4] L. A. Goddyn and B. Jackson, Removable circuits in binary matroids, Combin

Probab Comput 8 (1999), 539–545.

[5] B. Jackson, Removable cycles in 2-connected graphs of minimum degree at

least four, J London Math Soc (2), 21 (1980), 385–392.

[6] M. Lemos and J. Oxley, On removable circuits in graphs and matroids, J Graph

Theory 30 (1999), 51–66.

[7] W. Mader, Kreuzungfreie a; b-Wege in endlichen Graphe, Abh Math Sem

Univ Hamburg 42 (1974), 187–204.

[8] J. G. Oxley, Matroid Theory, Oxford University Press, New York, 1992.


On removable cycles through every edge

Documents

Transcript of On removable cycles through every edge