Dominating circuits in regular matroids

Advances in Applied Mathematics 53 (2014) 72–111

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Dominating circuits in regular matroids Sean McGuinness Thompson Rivers University, Kamloops, BC V2C5N3, Canada

a r t i c l e

i n f o

Article history: Received 30 October 2012 Accepted 9 October 2013 Available online 22 November 2013 MSC: 05B35 05D40 Keywords: Matroid Binary matroid Minor

a b s t r a c t In 1971, Nash-Williams proved that if G is a simple 2-connected graph on n vertices having minimum degree at least 13 (n + 2), then any longest cycle C in G is also edge-dominating; that is, each edge of G has at least one end-vertex incident with a vertex of C . We say that a circuit C in a matroid M is dominating if each component of M /C has rank at most one. In this paper, we show that an analogous theorem holds for regular matroids. More specifically, suppose M is a simple, connected, regular matroid and let C be a circuit in M. We show that if |C ∗ | > r (3M ) + 1, for all cocircuits C ∗ in M which are disjoint from C , then either C is a dominating circuit, or there is a circuit D such that the symmetric difference of C with D is a circuit which is strictly larger than C . © 2013 Elsevier Inc. All rights reserved.

1. Introduction There are numerous theorems in graph theory relating a minimum degree condition on vertices of a graph to the length of a longest cycle. The first such result is a seminal theorem of Dirac [4]: Theorem 1.1 (Dirac). Let G be a simple graph of order n  3 having minimum degree at least Hamilton cycle.

n . 2

Then G has a

Since Dirac’s theorem, there has been a steady flow of similar theorems about cycles in graphs. The concept of a cycle in a graph corresponds to the more general concept of a circuit in a matroid. While some properties of cycles in graphs easily extend to circuits in matroids, there are other properties for which there is no simple answer. With Dirac’s theorem in mind, Welsh [17] posed the following conjecture: Conjecture 1.2 (Welsh). If M is a simple connected regular matroid and every cocircuit has at least 1 (r ( M ) + 1) elements, then M has a circuit of size r ( M ) + 1. 2 0196-8858/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.aam.2013.10.001

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In two papers [8], Hochstättler and Jackson verified the above conjecture of Welsh by using a decomposition theorem of Seymour [18] for regular matroids. Various other authors have exploited Seymour’s theorem for the purpose of extending known theorems about graphs to regular matroids or proving new theorems (see [3,6,7,9–13]). With proofs of this type, one must consider the cases of graphic and cographic matroids separately. In contrast, the proof we give is purely matroid-theoretic. While no graph theory is used, many of the ideas introduced owe their intuitive basis to the properties of cycles and cocycles in graphs. Graph theorists have also been interested in conditions which guarantee the existence of a socalled dominating cycle, a generalization of a Hamilton cycle. A cycle in a graph is dominating if every edge of the graph is incident with at least one vertex of the cycle. The dominating cycle counterpart to Dirac’s theorem is the following result of Nash-Williams [16]: Theorem 1.3 (Nash-Williams, 1971). Let G be a simple 2-connected graph of order n  3. If G has minimum degree at least 13 (n + 2), then any longest cycle in G is a dominating cycle. Since 1971, there have been many other variations and strengthenings of Nash-Williams’ result for graphs (see Voss [19]), but none thus far for matroids. In this paper, I show, using only basic concepts and ideas involving matroids, that Nash-Williams’ theorem can be extended in a natural way to regular matroids. Throughout this paper, C ( M ) (respectively C ∗ ( M )) shall denote the set of circuits (respectively cocircuits) of a matroid M. Let M be a matroid and let X ⊆ E ( M ). A subset Y ⊆ E ( M )\ X is called an X -bridge if Y is a component of M / X . We define the order of Y to be o(Y ) = r M / X (Y ). We shall let K ( M / X ) denote the set of X -bridges of M. We say that a circuit C in M is dominating if o(Y )  1 for each C -bridge Y . Let M be a connected binary matroid with ground set E and let C ∈ C ( M ). We say that a circuit D is C -minimal if C ∩ D = ∅ and C , D, and C  D are exactly the circuits contained in C ∪ D. We observe that if D is a C -minimal circuit, then C  D is also a C -minimal circuit. We say that a circuit D is C -augmenting if it is a C -minimal circuit for which |C  D | > |C |. The goal of this paper is to prove the following theorem. Theorem 1.4. Let M be a simple, connected, regular matroid and let C be a circuit in M. Suppose that |C ∗ | > r(M ) + 1 for all cocircuits C ∗ in M which are disjoint from C . Then either C is a dominating circuit, or there 3 exists a C -augmenting circuit D. An immediate corollary of the above theorem is the following result which one can view as an extension of Nash-Williams’ result to matroids. Corollary 1.5. Let M be a simple, connected, regular matroid where |C ∗ | > r (3M ) + 1 for all cocircuits C ∗ in M. Then any longest circuit of M is dominating. If we apply Corollary 1.5 to cographic matroids, which are matroids whose circuits are minimal cutsets (or cocycles) of a graph, we obtain a theorem about graphs which is interesting in its own right. Observe that a cocycle C ∗ in a cographic matroid M ∗ (G ) is dominating if every block of G \C ∗ is either an edge or a cycle. For a graph G, we let g (G ) denote its girth (length of a shortest cycle) and let g ∗ (G ) denote its cogirth (size of a smallest cocycle). Then we have the following result: | E (G )|−n+1

Corollary 1.6. Let G be a 2-connected graph of order n. Suppose that g ∗ (G )  3 and g (G ) > Then any largest cocycle C ∗ of G has the property that every block of G \C ∗ is an edge or a cycle.

3

+ 1.

2. A brief overview The proof of the main result builds on the concepts and ideas introduced in [15] that uses the minimum counterexample approach. Before we examine the properties of a minimal counterexample,

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we shall introduce various other preliminary results. An important aspect of the proof is the way in which sets in a binary matroid with no F 7 -minor can intersect or “cross” each other. This is the topic of Section 3. In Section 4, we introduce the notion of a pseudowheel. An important result here is that for a circuit Ω in a regular matroid, any Ω -bridge of order one corresponds to the “spokes” of a pseudowheel with “rim” Ω . In Section 5, we introduce the concept of a chord-circuit. We show that the way in which chord-circuits can “cross” each other is restricted when the matroid has no F 7 -minor. This will play a significant role when constructing C -augmenting circuit. In Section 6, we shall define, in the case where Ω is a spanning circuit of a matroid, so-called Ω -cocircuits. Such cocircuits contain exactly two elements of Ω where the remaining elements are chords of Ω . We shall later show that, given a binary matroid has no F 7 -minor, one can find an Ω -cocircuit with the property that its elements “fan out” across Ω . In Section 7, we shall look at the properties as a minimum counterexample to the theorem. Assuming Theorem 1.4 is false, we shall choose among its counterexamples a matroid M for which | E ( M )| is minimum. There is a circuit C in M for which

• C is a non-dominating circuit of M for which |C ∗ | > C = ∅. • M contains no C -augmenting circuit.

r(M ) 3

+ 1 for all cocircuits C ∗ where C ∗ ∩

One important step is to show that there is a C -minimal circuit D where | D \C | = |C ∩ D | and Ω = C  D is a dominating circuit of N = M \(C ∩ D ). This is a key property which will be heavily exploited in the remaining sections of the paper and we shall briefly outline the work that follows. Let U = D \C , V = C \ D, and W = C ∩ D. Let μ = |U |, υ = | V |, and ω = | W |, noting that μ = ω . We may replace W by a chord w of Ω . If μ > υ , then D would be C -augmenting circuit. Thus μ  υ , and there exists δ  0 such that υ = μ + δ . Given that Ω is a dominating circuit of N, one can show that every Ω -bridge is either a chord of Ω or the spoke-set of a pseudowheel with rim Ω . We shall examine how these Ω -bridges “overlap”, and later use them to construct C -augmenting circuits. There are two main cases to consider: the case where Ω is a spanning circuit of N, and the case where it is not. In the case where Ω is a spanning circuit, all the Ω -bridges are chords of Ω . We show that one can find a C -augmenting circuit using these chords and Ω . The methods applied here are derived from both the graphic and cographic cases; C can be augmented as if it were a cycle or a cocycle in a graph, depending on the situation. The methods work largely due to the fact that the chords of Ω “cross” each other either like chords of cycle, or chords of a cocycle in j a graph. For each chord x of Ω , we let Ωx , j = 1, 2 be the circuits of Ω ∪ {x} which contain x. j

j

j

j

Furthermore, for j = 1, 2, we let U x = Ωx ∩ U and V x = Ωx ∩ V . The fact that M has no F 7 -minor can be used to find “nested” sets of chords of Ω . A set of chords X of Ω is w-nested if there exists j j j an ordering x1 , x2 , . . . , xm of X and j 1 , j 2 , . . . , jm ∈ {1, 2} such that ∅ =  U x11 ⊆ U x22 ⊆ · · · ⊆ U xmm ⊂ U j

j

j

 V x11 ⊂ V x22 ⊂ · · · ⊂ V xmm ⊂ V . In Section 9, we show that under certain conditions, if one has and ∅ =

two w-nested sets X i , i = 1, 2 such that | X 1 | + | X 2 | > 23 (δ + 1), or if one has three w-nested sets X i ,

i = 1, 2, 3 for which | X 1 | + | X 2 | + | X 3 | > 2δ , then M has a C -augmenting circuit. In Sections 10 and 11, we show how one can find such w-nested sets as described above. In the second part, we deal with the case where Ω is not a spanning circuit of N and there are Ω -bridges of order one. For every Ω -bridge X of order one, the set X ∪ Ω forms a pseudowheel, as is shown in Lemma 4.1. For each pair of elements x, y ∈ X , there are two circuits in Ω ∪ {x, y } which j contain x and y. These circuits are denoted by Ωxy , j = 1, 2. The pseudowheel determined by X ∪ Ω j

is said to be V -complete if for all x, y ∈ X we have Ωxy ∩ V = ∅, j = 1, 2. In Section 12, we first examine how chords and pseudowheels of Ω overlap. In Section 13, we show that if we can find a w-nested set of n chords and V -complete pseudowheel with m spokes, then M has a C -augmenting circuit if m + n > max{ 2δ + 2, 23 (δ + 1) + 1}. This in turn can be used to show that if one has two V -complete pseudowheels with m and n spokes, respectively, where m and 4 n are greater than δ+ , then M has a C -augmenting circuit. In Section 14, we show that if there are 3 two Ω -bridges of order one, then one can find two V -complete pseudowheels as described above. Later in the same section, we show that if there is only one Ω -bridge which of order one, then one can find a w-nested set of chords and a V -complete pseudowheel as described above.

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3. Crossing sets and F 7 -minors In this section, we shall describe a sufficient condition for the existence of an F 7 -minor in a binary matroid. We shall first define some concepts and notation for sets. Let A and B be subsets of a set X . ◦

If X is the disjoint union of A and B, then we shall write X = A ∪ B. We say that A and B cross in X if the sets A ∩ B, ( X \ A ) ∩ B, A ∩ ( X \ B ), and ( X \ A ) ∩ ( X \ B ) are all nonempty subsets. In shorthand, we write A  B, the set X being implicitly known in most cases. If A does not cross B, then we write A   B. If either A ⊆ B or B ⊆ A, then we shall write A B. If this is not the case, we write A  B. Furthermore, if A ∩ B = ∅ or A ∪ B = X , then we shall write A B. If this is not the case, we write A ∦ B. It is seen that A B (respectively, A B) if and only if ( X \ A ) ( X \ B ) (respectively, ( X \ A ) ( X \ B )). It is also straightforward to see that A  B if and only if A B or A B. A collection of sets S is non-crossing if each pair of sets in S is non-crossing. A k-circuit is defined to be a circuit having k elements. A 2-circuit will also be called a digon and a 3-circuit is also called a triangle. / C is called a chord Let M be a binary matroid. For a circuit C ∈ C ( M ), a non-loop element e ∈ of C if e ∈ cl(C ). There are exactly two circuits of C ∪ {e } which contain e; we call these circuits i i i the chord-circuits of e with respect to C and denote them by C M ,e , i = 1, 2. We let C M ,e = C M ,e \{e }, i = 1, 2. For convenience, we shall usually drop M from our notation when M is implicit. Additionally, C e in place of  C e1 . We say that a chord-circuit C ei is minimal we shall just write C e in place of C e1 and  j if for all chords f and for all j ∈ {1, 2} we have  C f ⊂  C ei .

Lemma 3.1. Let M be a binary matroid. Suppose C 1 , C 2 , C 3 are 3-circuits intersecting in a common element e, and M |C 1 ∪ C 2 ∪ C 3 is simple. If there exists a 3-circuit C 4 where e ∈ / C 4 and C 4 intersects each of C 1 , C 2 , C 3 , then M |C 1 ∪ C 2 ∪ C 3  F 7 . Proof. This follows from three applications of the fact that if 123, 345, and 561 are circuits in a binary matroid, then 246 is also a circuit. 2 Let M be a connected binary matroid and let C be a circuit. Let C 1 , C 2 , C 3 ∈ C ( M ) where e ∈ C 1 ∩ C 2 ∩ C 3 and let f i , i = 1, 2, 3 be distinct elements where C i \(C ∪ {e }) = { f i }, i = 1, 2, 3. In addition, assume that either e ∈ C or e , f 1 , f 2 , f 3 ∈ E ( M )\cl M (C ). If (i) C i C , i = 1, 2, 3 are circuits of M and (ii) there exist distinct elements g i ∈ C i ∩ C , i = 1, 2, 3 where g i ∈ C j if and only if i = j, →

then we write ( g 1 , g 2 , g 3 ) C ,e (C 1 , C 2 , C 3 ). →

Lemma 3.2. If ( g 1 , g 2 , g 3 ) C ,e (C 1 , C 2 , C 3 ), then M has an F 7 -minor. Proof. Let N = M |C ∪ C 1 ∪ C 2 ∪ C 3 , and let N  = N /(C \{e , g 1 , g 2 , g 3 }). Then N  is simple, and {e , f 1 , g 1 }, {e , f 2 , g 2 }, {e , f 3 , g 3 } are seen to be circuits of N  . If e ∈ C , then {e , g 1 , g 2 , g 3 } is also a circuit of N  and { f 1 , f 2 , f 3 } = {e , f 1 , g 1 }{e , f 2 , g 2 }{e , f 3 , g 3 }{e , g 1 , g 2 , g 3 } is also a circuit (because N  is simple) which intersects the circuits {e , f 1 , g 1 }, {e , f 2 , g 2 }, {e , f 3 , g 3 } containing e. It / C , then { g 1 , g 2 , g 3 } is a circuit of N  which intersects follows from Lemma 3.1 that N   F 7 . If e ∈ the 3-circuits {e , f 1 , g 1 }, {e , f 2 , g 2 }, and {e , f 3 , g 3 }. Again, Lemma 3.1 implies that N   F 7 . 2 4. Pseudowheels Let M be a simple binary matroid which has a circuit-hyperplane C . Let C ∗ = E ( M )\C , which is a cocircuit. Assume that |C ∗ |  2. For any two elements e , f ∈ C ∗ , there are exactly two circuits j of C ∪ {e , f } which contain e and f . We denote such circuits by C M ,e f , j = 1, 2. For j = 1, 2 we

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j j let  C M ,e f = C M ,e f \{e , f }. Normally, we shall drop M from our notation when it is implicit. Also, to further simplify our notation, we shall usually just denote C e1f and  C e1f by C e f and  C e f , respectively. ∗ An ordering x0 , x1 , x2 , . . . , xn of C is called a cyclic ordering if there exist j 0 , . . . , jn ∈ {1, 2} such that j the sets  C xii xi+1 , i = 0, 1, . . . , n form a partition of C (where xn+1 = x0 ). If there exists a cyclic ordering for C ∗ , then we say that M is a pseudowheel with rim C and spoke-set C ∗ . Note that our definition of a pseudowheel here is more general than the standard definition of the wheel Wr , the cycle matroid of the “wheel” graph on r + 1 vertices where one vertex (the “hub”) is joined to each vertex lying on an r-cycle (the “rim”). If one modifies this definition so that instead of requiring that the hub vertex to be joined to all vertices on the rim, one only requires that hub vertex be joined to at least two vertices, then the cycle matroid of this graph corresponds to our definition of a pseudowheel. For convenience, we shall always assume that when x0 , . . . , xn is a cyclic ordering of C ∗ , the sets  C xi xi+1 , i = 0, 1, . . . , n partition C . We shall denote such a pseudowheel M by the pair M = (x, C ) where x = (x0 , . . . , xn ). We shall refer to x as a cyclic ordering for M. For any positive integer m, xm shall denote the element xi ∈ {x0 , . . . , xn } for which m ≡ i (mod n + 1). We note that several cyclic orderings for the same pseudowheel are possible. For i , j ∈ {0, . . . , n}, i = j, we see that C xi xi+1 C xi+1 xx+2  · · · C x j−1 x j is one j

of the circuits C xi x j , j = 1, 2. We shall adopt the convention that C xi x j = C xi xi+1 C xi+1 xx+2  · · · C x j−1 x j . Observe that by this convention, C xi x j = C x2j xi . The main result of this section is the following lemma:

Lemma 4.1. Let M be a simple, connected, regular matroid. Let C be a circuit-hyperplane of M and let C ∗ be the cocircuit E ( M )\C where |C ∗ |  2. Then M is a pseudowheel having rim C and spoke-set C ∗ . Proof. Let C ∗ = {x0 , x1 , . . . , xn }. We shall show that there exists a cyclic ordering for C ∗ . Clearly, if n = 1, then this is true. So we may assume that n  2. For all i , j ∈ {0, . . . , n} where i = j and for all k k k ∈ {1, 2} let C ikj = C xki x j and  C ikj =  C xki x j . For i = 1, . . . , n and k = 1, 2 let C ik = C 0i and  C ik =  C 0i . Let

C = {C ik | i = 1, . . . , n; k = 1, 2} and C= { C ik | i = 1, . . . , n; k = 1, 2}. (i) C is non-crossing.

Proof. Suppose to the contrary that two sets of C cross. We may assume that  C1   C 2 . For i = 1, 2 j C 1i ∩  C 2 . Since  C1   C 2 we have that E i j = ∅ for all i , j ∈ {1, 2}. We note that and j = 1, 2 let E i j =  ◦

C 1 C 2 = ( C 1  C 2 ) ∪ {x1 , x2 } = ( E 12 ∪ E 21 ) ∪ {x1 , x2 }. Thus C 1 C 2 ∈ {C x1 x2 , C x21 x2 } and we may assume ◦

that C x1 x2 = C 1 C 2 . Then C x21 x2 = C x1 x2 C = ( E 11 ∪ E 22 )∪{x1 , x2 }. Let N = M |(C 1 ∪ C 2 ∪ C ). For i = 1, 2 and j = 1, 2 let e i j ∈ E i j . Let N  = N /( E ( N )\{e 11 , e 12 , e 21 , e 22 , x0 , x1 , x2 }). Then it is straightforward to show that N   F 7∗ . This contradicts the fact that M is regular. We conclude that C is non-crossing. 2 (ii) A maximum antichain in C has size two. Proof. Suppose to the contrary that C contains an antichain of size 3. We may assume that  Ci , i = 1, 2, 3 is such an antichain. Suppose first that the sets are pairwise disjoint. Then letting g 1 ∈  C1, → g2 ∈  C 2 , and g 3 ∈  C 3 , it is seen that ( g 1 , g 2 , g 3 ) C ,x (C 1 , C 2 , C 3 ) and M has an F 7 -minor. This yields 0 a contradiction. Suppose instead that at least one pair of these sets intersects. We may assume that  C1 ∩ C 2 = ∅. By (i),  C 1   C 2 , and thus either  C1  C 2 or  C1  C 2 . Given that C i , i = 1, 2, 3 is an antichain, we have  Ci   C j for all i = j. Thus  C1   C 2 and hence  C1  C 2 . Since  C1 ∩  C 2 = ∅, it follows that  C1 ∪  C 2 = C , and hence  C 3 ∩ ( C1 ∪  C 2 ) = ∅. We may assume that  C1 ∩  C 3 = ∅. Then  C1 ∪  C 3 = C as        well. Now C = C 1 ∪ C 2 = C 1 ∪ C 3 , which implies that C 2 ∩ C 3 = ∅. Thus C 2 ∪ C 3 = C as well. Now it is seen that  C 1  C2 ⊆  C 3 , and thus  C 32 ⊆  C1 ∩  C 2 . From this, we see that the sets  C i2 , i = 1, 2, 3 are → pairwise disjoint. Letting g i ∈  C i2 , i = 1, 2, 3 we see that ( g 1 , g 2 , g 3 ) C ,x (C 12 , C 22 , C 32 ) and this yields a 0 contradiction. We conclude that a maximum antichain has size at most two, and hence it has size 2 exactly two since  Ci ,  C i is an antichain for i = 1, . . . , n. 2

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By Dilworth’s theorem (see [1]), the maximum size of an antichain equals the minimum number of chains which partition a poset. Thus by (ii), C is the disjoint union of two chains. Given that no pair  Ci ,  C i2 , i = 1, . . . , n belongs to the same chain, each of these chains contains exactly one set from C1 ⊂  C2 ⊂ · · · ⊂  C n and  C n2 ⊂  C n2−1 ⊂ · · · ⊂  C 12 are two each such pair. By reindexing, we may assume 

C1 =  C x0 x1 and  C i \ C i −1 =  C i  C i −1 ∈ { C x i −1 x i ,  C x2i−1 xi } for i = 2, . . . , n. Thus for some chains. We have 

i 1 , i 2 , . . . , in ∈ {1, 2} the sets  C x0 x1 ,  C x11 x2 , . . . ,  C xnn−−11 xn ,  C xnn x0 form a partition of C , and x = (x0 , . . . , xn ) ∗ is a cyclic ordering of C . We conclude that M = (x, C ) is a pseudowheel. 2 i

i

i

Note that the above lemma does not hold for binary matroids in general. For example, consider the Fano plane F 7 . Clearly any circuit-hyperplane C (that is, any line) cannot be the rim of a pseudowheel since the complement C ∗ has four elements and |C ∗ | > |C |. 5. Chord-circuits In this section, we shall assume that M is a connected, regular matroid and Ω is a circuit of M. The purpose of this section is to investigate how the chords of Ω “cross”. First, we shall show that certain rules apply to chords which cross a fixed chord of Ω . Secondly, we shall investigate how three chords can cross a fixed chord of Ω . We shall establish certain properties which will later be used to find “nested” sets of chords. These in turn will be used to construct augmenting circuits. Let X M ,Ω denote the set of chords of Ω in M. Normally, when M is implicit, we shall simply x  Ω y . For x, y ∈ X Ω and i , j ∈ {1, 2} we let write X Ω . We say that two chords x, y ∈ X Ω cross if Ω ij

j

j

1j

Ωxy = Ωxi ∩ Ω y . When i = 1 we usually write Ωxy in place of Ωxy . Similarly, when i = j = 1, we 11 shall simply write Ωxy in place of Ωxy . For a ∈ X Ω , we let X M ,a denote the set of chords of X Ω which cross a. We shall just write X a when M is implicit. A subset X ⊆ X a is said to be a-nested if either it is empty, or in the case when it is not empty, there exists an ordering x1 , x2 , . . . , xm of X j j j 2 j1 2j a and ∅ = Ωax ⊂ Ωax22 ⊂ · · · ⊂ and j 1 , j 2 , . . . , jm ∈ {1, 2} such that ∅ = Ωax11 ⊆ Ωax22 ⊆ · · · ⊆ Ωaxmm ⊂ Ω 1 2j a2 . Letting x = (x1 , x2 , . . . , xm ) and j = ( j 1 , j 2 , . . . , jm ), we call the pair (x, j) an a-ordering Ωaxmm ⊂ Ω for X . 2 Let a ∈ X Ω and let y ∈ X a . We define [ y ] M ,a = { y  ∈ X a | Ωay  ∈ {Ωay , Ωay }}. When M is implicit, we shall simply write [ y ]a in place of [ y ] M ,a . For the remainder of this section, we shall assume a and let B = Ω a2 . To simplify our that Ωa is a minimal chord-circuit with respect to X Ω . Let A = Ω j

j

j

2j

j

notation, we shall let A x = Ωax and B x = Ωax for all x ∈ X a and j ∈ {1, 2}. Let A = { A x | x ∈ X a , j ∈ {1, 2}}. Let x1 , x2 , x3 ∈ X Ω . We say that the chords x1 , x2 , x3 3-cross a if X a = {x1 , x2 , x3 } and there exist j

◦

j

◦

3− j 1 ◦

j

j 1 , j 2 , j 3 ∈ {1, 2} such that A = A x11 ∪ A x22 ∪ A x33 and B = B x1

3− j 2 ◦

∪ B x2

j C xii ,

3− j

∪ B x3 3 . In this case, we say i = 1, 2, 3 intersect when x1 ,

that a is 3-crossed. Fig. 1 illustrates the way in which the circuits x2 , x3 3-cross a. One of the goals in this section is to examine how three chords of X a cross each other. Specifically, we shall show that if there exist xi ∈ X a , i = 1, 2, 3 and βi ∈ {1, 2}, i = 1, 2, 3 such β β that the sets A xii , i = 1, 2, 3 are pairwise disjoint and the sets B xii , i = 1, 2, 3 are not pairwise disjoint, then x1 , x2 , x3 3-cross a. To do this, we shall need the next lemma (which also appears in [15]). This lemma will be of central importance throughout the paper. Lemma 5.1. We have the following: (i) A is a non-crossing collection of subsets of A. j j (ii) For all x, y ∈ X a and for all i , j ∈ {1, 2} we have A ix A y iff B ix B y . Furthermore, if M | X Ω is simple and j

j

j

A ix = A y , then either B ix ⊂ B y or B y ⊂ B ix . j

j

(iii) For all x, y ∈ X a and for all i , j ∈ {1, 2}, we have A ix A y iff B ix B y . Proof of (i). Let x, y ∈ X a . It suffices to show that A x  A y . Suppose to the contrary that A x  A y . Then there exist elements a1 , a2 , a3 , a4 where a1 ∈ A x ∩ A y , a2 ∈ A x \ A y , a3 ∈ A y \ A x , and a4 ∈ A \( A x ∪ A y ).

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j

Fig. 1. The circuits C xii , i = 1, 2, 3 where x1 , x2 , x3 3-cross a.

Case 1. Suppose B x  B y . There exist b1 ∈ B x \ B y and b2 ∈ B y \ B x . We have that a1 ∈ Ωa ∩ Ωx ∩ Ω y , and a4 ∈ Ωa \(Ωx ∪ → Ω y ). We also have that b1 ∈ Ωx \(Ωa ∪ Ω y ) and b2 ∈ Ω y \(Ωa ∪ Ωx ). Consequently, (a4 , b1 , b2 ) Ω, a1 (Ωa , Ωx , Ω y ). By Lemma 4.1, M has an F 7 -minor, a contradiction. Case 2. Suppose B x B y . Without loss of generality, we may assume B x ⊆ B y . Let b1 ∈ B x , and b2 ∈ B y = B \ B y . Then a2 ∈ Ωx \(Ωa ∪ Ω y ) and a3 ∈ Ω y \(Ωa2 ∪ Ωx ). Furthermore, b1 ∈ Ωa2 ∩ Ωx ∩ Ω y and b2 ∈ Ωa2 \(Ωx ∪ Ω y ). → Consequently, (b2 , a2 , a3 ) Ω,b (Ωa2 , Ωx , Ω y ). Again this gives a contradiction. 1

From Cases 1 and 2 we conclude that A x  A y . j

2 j

j

From (i) it follows that for all A ix , A y ∈ A , either A ix A y or A ix A y . j

j

j

j

Proof of (ii). By contradiction. Suppose A ix ⊆ A y and B ix  B y . Then ∃b1 ∈ B ix \ B y and ∃b2 ∈ B y \ B ix . j

j

Let a1 ∈ A ix and a2 ∈ A \ A y ; a2 exists since y crosses a and hence A  A y . Now it is seen that j j ∩ a2 ∈ Ωa \(Ωxi ∪ Ω y ), b1 ∈ Ωxi \(Ωa ∪ Ω y ), and b2 ∈ C y \(Ωa ∪ Ωxi ). Consequently, a1 ∈ Ωa ∩ j j j → i (a2 , b1 , b2 ) Ω,a1 (Ωa , Ωx , Ω y ). This gives a contradiction. We conclude that if A ix A y , then B ix B y . j j j j A similar proof also works in reverse; that is, if B ix B y , then A ix A y . Thus A ix A y , iff B ix B y . If j j j i i i M | X Ω is simple and A x = A y , then B x = B y ; otherwise, Ωx Ω y = {x, y }, implying that {x, y } conj tains a circuit, contradicting the fact that M | X Ω is simple. Thus if A ix = A y , and M | X Ω is simple, then j j either B ix ⊂ B y or B y ⊂ B ix . 2

Ωxi

j Ωy ,

Proof of (iii). From (ii), we have the following implications: A ix A y ⇔ A 3x −i A y ⇔ B 3x −i B y ⇔ B ix j

j B y.

Thus

A ix

j Ay

⇔

B ix

j B y.

2

Lemma 5.1 has some useful consequences one of which is the following: Lemma 5.2. Suppose that M | X Ω is simple. Then [x]a is a-nested for all x ∈ X a .

j

j

S. McGuinness / Advances in Applied Mathematics 53 (2014) 72–111

79

j

Proof. Let x ∈ X a and let u , v ∈ [x]a . Then A x = A iu = A v for some i , j ∈ {1, 2}. By Lemma 5.1(ii), either B ui ⊂

j Bv

have ∅ =

or

j Bv

i A x11

⊂ B ui . Thus for some i 1 , i 2 , . . . , im ∈ {1, 2} and some ordering x1 , x2 , . . . , xm of [x]a we = A ix22 = · · · = A ixmm ⊂ A and ∅ = B i 1 ⊂ B ix22 ⊂ · · · ⊂ B ixmm ⊂ B. Thus [x]a is a-nested. 2 β

Let xi ∈ X a , i = 1, 2, 3 and let βi ∈ {1, 2}, i = 1, 2, 3 be such that the sets A xii are pairwise disjoint. In the remainder of this section, we shall examine how the three chords cross. 3−β1 ◦

β

Lemma 5.3. If the sets B xii , i = 1, 2, 3 are not pairwise disjoint, then B = B x1 β

◦

β

◦

3−β2 ◦

∪ B x2

3−β3

∪ B x3

and A =

β

A x11 ∪ A x22 ∪ A x33 . β

β

Proof. Assume that the sets A xii , i = 1, 2, 3 are pairwise disjoint and the sets B xii , i = 1, 2, 3 are not βj

β

βj

β

pairwise disjoint. Then A xii A x j for all i = j and it follows from Lemma 5.1(iii) that B xii B x j for 3−β j

β

all i = j. Thus we also have B xii B x j β

β

that B x11 ∩ B x22 = ∅. Since B x11 B x2 3−β B x3 3

3−β B x3 3

β1

β1

3−β2

Bxj

for all i = j. We may assume

β 3−β3 β , either B x11 ⊆ 3−β3 3−β2 3−β3 3−β2 3−β3 or ⊆ B x1 . If B x1 ⊆ B x3 , then B x2 ⊆ B x3 , contradicting the fact that B x2 B x3 . 3−β3 β 3−β 3−β β 3−β 3−β 3−β Thus B x3 ⊆ B x11 and hence B x2 3 ∪ B x3 3 ⊆ B x11 ⊂ B. Since B x2 2 B x3 3 it follows that B x2 2 ∩ 3−β3 3−β2 3−β3 β1 β1 3−β2 3−β3 β1 β2 β3 B x3 = ∅. Suppose B x2 ∪ B x3 ⊂ B x1 . Let e ∈ B x1 \( B x2 ∪ B x3 ). Then e ∈ B x1 ∩ B x2 ∩ B x3 . Now β

3−β2

3−β j

3−βi

for all i = j and B xi

β

⊆ B x11 . Since B x11 B x3

, it follows that B x2

→

β

β

β

β

3−β2

◦

letting ai ∈ A xii , i = 1, 2, 3, it is seen that (a1 , a2 , a3 ) Ω,e (Ωx11 , Ωx22 , Ωx33 ). This gives a contradiction. 3−β2

β

We conclude that B x11 = B x2 β1

◦

◦

3−β3

∪ B x3 ◦

β2

◦

3−β1

and hence B = B x1

∪ B x2

3−β B xi i , 3−βi B xi ,

β3

To prove that A = A x1 ∪ A x2 ∪ A x3 , we observe that the sets 3−βi

but the sets A xi

, i = 1, 2, 3 are not. Thus by having the sets β

◦

◦

β

β

i = 1, 2, 3 in the above, one can deduce that A = A x11 ∪ A x22 ∪ A x33 . β

3−β3

∪ B x3

.

i = 1, 2, 3 are pairwise disjoint β

i = 1, 2, 3 play the role of A xii ,

2

β

Lemma 5.4. If there exists x ∈ X a \{x1 , x2 , x3 } such that B x = B xii for all i ∈ {1, 2, 3} and for all β ∈ {1, 2}, βi

then the sets B xi , i = 1, 2, 3 are pairwise disjoint. β

β

Proof. By contradiction. Let x ∈ X a \{x1 , x2 , x3 } be such that B x = B xii for all i ∈ {1, 2, 3} and for all β

β ∈ {1, 2}. Suppose that the sets B xii , i = 1, 2, 3 are not pairwise disjoint. Without loss of generality, β1

β2

β

◦

◦

β

3−β1 ◦

β

we may assume B x1 ∩ B x2 = ∅. By Lemma 5.3, A = A x11 ∪ A x22 ∪ A x33 and B = B x1

3−β2 ◦

3−β

∪ B x3 3 . β β1 β β1 β β For some β ∈ {1, 2}, we have that A x ∩ A x1 = ∅. Since A x  A x1 (by Lemma 5.1(i)) either A x A x11 or β β1 β β1 3−β β1 β β1 β β1 A x A x1 . If A x A x1 , then A x A x1 . So we might as well assume that A x A x1 . Then B x B x1 . β

3−β2 ◦

β

Suppose first that B x11 ⊆ B x . Then B x2

3−β

β

∪ B x2

β

3−β

β

∪ B x3 3 = B x11 ⊆ B x and it follows that B x B xi i , β β β β β i = 2, 3 and A x A xii , i = 2, 3. If A x ∩ A x22 = ∅, then A x ∪ A x22 = A and i = 2, 3. Then A x β3 β β3 β β β2 consequently, A x3 ⊆ A x , yielding a contradiction (since A x3 A x ). Thus A x ∩ A x2 = ∅, and likewise β β β β β β β β A x ∩ A x33 = ∅. It now follows that the sets A x , A x22 , A x33 are pairwise disjoint but the sets B x , B x22 , B x33 β

3−β A xi i ,

β

3−β ◦

are not. Then by Lemma 5.3, we have B = B x our assumptions. β

β

3−β3

∪ B x3

3−β2 ◦

β

Suppose instead that B x ⊂ B x11 . Since B x11 = B x2

3−β B x ⊆ B xi i β 3−β B x B x3 3

β

3−β2 ◦

∪ B x2

3−β3

3−βi

β

∪ B x3

and B x  B xi

β

, i = 2, 3, it follows that 3−β2

β

for some i ∈ {2, 3}. Without loss of generality, we may assume that B x ⊆ B x2 3−β3

β

and hence A x A x3 β

◦

β

β

β

β

β

3−β

β

β

β

β

. Thus

β

and A x A x33 . Now since A x A x33 , A x A x11 , and A x11 ∩ A x33 = ∅, it β

β

3−β

follows that A x11 ∪ A x33 ⊆ A x . Thus the sets A x11 , A x33 , and A x β

β

and hence B x = B x11 . This contradicts

3−β

β

are pairwise disjoint but the sets B x11 , ◦

3−β3 ◦

are not. Now Lemma 5.3 implies that B = B x1 1 ∪ B x3 B x33 , and B x This contradicts our assumptions and the proof is complete. 2

β

β

3−β2

∪ B x and hence B x = B x2

.

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S. McGuinness / Advances in Applied Mathematics 53 (2014) 72–111

The next lemma will be needed in various places in this paper. j

β

Lemma 5.5. Suppose there exist x ∈ X a \{x1 , x2 , x3 }, i ∈ {1, 2, 3}, and j ∈ {1, 2} such that A x ⊆ A xii and j

β

β

B x ⊂ B xii . Then the sets B xii , i = 1, 2, 3 are pairwise disjoint. j

j

β

j

j

β

Proof. Without loss of generality, we may assume that A x ⊆ A x11 and B x ⊂ B x11 . Then A x ∩ A xii = ∅, i = 3− j

β

◦

j Bx

βi

β

◦

3−β1 ◦

β

A = A x11 ∪ A x22 ∪ A x33 and B = B x1 pairwise disjoint and the sets B=

j Bx

βi

βi

A xi , i = 2, 3. Consequently,

B xi , i = 2, 3 and hence = B xi , i = 2, 3. Suppose β β β β j = B x22 . Given that B x ⊂ B x11 , it follows that B x11 ∩ B x22 = ∅. Now Lemma 5.3 implies that

2, 3 and hence that B x

j Ax

3− j ◦ Bx ∪

3−β ◦ B x2 2 ∪

3−β B x3 3

j Bx,

β2

3−β2 ◦

∪ B x2

3−β3

∪ B x3

β3

j

β

β

. However, since the sets A x , A x22 , and A x33 are

B x2 , and B x3 are not pairwise disjoint, Lemma 5.3 also implies that 3− j

and consequently B x 3− j Bx

β2

3−β1

= B x1

j

β

. This implies that B x = B x11 , contradicting j

β3

β

= B x3 . Now we see that B x = B xii β for all i ∈ {1, 2, 3} and j ∈ {1, 2}. By Lemma 5.4, it follows that the sets B xii , i = 1, 2, 3 are pairwise disjoint. 2

our assumptions. We conclude that

j

◦

j

◦

= B x2 and likewise,

3− j Bx

3− j 1 ◦

j

Lemma 5.6. Suppose that A = A x11 ∪ A x22 ∪ A x33 and B = B x1 M | X a is simple. Then x1 , x2 , and x3 3-cross a.

3− j 2 ◦

∪ B x2

3− j 3

∪ B x3

. Furthermore, assume that

Proof. We need only show that X a = {x1 , x2 , x3 }. Suppose to the contrary that there exists x ∈ j j X a \{x1 , x2 , x3 }. Given that the sets A xii , i = 1, 2, 3 are pairwise disjoint and the sets B xii , i = 1, 2, 3 β

j

are not pairwise disjoint, Lemma 5.4 implies that B x = B xii for some β ∈ {1, 2} and i ∈ {1, 2, 3}. We ◦

β j 3− j 2 3− j 3− j ∪ B x3 3 , and it follows that B xi i ⊂ B x , i = 2, 3 β β β 3− j i ji j B x , i = 2, 3. Now A xi A x , i = 2, 3 and hence A xi A x , i = 2, 3. Also, A x11 A x β β j1 j1 If A x1 = A x , then Ωx1 Ωx = {x, x1 }, and x1 belongs to a digon. This contradicts β β β j j j assumption that M | X a is simple. Thus A x11 = A x and hence either A x11 ⊂ A x or A x ⊂ A x11 . β β β j1 j2 j3 j2 j Suppose A x ⊂ A x1 . Then A x , A x2 , and A x3 are pairwise disjoint. Now since B x , B x2 , and B x33 are ◦ ◦ ◦ ◦ β j j j j j not pairwise disjoint, Lemma 5.3 implies that A = A x ∪ A x22 ∪ A x33 . Given that A = A x11 ∪ A x22 ∪ A x33 , β β β j1 j1 ji it follows that A x = A x1 , yielding a contradiction. Thus A x1 ⊂ A x and consequently A x ∩ A xi = ∅ for β β β 3−β j j j j some i ∈ {2, 3}. If A x ∩ A x22 = ∅, then A x ∪ A x22 = A (since A x A x22 ). Now A x ⊂ A x22 and hence 3−β 3−β j j j j A x , A x11 , and A x33 are pairwise disjoint. However, we see that B x , B x11 , and B x33 are not pair3−β ◦ 3−β j1 ◦ j3 j2 wise disjoint, and hence Lemma 5.3 implies that A = A x1 ∪ A x ∪ A x3 . Now A x2 = A x , yielding a β j contradiction. A similar contradiction is reached if A x ∩ A x33 = ∅. From the above, we conclude that β

j

β

may assume that B x = B x11 . Then B x = B x11 = B x2 3− j and hence B xi i β j since B x = B x11 .

X a = {x1 , x2 , x3 }.

β

2

As a consequence of Lemmas 5.3 and 5.6 we have the following: j

Corollary 5.7. Let xi ∈ X a , i = 1, 2, 3 and suppose there exist j 1 , j 2 , j 3 ∈ {1, 2} such that A xii , i = 1, 2, 3 are j B xii ,

pairwise disjoint and i = 1, 2, 3 are not pairwise disjoint. Furthermore, suppose that M | X a is simple. Then x1 , x2 , and x3 3-cross a. 6. Ω -cocircuits In this section, we shall assume that M is a connected, regular matroid and Ω is a spanning ∗ ∗ ∗ circuit of M. For e , f ∈ Ω , let Ω M ,e f = E ( M )\cl M (Ω\{e , f }) and let Ω M ,e f = Ω M ,e f \{e , f }. Since Ω ∗ is a spanning circuit of M, the set Ω M ,e f is seen to be a cocircuit of M containing e and f . We

S. McGuinness / Advances in Applied Mathematics 53 (2014) 72–111

81

∗ ∗ ∗ refer to Ω M ,e f as an Ω -cocircuit of M. When M is implicit, we shall just write Ωe f and Ωe f in place ∗ ∗  of Ω M ,e f and Ω M ,e f . In this section, we shall show that, under certain conditions, one can find a ∗ are nested. That is, there exists an ordering pair of elements e , f ∈ Ω for which the chords in Ω ef

∗ and there exist j 1 , j 2 , . . . , jm ∈ {1, 2} such that Ω xj 1 ⊂ Ω xj 2 ⊂ x1 , x2 , . . . , xm of the elements of Ω 1 2 ef x m . In addition, we shall show that for some a ∈ X Ω where a is not 3-crossed, we can choose ··· ⊂ Ω m ◦ ∗ = [x]a for some x ∈ X a , or Ω ∗ = [x]a ∪ [ y ]a for some x, y ∈ X a . a such that either Ω e, f ∈ Ω j

ef

ef

∗ ⊆ X a . a . Then Ω Lemma 6.1. Let a ∈ X Ω where Ωa is minimal. Let e , f ∈ Ω ef ∗ . By definition, x ∈ Proof. Let x ∈ Ω / cl M (Ω\{e , f }). This implies that each of the chord-circuits Ωx , ef j

j

j = 1, 2 contains exactly one of the elements e, f . Thus Ωax = ∅, j = 1, 2. Furthermore, since Ωa 2j Ωax

is minimal, we have ∗ ⊆ X a . follows that Ω ef

= ∅, j = 1, 2. Thus 2

ij Ωax

= ∅, for i = 1, 2 and j = 1, 2 and hence x ∈ X a . It

∗ is a for which Ω The next theorem shows that for certain chords a ∈ X Ω there exist e , f ∈ Ω ef ◦

∗ = [x]a for some x ∈ X a or Ω ∗ = [x]a ∪ a-nested. Moreover, when a is not 3-crossed, then either Ω ef ef [ y ]a for some x, y ∈ X a . The first possibility occurs when M is graphic, and second possibility was derived from the case where M is graphic or cographic. Theorem 6.2. Let a ∈ X Ω where Ωa is minimal and a is not 3-crossed. Suppose that M | X a is simple. Then ∗ is a-nested and either a such that Ω there exist e , f ∈ Ω ef

∗ = [x]a for some x ∈ X a (i) Ω ef or ◦

∗ = [x]a ∪ [ y ]a for some x, y ∈ X a . (ii) Ω ef a and B = Ω a2 . For x ∈ X Ω and j ∈ {1, 2} we let A x = Ωax and Proof. As before, we shall let A = Ω j

j

j Bx B ix

2j j j j j = Ωax . Let A = { A x | x ∈ X a , j ∈ {1, 2}}. By Lemma 5.1, we have A ix A y iff B ix B y and A ix A y iff j j j j j

B y for all A ix , A y ∈ A . Furthermore, for all A ix , A y ∈ A , we have A ix  A y and thus either A ix A y j or A ix A y . For A ix ∈ A let ( A ix ) denote the maximum integer k such that there is a chain of sets i i i0 i i i1 i = max A ix ∈A ( A ix ) A x0 ⊂ A x1 ⊂ · · · ⊂ A xkk−−11 ⊂ A where A x00 , A x11 , . . . , A xkk−−11 ∈ A and A x00 = A ix . Let and let x0 ∈ X a where = max{ ( A x0 ), ( A 2x0 )}. Without loss of generality, we may assume that

α

α

α

α α α α = α ( A x0 ). Let e0 ∈ A x0 . By relabelling, we may assume that e0 ∈ A x for all x ∈ Xa . By our choice of x0 , it is evident that A ix ⊂ A x0 for all A ix ∈ A . The remainder of the proof will be divided into several subproofs. (6.3) A x0 ⊆ A x and B x0 B x for all x ∈ X a . Proof. Let x ∈ X a \{x0 }. We have that either A x0 A x or A x0 A x . Suppose A x0 A x . Since A x0 ∩ A x = ∅ (both sets contain e 0 ), we have A x0 ∪ A x = A and hence A 2x ⊂ A x0 . This gives a contradiction. Thus A x0 A x . Since A x ⊂ A x0 , it follows that A x0 ⊆ A x , A x0 A x , and B x0 B x . 2 By our choice of x0 , there is a chain of sets A x0 ⊂ A x1 ⊂ · · · ⊂ A xα−1 of length α . Since [x0 ]a is a-nested (by Lemma 5.2), we may assume that B x ⊂ B x0 for all x ∈ [x0 ]a \{x0 }. If α = 1, then A x0 = A x

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S. McGuinness / Advances in Applied Mathematics 53 (2014) 72–111

∗ = [x0 ]a and (i) holds in this case. for all x ∈ X a and [x0 ]a = X a . Letting f 0 ∈ A 2x0 we see that Ω e0 f 0 Thus for the remainder, we may assume that α  2 and | A |  3. Let f 0 ∈ A x1 \ A x0 . ∗ (6.4) Suppose f 0 ∈ / A 2x for all x ∈ X a where A 2x ⊆ A x1 \ A x0 . Then Ω e

0 f0

∗ is a-nested. = [x0 ]a and Ω e0 f 0

∗ Proof. Since e 0 ∈ A x and f 0 ∈ A 2x = A 2x0 for all x ∈ [x0 ]a , it is clear that [x0 ]a ⊆ Ω e

0 f0

. We shall show

∗ . Since e 0 ∈ A x , it follows that f 0 ∈ A 2x . Suppose x ∈ X a \[x0 ]a . By (6.3), that equality holds. Let x ∈ Ω e0 f 0 we have A x0 ⊂ A x . Thus α ( A x )  α − 1 = α ( A x1 ) and therefore A x ⊂ A x1 . Since f 0 ∈ A 2x ∩ A x1 , we also have A x1  A x . Thus A x  A x1 and consequently A x A x1 . Since A x ∩ A x1 = ∅ (both sets contain e 0 ), it follows that A x ∪ A x1 = A and A 2x ⊂ A x1 . Since A 2x ∩ A x0 = ∅, we have A 2x ⊆ A x1 \ A x0 . However, f 0 ∈ A 2x ∗ ⊆ [x0 ]a . From the and this contradicts our assumptions. We conclude that x ∈ [x0 ]a and hence Ω e0 f 0 ∗ ∗   above, it now follows that Ω = [x0 ]a and Ω is a-nested (by Lemma 5.2). 2 e0 f 0

e0 f 0

In light of (6.4), we may henceforth assume that there exists y 0 ∈ X a such that A 2y 0 ⊆ A x1 \ A x0

and f 0 ∈ A 2y 0 . Since [ y 0 ]a is a-nested, we may assume that B y 0 ⊂ B y for all y ∈ [ y 0 ]a \{ y 0 }. Since

α ( A x1 ) = α − 1, we see that α ( A 2y0 ) = α . We also observe that the sets A x0 , A 2x1 , A 2y0 are pairwise

disjoint and consequently A x0 A 2x1 , A x0 A 2y 0 , and A 2x1 A 2y 0 . Thus we also have B x0 B 2x1 , B x0 B 2y 0 , B 2x1 B 2y 0 , and B x0 B x1 , B x0 B y 0 , B x1 B y 0 .

Suppose B x1 ⊂ B x0 . Then B 2x1 ∩ B x0 = ∅, the sets A x0 , A 2x1 , A 2y 0 are pairwise disjoint, but the sets

B x0 , B 2x1 , B 2y 0 are not pairwise disjoint. In this case, Corollary 5.7 implies that X a = {x0 , x1 , y 0 } and a is 3-crossed, contradicting our assumptions. Thus B x0 ⊆ B x1 .

∗ (6.5) Ω e

◦

0 f0

∗ is a-nested. = [x0 ]a ∪ [ y 0 ]a and Ω e0 f 0

∗ Proof. We first observe that [x0 ]a ∪ [ y 0 ]a ⊆ Ω e

. To see this, let x ∈ [x0 ]a ∪ [ y 0 ]a . Then e 0 ∈ A x 0 f0 i  and f 0 ∈ This implies that Ωx  Ω\{e 0 , f 0 }, i = 1, 2 and hence x ∈ / cl M (Ω\{e 0 , f 0 }). Thus x ∈ ∗ ⊆ [x0 ]a ∪ [ y 0 ]a , after which equality will be E ( M )\cl M (Ω\{e 0 , f 0 }) = Ωe∗ f . We shall show that Ω e0 f 0 0 0 ∗ . Since e 0 ∈ A x , we have f 0 ∈ A 2x . Suppose x ∈ / [x0 ]a ∪ [ y 0 ]a . Then A x0 ⊂ seen to hold. Let x ∈ Ω e0 f 0 A x (by (6.3)). Following the arguments of the proof of (6.4), we deduce that A 2x ⊆ A x1 \ A x0 . Thus α ( A 2x )  α ( A x1 ) + 1 = α , and hence α ( A 2x ) = α . We see that A 2x ∩ A 2y0 = ∅ (both sets contain f 0 ) and A 2x ∪ A 2y 0 ⊂ A (since e 0 ∈ A x ∩ A y 0 ). Thus A 2x ∦ A 2y 0 and hence A 2x A 2y 0 . However, since x ∈ / [ y 0 ]a , 2 2 2 2 2 2 2 2 2 we have A x = A y 0 . We also have A x ⊂ A y 0 and A y 0 ⊂ A x since α ( A x ) = α ( A y 0 ) = α . Thus A x  A 2y 0 , ∗ ⊆ [x0 ]a ∪ [ y 0 ]a . We need only yielding a contradiction. We conclude that x ∈ [x0 ]a ∪ [ y 0 ]a and Ω e0 f 0 ∗  show that Ωe f is a-nested. Recall that B x0 B y 0 and B x1 B y 0 . Suppose B y 0 ⊆ B x0 . Then B y 0 ⊆ B x0 ⊆ 0 0 B x1 , implying that B x1 B y 0 . This yields a contradiction. Consequently, B x0 ⊂ B y 0 . Recall that by our choice of x0 and y 0 , B x ⊂ B x0 for all x ∈ [x0 ]a \{x0 } and B y 0 ⊂ B y for all y ∈ [ y 0 ]a \{ y 0 }. It now follows that B x ⊂ B x0 ⊂ B y 0 ⊂ B y for all x ∈ [x0 ]a \{x0 } and y ∈ [ y 0 ]a \{ y 0 }, and it is seen that [x0 ]a ∪ [ y 0 ]a is a-nested. 2 A 2x .

The proof of the theorem now follows by (6.4) and (6.5).

2

The following remarks are a direct consequence of the proof of Theorem 6.2.

∗ Remark 1. Let f 0 ∈ A x1 \ A x0 . Then Ω e

0 f0

∗ (a) Ω e

0 f0

= [x0 ]a

is a-nested and either

S. McGuinness / Advances in Applied Mathematics 53 (2014) 72–111

83

or

∗ (b) Ω e

0 f0

= [x0 ]a ∪ [ y 0 ]a for some y 0 ∈ X a such that A 2y0 ⊆ A x1 \ A x0 and f 0 ∈ A 2y0 .

Suppose α  2. Let e 0 ∈ A 2xα−1 and let x0 = xα −1 and x1 = xα −2 . By symmetry, the result in Remark 1 above has its counterpart for e 0 and x0 :

∗ Remark 2. Let f 0 ∈ A 2x \ A 2x . Then Ω e 1

∗ (a ) Ω e

0



0 f0

is a-nested and either



= [x0 ]a



= [x0 ]a ∪ [ y 0 ]a for some y 0 ∈ X a such that A 2y ⊆ A 2x \ A 2x and f 0 ∈ A 2y .

0 f0

or

∗ (b ) Ω e

0 f0

0

1

0

0

Using the above remarks, we shall prove the following theorem which is the second main result of this section. We remark that the first part (i) applies when M is graphic, and either (i) or (ii) applies when M is cographic.

a |  3. Let ξ = mine, f ∈Ω |Ω ∗ | and suppose Theorem 6.6. Let a ∈ X Ω and suppose that Ωa is minimal and |Ω ef a that ξ  1. Suppose that M | X a is simple. If a is not 3-crossed, then either i 3− i ∩ Ωay = ∅, and Ωax ∩ (i) there exist x, y ∈ X a and i , j ∈ {1, 2} such that |[x]a |  ξ , |[ y ]a |  ξ , Ωax j

3− j Ωay

= ∅

or j

i k (ii) there exist x, y , z ∈ X a and i , j , k ∈ {1, 2} such that |[x]a | + |[ y ]a | + |[ z]a |  32 ξ and Ωax , Ωay , Ωaz are pairwise disjoint. j

j

Proof. Assume that a is not 3-crossed. Let A, B, A x , B x , e 0 , x0 , y 0 , x1 etc. be as in the proof of Theorem 6.2. As before, we may assume that e 0 ∈ A x for all x ∈ X a and (6.3) holds. We shall first show that A x0 = {e 0 } and α  2. Suppose to the contrary that A x0 = {e 0 } and let f 0 ∈ A x0 \{e 0 }. Since ∗ |  ξ  1, there exists x ∈ Ω ∗ . Since e 0 ∈ A x , we have that f 0 ∈ |Ω / A x . However, A x0 ⊆ A x (by e f e f 0 0

0 0

(6.3)) and thus f 0 ∈ A x . This gives a contradiction. We conclude that A x0 = {e 0 } and | A 2x0 | = | A | −

∗ . We may assume | A x0 | = | A | − 1  2 (since | A |  3 by assumption). Let f 0 , g 0 ∈ A 2x0 and let x ∈ Ω f 0 g0 2 that f 0 ∈ A x and g 0 ∈ A x . Then A x0 ⊂ A x (since f 0 ∈ A x \ A x0 ) and hence α  2. Let f 0 ∈ A x1 \ A x0 and f 0 ∈ A 2x \ A 2x . By Remarks 1 and 2, we have that (a) or (b) holds and (a ) 1 0 ∗ |  ξ and |[x ]a | = |Ω ∗  |  ξ . We have or (b ) holds. Suppose (a) and (a ) hold. Then |[x0 ]a | = |Ω 0 e0 f 0 e f 0 0

 A x1 \ A x0 ⊆ A 2x0 ∩ A x . Then (i) is seen to hold with x = x0 , y = x0 , and i = 1, A x0 ∩ A 2x = ∅ and ∅ = 0

0

∗ j = 2. Suppose that (a) and (b ) hold. Then |[x0 ]a | = |Ω e

0 f0

∗ |  ξ and |[x0 ]a | + |[ y 0 ]a | = |Ω e



0 f0

|  ξ.

Consequently, |[x0 ]a | + |[x0 ]| + |[ y 0 ]a |  2ξ . Observing that A x0 , A 2x , and A 2y  are pairwise disjoint, 0

0

(ii) is seen to hold with x = x0 , y = y 0 , z = x0 , and i = 1, j = 2, k = 2. If (b) and (a ) hold, then one can show with similar reasoning that (ii) holds with x = x0 , y = y 0 , z = x0 , and i = 1, j = 2, k = 2. ∗ |  ξ , |[x ]a | + |[ y  ]a | = |Ω ∗  |  ξ , and Lastly, suppose (b) and (b ) hold. Then |[x0 ]a | + |[ y 0 ]a | = |Ω 0 0 e f e f 0 0

0 0

|[x0 ]a | + |[ y 0 ]a | + |[x0 ]a | + |[ y 0 ]a |  2ξ . If A 2y0 ∩ A 2y = ∅, then the sets A x0 , A 2x , A 2y0 , A 2y are pairwise 0

0

0

disjoint. In this case, there exist x, y , z ∈ {x0 , x0 , y 0 , y 0 } such that |[x]a | + |[ y ]a | + |[ z]a | 

3 4

· 2ξ = 32 ξ

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S. McGuinness / Advances in Applied Mathematics 53 (2014) 72–111 j

and A ix , A y , A kz are pairwise disjoint for some i , j , k ∈ {1, 2}. Thus (ii) holds in this case. Suppose instead that A 2y 0 ∩ A 2y  = ∅. Then x0 = x1 , x1 = x0 , and α ( A x0 ) = α ( A 2y 0 ) = α ( A 2y  ) = 2. We have A 2y 0 ∦ 0

0

A 2y  since A 2y 0 ∩ A 2y  = ∅ and A 2y 0 ∪ A 2y  ⊆ A x1 \ A x0 ⊂ A. Thus A 2y 0 A 2y  . Since 0

0

0

0

α ( A 2y0 ) = α ( A 2y ), it 0

follows that A 2y 0 ⊂ A 2y  and A 2y  ⊂ A 2y 0 . Consequently, A 2y 0 = A 2y  , [ y 0 ]a = [ y 0 ]a , and |[ y 0 ]a | + |[x0 ]a | = 0

0

0

|[ y 0 ]a | + |[x0 ]a |  ξ . We see that

◦ ◦    ◦      ∗ Ω ∗  = [x0 ]a ∪ [ y 0 ]a  [ y 0 ]a ∪ x0 = [x0 ]a ∪ x0 . ∗  = Ω Ω e0 f 0 e0 e f0e a a 0

0

∗  |  ξ and consequently, |[x]a | + |[ y ]a |  ξ for all x, y ∈ {x0 , y 0 , x }. SumThus |[x0 ]a | + |[x0 ]a | = |Ω 0 e e 0 0

ming these inequalities, we obtain that 2(|[x0 ]a | + |[ y 0 ]a | + |[ z0 ]a |)  3ξ . Thus |[x0 ]a | + |[ y 0 ]a | + |[ z0 ]a |  32 ξ . Now (ii) is seen to hold with x = x0 , y = y 0 , z = x0 , and i = 1, j = 2, k = 2. This completes the proof. 2 7. Properties of a minimum counterexample To prove Theorem 1.4, we shall use the minimum counterexample approach. We shall assume that the theorem is false. Among the counterexamples to the theorem, let M be a counterexample for which | E ( M )| is minimum. Then there is a circuit C of M where (i) C is not dominating. (ii) There is no C -augmenting circuit. r(M ) (iii) |C ∗ | > 3 + 1 for all cocircuits C ∗ ∈ C ∗ ( M ) where C ∗ ∩ C = ∅. In this section, we shall show that M and C satisfy various properties. In particular, we shall show that there is a C -minimal circuit D for which the circuit C  D is a dominating circuit of M \(C ∩ D ). (7.1) M /e is connected for all e ∈ C . Proof. Let e ∈ C . Suppose M /e is not connected. Then M is a 2-sum M = M 1 ⊕2 M 2 where {e } = E ( M 1 ) ∩ E ( M 2 ) and C ⊆ E ( M 1 ). Let D be a largest circuit in M 2 which contains e. Then D is a C -minimal circuit where C ∩ D = {e }. Also, D is C -augmenting since |C  D | = | D | + |C | − 2 > |C |. This contradicts our assumptions. Thus M /e is connected. 2 Let M 0 be the matroid obtained from M by deleting all C -bridges of order at most one. (7.2) There exists a C -minimal circuit D in M for which |C  D | = |C |. Proof. We observe that since C is not a dominating circuit of M, it is also not a dominating circuit of M 0 . We also note that if C ∗ ∈ C ∗ ( M 0 ) where C ∗ ∩ C = ∅, then C ∗ is also a cocircuit of M, and thus by assumption |C ∗ | > 13 r ( M ) + 1. Let e ∈ C and let C  = C \{e }. By (7.1), M /e is connected, and hence M 0 /e is also seen to be connected. Suppose that M 0 /e is not simple, and let { f , g } be a digon of M 0 /e. Then {e , f , g } is a triangle of M. Neither f nor g are chords of C since such chords are C -bridges of C which are deleted from M when we construct M 0 . Now {e , f , g }C = (C \{e }) ∪ { f , g } is seen to be a circuit of M 0 which has more elements than C . Thus {e , f , g } is C -augmenting and this gives a contradiction. We conclude that M 0 /e is simple. Now the conditions of Theorem 1.4 are seen to be met by M 0 /e. Given that | E ( M 0 )/e | < | E ( M )|, and C  is a circuit of M 0 /e which is not dominating, there exists a C  -augmenting circuit D  in M 0 /e. If D  is a circuit of M 0 , then it is seen to be a C -augmenting circuit. Thus D  is not a circuit of M 0 and hence D = D  ∪ {e } is a circuit of M. Now D is seen to be a C -minimal circuit in M 0 and hence it is also C -minimal in M as well. Since

S. McGuinness / Advances in Applied Mathematics 53 (2014) 72–111

85

    |C |  |C  D | = C   D   > C   = |C | − 1 it follows that |C  D | = |C |, and D is the desired circuit.

2

For the remainder of the section, let D be a C -minimal circuit in M where (a) |C  D | = |C |. (b) |C ∩ D | is minimum subject to (a). Let Ω = C  D, U = D \C , V = C \ D, and W = C ∩ D. Let that U and W are circuits of M / V . (7.3)

μ = |U |, υ = | V |, and ω = | W |. We observe

υ  μ and μ = ω  2.

Proof. Since |C | = |C  D |, it follows that ω + υ = μ + υ . Thus μ = ω . Clearly ω  2, for otherwise D would be a C -augmenting circuit. If μ > υ , then | D | > |C | and C  D would be a C -augmenting circuit, contradicting our assumptions. Thus υ  μ. 2 The next assertion will be instrumental in showing that Ω is a dominating circuit of M \ W . The proof follows along the lines of the proof of Lemma 2.2 in [14], with some modifications. (7.4) U and W belong to different components of M / V . Proof. We shall give a sketch of the proof. Let M  = M / V . Suppose that U and W belong to the same component of M  . One sees that U and W are circuits of M  . Furthermore, since C , D, and Ω are the only circuits contained in C ∪ D (because D is C -minimal), we see that r M  (U ∪ W ) = r M  (U ) + r M  ( W ). Since U and W belong to the same component of M  , there is a circuit in M  which intersects both U and W . Among such circuits, let A  be a circuit for which | A  \(U ∪ W )| is minimum. We note that A  \(U ∪ W ) = ∅ since r M  (U ∪ W ) = r M  (U ) + r M  ( W ). By Lemma 5.2 of [13], it follows that A  U , A   W , and A  U  W are circuits of M  . There is a circuit A in M where A  ⊆ A ⊆ A  ∪ V . Following similar arguments as given in the proof of Lemma 2.2 in [14], one can show that A is a C -minimal circuit. Moreover, A  D is also a C -minimal circuit. Suppose A ∩ V = ∅. Then A C and ( A  D )C are circuits such that

  | A C | + ( A  D )C  = μ + ω + 2υ + 2| A \ D |   = 2 |C | + | A \ D | > |C |. Thus at least one of the circuits A or A  D is C -augmenting. This gives a contradiction. If, on the other hand, A ∩ V = ∅, then one can show (using similar arguments as in [14]) that M has an F 7∗ -minor. This also gives a contradiction. We conclude that U and W must belong to different components of M  . 2 Since C is a non-dominating circuit of M, there exists a C -bridge of order at least two. Let K ∈

K ( M /C ) where o( K )  2.

(7.5) K ( M /C ) = { K }, U ⊆ K , and E ( M ) = C ∪ K . Proof. Suppose U  K . Let M  = M |(C ∪ K ). Then M  is simple and | E ( M  )| < | E ( M )|. Furthermore, C is a non-dominating circuit of M  since K is also a C -bridge of M  of order at least two. We also note that for any cocircuit C ∗ ∈ C ∗ ( M  ) where C ∗ ∩ C = ∅, C ∗ is also a cocircuit of M and r(M ) r(M  ) hence |C ∗ | > 3 + 1  3 + 1. Since | E ( M  )| < | E ( M )|, it follows by our assumptions that there

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is a C -augmenting circuit in M  . However, such a circuit is seen to be a C -augmenting circuit in M as well. This gives a contradiction. We conclude that U ⊆ K , and given that there is exactly one C -bridge which contains U , we must also have K ( M /C ) = { K }. It now follows immediately that E (M ) = C ∪ K . 2 By (7.4) and (7.5), it is not hard to see that K ( M / V ) = { K , W }. Let N = M |(Ω ∪ K ). Then N = M \ W . We shall now show that Ω is a dominating circuit of N, a property which will be heavily exploited in the proof of the main theorem. (7.6) Ω is a dominating circuit of N. Proof. By contradiction. Suppose Ω is not a dominating circuit of N. Then |Ω| < r ( N ) − 1. For all cocircuits C ∗ ∈ C ∗ ( N ) where C ∗ ∩ Ω = ∅, we have C ∗ ⊆ K and C ∗ is also a cocircuit of M. Thus for r(M ) r(N ) all such cocircuits C ∗ where C ∗ ∩ Ω = ∅ we have |C ∗ | > 3 + 1  3 + 1. Since | E ( N )| < | E ( M )|, Theorem 1.4 holds for N and hence there is an Ω -augmenting circuit D N in N. We note that D N is also a circuit of M. Case 1. Suppose D N ∩ U = ∅. Proof. Let D N = D N  D. Since D N ∩ W = ∅ and W ⊂ D, it follows that W ⊆ D N . Suppose D N ⊆ D N  is a circuit. If D N ∩ W = ∅, then D N contains a circuit D  N in M / V where D N ∩ W = ∅. Since W is . From this, we conclude that if a a V -bridge of M and also a circuit of M / V , we must have W = D  N circuit of D N of D N intersects W , then W ⊂ D N . We shall first show that D N is a circuit. Suppose it is not. Then D N is a disjoint union of circuits. Thus there exist disjoint circuits D N ,i ⊂ D N , i = 1, 2. If D N ,1 ∩ W = ∅, then D N ,1 ⊂ D N Ω . This would contradict the assumption that D N is Ω -minimal since clearly D N ,1 ∈ / { D N , Ω, D N Ω}. Thus D N ,1 ∩ W = ∅ and likewise, D N ,2 ∩ W = ∅. However, by the above, this means that W ⊂ D N ,i , i = 1, 2, contradicting the assumption that D N ,i , i = 1, 2 are disjoint. From this, we conclude that D N is a circuit. We shall show that D N is C -minimal. Suppose this is not the case. Then there exists a circuit  D N ⊂ D N ∪ C where D N ∈ / { D N , C , D N C }. If D N ∩ W = ∅, then D N ⊂ D N ∪ Ω and this contradicts the assumption that D N is Ω -minimal, since clearly D N ∈ / { D N , Ω, D N Ω}. Thus D N ∩ W = ∅, and     hence by the above, W ⊆ D N . Since W ⊆ D N , it follows that D N  D N ⊆ D N Ω . Let D  N ⊆ D N  D N be a circuit. Clearly D  ∈ / { D , Ω, D Ω} , and this contradicts the assumption that D is Ω -minimal. N N N N From the above, we conclude that D N must be C -minimal. To show that D N is C -augmenting, we note that μ = ω by (7.3). Consequently, |Ω| = |C | and |C  D N | = |Ω D N | > |Ω| = |C |. It follows that D N is C -augmenting, which contradicts our assumptions. This completes Case 1. 2 Case 2. Suppose D N ∩ Ω ⊂ V . Proof. In the same way D N was shown to be C -minimal in Case 1, one can show that D N is C -minimal. We have that

|C  D N | = ω + | D N  V | = μ + | D N  V | = |Ω D N | > |Ω| = |C |. In this case, D N is a C -augmenting circuit, contradicting our assumptions.

2

From Cases 1 and 2, we conclude that Ω is a dominating circuit of N.

2

For the remainder of this paper, M, C , D, Ω , U , V , W , μ, υ , ω , and N shall be as described above. Let δ  0 be such that υ = μ + δ . We observe that μ  3, for if μ = 2, then C would be a

S. McGuinness / Advances in Applied Mathematics 53 (2014) 72–111

87

dominating circuit of M. We also note that since M only has one C -bridge (by (7.5)), the circuit C w = U . has no chords in M. We shall add an element w to N where w is a chord of Ω such that Ω We may do this since such a chord can be achieved by contracting all but one element of W . To see that N will still have fewer elements than M, even after adding w, we note that ω  2, and | E ( N )| = | E ( M )| − | W |  | E ( M )| − 2. Thus after adding w to N, we will still have | E ( N )|  | E ( M )| − 1. We shall let X Ω = X N ,Ω where we will also assume that w ∈ X Ω . We shall also assume that for all

xj ⊆ U for some j ∈ {1, 2}, then j = 1. Furthermore, for all x ∈ X Ω and j ∈ {1, 2}, we let x ∈ X Ω , if Ω j

j

j

j

U x = Ωx ∩ U and V x = Ωx ∩ V . 8. Refinements Let X ⊆ X Ω . A subset X  ⊆ X is called a refinement of X if X  is w-nested; that is, either X  is empty or, in the case that X  is nonempty, there is an ordering x1 , . . . , xm for X  and there exist j j j j j j j 1 , . . . , jm ∈ {1, 2} such that ∅ = U x11 ⊆ U x22 ⊆ · · · ⊆ U xmm ⊂ U and ∅ = V x11 ⊂ V x22 ⊂ · · · ⊂ V xmm ⊂ V . Our basic objective in this section is to show how one can construct certain w-nested sets from a-nested a ⊆ U = Ω w . sets where a ∈ X Ω \{ w } and Ω

a ⊆ U . Let X ⊆ X a where X is a-nested. Then there exists a refinement X  ⊆ X Lemma 8.1. Let a ∈ X Ω where Ω a |. such that | X  |  | X | − μ + |Ω a , B = Ω a2 , and let A x = Ωax and B x = Ωax for all x ∈ X a and for all Proof. As before, let A = Ω  j ∈ {1, 2}. If | X |  μ − | A |, then X = ∅ will do. Thus we shall assume that | X | > μ − | A |. Let (x, j) j

j

j

2j

3− j m

j

be an a-ordering of X where x = (x1 , . . . , xm ) and j = ( j 1 , . . . , jm ). Let 1 = Ωx11 , m+1 = Ωxm for i = 2, . . . , m let i =

j j Ωxii Ωxii−−11 .

, and

Let S = {i | i ∩ V = ∅} and let s = | S |. Let T = {1, . . . , m + 1}\ S j

j

j

and let t = | T | = m + 1 − s. Since X is a-nested, we have ∅ =  B x11 ⊂ B x22 ⊂ · · · ⊂ B xmm ⊂ B. From this, it follows that i ∩ B = ∅, i = 1, 2, . . . , m + 1. We also observe that the sets i ∩ B, i = 1, . . . , m + 1 partition B. Thus the sets i ∩ ( B \ V ), i = 1, . . . , m + 1 partition B \ V . Since i ∩ V = ∅ for all i ∈ S it follows that i ∩ ( B \ V ) = ∅ for all i ∈ S. Thus

|B \V | =

m +1 

     i ∩ ( B \ V )   i ∩ ( B \ V )  s

i =1

i∈ S

and hence s  | B \ V | = |U \ A | = μ − | A |. Thus t = m + 1 − s  m + 1 − μ + | A |. Let T = {k1 , k2 , . . . , kt } jk

jk

jk

jk

jk

where k1 < k2 < · · · < kt . Then ∅ =  U xk 1 ⊆ U xk 2 ⊆ · · · ⊆ U xk t −1 ⊂ U and ∅ =  V xk 1 ⊂ V xk 2 ⊂ · · · ⊂ 1

jk

2

t −1

1

2

V xk t −1 ⊂ V . Thus X  = {xk1 , xk2 , . . . , xkt −1 } is w-nested and hence it is a refinement of X . Moreover, t −1

| X  | = t − 1  m − μ + | A |. 2

As will later be seen, we shall need to extend the above lemma to the case where we have three disjoint a-nested sets. The next lemma serves our purposes.

a ⊆ U . Let X i , i = 1, 2, 3 be three pairwise disjoint a-nested sets and let Lemma 8.2. Let a ∈ X Ω where Ω (xi , ji ), i = 1, 2, 3 be a-orderings for X i , i = 1, 2, 3 where xi = (xi1 , xi2 , . . . , ximi ), ji = ( j i1 , j i2 , . . . , j imi ), j im

i = 1, 2, 3. Let m = m1 + m2 + m3 and suppose that m  4. Suppose that Ωaximi , i = 1, 2, 3 are pairwise i

2 j im

disjoint. Then Ωaxim i , i = 1, 2, 3 are pairwise disjoint and there exist refinements X i ⊆ X i , i = 1, 2, 3 such that i

3     a | + 1; if μ  |Ω  X   m − μ + |Ωa |, i  − 1, if μ  |Ω a | + 2. m − μ + |Ω| i =1

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a , B = Ω a2 , and let A x = Ωax and B x = Ωax for all x ∈ X a and for all j ∈ Proof. As before, let A = Ω {1, 2}. Since m  4, we have that mi  2 for some i ∈ {1, 2, 3}. We may assume that m1  2. Now since j

j 1m

j

j

j

j

2j

j im

j 1m

11 ⊆ A x1m1 and B x11 ⊂ B x1m1 , Lemma 5.5 implies that the sets B ximi , i = 1, 2, 3 are pairwise disjoint. A x11 11 1

1

i

j

j

j i (k−1)

Let i1 = Ωxi1i1 , i = 1, 2, 3. For i = 1, 2, 3 if mi  2, then let ik = Ωxikik Ωxi(k−1) , k = 2, . . . , mi . Let j 1m

j 2m

j 3m

= ΩΩx1m 1 Ωx2m 2 Ωx3m 3 . The set ∩ B together with the sets i j ∩ B, i = 1, 2, 3; j = 1, . . . , mi 1 2 3 are seen to partition B. Let S i = { j | i j ∩ V = ∅} and T i = {1, . . . , mi }\ S i , i = 1, 2, 3. Let si = | S i | and

3 t i = | T i | = m − si , i = 1, 2, 3. Let s = s1 + s2 + s3 and t = t 1 + t 2 + t 3 . Then t = i =1 mi − si = m − s. Since X i , i = 1, 2, 3 are a-nested, we see that i j ∩ B = ∅, i = 1, 2, 3; j = 1, . . . , mi . Since i j ∩ V = ∅ for all j ∈ S i , i = 1, 2, 3, it follows that i j ∩ ( B \ V ) = ∅ for all j ∈ S i , i = 1, 2, 3. We also see that the set ∩ ( B \ V ) together with the sets i j ∩ ( B \ V ), i = 1, 2, 3; j = 1, . . . , mi partition B \ V . Thus we have 3          i j ∩ ( B \ V )   i j ∩ ( B \ V )  s1 + s2 + s3 = s. | B \ V | =  ∩ ( B \ V ) + i =1 j ∈ S i

i, j

Thus s  | B \ V | = |U \ A | = μ − | A | and t = m − s  m − μ + | A |. Let X i ⊆ X i , i = 1, 2, 3 where

X i =



if T i = ∅; ∅, {xi j | j ∈ T i }, otherwise.

If for all i, the set X i contains no set X  which contains V , then the sets X i , i = 1, 2, 3 are seen to be

3

 refinements of X i , i = 1, 2, 3 and i =1 | X i | = t  m − μ + | A |. In the case where, for some i, the set X i contains a set X  which contains V , the set X i is not a refinement. In this case, we also have that jkm

B xkmk ⊂ U for all k = i. However, by redefining X i with the set X  excluded, we see that the sets X j , k

j = 1, 2, 3 are refinements and furthermore, this case,

μ  | A | + 2 since

jkm B xkmk k

3



i =1 | X i |

= t − 1  m − μ + | A | − 1. We observe that in

⊂ U for all k = i. 2

9. C -augmenting circuits via w-nested sets Our ultimate goal is to show that certain kinds of circuits are C -augmenting. Before doing this, we shall introduce conditions which guarantee that a circuit is C -minimal. After this, we shall introduce a condition which guarantees that a C -minimal circuit is also C -augmenting. The main goal of this section is however, to show how one can find C -augmenting circuits given two or three disjoint sets of w-nested chords. This we shall do in Lemmas 9.4 and 9.7. j

Lemma 9.1. For all x ∈ X w and for all j ∈ {1, 2}, the circuit Ωx is C -minimal. Furthermore, suppose the

j j j sets U x11 , U x22 , . . . , U xmm are pairwise disjoint for some x1 , x2 , . . . , xm ∈ X w and j1 j2 j = ΩΩx1 Ωx2  · · · Ωxmm . If ∩ C = ∅, then is a C -minimal circuit.

j 1 , j 2 , . . . , jm ∈ {1, 2}. Let

Proof. The first assertion is straightforward and we leave its proof to the reader. We shall prove the second assertion. Let x1 , x2 , . . . , xm , and j 1 , j 2 , . . . , jm and be as in the statement of the lemma. Suppose ∩ C = ∅. To show that is C -minimal, suppose  is a circuit where  ⊆ ∪ C . If  ∩ ( \Ω) = ∅, then  ⊆ C ∪ D and given that D is C -minimal, it follows that  ∈ {C , D , Ω}. By

the definition of , we see that  ∩ U ⊆ U \{U x11 ∪ · · · ∪ U xmm } and  ∈ / { D , Ω}. In this case,  = C . Therefore, we may assume that  ∩ ( \Ω) = ∅. Suppose \Ω ⊆  . Then   ⊆ C ∪ D. If   = ∅, then =  . Otherwise, if   = ∅, then there is a circuit  ⊆   ⊆ C ∪ D. Since D is C -minimal, it follows that  ∈ {C , D , Ω}. Clearly  = D , Ω since D  and Ω  . Thus  = C , and given that this is the only choice, we j

j

S. McGuinness / Advances in Applied Mathematics 53 (2014) 72–111

89

must have that   =  = C . In this case,  = C . In each of the above cases, we see that  ∈ {C , , C }. Suppose \Ω   . Then there exist xi ∈  and x j ∈ \  for some i = j. Without loss of generality, we may assume x1 ∈  and x2 ∈ \  . Let e ∈ U x11 and f ∈ U x22 . There is a hyperplane H in M where cl M (Ω\{e , f }) ⊆ H and e , f ∈ / H . Let C ∗ = E ( M )\ H . Then C ∗ is a cocircuit of M containj

j

j xj i ⊆ Ω\{e , f }, ing e and f . Moreover, since U xii ∩ = ∅, i = 1, . . . , m, we have that e , f ∈ / and Ω i

i = 3, . . . , m. Thus xi ∈ H and xi ∈ / C ∗ for i = 3, . . . , m. Also, since e ∈ U x11 and f ∈ U x22 , we have that j

xj i Ω i

j

x3− j i Ω i

 Ω\{e , f } and  Ω\{e , f } for i = 1, 2. Thus xi ∈ / clM (Ω\{e , f }) and xi ∈ C ∗ for i = 1, 2.  ∗ Now we see that ∩ C = {x1 }, which is impossible since the nonempty intersection of a circuit and a cocircuit must contain at least two elements. This completes the proof. 2 The following lemma, which will be frequently used, gives a simple sufficient condition for the existence of a C -augmenting circuit. Recall that μ = |U |, υ = | V |, and υ = μ + δ . Lemma 9.2. Let 1 , . . . , k be C -minimal circuits where the sets i ∩ V , i = 1, . . . , k form a partition of V k

k and U ⊆ i =1 i ∩ U . If i =1 | i \Ω| > δ , then at least one of the circuits i , i = 1, . . . , k is C -augmenting. Proof. By contradiction. Suppose that none of the circuits i , i = 1, . . . , k are C -augmenting. Then | i C |  |C |, i = 1, . . . , k. Given that | i C | = | i ∩ U | + |C | − | i ∩ V | + | i \Ω|, i = 1, . . . , k, the previous inequalities imply that | i ∩ U | + | i \Ω|  | i ∩ V |, i = 1, . . . , k. Summing these inequalities, we obtain that k 

| i ∩ U | +

i =1

k 

| i \Ω| 

i =1

k 

| i ∩ V |.

(1)

i =1

k

∩ U and i ∩ V , i = 1, . . . , k form a partition of V , it follows that μ  ki=1 | i ∩ U |

k i =1 i

k and υ = i =1 | i ∩ V |. These inequalities together with (1) imply that μ + i =1 | i \Ω|  υ . Thus

k i =1 | i \Ω|  υ − μ = δ . This contradicts our assumptions. Thus at least one of i , i = 1, . . . , m must be C -augmenting. 2 Since U ⊆

We can now use the above lemma to show that if there is a w-nested set with more than 2δ chords, then one can find a C -augmenting circuit. Lemma 9.3. Let X be a w-nested set of chords of Ω . If | X | > 2δ , then M |Ω ∪ X has a C -augmenting circuit. j

Proof. Let (w, j) be a w-ordering for X where x = (x1 , . . . , xm ) and j = ( j 1 , . . . , jm ). Then ∅ = U x11 ⊆

j j j j j U x22 ⊆ · · · ⊆ U xmm ⊂ U and ∅ = V x11 ⊂ V x22 ⊂ · · · ⊂ V xmm ⊂ V . By assumption, m > 2δ . Let j 0 = 1, jm+1 = 1, j

j

Ωx10 = ∅, Ωx20 = Ω , Ωx1m+1 = Ω and Ωx2m+1 = ∅. For i = 1, 2, . . . , m + 1 let i = Ωxii Ωxii−−11 . Thus it is seen that i ∩ V = ∅, i = 1, 2, . . . , m + 1. Furthermore, since j

j

3− j i

i = Ωxii Ωxii−−11 = ΩΩxi

j

Ωxii−−11 ,

i = 1, 2, . . . , m + 1

Lemma 9.1 implies that the circuits i , i = 1, 2, . . . , m + 1 are C -minimal. By construction, the sets i ∩ Ω , i = 1, 2, . . . , m + 1 partition Ω . Thus the sets i ∩ U , i = 1, . . . , m + 1 partition U and the sets i ∩ V , i = 1, . . . , m + 1 partition V . We note that | 1 \Ω| = | m+1 \Ω| = 1 and | i \Ω| = 2 for all

m+1 i = 1, m + 1. Thus i =1 | i \Ω| = 2m > δ . By Lemma 9.2, at least one of the circuits 1 , . . . , m+1 must be C -augmenting. 2

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Our immediate goal now is to build on the results of the previous lemma. One important task in the paper is to show how one can find a C -augmenting circuit given two w-nested sets of chords. This we shall do in the next lemma. Lemma 9.4. Suppose X i , i = 1, 2 are two disjoint w-nested sets of chords of Ω having w-orderings (xi , ji ), i = j 1m

3− j 1m1

j 2m

1, 2 where xi = (xi1 , . . . , ximi ), ji = ( j i1 , . . . , j imi ), i = 1, 2. Suppose that U x1m 1 ∩ U x2m 2 = ∅ and U x1m 3− j 2m U x2m 2 2

1

= ∅. If | X 1 | + | X 2 | >

2 (δ 3

2

1

∩

+ 1), then M |Ω ∪ X 1 ∪ X 2 contains a C -augmenting circuit.

Proof. We shall prove the lemma in the case where j i1 = j i2 = · · · = j im1 = 1, i = 1, 2; the proof of the general case will be evident from this. Let m = m1 + m2 = | X 1 | + | X 2 |. Assume that m > 23 (δ + 1). We shall show that M |Ω ∪ X 1 ∪ X 2 has a C -augmenting circuit. Suppose to the contrary that it does not. Given that (xi , ji ), i = 1, 2 are w-orderings, we have (i) ∅ = U xi1 ⊆ U xi2 ⊆ · · · ⊆ U xim ⊂ U and ∅ = V xi1 ⊂ V xi2 ⊂ · · · ⊂ V xim ⊂ V , i = 1, 2. i

i

(ii) U x1m ∩ U x2m = ∅ and U x21m ∩ U x22m = ∅. 1

2

1

2

Since U x1m ∩ U x2m = ∅ (by (ii)), it follows that U x1i ∩ U x2 j = ∅ for all i, j. Thus U x1i U x2 j and 1

2

U x1i U x22 j for all i, j and consequently, V x1i V x22 j for all i, j. Therefore, the sets belonging to { V x11 , V x12 , . . . , V x1m } ∪ { V x221 , V x222 , . . . , V x22m } are totally ordered under inclusion. Thus we can 1

1

j

j

j

find an ordering x1 , x2 , . . . , xm of X 1 ∪ X 2 such that ∅ = V x11 ⊆ V x22 ⊆ · · · ⊆ V xmm ⊂ V where for all i ∈ {1, . . . , m} and for all k ∈ {1, 2}, j i = k iff xi ∈ X k . Let j 0 = 1, jm+1 = 1, Ωx10 = ∅, Ωx20 = Ω , j

j

Ωx1m+1 = Ω , and Ωx2m+1 = ∅. For i = 1, . . . , m + 1, let i = Ωxii Ωxii−−11 . By our choice of x1 , . . . , xm , 2 }, m+1 ∈ {Ωx21m , Ωx21 }, and we see that x1 ∈ {x11 , x2m2 } and xm ∈ {x1m1 , x21 }. Thus 1 ∈ {Ωx11 , Ω2m 2 1 hence 1 ∩ V = ∅ and m+1 ∩ V = ∅. Suppose that for some 1  i 1 < i 2  m and k ∈ {1, 2}, have ji

ji

xi 1 , xi 2 ∈ X k ; that is, j i 1 = j i 2 = k. Then by (i) V xi 1 \ V xi 2 = ∅. In particular, if xi −1 , xi ∈ X k for some 1 2 j

j

i ∈ {2, . . . , m} and k ∈ {1, 2}, then i ∩ V = V xii \ V xii−−11 = ∅. On the other hand, if i ∩ V = ∅, then xi −1 and xi belong to different sets X j , j = 1, 2. Let S = {1  i  m | i ∩ V = ∅} and s = | S |. Let T = {1, . . . , m}\ S and t = | T | = m − s, noting that 1 ∈ T . Suppose i ∈ S. Then xi −1 and xi belong to different sets X j , j = 1, 2. Without loss of generality, we may assume that xi −1 ∈ X 1 and xi ∈ X 2 . Suppose i + 1 ∈ S. Then i ∩ V = i +1 ∩ V = ∅ and furthermore, xi +1 ∈ X 1 . Now V xi+1 \ V xi−1 = ( V xi+1 \ V x2i ) ∪ ( V x2i \ V xi−1 ) = ( i +1 ∪ i ) ∩ V = ∅. However, V xi+1 \ V xi−1 = ∅ since xi −1 , xi +1 ∈ X 1 . This gives a contradiction. From this, we conclude that if i ∈ S, then i + 1 ∈ T . Consequently, s  m and 2 t =m−s m . We shall define β < β < · · · < β recursively in the following way: Let 1 2 t 2

β1 =

1, if x1 ∈ X 2 or 2 ∈ T ; 2, if x1 ∈ X 1 and 2 ∈ S .

We note that |{ j ∈ T | β1 < j  m}| = t − 1. Next, suppose that β1 , . . . , βk have already be defined for some 1  k < t. Suppose xβk ∈ X i for some i ∈ {1, 2} and |{ j ∈ T | βk < j  m}| = t − k. Let β = min{ j ∈ T | βk < j  m}. Now let

βk+1 =

if xβ ∈ X 3−i or β + 1 ∈ T ; β, β + 1, if xβ ∈ X i and β + 1 ∈ S .

By the definition of βk+1 , we have |{ j ∈ T | βk+1 < j  m}| = t − (k + 1). (9.5) If β + 1 ∈ S, then xβk+1 ∈ X 3−i .

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91

Proof. Suppose β + 1 ∈ S. If xβ ∈ X i then xβ+1 ∈ X 3−i . In addition, we also have xβk+1 = xβ+1 . Thus xβk+1 ∈ X 3−i . On the other hand, if xβ ∈ X 3−i , then βk+1 = β and xβk+1 = xβ . Thus xβk+1 ∈ X 3−i in this case as well. 2 Starting with β1 , we continue the process generating β1 , . . . , βt at which time the process stops since |{ j ∈ T | βt < j  m}| = 0. For i = 1, . . . , t let y i = xβi and let ki = j βi . Let k0 = 1, kt +1 = 1, k

Ω y10 = ∅ and Ω y1t +1 = Ω . For i = 1, . . . , t + 1 let i = Ω yii Ω yii−−11 . By our choice of y 1 , . . . , yt , we see that the sets i ∩ V , i = 1, . . . , t + 1 are nonempty and form a partition of V . By (ii), there exists t +1 u 0 ∈ U x21m ∩ U x22m . It is seen that U \{u 0 } ⊆ i =1 ( i ∩ (U \{u 0 })). Thus we have k

1

2

υ=

t +1 t +1           ∩ V  and μ − 1   ∩ U \{u 0 } . i i i =1

(2)

i =1

Using Lemma 9.1, the sets i , i = 1, . . . , t + 1 can be shown to be C -minimal circuits. Suppose that none of them are C -augmenting. Then | i C |  |C |, i = 1, . . . , t + 1 and hence | i ∩ U | + | i \Ω|  | i ∩ V |, i = 1, . . . , t + 1. Since | i \Ω| = 1 for i ∈ {1, t + 1} and | i \Ω| = 2 for all i ∈ {2, . . . , t }, we obtain the following inequalities:

















1 +  i ∩ U    i ∩ V ,

i = 1, t + 1

(3)

i = 2, . . . , t .

(4)

and

2 +  i ∩ U    i ∩ V , Summing the inequalities in (3) and (4), we obtain

2t +

t +1 t +1       ∩ U   | i ∩ V |. i i =1

(5)

i =1

Let 1 X be the indicator function for a set X . For i = 1, . . . , t + 1 we have | i ∩ U | = | i ∩ (U \{u 0 })| + 1  (u 0 ). Substituting the above into (5), we obtain i

2t +

t +1 t +1 t +1            ∩ U \{u 0 }  +  ∩ V . 1  (u 0 )  i i i =1

i =1

Now (2) and (6) imply that 2t + μ − 1 + δ + 1. Thus

t +1

2t +

i =1

t +1  i =1

i

1  (u 0 )  υ , and thus 2t + i

(6)

i =1

t +1 i =1

1  (u 0 )  υ − μ + 1 = i

1  (u 0 )  δ + 1.

(7)

i

k

We observe that if y i −1 and y i belong to different sets X j , j = 1, 2, then u 0 ∈ Ω yii Ω yii−−11 = i . We k

also note that if y 1 ∈ X 2 , then 1 = Ω y21 and u 0 ∈ 1 .

(9.6) Let i ∈ S. Then there exists β j such that β j = i − 1 or β j = i. Furthermore, y j −1 and y j belong to different sets X k , k ∈ {1, 2} and u 0 ∈ j .

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Proof. Let i ∈ S. Then i − 1 ∈ T . By our choice of β1 , . . . , βt , there exists β j such that β j = i − 1 or β j = i. If β j = i − 1, then xβ j−1 and xβ j belong to different sets X k , k = 1, 2 (since i ∈ S). If β j = i, then xi −1 and xβ j−1 belong to the same sets X k , k = 1, 2. Also, since β j = i ∈ S, it follows that xi −1 and xβ j belong to different sets X k , k = 1, 2. Therefore, xβ j−1 and xβ j belong to different sets X k , k = 1, 2. In either case, y j −1 = xβ j−1 and y j = xβ j belong to different sets X k , k = 1, 2. It now follows that u 0 ∈ j . 2 Let i ∈ S. By (9.6), there exists β j such that β j = i − 1 or β j = i and u 0 ∈ j . Since this holds for every value i ∈ S, it follows that there are at least s values j ∈ {1, . . . , t } for which 1  (u 0 ) = 1. Consequently,

t +1 i =1

j

1  (u 0 )  s and hence by (7), 2t + s  δ + 1. Since t = m − s, we have 2(m − s)+ s  i

m , it follows that 32 m  δ + 1 and hence m  23 (δ + 1). This 2 contradicts our assumptions. We conclude that at least one of the circuits i , i = 1, . . . , t + 1 must be C -augmenting. 2

δ + 1, and 2m − s  δ + 1. Since s 

The final goal of this section is to show how one can find a C -augmenting circuit in the case where one has three w-nested sets of chords. Lemma 9.7. Let X i , i = 1, 2, 3 be three pairwise disjoint w-nested sets of chords of Ω . Let (xi , ji ), i = 1, 2, 3 be w-orderings for X i , i = 1, 2, 3 where xi = (xi1 , . . . , ximi ), and ji = ( j i1 , . . . , j imi ), i = 1, 2, 3. Suppose that the j im

sets U xim i , i = 1, 2, 3 are pairwise disjoint. If i circuit.

3

i =1 | X i |

> 2δ , then M |Ω ∪ X 1 ∪ X 2 ∪ X 3 has a C -augmenting

3

Proof. Let m = m1 + m2 + m3 = i =1 | X i |. Assume that m > 2δ . For convenience, we may assume that j i1 = j i2 = · · · = j imi = 1, i = 1, 2, 3; the proof of the general case may be derived from this. Given that (xi , ji ), i = 1, 2, 3 are w-orderings, we have (i) ∅ = U xi1 ⊆ U xi2 ⊆ · · · ⊆ U xim ⊂ U and ∅ = V xi1 ⊂ V xi2 ⊂ · · · ⊂ V xim ⊂ V , i = 1, 2, 3. i

i

We also have that (ii) U xim , i = 1, 2, 3 are pairwise disjoint. i

Suppose m = 3. Then m1 = m2 = m3 = 1 and U xi1 , i = 1, 2, 3 are pairwise disjoint (by (ii)). Thus U xi1 U x j1 for all i = j, and hence V xi1 V x j1 for all i = j. Suppose V xi1 ∩ V x j1 = ∅ for some i = j. By ◦

◦

◦

◦

Lemma 5.3, it follows that U = U x11 ∪ U x21 ∪ U x31 and V = V x211 ∪ V x221 ∪ V x231 . For i = 1, 2, 3 let

|U xi1 | and βi =

| V x2i1 |.

Then

μ = |U | =

3

i =1

αi and υ

seen to be C -minimal. Suppose that none of them and hence 1 + |U x2i1 | + |C | − | V x2i1 |  |C |. From this, that |U x2i1 | =



j =i

i = 2 = | V | = i =1 βi . The circuits Ωxi1 , i = 1, 2, 3 are are C -augmenting. Then |C Ωx2i1 |  |C |, i = 1, 2, 3 we obtain that 1 + |U x2i1 |  | V x2i1 |, i = 1, 2, 3. Given

3

α

α j , this results in the inequalities:

1 + α2 + α3  β1 ,

1 + α1 + α3  β2 ,

1 + α1 + α2  β3 .

(8)

Summing the above inequalities, we obtain 2μ + 3 = 2(α1 + α2 + α3 ) + 3  β1 + β2 + β3 = υ . Thus μ+3 2μ + 3  υ . Given that υ = μ + δ , it follows that δ  μ + 3. Thus m > 2δ  2  3 (since μ  3). This gives a contradiction. We conclude that at least one of the circuits Ωx2i1 , i = 1, 2, 3 must be C -augmenting. From the above, we may assume that m  4. Now (i), (ii) and Lemma 5.5 imply that V xim , i = i 1, 2, 3 are pairwise disjoint. Let Ωxi0 = ∅, i = 1, 2, 3 and let Ψ = ΩΩx1m Ωx2m Ωx3m . For i = 1 2 3 1, 2, 3 and 1  j < mi let Ψi j = Ωxi j Ωxi( j−1) . For i = 1, 2, 3 let

S. McGuinness / Advances in Applied Mathematics 53 (2014) 72–111

Ψimi =

93

Ψ Ωximi Ωxi(mi −1) , if Ψ ∩ V = ∅; Ωximi Ωxi(mi −1) , if Ψ ∩ V = ∅.

By (i), Ψi j ∩ V = ∅, i = 1, 2, 3; j = 1, . . . , mi . Furthermore, by Lemma 9.1, Ψi j is seen to be a C -minimal circuit for all i, j and Ψ is also seen to be a C -minimal circuit. We shall consider two cases. Case 1. Ψ ∩ V = ∅. The set Ψ ∩ Ω together with the sets Ψi j ∩ Ω , i = 1, 2, 3; j = 1, . . . , mi form a partition of Ω . Thus Ψ ∩ U together with the sets Ψi j ∩ U , i = 1, 2, 3; j = 1, . . . , mi form a partition of U and Ψ ∩ V together with the sets Ψi j ∩ V , i = 1, 2, 3; j = 1, . . . , mi form a partition of V . Since |Ψ \Ω| = 3 and |Ψi1 \Ω| = 1, i = 1, 2, 3 and |Ψi j \Ω| = 2, i = 1, 2, 3; j  2, we have

|Ψ \Ω| +



|Ψi j \Ω|  3 +

i, j

3 

(2mi − 1) = 2

i =1

3 

mi = 2m > δ.

i =1

By Lemma 9.2, either Ψ or one of the circuits Ψi j , i = 1, 2, 3; j = 1, . . . , mi is C -augmenting. This completes Case 1. Case 2. Ψ ∩ V = ∅. The sets Ψi j ∩ V , i = 1, 2, 3; j = 1, . . . , mi form a partition of V , and U ⊆ see that

|Ψi1 \Ω| =

1, if mi  2; 2, if mi = 1,

|Ψimi \Ω| =

3, if mi  2; 2, if mi = 1,

and |Ψi j \Ω| = 2, i = 1, 2, 3; j = 2, . . . , mi − 1. From this it follows that



3

mi

j =1



i , j (Ψi j

∩ U ). We also

i = 1, 2, 3

|Ψi j \Ω| = 2mi , i = 1, 2, 3

and hence i =1 mi = 2m > δ . Now Lemma 9.2 implies that at least one of the Ψi j ’s i , j |Ψi j \Ω| = 2 must be C -augmenting. This completes Case 2. The proof of the lemma follows from Cases 1 and 2 above. 2 10. Ω is a spanning circuit of N : Part I In the next two sections, we shall complete the first part of the proof of Theorem 1.4. Namely, we shall show that if Ω is a spanning circuit of N, then we can find a C -augmenting circuit in M. j U Recall that by convention, for all x ∈ X Ω , if Ωx ⊆ U for some j ∈ {1, 2}, then j = 1. Let X Ω = {x ∈ X Ω | x ⊆ U }. In this section, we shall assume that Ω is a spanning circuit of N and Ωx is minimal and Ω U a |  3. Our goal is to show that, under these assumptions, M has a for which |Ω there exists a ∈ X Ω C -augmenting circuit. For the first part of this section, we shall assume that a is not 3-crossed. Our first important step is to show that either there are two disjoint w-nested sets of chords of chords X i , i = 1, 2 for which

2 | X  | > 23 (δ + 1) or there are three disjoint w-nested sets of chords X i , i = 1, 2, 3 for which

3i =1 i δ i =1 | X i | > 2 . By applying Lemmas 9.4 and 9.7, we can then find a C -augmenting circuit. 2j a and B = Ω a2 . For all x ∈ X a and j ∈ {1, 2}, let A xj = Ωaxj and let B xj = Ωax Let A = Ω . By assumption, | A |  3. Since N = M \ W and μ = ω (by (7.3)) we have r ( M ) = r ( N ∪ W ) = r (Ω ∪ W ) = μ + υ + ω − 2 = 3μ + δ − 2.

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For all e , f ∈ A, let Ωe∗f = Ω N∗ ,e f = E ( N )\cl N (Ω\{e , f }) be the Ω -cocircuit containing e and f (as defined in Section 6). One sees that Ωe∗f ∩ C = ∅ and Ωe∗f is a cocircuit of M. Thus by assumption,

|Ωe∗f | >

r(M ) 3

∗ | > μ + δ−2 − 1 for all e , f ∈ A. Let ξ = mine, f ∈ A |Ω ∗ |. + 1 = μ + δ−3 2 + 1 and hence |Ω 3 ef ef

2 − 1 and furthermore ξ > 1 since Then ξ > μ + δ− 3 implies that either

μ  | A |  3. Since a is not 3-crossed, Theorem 6.6

(i) there exist x, y ∈ X a and i , j ∈ {1, 2} such that |[x]a |  ξ , |[ y ]a |  ξ , A ix ∩ A y = ∅, and A 3x −i ∩ j

3− j

Ay

= ∅

or j

(ii) there exist x, y , z ∈ X a and i , j , k ∈ {1, 2} such that |[x]a | + |[ y ]a | + |[ z]a |  32 ξ and A ix , A y , A kz are pairwise disjoint. Suppose (i) holds. Let X 1 = [x]a , X 2 = [ y ]a , and let mi = | X i |, i = 1, 2. Let m = m1 + m2 . Then 2 mi  ξ , i = 1, 2 and consequently m  2ξ > 2(μ + δ− − 1) = 2μ + 23 (δ − 2) − 2. By (i), we also have 3 3− j

A ix ∩ A y = ∅ and A 3x −i ∩ A y = ∅ for some i , j ∈ {1, 2}. Without loss of generality, we may assume A x ∩ A y = ∅ and A 2x ∩ A 2y = ∅. By Lemma 8.1, there exist refinements X i ⊆ X i , i = 1, 2 such that | X i |  mi − μ + | A |, i = 1, 2. Thus j

2        2 X  + X   mi − μ + | A | = m − 2μ + 2| A | > 2| A | + (δ − 2) − 2. 1 2

(9)

3

i =1

The last inequality follows from that fact that m > 2μ + 23 (δ − 2) − 2. Since | A |  3, it follows that

2| A | + 23 (δ − 2) − 2 > 23 (δ + 1) and hence | X 1 | + | X 2 | > 23 (δ + 1). For i = 1, 2, let (xi , ji ) be a w-ordering for X i , where xi = (xi1 , xi2 , . . . , xini ) and ji = ( j i1 , j i2 , . . . , j ini ). Without loss of generality, we may assume j i1 = j i2 = · · · = j ini = 1, i = 1, 2. Thus for i = 1, 2 we have ∅ =  U xi1 ⊆ U xi2 ⊆ · · · ⊆ U xin ⊂ U i

and ∅ = V xi1 ⊂ V xi2 ⊂ · · · ⊂ V xin ⊂ V . Since A 2x ∩ A 2y = ∅, it follows that U x21n ∩ U x22n = ∅. Given that i

1

2

∅ = A x ⊆ U x1n1 \U x2n2 and ∅ = A y ⊆ U x2n2 \U x1n1 , if U x1n1 ∩ U x2n2 = ∅, then U x1n1  U x2n2 , yielding a contradiction. Thus U x1n ∩ U x2n = ∅. Now since | X 1 | + | X 2 | > 23 (δ + 1), by (9), it follows by Lemma 9.4 1 2

that M has a C -augmenting circuit. Suppose that (ii) holds. Without loss of generality, we may assume that A x , A y , and A z are pairwise disjoint. Let X 1 = [x]a , X 2 = [ y ]a , X 3 = [ z]a and let mi = | X i |, i = 1, 2, 3. Let m = m1 + m2 + m3 . 2 − 1) = 32 μ + 2δ − 52 . Suppose first that m  4. By Lemma 8.2, the cirThen m  32 ξ > 32 (μ + δ− 3 cuits Ωx , Ω y , and Ωz are pairwise disjoint and there exist refinements X i ⊆ X i , i = 1, 2, 3 such that

3



i =1 | X i |

 m − μ + | A | − 1. Now

3     X   m − μ + | A| − 1 > 3 μ + δ − 5 − μ + | A| − 1 > δ . i i =1

2

2

2

(10)

2

The last inequality follows from the fact that μ  | A |  3. Suppose X i = ∅, i = 1, 2, 3. For i = 1, 2, 3, let (xi , ji ) be a w-ordering for X i where xi = (xi1 , . . . , xini ) and ji = ( j i1 , . . . , j ini ). Without loss of generality, we may assume that j i1 = j i2 = · · · = j ini = 1, i = 1, 2, 3. We may also assume that A x = A x1 j , j = 1, . . . , n1 , A y = A x2 j , j = 1, . . . , n2 , and A z = A x3 j , j = 1, . . . , n3 . Since the circuits Ωx , Ω y , Ωz are pairwise disjoint, the sets U xini , i = 1, 2, 3 are pairwise disjoint. It now follows by (10) and Lemma 9.7 that M has a C -augmenting circuit. Suppose X i = ∅ for exactly one value i ∈ {1, 2, 3}. We may assume X 3 = ∅ and X i = ∅, i = 1, 2. Let (xi , ji ), i = 1, 2 be w-orderings of X i i = 1, 2 as before. We have U x1n ∩ U x2n = ∅ and V x1n ∩ V x2n = ∅, 1

2

1

2

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95

Ωx and Ω y are disjoint. We now see that the set X 1 ∪ X 2 is w-nested. Since | X 1 | + | X 2 | > 2δ by (10), Lemma 9.3 implies that M has a C -augmenting circuit. Suppose X i = ∅ for exactly two values i ∈ {1, 2, 3}. Without loss of generality, we may assume X 2 = ∅ and X 2 = ∅. Then X 1 = ∅. Let (w1 , j1 ) be a w-ordering of X 1 as before. Then by (10), | X 1 | > 2δ , and hence it follows by Lemma 9.3 that M has a C -augmenting circuit. Suppose m < 4. Then m = 3 and hence 3 > 32 μ + 2δ − 52 . This means that μ < 11 and hence 3 μ = | A | = m = 3 and δ  1. Thus A = U , B = V , A x = U x , A y = U y , and A z = U z . Now U x , U y , U z are pairwise disjoint. Since m = 3 > 12  2δ , it follows by Lemma 9.7 that M has a C -augmenting circuit. To finish off this section, we suppose that a is 3-crossed. Since a is 3-crossed, then there exist ◦

j ◦

◦

3− j ◦

x, y , z ∈ X Ω and i , j , k ∈ {1, 2} such that X a = {x, y , z}, A = A ix ∪ A y ∪ A kz , and B = B 3x −i ∪ B y ◦

◦

∪ B 3z −k . ◦

◦

Without loss of generality, we may assume i = j = k = 1. Then A = A x ∪ A y ∪ A z and B = B 2x ∪ B 2y ∪

∗ = {x, y }. It now follows that ξ  2 and B 2z . Let e ∈ A x and f ∈ A y . Then Ωe∗f = {e , f , x, y } and Ω ef

2ξ >μ+

δ−2 3

−12+

The above inequalities imply that ξ = 2, δ  1 and

3

.

μ  3. Thus μ = 3, A = U , B = V and A ix =

∈ X a and i ∈ {1, 2}. We see that the circuits Ωx2 , x ∈ X a are C -minimal, the sets V x2 , ◦ ◦ 

2 x ∈ X a partition V (since B = B 2x ∪ B 2y ∪ B 2z ), and U ⊆ x ∈ Xa U x2 . Since x ∈ X a |Ωx \Ω| = 3 > 1  δ , 2  Lemma 9.2 implies that at least one of the circuits Ωx , x ∈ X a is C -augmenting. This completes the U xi 

for all x

δ−2

case where a is 3-crossed.

11. Ω is a spanning circuit of N : Part II In this section, we shall complete the proof of Theorem 1.4 in the case where Ω is a spanning x | = 2 for all x ∈ X U ; that is, Ωx is a circuit of N. By the previous section, we may assume that |Ω Ω U U triangle for all x ∈ X Ω . Observe that w ∈ / X Ω since μ  3. Also, since Ω is a spanning circuit of N, we have r ( M ) = 3μ + δ − 2 (as in Section 10). We shall show that M has a C -augmenting circuit.

a ∩ Ω b = ∅ for some a, b ∈ X U . Then M has a C -augmenting circuit. (11.1) Suppose Ω Ω a = {e , f } and Ω b = { f , g }. Let x ∈ X a and y ∈ X b . Then it is seen that [x]a = X a = Ω ∗ Proof. Let Ω ef

∗ . Since Ω ∗ and Ω ∗ are cocircuits of M not intersecting C , we have |Ω ∗ | > and [ y ]b = X b = Ω fg ef fg ef

∗ | > r (M ) − 1 = μ + δ−2 − 1. Let X = [x]a , Y = [ y ]b , and let m = | X | − 1 = μ + δ−3 2 − 1 and |Ω 3 3 fg and n = |Y |. The sets X and Y are a-nested and b-nested, respectively (by Lemma 5.2). By the above, ∗ | > μ + δ−2 − 1 and n = |[ y ]b | = |Ω ∗ | > μ + δ−2 − 1. By Lemma 8.1, there exist m = |[x]a | = |Ω ef 3 fg 3    a | = m − μ + 2 and |Y  |  n − μ + |Ω b | = refinements X ⊆ X and Y ⊆ Y such that | X |  m − μ + |Ω n − μ + 2. Let m = | X  | and n = |Y  |. Then m  m − μ + 2 and n  m − μ + 2. We shall show i ac i = that X ∩ Y ∩ X w = ∅. Suppose to the contrary that there exists c ∈ X ∩ Y ∩ X w . Then Ω =Ω bc i i i { f } for some i ∈ {1, 2}. Since c ∈ X w , we have that Ωc ∩ V = V c = ∅. Let h ∈ V c . Then we see that → (e , g , h) Ω, (Ωa , Ωb , Ωci ), yielding a contradiction. Thus X ∩ Y ∩ X w = ∅. Given that X  ∩ Y  ⊆ X ∩ Y ∩ f X w , it follows that X  ∩ Y  = ∅. Let (x, i) and (y, j) be w-orderings for X  and Y  where x = (x1 , . . . , xm ), y = ( y 1 , . . . , yn ), i = (i 1 , . . . , im ), and j = ( j 1 , . . . , jn ). We may assume i 1 = i 2 = · · · = im = 1 and j 1 = j 2 = · · · = jn = 1. Furthermore, we may assume that e ∈ U xi , f ∈ U x2i , i = 1, . . . , m and g ∈ U y i , and f ∈ U 2y i , i = 1, . . . , n . Clearly g ∈ U x2i for i = 1, . . . , m since X ∩ Y ∩ X w = ∅, as was previously shown. Likewise, e ∈ U 2y i for i = 1, . . . , n . Since f ∈ U x2  ∩ U 2y  , it follows that U xm ∩ U yn = ∅; for otherwise, m n r(M ) 3

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U xm  U yn . Now

m  + n   m + n − 2μ + 4 > 2



μ+

δ−2 3

2 − 1 − 2μ + 4 = (δ + 1).

Thus Lemma 9.4 implies that M has a C -augmenting circuit.

3

2

x ∩ Ω a ⊆ U and y = ∅ for all x, y ∈ X U . Let a ∈ X Ω where Ω By (11.1), we may assume that Ω Ω a |  3. At least one such element exists since we could take a = w. Since |Ω a |  3, it follows by |Ω U a , B = Ω a2 and for all x ∈ X a , / XΩ , and hence Ωa is not minimal. Let A = Ω our assumptions that a ∈ j j j 2j A   and for all j ∈ {1, 2}, let A x = Ωax and B x = Ωax . Let X Ω = {x ∈ X Ω | Ωx ⊂ Ωa }. We may choose a U A A A  = X \ X A . For ; that is, X Ω ⊆ XΩ . Let N  = N \ X Ω and let X Ω such that Ωx is minimal for all x ∈ X Ω Ω Ω ∗ ∗ A  . Let each pair of elements e , f ∈ A, we see that Ω N  ,e f is a cocircuit of N and Ω N  ,e f = Ω N∗ ,e f \ X Ω δ− 2 ∗ ∗     |. As before, we have ξ > μ + ξ = mine, f ∈ A |Ω | and let ξ = mine, f ∈ A |Ω − 1. We note that N ,e f

since

N ,e f

μ  3, this inequality implies that ξ  2.

3

A (11.2) ξ   ξ − 2. Furthermore, if ξ  = ξ − 2, then | X Ω |  2 and | A |  4.

 , it is seen that x ∈ Ω ∗ x × Ω x2 or (e , f ) ∈ Proof. For all e , f ∈ A and for all x ∈ X Ω iff (e , f ) ∈ Ω N ,e f A 2 x × Ω x , x ∈ X are pairwise disjoint. Thus for any two elements x, y ∈ x . By (11.1), the sets Ω Ω Ω A ∗ , either (e , f ) ∈ Ω x × Ω y or ( f , e ) ∈ Ω x × Ω y . This means that if there are three elements XΩ ∩Ω N ,e f

A x, y , z ∈ X Ω ∩ ΩN∗ ,e f , then either e or f belongs to at least two of the sets Ωx , Ω y , Ωz . This contradicts A ∗ |  2 and it follows that ξ   ξ − 2. If ξ  = ξ − 2, then |Ω ∗  | = ξ − 2, the above. Thus | X Ω ∩Ω N ,e f N ,e f

∗ |  ξ , it follows that |Ω ∗ \Ω ∗  |  2. Thus | X A |  2 and hence for some e , f ∈ A. Given that |Ω Ω N ,e f N ,e f N ,e f | A |  4. 2 (11.3) ξ   1.

Proof. Suppose to the contrary that ξ  = 0. By (11.2), we have 0 = ξ   ξ − 2. Thus ξ  2. However, since ξ  2, it follows that ξ = 2 and ξ  = ξ − 2. Now (11.2) implies that | A |  4 and hence μ  | A |  4. However, this would mean that ξ > μ + δ−3 2 − 1 > 2, yielding a contradiction. We conclude that ξ   1. 2 A For all x ∈ X a , let [x]a = [x]a \ X Ω . We shall consider two cases.

Case 1. Suppose a is not 3-crossed in N  . We may apply Theorem 6.6 with [x]a , [ y ]x , [ z]a in place of [x]a , [ y ]a , [ z]a and ξ  in place of ξ . Thus either (i) there exist x, y ∈ X a and i , j ∈ {1, 2} such that |[x]a |  ξ  , |[ y ]a |  ξ  , A ix ∩ A y = ∅, and A 3x −i ∩ j

3− j

Ay

= ∅

or (ii) there exist x, y , z ∈ X a and i , j , k ∈ {1, 2} such that |[x]a | + |[ y ]a | + |[ z]a |  32 ξ  and A ix , A y , A kz are pairwise disjoint. j

We shall look at two subcases, depending on whether (i) or (ii) occurs.

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97

Case 1.1. Suppose (i) holds. We may assume that A x ∩ A y = ∅ and A 2x ∩ A 2y = ∅. Let X 1 = [x]a , X 2 = [ y ]a and let mi = | X i |, i = 1, 2. Let m = m1 + m2 . By (11.2), mi  ξ   ξ − 2, i = 1, 2. (11.4) If ξ   ξ − 1, then m > 2μ + 23 (δ − 2) − 4. Otherwise, if ξ  = ξ − 2, then m > 2μ + 23 (δ − 2) − 6. 2 Proof. Suppose ξ   ξ − 1. Then mi  ξ   ξ − 1 > μ + δ− − 2, i = 1, 2 and we have m > 2μ + 23 (δ − 3

2) − 4. On the other hand, if ξ  = ξ − 2, then m  2(ξ − 2) > 2μ + 23 (δ − 2) − 6.

2

The sets X i , i = 1, 2 are a-nested (by Lemma 5.2) and hence by Lemma 8.1 there exist refinements X i ⊆ X i , i = 1, 2 such that | X i |  mi − μ + | A |, i = 1, 2. Let ni = | X i |, i = 1, 2 and let n = n1 + n2 . (11.5) n > 23 (δ + 1). Proof. From the above, ni = | X i |  mi − μ + | A |, i = 1, 2. Thus n = n1 + n2  m − 2μ + 2| A |. Suppose

ξ   ξ − 1. Then by (11.4), m > 2μ + 23 (δ − 2) − 4. Thus n > 23 (δ − 2) + 2| A | − 4  23 (δ + 1). The last inequality follows from the fact that | A |  3. Suppose on the other hand that ξ  = ξ − 2. Then | A |  4 (by (11.2)), and m > 2μ + 23 (δ − 2) − 6 (by (11.4)). In this case, n > 23 (δ − 2) + 2| A | − 6  23 (δ + 1). 2 For i = 1, 2 let (xi , ji ) be a w-ordering for X i where xi = (xi1 , xi2 , . . . , xini ), i = 1, 2 and ji = ( j i1 , j i2 , . . . , j ini ), i = 1, 2. We may assume that j i1 = j i2 = · · · = j ini = 1, i = 1, 2. Furthermore, we may assume that A x1 j = A x , j = 1, . . . , n1 and A x2 j = A y , j = 1, . . . , n2 . We see that ∅ = A 2x1n ∩ A 2x2n ⊆ 1

2

U x21n ∩ U x22n . If U x1n ∩ U x2n = ∅, then U x1n  U x2n , yielding a contradiction. Thus U x1n ∩ U x2n = ∅. 1

2

1

2

1

2

1

2

Now since n > 23 (δ + 1), (by (11.5)), Lemma 9.4 implies that M has a C -augmenting circuit. This completes Case 1.1. Case 1.2. Suppose (ii) holds. Let X 1 = [x]a , X 2 = [ y ]a , X 3 = [ z]a and let mi = | X i |, i = 1, 2, 3. Let m = m1 + m2 + m3 . Then m  32 ξ  . We may assume that A x , A y , and A z are pairwise disjoint. (11.6) If m  4, then M has a C -augmenting circuit.

3

 Proof. Assume m  4. By Lemma 8.2, there exist refinements X i ⊆ X i , i = 1, 2, 3 such that i =1 | X i |  m − μ + | A | −  where  = 0 if μ  | A | + 1 and  = 1 otherwise. Let ni = | X i |, i = 1, 2, 3 and let

n = n1 + n2 + n3 . Then n  m − μ + | A | −   32 ξ  − μ + | A | −  . For i = 1, 2, 3, let (xi , ji ) be a w-ordering for X i where xi = (xi1 , xi2 , . . . , xini ), i = 1, 2, 3 and ji = ( j i1 , j i2 , . . . , j ini ), i = 1, 2, 3. We may assume j i1 = j i2 = · · · = j ini = 1, i = 1, 2, 3. We may also assume that A x1 j = A x , j = 1, . . . , n1 , A x2 j = A y , j = 1, . . . , n2 , and A x3 j = A z , j = 1, . . . , n3 . Since the sets A x , A y , and A z are pairwise disjoint, the sets A xin , i = 1, 2, 3 are also pairwise disjoint. Since U xin  U x jn for all i = j, it can be i

i

j

easily shown that the sets U xin , i = 1, 2, 3 are also pairwise disjoint. We shall show that n > 2δ . i

By (11.2), ξ   ξ − 2. Suppose ξ   ξ − 1. Then

n

=

3 2

(ξ − 1) − μ + | A | −  >

μ 2

δ > . 2

− 4 + | A| −  +

δ 2

3 2



μ+

δ−2 3

− 2 − μ + | A| − 

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The last inequality follows from the fact that μ  | A |  3 and when instead that ξ  = ξ − 2. Then | A |  4 by (11.2). We now have

n

=

3 2

(ξ − 2) − μ + | A | −  >

μ 2

−

11 2

+ | A| −  +

3



μ+

2

 = 1, μ  | A | + 2  5. Suppose

− 3 − μ + | A| − 

δ−2 3

δ 2

δ

> . 2

The last inequality follows from the fact that μ  | A |  4 and when  = 1, μ  3| A | + 2  6. Thus in both cases, n > 2δ . Now Lemma 9.7 implies that M has a C -augmenting circuit. 2 By (11.6), we may assume that m = 3. Then m1 = m2 = m3 = 1. Given that m  32 ξ  , it follows that

3  32 ξ  .

(11.7) If μ = | A |, then M has a C -augmenting circuit. Proof. Suppose μ = | A |. Then U = A, and A x = U x and B x = V x for all x ∈ {x, y , z} and j ∈ {1, 2}. If the sets V x , x ∈ {x, y , z} are not pairwise disjoint, then Corollary 5.7 implies that a is 3-crossed. This contradicts our assumptions. Thus the sets V x , x ∈ {x, y , z} are pairwise disjoint and now one can proceed as in the proof of (11.6) to show that M has a C -augmenting circuit. 2 j

From the above, we may assume that

j

j

j

μ > | A |. In particular, this means that μ  4.

(11.8) If ξ   ξ − 1, then M has as C -augmenting circuit. Proof. Suppose ξ   ξ − 1. Then

3

3  3 3 ξ  (ξ − 1) > 2 2 2



μ+

δ−2 3

δ 3 − 2 = μ + − 4. 2

2

From this we see that μ  4, and hence μ = 4. The above inequality now implies that 2δ < 1, and hence δ  1. Since 4 = μ > | A |  3, we have | A | = 3 and consequently, μ − | A | = 1. Now at least one of x, y, z belongs to X w , and may assume x ∈ X w . Given that {x} is a w-nested set and δ  1, Lemma 9.3 implies that either Ωx or Ωx2 is a C -augmenting circuit. 2 (11.9) If ξ  = ξ − 2, then M has as C -augmenting circuit. Proof. Suppose ξ  = ξ − 2. Then | A |  4 by (11.2), and hence have

3

3  3 3 ξ = (ξ − 2) > 2 2 2



2 Thus 2 > μ + δ− − 3 and 5 > μ + δ−3 2 . It follows that 3

μ+

μ > | A |  4 and μ  5. As before, we

δ−2 3

−3 .

μ  5 and hence μ = 5 and | A | = 4. The same υ  μ + 1. Proceeding as in the proof of

2 inequality now implies that δ− < 0, and hence δ  1 and 3 (11.8), it follows that M has a C -augmenting circuit. 2

The proof of Case 1.2 follows from (11.2)–(11.9) above, and the proof of Case 1 now follows from Cases 1.1 and 1.2 above.

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99

Case 2. Suppose a is 3-crossed in N  . ◦

j ◦

There exist x, y , z ∈ X Ω and i , j , k ∈ {1, 2} such that X a = {x, y , z}, A = A ix ∪ A y ∪ A kz and B = ◦

3− j ◦

B 3x −i ∪ B y ◦

◦

◦

◦

∪ B 3z −k . We may assume i = j = k = 1. Thus we have A = A x ∪ A y ∪ A z and B = B 2x ∪

B 2y ∪ B 2z . Suppose | A x |  2 and let e , f ∈ A x . We observe that x ∈ / Ωe∗f . Moreover, since the sets A x , A y ,

∗ . From this, it follows that x, y , z ∈ ∗ /Ω /Ω and A z are pairwise disjoint, we also have that y , z ∈ N ,e f N ,e f

A ∗ ⊆ X A . Since ∗ , we have that x ∈ and consequently, for all x ∈ Ω / X a and hence x ∈ X Ω . Thus Ω Ω ef N ,e f x , x ∈ X A are pairwise disjoint, it follows that |Ω ∗ |  2 and consequently ξ  |Ω ∗ |  2. the sets Ω Ω N ,e f ef

However, since

μ  | A | = | A x | + | A y | + | A z |  4, it follows that ξ > μ +

δ−2 3

− 1 > 2. This gives a

contradiction. We conclude that | A x | = 1, and likewise, | A y | = | A z | = 1. Thus | A | = 3 and | A 2x | = A | A 2y | = | A 2z | = 2. Let A = {e , f , g } where A x = {e }, A y = { f }, and A z = { g }. Let b ∈ X Ω . Then | A b | = 2, and we may assume that A b = {e , f }. Since A x = {e }, A y = { f }, and A z = { g }, we have that x, y ∈ ∗ and z ∈ ∗ . Furthermore, since A x ∩ A b = ∅ for all x ∈ X A \{b}, it follows that x ∈ ∗ for Ω /Ω /Ω Ω N ,e f N ,e f N ,e f

A ∗ = {x, y }. Thus ξ  |Ω ∗ | = 2. We now have that μ + δ−2 − 1 < ξ  2, , and hence Ω all x ∈ X Ω 3 N ,e f N ,e f and hence μ  3 and μ = | A | = 3. One can now show that at least one of the circuits Ωx2 , x ∈ {x, y , z} is C -augmenting (see Section 10). This completes Case 2.

12. Ω -bridges of order one A major obstacle in the proof of the main theorem is the case where Ω is not a spanning circuit of N. In this case, N has Ω -bridges of order one (since Ω is a dominating circuit of N). As we shall see, the presence of such bridges will enable us to find a C -augmenting circuit in M. In this section, we shall show that the way in which chords and Ω -bridges of order one can “overlap” is very restricted, due to the fact that N is regular. Let X be an Ω -bridge of order one. Then X is a cocircuit of M and moreover, Lemma 4.1 implies that M |( X ∪ Ω) is a pseudowheel with rim Ω and spoke-set X . For a pseudowheel Θ = (x, Ω) where x = (x1 , . . . , xm ) we let U xki x j = Ωxki x j ∩ U and V xki x j = Ωxki x j ∩ V , for all i , j ∈ {1, . . . , m} and for all k ∈ {1, 2} where i = j. We call each set U xi xi+1 , i = 1, . . . , m (respectively, V xi xi+1 , i = 1, . . . , m) a U -sector (respectively, V -sector) of Θ . We say that two distinct U -sectors U xi xi+1 and U x j x j+1 are adjacent if xi +1 = x j or x j +1 = xi . Otherwise, we say that they are non-adjacent. We define σ (Θ) to be the number of nonempty U -sectors of Θ and we define σ ∗ (Θ) to be the maximum number of pairwise non-adjacent nonempty U -sectors. We say that Θ is V -complete if each V -sector of Θ is nonempty. If Θ is V -complete, then a pseudowheel Φ = (y, Ω) where y = ( y 1 , . . . , yn ) and y j = xi j , j = 1, . . . , n for some 1  i 1 < i 2 < · · · < in  m, is called a refinement of Θ . Lemma 12.1. Let Θ = (x, Ω) be a pseudowheel where x = (x1 , . . . , xm ). If |Ωxi xi+1 | = 1 for some i, then M has a C -augmenting circuit.

xi xi+1 | = 1 for some i. Either |Ω xi xi+1 | = |U xi xi+1 | = 1 or |Ω xi xi+1 | = | V xi xi+1 | = 1. Proof. Suppose |Ω Suppose the former occurs. Let D  = D Ωxi xi+1 . Then | D  | = | D | + 1, D  ∩ C = D ∩ C = W and D  is seen to be a C -minimal circuit. Now,

     C  D   =  D   − ω + υ = | D | + 1 − ω + υ = μ + υ + 1 = ω + υ + 1 = | C | + 1. xi xi+1 | = | V xi xi+1 | = 1, then Ωxi xi+1 is seen to be a Therefore, D  is a C -augmenting circuit. If |Ω C -minimal circuit where |Ωxi xi+1 C | = |C | + 1. In this case, Ωxi xi+1 is C -augmenting. 2 By Lemma 12.1, we may assume for every pseudowheel Θ = (x, Ω) where x = (x1 , . . . , xm ) that

xi xi+1 |  2, i = 1, . . . , m. |Ω

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Lemma 12.2. Let Θ = (x, Ω) be a pseudowheel where x = (x1 , . . . , xm ) and σ (Θ)  2. Then there exists a μ refinement Φ = (y, Ω) of Θ where y = ( y 1 , . . . , yn ), σ (Φ)  2, and n  m − 2 . Proof. Let S = {1  i  m | V xi xi+1 = ∅} and let T = {1, 2, . . . , m}\ S. Let s = | S | and t = | T | = m − s. xi xi+1 |  2. Now we have Then for all i ∈ S we have |U xi xi+1 | = |Ω

2s 



| U x i x i +1 | 

m 

i∈ S

|U xi xi+1 | = |U | = μ.

i =1

μ

μ

Thus 2s  μ and hence s  2 . This also means that t = m − s  m − 2 . Let T = { j 1 , j 2 , . . . , jt } where j 1 < j 2 < · · · < jt . Let y i = x j i , i = 1, 2, . . . , t and let y = ( y 1 , . . . , yt ). Then Φ = (y, Ω) is seen to be a V -complete pseudowheel. Suppose U y i y i+1 ⊂ U for all i ∈ T . Then σ (Φ)  2 and Φ is the desired refinement of Θ . Suppose instead that U y i y i+1 = U for some i ∈ T . Without loss of generality, we may assume U yt y 1 = U (and U y 1 yt = ∅). Since σ (Θ)  2, there exists j ∈ {1, . . . , m}\{ j 1 , j 1 + 1, j 1 + ◦

2, . . . , jt } such that U x jt x j = ∅ and U x j x j = ∅. Since V yt y 1 = V x jt x j = V x jt x j ∪ V x j x j , it follows that 1 1 1 either V x jt x j = ∅ or V x j x j = ∅. If V x jt x j = ∅, then we can redefine y 1 , y 2 , . . . , yt so that y 1 = x j , 1 y 2 = x j 2 , . . . , yt = x jt . Then ∅ =  U y1 y2 and ∅ =  U yt y1 , and hence σ (Φ)  2. If V x j x j1 = ∅, then let y 1 = x j , y 2 = x j 1 , y 3 = x j 2 , . . . , yt = x jt −1 . Then σ (Φ)  2 as well. In either case, we have the desired refinement. 2 In the next lemma, we shall examine how chords and pseudowheels of Ω “overlap”. We first note that for a pseudowheel Θ , the condition that σ (Θ) + σ ∗ (Θ)  4 is equivalent to the condition that either σ (Θ)  3 or σ (Θ) = σ ∗ (Θ) = 2. Lemma 12.3. Let x ∈ X w and let Θ = (y, Ω) be a V -complete pseudowheel where y = ( y 1 , . . . , yn ), n  3,

y i y i+1 |  2 for all i = 1, . . . , n. Then we have the following: σ (Θ)  2, and |Ω

(i) For all i = j we have U x  U y i y j . Furthermore, U x U y i y j iff V x V y i y j , and U x U y i y j iff V x V y i y j . (ii) For some i ∈ {1, 2, . . . , n} either U x ⊆ U y i y i+1 or U x2 ⊆ U y i y i+1 . (iii) Suppose σ (Θ)  3. Then there exists i ∈ {1, 2, . . . , n} such that either U x ⊆ U y i y i+1 and V x ⊆ V y i y i+1 or U x2 ⊆ U y i y i+1 and V x2 ⊆ V y i y i+1 . (iv) Suppose σ ∗ (Θ) = σ (Θ) = 2 and U y i y i +1 and U y i y i +1 are nonempty, non-adjacent U -sectors for which U = U y i 3− j

Vx

1

◦

y 1 i 1 +1

1

2

2

j

j

∪ U y i2 y i1 +2 . If U x ⊆ U y i1 y i1 +1 for some j ∈ {1, 2}, then either V x ⊆ V y i1 y i1 +1 or

⊆ V y i2 y i2 +1 .

j

j

j

j

(v) Let y ∈ X w \{x}. Suppose that U x 1 ∩ U y2 = ∅, U x 1 ⊆ U y i y i+1 , and U y2 ⊆ U y i y i+1 for some i ∈ {1, 2, . . . , n} and j 1 , j 2 ∈ {1, 2}. If σ (Θ) + σ ∗ (Θ)  4, then

j V x1

⊆ V y i y i+1 ,

j V y2

j

j

⊆ V y i y i+1 , and V x 1 ∩ V y2 = ∅.

y i y i+1 |  2 for all i, we have |Ω y i y j |  2 for all i = j. Let i 0 , j 0 ∈ {1, . . . , n} where Proof of (i). Since |Ω i 0 = j 0 . If U y i y j ∈ {∅, U }, then it is clear that U x  U y i y j . Therefore, for the remainder we may as0

0

sume that ∅ = U y i

0

0

y j0

0

⊂ U . Let M  = M / y j 0 . We see that y i ∈ X M  ,Ω for all i ∈ {1, . . . , n}\{ j 0 } (where

j j X M  ,Ω is the set of chords of Ω in M  ). For all x ∈ X M  ,Ω and for all j ∈ {1, 2}, let U M  ,x = Ω M  ,x ∩ U j

j

and V M  ,x = Ω M  ,x ∩ V .

2 Then {U M  , y i , U M  , y } = {U y i 0

U yi

0

y j0

i0

0

2 y j0 , U y i y j } 0 0

2 and { V M  , y i , V M  , y } = { V yi

⊂ U , and Θ is V -complete, it is evident that

0

j U M, y i0

i0

0

2 y j 0 , V y i y j }. 0 0

= ∅, j = 1, 2 and

j V M, y i0

Since ∅ = 

= ∅, j = 1, 2.

Consequently, y i 0 crosses w in M  (as does x). It now follows by Lemma 5.1 that U M  ,x  U M  , y i , and 0 U M  ,x U M  , y i iff V M  ,x V M  , y i . Thus U x  U y i y j and U x U y i y j iff V x V y i y j (and U x U y i y j 0 0 0 0 0 0 0 0 0 0 iff V x V y i y j ). This completes the proof of (i). 2 0

0

S. McGuinness / Advances in Applied Mathematics 53 (2014) 72–111

101

Proof of (ii). Since σ (Θ)  2, it is evident that U y i y i+1 = ∅ or ∅ =  U y i y i+1 ⊂ U for all i. Let I = {1  i  n | U x ∩ U y i y i+1 = ∅} and let J = {1  j  n | U x2 ∩ U y j y j+1 = ∅}. Let i ∈ I . By (i), U x  U y i y i+1 and

consequently, U x U y i y i+1 or U x U y i y i+1 . If U x U y i y i+1 , then U x ∪ U y i y i+1 = U and hence U x2 ⊆ U y i y i+1 . Thus we may assume that U x U y i y i+1 . If U x ⊆ U y i y i+1 , then we are done. Thus we may 2 assume that U y i y i+1 ⊂ U x for all i ∈ I . Likewise, we may assume  that U y j y j+1 ⊂ Ux for all j ∈ J . From this one also deduces that | I |  2 and | J |  2. Now U x = i ∈ I U x ∩ U y i y i+1 = i ∈ I U y i y i+1 and  likewise, U x2 = j ∈ J U y j y j+1 . Let i 1 , i 2 ∈ I and j 1 , j 2 ∈ J where i 1 < i 2 and j 1 < j 2 . Without loss of generality, we may assume i 1 < j 1 . If i 1 < i 2 < j 1 < j 2 , then U x  U y i y j , which contradicts (i). If i 1 < 2 2 j 1 < i 2 < j 2 , then U x  U y i y i , giving a contradiction. Finally, if i 1 < j 1 < j 2 < i 2 , then U x  U y i y j , 1

2

1

2

yielding a contradiction. We conclude that for some i ∈ {1, . . . , n}, either U x ⊆ U y i y i+1 or U x2 ⊆ U y i y i+1 . This proves (ii). 2 Proof of (iii). Suppose σ (Θ)  3. By (ii), there exists i ∈ {1, . . . , n} such that either U x ⊆ U y i y i+1 or U x2 ⊆ U y i y i+1 . Without loss of generality, we may assume that U x ⊆ U y i y i+1 . Given that σ (Θ)  3, there exists k = i , i + 1 for which U y i+1 yk = ∅ and U yk y i = ∅. Since U x ⊆ U y i y i+1 , it follows by (i) that V x V y i y i+1 . If V x ⊆ V y i y i+1 , then we are done. Thus we may assume that V y i y i+1 ⊂ V x . Then ◦

V x ∩ V y i+1 y i = ∅ and given that V y i+1 y i = V y i+1 yk ∪ V yk y i , it follows that either V x ∩ V y i+1 yk = ∅ or V x ∩ V yk y i = ∅. Suppose first that V x ∩ V y i+1 yk = ∅. Since U y i+1 yk ∩ U x = ∅, it follows that U x U y i+1 yk and hence V x V y i+1 yk (by (i)). Thus V x ∪ V y i+1 yk = V and hence V yk y i ⊂ V x . Then V x V yk y i , and hence U x U yk y i (by (i)). However, this gives a contradiction since U x ∩ U yk y i = ∅. A similar contradiction is reached if V x ∩ V yk y i = ∅. We conclude that V x ⊆ V y i y i+1 . This proves (iii). 2 Proof of (iv). Without loss of generality, we may assume that U x ⊆ U y i y i +1 . Then U x U y i y i +1 1 1 1 1 and hence V x V y i y i +1 . If V x ⊆ V y i y i +1 , then we are done. Therefore, we may assume that 1 1 1 1 V y i y i +1 ⊂ V x . Then V x ∩ V y i +1 y i = ∅ and given that V y i +1 y i = V y i +1 y i +1 ∪ V y i y i , it follows that 1 1 1 1 1 1 1 2 2 1 either V x ∩ V y i y i = ∅ or V x ∩ V y i +1 y i +1 = ∅. Suppose V x ∩ V y i y i = ∅. Since U x ∩ U y i y i = ∅ and 2 1 1 2 2 1 2 1 U y i y i = ∅ (because ∅ = U y i y i +1 ⊆ U y i y i ), it follows that V x V y i y i (since U x U y i y i ). Thus 2 1 2 2 2 1 2 1 2 1 V x ∪ V y i y i = V and consequently, V y i y i ⊂ V x . Now V x ∩ V y i +1 y i +1 = ∅, since V y i+1 y i ⊂ V x . By the 2 1 1 2 1 2 2 same reasoning as above, we have V x ∪ V y i +1 y i +1 = V and hence V y i +1 y i ⊂ V x . It now follows that 1

V yi

y 2 +1 i 2

2

2

⊆ V x and hence V x2 ⊆ V y i2 y i2 +1 . This proves (iv). 2

1

Proof of (v). Since σ (Θ) + σ ∗ (Θ)  4, it follows that either σ (Θ) = 3 or σ ∗ (Θ) = σ (Θ) = 2. Thus there exists j ∈ {1, . . . , n}\{i } such that j = i − 1 (mod n) and U y j y j+1 = ∅. We observe that U y j y j+1 ⊆

k U y i+1 y i−1 . Let M  = M / y i +1 . Then y i −1 , y i ∈ X M  ,Ω . For all x ∈ X M  ,Ω , let U kM  ,x = Ω M  ,x ∩ U ,

k k k k = 1, 2 and let V M  ,x = Ω M  ,x ∩ V , k = 1, 2. We see that U x y

i +1

= ∅, k = 1, 2 and V xk y i+1 = ∅,

k  k = 1, 2 for x ∈ { y i −1 , y i }. Thus U kM  ,x = ∅, k = 1, 2 and V M  ,x = ∅, k = 1, 2 for x ∈ { y i −1 , y i }. Thus j

j

y i −1 , y i ∈ X M  , w (and x, y ∈ X M  , w ). Furthermore, we see that U M1 ,x ⊂ U y i y i+1 , U M2 , y ⊂ U y i y i+1 , and j

j

2 U M  , y i ∈ {U y i y i+1 , U 2y i y i+1 }. We may assume that U M  , y i = U y i y i+1 . Then U M1 x , U M2 , y , and U M  , y are i

pairwise disjoint. Since | X M  , w |  4, it follows that w is not 3-crossed in M  and hence Corollary 5.7 j

j

j

j

1 2 2 implies that V M1 ,x , V M2 , y , and V M  , y are pairwise disjoint. Consequently, the sets V x , V y , and i

j

j

j

j

V 2y i y i+1 are pairwise disjoint, and now it follows that V x 1 ⊂ V y i y i+1 , V y2 ⊂ V y i y i+1 , and V x 1 ∩ V y2 = ∅. This proves (v). 2 13. C -augmenting circuits via pseudowheels Our goal in this section, is to prove the following theorem: Theorem 13.1. Let X be a w-nested set of chords of Ω and let Θ = (y, Ω) be a V -complete pseudowheel where y = ( y 1 , . . . , yn ) and σ (Θ)  2. Let m = | X |.

102

S. McGuinness / Advances in Applied Mathematics 53 (2014) 72–111

(i) Suppose σ (Θ) + σ ∗ (Θ)  4. If m + n > 2δ + 2, then M |(Ω ∪ X ∪ { y 1 , . . . , yn }) has a C -augmenting circuit. (ii) Suppose σ (Θ) = 2 and σ ∗ (Θ) = 1. If m + n > 23 (δ + 1) + 1, then M |(Ω ∪ X ∪ { y 1 , . . . , yn }) has a C -augmenting circuit. The lemma below gives sufficient conditions for a circuit to be C -minimal. The proof is similar to the proof of Lemma 9.1, and we shall leave the details to the reader. j

j

j

Lemma 13.2. Let Θ = (y, Ω) be a pseudowheel where y = ( y 1 , y 2 , . . . , yn ). Suppose that U x11 , U x22 , . . . , U xmm are pairwise disjoint for some x1 , x2 , . . . , xm ∈ X w and j 1 , j 2 , . . . , jm ∈ {1, 2}. In addition, suppose that for j j j j some j ∈ {1, . . . , n} we have U xii ⊆ U y j y j+1 , i = 1, . . . , m. Let = Ω y j y j+1 Ωx11 Ωx22  · · · Ωxmm . If ∩ C = ∅, then is a C -minimal circuit. As a first step in the proof of Theorem 13.1, we shall prove the following lemma. Lemma 13.3. Let X be a w-nested set of chords and let (x, j) be a w-ordering for X where x = (x1 , . . . , xm ) and j = ( j 1 , . . . , jm ). Let Θ = (y, Ω) be a V -complete pseudowheel where y = ( y 1 , . . . , yn ) and σ (Θ) + σ ∗ (Θ)  4. Suppose that U xjmm ⊆ U y i y i+1 for some i ∈ {1, . . . , n}. If m + n > 2δ + 2, then M |(Ω ∪ X ∪ { y 1 , . . . , yn }) has a C -augmenting circuit. Proof. We shall assume that m + n > 2δ + 2. We may also assume that j 1 = j 2 = · · · = 1 and U xm ⊆ U y 1 y 2 ; the proof of the general case can be inferred from this special case. We have that ∅ =  U x1 ⊆ U x2 ⊆ · · · ⊆ U xm ⊆ U y 1 y 2 ⊂ U and ∅ = V x1 ⊂ V x2 ⊂ · · · ⊂ V xm ⊂ V . The proof is divided into two cases. Case 1.

σ (Θ)  3.

Since U xm ⊆ U y 1 y 2 , Lemma 12.3(iii) implies that V xm ⊆ V y 1 y 2 . Let

1 =



if m  2; Ωx1 , Ω y 1 y 2 , if m = 1,

and

m =

Ωxm−1 Ω y 1 y 2 , if m  2; Ω y1 y2 , if m = 1.

Let i = Ωxi Ωxi−1 , i = 2, . . . , m − 1 and let m+i = Ω y i+1 y i+2 , i = 1, . . . , n − 1. By construction, the sets i ∩ Ω , i = 1, . . . , m + n − 1 are seen to form a partition of Ω and consequently, the sets i ∩ U , i = 1, . . . , m + n − 1 and i ∩ V , i = 1, . . . , m + n − 1 form partitions of U and V , respectively. Since Θ is V -complete and ∅ = V x1 ⊂ V x2 ⊂ · · · ⊂ V xm ⊂ V , it follows that i ∩ V = ∅, i = 1, . . . , m + n − 1. Now Lemmas 9.1 and 13.2 imply that the sets i , i = 1, . . . , m + n − 1 are C -minimal circuits. We also see that

| 1 \Ω| =

1, if m  2; 2, if m = 1,

| m \Ω| =

3, if m  2; 2, if m = 1,

m+n−1

and | i \Ω| = 2, for all i = 2, . . . , m − 1. Thus | i \Ω| = 2(m + n − 1) > 2( 2δ + 1) = δ + 2. It i =1 follows by Lemma 9.2 that at least one of the circuits i , i = 1, . . . , m + n − 1 must be C -augmenting. This completes Case 1. Case 2.

σ ∗ (Θ) = σ (Θ) = 2.

We have that U is the disjoint union of two non-adjacent U -sectors, which we may assume to be U y 1 y 2 and U y i y i +1 . By Lemma 12.3(iv), we have V xi ⊆ V y 1 y 2 or V x2i ⊆ V y i y i +1 , for i = 1, . . . , m. In 0

0

particular, either V xm ⊆ V y 1 y 2 or V x2m ⊆ V y i

0

0

y i 0 +1 .

0

If V xm ⊆ V y 1 y 2 , then the proof proceeds as in Case 1.

S. McGuinness / Advances in Applied Mathematics 53 (2014) 72–111

Therefore, we may assume that V x2m ⊆ V y i all i < k and

V x2i

0

y i 0 +1 .

Let k = min{i | V x2i ⊆ V y i

0

⊆ V y i0 y i0 +1 for all i  k. We shall consider two subcases.

y i 0 +1 }.

103

Then V xi ⊆ V y 1 y 2 for

Case 2.1. k > 1. Let

1 =

if k  3; Ωx1 , Ω y 1 y 2 , if k = 2,

k−1 =

Ωxk−2 Ω y 1 y 2 , if k  3; Ω y1 y2 , if k = 2,

and let

 k =



Ωx2k+1 Ω y i0 y i0 +1 , if k < m; Ω y i 0 y i 0 +1 , if k = m,

m =

if k < m; Ωx2m , Ω y i0 y i0 +1 , if k = m.

For i = 2, . . . , k − 2 let i = Ωxi Ωxi−1 and for i = k + 1, . . . , m − 1 let i = Ωx2i Ωx2i+1 . For i = 1, . . . , i 0 − 2, let m+i = Ω y i+2 y i+3 and for i = i 0 − 1, . . . , n − 2 let m+i = Ω y i+2 y i+3 . The sets i , i = 1, . . . , m + n − 2 are seen to be  C -minimal circuits. By our construction, the sets i ∩ V , i = m+n−2 1, . . . , m + n − 2 partition V and U ⊆ i =1 ( i ∩ U ). We also see that

| 1 \Ω| =

1, if k  3; 2, if k = 2,

| k−1 \Ω| =

3, 2,

if k  3; if k = 2,

and

| k \Ω| =

3, 2,

if k < m; if k = m,

| m \Ω| =

1, if k < m; 2, if k = m.

m+n−2

Furthermore, | i \Ω| = 2 for all i ∈ / {1, k − 1, k, m}. We now see that i =1 | i \Ω| = 2(m + n − 2) > 2 · 2δ = δ . Thus by Lemma 9.2, at least one of the circuits i , i = 1, . . . , m + n − 2 is C -augmenting. This completes Case 2.1. Case 2.2. k = 1. Let

 1 =

Ω y i0 y i0 +1 Ωx22 , if m > 1; Ω y i 0 y i 0 +1 , if m = 1,

and

m =

Ωx2m , if m > 1; Ω y i0 y i0 +1 , if m = 1.

Let i = Ωx2i Ωx2i+1 , i = 2, . . . , m − 1 and let m+i = Ω y i +i y i +i+1 , i = 1, . . . , n − 1. The sets i , i = 0 0 1, . . . , m + n − 2 are seen to be C -minimal circuits. By construction, the sets i ∩ V , i = 1, . . . , m + n − 1 m+n−1 form a partition of V and U ⊆ i =1 ( i ∩ U ). We also see that

| 1 \Ω| =

3, 2,

if m > 1; if m = 1,

| m \Ω| =

m+n−1

1, if m > 1; 2, if m = 1,

and | i \Ω| = 2 for all i = 1, m. Thus we have | i \Ω| = 2(m + n − 1) > 2( 2δ + 1) = δ + 2. i =1 Thus at least one of the circuits i , i = 1, . . . , m + n − 1 is C -augmenting. This completes Case 2.2, and the proof of Case 2 follows from Cases 2.1 and 2.2. 2

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Proof of Theorem 13.1. We shall begin by proving (i). We shall assume that m + n > 2δ + 2. Let (x, j) be a w-ordering for X where w = (x1 , . . . , xm ) and j = ( j 1 , . . . , jm ). We may assume that j 1 = j 2 = · · · = jm = 1. Then ∅ = U x1 ⊆ U x2 ⊆ · · · ⊆ U xm ⊂ U and ∅ = V x1 ⊂ V x2 ⊂ · · · ⊂ V xm ⊂ V . By Lemma 12.3(ii), there exists j ∈ {1, . . . , n} such that U x1 ⊆ U y j y j+1 or U x21 ⊆ U y j y j+1 . Without loss of generality, we

 U x2m ⊆ U x2m−1 ⊆ · · · ⊆ may assume that U x1 ⊆ U y 1 y 2 or U x21 ⊆ U y 1 y 2 . Suppose U x21 ⊆ U y 1 y 2 . Then ∅ =

U x21 ⊂ U y 1 y 2 ⊂ U . Since m + n > 2δ + 2, Lemma 13.3 implies that M has a C -augmenting circuit. Thus we may assume that U x1 ⊆ U y 1 y 2 . By Lemma 12.3(ii), there exists j 0 ∈ {1, . . . , n} such that either U xm ⊆ U y j y j +1 or U x2m ⊆ U y j y j +1 . 0 0 0 0 Suppose U xm ⊆ U y j y j +1 . Then U x1 ⊆ U xm ⊆ U y j y j +1 and hence j 0 = 1 (since U x1 ⊆ U y 1 y 2 ). In this 0 0 0 0 case, ∅ =  U x1 ⊆ U x2 ⊆ · · · ⊆ U xm ⊆ U y1 y2 and now Lemma 13.3 implies that M has a C -augmenting circuit. Thus we may assume that U x2m ⊆ U y j y j +1 . We shall consider two cases. 0

0

Case 1. j 0 = 1. We have U x1 ⊆ U y 1 y 2 and U x2m ⊆ U y 1 y 2 . By Lemma 12.3(ii), for i = 2, . . . , m − 1, there exists j ∈

{1, . . . , n} such that either U xi ⊆ U y j y j+1 or U x2i ⊆ U y j y j+1 . Given that U x1 ⊆ U xi and U x2m ⊆ U x2i , i = 2, . . . , m − 1, it follows that either U xi ⊆ U y 1 y 2 or U x2i ⊆ U y 1 y 2 , i = 2, . . . , m − 1. By Lemma 12.3(v), we have V x1 ⊂ V y 1 y 2 and V x2m ⊂ V y 1 y 2 . Let i 0 = max{i | U xi ⊂ U y 1 y 2 }. Then U xi ⊂ U y 1 y 2 , i = 1, . . . , i 0 and U x2i ⊂ U y 1 y 2 , i = i 0 + 1, . . . , m. By Lemma 12.3(v), we have V xi ⊂ V y 1 y 2 , i = 1, . . . , i 0 and V x2i ⊂ V y 1 y 2 , i = i 0 + 1, . . . , m. Let Ωx2m+1 = ∅ and let





if i 0 > 1; Ωx1 , 1 = Ωx22 Ω y 1 y 2 , if i 0 = 1,

and

i 0 =

Ωxi0 −1 Ωx2i

0 +1

Ωx22 Ω y 1 y 2 ,

Ω y 1 y 2 , if i 0 > 1; if i 0 = 1.

Let i = Ωxi Ωxi−1 , i = 2, . . . , i 0 − 1, i = Ωx2i Ωx2i+1 , i = i 0 + 1, . . . , m and let m+i = Ω y i+1 y i+2 , i = 1, . . . , n − 1. By construction, we see that the sets i ∩ Ω , i = 1, . . . , m + n − 1 form a partition of Ω . This in turn implies that the sets i ∩ U , i = 1, . . . , m + n − 1 form a partition of U and the sets i ∩ V , i = 1, . . . , m + n − 1 form a partition of V . Moreover, by our choice of i , i = 1, . . . , m + n − 1, we have i ∩ V = ∅, i = 1, . . . , m + n − 1. Now Lemmas 9.1 and 13.2 imply that i , i = 1, . . . , m + n − 1 are C -minimal circuits. By construction,

| 1 \Ω| =



1, if i 0 > 1; 3, if i 0 = 1,

| i 0 \Ω| =

4, 3,

if i 0 > 1; if i 0 = 1,

| m \Ω| = 1

m+n−1

and | i \Ω| = 2 for all i = 1, i 0 , m. Thus we have | i \Ω| > 2(m + n − 1) > 2( 2δ + 1) = δ + 2. i =1 Now Lemma 9.2 implies that at least one of the circuits i , i = 1, . . . , m + n − 1 is C -augmenting. This completes Case 1. Case 2. j 0 = 1. We shall look at two subcases. Case 2.1.

σ (Θ)  3.

We have U x1 ⊆ U y 1 y 2 and U x2m ⊆ U y j U x2i U x2i V x2i

0

y j 0 +1 .

By Lemma 12.3(ii), it follows that U xi ⊆ U y 1 y 2 or

⊆ U y j0 y j0 +1 , i = 2, . . . , m − 1. Let i 0 = max{i | U xi ⊆ U y1 y2 }. Then U xi ⊆ U y1 y2 , i = 1, . . . , i 0 and ⊆ U y j0 y j0 +1 , i = i 0 + 1, . . . , m. By Lemma 12.3(ii), it follows that V xi ⊆ V y1 y2 , i = 1, . . . , i 0 and ⊆ V y i0 y i0 +1 , i = i 0 + 1, . . . , m. We shall define i , i = 1 . . . , m + n − 2 as follows:

S. McGuinness / Advances in Applied Mathematics 53 (2014) 72–111

1 =

105

Ω y 1 y 2 , if i 0 = 1; if i 0 > 1, Ωx1 ,

if i 0 = 1; Ωy y , i 0 = Ω 1 2 Ω x i 0 −1 y 1 y 2 , if i 0 > 1, i = Ωxi Ωxi−1 ,

i = 2, . . . , i 0 − 1

and

 i 0 +1 =

Ω y j0 y j0 +1 Ωx2i +2 , if i 0  m − 2; 0 if i 0 = m − 1, Ω y j 0 y j 0 +1 ,

m =

if i 0  m − 2; Ωx2m , Ω y j0 y j0 +1 , if i 0 = m − 1.

Let

i = Ωx2i Ωx2i+1 ,

i = i 0 + 2, . . . , m − 1 and

m+i =

Ω y i +1 y i +2 , i = 1, . . . , j 0 − 2; Ω y i +2 y i +3 , i = j 0 − 1, . . . , n − 2.

By construction, we see that the sets i ∩ Ω , i = 1, . . . , m + n − 2 form a partition of Ω . Thus the sets i ∩ U , i = 1, . . . , m + n − 2 form a partition of U and the sets i ∩ V , i = 1, . . . , m + n − 2 form a partition of V . The sets i , i = 1, . . . , m + n − 1 are seen to be C -minimal circuits. Furthermore, from the above definitions, we deduce that

| 1 \Ω| =

if i 0 = 1; if i 0 > 1,

2, 1,

| i 0 \Ω| =

2, 3,

if i 0 = 1; if i 0 > 1,

and

| i 0 +1 \Ω| =

3, if i 0  m − 2; 2, if i 0 = m − 1,

| m \Ω| =

1, if i 0  m − 2; 2, if i 0 = m − 1,

m+n−2

and | i \Ω| = 2 for all i = 1, i 0 , i 0 + 1, m. From the above, we deduce that | i \Ω| = 2(m + i =1 n − 2) > 2 · 2δ = δ . Thus at least one of the circuits i , i = 1, . . . , m + n − 2 is C -augmenting. This completes Case 2.1. Case 2.2.

σ (Θ) = σ ∗ (Θ) = 2.

We have that U is the disjoint union of two nonempty, non-adjacent U -sectors, one of which is U y 1 y 2 and the other which is U y j y j +1 . Since U x1 ⊆ U y 1 y 2 and U x2m ⊆ U y j y j +1 , Lemma 12.3(iv), 0

implies that V x1 ⊆ V y 1 y 2 or V x21 ⊆ V y j

0

0

y j 0 +1

and V xm ⊆ V y 1 y 2 or V x2m ⊆ V y j

0

0

0

y j 0 +1 .

Suppose V x1 ⊆ V y 1 y 2 . If V x2m ⊆ V y j y j +1 , then we can proceed as in Case 2.1. Therefore, we may 0 0 assume that V xm ⊆ V y 1 y 2 . Then ∅ = V x1 ⊂ V x2 ⊂ · · · ⊂ V xm ⊂ V y 1 y 2 . Let Ωx0 = ∅. For i = 1, . . . , m − 1 let i = Ωxi Ωxi−1 , and let m = Ωxm−1 Ω y 1 y 2 . For i = 1, . . . , n − 1 let m+i = Ω y i+1 y i+2 . By con− 1 form a partition of V where each of the sets are struction, the sets i ∩ V , i = 1, . . . , m + n  m+n−1 nonempty. Furthermore, it is seen that U ⊆ i =1 ( i ∩ U ). By Lemmas 9.1 and 13.2, the sets i , i = 1, . . . , m + n − 1 are seen to be C -minimal circuits. We also see that | 1 \Ω| = 1, | m \Ω| = 3, and +n−1 | i \Ω| = 2 for all i = 1, m. Thus m | i \Ω| = 2(m + n − 1) > 2( 2δ + 1) = δ + 2. Now Lemma 9.2, i =1 implies that at least one of the circuits i , i = 1, . . . , m + n − 1 is C -augmenting. Suppose V x21 ⊆ V y j y j +1 . Then V x2m ⊂ V x21 ⊂ V y j y j +1 and we can proceed in a similar fashion as 0 0 0 0 in the case when V x1 ⊆ V y 1 y 2 and V xm ⊆ V y 1 y 2 . This completes the proof of Case 2.2 and Case 2. The proof of (i) follows by Cases 1 and 2 above.

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For the sake of brevity, we shall only give a brief sketch of the proof of (ii). Many of the details (which are left to the reader) involve modifying the proofs of Lemmas 9.4 and 9.7 to suit our needs. Suppose σ (Θ) = 2, σ ∗ (Θ) = 1, and m + n > 23 (δ + 1) + 1. Then Θ has exactly two nonempty U -sectors and these are adjacent. Without loss of generality, we may assume that U y 1 y 2 and U yn y 1 are these two sectors. Since X is w-nested, there is an ordering x1 , x2 , . . . , xm of X where we may assume that ∅=  U x1 ⊆ U x2 ⊆ · · · ⊆ U xm ⊂ U and ∅ =  V x1 ⊆ V x2 ⊆ · · · ⊆ V xm ⊂ V . Lemma 12.3(ii) implies that for j

j

all x ∈ X , there exists j ∈ {1, 2} such that either U x ⊆ U yn y 1 or U x ⊆ U y 1 y 2 . Without loss of generality, we may assume that U x1 ⊆ U y 1 y 2 . We have three different scenarios to consider. If U xm ⊆ U y 1 y 2 , then we can find a C -augmenting circuit in much the same way as was done in the proof of Lemma 9.4. If U x2m ⊆ U y 1 y 2 , then we can use Lemma 9.7 to find a C -augmenting circuit by contracting yn and then breaking X ∪ { y 1 , . . . , yn−1 } into three disjoint w-nested sets of chords of Ω . The only remaining possibility is when U x2m ⊆ U yn y 1 . In this case, we can also use the proof of Lemma 9.4 to find a C -augmenting circuit. To see this, let M  = M / yn and let Y  = { y 1 , y 2 , . . . , yn−1 }. Then X ∪ Y  ⊆ X M  , w .

For all x ∈ X M  , w let U M  ,x = Ω M  ,x ∩ U , j = 1, 2 and let V M  ,x = Ω M  ,x ∩ V , j = 1, 2. We may assume that U M  ,x = U x for all x ∈ X and U M  , y  = U y 1 y 2 for all y  ∈ Y  . One sees that U M  ,x U M  , y  for all x , y  ∈ X ∪ Y  . This in turn implies that V M  ,x V M  , y  for all x , y  ∈ X ∪ Y  . Thus there is an ordering z1 , z2 , . . . , zm+n−1 of the elements of X ∪ Y  such that ∅ ⊂ V z1 ⊆ V z2 ⊆ · · · ⊆ V zm+n−1 ⊂ V . Now by using similar arguments as in the proof of Lemma 9.4, one can find a C -augmenting circuit. 2 j

j

j

j

As an important corollary to Theorem 13.1, the following result shows how one can find a C -augmenting circuit with the help of two pseudowheels. Corollary 13.4. Let Θ = (x, Ω) and Φ = (y, Ω) be two V -complete pseudowheels where x = (x1 , . . . , xm ), 4 4 y = ( y 1 , . . . , yn ), σ (Θ)  2, and σ (Φ)  2. If m > δ+ and n > δ+ , then M |(Ω ∪ {x1 , . . . , xm } ∪ 3 3 { y 1 , . . . , ym }) has a C -augmenting circuit.

xi xi+1 |  4, Proof. Let X = {x1 , . . . , xm } and Y = { y 1 , . . . , yn }. By Lemma 12.1, we may assume that |Ω i = 1, . . . , m and |Ω y i y i+1 |  4, i = 1, . . . , n. We may also assume that U xm x1 = ∅ and U yn y 1 = ∅. Since σ (Θ)  2 and σ (Φ)  2, there exist 1  i 0 < m and 1  j 0 < n such that U xi xi +1 = ∅ and 0 0 U y j y j +1 = ∅. We shall consider two cases. 0

0

Case 1. Either

σ (Θ) + σ ∗ (Θ)  4 or σ (Φ) + σ ∗ (Φ)  4.

Without loss of generality, we may assume that σ (Φ) + σ ∗ (Φ)  4. Let X 1 = {x1 , . . . , xi 0 }, X 2 = X \ X 1 , Y 1 = { y 1 , . . . , y j 0 }, and Y 2 = Y \Y 1 . We may assume that | X 1 |  | X 2 |. Then i 0 = | X 1 |  m > 2 δ+4 6

4 and | X 1 | + |Y | = i 0 + n > δ+ + δ+3 4 = 6 j U M  ,x

j = Ω M  ,x ∩ that Ω M  ,xi

For all x ∈ X M  ,Ω , let generality, we may assume · · · ⊆ U M  ,x1 ⊂ U and ∅ = V xi

x 0 i 0 +1

1 (δ 2

+ 4). Let M  = M /xm . We see that X \{xm } ⊆ X M  ,Ω . j

j

U , j = 1, 2 and V M  ,x = Ω M  ,x ∩ V , j = 1, 2. Without loss of = Ωxi xm \{xm }, i = 1, . . . , m − 1. Thus ∅ = U xi0 xi0 +1 ⊆ U M  ,xi0 ⊆ ⊂ V M  ,xi ⊂ · · · ⊂ V M  ,x1 ⊂ V . The set X 1 = {x1 , . . . , xi 0 } is now 0

seen to be a w-nested set of chords of M  . We note that M  = M  |(Ω ∪ X 1 ∪ Y ) is simple. We may apply Theorem 13.1(i) to M  , X 1 , and Φ . Since | X 1 | + |Y | > 12 (δ + 4) = 2δ + 2, it follows that M  has a C -augmenting circuit. Now M |(Ω ∪ X ∪ Y ) is seen to have a C -augmenting circuit as well. This completes Case 1. Case 2.

σ (Θ) + σ ∗ (Θ) < 4 and σ (Φ) + σ ∗ (Φ) < 4.

Since σ (Θ)  2 and σ (Φ)  2, it follows that σ (Θ) = σ (Φ) = 2 and σ ∗ (Θ) = σ ∗ (Φ) = 1. We may assume that U xm−1 xm and U xm x1 are the only nonempty U -sectors of Θ . Let M  = M /xm and let X  = {x1 , . . . , xm−1 }. Let M  = M  |(Ω ∪ X  ∪ Y ). We see that M  is simple, and given that m − 1 + n > 2 (δ + 4) − 1 = 23 (δ + 1) + 1, it follows from Theorem 13.1(ii) that M  has a C -augmenting circuit. 3 Therefore, M |(Ω ∪ X ∪ Y ) has a C -augmenting circuit. This completes Case 2. The proof of the corollary now follows from Cases 1 and 2 above. 2

S. McGuinness / Advances in Applied Mathematics 53 (2014) 72–111

107

14. The final steps In this section, we shall complete the proof of Theorem 1.4. Suppose that Ω is not a spanning circuit of N. Given that Ω is a dominating circuit of N, it follows that there are one or more Ω -bridges of order equal to one. We shall show that the presence of such bridges enables us to find a C -augmenting circuit. Suppose that there are at least two Ω -bridges of order one. Then r ( M )  r M (C ∪ Ω) + 2 = μ + ω + υ = 3μ + δ . Let X i , i = 1, 2 be two Ω -bridges of order one. The sets X i , i = 1, 2 are cocircuits r(M ) of M and N which are disjoint from C . Thus | X i | > 3 + 1  μ + 3δ + 1, i = 1, 2. Let Θi = (xi , Ω), i = 1, 2 represent pseudowheels where xi = (xi1 , xi2 , . . . , ximi ), i = 1, 2 and X i = {xi1 , xi2 , . . . , ximi }, i = 1, 2. Then mi > μ + 3δ + 1, i = 1, 2. By (7.5), there is exactly one C -bridge, and this C -bridge contains U . Thus the sets X i , i = 1, 2 are not C -bridges. It follows that U xi j xi( j+1) ⊂ U for all i, j and hence σ (Θi )  2, i = 1, 2. By Lemma 12.2, there exist refinements Φi = (yi , Ω), i = 1, 2 for Θi , μ μ i = 1, 2 where yi = ( y i1 , y i2 , . . . , y ini ), σ (Φi )  2, and ni  mi − 2 , i = 1, 2. Thus ni > μ + 3δ + 1 − 2 = μ

+

δ

+ 1, i = 1, 2. Since μ  2, these inequalities imply that ni >

δ

+2>

δ+4

. It now follows by 3 Corollary 13.4 that M has a C -augmenting circuit. For the remainder of this section, we shall assume that N has exactly one C -bridge of order one, which we denote by Y . We have that r ( M ) = r (C ∪ Ω) + 1 = μ + ω + υ − 1 = 3μ + δ − 1. The set Y is 1 + 1. Let Θ = (y, Ω) seen to be a cocircuit of M and N where Y ∩ C = ∅. Thus |Y | > r (3M ) + 1  μ + δ− 3 2

3

3

1 be a pseudowheel where y = ( y 1 , y 2 , . . . , yn ) and Y = { y 1 , y 2 , . . . , yn }. Then n > μ + δ− + 1. As 3 before, we have σ (Θ)  2. Let N Ω = N |Ω ∪ X w . Then Ω is a spanning circuit of N Ω . For e , f ∈ U , let ∗ = Ω ∗ \{e , f }. We observe that Ω ∗ ΩN∗ ,e f = E ( N )\cl N ((Ω\{e , f }) ∪ { yn }) and let Ω and Ω N∗ ,e f N ,e f N ,e f N Ω ,e f ∗ are cocircuits of N Ω and N respectively, and neither intersects C . Moreover, Ω N ,e f is also seen to be

r(M ) 1 ∗ | > μ + δ−1 − 1. + 1 and hence |Ω a cocircuit of M. It follows that |Ω N∗ ,e f | > 3 + 1 = μ + δ− 3 3 N ,e f We also observe that Ω N∗ ,e f = Ω N∗ ,e f \Y . We shall consider two cases. Ω

Case 1. Suppose Θ is V -complete.

a ⊆ U . Let A = Ω a , B = Ω a2 , and let A x = Ω ax , B x = Ω ax Let a ∈ X Ω where Ωa is minimal and Ω for all x ∈ X a . We shall consider two subcases. j

j

j

2j

Case 1.1. Suppose a is not 3-crossed in N Ω .

∗ ∗ is a-nested. Let X = Ω and let m = | X |. By Theorem 6.2, there exist e , f ∈ A such that Ω N Ω ,e f N Ω ,e f   By Lemma 8.1, there exists a refinement X ⊆ X such that | X |  m − μ + | A |. Let m = | X  |. Then m  m − μ + | A |. (14.1) If Ω N∗ ,e f ∩ Y = ∅, then M has a C -augmenting circuit.

∗ ∗ | > μ + δ−1 − 1. Proof. Suppose Ω N∗ ,e f ∩ Y = ∅. Then Ω N∗ ,e f = Ω N∗ ,e f and hence m = |Ω | = |Ω N Ω ,e f N ,e f 3 Ω δ− 1 Thus m  m − μ + | A | > | A | + 3 − 1. It now follows that m + n > | A | +

δ−1 3

−1+μ+

δ−1 3

2

+ 1 = μ + | A | + (δ − 1). 3

Since μ  | A |  2, it follows from the above that m + n > max{ 2δ + 2, 23 (δ + 1) + 1}. Thus by Theorem 13.1, M has a C -augmenting circuit. 2 (14.2) If Ω N∗ ,e f ∩ Y = ∅, then M has a C -augmenting circuit.

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Proof. Suppose Ω N∗ ,e f ∩ Y = ∅. We shall first assume that σ (Θ) + σ ∗ (Θ)  4. Let X 1 = Ω N∗ ,e f , X 2 = X 1 Y , and X 3 = Y . Then X i , i = 1, 2, 3 are seen to be cocircuits of M where X i ∩ C = ∅. Thus | X i | > r(M ) + 1 = μ + δ−3 1 + 1, i = 1, 2, 3. We now see that 3 3  i =1

 3        ∗      | X i | = 2 X i  = 2Ω N∗ Ω ,e f ∪ Y  = 2 Ω N Ω ,e f + 2 + n = 2(m + n + 2).   i =1

Thus

1 3

m+n+2=

2

| Xi | >

i =1

From this, we obtain that m + n >

3 2

3 2



μ+

δ−1 3

δ−1 3 3 +1 = μ+ + . 2

2

2

μ + 2δ − 1. Since m  m − μ + | A |, we have

m + n  m − μ + | A | + n >

3 2

μ+

δ 2

− 1 − μ + | A| =

μ 2

+ | A| +

δ 2

− 1.

Since μ  | A |  2, it follows that 2 + | A | + 2δ − 1  2δ + 2. Thus m + n > 2δ + 2, and hence Theorem 13.1(i) implies that M has a C -augmenting circuit. Suppose σ (Θ) + σ ∗ (Θ) < 4. Then σ (Θ) = 2, σ ∗ (Θ) = 1 and U is the disjoint union of two nonempty, adjacent U -sectors of Θ which we may assume to be U y 1 y 2 and U y 2 y 3 . Since Ω N∗ ,e f ∩ Y = ∅, it follows that e and f belong to different U -sectors (either U y 1 y 2 or U y 2 y 3 ). As such, ∗ ∗ | − 1 > μ + δ−1 − 2. Thus m  m − μ + | A | > ΩN∗ ,e f ∩ Y = { y 2 }. Thus m = |Ω | = |Ω 3 N ,e f N ,e f μ

Ω

+ | A | − 2, and hence m + n > μ + 23 (δ − 1) − 1 + | A |. Given that μ  | A |  2, it follows that  m + n > 3 + 23 (δ − 1) > 23 (δ + 1) + 1. It now follows by Theorem 13.1(ii) that M has a C -augmenting circuit. 2 δ−1 3

By (14.1) and (14.2), M contains a C -augmenting circuit. This completes Case 1.1. Case 1.2. Suppose a is 3-crossed in N Ω . ◦

j ◦

There exist x, y , z ∈ X Ω and i , j , k ∈ {1, 2} such that X a = {x, y , z}, A = A ix ∪ A y ∪ A kz , and B = ◦

3− j ◦

B 3x −i ∪ B y

◦

◦

◦

◦

∪ B 3z −k . We may assume that i = j = k = 1. Then A = A x ∪ A y ∪ A z and B = B 2x ∪ B 2y ∪ B 2z . ∗ ∗ = {x, y }, and |Ω | = 2. Let e ∈ A x and f ∈ A y . Then Ω N∗ ,e f = {e , f , x, y }, Ω N Ω ,e f N Ω ,e f Ω (14.3) If Ω N∗ ,e f ∩ Y = ∅, then M has a C -augmenting circuit.

∗ ∗ and 2 = |Ω ∗ ∗ | > μ + δ−1 − 1. Then Proof. Suppose Ω N∗ ,e f ∩ Y = ∅. Then Ω =Ω | = |Ω 3 N Ω ,e f N ,e f N Ω ,e f N ,e f μ  3 and hence 3  | A |  μ  3. Consequently, μ = | A | = 3. Now by substituting μ = 3 into the 1 above inequality, it follows that δ− < 0 and consequently δ = 0 and μ = υ . It follows that U = A 3 j

and hence x, y , z ∈ X w . It is now straightforward to show that at least one of the circuits Ωx , j = 1, 2 is C -augmenting. 2 (14.4) If Ω N∗ ,e f ∩ Y = ∅, then M has a C -augmenting circuit. Proof. Suppose Ω N∗ ,e f ∩ Y = ∅. Let X 1 = Ω N∗ ,e f and X 2 = X 1 Y . Then X i , i = 1, 2 are seen to be r(M ) 1 + 1, i = 1, 2. Let Y i = X i ∩ Y , i = 1, 2. cocircuits of M where X i ∩ C = ∅. Thus | X i | > 3 + 1 = μ + δ− 3

S. McGuinness / Advances in Applied Mathematics 53 (2014) 72–111

◦

109

◦

Then Y = Y 1 ∪ Y 2 , X i = Ω N∗ ,e f ∪ Y i , i = 1, 2, and |Y i | = | X i | − |Ω N∗ ,e f | = | X i | − 4, i = 1, 2. Thus |Y i | > Ω Ω μ + δ−3 1 − 3, i = 1, 2 and hence |Y i |  3δ , i = 1, 2. From this, we deduce that n = |Y | = |Y 1 | + |Y 2 |  23 δ . Let i = Ω y i y i+1 , i = 1, . . . , n. Then the sets i ∩ Ω , i = 1, . . . , n form a partition of Ω and i ∩ V = ∅,

n

n

4 i = 1, . . . , n (since Θ is V -complete). Since i =1 | i \Ω| = 2n  3 δ , it follows that i =1 | i \Ω| > δ . Now Lemma 9.2 implies that at least one of the circuits 1 , . . . , n is C -augmenting. 2

By (14.3) and (14.4), M contains a C -augmenting circuit. This completes Case 1.2 and Case 1. Case 2. Suppose Θ is not V -complete. Since Θ is not V -complete, there is a V -section which is empty. Without loss of generality, we y i y i+1 |  2 for all i, it follows that |U y1 y2 |  2. may assume that V y 1 y 2 = ∅. Given that |Ω a ⊆ U y1 y2 . Let To start with, we suppose that there exists a ∈ X Ω such that Ωa is minimal and Ω

a and B = Ω a2 . For all x ∈ X Ω and j ∈ {1, 2}, let A x = Ωx ∩ A and B x = Ωx ∩ B. For all e , f ∈ A, A=Ω j

j

j

Ω∗

is seen to be a cocircuit of N and M where Ω N∗ ,e f ∩ C = ∅. Thus N ,e f ∗ | > μ + δ−1 − 1. We shall consider two subcases. and hence |Ω 3 N ,e f

|Ω ∗

N ,e f

j

|>

r(M ) 3

+ 1 = μ + δ−3 1 + 1

Case 2.1. Suppose a is not 3-crossed in N.

∗ ∗ By Theorem 6.2, there exist e , f ∈ A such that Ω is a-nested. Let X = Ω and let m = N ,e f N ,e f δ− 1  ⊆ X such that N ,e f | > μ + | X |. Then m = |Ω − 1. By Lemma 8.1, there exists a refinement X 3

| X  |  m − μ + | A |. Let m = | X  |. Then m  m − μ + | A | > | A | + δ−3 1 − 1. By Lemma 12.2, there μ exists a refinement Θ  = (y , Ω) of Θ where y = ( y 1 , y 2 , . . . , yn  ), n  n − 2 , and σ (Θ  )  2. Since

1 + 1, it follows that n > μ2 + δ−3 1 + 1 and hence m + n > μ2 + | A | + n > μ + δ− 3

2 (δ 3

− 1). Since

μ  | A |  2, we see that m + n > max{ 2δ + 2, 23 (δ + 1) + 1}. Now Theorem 13.1 implies that M has a C -augmenting circuit. This completes Case 2.1.

Case 2.2. Suppose a is 3-crossed in N. ◦

j ◦

There exist x, y , z ∈ X Ω and i , j , k ∈ {1, 2} such that X a = {x, y , z}, A = A ix ∪ A y ∪ A kz and B = ◦

3− j ◦

∪ B 3z −k . We note that | A |  3. Let e ∈ A ix and f ∈ A y . Then it is seen that Ω N∗ ,e f = ∗ = {x, y }. Therefore, 2 = |Ω ∗ | > μ + δ−1 − 1, and hence μ  3. Since | A |  3, {e , f , x, y } and Ω N ,e f N ,e f 3 it follows that μ = | A | = 3 and A = U . However, since σ (Θ)  2 we also have A ⊆ U y 1 y 2 ⊂ U . This B 3x −i ∪ B y

j

yields a contradiction and Case 2.2 is complete. a ⊆ U y1 y2 . Let N  = N / y 2 . Then y 1 is seen Lastly, suppose that no chord a ∈ X Ω exists for which Ω to be a chord of Ω in N  . We note that while N  may not be simple, N  | X N  , y 1 is simple. We see that

N  , y1 = U y1 y2 . Now y 1 can play the j  for some j ∈ {1, 2} and we may assume that Ω U y1 y2 = Ω N , y1  N  , y1 = U y1 y2 for which Ω ∗  is role of a in N . As with a, one can show that there exist e , f ∈ Ω N ,e f

1 ∗  | > μ + δ−1 − 1. + 1 and |Ω y 1 -nested in N  . Since Ω N∗  ,e f ∩ C = ∅, it follows that |Ω N∗  ,e f | > μ + δ− 3 N ,e f 3 ∗  and let m = | X |. As before, there exists a refinement X  ⊂ X where | X  | > | A | + δ−1 − 1. Let X = Ω 3 N ,e f

We observe that X  is also a w-nested set in N. Let m = | X  | and let Θ  be the refinement of Θ as described in Case 2.1. Then m + n > max{ 2δ + 2, 23 (δ + 1) + 1}. Now Theorem 13.1 implies that M has a C -augmenting circuit.

15. Summary and discussion From Sections 10, 11, 14, we conclude that M has a C -augmenting circuit. Given that M was assumed to have no such circuits, we arrive at a contradiction. This completes the proof of Theorem 1.4.

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There are a number of different avenues one might take to extend and/or strengthen the main result of this paper. In particular, there are two results of Bondy [2] and Fraisse [5] which might offer some insight as to what

is possible for matroids. Let G be a simple graph and let k be a positive integer. We define σk = min x∈ X d G (x) where the minimum is taken over all sets X of k independent vertices. If G has no independent sets of k vertices, then we define σk = ∞. We have the following strengthening of Nash-Williams’ theorem due to Bondy [2]: Theorem 15.1 (Bondy). Let G be a simple, 2-connected graph on n vertices. If cycle of G is dominating.

σ3  n + 2, then any longest

A cycle C of a simple graph G is a D k -cycle if | V ( H )| < k for any component H of G \ V (C ). The next theorem due to Fraisse [5], extends Bondy’s result to the general case of k-connected graphs. Theorem 15.2 (Fraisse). Let k  2 be a positive integer and let G be a simple, k-connected graph on n vertices. If σk+1  n + k(k − 1), then G has a D k -cycle. The above concepts extend naturally Let M be a simple matroid. For every positive

to matroids. ∗ integer k we define σk ( M ) = minS C ∗ ∈S |C | where the minimum is taken over all sets S of k pairwise disjoint cocircuits. If M does not have k disjoint cocircuits, then we define σk ( M ) = ∞. In light of Bondy’s theorem, the following conjecture seems natural: Conjecture 15.3. Let M be a connected regular matroid. If σ3 ( M )  r ( M ) + 3, then any longest circuit in M is dominating. We can also generalize the concept of a D k -cycle to matroids. We say that a circuit C in a simple matroid M is a D k -circuit if o( X ) < k for all C -bridges X . By this definition, a D 1 -circuit is a spanning circuit, and a D 2 -circuit is a dominating circuit. In the case k = 3, Fraisse’s theorem says that any 3-connected simple graph on n vertices where σ3  n + 6 has a D 3 -circuit. We conjecture that an analogous result holds for regular matroids. Conjecture 15.4. Let M be a simple, 3-connected, regular matroid where D 3 -circuit.

σ3 ( M )  r ( M ) + 7. Then M has a

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