A Generalization of a Result of Catlin

Ronald J. GouldEmory UniversityAtlanta GA 30322

Emily A. HyndsSamford University

Birmingham AL 35229

August 3, 2005

Abstract

A 2-factor of a graph G consists of a spanning collection of vertex disjoint cycles.In particular, a hamiltonian cycle is an example of a 2-factor consisting of precisely onecycle. Harary and Nash-Williams characterized graphs with hamiltonian line graphs.Gould and Hynds generalized this result, characterizing those graphs whose line graphscontain a 2-factor with exactly k (k ≥ 1) cycles. With this tool we show that certainproperties of a graph G, that were formerly shown to imply the hamiltonicity of theline graph, L(G), are actually strong enough to imply that L(G) has a 2-factor with k

cycles for 1 ≤ k ≤ f(n), where n is the order of the graph G.

1 Introduction

All graphs considered in this paper are simple graphs. For terms or notation not definedhere, see [4]. For a graph G, let N(v) denote the neighborhood of vertex v. A set S ⊆ V (G)is said to be independent if uv 6∈ E(G) for every u, v ∈ S. The independence number of agraph G, denoted α(G), is the size of a largest independent set of vertices of G. For a setS ⊆ V (G) we use 〈S〉 to denote the subgraph induced by S.

A circuit of G is an alternating sequence C : v1, e1, v2, e2, ..., vm, em, v1 of vertices andedges of G, such that ei = vivi+1, i = 1, 2, ...,m − 1, em = vmv1, and ei 6= ej if i 6= j. Acircuit whose m vertices vi are distinct is called a cycle.

We define a dominating circuit of a graph G to be a circuit of G with the property thatevery edge of G either belongs to the circuit or is adjacent to an edge of the circuit.

A star is the complete bipartite graph K1,n.The vertex of degree n is termed the center of the star and the vertices of degree 1 are

the leaves. If a star has center w we often denote it as Sw. Further, if we wish to specify astar centered at w with some specific leaves, say a, b, c, we will denote it by Sw(a, b, c). Notethat there may be other leaves in Sw not specified.

The subgraph H of G is said to be a 2-factor of G if H spans G and for every v ∈ V (H),degH v = 2. A trivial consequence of the definition is that every 2-factor of a graph G consistsof a spanning collection of vertex disjoint cycles. In particular, a hamiltonian cycle is anexample of a 2-factor consisting of precisely one cycle.

Early studies of 2-factors centered on the question of existence, often of simply a hamil-tonian cycle. More recently the focus in the area of 2-factors has shifted from the problem of

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showing the existence of a 2-factor to that of showing the existence of 2-factors with specificstructural features. In 1978 Sauer and Spencer [8] made the following conjecture along thoselines.

Conjecture 1 Let H be any graph on n vertices with maximum degree ∆ ≤ 2. If G is agraph on n vertices with minimum degree δ(G) > 2n/3 then G contains an isomorphic copyof H.

In 1993 Aigner and Brandt [1] settled Conjecture 1 with a slight improvement.

Theorem 1 Let G be a graph of order n with δ(G) ≥ (2n−1)/3, then G contains any graphH of order at most n with ∆(H) ≤ 2.

In the above result the minimum degree must be very high to guarantee that a graphcontains all possible 2-factors or 2-factors with a particular structure. Thus, a more relaxedquestion would be: is there a lesser degree condition that will imply the existence of 2-factorswith k cycles for a range of k.

The following was shown in [2].

Theorem 2 Let k be a positive integer and let G be a graph of order n. If deg(x)+deg(y) ≥ nfor all x, y ∈ V (G) such that xy 6∈ E(G), then G contains a 2-factor with k cycles for all k,1 ≤ k ≤ ⌊n/4⌋.

Note that Theorem 2 is a generalization of the classic hamiltonian result of Ore [7] forthe case when n ≥ 4k. The complete bipartite graph Kn/2,n/2 shows that this result is bestpossible.

This type result naturally leads to the question of whether or not other hamiltonianresults can be extended in a similar manner.

The following is the well-known result of Harary and Nash-Williams [6] characterizinggraphs with hamiltonian line graphs.

Theorem 3 Let G be a graph without isolated vertices. Then L(G) is hamiltonian if, andonly if, G ≃ K1,n, for some n ≥ 3, or G contains a dominating circuit.

Given a graph G, we say that G contains a k-system that dominates if G contains acollection of k edge disjoint circuits and stars, (K1,ni

, ni ≥ 3), such that each edge of G iseither contained in one of the circuits or stars, or is adjacent to one of the circuits.

We will use a generalization of Theorem 3 that allows us to characterize those graphswhose line graphs contain a 2-factor with exactly k(k ≥ 1) cycles.

Theorem 4 (Gould, Hynds[5]) Let G be a graph with no isolated vertices. The graph L(G)contains a 2-factor with k (k ≥ 1) cycles if, and only if, G contains a k-system that domi-nates.

The following result gives specific conditions on a graph G that imply that the line graphL(G) is hamiltonian.

Theorem 5 (Catlin[3]) If G is a 2-edge-connected simple graph of order n such that δ(G) ≥n/5, then L(G) is hamiltonian.

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2 Extension

We now show that this same condition actually implies much more.

Theorem 6 Let G be a graph of order n and k be an integer 1 ≤ k ≤ ⌊cn⌋, for someconstant c > 0 . If |E(G)| ≥ cn2 and L(G) is hamiltonian, then L(G) contains a 2-factorwith k cycles.

Proof: Let G be a graph as in the theorem. We know by assumption that L(G) ishamiltonian and so we will proceed by induction. Suppose that L(G) has a 2-factor withk − 1 cycles for k ≤ cn, but L(G) does not have a 2-factor with k cycles. By Theorem 4 weknow two things about the graph G, that G does have a dominating (k − 1)-system and Gdoes not have a dominating k-system.

Consider a dominating (k − 1)-system in G. Every edge of G is either in a star in thissystem, a circuit in this system, or is dominated by a circuit in the system. We will let i bethe number of stars in the system and thus k − 1 − i is the number of circuits.

Claim 1 All stars in this system have at most 5 edges.

Proof: Suppose there is a star with six or more edges. Then we can separate the starinto two smaller stars with at least 3 edges each. This gives us a dominating k-system in Gand a contradiction. 2

Claim 2 The circuits in this system must be cycles.

Proof: Suppose there is a circuit in the system that is not a cycle. Then we can separatethe circuit into 2 edge disjoint circuits which again gives us a dominating k-system and acontradiction. 2

Claim 3 There can be at most n − 1 dominated edges.

Proof: Consider the subgraph of G induced by these dominated edges. No vertex inthis subgraph can have degree 3 or higher. Such a vertex would allow us to form anotherstar and thus a dominating k-system. Consequently this subgraph has maximum degree 2.Now suppose all of the vertices in the subgraph have degree 2. Then we can find a cycle inthe subgraph which when added to the (k − 1)-system gives us a dominating k-system inG. Hence, we have a subgraph which must be a collection of disjoint paths, and thus cancontribute at most n − 1 edges to our graph G. 2

Combining the results of these three claims we see that |E(G)| ≤ 5i+n(k−1−i)+(n−1).For n > 5 this is maximized at i = 0. So, |E(G)| ≤ 5i+n(k−1−i)+(n−1) ≤ nk−1 (n > 5).By our original assumption, |E(G)| ≥ cn2 which implies that cn2 ≤ nk − 1. On the otherhand, we know that k ≤ cn. But, k ≤ cn implies that k < cn + 1

nwhich in turn implies that

cn2 > nk − 1, a contradiction. Thus, it must be the case that L(G) has a 2-factor with kcycles. 2

Now Theorem 5 can be extended as an easy Corollary to Theorem 6.

Corollary 7 If G is a 2-edge-connected simple graph of order n such that δ(G) ≥ n/5 thenL(G) has a 2-factor with k cycles for k = 1, ..., ⌊n/10⌋.

Proof: From Theorem 5 we know that L(G) is hamiltonian. Now since δ(G) ≥ n/5 weknow that |E(G)| ≥ n2/10. Applying Theorem 6 we easily get the desired result. 2

3

3 Lemmas

In this section we state two technical lemmas that will be useful in Section 4. The proofsare not included as they are lengthy and the techniques are very similar to those found inthe proofs of the main theorems in Section 4.

In Lemma 1 we will consider four types of dominating (k − 2)-systems. For example, atype 1 dominating (k − 2)-system is one that can be formed from a dominating k-systemthat contains three stars which can be combined to form a cycle as pictured in Figure 1.Similarly, dominating (k− 2)-systems of types 2, 3, and 4 are those that can be formed fromdominating k-systems by combining three particular elements as pictured in Figures 2, 3,and 4.

Lemma 1 Let G be a graph of order n > 65, such that δ(G) ≥ n/5. If G contains adominating k-system and a dominating (k − 2)-system of type 1,2,3, or 4, then G containsa dominating (k − 1)-system.

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Lemma 2 Let G be a graph of order n > 65 with δ(G) ≥ n/5. If G contains at leastone dominating k-system , but does not contain a dominating k-system consisting entirelyof circuits, and G does not contain a dominating (k − 1)-system, then any two degree onevertices x and y in the same star in a dominating k-system of G cannot have a commonneighbor in G other than the center of the star.

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Figure 3: Type 3

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4 Catlin Generalization

In an effort to improve the result of Corollary 7 we continued our study of line graphsobtained from graphs G of order n with δ(G) ≥ n/5. We concentrated on such graphs withlarge independence number. Our goal is the following:

Theorem 8 If G is a 2-connected graph with δ(G) ≥ n/5 and α(G) = cn, for some c, 1/4 ≤c < 1, then L(G) contains a 2-factor with k cycles, for each k = 1, 2, . . . , {cn⌊ δ

3⌋−(1−c)n−δ}.

The proof of Theorem 8 is immediate from Corollary 7, and Theorems 9 and 10 thatfollow.

Theorem 9 Let G be a 2-edge-connected simple graph of order n > 65 such that δ(G) ≥n/5 and α(G) = cn for some c, 1/4 ≤ c < 1. Then L(G) has a 2-factor with at leastk = {cn⌊ δ

3⌋ − [(1 − c)n − δ]} cycles.

Proof: Since G has an independent set, say I, of size cn (c ≥ 1/4), let R = V (G)\ I and

x ∈ I. As deg(x) ≥ δ(G), we can form at least ⌊ δ(G)3⌋ stars centered at the vertex x. Doing

this for each vertex in I we build a collection of at least cn⌊ δ3⌋ stars. Place these stars in

S which represents a dominating k-system at all stages of the proof. Let |S| represent thenumber of stars and circuits in S. If there are any stars or circuits contained in 〈R〉G thenwe add them to S.

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Removing from G the edges that have already been used, we are left with at most somedisjoint paths in 〈R〉G. Let P1, P2, ..., Ps be the disjoint paths in R. We want to add all ofthe edges of these paths to S so that S dominates. Suppose some Pi contains two or morevertices that share a common neighbor z in I. Choose two such vertices xi 6= yi, such thatany other neighbors of z on Pi lie between xi and yi on Pi.

Form the cycle C : xi, z, yi, ..., xi where yi, ..., xi denotes the portion of the path Pi betweenyi and xi

The edges xiz and yiz are the only edges of C that were already in S. Because of the waywe are building S, xiz and yiz must either be dominated edges or edges of stars with centerz in S. If there exists any star with center z in S then move edges, if necessary, to form thestar Sz(xi, yi) without decreasing |S|. In S, replace Sz(xi, yi) with C which dominates theremaining edges of Sz(xi, yi). If there is no star with center z in S, then xi and yi are bothdominated and z must be on a circuit C∗ in S. We can then attach C to C∗ making a largercircuit. In either case, we have added edges from 〈R〉G without decreasing |S|. Repeat thisprocess until all that remains in 〈R〉G are disjoint paths which contain no two vertices thathave a common neighbor in I.

Now, we must incorporate the remaining edges into S without affecting |S| too much.Assume that if we can add one of the remaining edges in R to S ,we must lose at least 2from |S|. Choose an edge pq in 〈R〉G such that pq is not part of the system yet and it isthe only edge adjacent to p unused in our system. Consequently, neither p nor q can be thecenter of any star in S and neither can be found on any circuit, for otherwise we could easilyadd pq to S leaving |S|unchanged.

As I is an independent set of maximum size, there exists an x0 ∈ I such that px0 ∈ E(G)and a y0 ∈ I such that qy0 ∈ E(G). We know x0 6= y0 as none of the remaining pathshave 2 vertices with a common neighbor in I. Let N(p) = {q, x0, x1, ..., xr} and N(q) ={p, y0, y1, ..., ys} where r, s ≥ n/5 − 2.

Note that we can assume that S is made up of stars and cycles since any circuit can befactored into edge disjoint cycles, which only increases |S|.

Claim 1 The set N(p) ∩ N(q) = ∅.

Proof: Suppose there exists a z ∈ N(p)∩N(q). As p and q have no common neighborsin I, z ∈ R. We know that pz is already in our system, which means, because of theproperties of p discussed earlier, z is the center of a star or is on a circuit. Consequently, theedge qz must be in our system as well. Remember that the edges pz and qz can only be ina star with center z or dominated by a cycle containing z. If either is in a star with centerz then we can clearly rearrange our system, leaving the size the same, to form Sz(p, q). Wecan then add the edge pq to Sz(p, q) which forms a cycle p, z, q, p that dominates the otheredges of Sz(p, q) leaving the system size unchanged, a contradiction. So it must be thecase that pz and qz are both dominated by circuits. However, then we can form the cyclep, z, q, p, adding the edge pq to S and increasing |S| by 1, another contradiction. Hence,N(p) ∩ N(q) = ∅. 2

Claim 2 The set N(p) is independent.

Proof: Recall that N(p) = {q, x0, x1, ..., xr} and that pq is the edge we are trying to addto our system. We know from Claim 1 that qxi 6∈ E(G) for all i ∈ {0, 1, ..., r}. Supposethat xixj ∈ E(G) for some i 6= j. We want to move the edges pxi and pxj. As in the proof

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of Claim 1, pxi (pxj) must appear in the system as an edge of a star with center xi (or xj)or as an edge dominated by a cycle containing xi (or xj).

(i) Suppose that xixj is an edge in a star in S.

Without loss of generality let Sxi(xj) be this star. Now the cycle xi, xj, p, xi dominates

the remaining edges of Sxi(xj). If the edges pxi and pxj are both either from stars of

size 4 or larger or are dominated edges, then we have left the system size unchanged.However, we now have a cycle containing p in our system and so we can add the edgepq as a dominated edge without decreasing the system size, a contradiction. Thus, atleast one of pxi or pxj is from a star of size 3. If only one of them is in a star of size 3,then we can let our cycle dominate the remaining edges of that star, which decreasesthe system size by one. But, we can still add the edge pq to our system as an edgedominated by the cycle containing p, which is again a contradiction. Therefore theedges pxi and pxj must both be found in stars of size 3. If xixj is in a star of size 4 orlarger then we only destroy 2 stars when forming our new cycle. So again we are ableto add the edge pq while only reducing the system size by 1. Consequently, the edgexixj must also be in a star of size 3. If the edge xip is not in Sxi

(xj) then interchangeedges to form Sxi

(xj, p) without decreasing the size of the system S. Again we destroyat most 2 elements of S when forming the new cycle containing p, which means |S|has decreased by at most one. However, since p is now on a cycle in S, we can againadd the edge pq to S giving us another contradiction

In all possible cases we get a contradiction to the claim, hence, xixj cannot be in astar in S.

The proofs of (ii) and (iii) are similar to (i).

(ii) Suppose that xixj is an edge of a cycle xi, ..., xj , xi in our system.

(iii) Suppose that the edge xixj is either a dominated edge in our system or is not in thesystem at all.

In all possible cases we reach a contradiction implying that xixj cannot be an edge ofour graph G, completing the claim. 2

Claim 3 The set N(q) is independent.

Proof: The vertices p and q are virtually identical. The only possible difference wouldbe that the edge qys may not be in S whereas all of the edges adjacent to p except for pqare in S. However a proof similar to that in Claim 2 still applies. 2

Claim 4 The set N(a) ∩ N(q) \ {p} is empty for all a ∈ N(p) ∩ I.

Proof: Suppose there exists z ∈ N(a) ∩ N(q) \ {p}. Consider the edges pa, qz, and az,which may already be part of S. Note that |S| can only decrease if we take an edge froman element in S and what remains is not a valid element.

If we can use pa, az, and zq without affecting the system size then we form the cyclep, a, z, q, p and add it to the system, a contradiction. Therefore, the removal of at least oneof these three edges effects |S| . First consider the case when the removal of each of these

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edges effects |S| . Note, similar arguments can be used to reach a contradiction no matterwhich subset of these three edges affects the system when removed.

As before, it must be the case that Sa(p) and Sz(q) are K1,3’s in S. The edge az musteither be an edge of a K1,3 or a cycle edge. Suppose, without loss of generality, that Sa(z) is aK1,3 in S. We can then assume that Sa(p, z) ∈ S. Form the cycle p, a, z, q, p which dominatesthe edges from Sa(p, z) and Sz(q). Thus we have incorporated pq into S decreasing |S| atmost 1. Consequently, the edge az must be an edge of a cycle in S. From this cycle we formthe new cycle C : a, p, q, z, ..., a where z, ..., a represents the vertex sequence of the old cycle.Let C dominate the edges from Sz(q) and use the edge az to complete the star from whichap was removed. Again we have incorporated pq decreasing |S| at most 1, a contradiction.2

Claim 5 The set N(b) ∩ N(p) \ {q} is empty for all b ∈ N(q) ∩ I.

Proof: Similar to the proof of Claim 4. 2

Claim 6 The set N(a)∩N(b) is empty for all {a, b} such that a ∈ N(p)∩I and b ∈ N(q)∩I.

Proof: Suppose there does exist z ∈ N(a) ∩ N(b). We know that p and q do not shareany neighbors in I so it follows that z 6= p and z 6= q. Consider the edges ap, az, bz, andbq. We know all of these edges are in S. Now Sa(p) and Sb(q) are K1,3’s in S. The edges azand bz are either in a K1,3 with center other than z or are a cycle edge in S.

Suppose both az and bz are in a K1,3 in S. Then, based on the way we formed S thusfar, the centers must be a and b respectively. We can easily trade edges between stars, ifnecessary, so that Sa(p, z) and Sb(q, z) are in S. Now form the cycle a, z, b, q, p, a. This cycledominates the edges of Sa(p, z) and Sb(q, z). So we have formed a new cycle that containsthe edge pq while only decreasing |S| by 1, a contradiction.

Now suppose that only az is in a K1,3 and that bz is a cycle edge in our system. Againit must be the case that Sa(z) ∈ S so we can swap edges, if necessary, to ensure thatSa(z, p) ∈ S. Remove bz from the cycle that contains it and add it to Sb(q). Now add tothe cycle the path b, q, p, a, z to replace the edge bz, thus forming a larger cycle. We can usethe edge bq since we added the edge bz to the star which originally contained it. The edgespa and az are from Sa and the new cycle dominates the remaining edge of that star. So wehave again decreased |S| by only 1, a contradiction. The case that bz is in a K1,3 and az isa cycle edge in S is handled similarly.

Thus, az and bz must both be cycle edges in S. Begin by moving az to Sa(p) and bz toSb(q). Now use the edges ap and bq knowing that what remains after their removal is stilla star. If az and bz were on the same cycle in S we replace the path b, z, a with the pathb, q, p, a, which gives a new cycle containing the edge pq. We have left |S| unchanged. Soaz and bz must have been on different cycles in S. Both of these cycles contain z so we takewhat is left of them after the removal of az and bz and join them at the point z. We willthen add the path b, q, p, a, giving us a new cycle containing the edge pq. In this case wedecreased |S| by 1, a contradiction.

Thus, in all possible cases we reach a contradiction, proving the claim. 2

Claim 7 If there exists w ∈ N(p) \ {x0} such that w ∈ I, then N(x0) ∩ N(w) = p.

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Proof: Suppose there exists w ∈ N(p)\{x0} such that w ∈ I. Also suppose there existsz 6= p such that z ∈ N(x0)∩N(w). We consider the edges x0p, x0z, wp, and wz. We knowSx0

(p), Sw(p) ∈ S and the edges x0z and wz are either in a star with center other than z orare a cycle edge in S.

Suppose both x0z and wz are in stars. Then by the way we formed S, the centersmust be x0 and w respectively. We can easily trade edges between stars, if necessary, toform Sx0

(p, z) and Sw(p, z). Now form the cycle x0, z, w, p, x0. This cycle dominates theremaining edges of the stars with centers x0 and w which we used to form the cycle. So wehave formed a new cycle that contains the vertex p while decreasing |S| by 1, a contradiction.

Now suppose that only x0z is in a star and wz is a cycle edge in our system. Again itmust be the case that x0 is the center of the star containing x0z so we can swap edges, ifnecessary, to form Sx0

(z, p). Remove the edge wz from the cycle that contains it and add itto Sw(p). Add to the cycle the path w, p, x0, z to replace the edge wz, thus forming a largercycle. We can use the edge wp since we added the edge wz to the star in which it originallyoccured. The new cycle dominates the remaining edge of Sx0

(p, z) Thus, we have decreased|S| by 1, a contradiction. Similarly, we arrive at a contradiction if we assume instead thatonly wz is in a star and x0z is a cycle edge in S.

Thus, we have that x0z and wz are both cycle edges in S. Begin by moving x0z to Sx0(p)

and wz to Sw(p). We now use the edges x0p and wp, since what remains after their removalis still a star. If x0z and wz were on the same cycle in our system we replace the pathw, z, x0 that we removed with the path w, p, x0 which gives a new cycle containing the vertexp. Thus, we can incorporate pq and leave |S| unchanged. Thus, x0z and wz must have beenon different cycles in S. Both of these cycles contain z so take what is left of them after theremoval of x0z and wz and join them at the point z. We then add the path w, p, x0, givinga new cycle containing the vertex p. In this case we decreased |S| by 1, a contradiction.

Hence, in all of the possible cases we reach a contradiction, thus proving the claim. 2

Claim 8 If there exists u ∈ N(q) \ {y0} such that u ∈ I, then N(y0) ∩ N(u) = q.

Proof: Same as the proof of Claim 7. 2

We now consider the 4 cases that arise when we consider whether or not the hypothesesof Claims 7 and 8 hold.

Case 1 Suppose the hypotheses of both Claim 7 and Claim 8 hold.

Then R contains N(x0), N(w), N(y0), and N(u). By Claim 1 we know that u 6= w. Claim7 implies that N(x0)∩N(w) = p. Claim 6 implies that N(x0)∩N(y0) = ∅, N(x0)∩N(u) = ∅,N(w) ∩ N(u) = ∅, and N(w) ∩ N(y0) = ∅. Finally, Claim 8 implies that N(y0) ∩ N(u) = q.From these it follows that

|R| ≥ |N(x0) ∪ N(w) ∪ N(y0) ∪ N(u)| ≥4n

5− 2

which implies that |I| ≤ n/5 + 2. However, we originally assumed that |I| ≥ n/4 whichmeans that n ≤ 40, a contradiction. 2

Case 2 Suppose the hypothesis of Claim 7 holds but the hypothesis of Claim 8 does not.

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Then R contains N(x0), N(w), N(y0), and N(q) \ {y0}. Claim 7 implies that N(x0) ∩N(w) = p. Claim 6 implies that N(x0)∩N(y0) = ∅, and N(w)∩N(y0) = ∅. Claim 4 impliesthat N(x0) ∩ N(q) \ {y0} = p, and N(w) ∩ N(q) \ {y0} = p. Finally, Claim 3 implies thatN(y0) ∩ N(q) \ {y0} = ∅. From these it follows that

|R| ≥ |N(x0) ∪ N(w) ∪ N(y0) ∪ N(q) \ {y0}| ≥4n

5− 3

which implies that |I| ≤ n/5 + 3. However, we originally assumed that |I| ≥ n/4 whichmeans that n ≤ 60, a contradiction. 2

Case 3 Suppose the hypothesis of Claim 8 holds but the hypothesis of Claim 7 does not hold.

Then R contains N(x0), N(p) \ {x0}, N(y0), and N(u). Claim 8 implies that N(y0) ∩N(u) = q. Claim 6 implies that N(x0)∩N(y0) = ∅, and N(x0)∩N(u) = ∅. Claim 5 impliesthat N(y0) ∩ N(p) \ {x0} = p, and N(u) ∩ N(p) \ {x0} = q. Finally, Claim 2 implies thatN(x0) ∩ N(p) \ {x0} = ∅. From these it follows that

|R| ≥ |N(x0) ∪ N(p) \ {x0} ∪ N(y0) ∪ N(u)| ≥4n

5− 3

which implies that |I| ≤ n/5 + 3. However, we originally assumed that |I| ≥ n/4 whichmeans that n ≤ 60, a contradiction. 2

Case 4 Suppose neither of the hypotheses of Claims 7 and 8 hold.

Then R contains N(x0), N(p) \ {x0}, N(y0), and N(q) \ {y0}. Claim 6 implies thatN(x0) ∩ N(p) \ {x0} = ∅. Claim 5 implies that N(y0) ∩ N(p) \ {x0} = q. Claim 4 impliesthat N(x0)∩N(q)\{y0} = p. Claim 3 implies that N(y0)∩N(q)\{y0} = ∅. Claim 2 impliesthat N(x0) ∩ N(p) \ {x0} = ∅. Finally Claim 1 implies that N(p) \ {x0} ∩ N(q) \ {y0} = ∅.From these it follows that

|R| ≥ |N(x0) ∪ N(p) \ {x0} ∪ N(y0) ∪ N(q) \ {y0}| ≥4n

5− 2

which implies that |I| ≤ n/5 + 2. However, we originally assumed that |I| ≥ n/4 whichmeans that n ≤ 40, a contradiction. 2

In all cases we get a contradiction which means that if we can add one of the remainingedges to S then we can add it losing at most one from the size of our system. Of courseit is important to us that we have a dominating system. Suppose there is an edge pq thatwe cannot add to the system. Then each of the above claims would hold leading us to thesame contradictions. So we know we can add each of the remaining edges to S, decreasing|S| by at most 1 each time. We now know that we have a dominating system in G, but itstill remains to show that is the size we claimed it would be.

Recall that we started our process with at least cn⌊ δ3⌋ stars. It follows then that the size

of the system is at least cn⌊ δ3⌋ − M where M is the maximum number of edges that could

have been remaining in R.Choose x ∈ I and let d = deg(x) ≥ δ ≥ n/5. Let p be the number of vertices in N(x)

that are not on one of the remaining paths. The (d− p) vertices that are on paths must beon different paths because no two vertices on our remaining paths have a common neighbor

10

in I. Consider an edge from each of these different paths. Then we have at least 2(d − p)vertices and d− p edges so far. We have now accounted for p + 2(d− p) = 2d− p vertices ofR. Because |R| ≤ (1 − c)n we know that there are at most (1 − c)n − (2d − p) vertices leftwhich can each contribute at most 1 of the remaining edges. When we combine this withthe d− p edges from before we see that we have at most [(1− c)n− (2d− p)] + [d− p] edgescomprising the disjoint paths. This gives us that the number of edges remaining in R is atmost (1− c)n− δ. So we have a dominating system of size at least cn⌊ δ

3⌋− [(1− c)n− δ] in

G which means we have a 2-factor of size at least cn⌊ δ3⌋− [(1− c)n− δ] in L(G), as claimed.

2

Theorem 10 Let G be a 2-edge-connected simple graph of order n > 65 such that δ(G) ≥n/5 and α(G) = cn for some c, 1/4 ≤ c < 1. Then for any k ≥ ⌊n/10⌋ + 2, if L(G) has a2-factor with k cycles then L(G) has a 2-factor with k − 1 cycles.

Proof: Suppose by way of contradiction that for some k ≥ ⌊n/10⌋ + 2, that L(G) hasa 2-factor with k cycles but does not have a 2-factor with k − 1 cycles. Then we know ourgraph G has a dominating k-system but no dominating (k − 1)-system.

The following observations are easily seen:

• No vertex can appear on more than one circuit of the k-system or we could connectthe two circuits to form a dominating (k − 1)-system.

• No vertex can be the center of more than one star of the k-system.

• No vertex can be the center of a star and on a circuit of the k-system .

• If there is an edge e between two leaves of a star then it must be a dominated edge inour system.

• No two stars in our system can share a pair of leaves.

• Every star must contain at least one leaf that is not the center of any star and is noton any circuit. Otherwise we could distribute all edges of the star to other elementsin the system and form a dominating (k − 1)-system.

We now consider cases based on the structure of the k-system.

Case 1 The graph G contains a dominating k-system that consists entirely of circuits.

Among all such k-system in G, choose one, say Sk, that has the maximum number ofdistinct vertices on circuits. In Sk there must be a circuit with 9 or fewer distinct vertices.

Otherwise, we see that 10k ≤ n or k ≤n

10, a contradiction.

Consider the smallest circuit C in Sk. Now choose a pair of adjacent vertices x and y onC. Recall that neither x nor y can appear on another circuit and note that they can each

have at most 8 neighbors C. Thus x and y each dominate at leastn

5− 8 vertices off C in

Sk.

Let Nx = N(x) \ C and Ny = N(y) \ C. Then |Nx| ≥n

5− 8 and |Ny| ≥

n

5− 8.

Claim 1 The set Nx ∩ Ny is empty.

11

Suppose there exist z ∈ Nx ∩Ny. Replace the edge xy of C with the path x, z, y to forma new circuit C ′ that now dominates the edge xy. If z appears on another circuit then weattach this circuit to C ′ and form a dominating (k − 1)-system, a contradiction. If z is noton a circuit then we contradict the way we chose the original system. 2

Claim 2 The set Nx is independent.

Suppose this is not the case. Let u, v ∈ Nx, u 6= v, such that uv ∈ E(G). The edge uvmust appear somewhere in Sk. Note that xu and xv must be dominated edges.

(i) Suppose the edge uv is dominated. Without loss of generality assume v is on a circuit.Then form the cycle u, x, v, u and use it to connect the circuit containing x and thecircuit containing v. This produces a dominating (k − 1)-system , a contradiction.

(ii) Suppose uv is in a circuit. Then replace the edge uv with the path u, x, v to form anew circuit that dominates uv. We now have 2 circuits containing x, which means wecan again form a dominating (k − 1)-system, a contradiction. 2

Claim 3 The set Ny is independent.

The argument is similar to that for Claim 2. 2

Claim 4 The set Nx ∪ Ny is independent.

Because of Claims 2 and 3 we need only to show that there is not an edge from Nx toNy. Again by way of contradiction, suppose there exists u ∈ Nx and v ∈ Ny such thatuv ∈ E(G). The edge uv must appear somewhere in our system.

(i) If the edge uv is dominated, without loss of generality assume that the vertex v is ona circuit. Then in C replace xy with the path x, u, v, y to form a new circuit thatnow dominates xy. Now 2 circuits contain the vertex v which allows us to form adominating (k − 1)-system, a contradiction.

(ii) If uv is on a circuit, then we will use xu and yv to connect the circuits containinguv and xy to form one circuit that now dominates xy and uv, giving us a dominating(k − 1)-system , a contradiction. 2

From our claims we see that we have an independent set I = Nx ∪ Ny such that |I| ≥2n/5 − 16. Let a, b ∈ Nx, (a 6= b). Now suppose that there exists a vertex w ∈ N(a) ∩N(b)(w 6= x).

Subcase 1 The vertex w is on the circuit C that contains x.

Because a and b are not on the C and because w is not on any other circuit except C,it must be the case that the edges aw and bw are dominated edges in our system. We usethe edges aw, bw, xa, and xb to form the cycle x, a, w, b, x, which we attach to C. If eithera or b are on another circuit in our system then we have 2 circuits that share a vertex,hence we can form a dominating (k − 1)-system. Thus, neither a nor b was on any circuit inour original system which means by adding them to C we contradict our original choice ofsystem. Consequently, the vertex w cannot be on the circuit that contains x. 2

12

Subcase 2 The vertex w is not on the circuit containing x.

(i) If both of the edges aw and bw are dominated edges in our system, then again weform the cycle x, a, w, b, x which we attach to the circuit containing x. All of thevertices a, b, w must be on some circuit in our system, or by adding them to the circuitcontaining x, we contradict our original choice of system. But then we have two circuitswith a common vertex which allows us to form a dominating (k − 1)-system, also acontradiction.

(ii) If exactly one of aw and bw is a circuit edge in our system, then assume withoutloss of generality that aw is dominated and that bw is in a circuit of our system.Now we replace bw with the path b, x, a, w to form a new circuit that dominates bw.Now two different circuits contain the vertex x which allows us to form a dominating(k − 1)-system, a contradiction.

(iii) If both of aw and bw are circuit edges in our system, then they must be on the samecircuit, as w cannot appear on two different circuits. If the circuit containing aw andbw can be factored in such a way that aw and bw are on different cycles, then we canremove aw and bw from the circuit and attach the path a, x, b to form a new circuitthat now domintes aw and bw and contains the vertex x. We now have 2 differentcircuits that contain the vertex x which allows us to form a dominating (k−1)-system,a contradiction.

The only remaining possibility for aw and bw is that they are both edges of the samecircuit in our system and that this circuit cannot be factored as above. This is thesituation we get if any 2 elements of Nx have a common neighbor other than x.

Similar arguments show that if any two elements of Ny, say a and b, have a commonneighbor other than y, say w, then aw and bw are both edges of the same circuit in oursystem, which is not the circuit containing y, and this circuit cannot be factored in such away that the edges aw and bw are on different cycles of the circuit.

Now suppose that there exists a ∈ Nx and b ∈ Ny such that there exists a vertexw ∈ N(a) ∩ N(b) where w 6= x, y. Then again similar arguments show that the edges awand bw are both edges of the same circuit in our system, which is not the circuit containingx and y, and this circuit cannot be factored in such a way that the edges aw and bw are ondifferent cycles of the circuit. The only difference in the arguments in this case is that wherewe were attaching a new circuit either to x or y in the previous cases, we are now combiningthe circuit containing aw and bw with the circuit containing xy to form one large circuitthat in addition to dominating aw and bw as before, now dominates xy as well. As beforewe always get a dominating (k − 1)-system unless we have the situation described above in(iii).

Our conclusion is that if any two vertices of the set I have a common neighbor otherthan x or y we get the situation described above in (iii).

Now suppose that 3 vertices a, b, c of I have a common neighbor w such that w 6= x, y.By our previous arguments, aw and bw are both edges of the same circuit in our system,which is not the circuit containing x and y, and that this circuit cannot be factored in sucha way that the edges aw and bw are on different cycles of the circuit. But cw must also beon the same cycle of the circuit as aw and bw, which is impossible. So no 3 vertices of Ihave a common neighbor other than x or y.

13

Recall that our set I = Nx ∪ Ny is such that |I| ≥ 2n/5 − 16. Let R be the remainingvertices of V (G) that are not in the set I. Consider the set R \ {x, y} that has size at most3n/5 − 14. By our previous arguments no three vertices of I can have a common neighborin R \ {x, y}. Let e be the number of edges between the sets I and R \ {x, y} . Since I isindependent and each vertex of I is adjacent to only one of x and y, each vertex of I sendsat least n/5 − 1 edges to the set R \ {x, y}. Thus, there are at least (n/5 − 1)(2n/5 − 16)

edges from I to R \ {x, y}, that is, e ≥2n2

25−

18n

5+ 16.

Now there are at most 3n/5 + 14 vertices in R \ {x, y} that must receive these edgesbut each vertex in R \ {x, y} receives at most two edges from I. Hence, there are at most

2(3n

5+ 14) edges from R \ {x, y} to I. In other words, e ≤

6n

5+ 28.

It follows that2n2/25 − 18n/5 + 16 ≤ 6n/5 + 28,

hence, n2 − 60n − 150 ≤ 0 which implies n < 63, a contradiction. 2

Case 2 Every dominating k-system of G contains at least one star.

Choose a dominating k-system of G, say Sk, in such a way that the system contains amaximum number of circuits. By our assumption, Sk must contain at least one star. Supposethere exists a star Sz(x1, x2, x3, x4, x5). Note that for all i 6= j, the edge xixj cannot be inour graph. If so it would be a dominated edge and we could use it to form a k-system withone more circuit, which contradicts the way we chose our original system. Lemma 2 impliesthat for every i 6= j, N(xi) ∩ N(xj) = z. This implies that

n ≥ |{x1, x2, x3, x4, x5, z}| + |N(x1)| + |N(x2)| + |N(x3)| + |N(x4)| + |N(x5)|

and thus n ≥ n + 1, a contradiction. So it must be the case that each star in our system hasat most 4 edges.

Suppose we do have a star in our system with 4 edges, say Sz(x1, x2, x3, x4). Let N ={N(x1), N(x2), N(x3), N(x4)} \ {z}. By Lemma 2 we know that for every i 6= j, N(xi) ∩N(xj) = z which means that |N | ≥ 4n/5−4. Let X = {x1, x2, x3, x4}. We know that z 6∈ Xand z 6∈ N . Again, by the same argument as above, xixj 6∈ E(G) which means N ∩ X = ∅.This implies that we have accounted for a total of 4n/5 + 1 vertices in G.

Now z must have at least n/5− 4 neighbors in G−X. None of the other edges adjacentto z can be dominated edges in our system or we could add them to Sz giving us star with5 edges or more which we cannot have. The other edges also cannot be in a circuit becauseas a center of a star z cannot appear on a circuit. There is not another star with centerz so the remaining edges adjacent to z must be in stars with center other than z. LetC = {c1, c2, ..., cl} be the neighbors of z not in the set X. Note that l ≥ n/5 − 4. We knowthat each of the vertices in C is the center of a star with leaf z. Also we know that X∩C = ∅or we would have two stars in our system that shared a pair of vertices.

Claim 1 The set N ∩ C is empty.

Proof: Suppose cixj ∈ E(G), for some i ∈ {1, 2, ..., l} and j ∈ {1, 2, 3, 4}. This edgecannot appear in the star with center ci or we have two stars that share xj and z. It cannotbe in a circuit or dominated by a circuit containing ci as ci cannot be on a circuit. If it is

14

dominated by a circuit containing xj then we can use the edge cixj to combine the starscontaining zxj and ciz into a cycle that dominates any edges in the stars not a part of thenew cycle, thus forming a dominating (k − 1)-system. So it must be the case that the edgecixj is found in a star with center xj. Because we know the star containing zxj has 4 edgeswe can use the edge zxj to combine the stars containing the edges cixj and ciz into a cyclethat dominates the edges of the stars and thus forms a dominating (k − 1)-system, again acontradiction. It follows that cixj /∈ E(G), thus N ∩ C is empty. 2

What we have shown is that none of the vertices in the set C were included in the 4n/5+1vertices that we had accounted for before. So now, with the sets X, N , C, and the vertex zwe have accounted for n − 3 different vertices.

Consider one last set of vertices. Recall that for all i ∈ {1, 2, ..., l}, ci is the center of astar with z as a leaf. As z is the center of a star, in each of these stars there is a leaf otherthan z that is not the center of a star and is not on a circuit. For each i ∈ {1, 2, ..., l} let yi bethe leaf in the star with center ci that meets these qualifications and let Y = {y1, y2, ..., yl},l ≥ n/5 − 4. Hence, |Y | ≥ n/5 − 4. By our choice of each yi we know that z 6∈ Y . ThenY ∩ C is empty or we would have two different stars that contain the pair ci and z. Also,Y ∩ X is empty or we would have two different stars that contain the pair xi and z.

Claim 2 The set Y ∩ N is empty.

Proof: Suppose yixj ∈ E(G), for some i ∈ {1, 2, ..., l} and j ∈ {1, 2, 3, 4}. This edgecannot appear in a star with center yi, in a circuit, or as an edge dominated by a circuitcontaining yi by the way we chose yi. If yixj is an edge dominated by a circuit containing xj

then we can use it to combine the stars containing ciz and zxj into a cycle that dominatesthe edges of the stars and thus gives us a dominating (k − 1)-system. It remains then thatthe edge yixj must be found in a star with center xj. Because we know the star containingzxj has 4 edges we can use the edge zxj to combine the stars containing the edges yixj andciz into a cycle that dominates the edges of the stars and thus forms a dominating (k − 1)-system, again a contradiction. Thus, yixj /∈ E(G) and thus N ∩ Y is empty. 2

So, none of the vertices in the set Y were included in the n − 3 vertices that we hadpreviously accounted for in the sets X,N ,C, and the vertex z. We now have at least (n −3) + (n/5− 4) different vertices that we have accounted for which implies that n ≥ 6n/5− 7or n ≤ 35 which is a contradiction. Hence, it follows that we cannot have any stars in oursystem that contain 4 edges , thus every star in our system contains exactly 3 edges.

Choose any star Sz(x1, x2, x3) in S. Let N = {N(x1), N(x2), N(x3)} \ {z}. By Lemma2 we know that for every i 6= j, N(xi) ∩ N(xj) = z which means that |N | ≥ 3n/5 − 3. LetX = {x1, x2, x3}. We know that z 6∈ X and z 6∈ N . Also, by the same argument as before,N ∩ X = ∅. Otherwise we either get a (k − 1)-system or we contradict our choice of systemwith a maximum number of circuits. This implies that we already have accounted for a totalof 3n/5 + 1 of the vertices in our graph G.

Now the vertex z must have at least n/5 − 3 other neighbors in G besides those in theset X. None of the edges adjacent to z can be dominated edges in our system or we couldadd them to the star centered at z giving us star with 4 edges or more which we cannothave. The edges also cannot be in a circuit because as a center of a star the vertex z cannotappear on a circuit. There is not another star with center z so the remaining edges that areadjacent to z must be found in stars with center other than z. Let C = {c1, c2, ..., cl} be the

15

neighbors of z not in the set X. Note that l ≥ n/5 − 3. We know that each of the verticesin C is the center of a star with leaf z. Clearly the vertex z is not an element of C. Alsowe know that X ∩ C = ∅ or we would have two stars in our system that shared a pair ofvertices.

Claim 3 The set N ∩ C is empty.

Suppose that there does exist a vertex ci ∈ N(xj) for some i ∈ {1, 2, ..., l} and j ∈{1, 2, 3}. In other words, cixj is an edge of our graph G. This edge must appear somewherein our system. It cannot appear in Sci

or we have two stars in our system that share thepair of vertices xj and z. It cannot be in a circuit or dominated by a circuit containing ci

as the vertex ci cannot be on a circuit. If it is dominated by a circuit containing xj then wecan add the edge cixj to Sci

to form a star with 4 edges which we have shown we cannothave. So it must be the case that the edge cixj is found in a star with center xj. We cannow use the 3 stars that contain the edges xjci, ciz, and zxj to form a type 4 dominating(k − 2)-system, which by Lemma 1 implies that G contains a dominating (k − 1)-system. Itfollows that the edge cixj cannot be in our graph G and thus the set N ∩ C is empty. 2

What we have shown is that none of the vertices in the set C were included in the 3n/5+1vertices that we had accounted for before. So now, with the sets X, N , C, and the vertex zwe have accounted for 4n/5 − 3 different vertices.

We will now consider two more sets of vertices. Recall that for all i ∈ {1, 2, ..., l}, thevertex ci is the center of a star with vertex z as a leaf. We know that each star in our systemhas 3 degree one vertices which means that each of these stars centered at ci contains 2 othervertices besides ci and z. So for each i ∈ {1, 2, ..., l} let vi and yi be the degree one verticesin the star centered at ci other than z. And let V = {v1, v2, ..., vl} and Y = {y1, y2, ..., yl}.Recall that l ≥ n/5 − 3 which means that |V | ≥ n/5 − 3 and |Y | ≥ n/5 − 3. By our choiceof each vi and yi we know that z 6∈ V and z 6∈ Y . It must be the case that V ∩ X is emptyand Y ∩X is empty or we would have two different stars in our system that contain the pairof vertices xi and z. It also must be the case that V ∩ C is empty and Y ∩ C is empty orwe would have two different stars in our system that contain the pair of vertices ci and z.What remains for me to show is that the sets N ∩ Y , N ∩ V , and V ∩ Y are empty.

Claim 4 The set Y ∩ N is empty.

Proof: Suppose that there does exist a vertex yi ∈ N(xj) for some i ∈ {1, 2, ..., l} andj ∈ {1, 2, 3}. In other words, yixj is an edge of our graph G. This edge must appearsomewhere in our system and we will consider each possibility. In each case we will use thestars Sz(xj) and Sci

(z, yi).If the edge yixj is dominated in our system then we will use it to combine Sz and Sci

toform the cycle z, ci, yi, xj, z which dominates the edges of the stars giving us a dominating(k − 1)-system, a contradiction. If the edge yixj is in a star centered at xj then we cancombine this star with Sz and Sci

to form a type 2 (k−2)-system, which by Lemma 1 meansG contains a dominating (k − 1)-system. If the edge yixj is in a star centered at yi then wecan combine this star with Sz and Sci

to form a type 1 (k − 2)-system, which by Lemma 1means G contains a dominating (k − 1)-system. If the edge yixj is in a circuit then we cancombine this circuit with Sz and Sci

to form a type 3 (k − 2)-system, which by Lemma 1means G contains a dominating (k − 1)-system.

16

In each case we get a contradiction which means that the set Y ∩N must be empty. Bythe same arguments we also get that the set V ∩ N is empty. If the set Y ∩ V is not emptythat means that yi = vj for some i 6= j. But that means that the vertices z and vj appeartogether in two different stars which cannot happen so the set Y ∩ V must also be empty.2

Let us review what we have shown. We have defined five sets X,N,C, Y , and V and wehave shown that the intersection of any two of them is empty. We have also shown that thevertex z is not an element of any of the five sets. The result is that n ≥ 1 + |X| + |N | +|C| + |Y | + |V | or that

n ≥ 1 + 3 + 3n/5 − 3 + n/5 − 3 + n/5 − 3 + n/5 − 3 = 6n/5 − 8.

But this implies that n ≤ 40 which is a contradiction that arises from our original assumptionthat each star in our system has exactly 3 edges. This leads us to the conclusion that Gcannot contain a dominating k-system that contains any stars.

Let us review what we have done. We began by supposing that for some k ≥ ⌊n/10⌋+2,that L(G) has a 2-factor with k cycles but does not have a 2-factor with k − 1 cycles. Thatled us immediately to the assumption that our graph G has a dominating k-system but nodominating (k − 1)-system. Based on this assumption we then showed that G could notcontain a dominating k-system that consists entirely of circuits and that it also could notcontain a dominating k-system that contains any stars. The resulting conclusion then isthat G cannot contain a dominating k-system. This is obviously false which implies thatour supposition that G contains a dominating k-system but no dominating (k − 1)-systemcannot hold. So for every k ≥ ⌊n/10⌋+2 if L(G) has a 2-factor with k cycles then L(G) hasa 2-factor with k − 1 cycles, thus proving the theorem. 2

Finally, consider the complete bipartite graph G = Kn/2,n/2. We know that α(G) = n/2and δ(G) = n/2. From Theorem 8 we see that L(G) has a 2-factor with k cycles for eachk = 1, 2, ..., n2/12. Note that this is best possible since |E(G)| = n2/4 which implies that|V (L(G))| = n2/4 and so it would be impossible to have any more than n2/12 disjoint cyclesin L(G).

References

[1] M. Aigner and S. Brandt, “Embedding Arbitrary Graphs of Maximum Degree Two”,Journal of the London Mathematical Society (2), Vol. 48, pp. 39-51, 1993.

[2] S. Brandt, G. Chen, R.J. Faudree, R.J. Gould, L. Lesniak, “On the Number of Cycles ina 2-Factor”, Journal of Graph Theory, Vol. 24, No. 2, pp. 165-173, 1997.

[3] P. Catlin, “A Reduction Method to Find Spanning Eulerian Subgraphs, Journal of GraphTheory, Vol. 12, pp. 29-44, 1988.

[4] G. Chartrand, L. Lesniak, Graphs & Digraphs, Chapman and Hall, London, 1996.

[5] R.J. Gould, E.A. Hynds, “A Note on Cycles in 2-Factors of Line Graphs”, Bulletin ofthe Institute of Combinatorics and its Applications, Vol. 26, pp. 46-48, 1999.

[6] F. Harary and C. St. J. A. Nash-Williams, “On eulerian and hamiltonian graphs and linegraphs”, Canadian Mathematical Bulletin, pp. 701-710, 1965.

17

[7] O. Ore, “Note on hamiltonian circuits”, American Mathematical Monthly, Vol. 67, p.55,1960.

[8] N. Sauer and J. Spencer, “Edge Disjoint Placements of Graphs”, Journal of Combinato-rial Theory B, Vol. 25, pp. 295-302, 1978.

18

A Lemma Proofs

In Lemma 1 we will consider four types of dominating (k − 2)-systems. For example, atype 1 dominating (k − 2)-system is one that can be formed from a dominating k-systemthat contains three stars which can be combined to form a cycle as pictured in Figure 5.Similarly, dominating (k− 2)-systems of types 2, 3, and 4 are those that can be formed fromdominating k-systems by combining three particular elements as pictured in Figures 6, 7,and 8. For convenience we will say that for r ∈ {1, 2, 3, 4} a type r dominating k-system inG is one that leads to a type r dominating (k− 2)-system. For ease of referal we will denotea dominating l-system of type r as SYl(r).

Lemma 1 Let G be a graph of order n > 65, such that δ(G) ≥ n/5. If G contains adominating k-system and a dominating (k − 2)-system of type 1,2,3, or 4, then G containsa dominating (k − 1)-system.

u

x y z v

x

w

y

w

u

v

z

y

w

x

Figure 5: Type 1

u

x y z v

x

w

w

y

u

v

z

y

w

x

Figure 6: Type 2

19

u

x y z v

x

w

y

w

u

v

z

y

x

ww

Figure 7: Type 3

u

x y z

x

w

w

u

u

yz

x

w

Figure 8: Type 4

Proof: Assume G has a dominating k-system and a dominating (k − 2)-system of type1, 2, 3, or 4, but does not have a dominating (k − 1)-system. We will now make a fewobservations. In each of types 1, 2, and 3, the star in the k-system with center u, Su, canhave at most 4 leaves. A fifth leaf, v′, would allow us to form a dominating (k − 1)-systemfrom the (k − 2)-system, by producing another star Su(v, v′, z). Similarly, the remainingstars used to form the (k − 2)-system have exactly 3 leaves.

Case 1 Suppose the star Su has 4 leaves.

Note that u cannot be on a cycle of the k-system nor can it be the center of anotherstar or we can immediately form a (k − 1)-system. Also, the vertex u cannot be a vertex ofa dominated edge in the k-system or we could move the edge to make Su have 5 leaves, acontradiction. Consequently, the remaining edges adjacent to u must each be in a star withcenter other than u. Because d(u) ≥ n/5, there must be at least n/5 − 4 such stars. Ingeneral we will let ir denote the number of stars in the k-system in which the vertex r is aleaf of the star. So, by our previous observation we see that iu ≥ n/5 − 4 or d(u) ≤ 4 + iu.Similar counts hold for the degrees of the vertices which are leaves of Su. Let these verticesbe x, y, z, and v.

In each type k-system there is a star with center x, Sx, having exactly three leaves. Aswith the vertex u, the remaining edges adjacent to x must be found in stars with centerother than x. Because of the minimum degree condition, there must be at least n/5 − 3 ofthese stars which implies that d(x) ≤ 3 + ix. Now for the remaining leaves y, z, and v, wewill consider them based on the type of k-system we are working with.

20

Subcase 1 Suppose the k-system is type 1.

Then y and x have identical properties and so d(y) ≤ 3 + iy. At least one of the leavesof Su is not the center of any star and is not on any cycle, for otherwise we could distributethe edges of Su to other places in the system forming a (k − 1)-system. Assume, withoutloss of generality, that z is such a vertex. Then all of z’s adjacent edges are either in starswith center other than z or are dominated edges. There can be at most one dominated edgebecause otherwise, when we form the (k − 2)-system, we can use the dominated edges andthe edge uz to form a star making a (k − 1)-system. Hence, d(z) ≤ iz + 1. Finally, thevertex v is either the center of a star, on a cycle, or neither. If v is the center of a star thenthe star has at most 4 leaves or in the (k − 2)-system we can add the edge uv to the star,making a star with at least 6 leaves, and then split this star to form 2 smaller stars givingus a (k − 1)-system. Thus, in this case d(v) ≤ 4 + iv. If v is on a cycle then similar to thevertex z, v can be adjacent to at most 1 dominated edge which implies that d(v) ≤ 3+ iv. Ifv is neither the center of a star in our system nor on a cycle in our system then d(v) ≤ 1+ iv.In any case d(v) ≤ 4 + iv. Hence, if the k-system is type 1, then we know that d(y) ≤ 3 + iy,d(z) ≤ 1 + iz, and d(v) ≤ 4 + iv. 2

Subcase 2 Suppose the k-system is type 2 and y is not the center of any star and y is noton a cycle in the k-system.

Then the edges adjacent to y are either in a star in our system with center other that y,or are dominated edges in our system. At most 2 of these edges can be dominated becausethey remain dominated edges in the (k − 2)-system and if there are at least 3 dominatededges, we can form an extra star giving us a (k − 1)-system. So, d(y) ≤ 2 + iy. By similararguments to those for vertex v in Subcase 1, d(v) ≤ 4 + iv, and d(z) ≤ 4 + iz. 2

Subcase 3 Suppose the k-system is type 2 and either y is the center of a star or y is on acycle in the k-system.

There are at most 2 dominated edges adjacent to y and so if y is on a cycle then d(y) ≤4 + iy. If y is the center of a star, Sy, then Sy can have at most 5 leaves. This means that ifwe add the number of edges in Sy and the number of edges adjacent to y that are dominatedthe total can be at most 5. The star Sy is unaffected when we form the (k − 2)-system .Therefore we can add dominated edges to Sy so that it has 6 or more leaves, and then wecan form 2 stars from it in the (k − 2)-system and the result is a (k − 1)-system. So, in thiscase we can guarantee that whether y is the center of a star or on a cycle, d(y) ≤ 5 + iy.With the same type arguments as in Subcase 1, we get that d(z) ≤ 1+ iz, and d(v) ≤ 4+ iv.2

Subcase 4 Suppose the k-system is type 3.

The vertex y is on a cycle and has at most two dominated edges adjacent to it, and sod(y) ≤ 4 + iy. Now we assume, without loss of generality, that z is the vertex in Su that isnot the center of any star and is not on any cycle. Then, as in previous subcases, we getthat d(z) ≤ 1 + iz and d(v) ≤ 4 + iv.

21

Looking at the results from each subcase we see that in any case, d(y)+d(z)+d(v) ≤ 10+iy + iz + iv. Combining this with the previous results on d(u) and d(x), and the minimumdegree condition in the statement of the proof we see that

n ≤ d(u) + d(x) + d(y) + d(z) + d(v) ≤ 17 + iu + ix + iy + iz + iv.

The only star that could be counted more than once in the sum iu + ix + iy + iz + iv is thestar centered at u with leaves x, y, z, and v. No other star in our k-system can have any pairfrom these four vertices or we can combine it with Su and get a (k − 1)-system. The star Su

is counted 4 times in this sum, once in each of ix, iy, iz, and iv. This means that we have atleast ix + iy + iz + iv + iu − 3 different stars in our k-system. If i is the total number of starsin the k-system , then ix + iy + iz + iv + iu ≤ i + 3. Putting our results together we get thatn ≤ 17 + i + 3 = i + 20 or that i ≥ n − 20.

This means that in our k-system, by counting all of the stars with leaves u, x, y, z, or v,we have at least n − 20 different stars. Note that no two stars in our k-system can share apair of vertices, for otherwise we can form from the two stars a cycle with dominated edges,giving us a (k − 1)-system.

Now consider the collection of iu stars that have vertex u as a leaf.

Claim 1 Each of these iu stars has only 3 leaves.

Suppose one of them has center c, and more than 3 leaves. No matter which type k-system (1,2,or 3) we are working with, when we form the (k − 2)-system we can form thestar Su(v, z, c) and make a (k − 1)-system. 2

We have already noted that there are at least n/5 − 4 of these stars with leaf u, whichmeans we certainly have at least 5 of them. Choose any 5 and call the centers c1, c2, c3, c4,and c5. We want to show that, based on our assumption that we do not have a (k−1)-system,we have more stars in our k-system than is possible. We already have at least n − 20 ofthem.

Now consider the stars with c1 as a leaf. There must be at least n/5− 3 of them as c1 isalready the center of a star that can have at most 3 leaves. This means c1 is not on a cycleand is not adjacent to any dominated edges. The vertex c1 cannot be a leaf in a star whichhas leaf u as they are already in a star together. Also, at most one star with leaf x can alsohave leaf c1. The same holds for stars with leaves y, z, and v. Therefore, at most 4 of then/5 − 3 stars with leaf c1 are in the n − 20 stars we already had counted, which means wenow have at least (n− 20) + (n/5− 7) different stars in our k-system. Similar arguments forc2, c3, c4, and c5 show that we have at least

(n − 20) + (n/5 − 7) + (n/5 − 8) + (n/5 − 9) + (n/5 − 10) + (n/5 − 11) ≤ i

different stars in our k-system. Because we have no (k − 1)-system, all the stars in ourk-system must have different centers which means that i ≤ n. Combining these results weget

2n − 65 ≤ i ≤ n

which implies that n ≤ 65, a contradiction. 2

Case 2 Suppose the star with center u has 3 leaves, x, y, and z.

22

In this case we could be working with any of the types 1-4. By arguments similar tothose in Case 1 we get that d(u) ≤ 3 + iu and again that d(x) ≤ 3 + ix. Recall that it mustbe the case that one of the vertices that is a leaf of Su is not the center of any star and isnot on any cycle.

Subcase 1 Suppose the k-system is type 1.

Then the vertices y and x have identical properties and so d(y) ≤ 3 + iy. Now z is notthe center of any star, is not on any cycle, and is adjacent to at most one dominated edge. Ifz is adjacent to 2 or more dominated edges then when we look at the type 1 (k − 2)-systemwe see that we can use these dominated edges and the edge zu to form a star centered at zgiving us a (k − 1)-system. Hence, d(z) ≤ 1 + iz. 2

Subcase 2 Suppose the k-system is type 2 and y is not the center of any star and y is noton a cycle.

If we consider the type 2 (k − 2)-system then we see that y can be adjacent to at most 2dominated edges in the (k − 2)-system and thus y can be adjacent to at most 2 dominatededges in the k-system. If there are more than 2 edges adjacent to y in the k-system that aredominated edges, then after forming the (k − 2)-system we can use these edges to form anew star with center y giving us a (k − 1)-system. This means that d(y) ≤ 2 + iy.

If z is the center of a star, then the star has at most 4 leaves, since, if the star has 5 ormore leaves, then after we form the type 2 (k − 2)-system, we can add the edge zu to thestar making a star with 6 or more leaves in the (k − 2)-system. We can then break this starinto two stars giving us a (k − 1)-system and a contradiction. Hence if z is the center ofa star in our k-system, then d(z) ≤ 4 + iz. The vertex z can be adjacent to at most onedominated edge in our k-system. Otherwise we could use the dominated edges and the edgezu to form a (k − 1)-system from the type 2 (k − 2)-system. Consequently, if z is on a cyclethen d(z) ≤ 3+ iz and if z is not the center of a star and is not on a cycle then d(z) ≤ 1+ iz.In any case we know that d(z) ≤ 4 + iz. 2

Subcase 3 Suppose the k-system is type 2 and y is the center of a star or y is on a cycle.

As in Subcase 3 of Case 1, d(y) ≤ 5 + iy and d(z) ≤ 1 + iz. 2

Subcase 4 Suppose the k-system is type 3.

As in Subcase 4 of Case 1, d(y) ≤ 4 + iy and d(z) ≤ 1 + iz. 2

Subcase 5 Suppose the k-system is type 4.

Either the vertex y or the vertex z has the property that it is not the center of a starand it is not on a cycle. If y is this vertex then by arguments similar to those in previoussubcases we get that d(y) ≤ 1 + iy and d(z) ≤ 4 + iz. Similarly, if z is the vertex then weknow d(z) ≤ 1 + iz and d(y) ≤ 4 + iy. 2

Looking at the results from each subcase we see that in all cases d(y)+d(z) ≤ 6+ iy + iz.Combining this with the previous results on d(u) and d(x), and the minimum degree conditionin the statement of the proof we see that

4n/5 ≤ d(u) + d(x) + d(y) + d(z) ≤ 12 + iu + ix + iy + iz.

23

Again Su is the only star counted more than once in the sum iu + ix + iy + iz, in fact,Su is counted three times. So, we have at least iu + ix + iy + iz − 2 different stars in ourk-system. If i is the total number of stars in the k-system, then iu + ix + iy + iz ≤ i + 2.Putting our results together we get that 4n/5 ≤ 12 + i + 2 = i + 14 or that i ≥ 4n/5 − 14.This time, by counting the stars with leaves u, x, y, or z, we have shown that there are atleast 4n/5 − 14 stars in our k-system. Using the same strategy as in Case 1 we choose any6 of the stars with leaf u and call the centers c1, c2, c3, c4, c5, c6. Using the same argumentsas in Case 1, there are at least n/5 − 6 stars with c1 as a leaf that are not included in the4n/5 stars we already have. We again continue by adding stars with leaves c2, then c3, andso forth until we reach c6, each time only adding those we have not already counted. Bydoing this we see that in this case we have at least

(4n

5− 14) + (

n

5− 6) + (

n

5− 7) + (

n

5− 8) + (

n

5− 9) + (

n

5− 10) + (

n

5− 11) ≤ i

different stars in our k-system. Because we have no (k − 1)-system, all the stars in ourk-system must have different centers which means that i ≤ n. Combining these results wesee that

2n − 65 ≤ i ≤ n

which implies that n ≤ 65.In either case, our assumption that we have a k-system and resulting (k − 2)-system of

type 1,2,3 or 4 and no (k−1)-system leads us to the conclusion that n ≤ 65, a contradiction.2

For ease of referal we now classify the types of edges in a dominating system.

• Type E1x - an edge in a star with center x.

• Type E2x - an edge dominated by a circuit where x is the vertex shared by the domi-nated edge and the circuit.

• Type E3 - an edge in a circuit.

Lemma 2 Let G be a graph of order n > 65 with δ(G) ≥ n/5. If G contains at leastone dominating k-system , but does not contain a dominating k-system consisting entirelyof circuits, and G does not contain a dominating (k − 1)-system, then any two degree onevertices x and y in the same star in a dominating k-system of G cannot have a commonneighbor in G other than the center of the star.

Proof: Suppose we have a graph G of order n > 65, such that δ(G) ≥ n/5. Also supposethis graph G contains a dominating k-system, but does not contain a dominating k-systemthat consists entirely of circuits, and does not contain a dominating (k − 1)-system. Choosea dominating k-system S of G with a maximum number of circuits. Choose any star Su

∈ S with center u. Choose any two degree one vertices from this star and call them xand y. Now suppose, by way of contradiction, that there does exist a vertex w 6= u suchthat w ∈ N(x) ∩ N(y). Because S is dominating, the edges xw and yw must appear in Ssomewhere. We will consider all possible combinations of edge types for these two edges.

Case 1 Suppose the edge xw is type E1x, that is, Sx(w) ∈ S.

24

If yw is type E1y, then we can combine Sx(w) and Sy(w) with Su to get SYk−2(1), whichimplies by Lemma 1 that G has a dominating (k − 1)-system, a contradiction

If yw is type E1w, then we can combine Sx(w) and Sw(y) with Su to get SYk−2(2), andLemma 1 again implies that G has a dominating (k − 1)-system, a contradiction.

If yw is type E2y or type E2w, then we can use the edge yw, Sx(w), and Su to form thecycle u, y, w, x, u. This cycle dominates the edges from the two stars that are not a part ofthe cycle which means we have a dominating (k − 1)-system, a contradiction

If yw is type E3, then we can combine the cycle containing yw, Sx(w), and Su to getSYk−2(3), and Lemma 1 then implies that G has a dominating (k−1)-system, a contradiction

2

Case 2 Suppose the edge xw is type E1w, that is, Sw(x) ∈ S.

If yw is type E1y then we can combine Sw(x) and Sy(w) with Su to get SYk−2(2), andLemma 1 then implies that G contains a dominating (k − 1)-system, a contradiction.

If yw is type E1w then Sw(x, y) ∈ S. If not we have 2 stars with center w which we cancombine to get a dominating (k − 1)-system. If we combine Sw(x, y) with Su we can formthe cycle w, y, u, x, w which dominates the remaining edges of the stars not used to form thecycle giving us a dominating (k − 1)-system, a contradiction.

If yw is type E2 then we can use the edge yw, Sw(x), and Su to form the cycle u, y, w, x, u.This cycle dominates the edges from the two stars that are not a part of the cycle whichmeans we have a dominating (k − 1)-system, a contradiction

If yw is type E3 then we know that our system contains a circuit with the vertex w andSw(x). We can attach Sw(x) to the circuit to easily form a dominating (k − 1)-system, acontradiction. 2

Case 3 Suppose the edge xw is type E2.

If yw is type E1y or type E1w then we use the edge xw, the star containing yw, and Su

to form the cycle u, y, w, x, u. This cycle dominates the edges from the two stars that arenot a part of the cycle which means we have a dominating (k − 1)-system, a contradiction.

If yw is type E2 then we use the edges xw and yw and Su to form the cycle u, y, w, x, uwhich dominates the unused edges from Su. We know that since yw is type E2 that oursystem contains either a cycle containing the vertex y or a cycle containing the vertex w.We now attach our new cycle to whichever of these is in our system to form a dominating(k − 1)-system, a contradiction.

If yw is type E3 then extend the circuit containing the edge yw by replacing the edgewith the path y, u, x, w. Our new circuit now dominates the edge yw and any of the edgesother than ux and uy from Su. Thus, we have combined a star and a circuit to form onecircuit in our system which gives us a dominating (k − 1)-system, a contradiction.

Case 4 Suppose the edge xw is type E3.

If yw is type E1y, that is, Sy(w) ∈ S, then we can combine Sy(w), Su, and the circuitcontaining xw to get SYk−2(3), and Lemma 1 then implies that G contains a dominating(k − 1)-system, a contradiction.

If yw is type E1w, that is, Sw(y) ∈ S, then we know that our system contains a circuitwith the vertex w and Sw(y). We can attach the Sw(y) to the circuit to easily form adominating (k − 1)-system, a contradiction.

25

If yw is type E2 then we can extend the circuit containing the edge xw by replacing theedge with the path x, u, y, w. Our new circuit now dominates the edge xw and any of theedges other than ux and uy from Su. Thus, we have combined a star and a circuit to formone circuit in our system which gives us a dominating (k − 1)-system, a contradiction.

If yw is type E3 then it must be the case that xw and yw are edges of the same circuitC since the vertex w cannot appear on more than one circuit.

Subcase 1 Suppose C can be partitioned in such a way that xw and yw are on differentcycles of C.

Then we can remove the edges xw and yw from C and add the path x, u, y to form anew circuit. The new circuit dominates the edges xw and yw and the edges from Su whichmeans we have formed a dominating (k − 1)-system, a contradiction.

2

Subcase 2 Suppose that in every partition of C, the edges xw and yw are on the same cycleof C and, in at least one of the partitions, the vertex w appears on another cycle of C.

Consider a partitioning of the circuit into cycles. Let C∗

1 be the cycle in the partitionthat contains the vertex w and the edges xw and yw. Let C∗

2 6= C∗

1 be another cycle in thepartition containing the vertex w. Now we divide the circuit C into two smaller circuits C1

and C2, where C1 contains the cycle C∗

1 and C2 contains the cycle C∗

2 . We will do this insuch a way that every cycle of C is either in C1 or C2, and every edge that was dominatedby C in our k-system is now dominated by either C1 or C2. In this division we will placethe one stipulation that the vertices x and y are the only adjacencies of w in the circuit C1.These two new circuits, C1 and C2, when combined with the original dominating k-system,minus the circuit C, give us a dominating (k + 1)-system in G.

Now, we will combine C1 with Su to form a new circuit that dominates the edges xw andyw and the edges of Su other than ux and uw. As such we have formed a dominating k-systemof G that contains one more circuit than our original system did which is a contradiction ofour original choice. 2

Subcase 3 Suppose that in every partition of C, the edges xw and yw are on the same cycleof C and the vertex w is not on any other cycle of C.

If there are no dominated edges adjacent to the vertex w in our system or if all of thedominated edges adjacent to w can be moved elsewhere in the system then, after movingthem, we can combine C and Su to form a new circuit that dominates the edges xw andyw and the edges of Su other than ux and uw, giving us a dominating (k − 1)-system,a contradiction. Suppose then that there is a dominated edge vw that cannot be movedelsewhere in the system. We know then the following facts.

• The vertex w does not appear anywhere else on C, it is not on any other circuit, andit is not the center of any star.

• The vertex v is not on any circuits in our system and is not the center of any star inour system.

• The vertices v and w cannot appear in the same star.

26

Claim 1 From S we can form a new dominating k-system S ′ that has the following threeproperties:

• The star Su(x, y) ∈ S ′.

• There is a circuit C∗ in the system that contains the edges xw and yw on the samecycle of the circuit.

• We can move the edge vw elsewhere in the system.

Proof: Suppose we cannot form this new dominating k-system. Then we will let N(v) ={w, n1, n2, ..., ns,m1,m2, ...mt} where the ni’s are the neighbors of v that are the centers ofstars and the mi’s are the neighbors of v that are on circuits, hence miv is a dominated edge,1 ≤ i ≤ t. By the facts we noted above we know that these are the only types of neighborsthat v can have. We will let Nv = N(v) \ {w}. If the vertex x is in Nv then we can combineC∗, Su, and the edges vw and xv to form a circuit that dominates the edge yw and theedges of Su other than xu and yu which we used to form the new circuit. This gives us adominating (k − 1)-system, a contradiction. So it must be the case that x 6∈ Nv. Similarly,y 6∈ Nv. Note that |Nv| ≥ n/5 − 1

Also, we will let N(w) = {v, x, y, c1, ..., cp, r1, ..., rl} where the ci’s are the neighbors of wthat are the centers of stars and the ri’s are the neighbors of w that are not the centers ofstars. Now let Nw = N(w) \ {v, x, y}. Note that |Nw| ≥ n/5 − 3.

Subclaim 1 The set Nv is independent.

Proof: Suppose that this is not the case. If ninj is an edge in our graph we can formthe cycle ni, nj, v, ni which dominates the remaining edges of the stars centered at ni and nj

giving us a dominating (k − 1)-system. If mimj is an edge in our graph then it must eitherbe a dominated edge in our system or an edge of a cycle.

If mimj is a dominated edge, then we can form the cycle C : v,mi,mj, v and attach Cto a circuit containing mj. We now have a new dominating k-system that has the threeproperties listed in the claim, which is a contradiction. If the edge mimj is part of a circuitwe can remove it from the circuit and replace it with the path mivmj which forms a newcircuit that contains the vertex v and now dominates the edges mimj. Again we have formedthe type of system described in the claim, again a contradiction.

If nimj is an edge of our system, then it is either dominated or in a star Sni, since as

the center of a star ni cannot be on a circuit. However nimj appears in our system, we canplace it in Sni

. We can then form the cycle C1 : v, ni,mj, v which dominates the edges ofSni

. We know the vertex mj is on a circuit C2 so we can attach C1 to C2 at mj, giving us adominating (k − 1)-system. In each case we reach a contradiction which means the set Nv

must be independent.2

Subclaim2 The set Nw is independent.

Proof: Suppose this is not the case. If cicj is an edge of our graph G, then we canform the cycle ci, cj, w, ci which dominates the edges of Sci

and Scjgiving us a dominating

(k − 1)-system, a contradiction.

27

If rirj is an edge in our graph then it must either be a dominated edge in our system oran edge of a cycle. If it is a dominated edge then we can form the cycle w, ri, rj, w and attachthe cycle to C. This new k-system puts us back to Subcase 2 and a contradiction. If theedge rirj is part of a circuit we can remove it from the circuit and replace it with the pathriwrj which forms a new circuit that contains the vertex w and dominates the edge rirj.Now if the edge rirj was part of C already then we have created a circuit in which w is onmore than one cycle of the circuit which takes us back to Subcase 2 and a contradiction. Soit must be the case that the new circuit we have created did not already contain the vertexw. But now we have a dominating k-system in which two circuits contain a common vertex,w, which implies that we have a dominating (k − 1)-system, again a contradiction.

If cirj is an edge of our system it is either dominated or in Sci. It cannot be in a circuit

as ci is the center of a star and therefore cannot be on a circuit. However cirj appears in oursystem, we can place it in Sci

. We can then form the cycle C1 : w, ci, rj, w which dominatesthe edges of Sci

. We know the vertex rj is on a circuit C2 so we can attach C1 to C2 at rj

giving us a dominating (k − 1)-system. In each case we reach a contradiction which meansthe set Nw must be independent.2

Subclaim 3 The set Nv ∩ Nw is empty.

Proof: Suppose that there does exist a vertex z ∈ Nv ∩ Nw. Recall that the edges zvand zw must either be in a star with center z or be dominated edges. We know that v isnot on any circuit so if zv is a dominated edge it follows that z is on a circuit. The edgezw actually must be a dominated edge. This follows from the fact that the vertices v andw cannot appear in the same star so if zw was in a star then z would have to either bethe center of another star as well, or be on a circuit. Either way we can form a dominating(k − 1)-system and obtain a contradiction. Now if zv is in a star with center z, Sz, then wecan form the cycle z, v, w, z which dominates the edges of Sz. Now we have two circuits thatcontain w which means we can form a dominating (k − 1)-system, a contradiction. Finally,if zv is also a dominated edge then we again form the cycle z, w, v, z. We now attach thiscycle to the circuit that contains the vertex z. If z is in a different circuit than w then wehave two circuits now that contain w which allows us to form a dominating (k − 1)-system,a contradiction. But if z is in the same circuit as w then we now have a circuit in which w ison two different cycles of the circuit which takes us back to Subcase 2 and a contradiction.In each case we reach a contradiction which means that the set Nv ∩ Nw must be empty.2

Subclaim 4 The set Nv ∪ Nw is independent.

Proof: Recall that N(v) = {w, n1, n2, ..., ns,m1,m2, ...mt} where the ni’s are the neigh-bors of v that are the centers of stars and the mi’s are the neighbors of v that are on circuits,and that Nv = N(v) \ {w}. Also, N(w) = {v, x, y, c1, ..., cp, r1, ..., rl} where the ci’s are theneighbors of w that are the centers of stars and the ri’s are the neighbors of w that are notthe centers of stars, and Nw = N(w) \ {v, x, y}. Now if cinj ∈ E(G) we can form the cyclew, ci, nj, v, w which dominates the edges from Sci

and Snj. Whether the edge cinj was in a

star or was dominated we have made a dominating (k − 1)-system by creating this cycle. Ifcimj ∈ E(G), then we form the cycle w, ci,mj, v. We now have two circuits in our systemthat contain w which means we can form a dominating (k − 1)-system, a contradiction.Similarly, if rinj ∈ E(G) we reach a contradiction. Finally, if rimj ∈ E(G) we form the

28

cycle w, v,mj, ri, w. We now attach this cycle to the circuit that contains w which gives usa circuit in which w is on two different cycles of the circuit which takes us back to Subcase 2and a contradiction. In each case we reach a contradiction which means that the set Nv∪Nw

is independent. 2

From our claims we see that we have an independent set I = Nv ∪ Nw such that |I| ≥2n/5− 4. By definition and by previous observation we know that v, w, x, y 6∈ I. We will letR = V (G) \ {I, v, w, x, y} which implies that |R| ≤ 3n/5.

Now suppose a, b ∈ I have a common neighbor z in R. We now consider several casesbased on a and b.

• Suppose a = ni and b = nj, i 6= j. Then the cycle z, nj, v, ni, z dominates the rest ofSni

and Snjproducing a dominating (k − 1)-system, a contradiction. This is the case

whether the edges zni and znj are in stars or are dominated. Similarly we reach acontradiction if a = ci and b = cj for some i 6= j.

• Suppose a = mi and b = nj for some i, j. If the edge zmj is dominated then we canform the cycle z,mi, v, nj, z which dominates the rest of Snj

. This gives a dominatingk-system in G with one more cycle than the system we chose, contradicting our originalchoice of k-system with the maximum number of cycles. If zmi is a circuit edge thenwe can remove it from the circuit and replace it with the path mi, v, nj, z forming anew circuit that dominates the edges of Snj

and the edge zmi, giving us a dominating(k − 1)-system, a contradiction. Similarly we reach a contradiction if

• Suppose a = ni and b = cj for some i, j. Then we form the cycle z, cj, w, v, ni, zwhich dominates the edges of Sni

and Scjproducing a dominating (k − 1)-system, a

contradiction.

• Suppose a = ni and b = rj for some i, j. If zrj is in a star then we form the cyclez, rj, w, v, ni, z which dominates the edges of Sni

and Sz giving us a dominating (k−1)-system and a contradiction. If zrj is a dominated edge then we form the same cycle asabove which dominates the edges of Sni

. We then attach this new cycle to the circuitcontaining w again giving us a dominating (k−1)-system and a contradiction. Finally,if the edge zrj is a circuit edge then we remove it from the circuit and replace it withthe path z, ni, v, w, rj to form a new circuit that dominates the edges of Sni

and theedge zrj, once again forming a dominating (k−1)-system and reaching a contradiction.Similarly we reach a contradiction if a = ci and b = mj.

• Suppose a = mi and b = rj for some i, j.

(i) If both of the edges zmi and zrj are dominated edges in our system. We thenform the cycle z, rj, w, v,mi, z. Now at least one of the vertices mi, rj or z is on acircuit and we attach this new cycle to a circuit at one of these vertices. We nowhave a dominating k system that meets the 3 conditions of the claim which givesus a contradiction. 2

(ii) Suppose exactly one of the edges zmi and zrj is dominated in our system. Assumewithout loss of generality that the edge zmi is dominated. If the edge zrj is in astar then we form the same cycle as above which dominates the edges of Sz givingus a dominating (k − 1)-system, a contradiction. If zrj is a circuit edge then we

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remove it from the circuit and replace it with the path z,mi, v, w, rj creating anew circuit that dominates the edge zrj. We now have a dominating k systemthat meets the 3 conditions of the claim which gives us a contradiction. 2

(iii) Suppose both of the edges zmi and zrj are edges of a circuit in our system. Notethat it must be the same circuit as z cannot appear on 2 different circuits. Anycircuit can be factored into cycles. If the circuit containing zmi and zrj can befactored in such a way that the edges zmi and zrj are on different cycles of thecircuit then we can remove the edges zmi and zrj from the circuit and attach thepath rj, w, v,mi to form a new circuit that now domintes the edges zmi and zrj.We now have a dominating k system that meets the 3 conditions of the claimwhich gives us a contradiction. The only remaining possibility for the edges zmi

and zrj is that they are both edges of the same circuit in our system and thatthis circuit cannot be factored in such a way that the edges zmi and zmj are ondifferent cycles of the circuit. 2

Similar arguments show that if a = mi and b = mj or if a = ri and b = rj then we reacha contradiction unless the edges az and bz are both edges of the same circuit in our systemand this circuit cannot be factored in such a way that the edges az and bz are on differentcycles of the circuit. The only change in the argument is that where before we got a cycleor path with the sequence of vertices rj, w, v,mi we now get either mi, v,mj or ri, w, rj.

The conclusion then is that for any a, b in I if there exists a vertex z ∈ R such thatz ∈ N(a)∩N(b) then the edges az and bz are edges in the same circuit C in our system andC cannot be factored in such a way that az and bz are on different cycles of C. Consequentlyit is impossible for any 3 vertices of I to have a common neighbor in R. 2

Recall that I = Nv ∪Nw is such that |I| ≥ 2n/5−4 and R = V (G)\{I, w, v, x, y} is thensuch that |R| ≤ 3n/5. By our previous arguments no three vertices of I can have a commonneighbor in R. Let’s consider the edges between the sets I and R and let e be the numberof such edges. Because the set I is independent and each vertex of I is adjacent to only oneof v and w, each vertex of I sends at least n/5 − 3 edges to the set R. So there are at least

(n/5 − 3)(2n/5 − 4) edges from I to R. In other words, e ≥2n2

25−

10n

5+ 12.

Now there are at most 3n/5 vertices in R that must receive these edges, but each vertex

in R receives at most two edges from I. Thus, there are at most 2(3n

5) edges from R to I.

In other words, e ≤6n

5.

It follows that

2n2/25 − 10n/5 + 12 ≤ 6n/5,

hence n2 − 40n + 150 ≤ 0 which implies n < 36, a contradiction. 2

Let’s take a moment to recall what we have shown and what we are trying to show. Webegan by assuming that we have a graph G of order n > 65, such that δ(G) ≥ n/5 andsuch that G contains at least one dominating k-system but does not contain a dominatingk-system consisting entirely of circuits and does not contain a dominating (k − 1)-system.We then chose a dominating k-system in G with a maximum number of circuits. From thisdominating k-system we chose a star Su. Finally, from Su we chose two leaves which we calledx and y. We then supposed that there exists a vertex w 6= u such that w ∈ N(x) ∩ N(y).

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From this supposition we reached the conclusions that the edges xw and wy must beedges of the same circuit C in our system, that C cannot be factored in such a way that xwand yw are on different cycles of C, that the vertex w is not on any other cycle of C, andthat there is an edge vw that is dominated by w on C that cannot be moved elsewhere inthe system. However, our claim now essentially allows us to move the edge vw. We do notkeep exactly the same dominating k-system but we do keep one with a maximum numberof cycles, with Su intact, and with a circuit containing xw and yw on the same cycle. Themajor difference is that vw is no longer dominated by the vertex w. We can apply the claimas many times as we need until we arrive at the situation where there are no dominatededges adjacent to the vertex w in our system or all of the dominated edges adjacent to wcan be moved elsewhere in the system. We can then follow the procedure of Subcase 2 toarrive at a contradiction.

Thus there cannot be a vertex w 6= u such that w ∈ N(x) ∩ N(y) proving the Lemma.2

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