A Generalization of a result of Catlin: 2-factors in line graphs

AUSTRALASIAN JOURNAL OF COMBINATORICS Volume 72(2) (2018), Pages 164 184 A Generalization of a result of Catlin: 2-factors in line graphs Ronald J. Gould Emory University Atlanta, Georgia U.S.A. rg@mathcs.emory.edu Emily A. Hynds Samford University Birmingham, Alabama U.S.A. eahynds@samford.edu Abstract A 2-factor of a graph G consists of a spanning collection of verte disjoint cycles. In particular, a hamiltonian cycle is an eample of a 2-factor consisting of precisely one cycle. Harary and Nash-Williams described graphs ith hamiltonian line graphs. Gould and Hynds generalized this result, describing those graphs hose line graphs contain a 2-factor ith eactly k (k 1) cycles. With this tool, e sho that certain properties of a graph G, that ere formerly shon to imply the hamiltonicity of the line graph, L(G), are actually strong enough to imply that L(G) has a 2-factor ith k cycles for 1 k f(n), here n is the order of the graph G. 1 Introduction We present an etension and then a broader generalization of the folloing result of Catlin [3], hich gives specific conditions on a graph G that imply that the line graph L(G) is hamiltonian. Theorem 1.1 [3] If G is a 2-edge-connected simple graph of order n such that δ(g) n/5, then L(G) is hamiltonian. The subgraph H of G is said to be a 2-factor of G if for every v V (G), deg H v = 2. A trivial consequence of the definition is that every 2-factor of a ISSN: 2202-3518 c The author(s). Released under the CC BY-ND 4.0 International License

R.J. GOULD AND E.A. HYNDS / AUSTRALAS. J. COMBIN. 72 (2) (2018), 164 184 165 graph G consists of a spanning collection of verte disjoint cycles. In particular, a hamiltonian cycle is an eample of a 2-factor consisting of precisely one cycle. We begin by proving the folloing etension of Theorem 1.1. Theorem 1.2 If G is a 2-edge-connected simple graph of order n such that δ(g) n/5, then L(G) has a 2-factor ith k cycles for each k {1, 2,..., n/10 }. The remainder of the paper is dedicated to proving the folloing broader generalization for graphs ith large independence number. Theorem 1.3 If G is a 2-edge-connected simple graph of order n > 65 ith δ(g) n/5 and α(g) = cn, for some c, 1/4 c < 1, then L(G) contains a 2-factor ith k cycles, for each k = 1, 2,..., cn δ [(1 c)n δ]. 3 All graphs considered in this paper are simple graphs. For terms or notation not defined here, see [4]. For a graph G, let N(v) denote the neighborhood of verte v. A set S V (G) is said to be independent if for all u, v S, uv E(G). The independence number of a graph G, denoted α(g), is the size of a largest independent set of vertices of G. For a set S V (G) e use S to denote the subgraph induced by S. A circuit of G is an alternating sequence C : v 1, e 1, v 2, e 2,..., v m, e m, v 1 of vertices and edges of G, such that e i = v i v i+1, i = 1, 2,..., m 1, e m = v m v 1, and e i e j if i j. A circuit hose m vertices v i are distinct is called a cycle. We define a dominating circuit of a graph G to be a circuit of G ith the property that every edge of G either belongs to the circuit or is adjacent to an edge of the circuit. A star is a complete bipartite graph, K 1,n. The verte of degree n is termed the center of the star and the vertices of degree 1 are the leaves. If a star has center e often denote it as S. Further, if e ish to specify a star centered at ith some specific leaves, say a, b, c, e ill denote it by S (a, b, c). Note that there may be other leaves in S not specified. Early studies of 2-factors centered on the question of eistence, often of simply a hamiltonian cycle. More recently, the focus in the area of 2-factors has shifted from the problem of shoing the eistence of a 2-factor to that of shoing the eistence of 2-factors ith specific structural features. In 1978, Sauer and Spencer made the folloing conjecture along those lines. Conjecture 1.1 [8] Let H be any graph on n vertices ith maimum degree 2. If G is a graph on n vertices ith minimum degree δ(g) > 2n/3 then G contains an isomorphic copy of H. In 1993, Aigner and Brandt settled Conjecture 1.1 ith a slight improvement. Theorem 1.4 [1] Let G be a graph of order n ith δ(g) (2n 1)/3. Then G contains any graph H of order at most n ith (H) 2.

R.J. GOULD AND E.A. HYNDS / AUSTRALAS. J. COMBIN. 72 (2) (2018), 164 184 166 In the above result, the minimum degree must be very high to guarantee that a graph contains all possible 2-factors or 2-factors ith a particular structure. Thus, a more relaed question ould be: is there a lesser degree condition that ill imply the eistence of 2-factors ith k cycles for a range of k? The folloing as shon. Theorem 1.5 [2] Let k be a positive integer and let G be a graph of order n. If deg() + deg(y) n for all, y V (G) such that y E(G), then G contains a 2-factor ith k cycles for all k, 1 k n/4. Note that Theorem 1.5 is a generalization of the classic hamiltonian result of Ore [7] for the case hen n 4k. The complete bipartite graph K n/2,n/2 shos that this result is best possible. This type of result naturally leads to the question of hether or not other hamiltonian results can be etended in a similar manner. The folloing is the ell-knon result of Harary and Nash-Williams [6] describing graphs ith hamiltonian line graphs. Theorem 1.6 [6] Let G be a graph ithout isolated vertices. Then L(G) is hamiltonian if, and only if, G K 1,n, for some n 3, or G contains a dominating circuit. Given a graph G, e say that G contains a dominating k-system if G contains a collection of k edge disjoint circuits and stars (K 1,ni, n i 3) such that each edge of G is either contained in one of the circuits or stars, or is adjacent to one of the circuits. We ill use a generalization of Theorem 1.6 that allos us to describe those graphs hose line graphs contain a 2-factor ith eactly k (k 1) cycles. Theorem 1.7 [5] Let G be a graph ith no isolated vertices. The graph L(G) contains a 2-factor ith k (k 1) cycles if, and only if, G contains a dominating k-system. Recall that Catlin s result, stated in Theorem 1.1, gives specific conditions on a graph G that imply that the line graph L(G) is hamiltonian. We ill no sho that these same conditions actually imply much more, by proving the etension stated in Theorem 1.2. 2 Etension: Proof of Theorem 1.2 Recall that the hypothesis of Theorem 1.2 gives us a 2-edge-connected simple graph of order n such that δ(g) n/5. From Theorem 1.1, e kno that L(G) is hamiltonian. No since δ(g) n/5, e kno that E(G) n 2 /10. Therefore, by proving the folloing theorem, e achieve the desired result. Theorem 2.1 Let G be a graph of order n and k be an integer 1 k cn, for some constant c > 0. If E(G) cn 2 and L(G) is hamiltonian, then L(G) contains a 2-factor ith k cycles.

R.J. GOULD AND E.A. HYNDS / AUSTRALAS. J. COMBIN. 72 (2) (2018), 164 184 167 Proof: Let G be a graph as in the theorem. We kno by assumption that L(G) is hamiltonian and so e ill proceed by induction on the number of cycles in a 2-factor. Suppose that L(G) has a 2-factor ith k 1 cycles for k cn, but L(G) does not have a 2-factor ith k cycles. By Theorem 1.7 e kno that the graph G does have a dominating (k 1)-system and G does not have a dominating k-system. Consider a dominating (k 1)-system in G. Every edge of G is either in a star in this system, a circuit in this system, or is dominated by a circuit in the system (called a dominated edge). We ill let i be the number of stars in the system and thus k 1 i is the number of circuits. If there is a star in the system ith si or more edges, e can separate the star into to smaller stars ith at least 3 edges each. This gives us a dominating k-system in G, hich means that all stars in the system have at most 5 edges. If there is a circuit in the system that is not a cycle, then e can separate the circuit into 2 edge disjoint circuits. This again gives us a dominating k-system hich means all of the circuits in this system must be cycles. Finally, there can be at most n 1 dominated edges. Consider the subgraph of G induced by these dominated edges. No verte in this subgraph can have degree 3 or higher. Such a verte ould allo us to form another star and thus a dominating k-system. Consequently this subgraph has maimum degree 2. No suppose all of the vertices in the subgraph have degree 2. Then e can find a cycle in the subgraph hich, hen added to the (k 1)-system, gives us a dominating k -system in G. Hence, e have a subgraph hich must be a collection of disjoint paths, and thus can contribute at most n 1 edges to our graph G Combining these three results, e see that E(G) 5i + n(k 1 i) + (n 1). For n > 5 this is maimized at i = 0. So, E(G) 5i + n(k 1 i) + (n 1) nk 1 (n > 5). By our original assumption, E(G) cn 2 hich implies that cn 2 nk 1. On the other hand, e kno that k cn. But, k cn implies that k < cn + 1 hich in turn implies that n cn2 > nk 1, a contradiction. Thus, it must be the case that L(G) has a 2-factor ith k cycles. In an effort to improve the result of Theorem 1.2, e continued our study of line graphs obtained from graphs G of order n ith δ(g) n/5. We concentrated on such graphs ith large independence number. This is here e moved from an etension to a generalization. The proof of this generalization, stated in Theorem 1.3, ill be the focus of the remainder of the paper. 3 Generalization: Proof of Theorem 1.3 The proof of Theorem 1.3 ill follo from Theorem 1.2 and the to theorems e ill prove in this section. Before proving these to theorems, e state to technical lemmas that ill be useful. The proofs of the lemmas are not included as they are lengthy and the techniques are very similar to those found in the included proofs. Hoever, proofs are available upon request. In Lemma 1 e ill consider four types of dominating (k 2)-systems. For eample, a type 1 dominating (k 2)-system is one that can be formed from a dominating k-system that contains three stars hich can be combined to form a

R.J. GOULD AND E.A. HYNDS / AUSTRALAS. J. COMBIN. 72 (2) (2018), 164 184 168 cycle as pictured in Figure 1. Similarly, dominating (k 2)-systems of types 2, 3, and 4 are those that can be formed from dominating k-systems by combining three particular elements as pictured in Figures 2, 3, and 4. In types 1, 2, and 3, the star ith center u must have eactly three or four leaves, hence the dotted line for uv. u y y z v v u z y Figure 1: Lemma 1 Type 1 Lemma 1 Let G be a graph of order n > 65, such that δ(g) n/5. If G contains a dominating k-system and a dominating (k 2)-system of type 1, 2, 3 or 4, then G contains a dominating (k 1)-system. Lemma 2 Let G be a graph of order n > 65 ith δ(g) n/5. If G contains at least one dominating k-system, but does not contain a dominating k-system consisting entirely of circuits, and G does not contain a dominating (k 1)-system, then any to degree one vertices and y in the same star in a dominating k-system of G cannot have a common neighbor in G other than the center of the star. Recall that given a graph G, e say that G contains a dominating k-system if G contains a collection of k edge disjoint circuits and stars (K 1,ni, n i 3) such that each edge of G is either contained in one of the circuits or stars, or is adjacent to one of the circuits. An edge that is simply adjacent to one of the circuits is called a dominated edge. The term dominating system ill denote a collection of edge disjoint circuits and stars ith possible dominated edges, but unknon size. The term k system ill denote a collection of k edge disjoint circuits and stars ith possible dominated edges, but that doesn t necessarily dominate. The term system ill denote a collection of edge disjoint circuits and stars ith possible dominated edges, that doesn t necessarily dominate and hose size is unknon.

R.J. GOULD AND E.A. HYNDS / AUSTRALAS. J. COMBIN. 72 (2) (2018), 164 184 u y v z 169 y v z u y Figure 2: Lemma 1 Type 2 u y y z v v u z y Figure 3: Lemma 1 Type 3 Theorem 3.1 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 ith at least k = cnb 3δ c [(1 c)n δ] cycles. Proof: Since G has an independent set of vertices, say I, of size cn (c 1/4), let R = V (G) \ I and I. As deg() δ(g), e can form at least b δ(g) c stars 3

R.J. GOULD AND E.A. HYNDS / AUSTRALAS. J. COMBIN. 72 (2) (2018), 164 184 170 u y z u u y z Figure 4: Lemma 1 Type 4 centered at the verte, each ith at least 3 leaves. Doing this for each verte in I, e build a collection of at least cn δ stars that collectively use all edges beteen 3 the sets R and I. Place these stars in S, hich represents a system that e ill make into a dominating system. Let S represent the number of stars and circuits in S. If there are any stars or circuits contained in R G, then e add them at random to S. At this point, e ill say an edge is part of a system if it is in a star or circuit in the system or is a dominated edge. Removing from G the edges that are already part of the system, S, e are left ith at most some disjoint paths in R G, P 1, P 2,..., P s. We ant to add all of the edges of these paths to S so that S dominates. This ill be done through a finite number of steps, using an iterative process, during hich e modify S in such a ay that it remains a system but ultimately dominates. During the process e might reduce the size of S, but by at most one in each step. Suppose that in the original graph, one of the paths P i in R G contains to or more vertices that share a common neighbor z in I. Choose to such vertices i y i, such that any other neighbors of z on P i lie beteen i and y i on P i. Form the cycle C : i, z, y i,..., i here y i,..., i denotes the portion of the path P i beteen y i and i. The edges i z and y i z are the only edges of C that ere already in S. Because of the ay e are building S, i z and y i z must either be dominated edges or edges of stars ith center z in S. If there eists any star ith center z in S then modify S, if necessary, to form the star S z ( i, y i ) ithout decreasing S. In S, replace S z ( i, y i ) ith C hich dominates the remaining edges of S z ( i, y i ). If there is no star ith center z in S, then i z and y i z are both dominated and z must be on a circuit C in S. We can then attach C to C making a larger circuit. In either case, e have added edges from R G ithout decreasing S. Repeat this process until all that

R.J. GOULD AND E.A. HYNDS / AUSTRALAS. J. COMBIN. 72 (2) (2018), 164 184 171 remains in R G are disjoint paths hich contain no to vertices that have a common neighbor in I. No, e must incorporate the remaining edges into S ithout affecting S too much. We ill begin by shoing that any time e incorporate a ne edge, S decreases by at most one. By ay of contradiction, assume that if e can add one of the remaining edges in R G to S, e must lose at least 2 from S. Choose an edge pq in R G such that pq is not part of the system yet and it is the only edge adjacent to p unused in our system. Consequently, neither p nor q can be the center of any star in S and neither can be found on any circuit, for otherise e could easily add pq to S leaving S unchanged. As I is an independent set of maimum size, there eists an 0 I such that p 0 E(G) and a y 0 I such that qy 0 E(G). We kno 0 y 0 as none of the remaining paths have 2 vertices ith a common neighbor in I. Let N(p) = {q, 0, 1,..., r } and N(q) = {p, y 0, y 1,..., y s } here r, s n/5 2. Note that e can assume that S is made up of stars and cycles since any circuit can be divided into edge disjoint cycles, hich only increases S. Claim 3.1.1 The set N(p) N(q) =. Suppose there eists a z N(p) N(q). As p and q have no common neighbors in I, z R. We kno that pz is already in our system, hich means, because of the properties of p discussed earlier, z is the center of a star or is on a circuit. Consequently, the edge qz must be in our system as ell. Remember that the edges pz and qz can only be in a star ith center z or dominated by a cycle containing z. If either is in a star ith center z then e can clearly rearrange our system, leaving the size the same, to form S z (p, q). We can then add the edge pq to S z (p, q) hich forms a cycle p, z, q, p that dominates the other edges of S z (p, q) leaving the system size unchanged, a contradiction. So it must be the case that pz and qz are both dominated by circuits. Hoever, then e can form the cycle p, z, q, p, adding the edge pq to S and increasing S by 1, another contradiction. Hence, N(p) N(q) =. Claim 3.1.2 The sets N(p) and N(q) are independent. Recall that N(p) = {q, 0, 1,..., r } and that pq is the edge e are trying to add to our system. We kno from Claim 3.1.1 that q i E(G) for all i {0, 1,..., r}. Suppose, by ay of contradiction, that i j E(G) for some i j. As in the proof of Claim 3.1.1, p i (p j ) must appear in the system as an edge of a star ith center i (or j ) or as an edge dominated by a cycle containing i (or j ). There are four possible cases for here the edge i j could be located in relation to our system, S. It could be an edge in a star or cycle in S, a dominated edge in S, or not in S at all. In each case, similar arguments can be used to sho that the edge pq can be added ith a loss of at most one from S, contradicting our original assumption. Here is an eample of one such argument. Suppose that i j is an edge in a star in S. Without loss of generality let S i ( j ) be this star. No the cycle i, j, p, i dominates the remaining edges of S i ( j ). If

R.J. GOULD AND E.A. HYNDS / AUSTRALAS. J. COMBIN. 72 (2) (2018), 164 184 172 the edges p i and p j are both either from stars of size 4 or larger or are dominated edges, then e can add that cycle to our system ithout losing any edges or changing the size of the system. Then e can add the edge pq as a dominated edge ithout decreasing the system size, a contradiction. Thus, at least one of p i or p j is from a star of size 3. If only one of them is in a star of size 3, then e replace that star ith our cycle, hich dominates the remaining edges of that star, hich decreases the system size by one ithout losing any edges. But, e can still add the edge pq to our system as an edge dominated by the cycle containing p, hich is again a contradiction. Therefore the edges p i and p j must both be found in stars of size 3. If i j is in a star of size 4 or larger then e only lose to stars hen forming our ne cycle. So again, e are able to add the edge pq hile only reducing the system size by 1. Consequently, the edge i j must also be in a star of size 3. If the edge i p is not in S i ( j ) then interchange edges to form S i ( j, p) ithout decreasing the size of the system S. Again e lose at most to elements of S hen forming the ne cycle containing p, hich means S has decreased by at most one. Hoever, since p is no on a cycle in S, e can again add the edge pq to S giving us another contradiction As mentioned before, a similar argument shos that if the edge i j is in a cycle of S, is a dominated edge of S, or is not yet in S, the edge pq can be added to S, ithout losing any edges from S and losing at most one from S, a contradiction. Therefore, the edge i j cannot be in E(G). The independence of N(q) can be shon similarly. Claim 3.1.3 The set N(a) N(q) \ {p} is empty for all a N(p) I and the set N(b) N(p) \ {q} is empty for all b N(q) I. Suppose there eists z N(a) N(q) \ {p}. Consider the edges pa, qz, and az, hich may already be part of S. Note that S can only decrease if e take an edge from a dominating circuit or star in S and hat remains is no longer a circuit or star. If e can use pa, az, and zq ithout affecting the system size, as described above, then e form the cycle p, a, z, q, p and add it to the system, a contradiction. Therefore, at least one of these three edges as a vital part of a circuit or star in S, and thus the removal of at least one of these three edges must decrease S. First consider the case hen each of the three edges has the property that its removal decreases S. Note, similar arguments can be used to reach a contradiction no matter hich subset of these three edges affects the system hen removed. As before, it must be the case that S a (p) and S z (q) are K 1,3 s in S. The edge az must either be an edge of a K 1,3 or a cycle edge. Suppose, ithout loss of generality, that S a (z) is a K 1,3 in S. We can then assume that S a (p, z) S. Form the cycle p, a, z, q, p hich dominates the edges from S a (p, z) and S z (q). Thus e have incorporated pq into S decreasing S at most 1. Consequently, the edge az must be an edge of a cycle in S. From this cycle e form the ne cycle C : a, p, q, z,..., a here z,..., a represents the verte sequence of the old cycle. Let C dominate the edges from S z (q) and use the edge az to complete the star from hich ap as

R.J. GOULD AND E.A. HYNDS / AUSTRALAS. J. COMBIN. 72 (2) (2018), 164 184 173 removed. Again e have incorporated pq decreasing S at most 1, a contradiction. The arguement that N(b) N(p) \ {q} is empty for all b N(q) I is similar. Claim 3.1.4 The set N(a) N(b) is empty for all {a, b} such that a N(p) I and b N(q) I. Suppose there does eist z N(a) N(b). We kno that p and q do not share any neighbors in I so it follos that z p and z q. Consider the edges ap, az, bz, and bq. We kno all of these edges are in S. No S a (p) and S b (q) are K 1,3 s in S. The edges az and bz are either in a K 1,3 ith center other than z or are a cycle edge in S. Suppose both az and bz are in a K 1,3 in S. Then, based on the ay e formed S thus far, the centers must be a and b respectively. We can easily trade edges beteen stars, if necessary, so that S a (p, z) and S b (q, z) are in S. No form the cycle a, z, b, q, p, a. This cycle dominates the edges of S a (p, z) and S b (q, z). So e have formed a ne cycle that contains the edge pq hile only decreasing S by 1, a contradiction. No suppose that only az is in a K 1,3 and that bz is a cycle edge in our system. Again it must be the case that S a (z) S so e can sap edges, if necessary, to ensure that S a (z, p) S. Remove bz from the cycle that contains it and add it to S b (q). No add to the cycle the path b, q, p, a, z to replace the edge bz, thus forming a larger cycle. We can use the edge bq since e added the edge bz to the star hich originally contained it. The edges pa and az are from S a and the ne cycle dominates the remaining edge of that star. So e have again decreased S by only 1, a contradiction. The case that bz is in a K 1,3 and az is a cycle edge in S is handled similarly. Thus, az and bz must both be cycle edges in S. Begin by moving az to S a (p) and bz to S b (q). No use the edges ap and bq knoing that hat remains after their removal is still a star. If az and bz ere on the same cycle in S e replace the path b, z, a ith the path b, q, p, a, hich gives a ne cycle containing the edge pq. We have left S unchanged. So az and bz must have been on different cycles in S. Both of these cycles contain z so e take hat is left of them after the removal of az and bz and join them at the point z. We ill then add the path b, q, p, a, giving us a ne cycle containing the edge pq. In this case e decreased S by 1, a contradiction. Thus, in all possible cases e reach a contradiction, proving the claim. Claim 3.1.5 If there eists N(p)\{ 0 } such that I, then N( 0 ) N() = p. Suppose there eists N(p) \ { 0 } such that I. Also suppose there eists z p such that z N( 0 ) N(). We consider the edges 0 p, 0 z, p, and z. We kno S 0 (p), S (p) S and the edges 0 z and z are either in a star ith center other than z or are a cycle edge in S. Suppose both 0 z and z are in stars. Then by the ay e formed S, the centers must be 0 and respectively. We can easily trade edges beteen stars, if

R.J. GOULD AND E.A. HYNDS / AUSTRALAS. J. COMBIN. 72 (2) (2018), 164 184 174 necessary, to form S 0 (p, z) and S (p, z). No form the cycle 0, z,, p, 0. This cycle dominates the remaining edges of the stars ith centers 0 and hich e used to form the cycle. So e have formed a ne cycle that contains the verte p hile decreasing S by 1, a contradiction. No suppose that only 0 z is in a star and z is a cycle edge in our system. Again it must be the case that 0 is the center of the star containing 0 z so e can sap edges, if necessary, to form S 0 (z, p). Remove the edge z from the cycle that contains it and add it to S (p). Add to the cycle the path, p, 0, z to replace the edge z, thus forming a larger cycle. We can use the edge p since e added the edge z to the star in hich it originally occured. The ne cycle dominates the remaining edge of S 0 (p, z). Thus, e have decreased S by 1, a contradiction. Similarly, e arrive at a contradiction if e assume instead that only z is in a star and 0 z is a cycle edge in S. Thus, e have that 0 z and z are both cycle edges in S. Begin by moving 0 z to S 0 (p) and z to S (p). We no use the edges 0 p and p, since hat remains after their removal is still a star. If 0 z and z ere on the same cycle in our system e replace the path, z, 0 that e removed ith the path, p, 0 hich gives a ne cycle containing the verte p. Thus, e can incorporate pq and leave S unchanged. Thus, 0 z and z must have been on different cycles in S. Both of these cycles contain z so take hat is left of them after the removal of 0 z and z and join them at the point z. We then add the path, p, 0, giving a ne cycle containing the verte p. In this case e decreased S by 1, a contradiction. Hence, in all of the possible cases e reach a contradiction, thus proving the claim. The net claim is proved in the same manner. Claim 3.1.6 If there eists u N(q)\{y 0 } such that u I, then N(y 0 ) N(u) = q. We no eamine the four cases that arise hen e consider hether or not the hypotheses of Claims 3.1.5 and 3.1.6 hold. Case 1 Suppose the hypotheses of both Claim 3.1.5 and Claim 3.1.6 hold. Then R contains N( 0 ), N(), N(y 0 ), and N(u). By Claim 1 e kno that u. Claim 7 implies that N( 0 ) N() = p. Claim 6 implies that N( 0 ) N(y 0 ) =, N( 0 ) N(u) =, N() N(u) =, and N() N(y 0 ) =. Finally, Claim 8 implies that N(y 0 ) N(u) = q. From these it follos that R N( 0 ) N() N(y 0 ) N(u) 4n 5 2 hich implies that I n/5 + 2. Hoever, e originally assumed that I n/4 hich means that n 40, a contradiction. Case 2 Suppose the hypothesis of Claim 3.1.5 holds but the hypothesis of Claim 3.1.6 does not.

R.J. GOULD AND E.A. HYNDS / AUSTRALAS. J. COMBIN. 72 (2) (2018), 164 184 175 Then R contains N( 0 ), N(), N(y 0 ), and N(q) \ {y 0 }. Claim 7 implies that N( 0 ) N() = p. Claim 6 implies that N( 0 ) N(y 0 ) =, and N() N(y 0 ) =. Claim 4 implies that N( 0 ) N(q) \ {y 0 } = p, and N() N(q) \ {y 0 } = p. Finally, Claim 3 implies that N(y 0 ) N(q) \ {y 0 } =. From these it follos that R N( 0 ) N() N(y 0 ) N(q) \ {y 0 } 4n 5 3 hich implies that I n/5 + 3. Hoever, e originally assumed that I n/4 hich means that n 60, a contradiction. Case 3 Suppose the hypothesis of Claim 3.1.6 holds but the hypothesis of Claim 3.1.5 does not. Then R contains N( 0 ), N(p) \ { 0 }, N(y 0 ), and N(u). Claim 8 implies that N(y 0 ) N(u) = q. Claim 6 implies that N( 0 ) N(y 0 ) =, and N( 0 ) N(u) =. Claim 5 implies that N(y 0 ) N(p) \ { 0 } = p, and N(u) N(p) \ { 0 } = q. Finally, Claim 2 implies that N( 0 ) N(p) \ { 0 } =. From these it follos that R N( 0 ) N(p) \ { 0 } N(y 0 ) N(u) 4n 5 3 hich implies that I n/5 + 3. Hoever, e originally assumed that I n/4 hich means that n 60, a contradiction. Case 4 Suppose neither the hypothesis of Claim 3.1.5 nor the hypothesis of Claim 3.1.6 holds. Then R contains N( 0 ), N(p) \ { 0 }, N(y 0 ), and N(q) \ {y 0 }. Claim 6 implies that N( 0 ) N(p) \ { 0 } =. Claim 5 implies that N(y 0 ) N(p) \ { 0 } = q. Claim 4 implies that N( 0 ) N(q) \ {y 0 } = p. Claim 3 implies that N(y 0 ) N(q) \ {y 0 } =. Claim 2 implies that N( 0 ) N(p) \ { 0 } =. Finally Claim 1 implies that N(p) \ { 0 } N(q) \ {y 0 } =. From these it follos that R N( 0 ) N(p) \ { 0 } N(y 0 ) N(q) \ {y 0 } 4n 5 2 hich implies that I n/5 + 2. Hoever, e originally assumed that I n/4 hich means that n 40, a contradiction. In all cases e get a contradiction hich means that if e can add one of the remaining edges to S then e can add it losing at most one from the size of our system. Of course it is important to us that e have a dominating system. In other ords, e ant to ensure that e can incorporate all remaining edges into our system, S, creating a dominating system. Suppose, again by ay of contradiction, that there is an edge pq that e cannot add to the system. Then each of the above claims ould hold leading us to the same contradictions, shoing that it is impossible to have an edge e cannot incorporate. So e kno e can add each of the remaining edges to

R.J. GOULD AND E.A. HYNDS / AUSTRALAS. J. COMBIN. 72 (2) (2018), 164 184 176 S, decreasing S by at most 1 each time. We no kno that e have a dominating system in G, but it still remains to sho that is the size e claimed it ould be. Recall that e started our process ith at least cn δ stars. It follos then that 3 the size of the system is at least cn δ M here M is the maimum number of 3 edges that could have been remaining in R. Choose I and let d = deg() δ n/5. Let t be the number of vertices in N() that are not on one of the remaining paths. The (d t) vertices that are on paths must be on different paths because no to vertices on our remaining paths have a common neighbor in I. Consider an edge from each of these different paths. Then e have at least 2(d t) vertices and d t edges so far. We have no accounted for t + 2(d t) = 2d t vertices of R. Because R (1 c)n e kno that there are at most (1 c)n (2d t) vertices left hich can each contribute at most 1 of the remaining edges. When e combine this ith the d t edges from before e see that e have at most [(1 c)n (2d t)] + [d t] edges comprising the disjoint paths. This gives us that the number of edges remaining in R is at most (1 c)n δ. So e have a dominating system of size at least cn δ [(1 c)n δ] in G hich means 3 e have a 2-factor of size at least cn δ [(1 c)n δ] in L(G), as claimed. 3 Theorem 3.2 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 a 2-factor ith k cycles then L(G) has a 2-factor ith k 1 cycles. Proof: Suppose by ay of contradiction that, for some k n/10 + 2, L(G) has a 2-factor ith k cycles but does not have a 2-factor ith k 1 cycles. Then e kno our graph G has a dominating k-system but no dominating (k 1)-system. The folloing observations are easily seen: No verte can appear on more than one circuit of the dominating k-system or e could connect the to circuits to form a dominating (k 1)-system. No verte can be the center of more than one star of the dominating k-system. No verte can be the center of a star and on a circuit of the dominating k- system. If there is an edge e beteen to leaves of a star then it must be a dominated edge in our system. No to 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 not on any circuit. Otherise e could distribute all edges of the star to other elements in the system and form a dominating (k 1)-system. We no consider cases based on the structure of the dominating k-system. Case 1 The graph G contains a dominating k-system that consists entirely of circuits.

R.J. GOULD AND E.A. HYNDS / AUSTRALAS. J. COMBIN. 72 (2) (2018), 164 184 177 Among all such k-systems in G, choose one, say S k, that has the maimum number of distinct vertices on circuits. In S k there must be a circuit ith 9 or feer distinct vertices. Otherise, e see that 10k n or k n 10, a contradiction. Consider a circuit C in S k ith the feest number of distinct vertices. No choose a pair of adjacent vertices and y on C. Recall that neither nor y can appear on another circuit and note that they can each have at most 8 neighbors in C. Thus and y each dominate at least n 5 8 vertices off C in S k. Let N = N() \ C and N y = N(y) \ C. Then N n 5 8 and N y n 5 8. Claim 3.2.1 The set N N y is empty. Suppose there eist z N N y. Replace the edge y of C ith the path, z, y to form a ne circuit C that no dominates the edge y. If z appears on another circuit then e attach this circuit to C and form a dominating (k 1)-system, a contradiction. If z is not on a circuit then e contradict the ay e chose the original system. Claim 3.2.2 The sets N and N y are independent. Suppose this is not the case for N. Let u, v N, u v, such that uv E(G). The edge uv must appear somehere in S k. Note that u and v 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,, v, u and use it to connect the circuit containing and the circuit containing v. This produces a dominating (k 1)- system, a contradiction. (ii) Suppose uv is in a circuit. Then replace the edge uv ith the path u,, v to form a ne circuit that dominates uv. We no have 2 circuits containing, hich means e can again form a dominating (k 1)-system, a contradiction. The independence of N y is proved similarly. Claim 3.2.3 The set N N y is independent. Because N and N y are independent, e need only to sho that there is not an edge from N to N y. Again, by ay of contradiction, suppose there eists u N and v N y such that uv E(G). The edge uv must appear somehere in our system. (i) If the edge uv is dominated, ithout loss of generality assume that the verte v is on a circuit. Then in C replace y ith the path, u, v, y to form a ne circuit that no dominates y. No 2 circuits contain the verte v hich allos us to form a dominating (k 1)-system, a contradiction.

R.J. GOULD AND E.A. HYNDS / AUSTRALAS. J. COMBIN. 72 (2) (2018), 164 184 178 (ii) If uv is on a circuit, then e ill use u and yv to connect the circuits containing uv and y to form one circuit that no dominates y and uv, giving us a dominating (k 1)-system, a contradiction. From our claims e see that e have an independent set I = N N y such that I 2n/5 16. Let a, b N, (a b). No suppose that there eists a verte N(a) N(b)( ). Subcase 1 The verte is on the circuit C containing. Because a and b are not on C and because is not on any other circuit ecept C, it must be the case that the edges a and b are dominated edges in our system. We use the edges a, b, a, and b to form the cycle, a,, b,, hich e attach to C. If either a or b are on another circuit in our system then e have 2 circuits that share a verte, hence e can form a dominating (k 1)-system. Thus, neither a nor b as on any circuit in our original system hich means by adding them to C e contradict our original choice of system. Consequently, the verte cannot be on the circuit that contains. Subcase 2 The verte is not on the circuit containing. (i) If both of the edges a and b are dominated edges in our system, then again e form the cycle, a,, b, hich e attach to the circuit containing. All of the vertices a, b, must be on some circuit in our system, or by adding them to the circuit containing, e contradict our original choice of system. But then e have to circuits ith a common verte hich allos us to form a dominating (k 1)-system, also a contradiction. (ii) If eactly one of a and b is a circuit edge in our system, then assume ithout loss of generality that a is dominated and that b is in a circuit of our system. No e replace b ith the path b,, a, to form a ne circuit that dominates b. No to different circuits contain the verte hich allos us to form a dominating (k 1)-system, a contradiction. (iii) If both of a and b are circuit edges in our system, then they must be on the same circuit, as cannot appear on to different circuits. If the circuit containing a and b can be divided in such a ay that a and b are on different cycles, then e can remove a and b from the circuit and attach the path a,, b to form a ne circuit that no dominates a and b and contains the verte. We no have 2 different circuits that contain the verte hich allos us to form a dominating (k 1)-system, a contradiction. The only remaining possibility for a and b is that they are both edges of the same circuit in our system and that this circuit cannot be divided as above. This is the situation e get if any 2 elements of N have a common neighbor other than.

R.J. GOULD AND E.A. HYNDS / AUSTRALAS. J. COMBIN. 72 (2) (2018), 164 184 179 Similar arguments sho that if any to elements of N y, say a and b, have a common neighbor other than y, say, then a and b are both edges of the same circuit in our system, hich is not the circuit containing y, and this circuit cannot be divided in such a ay that the edges a and b are on different cycles of the circuit. No suppose that there eists a N and b N y such that there eists a verte N(a) N(b) here, y. Then, similar arguments sho that the edges a and b are both edges of the same circuit in our system, hich is not the circuit containing and y, and this circuit cannot be divided in such a ay that the edges a and b are on different cycles of the circuit. The only difference in the arguments in this case is that here e ere attaching a ne circuit either to or y in the previous cases, e are no combining the circuit containing a and b ith the circuit containing y to form one large circuit that in addition to dominating a and b as before, no dominates y as ell. As before e alays get a dominating (k 1)-system unless e have the situation described above in (iii). Our conclusion is that if any to vertices of the set I have a common neighbor other than or y e get the situation described above in (iii). No suppose that three vertices a, b, c of I have a common neighbor such that, y. By our previous arguments, a and b are both edges of the same circuit in our system, hich is not the circuit containing and y, and that this circuit cannot be factored in such a ay that the edges a and b are on different cycles of the circuit. But c must also be on the same cycle of the circuit as a and b, hich is impossible. So no three vertices of I have a common neighbor other than or y. Recall that our set I = N N y is such that I 2n/5 16. Let R be the remaining vertices of V (G) that are not in the set I. Consider the set R \ {, y} that has size at most 3n/5 14. By our previous arguments no three vertices of I can have a common neighbor in R \ {, y}. Let e be the number of edges beteen the sets I and R \ {, y}. Since I is independent and each verte of I is adjacent to only one of and y, each verte of I sends at least n/5 1 edges to the set R \ {, y}. Thus, there are at least (n/5 1)(2n/5 16) edges from I to R \ {, y}, that is, e 2n2 25 18n 5 + 16. No there are at most 3n/5 + 14 vertices in R \ {, y} that must receive these edges but each verte in R \ {, y} receives at most to edges from I. Hence, there are at most 2( 3n + 14) edges from R \ {, y} to I. In other ords, e 6n + 28. 5 5 It follos that 2n 2 25 18n 5 + 16 6n 5 + 28, hence, n 2 60n 150 0, hich implies n < 63, a contradiction. Case 2 Every dominating k-system of G contains at least one star. Choose a dominating k-system of G, say S k, in such a ay that the system contains a maimum number of circuits. By our assumption, S k must contain at

R.J. GOULD AND E.A. HYNDS / AUSTRALAS. J. COMBIN. 72 (2) (2018), 164 184 180 least one star. Suppose, for a contradiction, there eists a star S z ( 1, 2, 3, 4, 5 ) ith at least five leaves. Note that for all i j, the edge i j cannot be in our graph. If so, it ould be a dominated edge and e could use it to form a k-system ith one more circuit, hich contradicts the ay e chose our original system. Lemma 2 implies that for every i j, N( i ) N( j ) = z. This implies that n { 1, 2, 3, 4, 5, z} + N( 1 ) + N( 2 ) + N( 3 ) + N( 4 ) + N( 5 ) and thus n n + 1, a contradiction. So it must be the case that each star in our system has 3 or 4 edges. We ill sho a contradiction is reached in either case. While the setup for these cases is the same, the arguments are different so e ill handle them separatetly. Suppose e do have a star in our system ith four edges, say S z ( 1, 2, 3, 4 ). Let N = {N( 1 ) N( 2 ) N( 3 ) N( 4 )}\{z}. By Lemma 2 e kno that for every i j, N( i ) N( j ) = z hich means that N 4n/5 4. Let X = { 1, 2, 3, 4 }. We kno that z X and z N. Again, by the same argument as above, i j E(G) hich means N X =. This implies that e have accounted for a total of 4n/5 + 1 vertices in G. No z must have at least n/5 4 neighbors in G X. None of the other edges adjacent to z can be dominated edges in our system or e could add them to S z giving us a star ith 5 edges or more, hich e cannot have. The other edges also cannot be in a circuit because as a center of a star, z cannot appear on a circuit. There is not another star ith center z, so the remaining edges adjacent to z must be in stars ith center other than z. Let C = {c 1, c 2,..., c l } be the neighbors of z not in the set X. Note that l n/5 4. We kno that each of the vertices in C is the center of a star ith leaf z. Also e kno that X C = or e ould have to stars in our system that share a pair of vertices. We ill no argue that the set N C is empty. Suppose, by ay of contradiction, that c i j E(G), for some i {1, 2,..., l} and j {1, 2, 3, 4}. This edge cannot appear in the star ith center c i or e have to stars that share j and z. It cannot be in a circuit or dominated by a circuit containing c i as c i cannot be on a circuit. If it is dominated by a circuit containing j then e can use the edge c i j to combine the stars containing z j and c i z into a cycle that dominates any edges in the stars not a part of the ne cycle, thus forming a dominating (k 1)-system. So it must be the case that the edge c i j is found in a star ith center j. Because e kno the star containing z j has 4 edges e can use the edge z j to combine the stars containing the edges c i j and c i z into a cycle that dominates the edges of the stars and thus forms a dominating (k 1)-system, again a contradiction. It follos that c i j / E(G), thus N C is empty. What e have shon is that none of the vertices in the set C ere included in the 4n/5 + 1 vertices that e had accounted for before. So no, ith the sets X, N, C, and the verte z e have accounted for n 3 different vertices. Consider one last set of vertices. Recall that for all i {1, 2,..., l}, c i is the center of a star ith z as a leaf. As z is the center of a star, in each of these stars there is a leaf other than z that is not the center of a star and is not on a circuit.

R.J. GOULD AND E.A. HYNDS / AUSTRALAS. J. COMBIN. 72 (2) (2018), 164 184 181 For each i {1, 2,..., l} let y i be the leaf in the star ith center c i that meets these qualifications and let Y = {y 1, y 2,..., y l }, l n/5 4. Hence, Y n/5 4. By our choice of each y i e kno that z Y. Then Y C is empty or e ould have to different stars that contain the pair c i and z. Also, Y X is empty or e ould have to different stars that contain the pair i and z. No e ill sho that the set N Y is also empty. Suppose, by ay of contradiction, y i j E(G), for some i {1, 2,..., l} and j {1, 2, 3, 4}. This edge cannot appear in a star ith center y i, in a circuit, or as an edge dominated by a circuit containing y i by the ay e chose y i. If y i j is an edge dominated by a circuit containing j then e can use it to combine the stars containing c i z and z j into a cycle that dominates the edges of the stars and thus gives us a dominating (k 1)-system. It remains then that the edge y i j must be found in a star ith center j. Because e kno the star containing z j has 4 edges e can use the edge z j to combine the stars containing the edges y i j and c i z into a cycle that dominates the edges of the stars and thus forms a dominating (k 1)-system, again a contradiction. Thus, y i j / E(G) and thus N Y is empty. So, none of the vertices in the set Y ere included in the n 3 vertices that e had previously accounted for in the sets X, N, C, and the verte z. We no have at least (n 3) + (n/5 4) different vertices that e have accounted for hich implies that n 6n/5 7 or n 35 hich is a contradiction. Hence, it follos that e cannot have any stars in our system that contain edges edges; thus, every star in our system contains eactly three edges. We are no in the situation that e kno our system must have at least one star and that every star must have eactly three edges. Choose any star S z ( 1, 2, 3 ) in S. The set-up is similar to the case hen e assumed our system contained a star ith four edges. Let N = {N( 1 ) N( 2 ) N( 3 )} \ {z}. By Lemma 2 e kno that for every i j, N( i ) N( j ) = z hich means that N 3n/5 3. Let X = { 1, 2, 3 }. We kno that z X and z N. As in the previous case, N X =. This implies that e already have accounted for a total of 3n/5 + 1 of the vertices in our graph G. No the verte z must have at least n/5 3 other neighbors in G besides those in the set X. None of the edges adjacent to z can be dominated edges in our system or e could add them to the star centered at z giving us a star ith 4 edges or more hich e cannot have. The edges also cannot be in a circuit because as a center of a star the verte z cannot appear on a circuit. There is not another star ith center z so the remaining edges that are adjacent to z must be found in stars ith center other than z. Let C = {c 1, c 2,..., c l } be the neighbors of z not in the set X. Note that l n/5 3. We kno that each of the vertices in C is the center of a star ith leaf z. Clearly the verte z is not an element of C. Also e kno that X C = or e ould have to stars in our system that shared a pair of vertices. As before, e ill sho that the set N C is empty. Suppose, by ay of contradiction, that c i j E(G), for some i {1, 2,..., l} and j {1, 2, 3}. This edge cannot appear in the star ith center c i or e have to stars that share j and z. It cannot be in a circuit or dominated by a circuit

R.J. GOULD AND E.A. HYNDS / AUSTRALAS. J. COMBIN. 72 (2) (2018), 164 184 182 containing c i as c i cannot be on a circuit. If it is dominated by a circuit containing j then e can add the edge c i j to S ci to form a star ith four edges hich e have shon e cannot have. So it must be the case that the edge c i j is found in a star ith center j. We can no use the three stars hich contain the edges j c i, c i z, and z j to form a type 4 dominating (k 2)-system, hich by Lemma 1 implies that G contains a dominating (k 1)-system. It follos that the edge c i j cannot be in our graph G and thus the set N C is empty. What e have shon is that none of the vertices in the set C ere included in the 3n/5 + 1 vertices that e had accounted for before. So no, ith the sets X, N, C, and the verte z e have accounted for 4n/5 3 different vertices. We ill no consider to more sets of vertices. Recall that for all i {1, 2,..., l}, the verte c i is the center of a star ith verte z as a leaf. We kno that each star in our system has 3 degree one vertices hich means that each of these stars centered at c i contains 2 other vertices besides c i and z. So for each i {1, 2,..., l} let v i and y i be the degree one vertices in the star centered at c i other than z. And let V = {v 1, v 2,..., v l } and Y = {y 1, y 2,..., y l }. Recall that l n/5 3 hich means that V n/5 3 and Y n/5 3. By our choice of each v i and y i e kno that z V and z Y. It must be the case that V X is empty and Y X is empty or e ould have to different stars in our system that contain the pair of vertices i and z. It also must be the case that V C is empty and Y C is empty or e ould have to different stars in our system that contain the pair of vertices c i and z. What remains for us to sho is that the sets N Y, N V, and V Y are empty. Starting ith N Y, suppose, by ay of contradiction, that there does eist a verte y i N( j ) for some i {1, 2,..., l} and j {1, 2, 3}. In other ords, y i j is an edge of our graph G. This edge must appear somehere in our system and e ill consider each possibility. In each case e ill use the stars S z ( j ) and S ci (z, y i ). If the edge y i j is dominated in our system then e ill use it to combine S z and S ci to form the cycle z, c i, y i, j, z hich dominates the edges of the stars giving us a dominating (k 1)-system, a contradiction. If the edge y i j is in a star centered at j then e can combine this star ith S z and S ci to form a type 2 (k 2)-system, hich by Lemma 1 means G contains a dominating (k 1)-system. If the edge y i j is in a star centered at y i then e can combine this star ith S z and S ci to form a type 1 (k 2)-system, hich by Lemma 1 means G contains a dominating (k 1)-system. If the edge y i j is in a circuit then e can combine this circuit ith S z and S ci to form a type 3 (k 2)-system, hich by Lemma 1 means G contains a dominating (k 1)-system. In each case e get a contradiction, hich means that the set N Y must be empty. By the same argument e also get that the set N V is empty. If the set V Y is not empty that means that y i = v j for some i j. But that means that the vertices z and v j appear together in to different stars hich cannot happen, so the set V Y must also be empty. Let us revie hat e have shon. We have defined five sets, X, N, C, Y and V, and e have shon that the intersection of any to of them is empty. We have also shon that the verte z is not an element of any of the five sets. The result is that

R.J. GOULD AND E.A. HYNDS / AUSTRALAS. J. COMBIN. 72 (2) (2018), 164 184 183 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 hich is a contradiction that arises from our original assumption that each star in our system has eactly three edges. This leads us to the conclusion that G cannot contain a dominating k-system that contains any stars. We began by supposing that for some k n/10 + 2, that L(G) has a 2- factor ith k cycles but does not have a 2-factor ith k 1 cycles. That led us immediately to the assumption that our graph G has a dominating k-system but no dominating (k 1)-system. Based on this assumption e then shoed that G could not contain a dominating k-system that consists entirely of circuits and that it also could not contain a dominating k-system that contains any stars. The resulting conclusion then is that G cannot contain a dominating k-system. This is obviously false hich implies that our supposition that G contains a dominating k-system but no dominating (k 1)-system cannot hold. So for every k n/10 + 2 if L(G) has a 2-factor ith k cycles then L(G) has a 2-factor ith k 1 cycles, thus proving the theorem. 4 Conclusion Our main goal as to prove a generalization of Catlin s orginal theorem hich gives conditions on a graph G that imply its line graph is Hamiltonian. We began ith a small etension. Theorem 1.2 If G is a 2-edge-connected simple graph of order n such that δ(g) n/5 then L(G) has a 2-factor ith k cycles for k = 1,..., n/10. We then set out to prove the larger generalization. Theorem 1.3 If G is a 2-edge-connected simple graph of order n > 65 ith δ(g) n/5 and α(g) = cn, for some c, 1/4 c < 1, then L(G) contains a 2-factor ith k cycles, for each k = 1, 2,..., cn δ [(1 c)n δ]. 3 We finish ith a graph that shos our bound in Theorem 1.3 is the best possible. Consider the complete bipartite graph G = K n/2,n/2. We kno that α(g) = n/2 and δ(g) = n/2. From Theorem 1.3 e see that L(G) has a 2-factor ith k cycles for each k = 1, 2,..., n 2 /12. Note that this is best possible since E(G) = n 2 /4 hich implies that V (L(G)) = n 2 /4 and so it ould be impossible to have any more than n 2 /12 disjoint cycles in L(G).