On Vertex-Disjoint Chorded Cycles and Degree Sum Conditions

Ronald J. Gould1, Kazuhide Hirohata2, Ariel Keller Rorabaugh3
1Department of Mathematics, Emory University, Atlanta, GA 30322 USA.
2Department of Industrial Engineering, Computer Science, National Institute of Technology, Ibaraki College, Hitachinaka, Ibaraki 312-8508 Japan.
3Department of Electrical Engineering and Computer Science, University of Tennessee, Knoxville, TN 37996 USA.

Abstract

In this paper, we consider a degree sum condition sufficient to imply the existence of k vertex-disjoint chorded cycles in a graph G.
Let σ4(G) be the minimum degree sum of four independent vertices of G.
We prove that if G is a graph of order at least 11k+7 and σ4(G)12k3 with k1,
then G contains k vertex-disjoint chorded cycles.
We also show that the degree sum condition on σ4(G) is sharp.

Keywords: vertex-disjoint chorded cycles, minimum degree sum, degree sequence

1. Introduction

The study of cycles in graphs is a rich and an important area. One question of particular interest is to find conditions that guarantee the existence of k vertex-disjoint cycles. Corrádi and Hajnal [1] first considered a minimum degree condition to imply a graph must contain k vertex-disjoint cycles, proving that if |G|3k and the minimum degree δ(G)2k, then G contains k vertex-disjoint cycles. For an integer t1, let σt(G)=min{vXdG(v):X is an independent vertex setof G with |X|=t.}, and σt(G)= when the independence number α(G)<t. Enomoto [2] and Wang [3] independently extended the Corrádi and Hajnal result, requiring a weaker condition on the minimum degree sum of any two non-adjacent vertices. They proved that if |G|3k and σ2(G)4k1, then G contains k vertex-disjoint cycles. In 2006, Fujita et al. [4] proved that if |G|3k+2 and σ3(G)6k2, then G contains k vertex-disjoint cycles, and in [5], this result was extended to σ4(G)8k3.

An extension of the study of vertex-disjoint cycles is that of vertex-disjoint chorded cycles. A chord of a cycle is an edge between two non-adjacent vertices of the cycle. We say a cycle is chorded if it contains at least one chord. In 2008, Finkel proved the following result on the existence of k vertex-disjoint chorded cycles.

Theorem 1. (Finkel [6]) Let k1 be an integer. If G is a graph of order at least 4k and δ(G)3k, then G contains k vertex-disjoint chorded cycles.

In 2010, Chiba et al. proved Theorem 2. Since σ2(G)2δ(G), Theorem 2 is stronger than Theorem 1.

Theorem 2 (Chiba, Fujita, Gao, Li [7]). Let k1 be an integer. If G is a graph of order at least 4k and σ2(G)6k1, then G contains k vertex-disjoint chorded cycles.

Recently, Theorem 2 was extended as follows. Since σ3(G)3σ2(G)/2, when the order of G is sufficiently large, Theorem 3 is stronger than Theorem 2.

Theorem 3 (Gould, Hirohata, Rorabaugh [8]). Let k1 be an integer. If G is a graph of order at least 8k+5 and σ3(G)9k2, then G contains k vertex-disjoint chorded cycles.

Remark 1. We note if k=1 in Theorem 3, then Theorem 3 holds under the condition that |G|7.

In this paper, we consider a similar extension for chorded cycles, as, in [5], the existence of k vertex-disjoint cycles was proved under the condition σ4(G). In particular, we first show the following.

Theorem 4. If G is a graph of order at least 15 and σ4(G)9, then G contains a chorded cycle.

Remark 2. We consider the following graph G of order 14. (See Fig. 1.) Then G satisfies the σ4(G) condition in Theorem 4. However, G does not contain a chorded cycle. Thus |G|15 is necessary.

Figure 1. The Graph of G of Order 14. The White Vertex (o) shows Degree 2, and the Black Vertex (o) shows Degree 3

Theorem 5. Let k1 be an integer. If G is a graph of order n11k+7 and σ4(G)12k3, then G contains k vertex-disjoint chorded cycles.

Remark 3. Theorem 5 is sharp with respect to the degree sum condition. Consider the complete bipartite graph G=K3k1,n3k+1, where large n=|G|. Then σ4(G)=4(3k1)=12k4. However, G does not contain k vertex-disjoint chorded cycles, since any chorded cycle must contain at least three vertices from each partite set, in particular, from the 3k1 partite set. Thus σ4(G)12k3 is necessary.

For other related results on vertex-disjoint chorded cycles in graphs and bipartite graphs, we refer the reader to see [9-12].

Let G be a graph, H a subgraph of G and XV(G). For uV(G), the set of neighbors of u in G is denoted by NG(u), and we denote dG(u)=|NG(u)|. For uV(G), we denote NH(u)=NG(u)V(H) and dH(u)=|NH(u)|. Also we denote dH(X)=uXdH(u). If H=G, then dG(X)=dH(X). Furthermore, NG(X)=uXNG(u) and NH(X)=NG(X)V(H). Let A,B be two vertex-disjoint subgraphs of G. Then NG(A)=NG(V(A)) and NB(A)=NG(A)V(B). The subgraph of G induced by X is denoted by X. Let GX=V(G)X and GH=V(G)V(H). If X={x}, then we write Gx for GX. If there is no fear of confusion, then we use the same symbol for a graph and its vertex set. For two disjoint graphs G1 and G2, G1G2 denotes the union of G1 and G2. Let Q be a path or a cycle with a given orientation and xV(Q). Then x+ denotes the first successor of x on Q and x denotes the first predecessor of x on Q. If x,yV(Q), then Q[x,y] denotes the path of Q from x to y (including x and y) in the given direction. The reverse sequence of Q[x,y] is denoted by Q[y,x]. We also write Q(x,y]=Q[x+,y], Q[x,y)=Q[x,y] and Q(x,y)=Q[x+,y]. If Q is a path (or a cycle), say Q=x1,x2,,xt(,x1), then we assume an orientation of Q is given from x1 to xt (if Q is a cycle, then the orientation is clockwise). If P is a path connecting x and y of V(G), then we denote the path P as P[x,y]. If G is one vertex, that is, V(G)={x}, then we simply write x instead of G. For an integer r1 and two vertex-disjoint subgraphs A,B of G, we denote by (d1,d2,,dr) a degree sequence from A to B such that dB(vi)di and viV(A) for each 1ir. In this paper, since it is sufficient to consider the case of equality in the above inequality, when we write (d1,d2,,dr), we assume dB(vi)=di for each 1ir. For two disjoint X,YV(G), E(X,Y) denotes the set of edges of G connecting a vertex in X and a vertex in Y. For a graph G, comp(G) is the number of components of G. A cycle of length is called a -cycle. For terminology and notation not defined here, see [13].

2. Preliminaries

Definition 1. Suppose C1,,Cr are r vertex-disjoint chorded cycles in a graph G. We say {C1,,Cr} is minimal if G does not contain r vertex-disjoint chorded cycles C1,,Cr such that |i=1rV(Ci)|<|i=1rV(Ci)|.

Definition 2. Let C=v1,,vt,v1 be a cycle with chord vivj, i<j. We say a chord vvvivj is parallel to vivj if either v,vC[vi,vj] or v,vC[vj,vi]. Note if two distinct chords share an endpoint, then they are parallel. We say two distinct chords are crossing if they are not parallel.

Definition 3. Let uivj and uvm be two distinct edges between two vertex-disjoint paths P1=u1,,us and P2=v1,,vt. We say uivj and uvm are parallel if either i and jm, or i and mj. Note if two distinct edges between P1 and P2 share an endpoint, then they are parallel. We say two distinct edges between two vertex-disjoint paths are crossing if they are not parallel.

Definition 4. Let vivj and vvm be two distinct edges between vertices of a path P=v1,,vt, with ji+2 and m+2. We say vivj and vvm are nested if either i<mj or i<jm.

Definition 5. Let P=v1,,vt be a path. We say a vertex vi on P has a left edge if there exists an edge vivj for some j<i1, that is not an edge of the path. We also say vi has a right edge if there exists an edge vivj for some j>i+1, that is not an edge of the path.

3. Lemmas

The following lemmas will be needed.

Lemma 1 ([8]). Let r1 be an integer, and let C={C1,,Cr} be a minimal set of r vertex-disjoint chorded cycles in a graph G. If |Ci|7 for some 1ir, then Ci has at most two chords. Furthermore, if the Ci has two chords, then these chords must be crossing.

Lemma 2 ([8]). Let r1 be an integer, and let C={C1,,Cr} be a minimal set of r vertex-disjoint chorded cycles in a graph G. Then dCi(x)4 for any 1ir and any xV(G)i=1rV(Ci). Furthermore, for some CC and some xV(G)i=1rV(Ci), if dC(x)=4, then |C|=4, and if dC(x)=3, then |C|6.

Lemma 3 ([8]). Suppose there exist at least three mutually parallel edges or at least three mutually crossing edges connecting two vertex-disjoint paths P1 and P2. Then there exists a chorded cycle in P1P2.

Lemma 4 ([8]). Suppose there exist at least five edges connecting two vertex-disjoint paths P1 and P2 with |P1P2|7. Then there exists a chorded cycle in P1P2 not containing at least one vertex of P1P2.

Lemma 5 ([8]). Let P1,P2 be two vertex-disjoint paths, and let u1,u2 (u1u2) be in that order on P1. Suppose dP2(ui)2 for each i{1,2}. Then there exists a chorded cycle in P1[u1,u2]P2.

Lemma 6 ([8]). Let H be a graph containing a path P=v1,,vt (t3), and not containing a chorded cycle. If v1viE(H) for some i3, then dP(vj)3 for any ji1 and in particular, dP(vi1)=2. And if vtviE(H) for some it2, then dP(vj)3 for any ji+1 and in particular, dP(vi+1)=2.

Lemma 7 ([8]). Let H be a graph containing a path P=v1,,vt (t6), and not containing a chorded cycle. If dP(v1)=1, then dP(vi)=2 for some 3i5, and if v1v3E(H), then dP(vi)=2 for some 4i6.

Lemma 8 ([8]). Let H be a graph containing a path P=v1,,vt (t6), and not containing a chorded cycle. If dP(vt)=1, then dP(vi)=2 for some t4it2, and if vtvt2E(H), then dP(vi)=2 for some t5it3.

Lemma 9. Let H be a connected graph of order at least 6. Suppose H contains neither a chorded cycle nor a Hamiltonian path. Let H=P1P2, where P1=u1,,us (s5) is a longest path in H and P2=v1,,vt (t1) is a longest path in HP1. If uiV(P1) for some 2is3 is adjacent to an endpoint v of P2 and ujV(P1) for some i+2js1 is adjacent to an endpoint v of P2 (possibly, v=v), then dH(u)=2 for some {i+1,j1}.

Proof. Let v,v be as in the lemma, and we may assume v=v1 and v=vt (possibly, v=v). Suppose dH(u)3 for each {i+1,j1}. If ui+1 has a left edge, say ui+1uh with h<i, then P1[uh,ui],v1,P2[v1,vt],uj,P1[uj,ui+1],uh is a cycle with chord uiui+1, a contradiction. By symmetry, uj1 does not have a right edge. Since uiv1,ujvtE(H), NP2(u)= for each {i+1,j1}, otherwise, since consecutive vertices on P1 each have adjacencies on P2, there exists a longer path than P1 in H, a contradiction. Note that even if v=v, NP2(u)= for each {i+1,j1}. Since dH(u)3 for each {i+1,j1}, ui+1 has a right edge and uj1 has a left edge. No vertex in P1[ui,uj] can have an edge that does not lie on P1 to some other vertex in P1[ui,uj], otherwise, this edge is a chord of the cycle P1[ui,uj],vt,P2[vt,v1],ui. Thus we have edges ui+1uh with h>j, and uj1uh with h<i. Then P1[uh,ui],v1,P2[v1,vt],uj,P1[uj,uh],ui+1,P1[ui+1,uj1],uh is a cycle with chord uiui+1 (and uj1uj), a contradiction. Thus the lemma holds. ◻

Lemma 10 ([8]). Let H be a graph of order at least 13. Suppose H does not contain a chorded cycle. If H contains a Hamiltonian path, then there exists an independent set X of four vertices in H such that dH(X)8.

Lemma 11 ([8]). Let H be a connected graph of order at least 4. Suppose H contains neither a chorded cycle nor a Hamiltonian path. Let P1=u1,,us (s3) be a longest path in H, and let P2=v1,,vt (t1) be a longest path in HP1. Then the following statements hold.
(i) NHP1(ui)= for each i{1,s}.
(ii) dH(ui)=dP1(ui)2 for each i{1,s}.
(iii) NH(P1P2)(vj)= for each j{1,t}.
(iv) dP2(vj)2 for each j{1,t}.
(v) dPi(z)2 for each zV(H)V(Pi) and each i{1,2}.
(vi) dP1({v1,vt})3 for each t2.

Proofs of (v) and (vi). Note parts (i) to (iv) are from [8], hence we only prove parts (v) and (vi). Since H does not contain a chorded cycle, (v) holds. Suppose dP1({v1,vt})4. By (v), dP1(vj)=2 for each j{1,t}. Then, by Lemma 5, H has a chorded cycle, a contradiction. Thus (vi) holds. 0◻

Lemma 12. Let H be a connected graph of order at least 15. Suppose H contains neither a chorded cycle nor a Hamiltonian path. Let P1=u1,,us (s3) be a longest path in H, and let P2=v1,,vt (t1) be a longest path in HP1 such that dP1(v1)dP1(vt). Then there exists an independent set X of four vertices in H such that {u1,us,v1}X and dH(X)8.

Remark 4. Let H be a graph of order 14 shown in Fig. 1 (Remark 2, Theorem 4), P1=u1,,u11, and P2=v1,v2,v3. Then H satisfies all the conditions except for the order in Lemma 12. However, the conclusion does not hold. Thus |H|15 is necessary.

Proof. Suppose u1usE(H). Since H is connected and V(HP1), there exists a longer path than P1, a contradiction. Thus u1usE(H). Let R=H(P1P2). If t=1, that is, v1=vt, then dP1(v1)2 by Lemma 11(v). If t2, then dP1({v1,vt})3 by Lemma 11(vi). Then dP1(v1)1 by the assumption (dP1(v1)dP1(vt)), and dP1(vt)2 by Lemma 11(v).

Claim 6. If |P2|3, then H=P1P2.

Proof. Suppose HP1P2. Now we prove the following two subclaims.

Subclaim 7. For any vV(P2), NR(v)=.

Proof. By Lemma 11(iii), NR(vj)= for each j{1,t}. If |P2|2, then the subclaim holds. Thus we may assume |P2|=3. Suppose NR(v) for some vV(P2). Then v=v2. Let w1NR(v2). If v1v3E(H), then the subclaim holds, otherwise, there exists a longer path than P2 in HP1, a contradiction. Thus v1v3E(H). Since dP1(v1)1 and dP1(v3)2, we have dH(v1)2 and dH(v3)3. Suppose a vertex on P2 has a neighbor w1 in R. Then v2w1E(H). Recall u1usE(H), and note uivjE(H) for any i{1,s} and any j{1,3} by Lemma 11(i). We also note dH(ui)2 for any i{1,s} by Lemma 11(ii). If dH({v1,v3})4, then X={u1,us,v1,v3} is an independent set in H and dH(X)8, and X is the desired set. Thus we may assume dH({v1,v3})=5, that is, dH(v1)=2 and dH(v3)=3. Then dP1(v1)=1 and dP1(v3)=2. Recall w1NR(v2). Clearly, NR(w1)=, otherwise, there exists a longer path than P2 in HP1, a contradiction. If dH(w1)2, then X={u1,us,v1,w1} is the desired set. Thus dH(w1)3, that is, dP1(w1)2. Note w1 and v3 lie on a path P=w1,v2,v3, and w1,v3 send at least two edges each to P1. By Lemma 5, there exists a chorded cycle in P1P, a contradiction. ◻

Subclaim 8. For any uV(P1), NR(u)=.

Proof. We first prove dH(v1)2. Suppose not, that is, dH(v1)3. Recall dP1(v1)1. By Subclaim 7 and Lemma 11(iv), dP1(v1)=1 and dP2(v1)=2. Thus |P2|=3 and v1v3E(H). Since dP1(v1)dP1(v3) by the assumption, dP1(v3)1. Then P1P2 contains a cycle with chord v1v3, a contradiction. Thus dH(v1)2. Suppose there exists a vertex in P1 with a neighbor w1 in R. If dH(w1)2, then X={u1,us,v1,w1} is the desired set. Thus dH(w1)3.

First suppose dP1(w1)2. Then dP1(w1)=2 by Lemma 11(v), and dR(w1)1 by Subclaim 7. Let w2NR(w1). If dH(w2)2, then X={u1,us,v1,w2} is the desired set. Thus dH(w2)3. If dP1(w2)2, then we have two vertices on a path P=w1,w2, each sending at least two edges to another path P1, and by Lemma 5, a chorded cycle exists in P1P, a contradiction. Thus dP1(w2)1, and by Subclaim 7, dR(w2)2. Let w3NRw1(w2). If dH(w3)2, then X={u1,us,v1,w3} is the desired set. Thus dH(w3)3. Suppose dP1(w3)2. Then consider the path P=w1,w2,w3. Since w1 and w3 send at least two edges to another path P1, a chorded cycle exists in P1P by Lemma 5, a contradiction. Thus dP1(w3)1. Also, NR{w1,w2}(w3)=, otherwise, there exists a longer path than P2 in HP1, a contradiction. By Subclaim 7, NP2(w3)=. Thus dP1(w3)=1 and w1,w2NH(w3). Then P1P contains a cycle with chord w1w3, a contradiction.

Next suppose dP1(w1)=1. Then dR(w1)2 by Subclaim 7. Let w2,w3NR(w1). If dH(wi)2 for some i{2,3}, then X={u1,us,v1,wi} is the desired set. Thus dH(wi)3 for each i{2,3}. Suppose dR(wi)3 for some i{2,3}. Without loss of generality, we may assume i=2. Then w2 has a neighbor w4 in R distinct from w1 and w3, and hence w3,w1,w2,w4 is a longer path than P2 in HP1, a contradiction. Thus for each i{2,3}, dR(wi)2, and then dP1(wi)1 by Subclaim 7. Note wi for each i{2,3} does not have a neighbor in R distinct from w1,w2,w3, otherwise, there exists a longer path than P2 in HP1, a contradiction. Now suppose dR(wi)=2 for some i{2,3}. Then w2w3E(H). Let P=w2,w1,w3. Since dP1(wi)1 for each i{2,3}, there exists a cycle with chord w2w3 in P1P, a contradiction. Thus dR(wi)1 for each i{2,3}, and then dP1(wi)2 by Subclaim 7. By Lemma 5, a chorded cycle exists in P1P, a contradiction. ◻

Since H is connected, we get a contradiction by Subclaims 7 and 8. Thus Claim 6 holds. ◻

Claim 9. We have dP1(vt)1.

Proof. Suppose dP1(vt)=0. By the assumption (dP1(v1)dP1(vt)), we have dP1(v1)=0. Then we may assume |P2|=t3, otherwise, we get a contradiction by Claim 6 and the connectedness of H. Recall u1usE(H). By Lemmas 11(iii) and (iv), dH(vj)2 for each j{1,t}. If v1vtE(H), then X={u1,us,v1,vt} is the desired set. Thus v1vtE(H).

First suppose |P2|=t=3. By Claim 6, H=P1P2. Since v1v3E(H), consider P2=v2,v1,v3. Then v2 can be regarded as an endpoint of P2. Since dP1(v1)=0, we may assume dP1(v2)=0 by considering v2 instead of v1. Since NP1(P2)=, this contradicts the connectedness of H.

Next suppose |P2|=t4. Recall u1usE(H) and v1vtE(H). Consider P2=P2[vt1,v1],vt. Then vt1 can be regarded as an endpoint of P2. Thus NR(vt1)= by Lemma 11(iii), and dP2(vt1)2 by Lemma 11(iv). Since dP1(v1)=0, we may assume dP1(vt1)=0 by considering vt1 instead of v1. Thus dH(vt1)=2. Hence X={u1,us,v1,vt1} is the desired set, and Claim 9 holds. ◻

Now we consider the following three cases based on |P2|.

Case 1. Suppose |P2|=t=1.

Then P2=v1. By Claim 6, H=P1P2. Since |H|15, |P1|14. Recall dP1(v1)2 when t=1. By Claim 9, dP1(v1){1,2}. Note dH(v1)=dP1(v1).

First suppose dP1(v1)=2. Let ui,ujNP1(v1) with i<j. Note i2 and js1 by Lemma 11(i). If j=i+1, then H contains a Hamiltonian path, a contradiction. Thus ji+2. By Lemma 9, dH(u)=2 for some {i+1,j1}. Note uu1,uusE(H). Then X={u1,u,us,v1} is the desired set.

Next suppose dP1(v1)=1. Note dP1(u1)2. Assume u1uiE(H) for some 4is1. By Lemma 6, dP1(ui1)=2. If v1ui1E(H), then v1,ui1,P1[ui1,u1],ui,P1[ui,us] is a Hamiltonian path, a contradiction. Thus v1ui1E(H) and dH(ui1)=2. Then X={u1,ui1,us,v1} is the desired set. Thus if dP1(u1)=2, then u1u3E(H). Then dP1(ui)=2 for some 3i6 by Lemma 7. Similarly, either dP1(us)=1 or usus2E(H) by symmetry. Then dP1(uj)=2 for some s5js2 by Lemma 8. Note |P1|=s14. Since dP1(v1)=1 by our assumption, v1uE(H) for some {i,j}, and dH(u)=2. Thus X={u1,u,us,v1} is the desired set.

Case 2. Suppose |P2|=t{2,3}.

By Claim 6, H=P1P2. Recall dP1({v1,vt})3, dP1(v1)1, and dP1(vt)2. We also note dP1({v1,vt})1 by Claim 9. Since |H|15, |P1|=s12.

First suppose |NP1({v1,vt})|{2,3}. Let ui,ujNP1({v1,vt}) with i<j. Assume j=i+1. Then H contains a longer path than P1, a contradiction. Thus ji+2. Note i2 and js1 by Lemma 11(i). By Lemma 9, dH(u)=2 for some {i+1,j1}. Note uu1E(H) and uusE(H). If dH(v1)2, then X={u1,u,us,v1} is the desired set. Thus we may assume that dH(v1)3. Since dP1(v1)1 and dP2(v1)2, we have dP1(v1)=1 and dP2(v1)=2. Then t=3 and v1v3E(H). Since dP1(v1)dP1(vt)=dP1(v3) by the assumption, we have dP1(v3)1. Thus P1P2 contains a cycle with chord v1v3, a contradiction.

Next suppose |NP1({v1,vt})|=1. Assume u1uiE(H) for some 4is1. By Lemma 6, dP1(ui1)=2. Let P1=P1[ui1,u1],ui,P1[ui,us]. Then |P1|=|P1| and ui1 can be regarded as an endpoint of P1. By Lemma 11(i), dP2(ui1)=0. Then dH(ui1)=dP1(ui1)=2. If dH(v1)2, then X={u1,ui1,us,v1} is the desired set. Thus we may assume that dH(v1)3. Then dP1(v1)=1, and dP2(v1)=2, that is, t=3 and v1v3E(H). Also, dP1(v3)1. Thus P1P2 contains a cycle with chord v1v3, a contradiction. Hence, either dP1(u1)=1 or u1u3E(H). Then dP1(ui)=2 for some 3i6 by Lemma 7. Similarly, either dP1(us)=1 or usus2E(H) by symmetry. Then dP1(uj)=2 for some s5js2 by Lemma 8. Since |NP1({v1,vt})|=1 by our assumption, uNP1({v1,vt}) for some {i,j}. Suppose t=2. Then dH(v1)2 and dH(u)=dP1(u)=2. Thus X={u1,u,us,v1} is the desired set. Hence t=3. If v1v3E(H), then dH(v1)2 and dH(v3)2. Thus X={u1,us,v1,v3} is the desired set. Hence we may assume that v1v3E(H). Note dP1(v1)1. Suppose dP1(v1)=1. Since dP1(v3)1, P1P2 contains a cycle with chord v1v3, a contradiction. Suppose dP1(v1)=0. Then dH(v1)=2. If dH(u)=2, then X={u1,u,us,v1} is the desired set. Thus we may assume that dH(u)3. Then uv2E(H). Since dP1(v3)1, P1P2 contains a cycle with chord v2v3, a contradiction.

Case 3. Suppose |P2|=t4.

Recall dP1(v1)1 and dP1(vt)2. We consider two subcases as follows.

Subcase 1. Suppose dP1(v1)=1.

By Claim 9, dP1(vt)1. Then dP2(v1)=dP2(vt)=1, otherwise, there exists a cycle in P1P2 with chord adjacent to v1 or vt, a contradiction. Thus dH(v1)=2 by Lemma 11(iii). If dP1(vt)=1, then dH(vt)=2 by Lemma 11(iii). Then X={u1,us,v1,vt} is the desired set. Thus dP1(vt)=2. Let ui,ujNP1(vt) with i<j. Consider the vertex vt1. If dH(vt1)=2, then X={u1,us,v1,vt1} is the desired set. Thus dH(vt1)3. If dP2(vt1)3, then there exists a cycle in P1P2 with chord adjacent to vt1, a contradiction. Thus dP2(vt1)=2, and then NP1(vt1) or NR(vt1).

First suppose NP1(vt1). If v1 or vt1 has a neighbor in P1[u1,ui]P1[uj,us], then there exist three parallel edges between P1 and P2, and by Lemma 3, a chorded cycle exists in P1P2, a contradiction. Thus NP1(ui,uj)(v) for each {1,t1}. Then we again have three parallel edges or three crossing edges, and by Lemma 3, a chorded cycle exists in P1P2, a contradiction.

Next suppose NR(vt1). Let wNR(vt1). If dH(w)2, then X={u1,us,v1,w} is the desired set. Thus dH(w)3. Then dP1(w)1, otherwise, since dP1(vt)=2, there exists a chorded cycle in P1P2 by Lemma 5, a contradiction. Since P2 is a longest path in HP1, NR(w)=. Thus dP1(w)=1 and dP2(w)=2. Let upNP1(v1) and uqNP1(w). Without loss of generality, we may assume pq. By Lemma 11(iii), wv1,wvtE(H). Thus wvE(H) for some 2t2. Then w,vt1,P2[vt1,v1],up,P1[up,uq],w is a cycle with chord wv, a contradiction.

Subcase 2. Suppose dP1(v1)=0.

Suppose v1vtE(H). Then note dH(v1)=2. Now we consider the path P2=P2[vt1,v1],vt. Then vt1 can be regarded as an endpoint of P2. Since dP1(v1)=0 by the assumption, we may assume dP1(vt1)=0 by considering vt1 instead of v1. Thus dH(vt1)=2. Recall u1usE(H). Then X={u1,us,v1,vt1} is the desired set. Thus v1vtE(H). If dH(vt)2, then X={u1,us,v1,vt} is the desired set. Thus dH(vt)3. By Lemma 11(iii), (iv), and (v), we have dH(vt)4 and dP1(vt){1,2}.

First suppose dP1(vt)=2. Let ui,ujNP1(vt) with i<j. Note i2 and js1 by Lemma 11(i), and |P1||P2|4. If j=i+1, then there exists a longer path than P1, a contradiction. Thus ji+2. Therefore, |P1|5. If dH(u)=2 for some {i+1,j1}, then X={u1,u,us,v1} is the desired set. Thus dH(u)3 for each {i+1,j1}. By Lemma 9, we may assume HP1P2. Now we claim NR(u) for some {i+1,j1}. Assume not. Note NP2(u)= since P1 is a longest path in H. Since H does not contain a chorded cycle, there exist edges ui+1uh with h>j and uj1uh with h<i. Then P1[uh,ui],vt,uj,P1[uj,uh],ui+1,P1[ui+1,uj1],uh is a cycle with chord uiui+1 (and uj1uj), a contradiction. Thus the claim holds. If ji+3, then we may assume =j1, that is, NR(uj1), otherwise, consider P[us,u1]. Let w1NR(uj1), and let P3=w1,,wp(p1) be a longest path starting from w1 in R. If dH(wp)2, then X={u1,us,v1,wp} is the desired set. Thus dH(wp)3. If NP2(w) for some wV(P3), that is, vNP2(w) for some 1t, then P1[u1,uj1],w1,P3[w1,w],v,P2[v,vt],uj,P1[uj,us] is a longer path than P1, a contradiction. Thus NP2(w)= for any wV(P3). Since P3 is a longest path starting from w1 in R, NRP3(wp)=. Suppose |P3|=p=1. Since NR(w1)= and dH(wp)3, dP1(w1)3. This contradicts Lemma 11(v). Suppose |P3|=p=2. Then dH(w2)3, and by Lemma 11(v), dP1(w2)=2. If uNP1(w2) for some js, then P1[ui,uj1],w1,P3[w1,w2],u,P1[u,uj],vt,ui is a cycle with chord uj1uj, a contradiction. Thus u,uNP1(w2) for some 1<j1. Then P1[u,uj1],w1,P3[w1,w2],u is a cycle with chord w2u, a contradiction. Suppose |P3|=p3. Then dP3(wp)2. Assume dP3(wp)=2. Since dP1(wp)1, there exists a cycle in P1P3 with chord adjacent to wp, a contradiction. Thus dP3(wp)=1, and dP1(wp)=2. Then we have a chorded cycle in P1P3 as in the case where |P3|=2 by considering wp instead of w2, a contradiction.

Next suppose dP1(vt)=1. Let uiNP1(vt) with 1is. Note i{1,s} by Lemma 11(i). Since dH(vt)3, dP2(vt)=2 by Lemmas 11(iii) and (iv). Let vNP2(vt) with t2. Now we consider the path P2=P2[v1,v],vt,P2[vt,v+1]. Then v+1 can be regarded as an endpoint of P2. Since dP1(vt)=1, we may assume dP1(v+1)=1. Let ujNP1(v+1) with 1js. Note j{1,s} by Lemma 11(i). Then we may assume ji, otherwise, consider P[us,u1]. Suppose =t2, that is, vtvt2E(H). Then P1[uj,ui],vt,vt2,vt1,uj is a cycle with chord vt1vt, a contradiction. Thus t3. If j=i1, then there exists a longer path than P1, a contradiction.

Suppose j=i. Recall vtvE(H) with t3. If dH(vt1)=2, then X={u1,us,v1,vt1} is the desired set. Thus dH(vt1)3. Assume ujNP1(vt1) for some 1js. We may assume ji, otherwise, consider P[us,u1]. Then P1[uj,ui],vt,P2[v,vt1],uj is a cycle with chord vt1vt, a contradiction. Assume vNP2(vt1) for some t3. Since vtvE(H), we may assume <. Then P2[v,v],vt,ui,P2[v+1,vt1],v is a cycle with chord vv+1 (and vt1vt), a contradiction. Assume NR(vt1). Let wNR(vt1). Now we consider the path P2=P2[v1,vt1],w. Then w can be regarded as an endpoint of P2. Since dP1(vt)=1, we may assume dP1(w)=1. Let ujNP1(w) for some 1js. We may assume ji. Then P2[v,vt1],w,P1[uj,ui],vt,v is a cycle with chord vt1vt, a contradiction.

Suppose ji2. If dH(uh)=2 for some h{j+1,i1}, then X={u1,uh,us,v1} is the desired set. Thus dH(uh)3 for each h{j+1,i1}. Now we claim NR(uh) for some h{j+1,i1}. Assume not. Note NP2(uh)=, since P1 is a longest path in H. Since H does not contain a chorded cycle, there exist edges uj+1um with m>i and ui1um with m<j. Then P1[um,uj],v+1,P2[v+1,vt],ui,P1[ui,um],uj+1,P1[uj+1,ui1],um is a cycle with chord ujuj+1 (and ui1ui), a contradiction. Thus the claim holds. We also note that if ji3, then we may assume NR(ui1), otherwise, consider P[us,u1]. Let w1NR(ui1), and let P3=w1,,wp(p1) be a longest path in R. Then, as in the above case where dP1(vt)=2, there exists a chorded cycle in H, a contradiction. ◻

Lemma 13 ([8]). Let k2 be an integer, and let G be a graph. Suppose G does not contain k vertex-disjoint chorded cycles. Let C={C1,,Ck1} be a minimal set of k1 vertex-disjoint chorded cycles in G, and let H=GC and XV(H) with |X|=4. Suppose H contains a Hamiltonian path. Then dCi(X)12 for each 1ik1.

4. Proof of Theorem 4

Suppose G does not contain a chorded cycle.

Claim 10. G is connected.

Proof. Suppose not, then comp(G)2. Let G1,G2,,Gcomp(G) be the components of G.

First suppose comp(G)4. By Theorem 1, there exists xiV(Gi) for each 1i4 such that dGi(xi)2. Let X={x1,x2,x3,x4}. Then X is an independent set with dG(X)8. This contradicts the σ4(G) condition.

Next suppose comp(G)=3. Let |G1||G2||G3|. Since |G|15 by the assumption, we have |G1|5. If G1 is complete, then G1 contains a chorded cycle. Thus we may assume G1 is not complete. By Theorem 2, there exist non-adjacent x0,x1V(G1) such that dG1({x0,x1})4. Also, by Theorem 1, there exists xiV(Gi) for each i{2,3} such that dGi(xi)2. Then X={x0,x1,x2,x3} is an independent set with dG(X)8, a contradiction.

Finally, suppose comp(G)=2. Let |G1||G2|. Since |G|15, |G1|8. By Theorem 3 (Remark 1), G1 contains an independent set X0 of three vertices with dG1(X0)6. Also, by Theorem 1, there exists xV(G2) such that dG2(x)2. Then X=X0{x} is an independent set with dG(X)8, a contradiction. ◻

Let P1=u1,,us be a longest path in G. Note s3, since |G|15 and G is connected by Claim 10.

Claim 11. G contains a Hamiltonian path.

Proof. Suppose not, then P1 is not a Hamiltonian path in G, and V(GP1). Let P2=v1,,vt (t1) be a longest path in GP1 such that dP1(v1)dP1(vt). By Lemma 12, there exists an independent set X of four vertices in G such that dG(X)8. This contradicts the σ4(G) condition. ◻

Since |G|15, by Claim 11 and Lemma 10, there exists an independent set X of four vertices in G such that dG(X)8, a contradiction. This completes the proof of Theorem 4. 0◻

5. Proof of Theorem 5

By Theorem 4, we may assume k2. Suppose Theorem 5 does not hold. Let G be an edge-maximal counter-example. If G is complete, then G contains k vertex-disjoint chorded cycles. Thus we may assume G is not complete. Let xyE(G) for some x,yV(G), and define G=G+xy, the graph obtained from G by adding the edge xy. By the edge-maximality of G, G is not a counter-example. Thus G contains k vertex-disjoint chorded cycles C1,,Ck. Without loss of generality, we may assume xyi=1k1E(Ci), that is, G contains k1 vertex-disjoint chorded cycles. Over all sets of k1 vertex-disjoint chorded cycles, choose C1,,Ck1 with C=i=1k1Ci, H=GC, and with P1 a longest path in H, such that:

(A1) |C| is as small as possible,

(A2) subject to (A1), comp(H) is as small as possible, and

(A3) subject to (A1) and (A2), |P1| is as large as possible.

We may also assume H does not contain a chorded cycle, otherwise, G contains k vertex-disjoint chorded cycles, a contradiction.

Claim 12. H has an order at least 18.

Proof. Suppose to the contrary that |H|17. Next suppose |Ci|11 for each 1ik1. Since |G|11k+7 by assumption, it follows that |H|(11k+7)11(k1)=18, a contradiction. Thus |Ci|12 for some 1ik1. Without loss of generality, we may assume C1 is a longest cycle in C. Then |C1|12. By Lemma 1, C1 contains at most two chords, and if C1 has two chords, then these chords must be crossing. For integers t and r, let |C1|=4t+r, where t3 and 0r3.

Subclaim 13. Let t3 be an integer. The cycle C1 contains t vertex-disjoint sets X1,,Xt of four independent vertices each in G such that dC1(i=1tXi)8t+4.

Proof. For any 4t vertices of C1, their degree sum in C1 is at most 4t×2+4=8t+4, since C1 has at most two chords. Thus it only remains to show that C1 contains t vertex-disjoint sets of four independent vertices each. Recall |C1|=4t+r4t. Start anywhere on C1 and label the first 4t vertices of C1 with labels 1 through t in order, starting over again with 1 after using label t. If r1, then label the remaining r vertices of C1 with the labels t+1,,t+r. (See Fig. 2.) The labeling above yields t vertex-disjoint sets of four vertices each, where all the vertices labeled with 1 are one set, all the vertices labeled with 2 are another set, and so on. Given this labeling, since t3, any vertex in C1 has a different label than the vertex that precedes it on C1 and the vertex that succeeds it on C1. Let C0 be the cycle obtained from C1 by removing all chords. Then the vertices in each of the sets are independent in C0. Thus the only way vertices in the same set are not independent in C1 is if the endpoints of a chord of C1 were given the same label. Note any vertex labeled i is distance at least 3 in C0 from any other vertex labeled i. Thus even if we exchange the label of x in C0 for the one of x (or x+), the vertices in each of the resulting t sets are still independent in C0.

An Example When t=3, r=3

Case 1. No chord of C1 has endpoints with the same label.

Then there exist t vertex-disjoint sets of four independent vertices each in C1.

Case 2. Exactly one chord of C1 has endpoints with the same label.

Recall C1 contains at most two chords, and if C1 contains two chords, then these chords must be crossing. Since |C1|12, even if C1 has two chords, each chord has an endpoint x such that there exists a vertex x{x,x+} which is not an endpoint of the other chord. Choose such an endpoint x of the chord whose endpoints were assigned the same label, and exchange the label of x for the one of x. The vertices in each of the resulting t sets are independent in C1, and now no chord of C1 has endpoints with the same label. Thus there exist t vertex-disjoint sets of four independent vertices each in C1.

Case 3. Two chords of C1 each have endpoints with the same label.

Then the two chords are crossing. Since endpoints of a chord have the same label in this case, recall these endpoints have distance at least 3. First suppose there exists an endpoint x of one chord of C1 which is adjacent to an endpoint y(=x+) of the other chord on C1. (See Fig. 3(a).) Now we exchange the label of x for the one of y. Then no chord of C1 has endpoints with the same label, and the vertices in each of the resulting t sets are independent in C1. Thus there exist t vertex-disjoint sets of four independent vertices each in C1.

Next suppose no endpoint of one chord of C1 is adjacent to an endpoint of the other chord on C1. (See Fig. 3(b).) Let x1x2,y1y2 be the two distinct chords of C1. Since the two chords are crossing, without loss of generality, we may assume x1,y1,x2,y2 are in that order on C1. Now we exchange the labels of x1 and x1+, and next the ones of y2 and y2. Then no chord of C1 has endpoints with the same label, and the vertices in each of the resulting t sets are independent in C1. Thus there exist t vertex-disjoint sets of four independent vertices each in C1◻

Figure 3: Examples: (a) The Labels of x and y are 2 and 3; (b) The Labels of x1 and y2 are 2 and 1. ([i] Means i is a New Label for a Vertex after the Exchange)

Since |C1|12, dC1(v)2 for any vV(H) by Lemma 2 and (A1). Thus since |H|17 by our assumption, it follows that |E(H,C1)|34. Let X=i=1tXi be as in Subclaim 13. By the σ4(G) condition, dG(X)t(12k3). Suppose k=2. Then C has only one cycle C1. Since k=2 and t3, |E(C1,H)|dH(X)t(12k3)(8t+4)=13t435, a contradiction. Thus k3. Then we have |E(X,CC1)|=dG(X)dC1(X)dH(X)t(12k3)(8t+4)34=12kt11t38, and since t3, 12kt11t38=12t(k1)+t3812t(k1)35>12t(k1)12t=12t(k2). Thus |E(X,C)|>12t for some C in CC1, since CC1 contains k2 vertex-disjoint chorded cycles. Let h=max{dC(v)|vX}. Let v be a vertex of X such that dC(v)=h. Since |E(X,C)|>12t, if h3, then |E(X,C)|3×4t=12t, a contradiction. Thus we may assume h4. By the maximality of C1, |C||C1|=4t+r. It follows that h=dC(v)|C|4t+r. Recall t3 and 0r3. Then (1)|E(X{v},C)|(12t+1)dC(v)(12t+1)(4t+r)=8tr+122. Since h=dC(v)4, let v1,v2,v3,v4 be neighbors of v in that order on C. Note v1,v2,v3,v4 partition C into four intervals C[vi,vi+1) for each 1i4, where v5=v1. By (1), there exist at least 22 edges from C1v to C. Thus some interval C[vi,vi+1) contains at least six of these edges. Without loss of generality, we may assume this interval is C[v4,v1). Then by Lemma 4, (C1v)C[v4,v1) contains a chorded cycle not containing at least one vertex of (C1v)C[v4,v1). Also, v,C[v1,v3],v is a cycle with chord vv2, and it uses no vertices from C[v4,v1). Thus we have two shorter vertex-disjoint chorded cycles in C1C, contradicting (A1). Hence Claim 12 holds. ◻

Claim 14. H is connected.

Proof. Suppose not, then comp(H)2. Let H1,H2,,Hcomp(H) be the components of H. First we prove the following subclaim.

Subclaim 15. Suppose X is an independent set of four vertices in H such that dH(X)8. Then there exists some C in C such that the degree sequences from four vertices of X to C are (4,4,4,1), (4,4,3,2) or (4,3,3,3). Furthermore, then |C|=4.

Proof. By the σ4(G) condition, dC(X)(12k3)8=12k11>12(k1). Thus there exists some C in C such that dC(X)13. By Lemma 2, dC(x)4 for any xX. Now we consider degree sequences defined in Section 1 (Introduction) from four vertices of X to C. Recall that when we write (d1,d2,d3,d4), we assume dC(xj)=dj for each 1j4, since it is sufficient to consider the case of equality. It follows that the degree sequences from four vertices of X to C are (4,4,4,1), (4,4,3,2) or (4,3,3,3). Since each degree sequence contains a vertex with degree 4 in C, we have |C|=4 by Lemma 2. Thus the subclaim holds. ◻

Now we consider the following three cases based on comp(H).

Case 1. Suppose comp(H)4.

By Theorem 1, there exists xiV(Hi) for each 1i4 such that dHi(xi)2. Let X={x1,x2,x3,x4}. Then X is an independent set and dH(X)8. By Subclaim 15, the degree sequences from four vertices of X to some C in C are (4,4,4,1), (4,4,3,2) or (4,3,3,3) and |C|=4. Let C=v1,v2,v3,v4,v1. Without loss of generality, we may assume dC(x1)dC(x2)dC(x3)dC(x4). Then dC(x1)=4. Since |C|=4, for each degree sequence, x2,x3,x4 must all have a common neighbor in C, say v1. Since dC(x1)=4, C=x1,v2,v3,v4,x1 is a 4-cycle with chord x1v3. If x1 is not a cut-vertex of H1, then H1x1 is connected. Replacing C in C by C, we consider the new H. Then comp(H)comp(H)2. This contradicts (A2). Thus we may assume x1 is a cut-vertex of H1. Since dH1(x1)2, dH1(x1)=2. Thus comp(H1x1)=2, and comp(H)comp(H)1 for the new H. This contradicts (A2).

Case 2. Suppose comp(H)=3.

Without loss of generality, we may assume |H1||H2||H3|. Since |H|18 by Claim 12, we have |H1|6. Let P1=u1,,us be a longest path in H1. Note s3. By Theorem 1, there exists xjV(Hj) for each j{2,3} such that dHj(xj)2.

First suppose u1usE(G). Then P1[u1,us],u1 is a Hamiltonian cycle in H1, otherwise, since H1 is connected, there exists a longer path than P1, a contradiction. Since H1 does not contain a chorded cycle, we have u1u3E(H1). Note dH1(ui)=2 for each i{1,3}. Let X={u1,u3,x2,x3}. Then X is an independent set and dH(X)8. By Subclaim 15, the degree sequences from four vertices of X to some C in C are (4,4,4,1), (4,4,3,2) or (4,3,3,3) and |C|=4. Let C=v1,v2,v3,v4,v1. Without loss of generality, we may assume dC(u1)dC(u3). Then dC(u1)3 and NC(u3)NC(x2)NC(x3) by the degree sequences. Without loss of generality, we may assume v1NC(u3)NC(x2)NC(x3). Suppose dC(u1)=4. Then C=u1,v2,v3,v4,u1 is a 4-cycle with chord u1v3. Since H1 contains a Hamiltonian cycle, u1 is not a cut-vertex of H1. Thus H1u1 is connected. Replacing C in C by C, we consider the new H. Then comp(H)comp(H)2=32=1. This contradicts (A2). Thus dC(u1)=3 since dC(u1)3. Then the degree sequence is (4,4,3,2) or (4,3,3,3). Without loss of generality, we may assume dC(x2)dC(x3). In each degree sequence, it is sufficient to consider dC(u1)=3, dC(u3)=2, dC(x2)=3 and dC(x3)=4. Without loss of generality, we may assume vjNC(u1) for each 1j3. Then C=u1,v1,v2,v3,u1 is a 4-cycle with chord u1v2. If v4NC(x2), then v4NC(x2)NC(x3) since dC(x3)=4. Then comp(H)comp(H)1=31=2 for the new H, a contradiction. Thus NC(u1)=NC(x2). Note C has a chord. Suppose v1v3E(G). Then C=u1,v1,v4,v3,u1 is a 4-cycle with chord v1v3. Since dC(x3)=4, v2NC(x2)NC(x3). Then comp(H)comp(H)1=31=2 for the new H, a contradiction. Suppose v2v4E(G). Then C=u1,v1,v4,v2,u1 is a 4-cycle with chord v1v2. Since dC(x3)=4, v3NC(x2)NC(x3). Then comp(H)comp(H)1=31=2 for the new H, a contradiction.

Next suppose u1usE(G). Let X={u1,us,x2,x3}. Since H1 does not contain a chorded cycle, dH1(ui)2 for each i{1,s}. Then X is an independent set and dH(X)8. Replacing u3 by us in the above case where u1usE(G), we get a similar contradiction.

Case 3. Suppose comp(H)=2.

Let |H1||H2|. Since |H|18 by Claim 12, |H1|9. Let P1=u1,,us be a longest path in H1. Note s3. By Theorem 1, there exists x2V(H2) such that dH2(x2)2.

First suppose u1usE(H1). Note P1[u1,us],u1 is a Hamiltonian cycle in H1. Then X0={u1,u3,u5} is an independent set and dH1(X0)=6, and X=X0{x2} is an independent set and dH(X)8. By Subclaim 15, the degree sequences from four vertices of X to some C in C are (4,4,4,1), (4,4,3,2) or (4,3,3,3), and |C|=4. Let C=v1,v2,v3,v4,v1. Since X0 is on the Hamiltonian cycle, we may assume dC(u1)=max{dC(u)|u{u1,u3,u5}}. Then dC(u1)3 by the degree sequences. Suppose dC(u1)=4. Since NC(u3)NC(x2) by the degree sequences, without loss of generality, we may assume v4NC(u3)NC(x2). Since dC(u1)=4, viNC(u1) for each 1i3. Then C=u1,v1,v2,v3,u1 is a 4-cycle with chord u1v2. Since H1 contains a Hamiltonian cycle, u1 is not a cut-vertex of H1. Thus H1u1 is connected. Replacing C in C by C, we consider the new H. Then comp(H)comp(H)1=21=1 for the new H. This contradicts (A2). Now suppose dC(u1)=3. Then by the maximality of dC(u1), we have only to consider the case where dC(ui)=3 for each i{1,3,5}, and dC(x2)=4. Let viNC(u1) for each 1i3. Then we may assume NC(u1)=NC(u3)=NC(u5), otherwise, we get a contradiction by the same arguments as the case where dC(u1)=4. Note C has a chord. Suppose v1v3E(G). Then C=u1,v1,v4,v3,u1 is a 4-cycle with chord v1v3. Since dC(x2)=4, v2NC(u3)NC(x2). Then comp(H)comp(H)1=21=1 for the new H, a contradiction. Suppose v2v4E(G). Then C=u1,v1,v4,v2,u1 is a 4-cycle with chord v1v2. Since dC(x2)=4, v3NC(u3)NC(x2). Then comp(H)comp(H)1=21=1 for the new H, a contradiction.

Next suppose u1usE(H1). Without loss of generality, we may assume dC(u1)dC(us). Assume P1 is a Hamiltonian path in H1. Note s9 since |H1|9. Since P1 is a Hamiltonian path in H1, note dP1(u)=dH1(u) for any uV(P1). We also note dP1(ui)2 for each i{1,s}. Suppose dP1(u1)=1. By Lemma 7, dH1(ui)=2 for some 3i5. Since s9, X0={u1,ui,us} is an independent set and dH1(X0)6. Thus X=X0{x2} is an independent set and dH(X)8. Then we get a contradiction by the same arguments as the case where u1usE(G). Next suppose dP1(u1)=2. Now assume u1u3E(H1). By Lemma 7, dH1(ui)=2 for some 4i6. Since s9, X0={u1,ui,us} is an independent set and dH1(X0)6, and we get a contradiction by considering X=X0{x2} similar to the case where u1usE(H1). Thus u1u3E(H1), that is, u1uiE(H1) for some 4is1. By Lemma 6, dH1(ui1)=2. Since s9, X0={u1,ui1,us} is an independent set and dH1(X0)6, and we get a contradiction by considering X=X0{x2}.

Assume P1 is not a Hamiltonian path in H1. Then V(H1P1). Let P2=v1,,vt(t1) be a longest path in H1P1. Without loss of generality, we may assume dH1(v1)dH1(vt). If u1usE(H1), then since there exists a longer path than P1, we may assume u1usE(H1). Also we may assume dH1(v1)2, otherwise, since dP1(vi)1 for each i{1,t} by Lemma 11(iii) and (iv), there exists a cycle in P1P2 with chord incident to v1 or vt, a contradiction. Thus X0={u1,us,v1} is an independent set and dH1(X0)6. Then X=X0{x2} is an independent set and dH(X)8. By Subclaim 15, the degree sequences from four vertices of X to some C in C are (4,4,4,1), (4,4,3,2) or (4,3,3,3), and |C|=4. Let C=w1,w2,w3,w4,w1. Since dC(u1)dC(us) by our assumption, dC(u1)3 by the degree sequences. First suppose dC(u1)=4. Since NC(v1)NC(x2) by the degree sequences, without loss of generality, we may assume w4NC(v1)NC(x2). Since dC(u1)=4, wiNC(u1) for each 1i3. Then C=u1,w1,w2,w3,u1 is a 4-cycle with chord u1w2. Since u1 is an endpoint of the longest path P1, u1 is not a cut-vertex of H1. Thus H1u1 is connected. Then comp(H)comp(H)1=21=1 for the new H. This contradicts (A2). Suppose dC(u1)=3. Then we may assume the degree sequence is (4,4,3,2) or (4,3,3,3). In each degree sequence, it is sufficient to consider dC(u1)=3, dC(us)=2, and {dC(v1),dC(x2)}={3,4}. First assume dC(v1)=3 and dC(x2)=4. Without loss of generality, we may assume wiNC(u1) for each 1i3. Then C=u1,w1,w2,w3,u1 is a 4-cycle with chord u1w2. If w4NC(v1), then w4NC(v1)NC(x2) since dC(x2)=4. Then comp(H)comp(H)1=21=1 for the new H, a contradiction. Thus NC(u1)=NC(v1). Note C has a chord. Suppose w1w3E(G). Then C=u1,w1,w4,w3,u1 is a 4-cycle with chord w1w3. Since dC(x2)=4, w2NC(v1)NC(x2). Then comp(H)comp(H)1=21=1 for the new H, a contradiction. Suppose w2w4E(G). Then C=u1,w1,w4,w2,u1 is a 4-cycle with chord w1w2. Since dC(x2)=4, w3NC(v1)NC(x2). Then comp(H)comp(H)1=21=1 for the new H, a contradiction. If dC(v1)=4 and dC(x2)=3, then we get a contradiction, similarly. ◻

Claim 16. H contains a Hamiltonian path.

Proof. Suppose not, and let P1=u1,,us be a longest path in H. Note s3 since |H|18 and H is connected by Claim 14. Let P2=v1,,vt (t1) be a longest path in GP1 such that dP1(v1)dP1(vt). By Lemma 12, there exists an independent set X of four vertices in H such that {u1,us,v1}X and dH(X)8. Then the degree sequences from four vertices of X to some C in C are (4,4,4,1), (4,4,3,2) or (4,3,3,3), and |C|=4. Let C=x1,x2,x3,x4,x1. We may assume u1usE(H), otherwise, a path longer than P1 exists, a contradiction. Without loss of generality, we may assume dC(u1)dC(us). By the degree sequences, we have dC(u1)3.

Suppose dC(u1)=4. Since NC(us)NC(v1) by the degree sequences, without loss of generality, we may assume x4NC(us)NC(v1). Since dC(u1)=4, we have xiNC(u1) for each 1i3. Then C=u1,x1,x2,x3,u1 is a 4-cycle with chord u1x2. Since u1 is an endpoint of the longest path P1, u1 is not a cut-vertex of H. Thus Hu1 is connected. Replacing C in C by C, we consider the new H. Then P1[u2,us],x4,P2[v1,vt] is a longer path than P1 in H. This contradicts (A3).

Suppose dC(u1)=3. Then we may assume the degree sequence is (4,4,3,2) or (4,3,3,3). First assume the degree sequence is (4,4,3,2). Since dC(u1)dC(us), we have dC(u1)=3, dC(us)=2 and dC(v1)=4. Without loss of generality, we may assume xiNC(u1) for each 1i3. Then C=u1,x1,x2,x3,u1 is a 4-cycle with chord u1x2. Note u1 is not a cut-vertex of H. If x4NC(us), then since dC(v1)=4, there exists a longer path than P1 in the new H, a contradiction. Thus we may assume x4NC(us). Note C has a chord. Suppose x1x3E(G). Assume x2NC(us). Then C=u1,x3,x4,x1,u1 is a 4-cycle with chord x1x3. Since dC(v1)=4, x2NC(us)NC(v1), and there exists a longer path than P1 in the new H, a contradiction. Thus x2NC(us). Since dC(us)=2, x1,x3NC(us). Then C=us,x3,x4,x1,us is a 4-cycle with chord x1x3. Note us is not a cut-vertex of H. Since dC(v1)=4, x2NC(u1)NC(v1). Then P1[us1,u1],x2,P2[v1,vt] is a longer path than P1 in the new H, a contradiction. Suppose x2x4E(G). Assume x3NC(us). Then C=u1,x1,x4,x2,u1 is a 4-cycle with chord x1x2. Since dC(v1)=4, x3NC(us)NC(v1). Then there exists a longer path than P1 in the new H, a contradiction. Thus x3NC(us). By symmetry, x1NC(us). Thus dC(us)1. This contradicts dC(us)=2.

Next assume the degree sequence is (4,3,3,3). Since dC(u1)dC(us) and dC(u1)=3 by our assumption, we have dC(us)=3. Thus dC(v1)3. First assume dC(v1)=4. Let xiNC(u1) for each 1i3. Then C=u1,x1,x2,x3,u1 is a 4-cycle with chord u1x2. If x4NC(us), then since dC(v1)=4, there exists a longer path than P1 in the new H, a contradiction. Thus NC(u1)=NC(us). Suppose x1x3E(G). Then C=u1,x1,x4,x3,u1 is a 4-cycle with chord x1x3. Since dC(v1)=4, x2NC(us)NC(v1). Then there exists a longer path than P1 in the new H, a contradiction. Suppose x2x4E(G). Then C=u1,x1,x4,x2,u1 is a 4-cycle with chord x1x2. Since dC(v1)=4, x3NC(us)NC(v1). Then there exists a longer path than P1 in the new H, a contradiction.

Next assume dC(v1)=3. Recall dC(u1)=dC(us)=3. Then |NC(us)NC(v1)|2. Let xiNC(u1) for each 1i3. Suppose x1x3E(G). If xiNC(us)NC(v1) for some i{2,4}, then there exists a longer path than P1, a contradiction. Thus x1,x3NC(us)NC(v1). Suppose x4NC(us) and x2NC(v1). Then C=us,x4,x1,x3,us is a 4-cycle with chord x3x4, and P1[us1,u1],x2,P2[v1,vt] is a longer path than P1 in the new H, a contradiction. Suppose x2NC(us) and x4NC(v1). Let wX{u1,us,v1}. Since we assume the degree sequence is (4,3,3,3), we have dC(w)=4. Assume wV(P1). Then P1[u1,us],x2,u1 is a cycle with chord wx2, and v1,x1,x4,x3,v1 is the other cycle with chord x1x3. Thus we have two vertex-disjoint chorded cycles in HC, and G contains k vertex-disjoint chorded cycles, a contradiction. Assume wV(P1). Then C=us,x3,x4,x1,us is a 4-cycle with chord x1x3. Since dC(w)=4, w,x2,P1[u1,us1] is a longer path than P1 in the new H, a contradiction. Suppose x2x4E(G). Note |NC(us)NC(v1)|2. If xiNC(us)NC(v1) for some i{1,3,4}, then there exists a longer path than P1, a contradiction. Thus |NC(us)NC(v1)|1, a contradiction. ◻

By Claims 10, 16 and Lemma 10, H contains an independent set X of four vertices such that dH(X)8. By Claim 16 and Lemma 13, dG(X)=dC(X)+dH(X)12(k1)+8=12k4. This contradicts the σ4(G) condition. This completes the proof of Theorem 5. 0◻

Acknowledgments

The authors would like to thank referees for valuable suggestions and comments. The first author is supported by the Heilbrun Distinguished Emeritus Fellowship from Emory University. The second author is supported by JSPS KAKENHI Grant Number JP19K03610.

References:

  1. Corradi, K. and Hajnal, A., 1963. On the maximal number of independent circuits in a graph. Acta Mathematica Hungarica, 14(3-4), pp.423-439.
  2. Enomoto, H., 1998. On the existence of disjoint cycles in a graph. Combinatorica, 18(4), pp.487-492.
  3. Wang, H., 1999. On the maximum number of independent cycles in a graph. Discrete Mathematics, 205(1-3), pp.183-190.
  4. Fujita, S., Matsumura, H., Tsugaki, M. and Yamashita, T., 2006. Degree sum conditions and vertex-disjoint cycles in a graph. Australasian Journal of Combinatorics, 35, pp.237-251.
  5. Gould, R.J., Hirohata, K. and Keller, A., 2018. On vertex-disjoint cycles and degree sum conditions. Discrete Mathematics, 341(1), pp.203-212.
  6. Finkel, D., 2008. On the number of independent chorded cycles in a graph. Discrete Mathematics, 308(22), pp.5265-5268.

  7. Chiba, S., Fujita, S., Gao, Y. and Li, G., 2010. On a sharp degree sum condition for disjoint chorded cycles in graphs. Graphs and Combinatorics, 26, pp.173-186.
  8. Gould, R.J., Hirohata, K. and Rorabaugh, A.K., 2020. On independent triples and vertex-disjoint chorded cycles in graphs. Australas. J Comb., 77, pp.355-372.

  9. Chiba, S., Jiang, S. and Yan, J., 2020. Partitioning a graph into cycles with a specified number of chords. Journal of Graph Theory, 94(3), pp.463-475.
  10. Chiba, S. and Yamashita, T., 2018. Degree conditions for the existence of vertex-disjoint cycles and paths: A survey. Graphs and Combinatorics, 34, pp.1-83.

  11. Gao, Y., Lin, X. and Wang, H., 2019. Vertex-disjoint double chorded cycles in bipartite graphs. Discrete Mathematics, 342(9), pp.2482-2492.

  12. Molla, T., Santana, M. and Yeager, E., 2020. Disjoint cycles and chorded cycles in a graph with given minimum degree. Discrete Mathematics, 343(6), p.111837.

  13. Gould, R. J., (2012). Graph Theory, Dover Pub. Inc. Mineola, N.Y. 2012.