In this paper, we consider a degree sum condition sufficient to imply the existence of vertex-disjoint chorded cycles in a graph .
Let be the minimum degree sum of four independent vertices of .
We prove that if is a graph of order at least and with ,
then contains vertex-disjoint chorded cycles.
We also show that the degree sum condition on is sharp.
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 vertex-disjoint
cycles. Corrádi and Hajnal [1] first considered a minimum degree condition to
imply a graph must contain
vertex-disjoint cycles, proving that if and the minimum degree , then contains vertex-disjoint cycles. For an integer
, let and when the independence number . 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 and , then contains vertex-disjoint cycles. In 2006, Fujita
et al. [4] proved that if
and , then contains vertex-disjoint cycles, and in [5], this result was extended to
.
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 vertex-disjoint chorded cycles.
Theorem 1. Finkel [6] Let
be an integer. If is a graph of order at least and , then contains vertex-disjoint chorded cycles.
In 2010, Chiba et al. proved Theorem 2. Since , Theorem 2 is
stronger than Theorem 1.
Theorem 2 (Chiba, Fujita, Gao, Li [7]). Let be an integer. If is a graph of order at least
4k and , then contains
vertex-disjoint chorded cycles.
Recently, Theorem 2 was extended as follows. Since , when the
order of is sufficiently large,
Theorem 3 is stronger than Theorem 2.
Theorem 3 (Gould, Hirohata, Rorabaugh [8]). Let be an integer. If is a graph of order at least and , then contains vertex-disjoint chorded cycles.
Remark 1. We note if in Theorem 3, then Theorem 3
holds under the condition that .
In this paper, we consider a similar extension for chorded cycles,
as, in [5], the existence of
vertex-disjoint cycles was proved
under the condition . In
particular, we first show the following.
Theorem 4. If is a graph of order at least and , then contains a chorded cycle.
Remark 2. We consider the following graph of order 14. (See Fig. 1.) Then satisfies the condition in Theorem 4.
However, does not contain a
chorded cycle. Thus 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 be an integer. If is a
graph of order and
, then
contains vertex-disjoint chorded cycles.
Remark 3. Theorem 5 is sharp with
respect to the degree sum condition. Consider the complete bipartite
graph ,
where large . Then . However, does not contain vertex-disjoint chorded cycles, since
any chorded cycle must contain at least three vertices from each partite
set, in particular, from the
partite set. Thus 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 be a graph, a subgraph of and . For , the set of neighbors of in is denoted by , and we denote . For , we denote and . Also we denote . If , then . Furthermore, and . Let be two vertex-disjoint subgraphs of
. Then and . The subgraph of
induced by is denoted by . Let and . If , then we write for . If there is no fear of confusion,
then we use the same symbol for a graph and its vertex set. For two
disjoint graphs and , denotes the union of and . Let be a path or a cycle with a given
orientation and . Then
denotes the first successor of
on and denotes the first predecessor of
on . If , then denotes
the path of from to (including and ) in the given direction. The reverse
sequence of is denoted by
. We also write , and . If is a path (or a cycle), say , then we
assume an orientation of is given
from to (if is a cycle, then the orientation is
clockwise). If is a path
connecting and of , then we denote the path as . If is one vertex, that is, , then we simply write instead of . For an integer and two vertex-disjoint subgraphs
of , we denote by a degree
sequence from to such that and for each . In this paper, since it is
sufficient to consider the case of equality in the above inequality,
when we write , we assume for each . For two disjoint , denotes the set of edges of connecting a vertex in and a vertex in . For a graph , is the number of components of
. A cycle of length is called a -cycle. For terminology
and notation not defined here, see [13].
2. Preliminaries
Definition 1. Suppose are vertex-disjoint chorded cycles in a
graph . We say is minimal
if does not contain vertex-disjoint chorded cycles such that
.
Definition 2. Let be a cycle with chord
, . We say a chord is parallel
to if either or . 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 and be two distinct edges between
two vertex-disjoint paths and . We say and
are parallel
if either and , or and . Note if two distinct edges
between and 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 and be two distinct edges between
vertices of a path , with and
. We say and are nested if either
or .
Definition 5. Let be a path. We say a
vertex on has a left edge if there
exists an edge for some
, that is not an edge of
the path. We also say has a
right edge if there exists an edge for some , that is not an edge of the
path.
3. Lemmas
The following lemmas will be needed.
Lemma 1 ([8]). Let be an integer, and let be a
minimal set of vertex-disjoint
chorded cycles in a graph . If
for some , then has at most two chords. Furthermore,
if the has two chords, then
these chords must be crossing.
Lemma 2 ([8]). Let be an integer, and let be a
minimal set of vertex-disjoint
chorded cycles in a graph . Then
for any and any .
Furthermore, for some and some , if , then , and if
, then .
Lemma 3 ([8]). Suppose there exist at least three
mutually parallel edges or at least three mutually crossing edges
connecting two vertex-disjoint paths and . Then there exists a chorded cycle in
.
Lemma 4 ([8]). Suppose there exist at least five edges
connecting two vertex-disjoint paths and with . Then there exists a
chorded cycle in not containing at least one vertex of .
Lemma 5 ([8]). Let be two vertex-disjoint paths, and let be in that order on . Suppose for each . Then there exists a chorded
cycle in .
Lemma 6 ([8]). Let be a graph containing a path , and not containing a chorded
cycle. If for some
, then for any and in particular, . And if for some , then for any and in particular, .
Lemma 7 ([8]). Let be a graph containing a path , and not containing a chorded
cycle. If , then for some , and if , then for some .
Lemma 8 ([8]). Let be a graph containing a path , and not containing a chorded
cycle. If , then for some , and if , then for some .
Lemma 9. Let be a connected graph of order at least
. Suppose contains neither a chorded cycle nor a
Hamiltonian path. Let , where is a
longest path in and is a longest path in . If for some is adjacent to an endpoint of and for some is adjacent to an
endpoint of possibly, , then for some .
Proof. Let be
as in the lemma, and we may assume and (possibly, ). Suppose for each . If has a left edge, say with , then
is a cycle with chord ,
a contradiction. By symmetry, does not have a right edge. Since
, for each
, otherwise,
since consecutive vertices on
each have adjacencies on , there
exists a longer path than in
, a contradiction. Note that even
if , for each
. Since for each , has a right edge and has a left edge. No vertex in
can have an edge that
does not lie on to some other
vertex in , otherwise,
this edge is a chord of the cycle .
Thus we have edges with
, and with . Then
is a cycle with chord
(and ), a contradiction.
Thus the lemma holds.
Lemma 10 ([8]). Let be a graph of order at least . Suppose does not contain a chorded cycle. If
contains a Hamiltonian path, then
there exists an independent set
of four vertices in such that
.
Lemma 11 ([8]). Let be a connected graph of order at least
. Suppose contains neither a chorded cycle nor a
Hamiltonian path. Let be a longest
path in , and let be a longest path in . Then the following statements
hold. (i) for each . (ii) for each . (iii) for each . (iv)
for each . (v)
for each and each
. (vi) for each .
Proofs of (v) and
(vi). Note parts (i) to (iv) are from [8], hence we only prove parts
(v) and (vi). Since does not
contain a chorded cycle, (v) holds. Suppose . By (v), for each . Then, by Lemma 5, has a chorded cycle, a contradiction.
Thus (vi) holds. 0
Lemma 12. Let be a connected graph of order at least
. Suppose contains neither a chorded cycle nor a
Hamiltonian path. Let be a longest
path in , and let be a longest path in such that . Then there
exists an independent set of four
vertices in such that and .
Remark 4. Let be a graph of order 14 shown in Fig. 1 (Remark 2, Theorem 4), , and . Then satisfies all the conditions except for
the order in Lemma 12. However, the conclusion does not hold.
Thus is necessary.
Proof. Suppose . Since is connected
and , there
exists a longer path than , a
contradiction. Thus . Let . If , that is,
, then by Lemma 11(v). If , then by Lemma 11(vi). Then by the
assumption (), and by Lemma 11(v).
Claim 6. If , then .
Proof. Suppose . Now we prove the following two
subclaims.
Subclaim 7. For any , .
Proof. By Lemma 11(iii),
for each . If , then the subclaim holds. Thus
we may assume . Suppose
for some
. Then . Let . If , then the subclaim holds,
otherwise, there exists a longer path than in , a contradiction. Thus . Since and , we have and . Suppose a vertex on
has a neighbor in . Then . Recall , and note for any and any by Lemma 11(i). We also note for any by Lemma 11(ii). If , then is an independent
set in and , and is the desired set. Thus we may assume
, that is,
and . Then and . Recall . Clearly, , otherwise, there
exists a longer path than in
, a contradiction. If , then is the desired set.
Thus , that is,
. Note and lie on a path , and send at least two edges each to
. By Lemma 5, there exists a
chorded cycle in , a contradiction.
Subclaim 8. For any , .
Proof. We first prove . Suppose not, that is,
. Recall . By Subclaim 7 and Lemma 11(iv),
and . Thus and . Since by the
assumption, . Then
contains a cycle with chord ,
a contradiction. Thus . Suppose there exists a vertex in with a neighbor in . If , then is the desired set.
Thus .
First suppose .
Then by Lemma 11(v), and by Subclaim 7. Let . If , then is the desired set.
Thus . If , then we have two
vertices on a path , each
sending at least two edges to another path , and by Lemma 5, a chorded cycle
exists in , a contradiction. Thus , and by Subclaim 7, . Let . If , then is the desired set.
Thus . Suppose
. Then consider
the path . Since and send at least two edges to another
path , a chorded cycle exists in
by
Lemma 5, a contradiction.
Thus . Also,
,
otherwise, there exists a longer path than in , a contradiction. By Subclaim 7, . Thus and . Then contains a
cycle with chord , a
contradiction.
Next suppose .
Then by Subclaim 7. Let . If for some , then is the desired set.
Thus for each
. Suppose for some . Without loss of generality,
we may assume . Then has a neighbor in distinct from and , and hence is a longer path than
in , a contradiction. Thus for each
, , and then by Subclaim 7. Note for each does not have a neighbor in
distinct from , otherwise, there exists a
longer path than in , a contradiction. Now suppose for some . Then . Let . Since for each , there exists a cycle with
chord in , a
contradiction. Thus
for each , and then
by Subclaim 7. By Lemma 5, a chorded cycle
exists in , a contradiction.
Since is connected, we get a
contradiction by Subclaims 7 and 8. Thus Claim 6 holds.
Claim 9. We have .
Proof. Suppose . By the assumption (), we have . Then we may assume , otherwise, we get a
contradiction by Claim 6 and the connectedness of
. Recall . By Lemmas 11(iii) and (iv), for each . If , then is the desired set.
Thus .
First suppose . By
Claim 6, . Since
, consider . Then can be regarded as an endpoint of
. Since , we may assume by considering instead of . Since , this contradicts
the connectedness of .
Next suppose .
Recall and
. Consider . Then
can be regarded as an
endpoint of . Thus by Lemma 11(iii), and by Lemma 11(iv). Since , we may assume by considering instead of . Thus . Hence is the desired
set, and Claim 9 holds.
Now we consider the following three cases based on .
Case 1. Suppose .
Then . By Claim 6, .
Since , . Recall when . By Claim 9, . Note .
First suppose .
Let with
. Note and by Lemma 11(i). If
, then contains a Hamiltonian path, a
contradiction. Thus . By
Lemma 9, for some . Note .
Then is
the desired set.
Next suppose .
Note . Assume
for some . By Lemma 6, . If , then
is a Hamiltonian path, a contradiction. Thus and . Then is the desired
set. Thus if , then
. Then for some by Lemma 7. Similarly, either
or by symmetry. Then
for some by Lemma 8. Note . Since by our assumption, for some , and . Thus is the desired
set.
Case 2. Suppose .
By Claim 6, . Recall
, , and . We also note by Claim 9. Since , .
First suppose . Let with . Assume . Then contains a longer path than , a contradiction. Thus . Note and by Lemma 11(i). By
Lemma 9, for some . Note and . If , then is the desired
set. Thus we may assume that . Since and ,
we have and . Then and . Since by the assumption, we have . Thus contains a
cycle with chord , a
contradiction.
Next suppose . Assume for some . By Lemma 6, . Let .
Then and can be regarded as an endpoint of
. By Lemma 11(i), . Then . If
, then is the desired
set. Thus we may assume that . Then ,
and , that is, and . Also, . Thus contains a
cycle with chord , a
contradiction. Hence, either or . Then for some by Lemma 7. Similarly, either
or by symmetry. Then
for some by Lemma 8. Since by our
assumption, for some . Suppose .
Then and . Thus
is the
desired set. Hence . If , then and . Thus is the desired set.
Hence we may assume that . Note . Suppose .
Since , contains a
cycle with chord , a
contradiction. Suppose . Then . If , then is the desired
set. Thus we may assume that . Then . Since , contains a
cycle with chord , a
contradiction.
Case 3. Suppose .
Recall and
. We consider two
subcases as follows.
Subcase 1. Suppose .
By Claim 9, . Then , otherwise,
there exists a cycle in with chord adjacent to or , a contradiction. Thus by Lemma 11(iii). If , then by Lemma 11(iii). Then is the desired set.
Thus . Let with . Consider the vertex . If , then is the desired
set. Thus . If
, then there
exists a cycle in with chord adjacent to , a contradiction. Thus , and then or .
First suppose . If or has a neighbor in , then there
exist three parallel edges between and , and by Lemma 3, a chorded
cycle exists in , a contradiction. Thus
for each . Then we
again have three parallel edges or three crossing edges, and by Lemma 3, a chorded
cycle exists in , a contradiction.
Next suppose . Let . If , then is the desired set.
Thus . Then , otherwise, since , there exists a chorded
cycle in by Lemma 5, a contradiction.
Since is a longest path in
, . Thus and . Let and . Without loss of
generality, we may assume .
By Lemma 11(iii), . Thus for some . Then
is a cycle with chord , a
contradiction.
Subcase 2. Suppose .
Suppose . Then
note . Now we consider
the path . Then
can be regarded as an
endpoint of . Since by the assumption, we may
assume by
considering instead of
. Thus . Recall . Then is the desired
set. Thus . If
, then is the desired set.
Thus . By Lemma 11(iii), (iv), and (v), we have and .
First suppose .
Let with
. Note and by Lemma 11(i), and
. If , then there exists a longer path
than , a contradiction. Thus
. Therefore, . If for some , then is the desired
set. Thus for
each . By
Lemma 9, we may assume .
Now we claim for some . Assume not. Note since is a longest path in . Since does not contain a chorded cycle, there
exist edges with and with . Then
is a cycle with chord
(and ), a contradiction.
Thus the claim holds. If ,
then we may assume , that
is, ,
otherwise, consider .
Let , and let
be
a longest path starting from in
. If , then is the desired set.
Thus . If for some , that is, for some , then is a longer path than
, a contradiction. Thus for any . Since is a longest path starting from in , . Suppose . Since and , . This contradicts
Lemma 11(v). Suppose . Then , and by Lemma 11(v), . If for some , then
is a cycle with chord , a
contradiction. Thus
for some . Then
is a cycle with chord , a contradiction.
Suppose . Then . Assume . Since , there exists a cycle
in with
chord adjacent to , a
contradiction. Thus ,
and . Then we have a
chorded cycle in as in the case where by considering instead of , a contradiction.
Next suppose . Let
with . Note by Lemma 11(i). Since , by Lemmas 11(iii) and (iv). Let with . Now we consider the path
.
Then can be regarded as
an endpoint of . Since
, we may assume . Let with . Note by Lemma 11(i). Then we may assume , otherwise, consider . Suppose , that is, . Then is a
cycle with chord , a
contradiction. Thus .
If , then there exists a
longer path than , a
contradiction.
Suppose . Recall with . If , then is the desired
set. Thus .
Assume for
some . We may assume
, otherwise, consider . Then
is a cycle with chord , a
contradiction. Assume for some . Since , we may assume . Then
is a cycle with chord (and ), a contradiction. Assume
. Let
. Now we
consider the path . Then can be regarded as an endpoint of . Since , we may assume . Let for some . We may assume . Then
is a cycle with chord , a
contradiction.
Suppose . If for some , then is the desired set.
Thus for each . Now we claim for some . Assume not. Note , since is a longest path in . Since does not contain a chorded cycle, there
exist edges with and with . Then
is a cycle with chord
(and ), a contradiction.
Thus the claim holds. We also note that if , then we may assume , otherwise,
consider . Let , and let be a
longest path in . Then, as in the
above case where ,
there exists a chorded cycle in ,
a contradiction.
Lemma 13 ([8]). Let be an integer, and let
be a graph. Suppose does not
contain vertex-disjoint chorded
cycles. Let be a minimal set of vertex-disjoint chorded cycles in
, and let and with . Suppose contains a Hamiltonian path. Then for each .
4. Proof of Theorem 4
Suppose does not contain a
chorded cycle.
Claim 10.
is connected.
Proof. Suppose not, then . Let be the
components of .
First suppose . By
Theorem 1, there exists for each such that . Let . Then is an independent set with . This contradicts the condition.
Next suppose . Let
. Since
by the assumption, we
have . If is complete, then contains a chorded cycle. Thus we may
assume is not complete. By
Theorem 2, there exist non-adjacent such that . Also, by
Theorem 1, there exists for each such that . Then is an independent
set with , a
contradiction.
Finally, suppose . Let
. Since , . By Theorem 3 (Remark 1),
contains an independent set
of three vertices with . Also, by Theorem 1,
there exists such that
. Then is an independent set
with , a
contradiction.
Let be a
longest path in . Note , since and is connected by Claim 10.
Claim 11.
contains a Hamiltonian path.
Proof. Suppose not, then is not a Hamiltonian path in , and . Let () be a longest path in such that . By
Lemma 12, there exists an independent set of four vertices in such that . This contradicts the condition.
Since , by Claim 11 and Lemma 10,
there exists an independent set
of four vertices in such that
, a contradiction.
This completes the proof of Theorem 4. 0
5. Proof of Theorem 5
By Theorem 4, we may assume . Suppose Theorem 5 does not hold.
Let be an edge-maximal
counter-example. If is complete,
then contains vertex-disjoint chorded cycles. Thus we
may assume is not complete. Let
for some , and define , the graph obtained from
by adding the edge . By the edge-maximality of , is not a counter-example. Thus
contains vertex-disjoint chorded cycles . Without loss of
generality, we may assume , that is,
contains vertex-disjoint chorded cycles. Over
all sets of vertex-disjoint
chorded cycles, choose with , , and with a longest path in , such that:
(A1) is as small
as possible,
(A2) subject to (A1), is
as small as possible, and
(A3) subject to (A1) and (A2), is as large as possible.
We may also assume does not
contain a chorded cycle, otherwise, contains vertex-disjoint chorded cycles, a
contradiction.
Claim 12.
has an order at least .
Proof. Suppose to the contrary that . Next suppose for each . Since by assumption, it follows
that , a
contradiction. Thus
for some . Without
loss of generality, we may assume is a longest cycle in . Then . By Lemma 1, contains at most two chords, and if
has two chords, then these
chords must be crossing. For integers and , let , where and .
Subclaim 13. Let be an integer. The cycle contains vertex-disjoint sets of four independent
vertices each in such that .
Proof. For any
vertices of , their degree sum
in is at most , since has at most two chords. Thus it only
remains to show that contains
vertex-disjoint sets of four
independent vertices each. Recall . Start anywhere on and label the first vertices of with labels 1 through in order, starting over again with 1
after using label . If , then label the remaining vertices of with the labels . (See Fig. 2.)
The labeling above yields
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 , any vertex in
has a different label than the vertex that precedes it on and the vertex that succeeds it on
. Let be the cycle obtained from by removing all chords. Then the
vertices in each of the sets are independent in . Thus the only way vertices in the
same set are not independent in
is if the endpoints of a chord of were given the same label. Note any
vertex labeled is distance at
least 3 in from any other
vertex labeled . Thus even if we
exchange the label of in for the one of (or ), the vertices in each of the
resulting sets are still
independent in .
An Example When t=3, r=3
Case 1. No chord of has endpoints with the same
label.
Then there exist
vertex-disjoint sets of four independent vertices each in .
Case 2. Exactly one chord of has endpoints with the same
label.
Recall contains at most two
chords, and if contains two
chords, then these chords must be crossing. Since , even if has two chords, each chord has an
endpoint such that there exists a
vertex which
is not an endpoint of the other chord. Choose such an endpoint of the chord whose endpoints were
assigned the same label, and exchange the label of for the one of . The vertices in each of the
resulting sets are independent in
, and now no chord of has endpoints with the same label.
Thus there exist vertex-disjoint
sets of four independent vertices each in .
Case 3. Two chords of 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 of one chord of which is adjacent to an endpoint
of the other chord on
. (See Fig. 3(a).) Now
we exchange the label of for the
one of . Then no chord of has endpoints with the same label,
and the vertices in each of the resulting sets are independent in . Thus there exist vertex-disjoint sets of four
independent vertices each in .
Next suppose no endpoint of one chord of is adjacent to an endpoint of the
other chord on . (See Fig. 3(b).) Let be the two distinct chords
of . Since the two chords are
crossing, without loss of generality, we may assume are in that order on
. Now we exchange the labels of
and , and next the ones of and . Then no chord of has endpoints with the same label,
and the vertices in each of the resulting sets are independent in . Thus there exist vertex-disjoint sets of four
independent vertices each in .
Figure 3: Examples: (a) The Labels of and are 2 and 3;
(b) The Labels of and are 2 and 1.
([] Means is a New Label for a Vertex after the Exchange)
Since , for any by Lemma 2 and (A1). Thus since
by our assumption, it
follows that . Let
be as
in Subclaim 13. By the condition, . Suppose
. Then has only one cycle . Since and , , a contradiction. Thus . Then we have and since , Thus for some in , since contains vertex-disjoint chorded cycles. Let
. Let
be a vertex of such
that . Since
, if
, then , a contradiction. Thus we may assume . By the maximality of , . It follows that . Recall and
. Then Since , let be neighbors of in that order on . Note partition into four intervals for each , where . By (1), there exist at least 22 edges from to . Thus some interval contains at least six
of these edges. Without loss of generality, we may assume this interval
is . Then by
Lemma 4, contains a chorded cycle not containing
at least one vertex of . Also, is a cycle
with chord , and it uses no
vertices from . Thus
we have two shorter vertex-disjoint chorded cycles in ,
contradicting (A1). Hence Claim 12
holds.
Claim 14.
is connected.
Proof. Suppose not, then . Let be the
components of . First we prove the
following subclaim.
Subclaim 15. Suppose is an independent set of four vertices
in such that . Then there exists some
in such that the degree
sequences from four vertices of
to are , or . Furthermore, then .
Proof. By the condition, . Thus there exists some in such that . By Lemma 2, for any . Now we consider degree sequences
defined in Section 1 (Introduction) from four vertices of to . Recall that when we write , we assume for each , since it is sufficient to
consider the case of equality. It follows that the degree sequences from
four vertices of to are , or . Since each degree sequence
contains a vertex with degree 4 in , we have by Lemma 2. Thus the subclaim
holds.
Now we consider the following three cases based on .
Case 1. Suppose .
By Theorem 1, there exists for each such that . Let . Then is an independent set and . By Subclaim 15,
the degree sequences from four vertices of to some in are , or and . Let . Without loss of
generality, we may assume . Then . Since , for each degree sequence, must all have a common
neighbor in , say . Since , is a 4-cycle
with chord . If is not a cut-vertex of , then is connected. Replacing in by , we consider the new . Then . This
contradicts (A2). Thus we may assume is a cut-vertex of . Since , . Thus , and for the new
. This contradicts (A2).
Case 2. Suppose .
Without loss of generality, we may assume . Since by Claim 12, we have . Let be a longest path in
. Note . By Theorem 1, there exists
for each such that .
First suppose .
Then is a
Hamiltonian cycle in ,
otherwise, since is connected,
there exists a longer path than , a contradiction. Since does not contain a chorded cycle, we
have . Note
for each . Let . Then is an independent set and . By Subclaim 15,
the degree sequences from four vertices of to some in are , or and . Let . Without loss of
generality, we may assume . Then and by the degree sequences. Without loss of
generality, we may assume . Suppose . Then is a 4-cycle
with chord . Since contains a Hamiltonian cycle, is not a cut-vertex of . Thus is connected. Replacing in by , we consider the new . Then . This
contradicts (A2). Thus
since . Then the
degree sequence is or
. Without loss of
generality, we may assume . In each degree sequence, it is sufficient to consider
, , and . Without loss of generality,
we may assume for
each . Then is a 4-cycle
with chord . If , then since . Then for the
new , a contradiction. Thus
. Note has a chord. Suppose . Then is a 4-cycle
with chord . Since , . Then for the
new , a contradiction.
Suppose . Then is a 4-cycle
with chord . Since , . Then for the
new , a contradiction.
Next suppose .
Let . Since
does not contain a chorded
cycle, for each
. Then is an independent set and . Replacing by in the above case where , we get a similar
contradiction.
Case 3. Suppose .
Let . Since by Claim 12, . Let be a longest path in
. Note . By Theorem 1, there exists
such that .
First suppose .
Note is a
Hamiltonian cycle in . Then
is an
independent set and ,
and is an
independent set and .
By Subclaim 15, the degree sequences from four vertices
of to some in are , or , and . Let . Since is on the Hamiltonian cycle, we may
assume . Then by the degree sequences. Suppose . Since by
the degree sequences, without loss of generality, we may assume . Since
, for each . Then is a 4-cycle
with chord . Since contains a Hamiltonian cycle, is not a cut-vertex of . Thus is connected. Replacing in by , we consider the new . Then for the
new . This contradicts (A2).
Now suppose . Then by the
maximality of , we have
only to consider the case where for each , and . Let for each . Then we may assume , otherwise, we
get a contradiction by the same arguments as the case where . Note has a chord. Suppose . Then is a 4-cycle
with chord . Since , . Then for the
new , a contradiction.
Suppose . Then is a 4-cycle
with chord . Since , . Then for the
new , a contradiction.
Next suppose . Without loss of generality, we may assume . Assume is a Hamiltonian path in . Note since . Since is a Hamiltonian path in , note for any . We also note for each . Suppose . By Lemma 7, for some . Since , is an independent set
and . Thus is an independent set
and . Then we get a
contradiction by the same arguments as the case where . Next suppose . Now assume . By Lemma 7, for some . Since , is an independent set
and , and we get a
contradiction by considering similar to the case where . Thus , that is, for some . By Lemma 6, . Since , is an independent
set and , and we
get a contradiction by considering .
Assume is not a Hamiltonian
path in . Then . Let be a
longest path in . Without
loss of generality, we may assume . If , then since there exists
a longer path than , we may
assume . Also
we may assume ,
otherwise, since
for each by Lemma 11(iii) and (iv), there exists a cycle
in with
chord incident to or , a contradiction. Thus is an independent set
and . Then is an independent set
and . By Subclaim 15,
the degree sequences from four vertices of to some in are , or , and . Let . Since by our assumption,
by the degree
sequences. First suppose . Since by
the degree sequences, without loss of generality, we may assume . Since
, for each . Then is a 4-cycle
with chord . Since is an endpoint of the longest path
, is not a cut-vertex of . Thus is connected. Then for the
new . This contradicts (A2).
Suppose . Then we may
assume the degree sequence is or . In each degree sequence, it is
sufficient to consider ,
, and . First
assume and . Without loss of generality,
we may assume for
each . Then is a 4-cycle
with chord . If , then since . Then for the
new , a contradiction. Thus
. Note has a chord. Suppose . Then is a 4-cycle
with chord . Since , . Then for the
new , a contradiction.
Suppose . Then is a 4-cycle
with chord . Since , . Then for the
new , a contradiction. If
and , then we get a contradiction,
similarly.
Claim 16.
contains a Hamiltonian path.
Proof. Suppose not, and let be a longest path in
. Note since and is connected by Claim 14. Let () be a longest path in such that . By
Lemma 12, there exists an independent set of four vertices in such that and . Then the degree sequences
from four vertices of to some
in are , or , and . Let . We may assume
, otherwise, a
path longer than exists, a
contradiction. Without loss of generality, we may assume . By the degree
sequences, we have .
Suppose . Since by
the degree sequences, without loss of generality, we may assume . Since
, we have for each . Then is a 4-cycle
with chord . Since is an endpoint of the longest path
, is not a cut-vertex of . Thus is connected. Replacing in by , we consider the new . Then is a longer
path than in . This contradicts (A3).
Suppose . Then we may
assume the degree sequence is or . First assume the degree
sequence is . Since , we have , and . Without loss of generality,
we may assume for
each . Then is a 4-cycle
with chord . Note is not a cut-vertex of . If , then since , there exists a longer path
than in the new , a contradiction. Thus we may
assume . Note
has a chord. Suppose . Assume . Then is a 4-cycle
with chord . Since , , and there
exists a longer path than in
the new , a contradiction.
Thus . Since
, . Then is a 4-cycle
with chord . Note is not a cut-vertex of . Since , . Then is a
longer path than in the new
, a contradiction. Suppose
. Assume . Then is a 4-cycle
with chord . Since , . Then there
exists a longer path than in
the new , a contradiction.
Thus . By
symmetry, . Thus
. This contradicts
.
Next assume the degree sequence is . Since and by our assumption, we have
. Thus . First assume . Let for each . Then is a 4-cycle
with chord . If , then since , there exists a longer path
than in the new , a contradiction. Thus . Suppose . Then is a 4-cycle
with chord . Since , . Then there
exists a longer path than in
the new , a contradiction.
Suppose . Then is a 4-cycle
with chord . Since , . Then there
exists a longer path than in
the new , a
contradiction.
Next assume . Recall
. Then . Let for each . Suppose . If for some
, then there exists a
longer path than , a
contradiction. Thus . Suppose and . Then is a 4-cycle
with chord , and is a
longer path than in the new
, a contradiction. Suppose
and . Let . Since we assume
the degree sequence is ,
we have . Assume . Then is a cycle with
chord , and is the other cycle
with chord . Thus we have two
vertex-disjoint chorded cycles in , and
contains vertex-disjoint chorded
cycles, a contradiction. Assume . Then is a 4-cycle
with chord . Since , is a longer path
than in the new , a contradiction. Suppose . Note . If for some
, then there exists a
longer path than , a
contradiction. Thus , a contradiction.
By Claims 10, 16 and Lemma 10, contains an independent set of four vertices such that . By Claim 16 and Lemma 13, This contradicts the 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.
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