Chapter 14. Graphs with Cycles Containing Given Paths
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M . Grotschel
234
Proposition 1 (Bondy). Let G be a graph with degree sequence d,, . . ., d, such that
for some integer h < n the following holds:
d,
3
k
+h

1 forall 1 s k s n
Then G is hconnected.

dn,,+, 1.
(1)
0
A wellknown property of hconnected graphs is the following, cf. [ l , p. 1681:
Proposition 2. If G is h connected then the induced subgraph obtained by removing
one vertex is ( h  1)connected. 0
The next two theorems can also b e found in [1, p. 1691
Proposition 3. Let G = ( V ,E ) be hconnected. Let W = { w l , .. ., w h } be a set of
vertices, 1 W 1 = h. If v E V  W, there exist h vertexdisjoint paths (v, . . ., wn),
i = l , . . ., h, joining v and W. 0
Proposition 4. Let G be a h connected graph, h 3 2. Then there is a cycle passing
through an arbitrary set of two edges and h  2 vertices. 0
A frequently used theorem is the following, see 12, p. 1921:
Proposition 5 (MengerDirac). Let P = (ao,a,, . . ., a,) be a path. If G is 2connected then there exist two paths P' and P" with the following properties:
(a) the endpoints of P' and P" are a. and up,
( b ) P' and P" have no other points in common,
(c) if P' (or PI') contains vertices of P, then they appear in P' (or P " ) in the same
order as they do in P. 0
We now give an extension of Proposition 3 which will be of interest later.
Proposition 6. Let G be a 3connected graph and P = ( a o , .. .,a,) be a path, let
{as,a,,,} be an edge of this path. Then there exists a pair of paths P', P" with the
following properties :
(a) The endpoints of P' and P" are a. and ap,
(b) P' and P" have no other points in common,
(c) if P' (or P") contains vertices of P, then they appear in P' (or P") in the same
order as they do in P,
(d) P' contains {as,as+,}.
Proof. By induction.
(1) Let P = ( a o , a l ) ,i.e. P is an edge. Then necessarily s = 0. As G is 2connected, there is another path P" from a. t o a l . Take P' = P.
Graphs with cycles containing given paths
235
(2) P = (ao,a l ,az), s = 1. By Proposition 3 there are two vertexdisjoint paths
PI = ( a o , . ., a l ) and Pz = ( a o , . ., aZ).Define P' = (ao,P1,al, az), P" = P2. The case
s = 0 is similar.
(3) P = (ao,a ] ,a2,a3), s = 1. By Proposition 3 there are three vertexdisjoint
, = (ao, . . ., &), p3 = (ao,.. ., a3). Define
paths (G is 3connected): PI = (ao, . . ., a ~ )PZ
P' = ( a o ,PI,a ] ,az,a3) and P" = P3. All other cases are similar.
Now suppose the theorem is true for paths of length k . W e prove that it is true for
paths of length k + 1.
Let P = (ao,a l , . . ., ~ k + ~ ) PI, = (ao,P,a ) .
We may assume that s < k  1, otherwise we take the reverse p of P. By
assumption there exist paths Pi and PY connecting a. and an having the desired
properties with respect to P I . From G we now remove the vertex ak and add the
edge {ao,a t + l } ,if it does not already exist. By Proposition 2 the new graph G' is
2connected. By Proposition 4 there is a cycle in G' containing the edges {as,as+]}
and {ao,a k + l } . Thus there is a path Q = ( a o ,a : , . . ., ah, a k + l ) in G connecting a. and
a k + l , which contains the edge {as,
as+l}and does not contain the vertex ak.
Let x be the vertex of path Q which is as close as possible t o a k + l and is contained
in the union of the vertex sets P1,Pi,and P:. Clearly x lies between a,+land
on
the path Q as a,+lis in Q and in Pi. If x is in Pi then x lies between a,+]and ak in
P : . We now have to investigate several cases.
(i) x
=
ak+l
(a) x E Pi
(b)
(ii) x not in P
X
E P:
(a) x E P :
(b)
X
E P:'
P' = (ao,Pi, x),
P" = (ao,PY, ak7
P' = ( U o , P ; , Ua, & + I ) ,
P'' = (ao,ZJ:, x).
P' = (ao,P : , x, Q, ak+]),
P" = (ao,p:, aa, a*+1),
P' = (ao, p : , Ua, U a + i ) ,
P''= ( ~ o P',',
,
X, Q, ~ a + i ) .
(iii) x in P but x # @ + I , say x = a,, r 3 s + 1. Let p S r be the largest index such
that a, is contained in the union of the vertex sets of Pi and Py.
(a) a, E Pi
(b) a, E P','
P' = (ao,P i , a,, P, a,, Q,
P" = (ao,P'L a,a k + l ) ,
P' = (ao,Pi, at,
P" = (ao,P;,a,, P, a,, Q, at+$
These are all the cases which have to be considered and hence we are done. 0
Corollary 7 . Let G be ( r + 2)connected and P = ( a o , . ., a p ) be a path, r =sp , let
Q = (as,. . ., a,,,) be a path of length r contained in P. Then there exists a pair of paths
P', P" with the following properties:
(a) the endpoints of P' and P'' are a. and a,,
(b) P' and P" have no other points in common,
M. Grotschel
236
(c) if P' (or P") contains vertices of P, then they appear in P' ( P " ) in the same order
as they do in P,
(d) P' contains the path Q.
Proof. r = 0 : Then by definition Q is an empty path and Corollary 7 reduces to
Proposition 5 .
r = 1 : This is Proposition 6.
r > 1 : Remove the r  1 vertices a,+l, a,+z,.. ., a,,,. I and add the edge {as,a + , } .
The resulting graph G ' is 3connected by Proposition 2. The path PI =
(a",. .., a,, as+,,. .., a,), contains the edge {a,, as+,}.Application of Proposition 6
gives two paths Pi and P'i, and Pi contains {as,as+,}. T h e path P ' =
( a , ,PI, a,, Q, a,,,, P : , a,) is well defined in G. Define P" = P',', then the pair P ' , P"
has the desired properties. 0
3. The theorem and its corollaries
The following theorem establishes a sufficient condition in terms of the degree
sequencefor the following property of a graph: given any path of a specified
length, there exists a cycle containing this path and having a certain minimum
length. Formally the theorem is very like a theorem of Berge [l, p. 2041, which is an
extension of a theorem of ChvAtal [4] on hamiltonian graphs. The proof of case (i)
below is a slight variation of their proof whichin spirit is due to NashWilliams
[6]. Case (ii) of the proof was motivated by P6sa's proof of his own theorem [7]
which is also included in the following:
Theorem 8. Let d , , . . ., d, be the degree sequence of a graph G = (V, E ) . Let n
rn =G n, 0 r < m  3, and let the following condition be satisfied:
dkSk+r
d,.k,snk
forall O < k < ; ( m  r ) .
3
3,
(2)
Furthermore, let G be ( r + 2)connected if f ( m  r ) S n  dn..,,  1 holds and
dk > k + r holds for all 0 < k < $(in r). Then for each path Q of length r there exists
u cycle in G of length at least rn which contains Q.
Proof. (1) We prove: G is ( r + 2)connected. Let h = r
lent to
d, < k
+h

2
+ dnh+2k n
2

k
+ 2 < n, then
(2) is equiva
for all 0 < k < $ ( m  h
+ 2). (2')
j + h  2.
(a) Suppose there exists a j such that 0 < j < f (rn  h + 2 ) and d,
Condition (2') implies dnh+Z,
3 n  j . As d n  h + l 3 d ",,+*,,
we obtain j >
n  ( n  j )  1 3 n  dn,,+] 1. Thus if d k < k + h  1, then k > n  d n  h + l 1 .
Therefore the conditions of Proposition 1 are satisfied and G is hconnected.
(b) Suppose d, 3 k + h  1 for all 0 < k < 4 (rn  h + 2), then G is hconnected
237
Graphs with cycles containing given paths
by Proposition 1 if f ( r n  h + 2) > n  dn,,+,
 1. Otherwise hconnectedness
follows from the assumption. We note for the following that ( r + 2)connectedness
implies d , 3 r + 2.
(2) It is an easy exercise to see that a graph G’ obtained from G by adding any
new edge to G also satisfies (2) and the other conditions of the theorem.
( 3 ) Suppose now that G is a graph satisfying the required conditions but which
contains a path Q of length r such that Q is not contained in a cycle of length 3 rn.
By adding new edges to G we construct a “maximal” graph (also called G ) which
satisfies all the conditions of the theorem, contains a path Q of length r, has no
cycle of length 3 rn containing Q, and has the property that the addition of any
new edge to G creates a cycle of length 3 rn which contains Q. In the following we
shall deal with this maximal graph G.
(4) Let u, v E V be two nonadjacent vertices of G. T h e addition of the edge
{u, v } will create a cycle with the desired properties. Thus there exists a path
P : = ( u ,,..., u p ) , u ,= u, up = v , p 3 rn
of length
3 rn  1 connecting u
and v, and which contains
Q : = ( u s , .. ., us+,), where s E (1,. . . , p  r } .
Let
S : = { I E { l , .. . , p } :{u,,u , + ~E} E } n ((1,. . ., s  1) u { S
+ r, . . . , p } )
T : = { i E { l ) ...)p } : { u p , u , } E E } .
+.
(a) We prove: S n T =
Suppose i E S fl T, then [ul, u,+,,P, up,u,, p,u l ] is a
cycle with the desired properties. Contradiction!
(b) J S l + J T J < j P J ble c a u s e p t i f S U T .
( 5 ) The degree sequence of G necessarily has exactly one of the following
properties:
Case (i) there is a k o , 0 < k o < i(rn  r ) , such that db k o + r,
Case (ii) d, > k + r for all 0 < k < ( m  r ) .
These cases will be handled separately.
Case (i).
(6) As d , 3 r + 2 and as the degree sequence d,, . . ., d, is increasing there is a
j < k , such that d, = j + r. (2) implies d n  ,  r 3n  j , i.e. there are j + r + 1 vertices
of V having degree at least n  j . T h e vertex having degree j + r cannot be adjacent
to all of these. Thus there exist two nonadjacent vertices a, b E V such that
d ( a )+ d ( b ) 3 n + r.
(7) Among all nonadjacent vertices of G choose u, v such that d ( u ) + d ( u ) is as
large as possible. Define P, S, T, Q as in (4). We calculate d ( u ) + d ( v ) . Obviously
d ( u ) = I T I + a where a s ( V  P J
and
d ( u ) s J S I + r + P where
VPI.
M. Grotschel
238
Suppose there is a w E V  P which is adjacent to both u and u. Then
[ul, uz, . . ., up, w ] would b e a desired cycle. Therefore (Y + p S I V  P 1, which
leads, using (4) (a) and (b), to
d ( u ) + d ( v ) S ) TI+ a
+1S 1+r +p
sIPI+JVPI+~I
By (6) d ( u ) + d ( v ) cannot be maximal. Contradiction!
Case (ii).
(8) Among all longest paths in G containing Q choose a path such that the sum
of the degrees of the endpoints is as large as possible. As G is maximal, the length
of this path is at least rn  1, and the endpoints are not joined by an edge. Let this
path be P = ( u l , . . ., u p ) and Q, T,S be defined as in (4).
(9) We prove: d ( u J > l ( m r ) , d ( u p )ai(rn + r ) . Suppose d ( u l )< ! ( r n + r ) . A11
neighbours of u l and up are contained in P, otherwise P would not have maximal
length. As dl Z r + 2, we have d ( u , ) > r + 1 and therefore 1 S 13 d ( u l )  r > 1. All
vertices
u,,
i E S,
have
degree
at
most
d ( u l ) , otherwise
(u,, u ,  ~. ,. ., u l , u # + ~ ,
. ., u p ) would be a path of the same length as P and
d ( u , ) + d ( u p ) > d ( u l ) + d ( u p ) , contradicting the maximality assumption on the
endpoints of P. Let j o : = d(ul), then there are I S 12j o  r vertices of degree at most
jo. As we are in case (ii), dk > k + r holds for all 0 < k < i(m  r ) , which is
equivalent to d,, > j for all r < j < t ( r n + r ) . Therefore j o 3 ( r n + r ) . By similar
arguments d ( u p )3 !(rn + r ) .
(10) From (9) it follows that
+
IS I + r
+ I T l a d ( u l ) + d ( u p ) a m + r.
T h u s ~ S ~ + ~ T ~ ~ r n m , a n d f r o m ( 4 ) ( b ) w e h a v e ( P ~ ~ m + l . T hrne =r enf we
oreif
have n = I P 1 > n which is a contradiction, and in this case we are done.
(11) Let N : = N ( u J U N ( u p )U { u s , .. ., us+,}U { u , , u p } .W e prove: 1 N 12 rn. As
r 5 rn  3 , 1 { u s , .. ., us +,}n {ul, u p } /s 1.
(a) Suppose max{i E S } < min{j E T'}, where T ' : = T  {s, . . ., s + r } . This
means that the index of a neighbour of u 1which is not among us,.. ., us+,is less than
or equal to the smallest of the indices of the neighbours of up not among us,. . ., us+,.
Thus I(N(ul)n N ( u p ) )  { u s , ..., u s + , } 1 s1. Obviously
1 N l s l N ( u 1 )  { ~ ..,
~ , U~+,}I+
.
+ / { u s ,..., ZJ,+,)I+ I{u~,
 i{uS,.. ., U,+J
3
l N ( u p )  { u s ,.. ., u s + , } (
~ , } l  l ( N ( u l ) nN ( u p ) )  { u s , . .,
n {ul, up>i
I S 1  1+ 1 T'I + ( r + 1) + 2  1  1
a1 S 1 + ) T I 3 r n
U,+~}I
Graphs with cycles containing given paths
239
(b) Suppose max{i E S } 2 min{j E T’}. Let
d : = min{(i + 1)j : i E S, j E T’ such that i S j } ,
then we have d > 0. Now let io + 1  j o = d.
(b,) io + 1 s. By definition j o < s and n o vertex of the path P between u, and
u * + ~is linked t o u1 o r up by an edge. Thus
[Ul,
%+I,
%+2,
* .., u p , u,,
. ., U l ]
%I,.
is a cycle containing the path Q, all vertices u,, i E S, with the possible
exception of i = io, and all vertices u,, j E T’. It also contains u1and u,. Thus
the length of this cycle is at least:
(r +2)+)S
1
1 + 1 T’(31 S
1 + 1 TI 2 rn
which is impossible by assumption.
(b2) r + s G I . . Define the same cycle as in (b,) and by the same arguments we
obtain a contradiction.
(b,) j o < s, io > r + s. Define
i l :=max{i
jl :=min{j E T’}s j o ,
+ 1 : i E S } 3 r + s + 1.
The conditions of case (b3) imply the following:
UI
#
us, up# US+“
none of the vertices u,, jl < i s s, can be linked to u 1by an edge, none of the vertices
u,, i l < i S p , is a neighbour of u I , thus
N(u1) c{u27. . .>u J 1 }
u
{uS+l,.
.
.?
u,~}7
none of the vertices u,, 1 6 i < j , , is a neighbour of up, none of the vertices u,,
s r < i < i l , is a neighbour of up, thus
+
N(u,) c{u,,, . . ., u s + , }u {u,,,. . ., up,}.
Furthermore
I N(u1)  {us,. . ., u,+,) I = I s I,
I N ( u ,)  {u , ,..., u s + , } [ = I T ’ l .
The only vertices which might be neighbours of both u 1 and up are u,,, u,,and
us+,,. . ., u,+,.This implies
I ( N ( U Jn N ( u , ) ) {us,. . ., us +,}I
2.
Therefore
I N I 3 I N ( u J  {us,. . ., u,+,} 1 + I N ( u , )  {us,. . .,u s+ , 1} + ( r + 1)+ 2  2  1
aIS(+(T’(+r
z= 1 S
1 + I TI
m.
M. Groischel
240
These are all the cases which can occur, therefore I N 13 m is proved.
(12) Among all pairs of paths satisfying Corollary 7 with respect to P and Q
choose a pair P ' , P" such that the cycle K = [ u l ,P ' , up,P",ul] contains as many
bertices of P as possible.
(13) To show that K has length 3 rn, we will prove: K contains all vertices of N.
Suppose there is a vertex of N which is not contained in K. Trivially the vertex is
either in N ( u l ) { u s , .. ., u,,,} o r in N ( u , )  { u s , .. ., u,+,}. Without loss of generality
we assume that the vertex uk E N ( u l ) {u,, . . ., us+,}is not contained in K. Let
io= max{i 1 u, E Nn K, i < k } ,
ill=min{i 1 u, E N n K, i > k } .
(a) Suppose uq,, u,E P', then
p ; = (u1, P ' , u,, P, u,<,,P ' , u p ) ,
PI"= p " ,

is a pair of paths satisfying Corollary 7, and K , = [ u l ,P i , up,PY, ul] contains more
vertices of P then K does. Contradiction! If u,, u,,, E P" the contradiction follows
similarly.
(b) Suppose u,E P', u,E P". Let
PI = (u1, P, u,, P', up),
p:' = (u1, uk,p, ua, P''?u p ) .
If io s s, then Q is contained in (u,, P ' , u p ) ,otherwise Q is contained in ( u , , P, u,).
Therefore Pi and P: satisfy the conditions of Corollary 7, and K , contains more
vertices of P then K does. Contradiction!
(c) Suppose u,E P", u,E P ' .
(cl) io =Ss : this implies j o 6 s.
Take
pi = (u1, uk, p,u j o ,
up)
P:'= ( U l , P, ue7P", u p ) .
(cz) ill 3 r + s :
Let
P:
= ( u , , P, u,,
p : = ( u l , uk, p7
P", u p ) *
uj07
p', u p ) *
These pairs of paths satisfy Corollary 7. The contradiction follows as above.
Thus in Case (ii) we have constructed a cycle K of length 5 m containing the
path Q, which contradicts the assumption that G does not contain such a cycle, and
we are done.
Theorem 8 has some immediate Corollaries and also includes some of the
classical theorems on graphs containing cycles of a certain minimum length.
24 1
Graphs with cycles containing given paths
Corollary 9. Let d,, . . ., d, be the degree sequence of a graph G
q 2 2 and let the following condition be satisfied:
dkSkSql
=
(V,E ) . Let n
3 3,
(3)
d,,,>nk.
Furthermore, let G be 2connected i f q  1 < n  dn,  1 holds and dk > k holds for
all 1 c k s q  1. Then G contains a cycle of length at least min{n, 2q).
Proof. Take r = 0 in Theorem 8. 0
O ne of the wellknown theorems implied by Theorem 8 is the following due to
P6sa [ 7 ] , which generalizes results of Dirac [ 5 ] .
Corollary 10 (PBsa [ 7 ] ) . Let d,, . .., d, be the degree sequence of a 2connected
graph G. Let q 3 2, n 3 29. If
dk > k
for all k = 1 , . . ., 4  1,
(4)
then G contains a cycle of length at least 2q.
Proof. Immediate from Corollary 9. 0
For bipartite graphs a simple trick yields:
Corollary 11. Let G = (V, W, E ) be a bipartite graph with degree sequences
d ( v , ) S  . . S d ( v , ) a n d d ( w , ) ~ . . . s d ( w , ) ,n s m . If
d(w,)< k < n  1
+ d ( v n  k ) am

k
+ 1,
(5)
then G contains a cycle of length 2n.
Proof. Construct G * = ( V U W, E *) by adding all edges to E which have both
endpoints in V. Clearly G * contains a cycle of length 2n if and only if G does. If G
satisfies (5) then G * satisfies (3). As (5) implies that d ( w 1 ) > 2 and V defines a
clique in G*, G * is 2connected. 17
Standard theorems giving sufficient conditions for a graph to be hamiltonian can
also be derived from Theorem 8.
Corollary 12 (Berge, [l, p. 2041). Let G = ( V ,E ) be a graph with degree sequence
d,, . . ., d,. Let r be a n integer, 0 < r S n  3. If for every k with r < k < $ ( n + r ) the
following condition holds :
dk,S k
d,,,
3
n k
then for each subset Q of edges, I Q
cycle in G that contains Q.
+ r,
I = r, that forms a path
(6)
there is a hamiltonian
242
M. Grotschel
Proof. Clearly (6) is equivalent to (2) if rn = n. We have to prove that (6) implies
( r + 2)connectedness.
If there is a k with r < k < f ( n + r ) such that dk, S k , then by the arguments of
the proof of Theorem 8, Section (1) (a) ( r t 2)connectedness is assured.
If dk, > k for all r < k < 1 ( n t r ) , we have d,
Furthermore 2q
This implies
q
3
n

= 2q  q 3
2
q
+ r,
In 2
where q : = 
r and q < n  r  1 (as r < n  3), thus q t r < d,
n  ( r + q ) > n  ( q + r )  1 3 n  d,,,
dn,,.
1.
Thus condition (1) of Proposition 1 is satisfied and G is ( r + 2)connected.
Actually Berge proved a stronger theorem saying that Q only has to be a set of
edges of cardinality r such that the connected components of Q are paths.
Corollary 13 (Chvatal [4]). If the degree sequence d , , . . ., d, of a graph G, n
satisfies
dk < k < f n
dnk a n   ,
3,
(7)
then G contains a hamiltonian cycle.
Proof. Take r
=0
in Corollary 12. 0
Furthermore, Chvital showed that this theorem is best possible in the sense that
if there is a degree sequence of a graph not satisfying (7) then there exists a
nonhamiltonian graph having a degree sequence which majorizes the given one.
This proves that Theorem 8 is also best possible in this special case. Moreover
Chvital (see [4]) showed that most of the classical results on hamiltonian graphs are
contained in his theorem, and therefore are also implied by Theorem 8.
A trivial consequence of Corollary 13 which however is not too “workable” is
Corollary 14. Let G‘ be an induced subgraph of a graph G having m n vertices. If
the degree sequence d I,.. ., d A of G ‘ satisfies (7) then G contains a cycle of
length m . 0
4. Some examples
(a) We first show that the number m implied by Theorem 8 giving the minimum
length of a cycle containing a given path cannot be increased, i.e. we give an
example of a graph G with a path Q of length r such that the longest cycle
containing Q has length m .
Consider a graph with two disjoint vertex sets A and B. A is a clique of q
Graphs with cycles containing given paths
243
vertices, and B consists of p isolated vertices. Each vertex of A is linked to each
vertex of B by an edge. Suppose that 1 < q  r and p 3 q  r + 1. The degree
sequence of G is
q , q)...)q, n  1 ,
...) n  1 .
q times
p times
Hence we have
By Theorem 8 for each path Q of length r there is a cycle of length 2q  r
containing Q.
If we choose a path Q of length r such that all vertices of Q are contained in A it
is obvious that no longer cycle containing Q exists.
(b) We give an example showing that the assumption of ( r + 2)connectedness in
Theorem 8 under the specified conditions is necessary.
Consider the graph G consisting of three vertex sets A, B, C. A and B have k
vertices and are complete, C has r + 1 vertices and is complete. Each vertex of C is
joined to each vertex of A U B by an edge. Hence G is ( r + 1)connected but not
( r + 2)connected. Take a path Q of length r in C. Clearly the maximal length of a
cycle containing Q is k + r + 1. The degree sequence of this graph is
k + r , . . k + r , n  1 , ..., n  1
A'?
2k times
r
+ 1 times
+
We have di> i r for 0 < i 6 k  1, therefore Theorem 8 would imply the
existence of a cycle of length at least 2 k + r containing Q.
(c) We give an example showing that Corollary 14 is not stronger than
Corollary 9.
Consider a graph consisting of two disjoint cliques A, B, each having m vertices.
Link A and B by two disjoint edges. Obviously this graph is hamiltonian. The
degree sequence is
m  1 , ......, m  l , m , m , m , m .
P
2 m  4 times
Corollary 9 implies that there exists a cycle of length 3 2m  2, but Corollary 14
does not imply a cycle of length 3 2m  2.
(c,) Delete 2 vertices of A, both must necessarily be distinct from the two
vertices linking A to B. The degree sequence is
244
 M . Grotschel
m  3 ,..., m  3 , m  2 , m
2,m1,...,
ml,m,m
m  4 times
m  4 times
which does not satisfy (7).
(c2) Delete one vertex of A and one of B, again both must be distinct from the
vertices linking A to B. The degree sequence is
m  2 ,......, m  2 , m  l , m  l , m  l , m  1
2 ( m  3 ) times
which also does not satisfy (7)
It is clear that Corollary 9 does not imply Corollary 14.
(d) Bondy proved (see [3]) the following
Theorem (Bondy). Let G be a 2connected graph with degree sequence d , , . . ., d,. If
d, S j , dk S k (j#
k)
+ d, + d,
2
c,
(8)
then G has a cycle of length at least min (c, n ) . 0
Chvatal showed that in the case c = n his theorem (Corollary 13) implies Bondy’s
theorem, thus in the hamiltonian case Corollary 9 is stronger than the theorem of
Bondy. In general this is obviously not true, nor is the converse as the following
example shows: The graph has three vertex sets A, B, C. A = { a l , a z , a 3 } ,
B = { b , ,bZ,b,, b4}, I C 1 = m. T h e edges are the following: { a , , b J , {al,b J , {az,bl},
{ a 2 ,b3},{ a 3 ,b2},{ a 3 ,b3},{ a 3 ,b4},and all edges having both endpoints in B U C. The
degree sequence is
2,2,3, n  4
,...,It
4, n  3 , n  2 , n  2 , n  2
T
m times
d 2 < 2 and d 3< 3. By P6sa’s theorem there is a cycle of length 2 4, by Bondy’s
theorem there exists a cycle of length 3 5 . A s d,z* n  2 and dn3> n  3 and
d, > i, 4 < i < $ n, G is hamiltonian by Corollary 9.
(e) In [8] Woodall stated the following (to my knowledge unsettled)
Conjecture. Let d ] ,..., dn be the degree sequence of a 2connected graph G,
m < n  3, and let the following condition be satisfied:
[?,:.,>r
k
for 1 k < $ ( n  m  l),
if k = 5 ( n  m  1).
Then G contains a cycle of length at least n

(9)
m. 0
Obviously Corollary 9 does not imply Woodall’s Conjecture, but surprisingly nor