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Chapter 14. Graphs with Cycles Containing Given Paths

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 h-connected.



-



dn-,,+,- 1.



(1)



0



A well-known property of h-connected 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 h-connected. Let W = { w l , .. ., w h } be a set of

vertices, 1 W 1 = h. If v E V - W, there exist h vertex-disjoint 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 (Menger-Dirac). 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 3-connected 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 vertex-disjoint 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 vertex-disjoint

, = (ao, . . ., &), p3 = (ao,.. ., a3). Define

paths (G is 3-connected): 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

2-connected. 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 3-connected 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

sequence-for 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 which-in spirit -is due to Nash-Williams

[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-,sn-k



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



+ dn-h+2-k 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 dn-h+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 h-connected.

(b) Suppose d, 3 k + h - 1 for all 0 < k < 4 (rn - h + 2), then G is h-connected



237



Graphs with cycles containing given paths



by Proposition 1 if f ( r n - h + 2) > n - dn-,,+,

- 1. Otherwise h-connectedness

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



V-PI.



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+JV-PI+~-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:



dkSkSq-l



=



(V,E ) . Let n



3 3,



(3)



d,-,,>n-k.



Furthermore, let G be 2-connected 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 well-known 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 2-connected

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 2-connected. 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



dn-k 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

non-hamiltonian 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,m-1,...,



m-l,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 2-connected 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 dn-3> 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 2-connected 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



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Chapter 14. Graphs with Cycles Containing Given Paths

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