Edges and Cycles

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Chapter 7

• Edges and Cycles

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Line Graph

• The line graph of G, written L(G), is the simple
  graph whose vertices are edges of G, with ef
  ?E(L(G)) when e and f have a common endpoint in G.

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Eulerian circuit
Graph Theory

• An Eulerian circuit in G yields a spanning cycle
  in the line graph L(G)
• Eulerian circuit Passing through every edge in a
  graph
• Spanning cycle passing through every vertex in a
  graph
• An edge in G corresponds to a vertex in L(G)
• Hence an Eulerian circuit corresponds to a
  spanning cycle
• The converse need not hold

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Ch. 7. Edges and Cycles
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Eulerian circuit
Graph Theory
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Eulerian circuit d, e, g, h, f
Spanning cycle d, e, g, h, f
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Ch. 7. Edges and Cycles
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Matching

• A matching in G becomes an independent set in
  L(G)
• Thus ?(G) ?(L(G))
• ?( ) is the independence number
• ?( ) is the maximum size of matching
• Computing ? is harder for graphs than for line
  graphs

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Matching Red edges
Independent set Red vertices
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Edge-Coloring

• Edge-coloring problems arise when the objects
  being scheduled are pairs of underlying elements
• Example
• 2n teams, each pair of teams plays a game
• Each team plays at most once a week
• Each must play 2n-1 others, need 2n-1 weeks
• The games of each week must form a matching
• We can schedule the season in 2n-1 weeks if and
  only if we can partition E(K2n) into 2n-1
  matchings

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Example

• We can schedule the season in 2n-1 weeks if and
  only if we can partition E(K2n) into 2n-1
  matchings
• Graph modal

The games of a week a matching
Each pair A game
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k-edge-coloring

• A k-edge-coloring of G is a labeling fE(G)?S
  where S k (often S k colors), the edges
  of one color form a color class

Black edge class
Red edge class
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k-edge-coloring
Graph Theory

• A k-edge-coloring is proper if incident edges
  have different labels, that is, each color class
  is a matching
• A graph is k-edge-colorable if it has a proper
  k-edge-coloring
• The edge-chromatic number ?(G) of a loopless
  graph is the least k such that G is
  k-edge-colorable
• It is also called Chromatic index

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Ch. 7. Edges and Cycles
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Multiplicity

• In a graph G with multiple edges, we say that a
  vertex pair x, y is an edge of multiplicity m if
  there are m edges with endpoints x,y
• ?(x y) for the multiplicity of the pair x, y
• ?(G) for the maximum of the edge multiplicities
  in G

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If G is bipartite, then ?(G) ?(G) 7.1.7

• Proof
• Every regular bipartite graph H has a 1-factor
• By induction, H has a proper ?(H)-edge-coloring
• It is sufficient to show
• Given a G with ?(G) k, we have a k-regular
  graph H containing G
• 
  ?

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If G is bipartite, then ?(G) ?(G) 7.1.7

• Proof continued
• To construct such a graph, H
• Add vertices to the smaller partite set of G to
  equalize the sizes.
• If the resulting graph G is not regular, then
• Each partite set must have a vertex with degree
  less than k
• Add an edge with these two vertices as endpoints
• Continue adding such edges until the graph become
  k-regular

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If G is bipartite, then ?(G) ?(G) 7.1.7

• Proof continued
• Illustration of construction of H

?(G) 3
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Definition 7.1.8

• A decomposition of a regular graph G into
  1-factors is a 1-factorization of G
• A graph with a 1-factorization is 1-factorable
• An odd cycle is not 1-factorable
• ?(C2m1) 3 gt ?(C2m1)
• The Petersen graph also requires one and only one
  extra color

?(C5) 3, ?(C5) 2
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The Petersen graph is 4-edge-chromatic 7.1.9 1/2

• Petersen graph is 3-regular
• Deleting a perfect matching
• leaves a 2-factor
• All components are cycles
• 1-facorization can be completed only if they are
  all even cycles
• Need to show that every 2-factor ? 2 C5

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The Petersen graph is 4-edge-chromatic 7.1.9 2/2

• Consider the drawing consisting of two 5-cycles
  and a matching (the cross edges) between them
• Every cycles uses an even number m of cross edges
• m2 or m4 are infeasible
• m must be 0
• One 5-cycle needs 3 colors

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If G is a simple graph, then ?(G)??(G)1
?(G) 3 ?(G) 3
?(G) 4 ?(G) 3
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If G is a simple graph, then ?(G)??(G)1

• Proof (sketch)
• Let f be a proper ?(G)1-edge-coloring of a
  subgraph G of G
• If G ? G, then some edge uv is uncolored by f
• After recoloring some edges, we extend the
  coloring to include uv (called augmentation)
• After e(G) augmentations, we obtain a proper
  ?(G)1-edge-coloring of G

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Hamiltonian Cycles

• A Hamiltonian graph is a graph with a spanning
  cycle
• The spanning cycle is also called a Hamiltonian
  cycle
• Recall
• Eulerian cycle passes every edge
• Hamiltanion cycle passes every vertex
• We focus on simple graph

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Hamiltonian Cycle in Bipartite Graph

• A Hamiltonian Cycle in a bipartite graph visits
  the two partite sets alternatively
• Unless the partite sets have the same size, it
  has no Hamiltonian cycle

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If G has a Hamiltonian cycle, then for each
nonempty set S?V, the graph G-S has at most S
Componentsproposition 7.2.3

• Proof
• When Leaving a component of G-S, a Hamiltonian
  cycle can go only to S, and the arrivals in S
  must use distinct vertices of S.
• Hence S must have at least as many vertices as
  G-S has components

S
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Example of Proposition 7.2.3
The necessary condition holds, but it is not
sufficient
The necessary condition fails
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If G is a simple graph with at least three
vertices and ?(G)?n(G)/2, then G is Hamiltonian
7.2.8

• Proof
• The condition n(G) ? 3 is annoying but must be
  included
• Since K2 is not Hamiltonian but satisfies
• d(K2) n(K2)/2
• The proof uses contradiction and extremality
• 
  ?

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If G is a simple graph with at least three
vertices and ?(G)?n(G)/2, then G is Hamiltonian
7.2.8
Graph Theory

• Proof continue
• If there is a non-Hamiltonian graph satisfying
  the hypotheses, then adding edges cannot reduce
  the minimum degree
• Thus we may restrict our attention to maximal
  non-Hamiltonian graphs with minimum degree at
  least n/2
• "maximal" means that adding any edge joining
  nonadjacent vertices creates a spanning cycle
• 
  ?

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Ch. 7. Edges and Cycles
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If G is a simple graph with at least three
vertices and ?(G)?n(G)/2, then G is Hamiltonian
7.2.8
Graph Theory

• Proof continue
• When u ? v in G, the maximality of G implies
  that G has a spanning path v1, ... , vn. from u
  v1 to v vn., because every spanning cycle in G
  uv contains the new edge uv
• To prove the theorem, it suffices to make a small
  change in this cycle to avoid using the edge uv
  this will build a spanning cycle in G

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Ch. 7. Edges and Cycles
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If G is a simple graph with at least three
vertices and ?(G)?n(G)/2, then G is Hamiltonian
7.2.8
Graph Theory

• Proof continue
• If a neighbor of u directly follows a neighbor of
  v on the path, such as u ? vi1 and v ? vi then
  (u, vi1 vi2 .,v , vi, vi-1.. v2) is a
  spanning cycle
• To prove that such a cycle exists, we show that
  there is a common index in the sets S and T
  defined by S i u ? vi1 and T i v
  vi
• 
  ?

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Ch. 7. Edges and Cycles
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If G is a simple graph with at least three
vertices and ?(G)?n(G)/2, then G is Hamiltonian
7.2.8
Graph Theory

• Proof continue
• Summing the sizes of these sets yields
• S ? T S n T S T d(u) d(v)
  n
• Neither S nor T contains the index n.
• Thus S ? T lt n, and hence S n T 1
• We have established a contradiction by finding a
  spanning cycle in G
• Hence there is no (maximal) non-Hamiltonian graph
  satisfying the hypotheses

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Ch. 7. Edges and Cycles
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Lemma Let G be a simple graph. If u, v are
distinct non-adjacent vertices of G with d(u)
d(v) gt n(G), then G is Hamiltonian if and only
if Guv is Hamiltonian 7.2.9

• Proof
• The proof is the same as for Theorem 7.2.8

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Hamiltonian Closure

• The Hamiltonian closure of a graph G, denoted
  C(G), is the graph with vertex set V(G) obtained
  from G by iteratively adding edges joining pairs
  of non-adjacent vertices whose degree sum is at
  least n, until no such pair remains.

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A simple n-vertex graph is Hamiltonian if and
only if its closure is Hamiltonian 7.2.11

• It is seen that the graph above begins with
  vertices of degree 2 but its closure is K6
• According to Ores Lemma, it is proven

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Lemma The closure of G is well-defined. 7.2.12.

• Fortunately, the closure does not depend on the
  order in which we choose to add edges when more
  than one is available

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Lemma The closure of G is well-defined 7.2.12.

• Proof Let e1, .... er , and f1 .... , fs be
  sequences of edges added in forming C(G), the
  first yielding G1, and the second G2.
• If in either sequence nonadjacent vertices u and
  v acquire degree summing to at least n(G), then
  the edge uv' must be added before the sequence
  ends

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Lemma The closure of G is well-defined. 7.2.12.

• Proof continue
• Thus f1, being initially addable to G, must
  belong to G1
• Similarly, if f1 .... , fi-1 ? E(G1), then fi
  becomes addable to G1, and therefore belongs to
  G1
• Hence neither sequence contains a first edge
  omitted by the other sequence, and we have G1 ?
  G2 and G2 ? G1

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Condition to test for Hamiltonian cycles

• We now have a necessary and sufficient condition
  to test for Hamiltonian cycles in simple graphs
• It doesn't help much, because it requires us to
  test whether another graph is Hamiltonian!
• Nevertheless, it does furnish a method for
  proving sufficient conditions
• A condition that forces C(G) to be Hamiltonian
  also forces a Hamiltonian cycle in G

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Example

• The condition may imply C(G) Kn
• Chvatal used this method to prove the best
  possible degree sequence condition for
  Hamiltonian cycles.
• Some vertex degrees can be small if others are
  large enough.

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Theorem Let G be a simple graph with vertex
degrees d1 ... dn", where n3. If i lt n/2
implies that di gt i or dn-i n-i (Chvatal's
condition), then G is Hamiltonian. 7.2.13
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Theorem Let G be a simple graph with vertex
degrees d1 ... dn", where n3. If i lt n/2
implies that di gt i or dn-i n-i (Chvatal's
condition), then G is Hamiltonian. 7.2.13

• Proof Adding edges to form the closure reduces
  no entry in the degree sequence.
• Also, G is Hamiltonian if and only if C(G) is
  Hamiltonian.
• Thus it suffices to consider the case where C (G)
  G, which we describe by saying that G is
  closed.
• In this case, we prove that Chvatal's condition
  implies that G Kn.

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Theorem Let G be a simple graph with vertex
degrees d1 ... dn", where n3. If i lt n/2
implies that di gt i or dn-i n-i (Chvatal's
condition), then G is Hamiltonian. 7.2.13

• Proof continue
• We prove the contrapositive
• If G is a closed /I-vertex graph that is not a
  complete graph, then we construct a value of i
  less than n /2 for which Chvatal's condition is
  violated.
• Violation means that at least i vertices have
  degree at most i and at least n - i vertices have
  degree less than n - i.

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Theorem 7.2.13

• With G?Kn, we choose among the pairs of
  nonadjacent vertices a pair u, v with maximum
  degree sum.
• Because G is closed, u ? v implies that d(u)
  d(v) lt n.
• We choose the labels on u. v so that d(u) d(v)
• Since d(u) d(v) lt n, we thus have d(u) lt n/2.
  Let i d(u).

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Theorem 7.2.13

• We need to find i vertices with degree at most i
• Because we chose a nonadjacent pair with maximum
  degree sum, every vertex of V - v that is not
  adjacent to v has degree at most d(u), which
  equals i.
• There are n- 1 - d(v) such vertices, and d(u)
  d(v) n - 1 yields n - 1 - d(v) i.

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Theorem 7.2.13

• We also need n - i vertices with degree less than
  n - i.
• Every vertex of V - u that is not adjacent to u
  has degree at most d(v), and we have d(v) lt n -
  d(u) n - i.
• There are n -1- d(u) such vertices.
• Since d(u) d(v), we can also add u itself to the
  set of vertices with degree at most d(v).
• We thus obtain n - i vertices with degree less
  than n - i.
• We have proved that di i and dn-ilt n - i for
  this specially chosen I, which contradicts the
  hypothesis. _

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Example Non-Hamiltonian. graphs with "large"
vertex degrees 7.2.14

• Theorem 7.2.13 characterizes the degree
  sequences of simple graphs that force Hamiltonian
  cycles.
• If the degree sequence fails Chvatal's condition
  at i , then the largest we can make the terms in
  dl, ... dn is
• dji for j i,
• dj n - i-1 for i1 j n - i,
• dj n - 1 for j gt n - i.

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Example 7.2.14.

• Let G be a simple graph realizing this degree
  sequence (if it exists).
• The i vertices of degree n - 1 are adjacent to
  all others (the central clique in the figure).
• This already gives i neighbors to the i vertices
  of degree i, so they form an independent set and
  have no additional neighbors.

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Example 7.2.14.

• With degree n-i-1, each of the remaining n - 2i
  vertices must be adjacent to all vertices except
  itself and the independent set
• Thus these vertices form a clique
• The only possible realization is (Ki Kn-2i) V
  Ki , shown below

Kn-2i
Ki
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Example 7.2.14.

• This graph is not Hamiltonian because deleting
  the i vertices of degree n - 1 leaves a subgraph
  with i 1 components.
• If a simple graph H is nonHamiltonian and has
  vertex degrees d1 ... dn, then Chvátal's
  result implies that for some i the graph (Ki
  Kn-2i) v Ki with vertex degrees d1 ... dn
  satisfies dj dj for all i

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Definition A Hamiltonian path is a spanning path
7.2.15

• Every graph with a spanning cycle has a spanning
  path
• But Pn shows that the converse is not true
• We could make arguments like those above to prove
  sufficient conditions for Hamiltonian paths
• But it is easier to use our previous work and
  prove the new theorem by invoking a theorem about
  cycles
• To do this, we use a standard transformation

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Remark A graph G has a spanning path if and only
if the graph G v K1 has a spanning cycle 7.2.16

• This remark applies in several of the exercises
• Here we use it to derive the analogue for paths
  of Chvatal's condition for spanning cycles.

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Theorem Let G be a simple graph with vertex
degrees d1 ... dn. If i lt (n 1)/2 implies
(di i or dn1-i n i), then G has a
spanning path. 7.2.17

• Proof Let G' G v K1 let n' n 1, and let
  d1 .... , dn be the degree sequence of G'.
• Since a spanning cycle in G v K1 becomes a
  spanning path in G when the extra vertex is
  deleted,
• It suffices to show that G' satisfies Chvatal's
  sufficient condition for Hamiltonian cycles.

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Theorem Let G be a simple graph with vertex
degrees d1 ... dn. If i lt (n 1)/2 implies
(di i or dn1-i n i), then G has a
spanning path. 7.2.17

• Since the new vertex is adjacent to all of V(G),
  we have dn n and dj dj 1 for j lt n'. For
  i lt n' /2 (n 1)/2, the hypothesis on G yields
  
• di di 1 i 1 gt i or
• dn-i dn1-i n - i 1 n' - i.
• This is precisely Chvatal's sufficient condition,
  so G' has a spanning cycle, and deleting the
  extra vertex leaves a spanning path in G.

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Remark 7.2.18

• The degree requirements can be weakened under
  conditions such as regularity or high toughness.
• Every regular simple graph G with vertex degrees
  at least n(G)/3 is Hamiltonian (Jackson 1980).
• Only the Petersen graph prevents lowering the
  threshold to (n(G) - 1)/3 (Zhu-Liu-Yu 1985,
  partly simplified in Bondy-Kouider 1988 see
  also Exercise 13).

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Remark 7.2.18

• It may be possible to lower the degree condition
  further when connectivity is high.
• For example, Tutte 1971) conjectured that every
  3connecterl 3-regular bipartite graph is
  Hamiltonian.
• Horton 1982 found a counterexample with 96
  vertices, and the smallest known counterexample
  bas 50 vertices (Georges 1989), but stronger
  conditions of this sort may suffice.
• Our last sufficient condition for Hamiltonian
  cycles involves connectivity and independence,
  not degrees.
• The proof yields a good algorithm that constructs
  a Hamiltonian cycle or shows that the hypothesis
  is false.

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Th If ?(G) ?a(G), then G has a Hamiltonian cycle
(unless G K2) 7.2.19

• Proof
• With GK2 the conditions require ?(G) gt 1
• Suppose that ?(G)a(G). Let k ?(G), and let C
  be a longest cycle in G
• Since d(G) ?(G), and every graph with d(G) 2
  has a cycle of length at least d(G) 1
  (Proposition 1.2.28), C has at least k 1
  vertices

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Theorem If ?(G) a(G), then G has a Hamiltonian
cycle (unless G K2) 7.2.19

• Proof continue
• Let H be a component of G - V (C). The cycle C
  has at least k vertices with edges to H
  otherwise, deleting the vertices of C with edges
  to H contradicts K (G) k
• Let u1... uk be k vertices of C with edges to H,
  in clockwise order

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Theorem If ?(G) a(G), then G has a Hamiltonian
cycle (unless G K2) 7.2.19

• Proof continue
• For i 1, k, let ai be the vertex immediately
  following ui on C.
• If any two of these vertices are adjacent, say ai
  ? aj, then we construct a longer cycle by using
  ai aj, the portions of C from ai to uj and aj to
  ui, and a ui, uj-path through H (see
  illustration).

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Theorem If ?(G) a(G), then G has a Hamiltonian
cycle (unless G K2) 7.2.19

• Proof continue
• If ai has a neighbor in H, then we can detour to
  H between ui and ai, on C.
• Thus we also conclude that no ai has a neighbor
  in H.
• Hence a1.... ak plus a vertex of H forms an
  independent set of size k 1.
• This contradiction implies that C is a
  Hamiltonian cycle

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Remark 7.2.20

• Most sufficient conditions for Hamiltonian cycles
  generalize to conditions for long cycles.
• The circumference of a graph is the length of its
  longest cycle. A weaker form of a sufficient
  condition for spanning cycles may force a long
  cycle.

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Remark 7.2.20

• Dirac 1952b proved the first such result a
  2-connected graph with minimum degree k has
  circumference at least minn,2k.
• Proposition 1.2.28 only guarantees a cycle of
  length at least k1.
• Most long-cycle results are more difficult than
  the corresponding sufficient conditions for
  Hamiltonian cycles (see Lemma 8.4.36 -Theorem
  8.4.37).