cycle spectra of hamiltonian graphs - wvu math...
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Cycle Spectra of Hamiltonian Graphs
Kevin G. Milans ([email protected])University of South Carolina
Joint with F. Pfender, D. Rautenbach, F. Regen, and D. B. West
Fall 2011 Southeastern Section Meeting of the AMSWake Forest University
Winston-Salem, NC25 September 2011
Cycle spectrumI The cycle spectrum of a graph G is the set of lengths of
cycles in G .
I Let S(G ) denote the cycle spectrum of G .
I Let s(G ) denote |S(G )|.
Cycle spectrumI The cycle spectrum of a graph G is the set of lengths of
cycles in G .
I Let S(G ) denote the cycle spectrum of G .
I Let s(G ) denote |S(G )|.
Cycle spectrumI The cycle spectrum of a graph G is the set of lengths of
cycles in G .
I Let S(G ) denote the cycle spectrum of G .
I Let s(G ) denote |S(G )|.
Cycle spectrumI The cycle spectrum of a graph G is the set of lengths of
cycles in G .
I Let S(G ) denote the cycle spectrum of G .
I Let s(G ) denote |S(G )|.
Theorem (Bondy (1971))
If d(u) + d(v) ≥ n whenever u and v are non-adjacent, thenG = Kn/2,n/2 or S(G ) = 3, . . . , n.
Theorem (Gould–Haxell–Scott (2002))
∀ε > 0 ∃c: if G is a graph with δ(G ) ≥ εn and maximum evencycle length 2`, then S(G ) contains all even lengths up to 2`− c.
Conjecture
∃c : if G is a Hamiltonian subgraph of Kn,n with δ(G ) ≥ c√n, then
S(G ) = 4, 6, . . . , 2n.
Cycle spectrumI The cycle spectrum of a graph G is the set of lengths of
cycles in G .
I Let S(G ) denote the cycle spectrum of G .
I Let s(G ) denote |S(G )|.
Theorem (Bondy (1971))
If d(u) + d(v) ≥ n whenever u and v are non-adjacent, thenG = Kn/2,n/2 or S(G ) = 3, . . . , n.
Theorem (Gould–Haxell–Scott (2002))
∀ε > 0 ∃c: if G is a graph with δ(G ) ≥ εn and maximum evencycle length 2`, then S(G ) contains all even lengths up to 2`− c.
Conjecture
∃c : if G is a Hamiltonian subgraph of Kn,n with δ(G ) ≥ c√n, then
S(G ) = 4, 6, . . . , 2n.
Cycle spectrumI The cycle spectrum of a graph G is the set of lengths of
cycles in G .
I Let S(G ) denote the cycle spectrum of G .
I Let s(G ) denote |S(G )|.
Theorem (Bondy (1971))
If d(u) + d(v) ≥ n whenever u and v are non-adjacent, thenG = Kn/2,n/2 or S(G ) = 3, . . . , n.
Theorem (Gould–Haxell–Scott (2002))
∀ε > 0 ∃c: if G is a graph with δ(G ) ≥ εn and maximum evencycle length 2`, then S(G ) contains all even lengths up to 2`− c.
Conjecture
∃c : if G is a Hamiltonian subgraph of Kn,n with δ(G ) ≥ c√n, then
S(G ) = 4, 6, . . . , 2n.
Cycle spectrumI The cycle spectrum of a graph G is the set of lengths of
cycles in G .
I Let S(G ) denote the cycle spectrum of G .
I Let s(G ) denote |S(G )|.
Conjecture (Erdos)
If G has girth g and average degree k, then s(G ) ≥ Ω(kb(g−1)/2c).
I (Erdos–Faudree–Rousseau–Schelp 1999) True for g = 5.
I (Sudakov–Verstraete 2008) True for all g .
Question (Jacobson–Lehel)
I Lower bounds on s(G ) when G is Hamiltonian and k-regular.
I In particular, what about k = 3?
Cycle spectrumI The cycle spectrum of a graph G is the set of lengths of
cycles in G .
I Let S(G ) denote the cycle spectrum of G .
I Let s(G ) denote |S(G )|.
Conjecture (Erdos)
If G has girth g and average degree k, then s(G ) ≥ Ω(kb(g−1)/2c).
I (Erdos–Faudree–Rousseau–Schelp 1999) True for g = 5.
I (Sudakov–Verstraete 2008) True for all g .
Question (Jacobson–Lehel)
I Lower bounds on s(G ) when G is Hamiltonian and k-regular.
I In particular, what about k = 3?
Cycle spectrumI The cycle spectrum of a graph G is the set of lengths of
cycles in G .
I Let S(G ) denote the cycle spectrum of G .
I Let s(G ) denote |S(G )|.
Conjecture (Erdos)
If G has girth g and average degree k, then s(G ) ≥ Ω(kb(g−1)/2c).
I (Erdos–Faudree–Rousseau–Schelp 1999) True for g = 5.
I (Sudakov–Verstraete 2008) True for all g .
Question (Jacobson–Lehel)
I Lower bounds on s(G ) when G is Hamiltonian and k-regular.
I In particular, what about k = 3?
Cycle spectrumI The cycle spectrum of a graph G is the set of lengths of
cycles in G .
I Let S(G ) denote the cycle spectrum of G .
I Let s(G ) denote |S(G )|.
Conjecture (Erdos)
If G has girth g and average degree k, then s(G ) ≥ Ω(kb(g−1)/2c).
I (Erdos–Faudree–Rousseau–Schelp 1999) True for g = 5.
I (Sudakov–Verstraete 2008) True for all g .
Question (Jacobson–Lehel)
I Lower bounds on s(G ) when G is Hamiltonian and k-regular.
I In particular, what about k = 3?
Cycle spectrumI The cycle spectrum of a graph G is the set of lengths of
cycles in G .
I Let S(G ) denote the cycle spectrum of G .
I Let s(G ) denote |S(G )|.
Conjecture (Erdos)
If G has girth g and average degree k, then s(G ) ≥ Ω(kb(g−1)/2c).
I (Erdos–Faudree–Rousseau–Schelp 1999) True for g = 5.
I (Sudakov–Verstraete 2008) True for all g .
Question (Jacobson–Lehel)
I Lower bounds on s(G ) when G is Hamiltonian and k-regular.
I In particular, what about k = 3?
Cycle spectrumI The cycle spectrum of a graph G is the set of lengths of
cycles in G .
I Let S(G ) denote the cycle spectrum of G .
I Let s(G ) denote |S(G )|.
Example (Jacobson–Lehel)
I S(G ) = 4, 6 ∪23n,
23n + 2, 23n + 4, . . . , n
I s(G ) = n/6 + 3
I Generalizes to provide k-regular Hamiltonian graphs withs(G ) = k−2
2k n + k when 2k divides n.
Cycle spectrumI The cycle spectrum of a graph G is the set of lengths of
cycles in G .
I Let S(G ) denote the cycle spectrum of G .
I Let s(G ) denote |S(G )|.
Example (Jacobson–Lehel)
I S(G ) = 4, 6 ∪23n,
23n + 2, 23n + 4, . . . , n
I s(G ) = n/6 + 3
I Generalizes to provide k-regular Hamiltonian graphs withs(G ) = k−2
2k n + k when 2k divides n.
Cycle spectrumI The cycle spectrum of a graph G is the set of lengths of
cycles in G .
I Let S(G ) denote the cycle spectrum of G .
I Let s(G ) denote |S(G )|.
Example (Jacobson–Lehel)
I S(G ) = 4, 6 ∪23n,
23n + 2, 23n + 4, . . . , n
I s(G ) = n/6 + 3
I Generalizes to provide k-regular Hamiltonian graphs withs(G ) = k−2
2k n + k when 2k divides n.
Cycle spectrumI The cycle spectrum of a graph G is the set of lengths of
cycles in G .
I Let S(G ) denote the cycle spectrum of G .
I Let s(G ) denote |S(G )|.
Example (Jacobson–Lehel)
I S(G ) = 4, 6 ∪23n,
23n + 2, 23n + 4, . . . , n
I s(G ) = n/6 + 3
I Generalizes to provide k-regular Hamiltonian graphs withs(G ) = k−2
2k n + k when 2k divides n.
How small can the cycle spectrum be?
DefinitionLet fn(m) be the minimum size of the cycle spectrum of ann-vertex Hamiltonian graph with m edges.
Theorem (Bondy (1971))
If G is an n-vertex Hamiltonian graph with m edges and m > n2/4,then G is pancyclic (has cycles of all lengths from 3 to n).
Theorem (Entringer–Schmeichel (1988))
If G is an n-vertex bipartite Hamiltonian graph with m edges andm > n2/8, then G is bipancyclic (has cycles of all even lengthsfrom 4 to n).
How small can the cycle spectrum be?
DefinitionLet fn(m) be the minimum size of the cycle spectrum of ann-vertex Hamiltonian graph with m edges.
Theorem (Bondy (1971))
If G is an n-vertex Hamiltonian graph with m edges and m > n2/4,then G is pancyclic (has cycles of all lengths from 3 to n).
Theorem (Entringer–Schmeichel (1988))
If G is an n-vertex bipartite Hamiltonian graph with m edges andm > n2/8, then G is bipancyclic (has cycles of all even lengthsfrom 4 to n).
How small can the cycle spectrum be?
DefinitionLet fn(m) be the minimum size of the cycle spectrum of ann-vertex Hamiltonian graph with m edges.
Theorem (Bondy (1971))
If G is an n-vertex Hamiltonian graph with m edges and m > n2/4,then G is pancyclic (has cycles of all lengths from 3 to n).
Theorem (Entringer–Schmeichel (1988))
If G is an n-vertex bipartite Hamiltonian graph with m edges andm > n2/8, then G is bipancyclic (has cycles of all even lengthsfrom 4 to n).
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
ei ej
d(ei , ej) d(ei , ej)
e1 e2e1 e3e1 e4e1 e5e1 e2 ej
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
ei ej
d(ei , ej) d(ei , ej)
e1 e2e1 e3e1 e4e1 e5e1 e2 ej
I Let n be the length of P.
I Let e1, . . . , eh be the chords in G .
I Let Pi ,j be the x , y -path using ei , ej , and edges of P.
I The length of Pi ,j is n + 2− 2d(ei , ej).
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
ei ej
d(ei , ej) d(ei , ej)
e1 e2e1 e3e1 e4e1 e5e1 e2 ej
I Let n be the length of P.
I Let e1, . . . , eh be the chords in G .
I Let Pi ,j be the x , y -path using ei , ej , and edges of P.
I The length of Pi ,j is n + 2− 2d(ei , ej).
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.ei ej
d(ei , ej) d(ei , ej)
e1 e2e1 e3e1 e4e1 e5e1 e2 ej
I Let n be the length of P.
I Let e1, . . . , eh be the chords in G .
I Let Pi ,j be the x , y -path using ei , ej , and edges of P.
I The length of Pi ,j is n + 2− 2d(ei , ej).
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.ei ej
d(ei , ej) d(ei , ej)
e1 e2e1 e3e1 e4e1 e5e1 e2 ej
I Let n be the length of P.
I Let e1, . . . , eh be the chords in G .
I Let Pi ,j be the x , y -path using ei , ej , and edges of P.
I The length of Pi ,j is n + 2− 2d(ei , ej).
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.ei ej
d(ei , ej) d(ei , ej)
e1 e2e1 e3e1 e4e1 e5e1 e2 ej
I The length of Pi ,j is n + 2− 2d(ei , ej).
I Already, P1,2, . . . ,P1,h provide h − 1 lengths.
I Only h − 1 lengths: every length is realized by P1,j for some j .
I Length n is realized: e2 immediately follows e1.
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
ei ej
d(ei , ej) d(ei , ej)
e1 e2e1 e3e1 e4e1 e5e1 e2 ej
I The length of Pi ,j is n + 2− 2d(ei , ej).
I Already, P1,2, . . . ,P1,h provide h − 1 lengths.
I Only h − 1 lengths: every length is realized by P1,j for some j .
I Length n is realized: e2 immediately follows e1.
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
ei ej
d(ei , ej) d(ei , ej)
e1 e2
e1 e3e1 e4e1 e5e1 e2 ej
I The length of Pi ,j is n + 2− 2d(ei , ej).
I Already, P1,2, . . . ,P1,h provide h − 1 lengths.
I Only h − 1 lengths: every length is realized by P1,j for some j .
I Length n is realized: e2 immediately follows e1.
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
ei ej
d(ei , ej) d(ei , ej)
e1 e2
e1 e3
e1 e4e1 e5e1 e2 ej
I The length of Pi ,j is n + 2− 2d(ei , ej).
I Already, P1,2, . . . ,P1,h provide h − 1 lengths.
I Only h − 1 lengths: every length is realized by P1,j for some j .
I Length n is realized: e2 immediately follows e1.
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
ei ej
d(ei , ej) d(ei , ej)
e1 e2e1 e3
e1 e4
e1 e5e1 e2 ej
I The length of Pi ,j is n + 2− 2d(ei , ej).
I Already, P1,2, . . . ,P1,h provide h − 1 lengths.
I Only h − 1 lengths: every length is realized by P1,j for some j .
I Length n is realized: e2 immediately follows e1.
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
ei ej
d(ei , ej) d(ei , ej)
e1 e2e1 e3e1 e4
e1 e5
e1 e2 ej
I The length of Pi ,j is n + 2− 2d(ei , ej).
I Already, P1,2, . . . ,P1,h provide h − 1 lengths.
I Only h − 1 lengths: every length is realized by P1,j for some j .
I Length n is realized: e2 immediately follows e1.
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
ei ej
d(ei , ej) d(ei , ej)
e1 e2e1 e3e1 e4
e1 e5
e1 e2 ej
I The length of Pi ,j is n + 2− 2d(ei , ej).
I Already, P1,2, . . . ,P1,h provide h − 1 lengths.
I Only h − 1 lengths: every length is realized by P1,j for some j .
I Length n is realized: e2 immediately follows e1.
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
ei ej
d(ei , ej) d(ei , ej)
e1 e2e1 e3e1 e4e1 e5e1 e2 ej
I The length of Pi ,j is n + 2− 2d(ei , ej).
I Already, P1,2, . . . ,P1,h provide h − 1 lengths.
I Only h − 1 lengths: every length is realized by P1,j for some j .
I Length n is realized: e2 immediately follows e1.
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
ei ej
d(ei , ej) d(ei , ej)
e1 e2e1 e3e1 e4e1 e5
e1 e2
ej
I The length of Pi ,j is n + 2− 2d(ei , ej).
I Already, P1,2, . . . ,P1,h provide h − 1 lengths.
I Only h − 1 lengths: every length is realized by P1,j for some j .
I Length n is realized: e2 immediately follows e1.
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
ei ej
d(ei , ej) d(ei , ej)
e1 e2e1 e3e1 e4e1 e5
e1 e2 ej
I Consider a chord ej .
I The length of P2,j . . .
I . . . is also realized by P1,i .
I So, there is a chord immediately preceding ej .
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
ei ej
d(ei , ej) d(ei , ej)
e1 e2e1 e3e1 e4e1 e5
e1 e2 ej
I Consider a chord ej .
I The length of P2,j . . .
I . . . is also realized by P1,i .
I So, there is a chord immediately preceding ej .
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
ei ej
d(ei , ej) d(ei , ej)
e1 e2e1 e3e1 e4e1 e5
e1 e2 ej
I Consider a chord ej .
I The length of P2,j . . .
I . . . is also realized by P1,i .
I So, there is a chord immediately preceding ej .
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
ei ej
d(ei , ej) d(ei , ej)
e1 e2e1 e3e1 e4e1 e5
e1 e2 ej
I Consider a chord ej .
I The length of P2,j . . .
I . . . is also realized by P1,i .
I So, there is a chord immediately preceding ej .
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
ei ej
d(ei , ej) d(ei , ej)
e1 e2e1 e3e1 e4e1 e5
e1 e2 ej
I Consider a chord ej .
I The length of P2,j . . .
I . . . is also realized by P1,i .
I So, there is a chord immediately preceding ej .
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
ei ej
d(ei , ej) d(ei , ej)
e1 e2e1 e3e1 e4e1 e5
e1 e2 ej
I Consider a chord ej .
I The length of P2,j . . .
I . . . is also realized by P1,i .
I So, there is a chord immediately preceding ej .
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
ei ej
d(ei , ej) d(ei , ej)
e1 e2e1 e3e1 e4e1 e5
e1 e2 ej
I Consider a chord ej .
I The length of P2,j . . .
I . . . is also realized by P1,i .
I So, there is a chord immediately preceding ej .
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
ei ej
d(ei , ej) d(ei , ej)
e1 e2e1 e3e1 e4e1 e5
e1 e2 ej
I Consider a chord ej .
I The length of P2,j . . .
I . . . is also realized by P1,i .
I So, there is a chord immediately preceding ej .
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
ei ej
d(ei , ej) d(ei , ej)
e1 e2e1 e3e1 e4e1 e5
e1 e2 ej
I Lengths of paths: n, n − 2, . . . , n − 2(h − 2).
I Path with a single chord: length n − (`− 1).
I So `− 1 ∈ 0, 2, . . . , 2(h − 2).
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
ei ej
d(ei , ej) d(ei , ej)
e1 e2e1 e3e1 e4e1 e5
e1 e2 ej
I Lengths of paths: n, n − 2, . . . , n − 2(h − 2).
I Path with a single chord: length n − (`− 1).
I So `− 1 ∈ 0, 2, . . . , 2(h − 2).
Overlapping chords lemma
LemmaG: an x , y-path P plus h pairwise-overlapping chords of length `.Then G contains x , y-paths of h − 1 distinct lengths. Having onlyh − 1 lengths requires that
1. the chords are consecutive along P, and
2. ` is odd and h ≥ (`+ 3)/2.
Proof.
ei ej
d(ei , ej) d(ei , ej)
e1 e2e1 e3e1 e4e1 e5
e1 e2 ej
I Lengths of paths: n, n − 2, . . . , n − 2(h − 2).
I Path with a single chord: length n − (`− 1).
I So `− 1 ∈ 0, 2, . . . , 2(h − 2).
Greedy chord system
I G : Hamiltonian cycle C plus q chords of length `
I Find many distinct cycle lengths using a greedy chord system.
e1
e2
e3
e4 eα
e?
Greedy chord system
I G : Hamiltonian cycle C plus q chords of length `
I Find many distinct cycle lengths using a greedy chord system.
e1
e2
e3
e4 eα
e?
Greedy chord system
I G : Hamiltonian cycle C plus q chords of length `
I Find many distinct cycle lengths using a greedy chord system.
e1
e2
e3
e4 eα
e?
I Choose a forwarddirection along C .
I e1: chord with mostoverlapping chords goingforward.
I e2: first chord notoverlapping e1.
I e3: first chord notoverlapping e2 or e1.
I e4: first chord notoverlapping e3 or e1.
Greedy chord system
I G : Hamiltonian cycle C plus q chords of length `
I Find many distinct cycle lengths using a greedy chord system.
e1
e2
e3
e4 eα
e?
I Choose a forwarddirection along C .
I e1: chord with mostoverlapping chords goingforward.
I e2: first chord notoverlapping e1.
I e3: first chord notoverlapping e2 or e1.
I e4: first chord notoverlapping e3 or e1.
Greedy chord system
I G : Hamiltonian cycle C plus q chords of length `
I Find many distinct cycle lengths using a greedy chord system.
e1
e2
e3
e4 eα
e?
I Choose a forwarddirection along C .
I e1: chord with mostoverlapping chords goingforward.
I e2: first chord notoverlapping e1.
I e3: first chord notoverlapping e2 or e1.
I e4: first chord notoverlapping e3 or e1.
Greedy chord system
I G : Hamiltonian cycle C plus q chords of length `
I Find many distinct cycle lengths using a greedy chord system.
e1
e2
e3
e4 eα
e?
I Choose a forwarddirection along C .
I e1: chord with mostoverlapping chords goingforward.
I e2: first chord notoverlapping e1.
I e3: first chord notoverlapping e2 or e1.
I e4: first chord notoverlapping e3 or e1.
Greedy chord system
I G : Hamiltonian cycle C plus q chords of length `
I Find many distinct cycle lengths using a greedy chord system.
e1
e2
e3
e4 eα
e?
I Choose a forwarddirection along C .
I e1: chord with mostoverlapping chords goingforward.
I e2: first chord notoverlapping e1.
I e3: first chord notoverlapping e2 or e1.
I e4: first chord notoverlapping e3 or e1.
Greedy chord system
I G : Hamiltonian cycle C plus q chords of length `
I Find many distinct cycle lengths using a greedy chord system.
e1
e2
e3
e4 eα
e?
I Choose a forwarddirection along C .
I e1: chord with mostoverlapping chords goingforward.
I e2: first chord notoverlapping e1.
I e3: first chord notoverlapping e2 or e1.
I e4: first chord notoverlapping e3 or e1.
Greedy chord system
I G : Hamiltonian cycle C plus q chords of length `
I Find many distinct cycle lengths using a greedy chord system.
e1
e2
e3
e4
eα
e?
I Choose a forwarddirection along C .
I e1: chord with mostoverlapping chords goingforward.
I e2: first chord notoverlapping e1.
I e3: first chord notoverlapping e2 or e1.
I e4: first chord notoverlapping e3 or e1.
Greedy chord system
I G : Hamiltonian cycle C plus q chords of length `
I Find many distinct cycle lengths using a greedy chord system.
e1
e2
e3
e4 eα
e?
I The process ends with eα,when all remaining chordsin the forward directionoverlap eα or e1.
I For 1 ≤ j ≤ α, let Fjconsist of ej plus chordsoverlapping ej goingforward.
Greedy chord system
I G : Hamiltonian cycle C plus q chords of length `
I Find many distinct cycle lengths using a greedy chord system.
e1
e2
e3
e4 eα
e?
I The process ends with eα,when all remaining chordsin the forward directionoverlap eα or e1.
I For 1 ≤ j ≤ α, let Fjconsist of ej plus chordsoverlapping ej goingforward.
Greedy chord system
I G : Hamiltonian cycle C plus q chords of length `
I Find many distinct cycle lengths using a greedy chord system.
e1
e2
e3
e4 eα
e?
I The process ends with eα,when all remaining chordsin the forward directionoverlap eα or e1.
I For 1 ≤ j ≤ α, let Fjconsist of ej plus chordsoverlapping ej goingforward.
Greedy chord system
I G : Hamiltonian cycle C plus q chords of length `
I Find many distinct cycle lengths using a greedy chord system.
e1
e2
e3
e4 eα
e?
I The process ends with eα,when all remaining chordsin the forward directionoverlap eα or e1.
I For 1 ≤ j ≤ α, let Fjconsist of ej plus chordsoverlapping ej goingforward.
Greedy chord system
I G : Hamiltonian cycle C plus q chords of length `
I Find many distinct cycle lengths using a greedy chord system.
e1
e2
e3
e4 eα
e?
I The process ends with eα,when all remaining chordsin the forward directionoverlap eα or e1.
I For 1 ≤ j ≤ α, let Fjconsist of ej plus chordsoverlapping ej goingforward.
Greedy chord system
I G : Hamiltonian cycle C plus q chords of length `
I Find many distinct cycle lengths using a greedy chord system.
e1
e2
e3
e4 eα
e?
I The process ends with eα,when all remaining chordsin the forward directionoverlap eα or e1.
I For 1 ≤ j ≤ α, let Fjconsist of ej plus chordsoverlapping ej goingforward.
Greedy chord system
I G : Hamiltonian cycle C plus q chords of length `
I Find many distinct cycle lengths using a greedy chord system.
e1
e2
e3
e4 eα
e?
I The process ends with eα,when all remaining chordsin the forward directionoverlap eα or e1.
I For 1 ≤ j ≤ α, let Fjconsist of ej plus chordsoverlapping ej goingforward.
Greedy chord system
I G : Hamiltonian cycle C plus q chords of length `
I Find many distinct cycle lengths using a greedy chord system.
e1
e2
e3
e4 eα
e?
I Let F ? be the set ofremaining chords.
I When F ? 6= ∅, define e?
to be the first chord in F ?
after eα.
I F1, . . . ,Fα and F ? form apartition of the chords.
I Greedy choice of e1:|F1| ≥ |Fj | for 1 ≤ j ≤ α.
I Also: |F1| ≥ |F ?|.
Greedy chord system
I G : Hamiltonian cycle C plus q chords of length `
I Find many distinct cycle lengths using a greedy chord system.
e1
e2
e3
e4 eα
e?
I Let F ? be the set ofremaining chords.
I When F ? 6= ∅, define e?
to be the first chord in F ?
after eα.
I F1, . . . ,Fα and F ? form apartition of the chords.
I Greedy choice of e1:|F1| ≥ |Fj | for 1 ≤ j ≤ α.
I Also: |F1| ≥ |F ?|.
Greedy chord system
I G : Hamiltonian cycle C plus q chords of length `
I Find many distinct cycle lengths using a greedy chord system.
e1
e2
e3
e4 eα
e?
I Let F ? be the set ofremaining chords.
I When F ? 6= ∅, define e?
to be the first chord in F ?
after eα.
I F1, . . . ,Fα and F ? form apartition of the chords.
I Greedy choice of e1:|F1| ≥ |Fj | for 1 ≤ j ≤ α.
I Also: |F1| ≥ |F ?|.
Greedy chord system
I G : Hamiltonian cycle C plus q chords of length `
I Find many distinct cycle lengths using a greedy chord system.
e1
e2
e3
e4 eα
e?
I Let F ? be the set ofremaining chords.
I When F ? 6= ∅, define e?
to be the first chord in F ?
after eα.
I F1, . . . ,Fα and F ? form apartition of the chords.
I Greedy choice of e1:|F1| ≥ |Fj | for 1 ≤ j ≤ α.
I Also: |F1| ≥ |F ?|.
Greedy chord system
I G : Hamiltonian cycle C plus q chords of length `
I Find many distinct cycle lengths using a greedy chord system.
e1
e2
e3
e4 eα
e?
I Let F ? be the set ofremaining chords.
I When F ? 6= ∅, define e?
to be the first chord in F ?
after eα.
I F1, . . . ,Fα and F ? form apartition of the chords.
I Greedy choice of e1:|F1| ≥ |Fj | for 1 ≤ j ≤ α.
I Also: |F1| ≥ |F ?|.
Greedy chord system
I G : Hamiltonian cycle C plus q chords of length `
I Find many distinct cycle lengths using a greedy chord system.
e1
e2
e3
e4 eα
e?
I Let F ? be the set ofremaining chords.
I When F ? 6= ∅, define e?
to be the first chord in F ?
after eα.
I F1, . . . ,Fα and F ? form apartition of the chords.
I Greedy choice of e1:|F1| ≥ |Fj | for 1 ≤ j ≤ α.
I Also: |F1| ≥ |F ?|.
Spectrum bands
3 n
12. . .α
`− 1
Short Cycles Long Cycles
I We find many cycle lengths by dividing the space of possiblecycle lengths into bands.
I Let C [x , y ] denote the subpath of C from x to y along theforward direction.
I Let uv be a chord such that C [u, v ] has length `. ReplacingC [u, v ] with uv reduces the length of a cycle containingC [u, v ] by `− 1.
I We have α bands at the top, each of size `− 1.
Spectrum bands
3 n
12. . .α
`− 1
Short Cycles Long Cycles
I We find many cycle lengths by dividing the space of possiblecycle lengths into bands.
I Let C [x , y ] denote the subpath of C from x to y along theforward direction.
I Let uv be a chord such that C [u, v ] has length `. ReplacingC [u, v ] with uv reduces the length of a cycle containingC [u, v ] by `− 1.
I We have α bands at the top, each of size `− 1.
Spectrum bands
3 n
12. . .α
`− 1
Short Cycles Long Cycles
I We find many cycle lengths by dividing the space of possiblecycle lengths into bands.
I Let C [x , y ] denote the subpath of C from x to y along theforward direction.
I Let uv be a chord such that C [u, v ] has length `. ReplacingC [u, v ] with uv reduces the length of a cycle containingC [u, v ] by `− 1.
I We have α bands at the top, each of size `− 1.
Spectrum bands
3 n
12. . .α
`− 1
Short Cycles Long Cycles
I We find many cycle lengths by dividing the space of possiblecycle lengths into bands.
I Let C [x , y ] denote the subpath of C from x to y along theforward direction.
I Let uv be a chord such that C [u, v ] has length `. ReplacingC [u, v ] with uv reduces the length of a cycle containingC [u, v ] by `− 1.
I We have α bands at the top, each of size `− 1.
Spectrum bands
3 n
12. . .α
`− 1
Short Cycles Long Cycles
I The jth band: from n − j(`− 1) + 1 to n − (j − 1)(`− 1).
I The short cycles: lengths below the top α bands.
I The long cycles: lengths in the top 2 bands.
Spectrum bands
3 n
12. . .α
`− 1
Short Cycles
Long Cycles
I The jth band: from n − j(`− 1) + 1 to n − (j − 1)(`− 1).
I The short cycles: lengths below the top α bands.
I The long cycles: lengths in the top 2 bands.
Spectrum bands
3 n
12. . .α
`− 1
Short Cycles Long Cycles
I The jth band: from n − j(`− 1) + 1 to n − (j − 1)(`− 1).
I The short cycles: lengths below the top α bands.
I The long cycles: lengths in the top 2 bands.
Short cycles lemma
LemmaIf α ≥ 2, then G has short cycles of at least |F
?|−12 distinct lengths.
Short cycles lemma
LemmaIf α ≥ 2, then G has short cycles of at least |F
?|−12 distinct lengths.
e1
e2
e3
e4 eα
e?
v1
v`+1
vj
Short cycles lemma
LemmaIf α ≥ 2, then G has short cycles of at least |F
?|−12 distinct lengths.
e1
e?
v1
v`+1
vj
vk
Short cycles lemma
LemmaIf α ≥ 2, then G has short cycles of at least |F
?|−12 distinct lengths.
e1
e?
v1
v`+1
vj
vk
I We may assume |F ?| ≥ 2.
I Consider a chord e ∈ F ?
with e 6= e?.
I Cycle using e? and e haslength 2(k − j + 1).
I Cycle using e and e1 haslength 2(n − k + 2).
I Some cycle has length atmost n − j + 3.
I Note: j ≥ 1 + α`.
I This cycle has length atmost n − α`+ 2.
I α ≥ 2: this cycle is short.
Short cycles lemma
LemmaIf α ≥ 2, then G has short cycles of at least |F
?|−12 distinct lengths.
e1
e?
v1
v`+1
vj
vk
I We may assume |F ?| ≥ 2.
I Consider a chord e ∈ F ?
with e 6= e?.
I Cycle using e? and e haslength 2(k − j + 1).
I Cycle using e and e1 haslength 2(n − k + 2).
I Some cycle has length atmost n − j + 3.
I Note: j ≥ 1 + α`.
I This cycle has length atmost n − α`+ 2.
I α ≥ 2: this cycle is short.
Short cycles lemma
LemmaIf α ≥ 2, then G has short cycles of at least |F
?|−12 distinct lengths.
e1
e?
v1
v`+1
vj
vk
I We may assume |F ?| ≥ 2.
I Consider a chord e ∈ F ?
with e 6= e?.
I Cycle using e? and e haslength 2(k − j + 1).
I Cycle using e and e1 haslength 2(n − k + 2).
I Some cycle has length atmost n − j + 3.
I Note: j ≥ 1 + α`.
I This cycle has length atmost n − α`+ 2.
I α ≥ 2: this cycle is short.
Short cycles lemma
LemmaIf α ≥ 2, then G has short cycles of at least |F
?|−12 distinct lengths.
e1
e?
v1
v`+1
vj
vk
I We may assume |F ?| ≥ 2.
I Consider a chord e ∈ F ?
with e 6= e?.
I Cycle using e? and e haslength 2(k − j + 1).
I Cycle using e and e1 haslength 2(n − k + 2).
I Some cycle has length atmost n − j + 3.
I Note: j ≥ 1 + α`.
I This cycle has length atmost n − α`+ 2.
I α ≥ 2: this cycle is short.
Short cycles lemma
LemmaIf α ≥ 2, then G has short cycles of at least |F
?|−12 distinct lengths.
e1
e?
v1
v`+1
vj
vk
I We may assume |F ?| ≥ 2.
I Consider a chord e ∈ F ?
with e 6= e?.
I Cycle using e? and e haslength 2(k − j + 1).
I Cycle using e and e1 haslength 2(n − k + 2).
I Some cycle has length atmost n − j + 3.
I Note: j ≥ 1 + α`.
I This cycle has length atmost n − α`+ 2.
I α ≥ 2: this cycle is short.
Short cycles lemma
LemmaIf α ≥ 2, then G has short cycles of at least |F
?|−12 distinct lengths.
e1
e2
e3
e4 eα
e? v1
v`+1
vj
I We may assume |F ?| ≥ 2.
I Consider a chord e ∈ F ?
with e 6= e?.
I Cycle using e? and e haslength 2(k − j + 1).
I Cycle using e and e1 haslength 2(n − k + 2).
I Some cycle has length atmost n − j + 3.
I Note: j ≥ 1 + α`.
I This cycle has length atmost n − α`+ 2.
I α ≥ 2: this cycle is short.
Short cycles lemma
LemmaIf α ≥ 2, then G has short cycles of at least |F
?|−12 distinct lengths.
e1
e2
e3
e4 eα
e? v1
v`+1
vj
I We may assume |F ?| ≥ 2.
I Consider a chord e ∈ F ?
with e 6= e?.
I Cycle using e? and e haslength 2(k − j + 1).
I Cycle using e and e1 haslength 2(n − k + 2).
I Some cycle has length atmost n − j + 3.
I Note: j ≥ 1 + α`.
I This cycle has length atmost n − α`+ 2.
I α ≥ 2: this cycle is short.
Short cycles lemma
LemmaIf α ≥ 2, then G has short cycles of at least |F
?|−12 distinct lengths.
e1
e2
e3
e4 eα
e? v1
v`+1
vj
I We may assume |F ?| ≥ 2.
I Consider a chord e ∈ F ?
with e 6= e?.
I Cycle using e? and e haslength 2(k − j + 1).
I Cycle using e and e1 haslength 2(n − k + 2).
I Some cycle has length atmost n − j + 3.
I Note: j ≥ 1 + α`.
I This cycle has length atmost n − α`+ 2.
I α ≥ 2: this cycle is short.
Short cycles lemma
LemmaIf α ≥ 2, then G has short cycles of at least |F
?|−12 distinct lengths.
e1
e2
e3
e4 eα
e? v1
v`+1
vj
I We may assume |F ?| ≥ 2.
I Consider a chord e ∈ F ?
with e 6= e?.
I Cycle using e? and e haslength 2(k − j + 1).
I Cycle using e and e1 haslength 2(n − k + 2).
I Some cycle has length atmost n − j + 3.
I Note: j ≥ 1 + α`.
I This cycle has length atmost n − α`+ 2.
I α ≥ 2: this cycle is short.
Short cycles lemma
LemmaIf α ≥ 2, then G has short cycles of at least |F
?|−12 distinct lengths.
e1
e2
e3
e4 eα
e? v1
v`+1
vj
I We obtain |F ? − 1| shortcycles.
I Each length occurs atmost twice.
Short cycles lemma
LemmaIf α ≥ 2, then G has short cycles of at least |F
?|−12 distinct lengths.
e1
e2
e3
e4 eα
e? v1
v`+1
vj
I We obtain |F ? − 1| shortcycles.
I Each length occurs atmost twice.
Longer cycles
123α`− 1
Short Cycles Long Cycles
Longer cycles
123α`− 1
Short Cycles Long Cycles
Longer cycles
123α`− 1
Short Cycles Long Cycles
u
v
e?
e1
e2
e3
e4 eα
Longer cycles
123α`− 1
Short Cycles Long Cycles
u
ve?
e1
e2
e3
e4 eα
I A long cycle is good if itcontains C [u, v ].
I Let ρ be the number of lengthsof good cycles.
I Overlapping chords lemma:ρ ≥ |F1| − 1.
I First, suppose ρ ≥ |F1|.
Longer cycles
123α`− 1
Short Cycles Long Cycles
u
v
e?
e1
e2
e3
e4 eα
I A long cycle is good if itcontains C [u, v ].
I Let ρ be the number of lengthsof good cycles.
I Overlapping chords lemma:ρ ≥ |F1| − 1.
I First, suppose ρ ≥ |F1|.
Longer cycles
123α`− 1
Short Cycles Long Cycles
u
v
e?
e1
e2
e3
e4 eα
I A long cycle is good if itcontains C [u, v ].
I Let ρ be the number of lengthsof good cycles.
I Overlapping chords lemma:ρ ≥ |F1| − 1.
I First, suppose ρ ≥ |F1|.
Longer cycles
123α`− 1
Short Cycles Long Cycles
u
v
e?
e1
e2
e3
e4 eα
I A long cycle is good if itcontains C [u, v ].
I Let ρ be the number of lengthsof good cycles.
I Overlapping chords lemma:ρ ≥ |F1| − 1.
I First, suppose ρ ≥ |F1|.
Longer cycles
123α`− 1
Short Cycles Long Cycles
u
v
e?
e1
e2
e3
e4 eα
I A long cycle is good if itcontains C [u, v ].
I Let ρ be the number of lengthsof good cycles.
I Overlapping chords lemma:ρ ≥ |F1| − 1.
I First, suppose ρ ≥ |F1|.
Longer cycles
123α`− 1
Short Cycles Long Cycles
u
v
e?
e1
e2
e3
e4 eα
I A long cycle is good if itcontains C [u, v ].
I Let ρ be the number of lengthsof good cycles.
I Overlapping chords lemma:ρ ≥ |F1| − 1.
I First, suppose ρ ≥ |F1|.
Longer cycles
123α`− 1
Short Cycles Long Cycles
u
v
e?
e1
e2
e3
e4 eα
I A long cycle is good if itcontains C [u, v ].
I Let ρ be the number of lengthsof good cycles.
I Overlapping chords lemma:ρ ≥ |F1| − 1.
I First, suppose ρ ≥ |F1|.
Longer cycles
123α`− 1
Short Cycles Long Cycles
u
v
e?
e1
e2
e3
e4 eα
I Using a chord shifts theselengths down by `− 1.
I This yields α− 1 sets of ρlengths. Each length occurs atmost twice.
I Add one more set.
I Now: α sets of ρ lengths; eachlength appears at most once.
Longer cycles
123α`− 1
Short Cycles Long Cycles
u
v
e?
e1
e2
e3
e4 eα
I Using a chord shifts theselengths down by `− 1.
I This yields α− 1 sets of ρlengths. Each length occurs atmost twice.
I Add one more set.
I Now: α sets of ρ lengths; eachlength appears at most once.
Longer cycles
123α`− 1
Short Cycles Long Cycles
u
v
e?
e1
e2
e3
e4 eα
I Using a chord shifts theselengths down by `− 1.
I This yields α− 1 sets of ρlengths. Each length occurs atmost twice.
I Add one more set.
I Now: α sets of ρ lengths; eachlength appears at most once.
Longer cycles
123α`− 1
Short Cycles Long Cycles
u
v
e?
e1
e2
e3
e4 eα
I Using a chord shifts theselengths down by `− 1.
I This yields α− 1 sets of ρlengths. Each length occurs atmost twice.
I Add one more set.
I Now: α sets of ρ lengths; eachlength appears at most once.
Longer cycles
123α`− 1
Short Cycles Long Cycles
u
v
e?
e1
e2
e3
e4 eα
I Using a chord shifts theselengths down by `− 1.
I This yields α− 1 sets of ρlengths. Each length occurs atmost twice.
I Add one more set.
I Now: α sets of ρ lengths; eachlength appears at most once.
Longer cycles
123α`− 1
Short Cycles Long Cycles
u
v
e?
e1
e2
e3
e4 eα
I Using a chord shifts theselengths down by `− 1.
I This yields α− 1 sets of ρlengths. Each length occurs atmost twice.
I Add one more set.
I Now: α sets of ρ lengths; eachlength appears at most once.
Longer cycles
123α`− 1
Short Cycles Long Cycles
u
v
e?
e1
e2
e3
e4 eα
I Using a chord shifts theselengths down by `− 1.
I This yields α− 1 sets of ρlengths. Each length occurs atmost twice.
I Add one more set.
I Now: α sets of ρ lengths; eachlength appears at most once.
Longer cycles
123α`− 1
Short Cycles Long Cycles
u
v
e?
e1
e2
e3
e4 eα
I So we have αρ2 longer cycle
lengths, plus |F?|−12 short cycle
lengths.
I Since ρ ≥ |F1|,
s(G ) ≥ α
2|F1|+
|F ?| − 1
2
≥ α
2
q − |F ?|α
+|F ?| − 1
2
≥ q − 1
2
Longer cycles
123α`− 1
Short Cycles Long Cycles
u
v
e?
e1
e2
e3
e4 eα
I So we have αρ2 longer cycle
lengths, plus |F?|−12 short cycle
lengths.
I Since ρ ≥ |F1|,
s(G ) ≥ α
2|F1|+
|F ?| − 1
2
≥ α
2
q − |F ?|α
+|F ?| − 1
2
≥ q − 1
2
Longer cycles
123α`− 1
Short Cycles Long Cycles
u
v
e?
e1
e2
e3
e4 eα
I Otherwise ρ = |F1| − 1.
I The Overlapping cycles lemmaimplies:
1. ` is odd2. Chords in F1 are consecutive3. |F1| ≥ (`+ 3)/2
I We exploit the structure in twocases to show
s(G ) ≥(q − 1− q
`
)/2.
Longer cycles
123α`− 1
Short Cycles Long Cycles
u
v
e?
e1
e2
e3
e4 eα
I Otherwise ρ = |F1| − 1.I The Overlapping cycles lemma
implies:
1. ` is odd2. Chords in F1 are consecutive3. |F1| ≥ (`+ 3)/2
I We exploit the structure in twocases to show
s(G ) ≥(q − 1− q
`
)/2.
Longer cycles
123α`− 1
Short Cycles Long Cycles
u
v
e?
e1
e2
e3
e4 eα
I Otherwise ρ = |F1| − 1.I The Overlapping cycles lemma
implies:
1. ` is odd2. Chords in F1 are consecutive3. |F1| ≥ (`+ 3)/2
I We exploit the structure in twocases to show
s(G ) ≥(q − 1− q
`
)/2.
Longer cycles
123α`− 1
Short Cycles Long Cycles
u
v
e?
e1
e2
e3
e4 eα
I Otherwise ρ = |F1| − 1.I The Overlapping cycles lemma
implies:
1. ` is odd2. Chords in F1 are consecutive3. |F1| ≥ (`+ 3)/2
I We exploit the structure in twocases to show
s(G ) ≥(q − 1− q
`
)/2.
Summary and Open Problems
TheoremIf G is an n-vertex Hamiltonian graph with p chords, thens(G ) ≥ √p − 1
2 ln p − 1.
Open Problems
I What is the maximum number of edges in an n-vertexbipartite Hamiltonian graph that is not bipancyclic? Theanswer lies between (1 + o(1)) n
2
16 and (1 + o(1))n2
8 .
I What is the maximum number of edges in an n-vertexHamiltonian graph with s(G ) < n/2− 1? The answer lies
between (1 + o(1)) n2
16 and (1 + o(1))n2
4 .
I Obtain better bounds on fn(m).
I Is a constant c such that s(G ) ≥ cn for every Hamiltoniangraph G with δ(G ) ≥ 3?
Thank You
Summary and Open Problems
TheoremIf G is an n-vertex Hamiltonian graph with p chords, thens(G ) ≥ √p − 1
2 ln p − 1.
Open Problems
I What is the maximum number of edges in an n-vertexbipartite Hamiltonian graph that is not bipancyclic? Theanswer lies between (1 + o(1)) n
2
16 and (1 + o(1))n2
8 .
I What is the maximum number of edges in an n-vertexHamiltonian graph with s(G ) < n/2− 1? The answer lies
between (1 + o(1)) n2
16 and (1 + o(1))n2
4 .
I Obtain better bounds on fn(m).
I Is a constant c such that s(G ) ≥ cn for every Hamiltoniangraph G with δ(G ) ≥ 3?
Thank You
Summary and Open Problems
TheoremIf G is an n-vertex Hamiltonian graph with p chords, thens(G ) ≥ √p − 1
2 ln p − 1.
Open Problems
I What is the maximum number of edges in an n-vertexbipartite Hamiltonian graph that is not bipancyclic? Theanswer lies between (1 + o(1)) n
2
16 and (1 + o(1))n2
8 .
I What is the maximum number of edges in an n-vertexHamiltonian graph with s(G ) < n/2− 1? The answer lies
between (1 + o(1)) n2
16 and (1 + o(1))n2
4 .
I Obtain better bounds on fn(m).
I Is a constant c such that s(G ) ≥ cn for every Hamiltoniangraph G with δ(G ) ≥ 3?
Thank You
Summary and Open Problems
TheoremIf G is an n-vertex Hamiltonian graph with p chords, thens(G ) ≥ √p − 1
2 ln p − 1.
Open Problems
I What is the maximum number of edges in an n-vertexbipartite Hamiltonian graph that is not bipancyclic? Theanswer lies between (1 + o(1)) n
2
16 and (1 + o(1))n2
8 .
I What is the maximum number of edges in an n-vertexHamiltonian graph with s(G ) < n/2− 1? The answer lies
between (1 + o(1)) n2
16 and (1 + o(1))n2
4 .
I Obtain better bounds on fn(m).
I Is a constant c such that s(G ) ≥ cn for every Hamiltoniangraph G with δ(G ) ≥ 3?
Thank You
Summary and Open Problems
TheoremIf G is an n-vertex Hamiltonian graph with p chords, thens(G ) ≥ √p − 1
2 ln p − 1.
Open Problems
I What is the maximum number of edges in an n-vertexbipartite Hamiltonian graph that is not bipancyclic? Theanswer lies between (1 + o(1)) n
2
16 and (1 + o(1))n2
8 .
I What is the maximum number of edges in an n-vertexHamiltonian graph with s(G ) < n/2− 1? The answer lies
between (1 + o(1)) n2
16 and (1 + o(1))n2
4 .
I Obtain better bounds on fn(m).
I Is a constant c such that s(G ) ≥ cn for every Hamiltoniangraph G with δ(G ) ≥ 3?
Thank You
Summary and Open Problems
TheoremIf G is an n-vertex Hamiltonian graph with p chords, thens(G ) ≥ √p − 1
2 ln p − 1.
Open Problems
I What is the maximum number of edges in an n-vertexbipartite Hamiltonian graph that is not bipancyclic? Theanswer lies between (1 + o(1)) n
2
16 and (1 + o(1))n2
8 .
I What is the maximum number of edges in an n-vertexHamiltonian graph with s(G ) < n/2− 1? The answer lies
between (1 + o(1)) n2
16 and (1 + o(1))n2
4 .
I Obtain better bounds on fn(m).
I Is a constant c such that s(G ) ≥ cn for every Hamiltoniangraph G with δ(G ) ≥ 3?
Thank You