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 )|.

cycles in 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.

cycles in G .

Conjecture

S(G ) = 4, 6, . . . , 2n.

cycles in G .

Conjecture

S(G ) = 4, 6, . . . , 2n.

cycles in 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?

cycles in G .

Conjecture (Erdos)

cycles in G .

Conjecture (Erdos)

cycles in G .

Conjecture (Erdos)

cycles in G .

Conjecture (Erdos)

cycles in 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.

cycles in G .

I S(G ) = 4, 6 ∪23n,

23n + 2, 23n + 4, . . . , n

I s(G ) = n/6 + 3

cycles in G .

I S(G ) = 4, 6 ∪23n,

23n + 2, 23n + 4, . . . , n

I s(G ) = n/6 + 3

cycles in G .

I S(G ) = 4, 6 ∪23n,

23n + 2, 23n + 4, . . . , n

I s(G ) = n/6 + 3

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.

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.

2. ` is odd and h ≥ (`+ 3)/2.

Proof.

2. ` is odd and h ≥ (`+ 3)/2.

Proof.

2. ` is odd and h ≥ (`+ 3)/2.

Proof.

d(ei , ej) d(ei , ej)

e1 e2e1 e3e1 e4e1 e5e1 e2 ej

2. ` is odd and h ≥ (`+ 3)/2.

Proof.

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).

2. ` is odd and h ≥ (`+ 3)/2.

Proof.

2. ` is odd and h ≥ (`+ 3)/2.

Proof.ei ej

2. ` is odd and h ≥ (`+ 3)/2.

Proof.ei ej

2. ` is odd and h ≥ (`+ 3)/2.

Proof.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.

2. ` is odd and h ≥ (`+ 3)/2.

Proof.

2. ` is odd and h ≥ (`+ 3)/2.

Proof.

e1 e3e1 e4e1 e5e1 e2 ej

2. ` is odd and h ≥ (`+ 3)/2.

Proof.

e1 e4e1 e5e1 e2 ej

2. ` is odd and h ≥ (`+ 3)/2.

Proof.

e1 e2e1 e3

e1 e5e1 e2 ej

2. ` is odd and h ≥ (`+ 3)/2.

Proof.

e1 e2e1 e3e1 e4

e1 e2 ej

2. ` is odd and h ≥ (`+ 3)/2.

Proof.

e1 e2e1 e3e1 e4

e1 e2 ej

2. ` is odd and h ≥ (`+ 3)/2.

Proof.

2. ` is odd and h ≥ (`+ 3)/2.

Proof.

e1 e2e1 e3e1 e4e1 e5

2. ` is odd and h ≥ (`+ 3)/2.

Proof.

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 .

2. ` is odd and h ≥ (`+ 3)/2.

Proof.

e1 e2 ej

2. ` is odd and h ≥ (`+ 3)/2.

Proof.

e1 e2 ej

2. ` is odd and h ≥ (`+ 3)/2.

Proof.

e1 e2 ej

2. ` is odd and h ≥ (`+ 3)/2.

Proof.

e1 e2 ej

2. ` is odd and h ≥ (`+ 3)/2.

Proof.

e1 e2 ej

2. ` is odd and h ≥ (`+ 3)/2.

Proof.

e1 e2 ej

2. ` is odd and h ≥ (`+ 3)/2.

Proof.

e1 e2 ej

2. ` is odd and h ≥ (`+ 3)/2.

Proof.

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).

2. ` is odd and h ≥ (`+ 3)/2.

Proof.

e1 e2 ej

I So `− 1 ∈ 0, 2, . . . , 2(h − 2).

2. ` is odd and h ≥ (`+ 3)/2.

Proof.

e1 e2 ej

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.

e4 eα

Greedy chord system

e4 eα

Greedy chord system

e4 eα

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.

Greedy chord system

e4 eα

Greedy chord system

e4 eα

Greedy chord system

e4 eα

Greedy chord system

e4 eα

Greedy chord system

e4 eα

Greedy chord system

e4 eα

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

e4 eα

Greedy chord system

e4 eα

Greedy chord system

e4 eα

Greedy chord system

e4 eα

Greedy chord system

e4 eα

Greedy chord system

e4 eα

Greedy chord system

e4 eα

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

e4 eα

after eα.

I Also: |F1| ≥ |F ?|.

Greedy chord system

e4 eα

after eα.

I Also: |F1| ≥ |F ?|.

Greedy chord system

e4 eα

after eα.

I Also: |F1| ≥ |F ?|.

Greedy chord system

e4 eα

after eα.

I Also: |F1| ≥ |F ?|.

Greedy chord system

e4 eα

after eα.

I Also: |F1| ≥ |F ?|.

Spectrum bands

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

12. . .α

`− 1

Spectrum bands

12. . .α

`− 1

Spectrum bands

12. . .α

`− 1

Spectrum bands

12. . .α

`− 1

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

12. . .α

`− 1

Short Cycles

Long Cycles

Spectrum bands

12. . .α

`− 1

Short cycles lemma

LemmaIf α ≥ 2, then G has short cycles of at least |F

?|−12 distinct lengths.

Short cycles lemma

e4 eα

Short cycles lemma

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

with e 6= e?.

I Note: j ≥ 1 + α`.

Short cycles lemma

with e 6= e?.

I Note: j ≥ 1 + α`.

Short cycles lemma

with e 6= e?.

I Note: j ≥ 1 + α`.

Short cycles lemma

with e 6= e?.

I Note: j ≥ 1 + α`.

Short cycles lemma

e4 eα

with e 6= e?.

I Note: j ≥ 1 + α`.

Short cycles lemma

e4 eα

with e 6= e?.

I Note: j ≥ 1 + α`.

Short cycles lemma

e4 eα

with e 6= e?.

I Note: j ≥ 1 + α`.

Short cycles lemma

e4 eα

with e 6= e?.

I Note: j ≥ 1 + α`.

Short cycles lemma

e4 eα

I We obtain |F ? − 1| shortcycles.

I Each length occurs atmost twice.

Short cycles lemma

e4 eα

I We obtain |F ? − 1| shortcycles.

I Each length occurs atmost twice.

Longer cycles

123α`− 1

Longer cycles

123α`− 1

Longer cycles

123α`− 1

e4 eα

Longer cycles

123α`− 1

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

e4 eα

Longer cycles

123α`− 1

e4 eα

Longer cycles

123α`− 1

e4 eα

Longer cycles

123α`− 1

e4 eα

Longer cycles

123α`− 1

e4 eα

Longer cycles

123α`− 1

e4 eα

Longer cycles

123α`− 1

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

e4 eα

I Add one more set.

Longer cycles

123α`− 1

e4 eα

I Add one more set.

Longer cycles

123α`− 1

e4 eα

I Add one more set.

Longer cycles

123α`− 1

e4 eα

I Add one more set.

Longer cycles

123α`− 1

e4 eα

I Add one more set.

Longer cycles

123α`− 1

e4 eα

I Add one more set.

Longer cycles

123α`− 1

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

≥ α

q − |F ?|α

+|F ?| − 1

≥ q − 1

Longer cycles

123α`− 1

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

≥ α

q − |F ?|α

+|F ?| − 1

≥ q − 1

Longer cycles

123α`− 1

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

Longer cycles

123α`− 1

e4 eα

I Otherwise ρ = |F1| − 1.I The Overlapping cycles lemma

implies:

s(G ) ≥(q − 1− q

Longer cycles

123α`− 1

e4 eα

implies:

s(G ) ≥(q − 1− q

Longer cycles

123α`− 1

e4 eα

implies:

s(G ) ≥(q − 1− q

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

16 and (1 + o(1))n2

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

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

2 ln p − 1.

Open Problems

16 and (1 + o(1))n2

Thank You

2 ln p − 1.

Open Problems

16 and (1 + o(1))n2

Thank You

2 ln p − 1.

Open Problems

16 and (1 + o(1))n2

Thank You

2 ln p − 1.

Open Problems

16 and (1 + o(1))n2

Thank You

2 ln p − 1.

Open Problems

16 and (1 + o(1))n2

Thank You

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