Chapter 5: Concentration The Probabilistic Method Summer 2020 - - PowerPoint PPT Presentation

Chapter 5: Concentration

The Probabilistic Method Summer 2020 Freie Universität Berlin

Chapter Overview

• Prove some strong concentration inequalities
• Improve bounds on Ramsey numbers
• Study Hamiltonicity and chromatic number of 𝐻 𝑜, 𝑞

§1 Chernoff Bounds

Chapter 5: Concentration The Probabilistic Method

Domination vs Minimum Degree

Homework exercise

• Show bound is tight by consider 𝐻 𝑜, 𝑞

Degrees in 𝐻 𝑜, 𝑞

• Degree of a vertex ∼ Bin 𝑜 − 1, 𝑞
• ⇒ expected degree is 𝑜 − 1 𝑞
• Minimum degree: need to show not far from mean
• Suppose ℙ | deg 𝑤 − 𝑜 − 1 𝑞| ≥ 𝑏 <

1 2𝑜

• Union bound ⇒ ℙ 𝜀 𝐻 𝑜, 𝑞

≥ 𝑜 − 1 𝑞 − 𝑒 >

1 2

Corollary 2.2.5 Let 𝐻 be an 𝑜-vertex graph with 𝜀 𝐻 ≥ 𝜀. Then 𝐻 has a dominating set 𝑇 ⊆ 𝑊 𝐻 with 𝑇 ≤

ln 𝜀+1 +1 𝜀+1

𝑜.

Comparing Bounds

Concentration inequalities

• Let 𝑌 ∼ Bin 𝑜 − 1, 𝑞 , 𝑞 ≤

1 2

• 𝔽 𝑌 = 𝑜 − 1 𝑞 , Var 𝑌 = 𝑜 − 1 𝑞 1 − 𝑞 = Θ 𝑜𝑞
• Markov:

ℙ 𝑌 ≥ 𝑏 ≤

𝔽 𝑌 𝑏

• ⇒ error probability <

1 2𝑜 for 𝑏 = Ω 𝑜

• Chebyshev:

ℙ 𝑌 − 𝔽 𝑌 ≥ 𝑏 ≤

Var 𝑌 𝑏2

• ⇒ error probability <

1 2𝑜 for 𝑏 = Ω 𝑜 𝑞

• Central Limit Theorem:

ℙ 𝑌 − 𝔽 𝑌 ≥ 𝑏 ≤ exp

−𝑑𝑏2 Var 𝑌

• ⇒ error probability <

1 2𝑜 for 𝑏 = Ω

𝑜𝑞 log 𝑜

Definition 5.1.1 Let 𝑇𝑜 = σ𝑗=1

𝑜

𝑌𝑗, where the 𝑌𝑗 are independently and uniformly distributed on −1,1 .

The Problem With CLT

Asymptotics

• Central Limit Theorem is asymptotic, valid as 𝑜 → ∞
• We would like a quantitative bound for some given 𝑜

Binomial connection

• We have 𝑇𝑜 ∼ 2 Bin 𝑜,

1 2 − 𝑜 2

• Convenient to translate so mean is zero

Goal

• Show 𝑇𝑜 is exponentially unlikely to be far from zero

Theorem 5.1.2 (Symmetric Chernoff Bound) For every 𝑏 > 0, we have ℙ 𝑇𝑜 ≥ 𝑏 ≤ exp −

𝑏2 2𝑜 .

Chernoff Bounds

Remarks

• Concrete bounds for all 𝑜, 𝑏
• Symmetry: same bound for ℙ(𝑇𝑜 ≤ −𝑏)
• Bin 𝑜,

1 2 = 1 2 𝑇𝑜 + 𝑜

• ⇒ concentration for binomial random variables

Corollary 5.1.3 For every 𝑏 > 0, we have ℙ Bin 𝑜,

1 2 − 𝑜 2 ≥ 𝑏 ≤ 2 exp −2𝑏2 𝑜

.

Proving Chernoff

Proof

• Exponential conversion
• 𝑇𝑜 ≥ 𝑏 = 𝑓𝑇𝑜 ≥ 𝑓𝑏 = 𝑓𝜇𝑇𝑜 ≥ 𝑓𝜇𝑏
• Concentration
• 𝑓𝜇𝑇𝑜 a non-negative random variable
• Markov: ℙ 𝑓𝜇𝑇𝑜 ≥ 𝑓𝜇𝑏 ≤ 𝔽 𝑓𝜇𝑇𝑜 𝑓−𝜇𝑏
• Expectation
• Recall 𝑇𝑜 = σ𝑗=1

𝑜

𝑌𝑗

• ⇒ 𝑓𝜇𝑇𝑜 = ς𝑗=1

𝑜

𝑓𝜇𝑌𝑗

• Independence ⇒ 𝔽 𝑓𝜇𝑇𝑜 = ς𝑗=1

𝑜

𝔽 𝑓𝜇𝑌𝑗 =

1 2 𝑓𝜇 + 𝑓−𝜇 𝑜

= cosh𝑜 𝜇

Theorem 5.1.2 (Symmetric Chernoff Bound) For every 𝑏 > 0, we have ℙ 𝑇𝑜 ≥ 𝑏 ≤ exp −

𝑏2 2𝑜 .

Over the Cosh

Recall

• ℙ 𝑇𝑜 ≥ 𝑏 ≤ 𝔽 𝑓𝜇𝑇𝑜 𝑓−𝜇𝑏
• 𝔽 𝑓𝜇𝑇𝑜 = cosh𝑜 𝜇

A little calculus

• cosh 𝑦 =

1 2 𝑓𝑦 + 𝑓−𝑦

• Taylor series: 𝑓𝑦 = 1 + 𝑦 +

𝑦2 2 + 𝑦3 6 + 𝑦4 24 + ⋯

• ⇒ cosh 𝑦 = 1 +

𝑦2 2 + 𝑦4 24 + 𝑦6 720 + ⋯ ≤ 1 + 𝑦2 2 + 𝑦4 8 + 𝑦6 48 + ⋯ = 𝑓

𝑦2 2

Finishing the proof

• ∴ ℙ 𝑇𝑜 ≥ 𝑏 ≤ exp

1 2 𝑜𝜇2 − 𝜇𝑏

• Minimise: 𝜇 =

𝑏 𝑜 ⇒ ℙ 𝑇𝑜 ≥ 𝑏 ≤ exp − 𝑏2 2𝑜

∎

The General Setting

Shortcomings

• Required each 𝑌𝑗 to be uniform on −1,1

Wider Framework

• 𝑞1, 𝑞2, … , 𝑞𝑜 ∈ 0,1 , and 𝑞 = 𝑜−1 σ𝑗=1

𝑜

𝑞𝑗

• 𝑌𝑗 independent with ℙ 𝑌𝑗 = 1 − 𝑞𝑗 = 𝑞𝑗 and ℙ 𝑌 = −𝑞𝑗 = 1 − 𝑞𝑗
• 𝑌 = σ𝑗=1

𝑜

𝑌𝑗

Theorem 5.1.4 (Asymmetric Chernoff Bound) Let 𝑏 > 0 and let 𝑌 and 𝑞 be as above. Then ℙ 𝑌 ≤ −𝑏 ≤ exp −

𝑏2 2𝑞𝑜

and ℙ 𝑌 ≥ 𝑏 ≤ exp

−𝑏2 2𝑞𝑜 + 𝑏3 2 𝑞𝑜 2 .

Theorem 5.1.4 (Asymmetric Chernoff Bound) Let 𝑏 > 0 and let 𝑌 and 𝑞 be as above. Then ℙ 𝑌 ≤ −𝑏 ≤ exp

−𝑏2 2𝑞𝑜

and ℙ 𝑌 ≥ 𝑏 ≤ exp

−𝑏2 2𝑞𝑜 + 𝑏3 2 𝑞𝑜 2 .

An Asymmetric Chernoff Bound

Special case

• 𝑞𝑗 = 𝑞 for all 𝑗 ⇒ 𝑌 + 𝑜𝑞 ~ Bin 𝑜, 𝑞
• ⇒ ℙ Bin 𝑜, 𝑞 − 𝑜𝑞 ≥ 𝑏 ≤ 2 exp

−𝑏2 2𝑞𝑜 + 𝑏3 2 𝑞𝑜 2

Corollary 5.1.5 For every 𝜁 > 0 there is some 𝑑𝜁 > 0 such that, if 𝑍 is the sum of mutually independent indicator random variables and 𝜈 = 𝔽 𝑍 , then ℙ 𝑍 − 𝜈 ≥ 𝜁𝜈 ≤ 2 exp −𝑑𝜁𝜈 .

Any questions?

§2 Returning to Ramsey

Chapter 5: Concentration The Probabilistic Method

The Story So Far

Goal

• Determine the order of magnitude of 𝑆(3, 𝑙)

Upper bound

• Erdős-Szekeres (1935): 𝑆 3, 𝑙 ≤

𝑙+1 2

= O 𝑙2

Lower bounds

• First moment, Mantel: 𝑆 3, 𝑙 = Ω(𝑙)
• Alterations:

𝑆 3, 𝑙 = Ω

𝑙 log 𝑙 3/2

• Lovász Local Lemma:

𝑆 3, 𝑙 = Ω

𝑙 log 𝑙 2

Alterations Revisited

Proof sketch

• Take 𝐻 ∼ 𝐻 𝑜, 𝑞
• Remove one vertex from each triangle, independent set of size 𝑙
• Resulting graph 𝐻′ is Ramsey
• First moment ⇒ with positive probability 𝐻′ has many vertices

Optimisation

• Largest right-hand side can be is O

𝑙 log 𝑙 3/2

Theorem 2.1.2 (ℓ = 3) For every 𝑜, 𝑙 ∈ ℕ and 𝑞 ∈ [0,1], we have 𝑆 3, 𝑙 > 𝑜 − 𝑜 3 𝑞3 − 𝑜 𝑙 1 − 𝑞

𝑙 2 .

Alternative Alterations

Vertex removal

• Wasteful operation
• To fix a single, small triangle, we make Ω(𝑜) changes to the graph
• Shrinks our resulting Ramsey graph too much

Edge removal

• More efficient fix
• To fix a triangle, need only remove a single edge
• Problematic
• Being triangle-free and having small independence numbers are in conflict
• Need to ensure we can destroy all triangles without creating large independent sets
• A new hope
• Can our more advanced probabilistic tools help?

Plan of Attack

Detriangulation

• Need to remove at least one edge from each triangle
• Let 𝒰 be a maximal set of edge-disjoint triangles in 𝐻
• If 𝑈 is a triangle in 𝐻, maximality ⇒ must share an edge with some 𝑈′ ∈ 𝒰
• Remove all edges of all triangles in 𝒰
• Removes 3 𝒰 edges
• Need to remove at least 𝒰 edges

Independent sets

• Cannot let a set 𝑇 of 𝑙 vertices become independent
• Would help if 𝐻 𝑇 had many edges to begin with
• Expect to see 𝑙

2 𝑞 edges

• Chernoff ⇒ very unlikely to see many fewer
• Can afford a union bound over all such sets 𝑇

Theorem 5.1.4 (Asymmetric Chernoff Bound) Let 𝑏 > 0 and let 𝑌 and 𝑞 be as before. Then ℙ 𝑌 ≤ −𝑏 ≤ exp

−𝑏2 2𝑞𝑜 .

Local Edge Distribution

Local edge counts

• Fix a set 𝑇 of 𝑙 vertices
• 𝑓 𝐻 𝑇

∼ Bin

𝑙 2 , 𝑞

• Expect 𝑙

2 𝑞 edges, how likely are we to see at least half of that?

Applying Chernoff

• Set 𝑌 = Bin

𝑙 2 , 𝑞 − 𝑙 2 𝑞, and let 𝑏 = 1 2 𝑙 2 𝑞

• ⇒ ℙ 𝑓 𝐻 𝑇

≤

1 2 𝑙 2 𝑞 ≤ exp − 1 8 𝑙 2 𝑞

Local Properties Globally

Recall

• ℙ 𝑓 𝐻 𝑇

≤

1 2 𝑙 2 𝑞 ≤ exp − 1 8 𝑙 2 𝑞

Union bound

• We need every 𝑙-set to have many edges
• Apply a union bound over choice of 𝑇
• ℙ ∃𝑇: 𝑓 𝐻 𝑇

≤

1 2 𝑙 2 𝑞 ≤ 𝑜 𝑙 exp − 1 8 𝑙 2 𝑞 ≤ exp 𝑙 ln 𝑜 − 1 8 𝑙 2 𝑞

Setting parameters

• Small if 𝑙 ln 𝑜 ≤

1 10 𝑙 2 𝑞, say

• ⇔ 𝑞 ≥

20 ln 𝑜 𝑙−1

• To avoid too many triangles, take equality above
• Then with high probability each 𝑙-set spans at least

1 2 𝑙 2 𝑞 = 5𝑙 ln 𝑜 edges

A Tangle of Triangles

Recall

• Setting 𝑞 =

20 ln 𝑜 𝑙−1 ⇒ almost surely, every 𝑙-set has at least 5𝑙 ln 𝑜 edges

New independent sets

• Remove all edges from a maximal set 𝒰 of edge-disjoint triangles
• Need to avoid creating an independent set of 𝑙 vertices
• Fix a set 𝑇 of 𝑙 vertices

How many edges do we lose?

• Only triangles with an edge in 𝑇 are relevant
• Number of potential such triangles:
• 𝑙

3 + 𝑙 2

𝑜 − 𝑙

• Expected number of relevant triangles
• 𝑙

3 + 𝑙 2

𝑜 − 𝑙 𝑞3 ≈

4000 3

ln3 𝑜 + 4000

𝑜−𝑙 𝑙 ln3 𝑜 ≈ 4000 𝑜 𝑙 ln3 𝑜

Accounting for Triangles

Recall

• With high probability, each 𝑙-set 𝑇 spans at least 5𝑙 ln 𝑜 edges
• Expect there to be at most 4000

𝑜 𝑙 ln3 𝑜 triangles with an edge in 𝑇

Setting more parameters

• In order to ensure 𝑇 does not become independent, need

𝑜 𝑙 ln3 𝑜 ≤ 𝑑𝑙 ln 𝑜

• 𝑑 > 0 some small constant
• Solving: 𝑜 ≤ 𝑑

𝑙 ln 𝑜 2

= 𝑑′

𝑙 log 𝑙 2

Large deviations

• Need to ensure that no set 𝑇 sees too many triangles
• Union bound over 𝑜

𝑙 many sets

• ⇒ need the probability that we get more triangles than expected to be small

Lemma 5.2.1 (Erdős-Tetali, 1990) Let 𝐹1, 𝐹2, … , 𝐹𝑛 be a collection of events and set 𝜈 = σ𝑗=1

𝑛 ℙ 𝐹𝑗 . For

any 𝑡, ℙ 𝐹𝑗1 ∩ 𝐹𝑗2 ∩ ⋯ ∩ 𝐹𝑗𝑡 for some independent 𝐹𝑗1, 𝐹𝑗2, … , 𝐹𝑗𝑡 ≤ 𝜈𝑡 𝑡! .

Too Many Triangles

Concentration inequalities

• Chernoff: probability of seeing too many triangles is exponentially small
• Problem: indicator variables for triangles are not independent
• Chebyshev: error probabilities only polynomially small
• Not enough to make up for 𝑜

𝑙 summands in union bound

Saving grace

• We only remove edges of triangles in 𝒰, edge-disjoint set of triangles

The Erdős-Tetali Lemma

Proof

• Take a union bound over all such 𝑡-sets of events
• ℙ 𝐹𝑗1 ∩ ⋯ ∩ 𝐹𝑗𝑡 for some independent events

≤ σ 𝑗1,…,𝑗𝑡 ind ℙ 𝐹𝑗1 ∩ ⋯ ∩ 𝐹𝑗𝑡 =

1 𝑡! σ 𝑗1,…,𝑗𝑡 ind ℙ 𝐹𝑗1 ∩ ⋯ ∩ 𝐹𝑗𝑡

=

1 𝑡! σ 𝑗1,…,𝑗𝑡 ind ς𝑘=1 𝑡

ℙ 𝐹𝑗𝑘 ≤

1 𝑡! σ 𝑗1,…,𝑗𝑡 ∈ 𝑛 𝑡 ς𝑘=1 𝑡

ℙ 𝐹𝑗𝑘 =

1 𝑡! σ𝑗∈ 𝑛 ℙ 𝐹𝑗 𝑡

=

𝜈𝑡 𝑡!

∎

Lemma 5.2.1 (Erdős-Tetali, 1990) Let 𝐹1, 𝐹2, … , 𝐹𝑛 be a collection of events and set 𝜈 = σ𝑗=1

𝑛 ℙ 𝐹𝑗 . For

any 𝑡, ℙ 𝐹𝑗1 ∩ 𝐹𝑗2 ∩ ⋯ ∩ 𝐹𝑗𝑡 for some independent 𝐹𝑗1, 𝐹𝑗2, … , 𝐹𝑗𝑡 ≤ 𝜈𝑡 𝑡! .

Handling Triangle Errors

Recall

• With high probability, each 𝑙-set 𝑇 has at least 5𝑙 ln 𝑜 edges
• Expected number of triangles with an edge in 𝑇 at most 𝑑𝑙 ln 𝑜 for small 𝑑

Erdős-Tetali

• Events 𝐹𝑗: 𝑗th triangle meeting 𝑇 is present in 𝐻 𝑜, 𝑞
• 𝜈 ≤ 𝑑𝑙 ln 𝑜
• Let 𝑡 = 𝑙 ln 𝑜
• Lemma 5.2.1 ⇒ ℙ 𝑇 sees edges of 𝑡 disjoint triangles ≤

𝜈𝑡 𝑡!

Calculation

• Stirling: 𝑡! ≥

𝑡 𝑓 𝑡

• ⇒

𝜈𝑡 𝑡! ≤ 𝜈𝑓 𝑡 𝑡

≤ 𝑑𝑓 𝑙 ln 𝑜 < 𝑜−𝑙 if 𝑑 < 𝑓−2

Theorem 5.2.2 (Erdős, 1961; Krivelevich, 1995) As 𝑙 → ∞, 𝑆 3, 𝑙 = Ω

𝑙 log 𝑙 2

.

Completing the Proof

Union bound

• Union bound over all n

𝑙 < 𝑜𝑙 sets ⇒ with high probability, every 𝑙-set:

• Spans at least 5𝑙 ln 𝑜 edges and meets at most 𝑙 ln 𝑜 edge-disjoint triangles

Alteration

• Given 𝐻 ∼ 𝐻 𝑜, 𝑞 , where 𝑜 = 𝑑′

𝑙 log 𝑙 2

and 𝑞 =

20 ln 𝑜 𝑙−1

• Let 𝒰 be a maximal set of edge-disjoint triangles, and remove all edges in 𝒰
• Each 𝑙-set loses at most 3𝑙 ln 𝑜 edges ⇒ doesn’t become independent
• Resulting graph is therefore Ramsey.

∎

Theorem 1.5.5 (Erdős-Szekeres, 1935) For all ℓ, 𝑙 ∈ ℕ, 𝑆 ℓ, 𝑙 ≤ ℓ + 𝑙 − 2 ℓ − 1 = 𝑃 𝑙ℓ−1 . In particular, 𝑆 3, 𝑙 = 𝑃(𝑙2).

Closing In

Lower bounds

• Edge-alteration gave same bound as Lovász Local Lemma
• 𝑆 3, 𝑙 = Ω

𝑙 log 𝑙 2

• Could this be the truth? What can we say in the other direction?

Narrowing the gap

• Left with a log2 𝑙 gap to close

Independent Sets in Triangle-Free Graphs

Proof

• Key observation: 𝐻 triangle-free ⇒ every neighbourhood is independent
• ∴ if 𝐻 has a vertex of degree 𝑜 − 1, we are done
• Otherwise Δ 𝐻 <

𝑜 − 1

• Greedy algorithm:
• 𝛽 𝐻 ≥

𝑜 Δ 𝐻 +1 ≥

𝑜 ∎

Ramsey numbers

• Implies 𝑆 3, 𝑙 = 𝑃 𝑙2

Proposition 5.2.3 If 𝐻 is an 𝑜-vertex triangle-free graph, 𝛽 𝐻 ≥ 𝑜 − 1.

Theorem 5.2.4 (Ajtai, Komlós, Szemerédi, 1980; Shearer, 1995) If 𝐻 is an 𝑜-vertex triangle-free graph with maximum degree Δ, then 𝛽 𝐻 ≥ 𝑜 log Δ 8Δ .

Room for Improvement

Greedy algorithm

• Order vertices arbitrarily
• Add first vertex 𝑤 to independent set
• Remove all its ≤ Δ neighbours, and repeat
• Bound is sharp only if 𝑤 never has any neighbours previously removed
• Only true for disjoint union of cliques
• ⇒ cannot be sharp for triangle-free graphs

An Improved Upper Bound

Proof

• Let 𝑜 =

8𝑙2 log 𝑙 and let 𝐻 be an 𝑜-vertex triangle-free graph

• If Δ 𝐻 ≥ 𝑙
• Let 𝑤 be a vertex of maximum degree
• 𝑂 𝑤 is an independent set of size ≥ 𝑙
• If Δ 𝐻 < 𝑙
• Theorem 5.2.4 ⇒ 𝛽 𝐻 ≥

𝑜 log Δ 8Δ

≥

𝑜 log 𝑙 8𝑙

= 𝑙 ∎

Corollary 5.2.5 As 𝑙 → ∞, 𝑆 3, 𝑙 ≤

8𝑙2 log 𝑙.

The Big Picture

Randomness

• We show that a random independent set 𝐽 of 𝐻 has this size
• If we let 𝑍

𝑤 = 1 𝑤∈𝐽 , then 𝐽 = σ𝑤 𝑍 𝑤

• Would suffice to compute 𝔽 𝐽

= σ𝑤 𝔽[𝑍

𝑤] = σ𝑤 ℙ 𝑤 ∈ 𝐽

• Computing ℙ(𝑤 ∈ 𝐽) not straightforward – depends on neighbourhood

Neighbourhoods

• How does 𝐽 meet the neighbourhood 𝑂 𝑤 ?
• If 𝑤 ∈ 𝐽:
• Must have 𝐽 ∩ 𝑂 𝑤 = ∅
• If 𝑤 ∉ 𝐽:
• Can have 𝐽 ∩ 𝑂 𝑤 ≠ ∅
• Since 𝑂(𝑤) is independent, intersection can be arbitrary
• ⇒ might expect intersection to be large

Lemma 5.2.6 If Δ ≥ 16, we have 𝔽 𝑌𝑤 ≥

log Δ 4

for every 𝑤.

New Random Variables

Local variables

• Define new variables to account for local information
• Let 𝑌𝑤 = Δ ⋅ 1 𝑤∈𝐽 + 𝐽 ∩ 𝑂 𝑤
• Heuristic justification
• Regularise contribution of 𝑤
• When 𝑤 ∈ 𝐽, have 𝑌𝑤 = Δ
• When 𝑤 ∉ 𝐽, can still have 𝑌𝑤 = Θ Δ
• Easier to get useful bounds on 𝑌𝑤

Deducing the Theorem

Proof

• If Δ ≤ 15, done by 𝛽 𝐻 ≥

𝑜 Δ+1

• Otherwise, let 𝐽 be a uniformly random independent set of 𝐻
• For each vertex 𝑤, let 𝑌𝑤 = Δ ⋅ 1 𝑤∈𝐽 + 𝐽 ∩ 𝑂 𝑤
• Let 𝑌 = σ𝑤 𝑌𝑤
• Observe: 𝑌 ≤ 2Δ 𝐽
• Each 𝑤 ∈ 𝐽 contributes at most 2Δ: Δ via 𝑌𝑤, and 1 via 𝑌𝑣 for each neighbour 𝑣
• Lemma 5.2.6 ⇒ 𝔽 𝑌 ≥

𝑜 log Δ 4

∎

Theorem 5.2.4 (Ajtai, Komlós, Szemerédi, 1980; Shearer, 1995) If 𝐻 is an 𝑜-vertex triangle-free graph with maximum degree Δ, then 𝛽 𝐻 ≥ 𝑜 log Δ 8Δ .

Proving the Lemma

Proof

• 𝑌𝑤 = Δ ⋅ 1 𝑤∈𝐽 + 𝐽 ∩ 𝑂 𝑤
• Which 𝑣 ∈ 𝑂 𝑤 could be in 𝐽?
• Need to know 𝐽 ∩ 𝑂 𝑂 𝑤
• Idea: condition on how 𝐽 meets the rest of the graph
• Let 𝐼 = 𝐻 ∖

𝑤 ∪ 𝑂 𝑤

• 𝔽 𝑌𝑤 = 𝔽 𝔽 𝑌𝑤 𝐽 ∩ 𝑊 𝐼 = 𝐾
• Suffices to show 𝔽 𝑌𝑤 𝐽 ∩ 𝑊 𝐼 = 𝐾 ≥

log Δ 4

for every independent 𝐾 in 𝐼

Lemma 5.2.5 If Δ ≥ 16, we have 𝔽 𝑌𝑤 ≥

log Δ 4

for every 𝑤.

Extending Independent Sets

Goal

• 𝔽 𝑌𝑤 𝐽 ∩ 𝑊 𝐼 = 𝐾 ≥

log Δ 4

Available neighbours

• Let 𝐵 = 𝑂 𝑤 ∖ 𝑂 𝐾
• Those neighbours of 𝑤 that could be added to 𝐾
• Let 𝑏 = 𝐵

Independent extensions

• Two types of extensions of 𝐾 to 𝐽:
• 𝐽 = 𝐾 ∪ 𝑤
• 𝐽 = 𝐾 ∪ 𝑇, some 𝑇 ⊆ 𝐵
• 𝐽 is chosen uniformly at random from 2𝑏 + 1 optoins

Computing Conditional Expectations

Recall

• 𝑌𝑤 = Δ ⋅ 1 𝑤∈𝐽 + 𝐽 ∩ 𝑂 𝑤
• Want to show 𝔽 𝑌𝑤 𝐽 ∩ 𝑊 𝐼 = 𝐾 ≥

log Δ 4

Conditional Expectation

• Case: 𝑤 ∈ 𝐽
• Probability:

1 2𝑏+1

• 𝑌𝑤 = Δ
• Case: 𝑤 ∉ 𝐽
• Probability:

2𝑏 2𝑏+1

• 𝔽 𝑌𝑤 𝑤 ∉ 𝐽, 𝐽 ∩ 𝑊 𝐼 = 𝐾 = 𝔽 𝑇

=

𝑏 2

• ⇒ 𝔽 𝑌𝑤 𝐽 ∩ 𝑊 𝐼 = 𝐾 =

Δ 2𝑏+1 + a2a−1 2a+1

Concluding Calculations

Recall

• 𝔽 𝑌𝑤 𝐽 ∩ 𝑊 𝐼 = 𝐾 =

Δ 2𝑏+1 + 𝑏2𝑏−1 2𝑏+1

• Want to show 𝔽 𝑌𝑤 𝐽 ∩ 𝑊 𝐼 = 𝐾 ≥

log Δ 4

Contradiction

• If not,

log Δ 4

>

Δ 2𝑏+1 + 𝑏2𝑏−1 2𝑏+1

• ⇒ 2𝑏 + 1 log Δ > 4Δ + 2𝑏2𝑏
• ⇒ log Δ − 2𝑏 2𝑏 > 4Δ − log Δ
• Also ⇒ 𝑏 ≥ 1
• Must have 2𝑏 < log Δ
• ⇒ 2𝑏 <

Δ

• ⇒ log Δ − 2

Δ > 4Δ − log Δ

• False for Δ ≥ 16

∎

Theorem 5.2.7 (Kim, 1995) As 𝑙 → ∞, 𝑆 3, 𝑙 = Ω

𝑙2 log 𝑙 .

Epilogue

What we know

• Ω

𝑙2 log2 𝑙 = 𝑆 3, 𝑙 = 𝑃 𝑙2 log 𝑙

Remarks

• Kim’s proof a “tour de force”
• Lower bound recently sharpened via analysis of triangle-free process
• Asymptotics of 𝑆(𝑡, 𝑙), 𝑡 ≥ 4 fixed and 𝑙 → ∞, unknown

Any questions?

§3 Hamiltonicity

Chapter 5: Concentration The Probabilistic Method

Definition 5.3.1 A Hamiltonian cycle in a graph 𝐻 is a cycle passing through every vertex

• f 𝐻. A graph is called Hamiltonian if it contains a Hamiltonian cycle.

Setting the Scene

Questions

• Are there easy ways to recognise Hamiltonian graphs?
• What happens for the average graph?

Theorem 5.3.2 (Karp, 1972) Deciding whether a graph is Hamiltonian is NP-Complete.

Theorem 5.3.3 (Dirac, 1952) Every 𝑜-vertex graph 𝐻 with minimum degree 𝜀 𝐻 ≥

𝑜 2 is Hamiltonian.

A Sufficient Condition

Optimal bound

• 𝑜 even: two disjoint cliques
• 𝑜 odd: two cliques sharing one vertex

Corollary 5.3.4 For every 𝜁 > 0 and 𝑞 ≥

1 2 + 𝜁 𝑜, 𝐻 𝑜, 𝑞 is Hamiltonian w.h.p.

Proposition 5.3.5 For every 𝜁 > 0 and 𝑞 ≤

1−𝜁 log 𝑜 𝑜

, 𝐻 𝑜, 𝑞 is w.h.p. not Hamiltonian.

Threshold Lower Bound

First moment

• There are

𝑜−1 ! 2

=

𝑜 1+𝑝 1 𝑓 𝑜

possible Hamiltonian cycles

• Each appears in 𝐻 𝑜, 𝑞 with probability 𝑞𝑜
• ⇒ expected number of cycles is

𝑜𝑞 1+𝑝 1 𝑓 𝑜

• ⇒ if 𝑞 ≤

𝑓−𝜁 𝑜 , then 𝐻 𝑜, 𝑞 has no Hamiltonian cycles w.h.p.

Connectivity

• 𝐻(𝑜, 𝑞) Hamiltonian ⇒ 𝐻 𝑜, 𝑞 connected

Dirac’s Theorem

Proof

• 𝐻 is connected
• If not, smaller component would not support minimum degree
• Let 𝑄 = 𝑤0𝑤1𝑤2 … 𝑤𝑢 be a longest path
• 𝑂

𝑤0, 𝑤𝑢 ⊆ 𝑄, as otherwise path could be extended

• Pigeonhole: ∃𝑗 such that 𝑤𝑗, 𝑤𝑢 , 𝑤0, 𝑤𝑗+1 ∈ 𝐹 𝐻
• We have a cycle 𝐷 = 𝑤0𝑤1𝑤2 … 𝑤𝑗𝑤𝑢𝑤𝑢−1𝑤𝑢−2 … 𝑤𝑗+1𝑤0
• If 𝑢 = 𝑜, this is a Hamiltonian cycle
• If 𝑢 < 𝑜, connectivity ⇒ edge from 𝐷 to 𝐻 ∖ 𝐷
• Gives a longer path, contradiction.

∎

Theorem 5.3.3 (Dirac, 1952) Every 𝑜-vertex graph 𝐻 with minimum degree 𝜀 𝐻 ≥

𝑜 2 is Hamiltonian.

Dirac’s Algorithm

More than existential

• Proof shows us how to find a Hamiltonian cycle
• Start with any path
• If there are edges out from the endpoints, extend path
• Otherwise by pigeonhole turn path into cycle
• Use external edge to extend path
• Repeat until cycle is Hamiltonian

Random setting

• Extremal problem:
• Need to assume worst-case graph
• Used large degree, pigeonhole to rotate path into cycle
• Can we use properties of 𝐻 𝑜, 𝑞 to do this more efficiently?

Definition 5.3.6 (Booster) Given a graph 𝐻, a booster is a potential edge 𝑓 such that 𝐻 ∪ 𝑓 contains a longer path or a Hamiltonian cycle.

Pósa Rotations

Goal

• Given path 𝑄 = 𝑤0𝑤1 … 𝑤𝑢 in a graph 𝐻
• Want to find a longer path or a Hamiltonian cycle

Rotations

• If 𝐻 is connected, the pair 𝑤0, 𝑤𝑢 is a booster
• Suppose 𝑤𝑗, 𝑤𝑢 ∈ 𝐹 𝐻 , 1 ≤ 𝑗 ≤ 𝑢 − 2
• Rotation along 𝑤𝑗, 𝑤𝑢 : 𝑄′ = 𝑤0𝑤1 … 𝑤𝑗𝑤𝑢𝑤𝑢−1 … 𝑤𝑗+1 also a path of length 𝑢
• ⇒ the pair 𝑤0, 𝑤𝑗+1 is also a booster

Endpoint Neighbourhoods

Proof

• After rotating along {𝑤𝑗, 𝑤𝑢}, only 𝑤𝑗, 𝑤𝑢 get new neighbours on the path
• Let 𝑤 ∈ 𝑆
• Rotate to path 𝑄′ with 𝑤 as an endpoint
• Let 𝑧 ∈ 𝑂 𝑤 ∖ 𝑆
• If 𝑧 ∉ 𝑊(𝑄), extend 𝑄′ to 𝑧 ⇒ longer path than 𝑄
• If 𝑧 ∈ 𝑊 𝑄 , rotate 𝑄′ along the edge 𝑤, 𝑧
• ⇒ a neighbour 𝑦 of 𝑧 on 𝑄′ is an endpoint of the new path, so 𝑦 ∈ 𝑆
• If 𝑦 also a neighbour of 𝑧 on 𝑄, then 𝑧 ∈ 𝑂𝑄(𝑆)
• Otherwise must have rotated along an edge incident to 𝑧 ⇒ 𝑧 ∈ 𝑂𝑄 𝑆

∎

Lemma 5.3.7 Let 𝑄 = 𝑤0𝑤1 … 𝑤𝑢 be a longest path in a graph 𝐻, and let 𝑆 be the set

• f endpoints reachable from 𝑤0 by sequences of rotations. Then

𝑂𝐻 𝑆 ⊆ 𝑂𝑄 𝑆 .

Corollary 5.3.8 Let 𝑄 be a longest path in 𝐻, and let 𝑆 be the set of endpoints following sequences of rotations. Then 𝑂𝐻 𝑆 ≤ 2 𝑆 − 1.

Expanders

Proof

• Lemma 5.3.7 ⇒ 𝑂𝐻 𝑆 ⊆ 𝑂𝑄 𝑆
• Each vertex in 𝑆 contributes at most two neighbours to 𝑂𝑄 𝑆
• Final vertex 𝑤𝑢 only contributes one
• ⇒ 𝑂𝑄 𝑆

≤ 2 𝑆 − 1 ∎

Definition 5.3.9 (Expander) A graph 𝐻 is a 𝑙, 2 -expander if, for every 𝑇 ⊆ 𝑊 𝐻 with 𝑇 ≤ 𝑙, we have 𝑂𝐻 𝑇 ≥ 2 𝑇 .

Expanders Have Many Boosters

Proof

• If 𝐻 is Hamiltonian, every edge is a booster.
• Otherwise let 𝑄 = 𝑤0𝑤1 … 𝑤𝑢 be a longest path
• Fix 𝑤0, and let 𝑆0 be the endpoints after rotations
• Corollary 5.3.8 ⇒ 𝑂𝐻 𝑆0

≤ 2 𝑆0 − 1

• 𝐻 a 𝑙, 2 -expander ⇒ 𝑆0 ≥ 𝑙 + 1
• Given any 𝑧 ∈ 𝑆0, rotate to a 𝑤0-𝑧 path 𝑄′
• Fix 𝑧, and let 𝑆𝑧 be the endpoints of paths from 𝑧 after rotating 𝑄′
• Again, 𝑆𝑧 ≥ 𝑙 + 1
• For each 𝑨 ∈ 𝑆𝑧, 𝑧, 𝑨 is a booster, counted at most twice

∎

Corollary 5.3.10 If 𝐻 is a connected (𝑙, 2)-expander, then 𝐻 has at least

1 2 𝑙2 boosters.

Dirac’s Algorithm in Random Graphs

Assumptions

• 𝐻 𝑜, 𝑞 is connected – know to be true for 𝑞 ≥

1+𝜁 log 𝑜 𝑜

• 𝐻 𝑜, 𝑞 is a 𝑙, 2 -expander for 𝑙 large

Rotation-Extension process

• Start with a longest path 𝑄
• Corollary 5.3.10 ⇒ gives rise to Ω 𝑙2 boosters
• Each booster is an edge of 𝐻 𝑜, 𝑞 independently with probability 𝑞
• ⇒ Probability none of the boosters appear is 1 − 𝑞 𝑙2
• ⇒ if 𝑞 = 𝜕 𝑙−2 , then w.h.p. one of the boosters should be in 𝐻 𝑜, 𝑞
• Use it to extend path, repeat until Hamiltonian

Multiple Exposures

Recall

• Longest path gave rise to Ω 𝑙2 boosters
• Want to show w.h.p. a booster appears in 𝐻 𝑜, 𝑞

Problem

• To find the boosters, we needed to expose edges in 𝑊 𝑄

2

• Might already have found boosters are not edges
• They do not appear independently with probability 𝑞

Solution

• Split the random graph into independent subgraphs
• Let 𝑞0, 𝑟 satisfy 1 − 𝑞 = 1 − 𝑞0

1 − 𝑟

• Then 𝐻 𝑜, 𝑞 ∼ 𝐻 𝑜, 𝑞0 ∪ 𝐻 𝑜, 𝑟
• Use 𝐻 𝑜, 𝑞0 to obtain connectivity, expansion properties, find boosters
• Use 𝐻 𝑜, 𝑟 to show boosters appear in the random graph w.h.p.

Random Graphs are Expanders

Proof

• If not, there is some set 𝑇 of size 𝑡 ≔ 𝑇 ≤

𝑜 6 such that 𝑂 𝑇

< 2𝑡

• ⇒ ∃𝑋 ⊂ 𝑊 𝐻 ∖ 𝑇, 𝑋 = 2𝑡, such that we have no edges from 𝑇 to 𝑊 𝐻 ∖ 𝑇 ∪ 𝑋
• Probability these edges are missing is 1 − 𝑞 𝑡 𝑜−3𝑡 ≤ 𝑓−𝑞𝑡 𝑜−3𝑡 ≤ 𝑓−𝑞𝑡𝑜/2
• Count number of pairs 𝑇, 𝑋
• 𝑜

𝑡

≤

𝑜𝑓 𝑡 𝑡

choices for 𝑇,

𝑜−𝑡 2𝑡

≤

𝑜 2𝑡 ≤ 𝑜𝑓 2𝑡 2𝑡

choices for 𝑋

• Union bound
• ℙ 𝐻 𝑜, 𝑞 bad ≤ σ𝑡=1

𝑜 6

𝑜3𝑓3 4𝑡3 𝑓−𝑞𝑜/2 𝑡

≤ σ𝑡=1

𝑜 6

𝑓3 4 𝑜 𝑡

= 𝑝 1 ∎

Lemma 5.3.11 If 𝑞 ≥

7 log 𝑜 𝑜

, then 𝐻 𝑜, 𝑞 is w.h.p. an

𝑜 6 , 2 -expander.

The Hamiltonicity Threshold

Proof

• Let 𝑞0 =

7 log 𝑜 𝑜

and 𝑟 =

73 log 𝑜 𝑜2

• Let 𝐻0 ∼ 𝐻 𝑜, 𝑞0 , and for 𝑗 ∈ [𝑜], let 𝐻𝑗 ∼ 𝐻 𝑜, 𝑟 be independent
• If 𝐻 = 𝐻0 ∪ (∪𝑗 𝐻𝑗), then 𝐻 ∼ 𝐻 𝑜, 𝑞 for 𝑞 = 1 − 1 − 𝑞0

1 − 𝑟 𝑜 ≤

80 log 𝑜 𝑜2

• Lemma 5.3.11 ⇒ 𝐻0 is w.h.p. a connected

𝑜 6 , 2 -expander

• Corollary 5.3.10 ⇒ any supergraph of 𝐻0 has at least

𝑜2 72 boosters

• ⇒ probability 𝐻𝑗 does not contain one of the boosters ≤ 1 − 𝑟

𝑜2 72 ≤ 𝑓 −𝑟𝑜2 72 = 𝑝

1 𝑜

• ⇒ Grow a longest path, using 𝐻𝑗 to find a booster in the 𝑗th step

∎

Theorem 5.3.12 (Pósa, 1976) If 𝑞 ≥

80 log 𝑜 𝑜

, then 𝐻 𝑜, 𝑞 is w.h.p. Hamiltonian.

Theorem 5.3.13 (Komlós-Szemerédi, 1983) For 𝜁 > 0 and 𝑞 ≥

1+𝜁 log 𝑜 𝑜

, 𝐻 𝑜, 𝑞 is w.h.p. Hamiltonian.

Epilogue

• Even sharper results were later proven

Theorem 5.3.14 (Bollobás, 1984; Ajtai-Komlós-Szemerédi, 1985) In the random graph process, w.h.p. the graph becomes Hamiltonian precisely when the minimum degree is at least two.

• Hamiltonicity displays a very sharp threshold

Any questions?

§4 Martingales

Chapter 5: Concentration The Probabilistic Method

Threshold for Triangles

Triangular case

• ℓ = 3: threshold for containing triangles is 𝑜−1

Upper tail

• When 𝑞 ≫ 𝑜−1, how unlikely is 𝐻 𝑜, 𝑞 to be triangle-free?
• Proof of Theorem 3.3.1
• Used Chebyshev’s Inequality
• Gives polynomial error bounds

Theorem 3.3.1 For ℓ ≥ 2, the threshold for 𝐿ℓ ⊆ 𝐻 𝑜, 𝑞 is 𝑞0 𝑜 = 𝑜−2/(ℓ−1).

Exponential Dreams

Indicator random variables

• Let 𝑌 denote the number of triangles in 𝐻 ∼ 𝐻 𝑜, 𝑞
• Given 𝑈 ∈

𝑜 3 , let 𝑌𝑈 be the indicator that 𝐻 𝑈 ≅ 𝐿3

• Then ℙ 𝑌𝑈 = 1 = 𝑞3
• Also 𝑌 = σ𝑈 𝑌𝑈

Stronger concentration

• Using Chernoff would give ℙ 𝑌 = 0 ≤ exp −

1 2 𝑜 3 𝑞3

• Exponentially small error bound
• Problem: summands 𝑌𝑈 not independent
• 𝑌𝑈, 𝑌𝑈′ positively correlated when 𝑈 ∩ 𝑈′ = 2

Lemma 5.4.1 There exists a family of

1 3 𝑜−1 2

pairwise edge-disjoint triangles in 𝐿𝑜.

Sparse Independence

Cheap fix

• Restrict our attention to mutually independent events
• Equivalently: consider a family of edge-disjoint triangles

Proof

• Colour each triangle 𝑗, 𝑘, 𝑙 with the colour 𝑑 ≡ 𝑗 + 𝑘 + 𝑙 mod 𝑜
• Each colour class is edge-disjoint
• Given vertices 𝑗, 𝑘, third vertex 𝑙 ≡ 𝑑 − 𝑗 − 𝑘 determined
• Large colour class
• For some 𝑑, number of 𝑑-coloured triangles is at least

1 𝑜 𝑜 3 = 1 3 𝑜−1 2

∎

Don’t Let Your Dreams Be Dreams

Proof

• Let 𝒰 be the collection of triangles from Lemma 5.4.1
• If 𝐻 𝑜, 𝑞 is triangle-free, no triangle in 𝒰 appears
• These appear independently
• Probability none appear is 1 − 𝑞3

𝒰 ≤ exp − 𝒰 𝑞3

∎

Good news

• Exponential bound on error probability

Bad news

• Exponent 𝑜−1

2

𝑞3 = Θ 𝑜2𝑞3 is of lower order than expected

Corollary 5.4.2 𝐻 𝑜, 𝑞 is triangle-free with probability at most exp −

1 3 𝑜−1 2

𝑞3 .

Postmortem of a Proof

Improving the exponent

• Need to consider all 𝑜

3 possible triangles

• Dependencies are limited – can we recover Chernoff-type bounds?

Revisiting Chernoff

• 𝑇𝑜 = σ𝑗=1

𝑜

𝑌𝑗

• Properties of 𝑌𝑗:
• Bounded, −1,1 -variables
• 𝔽 𝑌𝑗 = 0
• 𝑌𝑗 mutually independent
• Using independence:
• Applied Markov to 𝑓𝑇𝑜
• Independence ⇒ 𝔽 𝑓𝑇𝑜 = 𝔽 𝑓σ𝑗 𝑌𝑗 = ς𝑗 𝔽 𝑓𝑌𝑗

Definition 5.4.3 (Martingale) A martingale is a sequence 𝑎0, 𝑎1, … , 𝑎𝑛 of random variables such that, for each 1 ≤ 𝑗 ≤ 𝑛, we have 𝔽 𝑎𝑗 𝑎

𝑘: 𝑘 < 𝑗

= 𝑎𝑗−1. Loosely speaking, given what has previously transpired, we expect nothing to change in the 𝑗th step.

Martingales

Conditional independence

• What if the 𝑌𝑗 are not independent?
• Recover independence by conditioning on previous variables
• Product rule: 𝔽 𝑓σ𝑗 𝑌𝑗 = ς𝑗 𝔽 𝑓𝑌𝑗 𝑌

𝑘: 𝑘 < 𝑗

• ⇒ if 𝑌𝑗 𝑌

𝑘: 𝑘 < 𝑗

has the right properties, can prove Chernoff-type bounds

Martingales, Tame and Wild

Boring mathsy example

• Let 𝑌𝑗 be independent and uniform on −1,1 , for 1 ≤ 𝑗 ≤ 𝑛
• Let 𝑎𝑗 = σ𝑘≤𝑗 𝑌

𝑘

• 𝔽 𝑎𝑗 𝑎

𝑘: 𝑘 < 𝑗

= 𝔽 𝑎𝑗−1 + 𝑌𝑗 𝑎

𝑘: 𝑘 < 𝑗

= 𝑎𝑗−1 + 𝔽 𝑌𝑗| 𝑎

𝑘: 𝑘 < 𝑗

• 𝔽 𝑌𝑗 𝑎

𝑘: 𝑘 < 𝑗

= 𝔽 𝑌𝑗 = 0

• ⇒ Zi: 0 ≤ 𝑗 ≤ 𝑛 is a martingale

Fun real-world example

• Gambling on (fair) coin tosses
• 𝑎𝑗 = cumulative profit/loss after 𝑗th toss
• Bet 𝑐𝑗 = 𝑐𝑗 𝑎0, 𝑎1, … , 𝑎𝑗−1 on the 𝑗th toss, depending on previous outcomes
• 𝔽 𝑎𝑗 𝑎

𝑘: 𝑘 < 𝑗

=

1 2 Zi−1 + bi + 1 2 Zi−1 − bi = Zi−1

• ⇒ 𝑎𝑗: 0 ≤ 𝑗 ≤ 𝑛 is a martingale

Disclaimer: gambling can be addictive and bad for your bank balance

Martingale Concentration

Proof

• Set 𝑌𝑗 = 𝑎𝑗 − 𝑎𝑗−1
• ⇒ 𝑌𝑗 ≤ 1 and 𝑎𝑛 = σ𝑗=1

𝑛 𝑌𝑗

• Martingale ⇒ 𝔽 𝑌𝑗 𝑎

𝑘: 𝑘 < 𝑗

= 0

• For any 𝜇 > 0, we have 𝑎𝑛 ≥ 𝑏 ⇔ e𝜇Zm ≥ 𝑓𝜇𝑏
• ℙ 𝑓𝜇𝑎𝑛 ≥ 𝑓𝜇𝑏 ≤ 𝔽 𝑓𝜇𝑎𝑛 𝑓−𝜇𝑏
• 𝔽 𝑓𝜇𝑎𝑛 = ς𝑗=1

𝑛 𝔽 𝑓𝜇𝑌𝑗 𝑎 𝑘: 𝑘 < 𝑗

Theorem 5.4.4 (Azuma’s Inequality) Let 𝑎0, 𝑎1, … , 𝑎𝑛 be a martingale with 𝑎0 = 0 and 𝑎𝑗 − 𝑎𝑗−1 ≤ 1 for all 1 ≤ 𝑗 ≤ 𝑛. Then, for any a > 0, we have ℙ 𝑎𝑛 ≥ 𝑏 ≤ exp −𝑏2/2𝑛 .

A Little Calculus

Proof

• Let 𝑔 𝑧 =

𝑓𝜇+𝑓−𝜇 2

+

𝑓𝜇−𝑓−𝜇 2

𝑧 = 𝑓𝜇

1 2 + 𝑧 2 + 𝑓−𝜇 1 2 − 𝑧 2

• ⇒ 𝑔 represents chord between 𝑕 𝑧 = 𝑓𝜇𝑧 between 𝑧 = −1 and 𝑧 = 1
• Convexity ⇒ 𝑕 𝑧 ≤ 𝑔 𝑧 for all 𝑧 ∈ −1,1
• Thus 𝔽 𝑓𝜇𝑍 = 𝔽 𝑕 𝑍

≤ 𝔽 𝑔 𝑍 =

𝑓𝜇+𝑓−𝜇 2

+

𝑓𝜇−𝑓−𝜇 2

𝔽 𝑍 = cosh 𝜇 ∎

Lemma 5.4.5 If 𝜇 > 0 and 𝑍 is a random variable with 𝔽 𝑍 = 0 and 𝑍 ≤ 1, then 𝔽 𝑓𝜇𝑍 ≤ cosh 𝜇 .

Completing the Proof

Proof (cont’d)

• ℙ 𝑓𝜇𝑎𝑛 ≥ 𝑓𝜇𝑏 ≤ 𝔽 𝑓𝜇𝑎𝑛 𝑓−𝜇𝑏
• 𝔽 𝑓𝜇𝑎𝑛 = ς𝑗=1

𝑛 𝔽 𝑓𝜇𝑌𝑗 𝑎 𝑘: 𝑘 < 𝑗

• By Lemma 5.4.5, 𝔽 𝑓𝜇𝑌𝑗 𝑎

𝑘: 𝑘 < 𝑗

≤ cosh 𝜇 ≤ 𝑓𝜇2/2

• ∴ ℙ 𝑎𝑛 ≥ 𝑏 ≤ exp

𝜇2𝑛 2 − 𝜇𝑏

• Substitute 𝜇 =

𝑏 𝑛

∎

Theorem 5.4.4 (Azuma’s Inequality) Let 𝑎0, 𝑎1, … , 𝑎𝑛 be a martingale with 𝑎0 = 0 and 𝑎𝑗 − 𝑎𝑗−1 ≤ 1 for all 1 ≤ 𝑗 ≤ 𝑛. Then, for any a > 0, we have ℙ 𝑎𝑛 ≥ 𝑏 ≤ exp −𝑏2/2𝑛 .

Graph Martingales

Upper tail for triangles

• Sample 𝐻 ∼ 𝐻 𝑜, 𝑞 , 𝑌 = # triangles in 𝐻
• 𝑌 = σ𝑈∈

𝑜 3 𝑌𝑈, with 𝑌𝑈 the indicator that 𝐻 𝑈 ≡ 𝐿3

Where is the martingale?

• Natural candidate
• Order sets 𝑈

1, 𝑈 2, … , 𝑈 𝑛

• Let 𝑎𝑗 = σ𝑘≤𝑗 𝑌𝑈𝑘
• Problem
• Positive correlation ⇒ cannot make 𝔽 𝑌𝑈𝑗 𝑎

𝑘: 𝑘 < 𝑗

= 0 for all choices of 𝑎

𝑘

• Solution
• Reveal information about 𝐻 in stages
• Let 𝑎𝑗 be the expected value of 𝑌 given the information after 𝑗 rounds

The Doob Martingale

General framework

• Sample 𝐻 ∼ 𝐻 𝑜, 𝑞 , interested in graph parameter 𝑔 𝐻 ∈ ℝ
• Example: 𝑔 𝐻 = # triangles in 𝐻

Revealing 𝐻

• Order the possible edges

𝑜 2

= 𝑓1, 𝑓2, … , 𝑓𝑛 for 𝑛 =

𝑜 2

• Let 𝑇𝑗 = 𝑓

𝑘: 𝑘 ≤ 𝑗

The martingale

• 𝑎𝑗 = 𝔽 𝑔 𝐻 𝐹 𝐻 ∩ 𝑇𝑗 − 𝔽 𝑔 𝐻
• Expected value of parameter given the previously revealed edges
• 𝑎0 = 𝔽 𝑔 𝐻

− 𝔽 𝑔 𝐻 = 0

• 𝑎𝑛 = 𝔽 𝑔 𝐻 𝐹 𝐻 ∩

𝑜 2

− 𝔽 𝑔 𝐻 = 𝑔 𝐻 − 𝔽 𝑔 𝐻

A Small Example

Framework

• 𝐻 ∼ 𝐻 3,

1 2 and 𝑔 𝐻 = 𝜕 𝐻

𝐻 𝑔 𝐻 1 2 2 2 2 2 2 3 1.5 2 2 2.5 1.75 2.25 2

Verifying Martingale-ness

Recall

• 𝐻 ∼ 𝐻(𝑜, 𝑞), and we are exploring a graph parameter 𝑔(𝐻)
• 𝑇𝑗 = 𝑓

𝑘: 𝑘 ≤ 𝑗

• 𝑎𝑗 = 𝔽 𝑔 𝐻 𝐹 𝐻 ∩ 𝑇𝑗

Conditional expectations

• 𝔽 𝑎𝑗+1 𝐹 𝐻 ∩ 𝑇𝑗 = 𝔽 𝔽 𝑔 𝐻 𝐹 𝐻 ∩ 𝑇𝑗+1 𝐹 𝐻 ∩ 𝑇𝑗

= 𝔽 𝑔 𝐻 𝐹 𝐻 ∩ 𝑇𝑗 = 𝑎𝑗

• ⇒ this is a martingale

Definition 5.4.6 (𝑑-Lipschitz) Let 𝑑 > 0. A graph parameter 𝑔 is 𝑑-(edge-)Lipschitz if, for any edge 𝑓, 𝑔 𝐻 − 𝑔 𝐻 △ 𝑓 ≤ 𝑑.

Lipschitz Properties

Bounded differences

• To apply Azuma’s Inequality, we need 𝑎𝑗 − 𝑎𝑗−1 ≤ 1 for all 𝑗
• Intuitively: changing one edge should not change 𝑔 𝐻 much

Fact 5.4.7 Given a 𝑑-Lipschitz parameter 𝑔, we have 𝑎𝑗 − 𝑎𝑗−1 ≤ 1 for the normalised Doob martingale 𝑎𝑗 =

1 𝑑 𝔽 𝑔 𝐻 𝐹 𝐻 ∩ 𝑇𝑗 − 𝔽 𝑔 𝐻

.

Theorem 5.4.4 (Azuma’s Inequality) Let 𝑎0, 𝑎1, … , 𝑎𝑛 be a martingale with 𝑎0 = 0 and 𝑎𝑗 − 𝑎𝑗−1 ≤ 1 for all 1 ≤ 𝑗 ≤ 𝑛. Then, for any a > 0, we have ℙ 𝑎𝑛 ≥ 𝑏 ≤ exp −𝑏2/2𝑛 .

Summary

Remarks

• Same bound holds for ℙ 𝑔 𝐻 ≤ 𝜈 − 𝑏
• Can also use a vertex-exposure martingale
• 𝑎𝑗 is the expected value of 𝑔 𝐻 after exposing induced subgraph of 𝐻 on first 𝑗 vertices

Corollary 5.4.8 Let 𝑔 be a 𝑑-Lipschitz graph parameter, 𝐻 ∼ 𝐻 𝑜, 𝑞 , 𝜈 = 𝔽 𝑔 𝐻 , and 𝑏 > 0. Then ℙ 𝑔 𝐻 ≥ 𝜈 + 𝑏 ≤ exp −𝑏2/𝑜2𝑑2 .

Any questions?

§5 Triangle-free Graphs

Chapter 5: Concentration The Probabilistic Method

Theorem 3.3.1 For ℓ ≥ 2, the threshold for 𝐿ℓ ⊆ 𝐻 𝑜, 𝑞 is 𝑞0 𝑜 = 𝑜−2/(ℓ−1).

A Quick Review

Triangle-freeness

• ⇒ when 𝑞 = 𝜕 𝑜−1 , ℙ 𝐿3 ⊈ 𝐻 𝑜, 𝑞

= 𝑝 1

• Error bound from Chebyshev ⇒ only polynomially small

Exponential error bounds

• Sharper estimates by considering edge-disjoint triangles

Corollary 5.4.2 𝐻 𝑜, 𝑞 is triangle-free with probability at most exp −

1 3 𝑜−1 2

𝑞3 .

Corollary 5.4.8’ Let 𝑔 be a 𝑑-Lipschitz graph parameter, 𝐻 ∼ 𝐻 𝑜, 𝑞 , 𝜈 = 𝔽 𝑔 𝐻 , and 𝑏 > 0. Then ℙ 𝑔 𝐻 ≤ 𝜈 − 𝑏 ≤ exp −𝑏2/𝑜2𝑑2 .

Applying Azuma

Counting triangles

• 𝑔 𝐻 = # triangles in 𝐻
• 𝜈 =

𝑜 3 𝑞3 = a

• 𝑑 = 𝑜 − 2

Corollary 5.5.1 𝐻 𝑜, 𝑞 is triangle-free with probability at most exp

− 𝑜−1 2𝑞6 36

.

Corollary 5.5.2 Let 𝑔 be a 𝑑𝑤-vertex-Lipschitz parameter, 𝜈 = 𝔽 𝑔 𝐻 , and 𝑏 > 0. Then, for 𝐻 ∼ 𝐻(𝑜, 𝑞), ℙ 𝑔 𝐻 ≤ 𝜈 − 𝑏 ≤ exp −𝑏2/2𝑜𝑑𝑤

2 .

Immeasurable Disappointment

Worse exponent

• Exponent

1 36 𝑜 − 1 2𝑞6 is worse than the 1 3 𝑜−1 2

𝑞3 from before

• Problems
• Long martingale, 𝑜

2 , and large Lipschitz constant, 𝑜 − 2

• What if we apply vertex-exposure instead?

Vertex-exposure martingale

• Shorter martingale, 𝑜, but worse Lipschitz constant, 𝑜−1

2

• Yields a worse exponent, Θ 𝑜𝑞6

Corollary 5.4.8 Let 𝑔 be a 𝑑-Lipschitz graph parameter, 𝐻 ∼ 𝐻 𝑜, 𝑞 , 𝜈 = 𝔽 𝑔 𝐻 , and 𝑏 > 0. Then ℙ 𝑔 𝐻 ≥ 𝜈 + 𝑏 ≤ exp −𝑏2/𝑜2𝑑2 .

A Judicious Parameter

Reducing the Lipschitz constant

• Need to decrease the influence a single edge can have
• Idea: edge-disjoint triangles
• Let 𝑔 𝐻 = maximum number of pairwise edge-disjoint triangles

New bound

• This choice of 𝑔 is 1-Lipschitz
• Still have 𝐻 triangle-free ⇔ 𝑔 𝐻 = 0, so take 𝑏 = 𝔽 𝑔 𝐻
• ⇒ ℙ 𝐻 triangle−free ≤ exp −𝔽 𝑔 𝐻

2/𝑜2

• How do we bound this expectation?

Edge-Disjoint Triangles

Proof

• Let 𝒰 be the collection of all 𝑌 triangles in 𝐻
• Let ℛ′ ⊆ 𝒰 be a 𝑟-random subcollection
• Triangle 𝑈 ∈ ℛ′ with probability 𝑟, independent of all other triangles
• Let 𝑍′ be the number of pairs of overlapping triangles in ℛ′
• From each pair in ℛ′ sharing an edge, remove one of the triangles
• ⇒ resulting ℛ ⊆ ℛ′ is pairwise edge-disjoint
• 𝔽 ℛ

≥ 𝔽 ℛ′ − 𝑍′ = 𝑟𝑌 − 𝑟2𝑍 ∎

Lemma 5.5.3 Let 𝑟 ∈ 0,1 , and let 𝐻 be a graph with 𝑌 triangles and 𝑍 pairs of triangles sharing an edge. Then 𝐻 has a collection of 𝑛 pairwise edge- disjoint triangles, for some 𝑛 ≥ 𝑟𝑌 − 𝑟2𝑍.

Corollary 5.5.4 Let 𝐻 ∼ 𝐻 𝑜, 𝑞 for 𝑞 ≥

1 3𝑜. Then 𝔽 𝑔 𝐻

≥

1 36 − 𝑝 1

𝑜2𝑞.

Random Edge-Disjoint Triangles

Random graph setting

• Let 𝐻 ∼ 𝐻 𝑜, 𝑞 , 𝑌 = # triangles, 𝑍 = # overlapping pairs of triangles
• Lemma 5.5.3 ⇒ 𝑔 𝐻 ≥ 𝑟𝑌 − 𝑟2𝑍 for all 𝑟 ∈ 0,1
• ⇒ 𝔽 𝑔 𝐻

≥ 𝑟𝔽 𝑌 − 𝑟2𝔽 𝑍

Choosing values

• We have 𝔽 𝑌 =

𝑜 3 𝑞3, 𝔽 𝑍 = 𝑜 2 𝑜−2 2

𝑞5

• Calculus ⇒ optimal 𝑟 =

1 3𝑜𝑞2

Theorem 5.5.5 Let 𝑞 ≥

1 3𝑜 and let 𝐻 ∼ 𝐻 𝑜, 𝑞 . Then

ℙ 𝐿3 ⊈ 𝐻 ≤ exp −Ω 𝑜2𝑞2 .

Immeasurable Joy

Recall

• 𝐻 ∼ 𝐻 𝑜, 𝑞
• 𝑔 𝐻 = maximum number of pairwise edge-disjoint triangles in 𝐻
• Corollary 5.5.4 ⇒ if 𝑞 ≥ 1/ 3𝑜, then 𝔽 𝑔 𝐻

≥ Ω 𝑜2𝑞

• Corollary 5.4.8 ⇒ ℙ 𝐻 triangle−free ≤ exp −𝔽 𝑔 𝐻

2/𝑜2

• Improves previous exponent when 𝑑𝑜−1/2 ≤ 𝑞 ≤ c′n

Any questions?

§6 Chromatic Number

Chapter 5: Concentration The Probabilistic Method

Introducing the Problem

General bounds

• What makes the chromatic number large?
• 𝜓 𝐻 ≥ 𝜕 𝐻
• 𝜓 𝐻 ≥

𝑜 𝛽 𝐻

Complexity

• Determining chromatic number of graphs is NP-Complete
• Even deciding if a graph is 3-colourable is NP-Complete

Typical behaviour

• What can we say about 𝜓 𝐻 𝑜,

1 2

?

Colouring Random Graphs

Question

• What is 𝜓 𝐻 𝑜,

1 2

?

Applying general bounds

• Homework: with high probability, 𝜕 𝐻 𝑜,

1 2

∼ 2 log 𝑜

• ⇒ 𝜓 𝐻 𝑜,

1 2

≥ 2 + 𝑝 1 log 𝑜

• Symmetry ⇒ 𝛽 𝐻 𝑜,

1 2

∼ 2 log 𝑜

• ⇒ 𝜓 𝐻 𝑜,

1 2

≥

1+𝑝 1 𝑜 2 log 𝑜

• Homework: will show this bound is sharp

Lemma 5.6.1 The parameter 𝜓 𝐻 is 1-vertex-Lipschitz.

Honing In

• Can we further narrow down the likely values of 𝜓 𝐻 𝑜,

1 2

? Proof

• Let 𝑤 ∈ 𝑊 𝐻 be arbitrary, and let 𝐼 = 𝐻 𝑊 ∖ 𝑤
• Chromatic number is monotone increasing
• ⇒ 𝜓 𝐻 ≥ 𝜓 𝐼
• Can always assign 𝑤 a new colour
• ⇒ 𝜓 𝐻 ≤ 𝜓 𝐼 + 1
• ⇒ changing 𝐻 at 𝑤 can change 𝜓(𝐻) by at most one

∎

Colouring with Martingales

Proof

• Apply the vertex-exposure martingale to the parameter 𝜓 𝐻
• 𝑎𝑗 = 𝔽 𝜓 𝐻 𝐻 𝑗

− 𝔽 𝜓 𝐻 , 0 ≤ 𝑗 ≤ 𝑜

• Lemma 5.6.1: 𝜓 𝐻 is 1-vertex-Lipschitz
• Azuma’s inequality: ℙ 𝑎𝑗 ≥ 𝑏 ≤ 2 exp −𝑏2/2𝑜
• If 𝑏 =

2𝑜 ln

2 𝜁, right-hand size is 𝜁

• ⇒ can take 𝐽𝑜 = 𝜈 − 𝑏, 𝜈 + 𝑏 , where 𝜈 = 𝔽 𝐻

∎

Theorem 5.6.2 For 𝜁 > 0 there is a constant 𝐷 = 𝐷 𝜁 such that for every 𝑜 there is an interval 𝐽𝑜 ⊆ 𝑜 of length 𝐷 𝑜 such that, for 𝐻 ∼ 𝐻 𝑜,

1 2 ,

ℙ 𝜓 𝐻 ∉ 𝐽𝑜 ≤ 𝜁.

Reflections on our Results

Narrow window

• Previously saw that 𝜓 𝐻 ≥

1+𝑝 1 𝑜 2 log 𝑜

= 𝑜1−𝑝 1 almost surely

• ⇒ margin of error of 𝑃

𝑜 is relatively small

• Theorem doesn’t say anything about where this interval is

Sparse random graphs

• Never used that 𝐻 ∼ 𝐻 𝑜,

1 2

• Proof applies to 𝐻 ∼ 𝐻 𝑜, 𝑞 for any 𝑞 = 𝑞 𝑜
• However, result is trivial for sparse graphs
• e.g.: if 𝑞 = 𝑝

1 𝑜 , then 𝐻 is bipartite with high probability

• If 𝑞 ≤

𝑑 𝑜, then with high probability Δ 𝐻 ≤ 𝐷 𝑜 ⇒ 𝜓 𝐻 ≤ 𝐷 𝑜 + 1

Colouring Subgraphs of Sparse Graphs

Proof

• If 𝐼 is 𝑒-degenerate, then 𝜓 𝐼 ≤ 𝑒 + 1
• ⇒ if 𝜓 𝐻 𝑇

> 3 for some 𝑇, 𝐻 𝑇 is not 2-degenerate

• ⇒ 𝐻 contains some subgraph 𝐼 with 𝑤 𝐼 ≤ 𝑑 𝑜 and 𝜀 𝐼 ≥ 3
• ⇒ 𝑓 𝐼 ≥

3 2 𝑤 𝐼

• Hence it suffices to show 𝐻 is unlikely to contain such a subgraph

Proposition 5.6.3 Fix 𝛽 >

5 6 and 𝑑 > 0. Then, if 𝑞 = 𝑜−𝛽 and 𝐻 ∼ 𝐻 𝑜, 𝑞 , with high

probability 𝐻 has the property that, for every set 𝑇 of 𝑑 𝑜 vertices, 𝜓 𝐻 𝑇 ≤ 3.

Subgraphs of Sparse Random Graphs are Sparse

Goal

• Every subgraph 𝐼 ⊆ 𝐻 on at most 𝑑 𝑜 vertices has at most

3𝑤 𝐼 2

edges

Proof (cont’d)

• Number of choices for 𝑊 𝐼 :

𝑜 𝑢

≤

𝑜𝑓 𝑢 𝑢

• Number of choices of

3𝑤 𝐼 2

edges of 𝐼:

𝑢 2 3𝑢 2

≤

𝑢 2 𝑓 3𝑢 2 3𝑢 2

≤

𝑢𝑓 3

3𝑢 2

• ⇒ ℙ ∃ bad 𝐼 on 𝑢 vertices ≤

𝑜𝑓 𝑢 𝑢 𝑢𝑓 3

3𝑢 2 𝑞 3𝑢 2 ≤ 𝑓𝑜1−3𝛽/2𝑢1/2 𝑢

• Since 𝑢 < 𝑑 𝑜, this is at most 𝑑′𝑜5/4−3𝛽/2 𝑢
• As 𝛽 >

5 6, exponent of 𝑜 is negative

• ⇒ summing over all 𝑢, ℙ ∃ bad 𝐼 = 𝑝 1

∎

Wow, Much Precise

Proof idea

• Enough to focus on likely values of 𝜓 𝐻
• Consider the smallest 𝑣 such that ℙ 𝜓 𝐻 ≤ 𝑣 = Ω 1
• Show that one can colour most vertices of 𝐻 with 𝑣 colours
• Use Proposition 5.6.3 for the rest

Theorem 5.6.4 (Shamir-Spencer, 1987) Fix 𝛽 >

5 6 and set 𝑞 = 𝑜−𝛽. There is some 𝑣 = 𝑣 𝑜, 𝑞 such that if 𝐻 ∼

𝐻 𝑜, 𝑞 , then almost surely 𝑣 ≤ 𝜓 𝐻 ≤ 𝑣 + 3.

A Wise Choice of Graph Parameter

Proof

• Suffices to show that for any 𝜁 > 0, there is 𝑣 = 𝑣 𝑜, 𝑞, 𝜁 such that

ℙ 𝑣 ≤ 𝜓 𝐻 ≤ 𝑣 + 3 ≥ 1 − 3𝜁

• Define 𝑣 = 𝑣 𝑜, 𝑞, 𝜁 to be smallest 𝑣 such that ℙ 𝜓 𝐻 ≤ 𝑣 ≥ 𝜁
• ⇒ ℙ 𝜓 𝐻 ≤ 𝑣 − 1 < 𝜁
• Now wish to show that most vertices can be 𝑣-coloured
• Define 𝑔 𝐻 = minimum size of 𝑇 ⊆ 𝑊 𝐻 such that 𝜓 𝐻 𝑊 ∖ 𝑇

≤ 𝑣

• ℙ 𝑔 𝐻 = 0 = ℙ 𝜓 𝐻 ≤ 𝑣 ≥ 𝜁

Theorem 5.6.4 Fix 𝛽 >

5 6 and set 𝑞 = 𝑜−𝛽. There is some 𝑣 = 𝑣 𝑜, 𝑞 such that if 𝐻 ∼

𝐻 𝑜, 𝑞 , then almost surely 𝑣 ≤ 𝜓 𝐻 ≤ 𝑣 + 3.

Setting Up Azuma

Recall

• 𝑣: least integer such that ℙ 𝜓 𝐻 ≤ 𝑣 ≥ 𝜁
• 𝑔(𝐻): minimum size of 𝑇 such that 𝜓 𝐻 𝑊 ∖ 𝑇

≤ 𝑣

Lipschitz

• Fix a vertex 𝑤 ∈ 𝑊 𝐻
• Choose a minimum set 𝑇′ whose removal from 𝐻 𝑊 ∖ 𝑤

⇒ 𝜓 ≤ 𝑣

• Worst-case: can always take 𝑇 = 𝑇′ ∪ 𝑤
• ⇒ 𝑔 is 1-vertex-Lipschitz

Martingale

• Run the vertex-exposure martingale on 𝑔 𝐻

Completing the Proof

Recall

• 𝑔(𝐻): minimum size of 𝑇 such that 𝜓 𝐻 𝑊 ∖ 𝑇

≤ 𝑣; let 𝜈 = 𝔽 𝑔 𝐻

• ℙ 𝑔 𝐻 = 0 ≥ 𝜁

Concentration

• Azuma’s Inequality ⇒ ℙ 𝑔 𝐻 ≤ 𝜈 − 𝑏 ≤ exp −𝑏2/2𝑜
• ⇒ 𝜁 ≤ ℙ 𝑔 𝐻 = 0 ≤ exp −𝜈2/2𝑜
• ⇒ 𝜈 ≤

2𝑜 ln 1/𝜁

• Azuma’s Inequality ⇒ ℙ 𝑔 𝐻 ≥ 𝜈 + 𝑏 ≤ exp −𝑏2/2𝑜
• ⇒ ℙ 𝑔 𝐻 ≥ 𝜈 +

2𝑜 ln 1/𝜁 ≤ 𝜁

• ⇒ ℙ 𝑔 𝐻 ≥ 2 2𝑜 ln 1/𝜁 ≤ 𝜁

And voila

• ⇒ can remove 𝑑 𝑜 vertices and 𝑣 colour the rest
• Proposition 5.6.3 ⇒ can 3-colour removed vertices with probability 1 − 𝜁

∎

Epilogue

Location of interval

• Again, proof only shows concentration
• Actual value of chromatic number not needed
• Concern: didn’t our choice of 𝑣 depend on 𝜁?
• Suppose 𝑣 = 𝑣(𝑜, 𝑞, 𝜁) and 𝑣′ = 𝑣 𝑜, 𝑞, 𝜁′
• We proved ℙ 𝜓 𝐻 ∈ 𝑣, 𝑣 + 3

≥ 1 − 𝜁, ℙ 𝜓 𝐻 ∈ 𝑣′, 𝑣′ + 3 ≥ 1 − 𝜁′

• ⇒ ℙ 𝜓 𝐻 ∈ 𝑣, 𝑣 + 3 ∩ 𝑣′, 𝑣′ + 3

≥ 1 − 𝜁 − 𝜁′

• ⇒ Different 𝑣’s give an even stronger concentration inequality

Further results

• Alon-Krivelevich (1997): if 𝛽 >

1 2 and 𝑞 = 𝑜−𝛽, there is some 𝑣 = 𝑣 𝑜, 𝑞 such

that 𝜓 𝐻 𝑜, 𝑞 ∈ 𝑣, 𝑣 + 1 with high probability

• Heckel-Riordan (2020+): if 𝐽 ⊆ 𝑜 is an interval such that 𝜓 𝐻 𝑜, 1/2

∈ 𝐽 with high probability, then 𝐽 = 𝑜1/2−𝑝 1

Any questions?