Chapter 2: Method of Alterations The Probabilistic Method Summer - - PowerPoint PPT Presentation

Chapter 2: Method of Alterations

The Probabilistic Method Summer 2020 Freie Universität Berlin

Chapter Overview

• Introduce the method of alterations

§1 Ramsey Revisited

Chapter 2: Method of Alterations The Probabilistic Method

Definition 1.5.4 (Asymmetric Ramsey numbers) Given ℓ, 𝑙 ∈ ℕ, 𝑆(ℓ, 𝑙) is the minimum 𝑜 for which any 𝑜-vertex graph contains either a clique on ℓ vertices or an independent set on 𝑙 vertices.

Asymmetric Ramsey Bounds

Obtained lower bounds by considering the random graph 𝐻 𝑜, 𝑞 Corollary 1.5.8 For fixed ℓ ∈ ℕ and 𝑙 → ∞, we have Ω 𝑙 2 ln 𝑙

ℓ−1 2

= 𝑆 ℓ, 𝑙 = 𝑃 𝑙ℓ−1 .

Triangle-free Graphs

The case ℓ = 3

• General lower bound gives Ω

𝑙 ln 𝑙

• 𝑆 3, 𝑙 ≥ 𝑙 is utterly trivial
• Complete bipartite graph gives 𝑆 3, 𝑙 ≥ 2𝑙 − 1
• Can improve lower bound by more careful computation

Corollary 1.5.8 For fixed ℓ ∈ ℕ and 𝑙 → ∞, we have Ω 𝑙 2 ln 𝑙

ℓ−1 2

= 𝑆 ℓ, 𝑙 = 𝑃 𝑙ℓ−1 .

Sharper Analysis

Better estimates

• 𝑜

3 𝑞3 ≈ 𝑜𝑞 3 6

• 𝑜

𝑙 ≈ 2𝐼 𝑙

𝑜 𝑜

• 1 − 𝑞

𝑙 2 ≈ 𝑓 −𝑞𝑙2 2

⇒ 𝑆 3, 𝑙 > 𝑑𝑙 for 𝑑 ≈ 1.298 Theorem 1.5.7 (ℓ = 3) Given 𝑙, 𝑜 ∈ ℕ and 𝑞 ∈ 0,1 , if 𝑜 3 𝑞3 + 𝑜 𝑙 1 − 𝑞

𝑙 2 < 1,

then 𝑆 3, 𝑙 > 𝑜.

Lemma 2.1.1 There is some constant 𝐷 such that, if 𝑜 > 𝐷𝑙, then 𝑜 3 𝑞3 + 𝑜 𝑙 1 − 𝑞

𝑙 2 > 1.

Can We Go Further?

What happens for larger 𝑜?

• 𝑜

𝑙 ≥ 𝑜 𝑙 𝑙

= 𝑓𝑙 ln𝑜

𝑙

• 1 − 𝑞

𝑙 2 > 𝑓−2𝑞 𝑙 2 > 𝑓−𝑞𝑙2

• ⇒ need 𝑞 = Ω 𝑙−1 ln

𝑜 𝑙 , otherwise 𝑜 𝑙

1 − 𝑞

𝑙 2 grows exponentially

• But then 𝑜

3 𝑞3 = Θ

𝑜𝑞 3 = Θ

𝑜 𝑙 ln 𝑜 𝑙 3

• Bigger than 1 if 𝑜 > 𝐷𝑙 for some constant 𝐷

Reinterpreting the Proof

Proof we saw

• ℙ

𝐻 𝑜, 𝑞 not Ramsey ≤

𝑜 3 𝑞3 + 𝑜 𝑙

1 − 𝑞

𝑙 2

• If this is less than 1, we get a Ramsey graph with positive probability
• If this is more than 1, we get no useful information

Linearity of expectation

• 𝑜

3 𝑞3: expected number of triangles in 𝐻 𝑜, 𝑞

• 𝑜

𝑙

1 − 𝑞

𝑙 2 : expected number of independent sets of size 𝑙 in 𝐻 𝑜, 𝑞

• ⇒

𝑜 3 𝑞3 + 𝑜 𝑙

1 − 𝑞

𝑙 2 is the expected number of bad subgraphs

• If this is less than 1, then with positive probability we have no bad subgraphs
• ⇒ we get a Ramsey graph

Method of Alterations Goal: existence of an object with property 𝒬

• 1. Show random object is with positive probability close to having 𝒬
• 2. Make deterministic changes to the random object to achieve 𝒬

Shades of Grey

Great expectations

• What does 𝔽 # bad subgraphs ≥ 1 mean?
• Do we have to have bad subgraphs?
• Not necessarily; see Chapter 3 for details
• Gives some guarantee of goodness
• There is a graph with at most 𝑜

3 𝑞3 + 𝑜 𝑙

1 − 𝑞

𝑙 2 bad subgraphs

• If this is small, perhaps we can fix it

Graph Surgery

Given

• Graph with few triangles/large independent sets

Graph Surgery

Given

• Graph with few triangles/large independent sets

Graph Surgery

Given

• Graph with few triangles/large independent sets

Graph Surgery

Given

• Graph with few triangles/large independent sets

Graph Surgery

Given

• Graph with few triangles/large independent sets

Graph Surgery

Given

• Graph with few triangles/large independent sets

Graph Surgery

Given

• Graph with few triangles/large independent sets

Goal

• Edit graph to obtain a Ramsey graph

Idea: remove an edge from each triangle

Graph Surgery

Given

• Graph with few triangles/large independent sets

Goal

• Edit graph to obtain a Ramsey graph

Idea: remove an edge from each triangle

Graph Surgery

Given

• Graph with few triangles/large independent sets

Goal

• Edit graph to obtain a Ramsey graph

Idea: remove an edge from each triangle Problem: creates new independent sets

Graph Surgery

Given

• Graph with few triangles/large independent sets

Goal

• Edit graph to obtain a Ramsey graph

Solution: remove a vertex from each triangle/independent set

Graph Surgery

Given

• Graph with few triangles/large independent sets

Goal

• Edit graph to obtain a Ramsey graph

Solution: remove a vertex from each triangle/independent set Result: a Ramsey graph, albeit on fewer vertices

An Altered Theorem

Proof

• Let 𝐻 ∼ 𝐻 𝑜, 𝑞
• 𝜈 ≔

𝑜 ℓ 𝑞

ℓ 2 +

𝑜 𝑙

1 − 𝑞

𝑙 2 is the expected number of 𝐿ℓ and 𝐿𝑙

• ⇒ there is an 𝑜-vertex graph with at most 𝜈 bad subgraphs
• Delete one vertex from each bad subgraph
• Obtain a Ramsey subgraph on at least 𝑜 − 𝜈 vertices

∎

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

ℓ 2 − 𝑜

𝑙 1 − 𝑞

𝑙 2 .

𝑆(3, 𝑙): A New Bound

Goal

• Choose 𝑜, 𝑞 to maximise 𝑜 −

𝑜 3 𝑞3 − 𝑜 𝑙

1 − 𝑞

𝑙 2

Choosing 𝑞

• Small 𝑞 makes the second term small
• Recall: need 𝑞 = Ω 𝑙−1 ln

𝑜 𝑙 , otherwise third term exponentially large

• When 𝑞 is this large, third term exponentially small – insignificant

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 .

𝑆(3, 𝑙): A New Bound

Recall

• Maximising 𝑜 −

𝑜 3 𝑞3 − 𝑜 𝑙

1 − 𝑞

𝑙 2

• Take 𝑞 = Θ 𝑙−1 ln

𝑜 𝑙

Choosing 𝑜

• Want to maximise 𝑜 − Θ

𝑜 𝑙 ln 𝑜 𝑙 3

• At maximum:

𝑜 𝑙 ln 𝑜 𝑙 3

= Θ 𝑜

• ⇒ 𝑜 = Θ

𝑙 ln𝑜

𝑙 3 2

= Θ

𝑙 ln 𝑙

3 2

Where We Stand

Lower bound

• Superlinear lower bound
• Beats Turán

Upper bound

• Erdős-Szekeres: 𝑆 3, 𝑙 = 𝑃(𝑙2)
• Can we narrow the gap? Stay tuned!

Corollary 2.1.3 As 𝑙 → ∞, we have R 3, k = Ω 𝑙 ln 𝑙

3 2

.

Any questions?

§2 Dominating Sets

Chapter 2: Method of Alterations The Probabilistic Method

BER: A Modern Tragicomedy

Sep 2006 Berlin-Brandenburg Airport to open Oct 2011 Jun 2010 Opening postponed to Jun 2012 May 2012 Fire detection systems do not work! Solution

• Hire people to stand around the airport looking for signs of fire

Problem

• Already overbudget
• ⇒ want to hire as few people as possible

Combinatorics to the Rescue

The airport is a graph

• Vertices: areas where fire could break out
• Edges:

lines of sight between areas

Objective

• Find a set of vertices that “see” all other vertices

Combinatorics to the Rescue

The airport is a graph

• Vertices: areas where fire could break out
• Edges:

lines of sight between areas

Objective

• Find a set of vertices that “see” all other vertices

Combinatorics to the Rescue

The airport is a graph

• Vertices: areas where fire could break out
• Edges:

lines of sight between areas

Objective

• Find a set of vertices that “see” all other vertices

Combinatorics to the Rescue

The airport is a graph

• Vertices: areas where fire could break out
• Edges:

lines of sight between areas

Objective

• Find a set of vertices that “see” all other vertices

Small Dominating Sets

Extremal problem

• How large can the smallest dominating set of an 𝑜-vertex graph 𝐻 be?

Answer

• 𝑜 (!)
• Isolated vertices must be in any dominating set

Avoiding trivialities

• What if we require 𝐻 to have minimum degree 𝜀?

Definition 2.2.1 Given a graph 𝐻 = 𝑊, 𝐹 , a set 𝑇 ⊆ 𝑊 of vertices is a dominating set if, for every 𝑤 ∈ 𝑊 ∖ 𝑇, there is some 𝑡 ∈ 𝑇 with 𝑡, 𝑤 ∈ 𝐹.

Do Random Sets Dominate?

Random set

• Let 𝑇 ⊆ 𝑊 be a random set
• 𝑤 ∈ 𝑇 with probability 𝑞, independently

Undominated vertices

• For 𝑣 ∈ 𝑊, define the event 𝐹𝑣 = 𝑣 not dominated by 𝑇
• For 𝐹𝑣 to hold, need:
• 𝑣 ∉ 𝑇
• 𝑤 ∉ 𝑇 for all neighbours 𝑤 of 𝑣
• ⇒ ℙ 𝐹𝑣 = 1 − 𝑞 𝑒 𝑣 +1

Problem Given 𝐻 on 𝑜 vertices with 𝜀 𝐻 ≥ 𝜀, how large can its smallest dominating set be?

Calculations Continued

Failure probability

• 𝑇 not dominating = ∪𝑣∈𝑊 𝐹𝑣
• ℙ ∪𝑣∈𝑊 𝐹𝑣 < σ𝑣∈𝑊 ℙ 𝐹𝑣 = σ𝑣∈𝑊 1 − 𝑞 𝑒 𝑣 +1
• σ𝑣∈𝑊 1 − 𝑞 𝑒 𝑣 +1 ≤ 𝑜 1 − 𝑞 𝜀+1 ≤ 𝑜𝑓−𝑞 𝜀+1
• ⇒ if 𝑞 =

ln 𝑜 𝜀+1, then ℙ 𝑇 not dominating < 1

• ⇒ 𝑇 is dominating with positive probability

Size of the dominating set

• 𝑇 ∼ Bin 𝑜, 𝑞
• ⇒ 𝔽 𝑇

= 𝑜𝑞 =

𝑜 ln 𝑜 𝜀+1

• ⇒ with positive probability, 𝑇 ≤

𝑜 ln 𝑜 𝜀+1

Proposition 2.2.3 Let 𝐻 be an 𝑜-vertex graph with 𝜀 𝐻 ≥ 𝜀 ≥ ln 2𝑜. Then 𝐻 has a dominating set 𝑇 ⊆ 𝑊(𝐻) with 𝑇 ≤

(𝑜+1) ln 2𝑜 𝜀+1

.

Putting Things Together

Concurrence of events

• Want events {𝑇 is dominating} and 𝑇 is small to hold simultaneously
• Suffices to have ℙ 𝑇 not dominating , ℙ 𝑇 large <

1 2

Non-domination

• ℙ 𝑇 not dominating < 𝑜𝑓−𝑞 𝜀+1 =

1 2 if 𝑞 = ln 2𝑜 𝜀+1

Large sets

• Binomial distribution ⇒ ℙ 𝑇 > 𝑜 + 1 𝑞 ≤

1 2

Observation For any T ⊆ 𝑊, the set T ∪ 𝑉(𝑈) is a dominating set.

Altering Our Approach

Reduced requirements

• Need large probability 𝑞 for the random set 𝑇 to be dominating
• What if we instead only want it to be close to dominating?
• Set 𝑇 should dominate most vertices of 𝐻

Undominated vertices

• Given a graph 𝐻, set of vertices T ⊆ 𝑊(𝐻)
• Let 𝑉 𝑈 = 𝑤 ∈ 𝑊 ∖ 𝑈: 𝑂 𝑤 ∩ 𝑈 = ∅ be the vertices not dominated by 𝑈

Altering Our Results

Proof

• Let 𝑈 be a random set of vertices, chosen independently with probability 𝑞
• ⇒ 𝔽 𝑈

= 𝑜𝑞

• Recall: ℙ 𝑣 not dominated by 𝑈 = 1 − 𝑞 𝑒 𝑣 +1 ≤ 𝑓−𝑞 𝜀+1
• Linearity of expectation ⇒ 𝔽 𝑉 𝑈

≤ 𝑜𝑓−𝑞 𝜀+1

• Let 𝑇 = 𝑈 ∪ 𝑉(𝑈)
• 𝑇 is a dominating set
• 𝔽 𝑇

= 𝔽 𝑈 ∪ 𝑉 𝑈 = 𝔽 𝑈 + 𝔽 𝑉 𝑈 ≤ 𝑜𝑞 + 𝑜𝑓−𝑞 𝜀+1

• ⇒ existence of a dominating set of at most this size

∎

Theorem 2.2.4 Let 𝐻 be an 𝑜-vertex graph with 𝜀 𝐻 ≥ 𝜀, and let 𝑞 ∈ [0,1]. Then 𝐻 has a dominating set 𝑇 ⊆ 𝑊 𝐻 with 𝑇 ≤ 𝑜𝑞 + 𝑜𝑓−𝑞 𝜀+1 .

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.

Running the Numbers

Goal

• Minimise 𝑜𝑞 + 𝑜𝑓−𝑞 𝜀+1 = 𝑜 𝑞 + 𝑓−𝑞 𝜀+1

A little calculus

• Let 𝑔 𝑞 = 𝑞 + 𝑓−𝑞 𝜀+1
• 𝑔′ 𝑞 = 1 − 𝜀 + 1 𝑓−𝑞 𝜀+1
• 𝑔′ 𝑞0 = 0 ⇔ 𝑞0 =

ln 𝜀+1 𝜀+1

• 𝑔 𝑞0 =

ln 𝜀+1 +1 𝜀+1

BER: Epilogue

May 2012 Firewatch plan rejected, opening set for Mar 2013 Sep 2012 Opening postponed further to Oct 2013 2013-2019 Series of delays, no new opening date set Apr 2020 Building authority approval! Opening 31 Oct 2020 Total delays 3072 days (and counting?) Original budget €2.3 billion Actual cost €7.3 billion (and counting?)

Any questions?

§3 Dependent Random Choice

Chapter 2: Method of Alterations The Probabilistic Method

Definition 2.3.1 Given a graph 𝐼 and 𝑜 ∈ ℕ, the Turán number ex(𝑜, 𝐼) is the maximum number of edges in an 𝐼-free 𝑜-vertex graph.

Turán Numbers

Theorem 2.3.2 (Erdős-Stone-Simonovits, 1966) For any graph 𝐼, ex 𝑜, 𝐼 = 1 −

1 𝜓 𝐼 −1 + 𝑝 1 𝑜 2 .

Theorem 1.5.6 (Turán, 1941) For ℓ ≥ 2, ex 𝑜, 𝐿ℓ = 1 −

1 ℓ−1 + 𝑝 1 𝑜 2 .

Theorem 2.3.2 (Erdős-Stone-Simonovits, 1966) For any graph 𝐼, ex 𝑜, 𝐼 = 1 −

1 𝜓 𝐼 −1 + 𝑝 1 𝑜 2 .

Bipartite Turán Numbers

• Determines ex(𝑜, 𝐼) asymptotically when 𝜓 𝐼 ≥ 3
• 𝐼 bipartite: only shows ex 𝑜, 𝐼 = 𝑝 𝑜2

Theorem 2.3.3 (Kővári-Sós-Turán, 1954) If 𝐼 is bipartite with at most 𝑢 vertices in one part, then ex 𝑜, 𝐼 = 𝑃 𝑜2−1/𝑢 .

Tightness of Kővári-Sós-Turán

Complete bipartite graphs

• Tight for 𝐼 = 𝐿𝑢,𝑡 when 𝑡 > 𝑢 − 1 ! [Alon-Rónyai-Szabó, 1999]

Even cycles

• Far from tight for cycles
• ex 𝑜, 𝐷2𝑙 = 𝑃(𝑜1+1/𝑙) [Bondy-Simonovits, 1974]

General graphs

• Can we find a sharper general bound?

Theorem 2.3.3 (Kővári-Sós-Turán, 1954) If 𝐼 is bipartite with at most 𝑢 vertices in one part, then ex 𝑜, 𝐼 = 𝑃 𝑜2−1/𝑢 .

Lemma 2.3.4 (Dependent Random Choice) Let 𝑏, 𝑒, 𝑛, 𝑜, 𝑢 ∈ ℕ. Let 𝐻 be an 𝑜-vertex graph with average degree 𝑒. If there is some 𝑡 ∈ ℕ with 𝑒𝑡 𝑜𝑡−1 − 𝑜 𝑢 𝑛 𝑜

𝑡

≥ 𝑏, then 𝐻 contains a subset 𝐵 of at least 𝑏 vertices, any 𝑢 of which have more than 𝑛 common neighbours.

Dependent Random Choice

Embedding 𝐼

• Need to show any sufficiently dense graph 𝐻 must contain a copy of 𝐼
• We know nothing about 𝐻 apart from its density
• This is enough to extract some structure

A Turánnical Application

Kővári-Sós-Turán

• Immediate consequence of the above theorem
• Same examples show bound can be tight

Wider class of graphs

• Gives reasonable bounds for graphs of arbitrary order
• e.g. even subdivisions 𝐺

𝑡𝑣𝑐 of a graph 𝐺

• Each edge of 𝐺 replaced by an even path
• Can apply Theorem 2.3.5 with 𝑢 = 2 ⇒ 𝑓𝑦 𝑜, 𝐺

𝑡𝑣𝑐 = 𝑃(𝑜3/2)

Theorem 2.3.5 (Alon-Krivelevich-Sudakov, 2003) Let 𝐼 be a bipartite graph with maximum degree 𝑢 in one part. Then 𝑓𝑦 𝑜, 𝐼 = 𝑃 𝑜2−1/𝑢 .

Setting Up the Proof

Given

• Bipartite 𝐼 with vertex classes 𝑉 ∪ 𝑋
• Maximum degree in 𝑋 is 𝑢

Objective

• Given 𝑜-vertex graph 𝐻 with 𝑓 𝐻 ≥ Ω(𝑜2−1/𝑢)
• Need to show 𝐼 ⊆ 𝐻

Theorem 2.3.5 (Alon-Krivelevich-Sudakov, 2003) Let 𝐼 be a bipartite graph with maximum degree 𝑢 in one part. Then 𝑓𝑦 𝑜, 𝐼 = 𝑃 𝑜2−1/𝑢 .

Applying Dependent Random Choice

Idea

• Embed 𝑉 in 𝐵 arbitrarily
• Each 𝑥 ∈ 𝑋 has at most 𝑢 neighbours in 𝑉
• Corresponding set of 𝑢 vertices in 𝐵 has at

least 𝑛 common neighbours in 𝐻

• May have used some on earlier vertices, but

if 𝑛 ≥ 𝑤(𝐼), one is free to embed 𝑥

• ⇒ can embed 𝑋 one vertex at a time

Lemma 2.3.4 (Dependent Random Choice) Let 𝑏, 𝑒, 𝑛, 𝑜, 𝑢 ∈ ℕ. Let 𝐻 be an 𝑜-vertex graph with average degree 𝑒. If there is some 𝑡 ∈ ℕ with 𝑒𝑡 𝑜𝑡−1 − 𝑜 𝑢 𝑛 𝑜

𝑡

≥ 𝑏, then 𝐻 contains a subset 𝐵 of at least 𝑏 vertices, any 𝑢 of which have more than 𝑛 common neighbours.

A Little Arithmetic

Target

• 𝑒𝑡

𝑜𝑡−1 − 𝑜 𝑢 𝑛 𝑜 𝑡

≥ 𝑏 where

• 𝑏 = 𝑉 ≤ 𝑤 𝐼 =: ℎ
• 𝑛 = ℎ
• 𝑒 = C𝐼𝑜1−1/𝑢 for some constant 𝐷𝐼 we can choose
• we can choose 𝑡 ∈ ℕ

Simplify

• 𝑜

𝑢

≤ 𝑜𝑢

• Sufficient to have 𝐷𝐼

𝑡𝑜1−𝑡/𝑢 − ℎ𝑡𝑜𝑢−𝑡 ≥ ℎ

• ⇒ need to take 𝑡 = 𝑢
• Sufficient to have 𝐷𝐼

𝑢 ≥ ℎ𝑢 + ℎ

• Satisfied by taking 𝐷𝐼 = 21/𝑢ℎ, completing the proof

∎

Proving Dependent Random Choice

Does a random set work for 𝐵?

• No - 𝐻 could be bipartite
• Then a random set will intersect both parts
• Subsets meeting both parts have no common neighbours

Lemma 2.3.4 (Dependent Random Choice) Let 𝑏, 𝑒, 𝑛, 𝑜, 𝑢 ∈ ℕ. Let 𝐻 be an 𝑜-vertex graph with average degree 𝑒. If there is some 𝑡 ∈ ℕ with 𝑒𝑡 𝑜𝑡−1 − 𝑜 𝑢 𝑛 𝑜

𝑡

≥ 𝑏, then 𝐻 contains a subset 𝐵 of at least 𝑏 vertices, any 𝑢 of which have at least 𝑛 common neighbours.

An Indirect Selection

Idea

• We choose a small random set 𝑇 of vertices
• Let their common neighbourhood 𝐶 be our candidate for 𝐵

Intuition

• 𝐻 has large average degree 𝑒
• ⇒ 𝑇 should have many common neighbours
• If a set of vertices has few neighbours, unlikely that 𝑇 was chosen from them
• ⇒ will not see these 𝑢 vertices in 𝐶
• Can expect 𝑢-subsets of 𝐶 to have large common neighbourhood

Fleshing Out the Details

Choosing 𝑇

• Sample 𝑡 vertices from 𝑊(𝐻), independently (with repetition!)
• Let 𝑇 be the set of vertices selected

Common neighbourhood 𝐶

• Let 𝐶 = 𝑤 ∈ 𝑊 𝐻 : ∀𝑡 ∈ 𝑇, {𝑤, 𝑡} ∈ 𝐹 𝐻
• For 𝑤 ∈ 𝐶, all vertices in 𝑇 had to be neighbours of 𝑤
• ⇒ ℙ 𝑤 ∈ 𝐶 =

𝑒 𝑤 𝑜 𝑡

• ⇒ 𝔽 𝐶

= σ𝑤

𝑒 𝑤 𝑜 𝑡

= 𝑜−𝑡 σ𝑤 𝑒 𝑤 𝑡

• 𝑦 ↦ 𝑦𝑡 is a convex function
• ⇒ 𝔽 𝐶

= 𝑜−𝑡 σ𝑤 𝑒 𝑤 𝑡 ≥ 𝑜1−𝑡

σ𝑤 𝑒 𝑤 𝑜 𝑡

=

𝑒𝑡 𝑜𝑡−1

Fixing the Set

Bad subsets

• Let 𝑈 be a set of 𝑢 vertices with at most 𝑛 common neighbours
• To have 𝑈 ⊆ 𝐶, need to select 𝑇 from these common neighbours
• ⇒ ℙ 𝑈 ⊆ 𝐶 ≤

𝑛 𝑜 𝑡

• Linearity of expectation ⇒ 𝔽 # bad subsets ≤

𝑜 𝑢 𝑛 𝑜 𝑡

Alteration

• Remove one vertex from each bad subset
• Let 𝐵 be the remaining set
• Every 𝑢-subset of 𝐵 has more than 𝑛 common neighbours
• 𝔽 𝐵

≥ 𝔽 𝐶 − 𝔽 # bad subsets ≥

𝑒𝑡 𝑜𝑡−1 − 𝑜 𝑢 𝑛 𝑜 𝑡

≥ 𝑏

• ⇒ there exists such a set 𝐵 of size at least 𝑏

∎

Any questions?