Chapter 1: Getting Started The Probabilistic Method Summer 2020 - - PowerPoint PPT Presentation

Chapter 1: Getting Started

The Probabilistic Method Summer 2020 Freie Universität Berlin

Chapter Overview

• Survey quick applications of the basic method to different areas

§1 Unsatisfiable Formulae

Chapter 1: Getting Started The Probabilistic Method

Boolean Logic

Binary values

• Computers can only talk in 0s and 1s
• In logical applications, we map those to 𝐺𝑏𝑚𝑡𝑓 and 𝑈𝑠𝑣𝑓

Logical operators

• Can obtain new truth values from old ones

Not: ¬ Or: ∨ And: ∧

Boolean formulae

• Can build any 𝑈𝑠𝑣𝑓/𝐺𝑏𝑚𝑡𝑓 expression using these operations
• Such a formula is a function 𝑔: 0,1 𝑜 → 0,1

Anatomy of a Formula

Every Boolean formula can be written in Conjunctive Normal Form: Variables

• 𝑦𝑗 ∈ 0,1

Literals

• Variable 𝑦𝑗 or its negation ¬𝑦𝑗

Clauses

• ‘OR’ of literals
• e.g.: 𝑦1 ∨ ¬𝑦2 ∨ 𝑦3

CNF Formula

• ‘AND’ of several clauses
• e.g.: 𝑦1 ∨ ¬𝑦2 ∨ 𝑦3 ∧ ¬𝑦1 ∨ 𝑦2 ∧ (𝑦2 ∨ 𝑦3 ∨ 𝑦4 ∨ ¬𝑦5)

A Little Complexity

Satisfiability Problem (SAT)

• Given a Boolean formula 𝑔, can 𝑔 ever evaluate to 𝑈𝑠𝑣𝑓?
• If not, say 𝑔 is unsatisfiable

Theorem 1.1.1 (Cook, 1971; Levin, 1973) SAT is 𝑂𝑄-Complete, i.e. is probably very difficult. A universal model

• Most interesting problems can be reduced to SAT instances
• e.g.: Travelling Salesman Problem, Subgraph Isomorphism, Largest Clique

Restricted Formulae

Simplifying the problem

• Perhaps the problem is easier for ‘nice’ formulae
• 𝑙-SAT: each clause must have exactly 𝑙 literals from distinct variables

Theorem 1.1.2 (Karp, 1972) For all 𝑙 ≥ 3, 𝑙-SAT is still 𝑂𝑄-Complete. Does size matter?

• Karp ⇒ unsatisfiability does not require long clauses
• Does it at least require many clauses? Are short formulae always satisfiable?

Minimum Unsatisfiability

Extremal problem

• How small can an unsatisfiable instance of 𝑙-SAT be?

Definition 1.1.3 Given 𝑙 ∈ ℕ, let 𝑛0 𝑙 be the minimum 𝑛 ∈ ℕ for which there is an unsatisfiable instance of 𝑙-SAT with 𝑛 clauses. Small 𝑙-SAT is easy

• Can solve instances of 𝑙-SAT with 𝑛 < 𝑛0 𝑙 clauses in constant time!
• (Existential) answer is always: yes (satisfiable)

A First Lower Bound

Lower bounds

• Given any instance of 𝑙-SAT with few clauses, need to show it is satisfiable
• First idea: build a satisfying argument greedily

A worked example (𝑙 = 3) 𝑦1 ∨ ¬𝑦2 ∨ 𝑦3 ∧ ¬𝑦1 ∨ 𝑦2 ∨ ¬𝑦4 ∧ ¬𝑦2 ∨ ¬𝑦3 ∨ ¬𝑦5

𝒚𝟐 𝒚𝟑 𝒚𝟒 𝒚𝟓 𝒚𝟔 ? ? ? ? ?

A First Lower Bound

Lower bounds

• Given any instance of 𝑙-SAT with few clauses, need to show it is satisfiable
• First idea: build a satisfying argument greedily

A worked example (𝑙 = 3) 𝑦1 ∨ ¬𝑦2 ∨ 𝑦3 ∧ ¬𝑦1 ∨ 𝑦2 ∨ ¬𝑦4 ∧ ¬𝑦2 ∨ ¬𝑦3 ∨ ¬𝑦5 Step 1

• Select 𝑦1 as the designated variable for the first clause

𝒚𝟐 𝒚𝟑 𝒚𝟒 𝒚𝟓 𝒚𝟔 ? ? ? ? ?

A First Lower Bound

Lower bounds

• Given any instance of 𝑙-SAT with few clauses, need to show it is satisfiable
• First idea: build a satisfying argument greedily

A worked example (𝑙 = 3) 1 ∨ ¬𝑦2 ∨ 𝑦3 ∧ 0 ∨ 𝑦2 ∨ ¬𝑦4 ∧ ¬𝑦2 ∨ ¬𝑦3 ∨ ¬𝑦5 Step 1

• Select 𝑦1 as the designated variable for the first clause
• Set 𝑦1 = 1 to satisfy the clause

𝒚𝟐 𝒚𝟑 𝒚𝟒 𝒚𝟓 𝒚𝟔 1 ? ? ? ?

A First Lower Bound

Lower bounds

• Given any instance of 𝑙-SAT with few clauses, need to show it is satisfiable
• First idea: build a satisfying argument greedily

A worked example (𝑙 = 3) 1 ∨ ¬𝑦2 ∨ 𝑦3 ∧ 0 ∨ 𝑦2 ∨ ¬𝑦4 ∧ ¬𝑦2 ∨ ¬𝑦3 ∨ ¬𝑦5 Step 2

• The second clause is still unsatisfied

𝒚𝟐 𝒚𝟑 𝒚𝟒 𝒚𝟓 𝒚𝟔 1 ? ? ? ?

A First Lower Bound

Lower bounds

• Given any instance of 𝑙-SAT with few clauses, need to show it is satisfiable
• First idea: build a satisfying argument greedily

A worked example (𝑙 = 3) 1 ∨ 0 ∨ 𝑦3 ∧ 0 ∨ 1 ∨ ¬𝑦4 ∧ 0 ∨ ¬𝑦3 ∨ ¬𝑦5 Step 2

• The second clause is still unsatisfied
• Select 𝑦2 as its designated variable, and set 𝑦2 = 1

𝒚𝟐 𝒚𝟑 𝒚𝟒 𝒚𝟓 𝒚𝟔 1 1 ? ? ?

A First Lower Bound

Lower bounds

• Given any instance of 𝑙-SAT with few clauses, need to show it is satisfiable
• First idea: build a satisfying argument greedily

A worked example (𝑙 = 3) 1 ∨ 0 ∨ 0 ∧ 0 ∨ 1 ∨ ¬𝑦4 ∧ 0 ∨ 1 ∨ ¬𝑦5 Step 3

• The third clause is still unsatisfied, so we set 𝑦3 = 0

𝒚𝟐 𝒚𝟑 𝒚𝟒 𝒚𝟓 𝒚𝟔 1 1 ? ?

A First Lower Bound

Lower bounds

• Given any instance of 𝑙-SAT with few clauses, need to show it is satisfiable
• First idea: build a satisfying argument greedily

A worked example (𝑙 = 3) 1 ∨ 0 ∨ 0 ∧ 0 ∨ 1 ∨ ¬𝑦4 ∧ 0 ∨ 1 ∨ ¬𝑦5 Step 3

• The third clause is still unsatisfied, so we set 𝑦3 = 0
• This satisfies the formula, so we are done

𝒚𝟐 𝒚𝟑 𝒚𝟒 𝒚𝟓 𝒚𝟔 1 1 ? ?

A First Lower Bound

Proof

• Let 𝑔 be an arbitrary 𝑙-SAT formula with 𝑛 ≤ 𝑙 clauses
• We use the greedy algorithm, satisfying each clause one at a time
• When dealing with the 𝑗th clause, for 1 ≤ 𝑗 ≤ 𝑛, either:
• it is already satisfied by our previous assignments, or
• we have set at most 𝑗 − 1 ≤ 𝑛 − 1 < 𝑙 variables, so there is a free variable to choose
• Hence we can satisfy all the clauses
• Thus 𝑔 is satisfiable

∎

Proposition 1.1.4 For all 𝑙 ∈ ℕ, 𝑛0 𝑙 > 𝑙.

Being Greedy Doesn’t Always Pay

What if we have more clauses?

• This greedy algorithm can get stuck

Extending our example

𝑦1 ∨ ¬𝑦2 ∨ 𝑦3 ∧ ¬𝑦1 ∨ 𝑦2 ∨ ¬𝑦4 ∧ ¬𝑦2 ∨ ¬𝑦3 ∨ ¬𝑦5 ∧ ¬𝑦1 ∨ ¬𝑦2 ∨ 𝑦3

𝒚𝟐 𝒚𝟑 𝒚𝟒 𝒚𝟓 𝒚𝟔 ? ? ? ? ?

Being Greedy Doesn’t Always Pay

What if we have more clauses?

• This greedy algorithm can get stuck

Extending our example

1 ∨ 0 ∨ 0 ∧ 0 ∨ 1 ∨ ¬𝑦4 ∧ 0 ∨ 1 ∨ ¬𝑦5 ∧ 0 ∨ 0 ∨ 0

Steps 1-3

• Proceed as before, with the same assignments
• Now the final clause is unsatisfiable

𝒚𝟐 𝒚𝟑 𝒚𝟒 𝒚𝟓 𝒚𝟔 1 1 ? ?

Being Greedy Doesn’t Always Pay

What if we have more clauses?

• This greedy algorithm can get stuck

Extending our example

1 ∨ 0 ∨ 1 ∧ 0 ∨ 1 ∨ ¬𝑦4 ∧ 0 ∨ 0 ∨ 1 ∧ 0 ∨ 0 ∨ 1

Formula is still satisfiable

• Could have satisfied earlier clauses with different variables

𝒚𝟐 𝒚𝟑 𝒚𝟒 𝒚𝟓 𝒚𝟔 1 1 1 ?

Proposition 1.1.5 For all 𝑙 ∈ ℕ, 𝑛0 𝑙 ≤ 2𝑙.

An Unsatisfiable Formula

Intuition

• Clauses with unique variables are can always be satisfied
• Maybe hardest when all clauses share the same variables

Building an unsatisfiable formula

• With 𝑙 variables, there are 2𝑙 possible inputs and 2𝑙 possible clauses
• Each clause is unsatisfied by a unique input
• e.g.: 𝑦1 ∨ ¬𝑦2 ∨ ¬𝑦3 is not satisfied by 𝑦1, 𝑦2, 𝑦3 = 0,1,1
• ⇒ The formula with all possible clauses is unsatisfiable

A Tight Result

Upper bound

• Previous construction

Lower bound

• Need to show every 𝑙-SAT instance with 𝑛 < 2𝑙 clauses is satisfiable

Existential reformulation

• Given: 𝑙-SAT formula 𝑔 with 𝑜 variables and 𝑛 < 2𝑙 clauses
• Goal: show there is some Ԧ

𝑦 ∈ Ω = {0,1}𝑜 with the property 𝑔 Ԧ 𝑦 = 1

Theorem 1.1.6 For all 𝑙 ∈ ℕ, 𝑛0 𝑙 = 2𝑙.

Randomness to the Rescue

Existential reformulation

• Given: 𝑙-SAT formula 𝑔 with 𝑜 variables and 𝑛 < 2𝑙 clauses
• Goal: show there is some Ԧ

𝑦 ∈ Ω = {0,1}𝑜 with the property 𝑔 Ԧ 𝑦 = 1

The probabilistic method

• Choose Ԧ

𝑦 ∈ {0,1}𝑜 uniformly at random

• Show ℙ 𝑔 Ԧ

𝑦 = 1 > 0

Theorem 1.1.6 For all 𝑙 ∈ ℕ, 𝑛0 𝑙 = 2𝑙.

Union Bound Given arbitrary events 𝐹𝑗, we have ℙڂ𝑗 𝐹𝑗 ≤ σ𝑗 ℙ 𝐹𝑗 .

Bounding Probabilities

Setting

• Given: 𝑔, a 𝑙-SAT formula with 𝑜 variables and 𝑛 < 2𝑙 clauses
• Given: uniformly random Ԧ

𝑦 ∈ {0,1}𝑜

• Goal: show ℙ 𝑔 Ԧ

𝑦 = 1 > 0

Bad events

• Equivalently, want to show ℙ 𝑔 Ԧ

𝑦 = 0 < 1

• Let 𝐹𝑗 be the event that the 𝑗th clause is not satisfied by Ԧ

𝑦

• 𝑔 Ԧ

𝑦 = 0 =ڂ𝑗=1

𝑛 𝐹𝑗

• ⇒ ℙ 𝑔 Ԧ

𝑦 = 0 = ℙڂ𝑗=1

𝑛 𝐹𝑗

Completing the Proof

Individual clauses

• Recall: 𝐹𝑗 is the event that the 𝑗th clause is not satisfied by Ԧ

𝑦

• 𝐹𝑗 only depends on the values of the 𝑙 variables it contains
• Exactly one of the 2𝑙 possible values does not satisfy the clause
• ⇒ ℙ 𝐹𝑗 = 2−𝑙

ℙ 𝑔 Ԧ 𝑦 = 0 = ℙ ራ

𝑗=1 𝑛

𝐹𝑗 ≤ ෍

𝑗=1 𝑛

ℙ 𝐹𝑗 = 𝑛2−𝑙 < 1 Conclusion

• Therefore ℙ 𝑔 Ԧ

𝑦 = 1 = 1 − ℙ Ԧ 𝑦 = 0 > 0

• Hence there is some Ԧ

𝑦 ∈ {0,1}𝑜 for which 𝑔 Ԧ 𝑦 = 1 ∎

Trivial unsatisfiability

• In our construction to show 𝑛0 𝑙 ≤ 2𝑙, each clause had the same variables
• Clauses are then forced to be in conflict with one another

Non-repetitive formulae

• A 𝑙-SAT formula is non-repetitive if each clause has a distinct set of variables
• e.g.: cannot have both (𝑦1 ∨ ¬𝑦2 ∨ 𝑦4) and (¬𝑦1 ∨ ¬𝑦2 ∨ ¬𝑦4) as clauses

Extremal problem

• How many variables must an unsatisfiable non-repetitive 𝑙-SAT formula have?

Definition 1.1.7 Given 𝑙 ∈ ℕ, let 𝑜0 𝑙 be the minimum 𝑜 ∈ ℕ for which there is an unsatisfiable non-repetitive 𝑙-SAT formula with 𝑜 variables.

Is Repetition Necessary?

Theorem 1.1.6 For all 𝑙 ∈ ℕ, 𝑛0 𝑙 = 2𝑙.

A Lower Bound

Observation

• A non-repetitive 𝑙-SAT formula with 𝑜 variables can have at most 𝑜

𝑙 clauses

Corollary 1.1.8 For all 𝑙, 𝑜 ∈ ℕ, if 𝑜

𝑙 < 2𝑙, then 𝑜0 𝑙 > 𝑜.

An Upper Bound

Existential formulation

• Set of objects Ω: non-repetitive 𝑙-SAT formulae with 𝑜 variables
• Desired property 𝒬: ∀ Ԧ

𝑦 ∈ {0,1}𝑜, 𝑔 Ԧ 𝑦 = 0

Probabilistic approach

• There are 𝑜

𝑙 sets of 𝑙 variables:

• For each variable 𝑦𝑗, there are two possible literals: 𝑦𝑗 and ¬𝑦𝑗
• Total of 2𝑙 possible clauses for this set of variables
• Choose one uniformly at random
• Make these choices independently
• This gives us a random 𝑔 ∈ Ω
• Want to show ℙ ∀ Ԧ

𝑦 ∈ {0,1}𝑜, 𝑔 Ԧ 𝑦 = 0 > 0

Analysing the Bad Events

Satisfying assignments

• For each Ԧ

𝑦 ∈ {0,1}𝑜, let 𝐹 Ԧ

𝑦 be the event that 𝑔 Ԧ

𝑦 = 1

• We want to show ℙڂ Ԧ

𝑦 𝐹 Ԧ 𝑦 < 1

• Union bound:
• There are 2𝑜 possible Ԧ

𝑦

• ℙڂ Ԧ

𝑦 𝐹 Ԧ 𝑦 ≤ σ Ԧ 𝑦 ℙ 𝐹 Ԧ 𝑦

• Suffices to have ℙ 𝐹 Ԧ

𝑦 < 2−n for all Ԧ

𝑦

Fix an assignment Ԧ 𝑦

• For 𝑔 Ԧ

𝑦 = 1, Ԧ 𝑦 must satisfy each of the 𝑜

𝑙 clauses

• Let 𝐺

𝑗 be the event that Ԧ

𝑦 satisfies the 𝑗th clause

• Then 𝐹 Ԧ

𝑦 =∩𝑗 𝐺 𝑗

Computing Probabilities

Recall

• 𝑔 formed by choosing a random clause for each set of variables
• 𝐹 Ԧ

𝑦: event that 𝑔 Ԧ

𝑦 = 1; 𝐺

𝑗: event that Ԧ

𝑦 satisfies the 𝑗th clause of 𝑔

• Suffices to show ℙ 𝐹 Ԧ

𝑦 = ℙ ∩𝑗 𝐺 𝑗 < 2−𝑜

Independence

• Clauses are chosen independently ⇒ events 𝐺

𝑗 are independent

• ⇒ ℙ ∩𝑗 𝐺

𝑗 = ς𝑗 ℙ 𝐺 𝑗

Satisfying a single clause

• Given 𝑗 and our fixed Ԧ

𝑦, unique choice of literals such that 𝐺

𝑗 doesn’t hold

• ⇒ ℙ 𝐺

𝑗 = 1 − 2−𝑙

Theorem 1.1.9 For all 𝑙 ∈ ℕ, if 𝑜

𝑙 < 2𝑙, then 𝑜0 𝑙 > 𝑜, and if 𝑜 𝑙 > 2𝑙𝑜 ln 2, then

𝑜0 𝑙 ≤ 𝑜.

Putting it all together

A final calculation

• We therefore have ℙ 𝐹 Ԧ

𝑦 = ς𝑗 ℙ 𝐺 𝑗 = 1 − 2−𝑙

𝑜 𝑙

Exponential bound For all 𝑦 ∈ ℝ, 1 + 𝑦 ≤ 𝑓𝑦

• ⇒ ℙ 𝐹 Ԧ

𝑦 = 1 − 2−𝑙

𝑜 𝑙 ≤ 𝑓−2−𝑙 𝑜 𝑙

• This is less than 2−𝑜 if 𝑜

𝑙 > 2𝑙𝑜 ln 2

Just Kidding, There’s One More Calculation

Binomial estimates

• For all 1 ≤ 𝑙 ≤ 𝑜, we have

𝑜 𝑙 𝑙

≤

𝑜 𝑙 ≤ 𝑜𝑓 𝑙 𝑙

• If 𝑙 = 𝛽𝑜, then 𝑜

𝑙 = 2 1+𝑝 1 𝐼 𝛽 𝑜 as 𝑜 → ∞

• Binary entropy: 𝐼 𝛽 = −𝛽 log 𝛽 − 1 − 𝛽 log 1 − 𝛽
• For more estimates:

http://page.mi.fu-berlin.de/shagnik/notes/binomials.pdf

Theorem 1.1.9 For all 𝑙 ∈ ℕ, if 𝑜

𝑙 < 2𝑙, then 𝑜0 𝑙 > 𝑜, and if 𝑜 𝑙 > 2𝑙𝑜 ln 2, then

𝑜0 𝑙 ≤ 𝑜.

Just Kidding, There’s One More Calculation

Lower bound

• We use

𝑜 𝛽𝑜 = 2 1+𝑝 1 𝐼 𝛽 𝑜

• Therefore 𝑜

𝑙 ∼ 2𝑙 if 𝑙 = 𝛽𝑜 and 𝐼 𝛽 = 𝛽

• This happens for 𝛽 = 0.7729 …

Upper bound

• Binomial coefficient grows very fast
• 𝑜+1

𝑙

=

𝑜+1 𝑜−𝑙+1 𝑜 𝑙 ∼ 1 1−𝛽 𝑜 𝑙

• ⇒ for some constant 𝑑, if 𝑜′ = 𝛽−1𝑙 + 𝑑 log 𝑙, then 𝑜′

𝑙

∼ 2𝑙𝑜 ln 2

Corollary 1.1.10 As 𝑙 → ∞, 𝑜0 𝑙 = 1.2938 … + 𝑝 1 𝑙.

Any questions?

§2 Prefix-free Codes

Chapter 1: Getting Started The Probabilistic Method

😢 😣 😑 🙃 😵

A Motivating Example

Evolutionary pitfalls

• Imagine in some parallel universe a species evolves so that:
• they develop binary computers, and
• they communicate their emotional states through a set of five emojis

Technical problem

• How should a binary computer transmit these states?

A First Attempt

Binary encoding

• We can index the emojis with integers 0 − 4
• The integers 0 − 7 can be written as binary strings of length 3
• Computers can send these strings to represent the emojis

Examples

😣😢😵 → 001 | 000 | 100 → 001000100 011000 → 011 | 000 → 🙃😢

😢 😣 😑 🙃 😵

000 001 010 011 100

A Problem and a Fix

Wasteful encoding

• This is a bit costly – it takes three bits per emoji
• Can we reduce the bandwidth by using a shorter encoding?

Idea

• We need to encode five emojis
• There are six non-empty binary strings of length at most two
• Five is less than six

😢 😣 😑 🙃 😵

1 00 01 10

The Problem in the “Fix”

Encoding is simple

🙃🙃🙃 → 01 | 01 | 01 → 010101

But how do we decode?

010101 → 01 | 01 | 01 → 🙃🙃🙃 ? 010101 → 0 | 1 | 0 | 1 | 0 | 1 → 😢😣😢😣😢😣? 010101 → 01 | 0 | 10 | 1 → 🙃😢😵😣?

😢 😣 😑 🙃 😵

1 00 01 10

Coding: General Framework

Set-up

• Have an alphabet 𝐵 = 𝑏1, 𝑏2, … , 𝑏𝑜 of size 𝑜
• Want to encode the letters of the alphabet as binary strings

Encoding

• Represent each 𝑏𝑗 with a word 𝑥𝑗 ∈ 0,1 ∗, a non-empty finite binary string
• Let ℓ𝑗 = 𝑥𝑗 be the length of the word 𝑥𝑗

Objectives

• Decipherability: given a concatenation of words, should be able to recover the
• riginal words uniquely
• Efficiency: would like to make the lengths ℓ𝑗 as small as possible

Prefix-free Codes

Prefixes

• Given a word 𝑥 ∈ 0,1 ∗, the ℓ-prefix of 𝑥 is the subword of the first ℓ bits
• e.g. the non-empty prefixes of 𝑥 = 00101 are 0, 00, 001, 0010 and 00101
• but the substring 010 is not a prefix

Prefix-free codes

• We say a code from an alphabet 𝐵 to 0,1 ∗ is prefix-free if no codeword 𝑥𝑗 is

a prefix of any other codeword 𝑥

𝑘, 𝑘 ≠ 𝑗

• Equivalently, if we place the codewords in the (infinite) binary tree of 0,1 ∗,

no codeword is an ancestor of another

Decipherability

Proof

• Want to show that the concatenation 𝑥 = 𝑥𝑗1𝑥𝑗2 … 𝑥𝑗𝑡 can be decoded
• Base case: 𝑡 = 0
• In this case, 𝑥 is the empty string ⇒ no codewords
• Induction step: 𝑡 ≥ 1
• Start from the beginning of 𝑥, and read until the prefix is some codeword 𝑥

𝑘

• Must terminate, as 𝑥𝑗1 is a prefix of 𝑥
• Cannot terminate on another codeword, as otherwise 𝑥

𝑘 would be a prefix of 𝑥𝑗1

• Thus we know 𝑏𝑗1 is the first letter
• Remove 𝑥𝑗1, and decode 𝑥′ = 𝑥𝑗2𝑥𝑗3𝑥𝑗4 … 𝑥𝑗𝑡 (induction hypothesis)

∎

Proposition 1.2.1 All prefix-free codes are decipherable.

Examples

Uniform codes Given ℓ, any injection 𝐵 → {0,1}ℓ is prefix- free

∅ 1 00 01 10 11 000 001 010 011 100 101 110 111 😢 😣 😑 🙃 😵 ___

Length We must have 𝐵 ≤ 0,1 ℓ = 2ℓ ⇒ ℓ ≥ log 𝐵 ⇒ ℓ ≥ log 𝐵

Examples - II

Improvements Can sometimes find prefix-free codes with a shorter average codeword length

∅ 1 00 01 10 11 😢 😣 😑____________ 000 001 010 011 100 101 110 111 🙃 😵

Theorem 1.2.2 (Kraft, 1949) Given an alphabet 𝐵 of size 𝑜, any prefix-free code with codeword lengths ℓ1, ℓ2, … , ℓ𝑜 must satisfy σ𝑗=1

𝑜

2−ℓ𝑗 ≤ 1.

Short Prefix-free Codes

Extremal problem

• How small can the average length of a codeword of a prefix-free code be?

Corollary 1.2.3 (Convexity) Given an alphabet 𝐵 of size 𝑜, the average length of the codewords in any prefix-free code is at least log 𝑜.

Proof Idea

Existential reformulation

• Want to show that an encoding with shorter codewords is not prefix-free
• Given:
• an encoding 𝑥1, 𝑥2, … , 𝑥𝑜 of 𝐵 with lengths ℓ1, ℓ2, … , ℓ𝑜 such that σ𝑗=1

𝑜

2−ℓ𝑗 > 1

• Seek:
• Codewords 𝑥𝑗, 𝑥

𝑘, 𝑗 ≠ 𝑘, such that 𝑥𝑗 is a prefix of 𝑥 𝑘

Key observation

• Suppose we have a string 𝑥 ∈ 0,1 ∗ such that both 𝑥𝑗, 𝑥

𝑘 are prefixes of 𝑥

• Then 𝑥𝑗 is a prefix of 𝑥

𝑘 or wj is a prefix of 𝑥𝑗

New objective

• Find a string 𝑥 ∈ 0,1 ∗ that contains at least two codewords as prefixes

Basic Fact For any random variable 𝑌, the events 𝑌 ≥ 𝔽 𝑌 and 𝑌 ≤ 𝔽 𝑌 must occur with positive probability.

Probabilistic Framework

Probability space

• Let 𝑀 = max 𝑚𝑗: 𝑗 ∈ 𝑜

be the length of the longest codeword

• Let 𝑥 ∈ 0,1 𝑀 be a uniformly random string of length 𝑀

Random variables

• Let 𝑌 =

𝑗: 𝑥𝑗 is a prefix of 𝑥 count the number of codeword prefixes

Simpler objective

• Since 𝑌 is integer-valued, it suffices to show 𝔽 𝑌 > 1

Linearity of Expectation For any sequence 𝑌1, 𝑌2, … , 𝑌𝑜 of random variables, and any sequence 𝑑1, 𝑑2, … , 𝑑𝑜 of constants, if 𝑌 = 𝑑1𝑌1 + 𝑑2𝑌2 + ⋯ + 𝑑𝑜𝑌𝑜, then 𝔽 𝑌 = 𝑑1𝔽 𝑌1 + 𝑑2𝔽 𝑌2 + ⋯ + 𝑑𝑜𝔽 𝑌𝑜 .

Computing the Expectation

Indicator random variables

• For each 𝑗 ∈ 𝑜 , let 𝐹𝑗 be the event that 𝑥𝑗 is a prefix of the random string 𝑥
• Let 𝑌𝑗 = 1𝐹𝑗 be the indicator function of this event
• Then 𝑌 = σ𝑗=1

𝑜

𝑌𝑗

Reduction to probabilities

• We therefore have 𝔽 𝑌 = σ𝑗=1

𝑜

𝔽[𝑌𝑗] = σ𝑗=1

𝑜

ℙ(𝐹𝑗)

Finishing the Proof

Recall

• 𝑥 is a uniformly random string
• 𝑌 is the number of codewords that are prefixes of 𝑥
• 𝐹𝑗 is the event that the codeword 𝑥𝑗 is a prefix of 𝑥
• 𝔽 𝑌 = σ𝑗=1

𝑜

ℙ(𝐹𝑗)

Computing probabilities

• The event 𝐹𝑗 only depends on the first ℓ𝑗 bits of 𝑥
• This is a uniformly random string in 0,1 ℓ𝑗
• ⇒ ℙ 𝐹𝑗 = 2−ℓ𝑗

A grand finale

• ⇒ if 𝔽 𝑌 = σ𝑗=1

𝑜

2−ℓ𝑗 > 1, there is some 𝑥 ∈ 0,1 ∗ with two codewords as prefixes

• Hence, in any prefix-free code, σ𝑗=1

𝑜

2−ℓ𝑗 ≤ 1 ∎

Linearity of Expectation

Union bound revisited

• In the previous calculation, we saw the expression σ𝑗 ℙ 𝐹𝑗
• Union bound: ℙ ∪𝑗 𝐹𝑗 < σ𝑗 ℙ 𝐹𝑗
• ∴ σ𝑗 ℙ 𝐹𝑗 < 1 ⇒ with positive probability, none of the events 𝐹𝑗 occur

Using linearity instead

• σ𝑗 ℙ 𝐹𝑗 is the expectation of the number 𝑌 of events 𝐹𝑗 that occur
• ∴ σ𝑗 ℙ 𝐹𝑗 < 1 ⇒ with positive probability, 𝑌 = 0
• With linearity, we get information when σ𝑗 ℙ 𝐹𝑗 ≥ 1 as well

Any questions?

§3 Sum-free Subsets

Chapter 1: Getting Started The Probabilistic Method

Sum Theorems

Definition 1.3.1 A set 𝐵 is sum-free if there are no 𝑦, 𝑧, 𝑨 ∈ 𝐵 with 𝑦 + 𝑧 = 𝑨. Theorem 1.3.3 (Schur, 1912) ℕ cannot be partitioned into finitely many sum-free sets. Theorem 1.3.2 (Fermat, 1637; Wiles, 1995) For all 𝑜 ≥ 3, the set 𝑦𝑜: 𝑦 ∈ ℕ is sum-free.

Sum-free Subsets of [𝑜]

Answer

• If 𝐵 is sum-free, then A ≤

𝑜 2

• Odd integers: 𝑃 = 𝑦 ∈ 𝑜 : 𝑦 ≡ 1 mod 2
• Large integers: 𝑀 = 𝑦 ∈ 𝑜 : 𝑦 >

𝑜 2

• These are the only two maximum sum-free subsets of [𝑜]

Question How large can a sum-free subset of [𝑜] be?

Sum-free Subsets of Sets

Extremal function

• Given a set S ⊆ ℕ, let 𝑔 𝑇 = max

𝐵 : 𝐵 ⊆ 𝑇, 𝐵 sum−free

• Let 𝑔 𝑜 = min 𝑔 𝑇 : 𝑇 ⊂ ℕ, 𝑇 = 𝑜
• Question: how quickly does 𝑔(𝑜) grow?

Question (Erdős, 1965) Does every set of 𝑜 natural numbers have a large sum-free subset? Theorem 1.3.4 (Deshouillers, Freiman, Sós) If 𝐵 ⊆ [𝑜] is sum-free, then either 𝐵 ⊆ 𝑃, 𝐵 ⊆ 𝑀, or 𝐵 <

2 5 𝑜 + 1.

Upper Bounds

Beating trivial

• Recall: biggest sum-free subsets have odd or large integers
• Let 𝑈 ⊆ 𝑜 be a set of

𝑜 10 large odd integers, take 𝑇 = 𝑜 ∖ 𝑈

• If 𝐵 ⊆ 𝑇 is sum-free, then either 𝐵 ⊆ 𝑃 ∖ 𝑈, 𝐵 ⊆ 𝑀 ∖ 𝑈 or 𝐵 <

2 5 𝑜 + 1

• Thus 𝑔

9 10 𝑜 ≤ 𝑔 𝑇 < 2 5 𝑜 + 1

• ⇒ 𝑔 𝑜 ≤

4 9 𝑜 + 10 9

A trivial bound

• 𝑔 𝑜 ≤ 𝑔

𝑜 =

𝑜 2

• Any good set should have lots of (well-distributed) sums
• [𝑜] has lots of sums – could this be best possible?

Lower Bounds

Goal

• Given a set 𝑇 of 𝑜 natural numbers, find a large sum-free 𝐵 ⊆ 𝑇

Greedy approach

• Start with 𝐵 = ∅, and add elements one-by-one, keeping 𝐵 sum-free
• If 𝐵 = 𝑏, 𝐵 defines at most 𝑏+1

2

sums

• If 𝑏+1

2

< 𝑜 − 𝑏, there is an element of 𝑇 ∖ 𝐵 that can be added to 𝐵

• ⇒ 𝑔 𝑜 >

2𝑜 − 2

Theorem 1.3.5 (Erdős, 1965) For all 𝑜 ∈ ℕ, 𝑔 𝑜 ≥

1 3 (𝑜 + 1).

A Cyclic Digression

The problem with 𝑜

• 𝑜 does have large sum-free sets, 𝑃 and 𝑀
• But 𝑇 might be far away from these

The cyclic group has more symmetry

• Largest sum-free set in ℤ𝑞, 𝑞 prime?
• 𝑁 = 𝑦:

1 3 𝑞 < 𝑦 < 2 3 𝑞 is sum-free

• Cauchy-Davenport: if 𝐵 ⊆ ℤ𝑞, then 𝐵 + 𝐵 ≥ min 2 𝐵 − 1, 𝑞
• Since 𝐵 ∩ 𝐵 + 𝐵 = ∅, 𝐵 ≤

𝑞 3

• ℤ𝑞 has many large sum-free sets
• For any 𝛽 ∈ ℤ𝑞 ∖ 0 , 𝛽𝑁 = {𝛽𝑦: 𝑦 ∈ 𝑁} is also sum-free

Finding Large Sum-free Subsets

Proof idea

• Given a set 𝑇 ⊂ ℕ of size 𝑜, embed 𝑇 in ℤ𝑞 for some suitable 𝑞
• ℤ𝑞 has many large sum-free subsets
• Find one that intersects 𝑇 significantly

Randomness to the rescue

• A random sum-free subset works!

Theorem 1.3.5 (Erdős, 1965) For all 𝑜 ∈ ℕ, 𝑔 𝑜 ≥

1 3 (𝑜 + 1).

Setting Up

Choosing a prime

• Let 𝑞 = 3𝑙 + 2 be prime with 𝑞 > max 𝑇
• Then 𝑁 = {𝑙 + 1, 𝑙 + 2, … , 2𝑙 + 1} is sum-free with size 𝑙 + 1
• Embed 𝑇 ⊆ ℤ𝑞

Choosing a sum-free subset

• Let 𝛽 ∈ ℤ𝑞 ∖ 0 be chosen uniformly at random
• Let 𝑇𝛽 = 𝑇 ∩ 𝛽𝑁
• 𝑇𝛽 ⊆ 𝑇 is sum-free:
• If 𝑦 + 𝑧 = 𝑨 in 𝑇𝛽, then 𝑦 + 𝑧 ≡ 𝑨 mod 𝑞 , so this would be a sum in 𝛽𝑁

No Devil in the Details

Using linearity

• 𝑇𝛽 = σ𝑡∈𝑇 1 𝑡∈𝛽𝑁
• ⇒ 𝔽 𝑇𝛽

= 𝔽 σ𝑡∈𝑇 1 𝑡∈𝛽𝑁 = σ𝑡∈𝑇 𝔽 1 𝑡∈𝛽𝑁 = σ𝑡∈𝑇 ℙ 𝑡 ∈ 𝛽𝑁

Computing probabilities

• 𝑡 ∈ 𝛽𝑁 ⇔ 𝛽−1𝑡 ∈ 𝑁
• 𝛽 uniform over ℤ𝑞 ∖ 0 ⇒ 𝛽−1 uniform ⇒ 𝛽−1𝑡 uniform
• ⇒ ℙ 𝛽−1𝑡 ∈ 𝑁 =

𝑁 𝑞−1 = 𝑙+1 3𝑙+1 > 1 3

Finishing the proof

• ⇒ 𝔽 𝑇𝛽

>

1 3 𝑇 ⇒ for some 𝛽, 𝑇𝛽 ≥ 1 3 (𝑜 + 1)

• This gives a sum-free subset of 𝑇 of the desired size

∎

Finishing the Story

Improving the lower bound

• Using Fourier analysis, Bourgain (1997) proved 𝑔 𝑜 ≥

1 3 (𝑜 + 2) for 𝑜 ≥ 3

• Best-known bound to date

Upper bounds

• Blow-ups of small constructions: several improvements over the years
• Until Eberhard, Green and Manners (2014) proved 𝑔 𝑜 ≤

1 3 + 𝑝 1

𝑜

• Construction randomised, but intricate

Any questions?

§4 Schütte Tournaments

Chapter 1: Getting Started The Probabilistic Method

War by Proxy

Rival superpowers

• Two powerful nations go to war
• Hire private military companies to do the actual fighting

Objectives

• Have enough power
• Need to ensure that hired companies can defeat any of the companies the enemy hires
• Be economical
• Hire as few companies as possible

Problem

• How many companies must be hired?

A Graph Theoretic Representation

Tournaments

• Build a directed graph
• Vertices: private military companies
• Arcs: edge 𝑦 → 𝑧 if 𝑦 would defeat 𝑧 in battle
• For every pair {𝑦, 𝑧}, exactly one of the arcs 𝑦 → 𝑧 or 𝑧 → 𝑦 is in the graph
• Such a graph is called a tournament

Objectives

• Dominating set
• A subset of vertices 𝑇 such that, for every 𝑦 ∈ 𝑊 ∖ 𝑇, there is some 𝑡 ∈ 𝑇 with 𝑡 → 𝑦
• Then we can always defeat the enemy’s army, regardless of their choice
• Economical
• Want to choose a small dominating set

Small Examples

Easy case

1 2 3

Small Examples

Easy case

• One vertex beats all others

1 2 3

Small Examples

Easy case

• One vertex beats all others
• Defeats any choice the enemy makes

1 2 3

Small Examples

Easy case

• One vertex beats all others
• Defeats any choice the enemy makes

1 2 3

Small Examples

Easy case

• One vertex beats all others
• Defeats any choice the enemy makes
• Dominating set of size one

1 2 3

Small Examples

Easy case

• One vertex beats all others
• Defeats any choice the enemy makes
• Dominating set of size one

1 2 3

Harder case

• No such vertex

1 2 3

Small Examples

Easy case

• One vertex beats all others
• Defeats any choice the enemy makes
• Dominating set of size one

1 2 3

Harder case

• No such vertex
• Any vertex we choose loses to another

1 2 3

Small Examples

Easy case

• One vertex beats all others
• Defeats any choice the enemy makes
• Dominating set of size one

1 2 3

Harder case

• No such vertex
• Any vertex we choose loses to another

1 2 3

Small Examples

Easy case

• One vertex beats all others
• Defeats any choice the enemy makes
• Dominating set of size one

1 2 3

Harder case

• No such vertex
• Any vertex we choose loses to another

1 2 3

Small Examples

Easy case

• One vertex beats all others
• Defeats any choice the enemy makes
• Dominating set of size one

1 2 3

Harder case

• No such vertex
• Any vertex we choose loses to another

1 2 3

Small Examples

Easy case

• One vertex beats all others
• Defeats any choice the enemy makes
• Dominating set of size one

1 2 3

Harder case

• No such vertex
• Any vertex we choose loses to another
• Dominating set of size two exists

1 2 3

An Extremal Reformulation

Worst-case scenario

• How large can the smallest dominating set in an 𝑜-vertex tournament be?
• Inverse formulation
• Say 𝑈 has the Schütte property 𝑇𝑙 if it has no dominating set of size at most 𝑙
• Let 𝜏(𝑙) be the minimum number of vertices in a tournament with the property 𝑇𝑙
• ⇒ if 𝑜 < 𝜏(𝑙), then 𝑈 has a dominating set of size ≤ 𝑙

Proving bounds on 𝜏(𝑙)

• Lower bound: 𝜏 𝑙 > 𝑜
• Prove that any tournament on 𝑜 vertices has a dominating set of size ≤ 𝑙
• Upper bound: 𝜏 𝑙 ≤ 𝑜
• Prove there is a tournament on 𝑜 vertices without a dominating set of size ≤ 𝑙

The Greedy Lower Bound

A recursive algorithm

• Given an optimal tournament 𝑈, let 𝑤 ∈ 𝑊(𝑈)
• Let 𝐵 be the vertices dominating 𝑤, and 𝐶 the vertices 𝑤 dominates
• Thus 𝑊 𝑈 = 𝐵 ∪ 𝐶 ∪ 𝑤 , with 𝐵 → 𝑤 → 𝐶
• Let 𝑇′ be a dominating set in 𝑈[𝐵], and set 𝑇 = 𝑇′ ∪ {𝑤}
• If 𝑦 ∈ 𝑊 𝑈 ∖ 𝑇:
• If 𝑦 ∈ 𝐵, then 𝑦 is dominated by 𝑇′, so there is an 𝑡 ∈ 𝑇′ ⊆ 𝑇 with 𝑡 → 𝑦
• If 𝑦 ∉ 𝐵, then 𝑦 ∈ 𝐶, so 𝑤 → 𝑦
• Thus 𝑇 is a dominating set for 𝑈

Proposition 1.4.1 For all 𝑙 ∈ ℕ, 𝜏 𝑙 ≥ 2𝑙+1 − 1.

Choosing the Right Vertex

Large out-degree

• If 𝐵 is small, then it has a small dominating set
• Thus we should choose 𝑤 to make 𝐵 as small as possible
• ⇒ choose a vertex of maximum out-degree
• Average out-degree is

1 𝑜 𝑜 2 = 1 2 (𝑜 − 1)

• ⇒ by choosing 𝑤 of maximum out-degree, we ensure 𝐵 ≤

1 2 (𝑜 − 1)

Induction

• Since 𝑈 has the property 𝑇𝑙, 𝑈[𝐵] must have the property 𝑇𝑙−1
• ⇒

1 2 𝑜 − 1 ≥ 𝐵 ≥ 𝜏 𝑙 − 1 ≥ 2𝑙 − 1 (induction hypothesis)

• Solving gives 𝜏 𝑙 = 𝑜 ≥ 2𝑙+1 − 1

∎

An Indomitable Tournament

Goal

• Need to construct a tournament with no dominating set of size 𝑙
• Greedy argument: tournament should be close to regular
• Idea: try a random tournament 𝑈

Random tournament

• Vertex set: 𝑊 = 𝑜
• For every pair 𝑦, 𝑧 ∈ [𝑜], choose 𝑦 → 𝑧 or 𝑧 → 𝑦 uniformly at random

Theorem 1.4.2 (Erdős, 1963) If 𝑜

𝑙

1 − 2−𝑙 𝑜−𝑙 < 1, then there is an 𝑜-vertex tournament with the property 𝑇𝑙.

Disproving Domination

Bad events

• Given a set 𝑇 ∈

𝑜 𝑙 , let 𝐹𝑇 be the event that 𝑇 is a dominating set

• Then ℙ 𝑈 has property 𝑇𝑙 = 1 − ℙ ∪𝑇 𝐹𝑇 ≥ 1 − σ𝑇 ℙ 𝐹𝑇
• Suffices to show σ𝑇 ℙ 𝐹𝑇 < 1

Computing probabilities

• Fix 𝑇 ∈

𝑜 𝑙

• For 𝑇 to dominate a fixed vertex 𝑤, cannot have all edges 𝑤 → 𝑇
• 𝑙 edges, chosen independently ⇒ probability is 1 − 2−𝑙
• This must be true for all vertices in 𝑊 ∖ 𝑇
• Edges again independent ⇒ ℙ 𝐹𝑇 = 1 − 2−𝑙 𝑜−𝑙
• ⇒ σ𝑇 ℙ 𝐹𝑇 =

𝑜 𝑙

1 − 2−𝑙 𝑜−𝑙 < 1. ∎

Corollary 1.4.3 As 𝑙 → ∞, 𝜏 𝑙 ≤ 𝑙22𝑙 ln 2 + 𝑝 1 .

Computing the Bound

Find the smallest 𝑜 for which 𝑜

𝑙

1 − 2−𝑙 𝑜−𝑙 < 1

• Estimates:
• 𝑜

𝑙 ≤ 𝑜𝑙 and 1 − 2−𝑙 < 𝑓−2−𝑙

• ⇒ suffices to have 𝑜𝑙𝑓−2−𝑙 𝑜−𝑙 < 1
• ⇔ 𝑙 ln 𝑜 < 𝑜 − 𝑙 2−𝑙 ∗
• ∗ ⇒ 𝑜 > 2𝑙
• ⇒ ln 𝑜 > 𝑙 ln 2
• ∗ ⇒ 𝑜 > 𝑙22𝑙 ln 2
• ⇒ ln 𝑜 > 𝑙 ln 2 + 𝑝 1

, so this suffices

Any questions?

§5 Ramsey Numbers

Chapter 1: Getting Started The Probabilistic Method

Theorem 1.5.2 (Erdős, 1947) As 𝑙 → ∞, we have 𝑆 𝑙 ≥ 1 𝑓 2 + 𝑝 1 𝑙 2

𝑙.

Reviewing the Classics

Definition 1.5.1 (Ramsey number) Given 𝑙 ∈ ℕ, 𝑆 𝑙 is the minimum 𝑜 for which any 𝑜-vertex graph has either a clique or independent set on 𝑙 vertices. Proof idea

• Show that a uniformly random graph on this many vertices works

Ramsey Upper Bounds

Proof by induction

• Introduce the asymmetric Ramsey numbers

Theorem 1.5.3 (Erdős-Szekeres, 1935) For all 𝑙 ∈ ℕ, we have 𝑆 𝑙 ≤

2𝑙−2 𝑙−1 . In particular, as 𝑙 → ∞,

𝑆 𝑙 ≤

1+𝑝 1 4 𝜌𝑙 4𝑙.

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.

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

Asymmetric Ramsey Bounds

Problem

• For fixed ℓ ∈ ℕ, how does 𝑆 ℓ, 𝑙 grow as 𝑙 → ∞?
• Asymmetric Ramsey numbers grow at most polynomially
• Can we find matching lower bounds?

Theorem 1.5.6 (Turán, 1941) An 𝑜-vertex 𝐿ℓ-free graph can have at most 1 −

1 ℓ−1 𝑜 2 edges.

Turán’s Lower Bound

Goal

• Find a 𝐿ℓ-free graph with no large independent sets

Intuition

• More edges ⇒ fewer independent sets
• How dense can a 𝐿ℓ-free graph be?

Construction

• Complete (ℓ − 1)-partite graph 𝑈𝑜,ℓ−1
• 𝛽 𝑈𝑜,ℓ−1 =

𝑜 ℓ−1 ⇒ 𝑆 ℓ, 𝑙 > ℓ − 1

𝑙 − 1

What About Randomness?

𝑆 𝑙 lower bound

• Symmetric situation – can switch edges and non-edges
• Used a uniformly random graph
• Equivalently: each edge appears independently with probability

1 2

𝑆(ℓ, 𝑙) for fixed ℓ ∈ ℕ, 𝑙 → ∞

• Situation far from symmetric
• “easier” to make clique on ℓ vertices than an independent set on 𝑙 vertices
• Should focus on graphs with fewer edges

Erdős-Rényi model

• 𝐻 𝑜, 𝑞 : 𝑜 vertices, each edge appears independently with probability 𝑞
• Allows us to “see” sparser graphs

A Random Lower Bound

Proof idea

• Sample the random graph 𝐻 ∼ 𝐻(𝑜, 𝑞)
• What could go wrong?
• Could find a clique on ℓ vertices
• Could find an independent set on 𝑙 vertices

Theorem 1.5.7 Given ℓ, 𝑙, 𝑜 ∈ ℕ and 𝑞 ∈ 0,1 , if 𝑜 ℓ 𝑞

ℓ 2 + 𝑜

𝑙 1 − 𝑞

𝑙 2 < 1,

then 𝑆 ℓ, 𝑙 > 𝑜.

Analysing Bad Events

Bad cliques

• Given a set 𝑇 ∈

[𝑜] ℓ , let 𝐹𝑇 be the event that 𝐻[𝑇] is a clique

• ℓ

2 pairs, each an edge independently with probability 𝑞

• ⇒ ℙ 𝐹𝑇 = 𝑞

ℓ 2

Bad independent sets

• Given a set 𝑈 ∈

𝑜 𝑙 , let 𝐺𝑈 be the event that 𝐻[𝑈] is an independent set

• 𝑙

2 pairs, each a non-edge independently with probability 1 − 𝑞

• ⇒ ℙ 𝐺𝑈 = 1 − 𝑞

𝑙 2

Completing the Proof

Recall

• 𝐹𝑇 = 𝐻 𝑇 is an ℓ−clique , ℙ 𝐹𝑇 = 𝑞

ℓ 2

• 𝐺𝑈 = 𝐻 𝑈 is an independent 𝑙−set , ℙ 𝐺𝑈 = 1 − 𝑞

𝑙 2

Union bound does the job

• 𝐻 not Ramsey = ∪𝑇 𝐹𝑇 ∪ ∪𝑈 𝐺𝑈
• ∴ ℙ 𝐻 not Ramsey = ℙ (∪𝑇 𝐹𝑇) ∪ ∪𝑈 𝐺𝑈

≤ σ𝑇 ℙ 𝐹𝑇 + σ𝑈 ℙ(𝐺𝑈)

• σ𝑇 ℙ(𝐹𝑇) + σ𝑈 ℙ(𝐺𝑈) =

𝑜 ℓ 𝑞

ℓ 2 +

𝑜 𝑙

1 − 𝑞

𝑙 2 < 1

⇒ ℙ 𝐻 Ramsey = 1 − ℙ(𝐻 not Ramsey) > 0 ∎

An Actual Bound

What does this tell us about 𝑆(ℓ, 𝑙)? Goal

• Maximise 𝑜
• Subject to 𝑜

ℓ 𝑞

ℓ 2 +

𝑜 𝑙

1 − 𝑞

𝑙 2 < 1 for some 𝑞 ∈ [0,1]

Theorem 1.5.7 Given ℓ, 𝑙, 𝑜 ∈ ℕ and 𝑞 ∈ 0,1 , if 𝑜 ℓ 𝑞

ℓ 2 + 𝑜

𝑙 1 − 𝑞

𝑙 2 < 1,

then 𝑆 ℓ, 𝑙 > 𝑜.

Computing a Lower Bound

Goal

• Maximise 𝑜
• Subject to 𝑜

ℓ 𝑞

ℓ 2 +

𝑜 𝑙

1 − 𝑞

𝑙 2 < 1 for some 𝑞 ∈ [0,1]

Varying 𝑞

• As 𝑞 increases, 𝑜

ℓ 𝑞

ℓ 2 increases and 𝑜

𝑙

1 − 𝑞

𝑙 2 decreases

• ⇒ at optimum, expect both quantities to be comparable

Simplification

• Instead solve 𝑜

ℓ 𝑞

ℓ 2 <

1 2 and 𝑜 𝑙

1 − 𝑞

𝑙 2 <

1 2

Computing Some More

𝑜 ℓ 𝑞

ℓ 2 <

1 2:

• Bound 𝑜

ℓ ≤ 𝑜ℓ, so 𝑜 ℓ 𝑞

ℓ 2 ≤ 𝑜𝑞 ℓ−1 2

ℓ

• Sufficient to have 𝑞 ≤ 1 − 𝑝 1

𝑜−2/(ℓ−1)

𝑜 𝑙

1 − 𝑞

𝑙 2 <

1 2:

• Bound 𝑜

𝑙 ≤ 𝑜𝑙 and 1 − 𝑞 ≤ 𝑓−𝑞, so 𝑜 𝑙

1 − 𝑞

𝑙 2 ≤ 𝑜𝑓−𝑞 𝑙−1 /2 𝑙

• Suffices to have 𝑜𝑓−𝑞 𝑙−1 /2 < 1 ⇒ 𝑞 𝑙 − 1 > 2 ln 𝑜
• Substitute 𝑞 ≈ 𝑜−2/(ℓ−1)
• ⇒ 𝑙 > 2𝑜2/(ℓ−1) ln 𝑜

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 .

Concluding the Computations

Recall

• 𝑙 ≈ 2𝑜2/(ℓ−1) ln 𝑜

Solve for 𝑜

• 𝑜 ≈

𝑙 2 ln 𝑜

ℓ−1 2 ≈

𝑙 2 ln 𝑙

ℓ−1 2

Any questions?