Chapter 4: The Lovsz Local Lemma The Probabilistic Method Summer - - PowerPoint PPT Presentation

Chapter 4: The Lovász Local Lemma

The Probabilistic Method Summer 2020 Freie Universität Berlin

Chapter Overview

• Introduce the Lovász Local Lemma and some variants
• Survey some applications, including to 𝑆 3, 𝑙

§1 Introducing the Lemma

Chapter 4: The Lovász Local Lemma The Probabilistic Method

Avoiding Bad Events

Second moment set-up

• Have a collection of good events 𝐹1, 𝐹2, … , 𝐹𝑛
• e.g.: 𝐹𝑗 = 𝑗th copy of 𝐼 appears in 𝐻 𝑜, 𝑞
• Goal: show that with positive probability, at least one event occurs
• Usually show this happens with probability 1 − 𝑝 1

Opposite situation

• Have a collection of bad events 𝐹1, 𝐹2, … , 𝐹𝑛
• e.g.: 𝐹𝑗 = 𝑗th clause in 𝑙−SAT formula not satisfied
• Goal: show that with positive probability, none of these events occur
• i.e.: ℙ ∩𝑗 𝐹𝑗

𝑑 > 0

• Union bound: ℙ ∩𝑗 𝐹𝑗

𝑑 = 1 − ℙ ∪𝑗 𝐹𝑗 ≥ 1 − σ𝑗 ℙ 𝐹𝑗

• Tight when events 𝐹𝑗 are disjoint
• In general, need either 𝑛 or ℙ 𝐹𝑗 to be small enough for effective bounds

Independence to the Rescue

Independent events

• If the events 𝐹1, 𝐹2, … , 𝐹𝑛 are mutually independent, we are in business
• ℙ ∩𝑗 𝐹𝑗

𝑑 = ς𝑗 ℙ 𝐹𝑗 𝑑 = ς𝑗 1 − ℙ 𝐹𝑗

• Might tend to zero, but is still positive (provided ℙ 𝐹𝑗 < 1 for all 𝑗)
• Doesn’t matter how many bad events there are, or how likely they are

A real-world example

• Work for the Bundesdruckerei
• Job: printing 𝑛 passports
• Bad event: 𝐹𝑗 = misprint in the 𝑗th passport
• Say ℙ 𝐹𝑗 =

1 2 for each 𝑗

• ℙ ∩𝑗 𝐹𝑗

𝑑 = 1 2 𝑛

> 0

• ⇒ it is possible to have a successful day

The Struggle for Independence

Do we need independence?

• In practice, true independence of events is rare
• Could hope to replace it with something weaker
• Most events being independent? Pairwise independence?

We might

• Bundesdruckerei example: suppose our passport printer is odd
• Never makes an even number of misprints
• Same marginal distributions
• ℙ 𝐹𝑗 =

1 2 for all 𝑗

• Almost complete independence
• Any 𝑛 − 1 of the 𝑛 events are mutually independent
• However, ℙ ∩𝑗 𝐹𝑗

𝑑 ≤ ℙ # misprints even = 0

Lovász to the Rescue

The Bundesdruckerei problem

• ℙ 𝐹𝑗 =

1 2 is a large probability for the bad event

• If ℙ 𝐹𝑗 <

1 2, then we lose even pairwise independence

• ℙ 𝐹𝑗 𝐹

𝑘 < ℙ 𝐹𝑗

The good news

• Suppose the bad events
• are independent of most other events
• occur with reasonably small probability
• Lovász Local Lemma ⇒ events behave as if independent
• Can show that with positive probability none occur

The Local Lemma – Symmetric Setting

“Local” Lemma

• Bound on 𝑞 independent of number of events (global property)
• Only depends on number of dependencies (local property)

Conclusion

• Only assert that with positive probability, none of the events occur
• This probability can depend on the number of events

Theorem 4.1.1 (Symmetric Lovász Local Lemma; Erdős-Lovász, 1975) Let 𝐹1, 𝐹2, … , 𝐹𝑛 be events such that each event 𝐹𝑗 is mutually independent of all but at most 𝑒 of the other events, and ℙ 𝐹𝑗 ≤ 𝑞 for all 𝑗. If 𝑓𝑞 𝑒 + 1 ≤ 1, then ℙ ∩𝑗 𝐹𝑗

𝑑 > 0.

Theorem 4.1.2 Any 𝑙-SAT formula in which each variable appears at most

2𝑙 𝑓𝑙 times is

satisfiable.

Re-restricted 𝑙-SAT

Recall

• Any 𝑙-SAT formula with fewer than 2𝑙 clauses is satisfiable
• Bound is best possible: take formula with all clauses on 𝑙 variables

Restricted 𝑙-SAT

• Previously: each 𝑙-set of variables appears in at most one clause
• What if we bound individual variable appearances instead?
• Applies to 𝑙-SAT formulae with any number of clauses!

Proof by Local Lemma

Proof

• Set each variable to true/false independently with probability

1 2

• Events
• 𝐹𝑗 = 𝑗th clause not satisfied
• ℙ 𝐹𝑗 = 𝑞 ≔ 2−𝑙 for all 𝑗
• Dependencies
• A clause is independent of any clauses with disjoint sets of variables
• Clause has 𝑙 variables, each in ≤

2𝑙 𝑓𝑙 − 1 other clauses

• ⇒ each event independent of all but 𝑒 ≔

2𝑙 𝑓 − 1 other events

• 𝑓𝑞 𝑒 + 1 = 1 ⇒ ℙ ∩𝑗 𝐹𝑗 > 0 ⇒ formula is satisfiable!

∎ Theorem 4.1.2 Any 𝑙-SAT formula in which each variable appears at most

2𝑙 𝑓𝑙 times is satisfiable.

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

𝑙.

Recalling Ramsey

• Disjoint sets of edges are independent
• Can improve bound with the local lemma

Theorem 4.1.3 As 𝑙 → ∞, we have 𝑆 𝑙 ≥ 2 𝑓 + 𝑝 1 𝑙 2

𝑙.

Setting Up the Proof

Events

• We take 𝐻 ∼ 𝐻 𝑜,

1 2 as before

• For 𝐽 ∈

𝑜 𝑙 , event 𝐹𝐽 = 𝐻 𝐽 ≅ 𝐿𝑙 or 𝐿𝑙 𝑑

• ℙ 𝐹𝐽 = 21− 𝑙

2 for all 𝐽

Dependencies

• 𝐹𝐽 independent of all events with disjoint edge-sets
• ⇒ 𝐹𝐽 depends on at most 𝑙

2 𝑜−2 k−2 − 1 other events

• ⇒ 𝑒 + 1 ≤

𝑙 2 𝑜−2 k−2

Lovász Local Lemma

• ⇒ suffices to show 𝑓21− 𝑙

2

𝑙 2 𝑜−2 𝑙−2 ≤ 1

Running the Calculations

Estimates

• 𝑙

2 ≤ 𝑙2 2

• 𝑜−2

𝑙−2 ≤ 𝑙2 𝑜2 𝑜 𝑙 ≤ 𝑙2 𝑜2 𝑜𝑓 𝑙 𝑙

Bounding 𝑜

• ⇒ 𝑓21− 𝑙

2

𝑙 2 𝑜−2 𝑙−2 ≤ 𝑓2− 𝑙

2 𝑙4

𝑜2 𝑜𝑓 𝑙 𝑙

=

𝑓𝑙4 𝑜2 𝑜𝑓 2 𝑙 2

𝑙

𝑙

• If 𝑜 =

1 𝑓 2 𝑙 2 𝑙, parenthetical term is 1

• Leading coefficient is then 𝑓2𝑙421−𝑙
• ⇒ can afford for the parenthetical term to be 2 + 𝑝(1)
• ⇒ can take 𝑜 =

2 𝑓 + 𝑝 1

𝑙 2

𝑙

∎

Any questions?

§2 The Ramsey Number 𝑆 3, 𝑙

Chapter 4: The Lovász Local Lemma The Probabilistic Method

Returning to 𝑆 3, 𝑙

Lower bound

• Proven using 𝐻 ∼ 𝐻(𝑜, 𝑞) and alterations

Limited dependence

• Again, disjoint sets of edges are independent
• What does the Local Lemma give?

Corollary 2.1.3 As 𝑙 → ∞, we have Ω 𝑙 ln 𝑙

3 2

= 𝑆 3, 𝑙 = 𝑃 𝑙2 .

Analysing the Events

Two classes of events

• For 𝐽 ∈

𝑜 3 , let 𝐹𝐽 = 𝐻 𝐽 ≅ 𝐿3

• For 𝐾 ∈

𝑜 𝑙 , let 𝐺 𝐾 = 𝐻 𝐾 ≅ 𝐿𝑙 𝑑

Probabilities

• For each 𝐽 ∈

𝑜 3 , 𝑞1 ≔ ℙ 𝐹𝐽 = 𝑞3

• For each 𝐾 ∈

𝑜 𝑙 , 𝑞2 ≔ ℙ 𝐺 𝐾 = 1 − 𝑞

𝑙 2 ≈ 𝑓−𝑞 𝑙 2

• ⇒ in Lovász Local Lemma, should take 𝑞′ = max {𝑞1, 𝑞2}
• ⇒ optimal to have 𝑞1 = 𝑞2
• ⇒ 𝑞 ≈

12 ln 𝑙 𝑙2

Analysing the Events Further

Edge involvements

• Each edge appears in 𝑜 − 2 events 𝐹𝐽 and 𝑜−2

k−2 events 𝐺 𝐾

Dependencies

• ⇒ each 𝐹𝐽 depends on fewer than 𝑒1 ≔ 3 𝑜 − 2 +

𝑜−2 𝑙−2

• ther events
• ⇒ each 𝐺

𝐾 depends on fewer than 𝑒2 ≔ 𝑙 2

n − 2 +

n−2 k−2

events

• ⇒ need to take d = max 𝑒1, 𝑒2 = 𝑒2 in the Local Lemma

Bounding 𝑜

• Thus 𝑓𝑞′ 𝑒 + 1 ≤ 𝑓 𝑞′ 3𝑒 ∼ 𝑓

12 ln 𝑙 𝑙2 3 𝑙 2

𝑜 − 2 +

𝑜−2 k−2

≤

123𝑓 ln3 𝑙 𝑙4 𝑜−2 𝑙−2 ≤ 123𝑓 ln3 𝑙 𝑜2𝑙2 𝑜 𝑙 ≤ 123𝑓 ln3 𝑙 𝑜2𝑙2 𝑜𝑓 𝑙 𝑙

• For this to be less than 1, need 𝑜 = 𝑃 𝑙

Post Mortem

Different kinds of events

• Triangle events 𝐹𝐽:
• Probability 𝑞1 = 𝑞3
• Depend on relatively few other events
• Independent set events 𝐺

𝐾:

• Probability 𝑞2 = 1 − 𝑞

𝑙 2

• Depend on many other events

A possible remedy

• Wasteful to use same probability, dependency bounds for all events
• Triangle events are “more independent”
• Could afford to let them occur with higher probability
• Ideally – track each event’s individual probability and dependencies

Definition 4.2.1 (Dependency digraph) Given events 𝐹1, 𝐹2, … , 𝐹𝑛, a directed graph 𝐸 on the vertices 𝑛 is a dependency digraph if, for each 𝑗 ∈ 𝑛 , the event 𝐹𝑗 is mutually independent of the set of events 𝐹

𝑘: 𝑗, 𝑘 ∉ 𝐸 .

Tracking Dependencies

Representing dependence

• Keep track of dependencies using a directed graph
• Events are independent of their non-neighbours

Why a digraph?

• In most applications, digraph will be symmetric
• 𝑗, 𝑘 ∈ 𝐸 ⇔ 𝑘, 𝑗 ∈ 𝐸
• Can sometimes help to have flexibility

The Lovász Local Lemma

Special case: independent events

• Can take 𝐸 to be edge-less
• ⇒ suffices to have 𝑦𝑗 = ℙ 𝐹𝑗 , and done

General case

• Dependencies → correction factor ς 𝑗,𝑘 1 − 𝑦𝑘
• The more dependencies, the smaller this factor
• ⇒ need probability of these events to shrink

Theorem 4.2.2 (Lovász Local Lemma; Erdős-Lovász, 1975) Let 𝐹1, 𝐹2, … , 𝐹𝑛 be events with a dependency digraph 𝐸. If there are 𝑦𝑗 ∈ 0,1 such that ℙ 𝐹𝑗 ≤ 𝑦𝑗 ς 𝑗,𝑘 ∈𝐸 1 − 𝑦𝑘 for all 𝑗 ∈ [𝑛], then ℙ ∩𝑗 𝐹𝑗

𝑑 ≥ ς𝑗 1 − 𝑦𝑗 .

Returning to 𝑆 3, 𝑙 Once Again

Symmetries

• All triangle events 𝐹𝐽 have the same probability and dependencies
• ⇒ should set 𝑦𝐽 = 𝑦 for some common 𝑦
• All independent set events 𝐺

𝐾 also share the same parameters

• ⇒ set 𝑦𝐾 = 𝑧 for some common 𝑧

Probability conditions

• Triangle events
• Depend on at most 3 𝑜 − 2 < 3𝑜 other triangle events
• Depend on at most 𝑜

𝑙 independent set bounds

• ⇒ suffices to have ℙ 𝐹𝐽 = 𝑞3 ≤ 𝑦 1 − 𝑦 3𝑜 1 − 𝑧

𝑜 𝑙

• Independent set events
• Depend on at most 𝑙

2

𝑜 − 2 <

𝑙 2 𝑜 triangle events, 𝑜 𝑙 independent set events

• ⇒ suffices to have ℙ 𝐺

𝐾 = 1 − 𝑞

𝑙 2 ≤ 𝑧 1 − 𝑦 𝑙 2 𝑜 1 − 𝑧 𝑜 𝑙

A Conditional Result

Proof

• Follows immediately from Lovász Local Lemma and previous calculations

∎

Optimisation

• Want to maximise 𝑜 while satisfying the two inequalities

Theorem 4.2.3 (Spencer, 1977) Let 𝑙, 𝑜 ∈ ℕ. If there are 𝑞, 𝑦, 𝑧 ∈ 0,1 such that 𝑞3 ≤ 𝑦 1 − 𝑦 3𝑜 1 − 𝑧

𝑜 𝑙

and 1 − 𝑞

𝑙 2 ≤ 𝑧 1 − 𝑦 𝑙 2 𝑜 1 − 𝑧 𝑜 k ,

then 𝑆 3, 𝑙 > 𝑜.

Some Heuristics

Maximise 𝑜 subject to

(1) 𝑞3 ≤ 𝑦 1 − 𝑦 3𝑜 1 − 𝑧

𝑜 𝑙

(2) 1 − 𝑞

𝑙 2 ≤ 𝑧 1 − 𝑦 𝑙 2 𝑜 1 − 𝑧 𝑜 𝑙

Setting 𝑧

• Do not want 1 − 𝑧

𝑜 𝑙 ≈ 𝑓−𝑧(𝑜 𝑙) to be exponentially small

• ⇒ set 𝑧 =

𝑜 𝑙 −1 ⇒ 1 − 𝑧

𝑜 k is constant

Understanding 𝑦

• From (2), we need 1 − 𝑦 𝑜 > 1 − 𝑞
• ⇒ 𝑜𝑦 < 𝑞, and 1 − 𝑦 𝑜 is constant

More Heuristics

Maximise 𝑜 subject to

(1’) 𝑞3 ≤ 𝑦 (2’) 1 − 𝑞

𝑙 2 ≤ 𝑧

(3) 𝑧 =

𝑜 𝑙 −1 and 𝑜𝑦 < 𝑞

Setting 𝑞

• From (1’) and (3), 𝑜𝑞3 ≤ 𝑜𝑦 < 𝑞 ⇒ 𝑞 < 𝑜−1/2

Fixing 𝑜

• From (2’) and (3), 1 − 𝑞

𝑙 2 ≈ 𝑓−𝑞 𝑙 2 ≤

𝑜 𝑙 −1 ≈ 𝑙 𝑜 𝑙

• ⇒ 𝑓−𝑞𝑙/2 ≤

𝑙 𝑜 ≤ 𝑙−3/2, since 𝑜 = 𝑃 𝑙2

• ⇒ 𝑙 ≥ 𝑞−1 ln 𝑙 = 𝑜1/2 ln 𝑙 ⇒ 𝑜 = 𝑃

𝑙 ln 𝑙 2

Wrapping Things Up Neatly

Proof

• Upper bound from Erdős-Szekeres
• Lower bound:
• Choose 𝑙 ≥ 20 𝑜 ln 𝑜, 𝑧 =

𝑜 𝑙 −1, 𝑦 = 1 9𝑜−3/2 and 𝑞 = 1 3 𝑜

• Substitute values into Theorem 4.2.3

∎

Corollary 4.2.4 As 𝑙 → ∞, we have Ω 𝑙 ln 𝑙

2

= 𝑆 3, 𝑙 = 𝑃 𝑙2 .

Any questions?

§3 Proving the Local Lemma

Chapter 4: The Lovász Local Lemma The Probabilistic Method

Theorem 4.1.1 (Symmetric Lovász Local Lemma; Erdős-Lovász, 1975) Let 𝐹1, 𝐹2, … , 𝐹𝑛 be events such that each event 𝐹𝑗 is mutually independent of all but at most 𝑒 of the other events, and ℙ 𝐹𝑗 ≤ 𝑞 for all 𝑗. If 𝑓𝑞 𝑒 + 1 ≤ 1, then ℙ ∩𝑗 𝐹𝑗

𝑑 > 0.

The Local Lemmas

• This follows easily from the general statement

Theorem 4.2.2 (Lovász Local Lemma; Erdős-Lovász, 1975) Let 𝐹1, 𝐹2, … , 𝐹𝑛 be events with a dependency digraph 𝐸. If there are 𝑦𝑗 ∈ 0,1 such that ℙ 𝐹𝑗 ≤ 𝑦𝑗 ς 𝑗,𝑘 ∈𝐸 1 − 𝑦𝑘 for all 𝑗 ∈ [𝑛], then ℙ ∩𝑗 𝐹𝑗

𝑑 ≥ ς𝑗 1 − 𝑦𝑗 .

• Recall the symmetric version

Deducing the Symmetric Statement

Proof

• For each event 𝐹𝑗, set 𝑦𝑗 =

1 𝑒+1

• Then 𝑦𝑗 ς 𝑗,𝑘 ∈𝐸 1 − 𝑦𝑘 =

1 𝑒+1 ς 𝑗,𝑘 ∈𝐸 1 − 1 𝑒+1 ≥ 1 𝑒+1 1 − 1 𝑒+1 𝑒

• For all 𝑒 ≥ 1, 1 −

1 𝑒+1 𝑒

≥ 𝑓−1

• ⇒ 𝑦𝑗 ς 𝑗,𝑘 ∈𝐸 1 − 𝑦𝑘 ≥

1 𝑓 𝑒+1 ≥ 𝑞

• ⇒ ℙ 𝐹𝑗 ≤ 𝑞 ≤ 𝑦𝑗 ς 𝑗,𝑘 ∈𝐸 1 − 𝑦𝑘 ⇒ ℙ ∩𝑗 𝐹𝑗 ≥ 1 −

1 𝑒+1 𝑛

> 0 ∎

Theorem 4.1.1 (Symmetric Lovász Local Lemma; Erdős-Lovász, 1975) Let 𝐹1, 𝐹2, … , 𝐹𝑛 be events such that each event 𝐹𝑗 is mutually independent of all but at most 𝑒 of the other events, and ℙ 𝐹𝑗 ≤ 𝑞 for all 𝑗. If 𝑓𝑞 𝑒 + 1 ≤ 1, then ℙ ∩𝑗 𝐹𝑗

𝑑 > 0.

Proving the General Statement

Chain rule

• ℙ ∩𝑗=1

𝑛

𝐹𝑗

𝑑 = ς𝑗=1 𝑛 ℙ 𝐹𝑗 𝑑 ∩𝑘=1 𝑗−1 𝐹 𝑘 𝑑

= ς𝑗=1

𝑛

1 − ℙ 𝐹𝑗 ∩𝑘=1

𝑗−1 𝐹 𝑘 𝑑

New objective

• Suffices to show ℙ 𝐹𝑗

𝑑 ∩𝑘=1 𝑗−1 𝐹 𝑘 𝑑 ≥ 1 − 𝑦𝑗 for each 𝑗 ∈ 𝑛

Theorem 4.2.2 (Lovász Local Lemma; Erdős-Lovász, 1975) Let 𝐹1, 𝐹2, … , 𝐹𝑛 be events with a dependency digraph 𝐸. If there are 𝑦𝑗 ∈ 0,1 such that ℙ 𝐹𝑗 ≤ 𝑦𝑗 ς 𝑗,𝑘 ∈𝐸 1 − 𝑦𝑘 for all 𝑗 ∈ [𝑛], then ℙ ∩𝑗 𝐹𝑗

𝑑 ≥ ς𝑗 1 − 𝑦𝑗 .

The Irrelevance of Order

Objective

• For each 𝑗 ∈ 𝑛 , ℙ 𝐹𝑗

𝑑 ∩𝑘=1 𝑗−1 𝐹 𝑘 𝑑 ≥ 1 − 𝑦𝑗

Reordering events

• If we reorder the events, the conditions do not change
• ⇒ the event 𝐹𝑗 could be preceded by any subset of the other events
• ⇒ we can hope that more is true

Newer objective

• For each 𝑗 ∈ 𝑛 and 𝑇 ⊆ 𝑛 ∖ 𝑗 , ℙ 𝐹𝑗

𝑑 ∩𝑘∈𝑇 𝐹 𝑘 𝑑 ≥ 1 − 𝑦𝑗

Conditional Probabilities

Objective

• For each 𝑗 ∈ 𝑛 and 𝑇 ⊆ 𝑛 ∖ 𝑗 , ℙ 𝐹𝑗

𝑑 ∩𝑘∈𝑇 𝐹 𝑘 𝑑 ≥ 1 − 𝑦𝑗

Independence

• We know 𝐹𝑗 is independent of some of the 𝐹

𝑘

• Conditioning on these events should be irrelevant
• Partition events
• Let 𝑇1 = 𝑘 ∈ 𝑇: 𝑗, 𝑘 ∈ 𝐸
• Let 𝑇2 = 𝑘 ∈ 𝑇: 𝑗, 𝑘 ∉ 𝐸

Rewriting the probability

• ℙ 𝐹𝑗

𝑑 ∩𝑘∈𝑇 𝐹 𝑘 𝑑 = 1 − ℙ 𝐹𝑗 ∩𝑘∈𝑇 𝐹 𝑘 𝑑 = 1 − ℙ 𝐹𝑗 ∩ ∩ℓ∈𝑇1 𝐹ℓ 𝑑

∩𝑘∈𝑇2 𝐹

𝑘 𝑑 ℙ ∩ℓ∈𝑇1 𝐹ℓ 𝑑 ∩𝑘∈𝑇2 𝐹 𝑘 𝑑

Simplifying the Numerator

Recall

• 𝑇1 = 𝑘 ∈ 𝑇: 𝑗, 𝑘 ∈ 𝐸
• 𝑇2 = 𝑘 ∈ 𝑇: 𝑗, 𝑘 ∉ 𝐸
• ℙ 𝐹𝑗 ∩𝑘∈𝑇 𝐹

𝑘 𝑑) = ℙ 𝐹𝑗 ∩ ∩ℓ∈𝑇1 𝐹ℓ 𝑑

∩𝑘∈𝑇2 𝐹

𝑘 𝑑 ℙ ∩ℓ∈𝑇1 𝐹ℓ 𝑑 ∩𝑘∈𝑇2 𝐹 𝑘 𝑑

Numerator

• 𝐹𝑗 ∩ ∩ℓ∈𝑇1 𝐹ℓ

𝑑 ⊆ 𝐹𝑗

• ⇒ ℙ 𝐹𝑗 ∩ ∩ℓ∈𝑇1 𝐹ℓ

𝑑 | ∩𝑘∈𝑇2 𝐹 𝑘 𝑑 ≤ ℙ 𝐹𝑗 ∩𝑘∈𝑇2 𝐹 𝑘 𝑑

• 𝐹𝑗 is mutually independent of the events in 𝑇2
• ⇒ ℙ 𝐹𝑗 ∩𝑘∈𝑇2 𝐹

𝑘 𝑑 = ℙ 𝐹𝑗

• Assumption: ℙ 𝐹𝑗 ≤ 𝑦𝑗 ς𝑘: 𝑗,𝑘 ∈𝐸 1 − 𝑦𝑘

Simplifying the Denominator

Objective

• For each 𝑗 ∈ 𝑛 and 𝑇 ⊆ 𝑛 ∖ 𝑗 , ℙ 𝐹𝑗

𝑑 ∩𝑘∈𝑇 𝐹 𝑘 𝑑 ≥ 1 − 𝑦𝑗

Denominator: ℙ ∩ℓ∈𝑇1 𝐹ℓ

𝑑 ∩𝑘∈𝑇2 𝐹 𝑘 𝑑

• Chain rule
• ℙ ∩ℓ∈𝑇1 𝐹ℓ

𝑑 ∩𝑘∈𝑇2 𝐹 𝑘 𝑑 = ςℓ∈𝑇1 ℙ 𝐹ℓ 𝑑

∩𝑠∈𝑇1,𝑠<ℓ 𝐹𝑠

𝑑 ∩ ∩𝑘∈𝑇2 𝐹 𝑘 𝑑

• Let 𝑈

ℓ = 𝑠 ∈ 𝑇1: 𝑠 < ℓ ∪ 𝑇2

• Apply the objective
• ⇒ ℙ 𝐹ℓ

𝑑 ∩𝑘∈𝑈ℓ 𝐹 𝑘 𝑑 ≥ 1 − 𝑦ℓ

• Substitute in
• ⇒ ℙ ∩ℓ∈𝑇1 𝐹ℓ

𝑑 ∩𝑘∈𝑇2 𝐹 𝑘 𝑑 ≥ ςℓ∈𝑇1 1 − 𝑦ℓ

• 𝑇1 ⊆ 𝑘: 𝑗, 𝑘 ∈ 𝐸
• ⇒ ℙ ∩ℓ∈𝑇1 𝐹ℓ

𝑑 ∩𝑘∈𝑇2 𝐹 𝑘 𝑑 ≥ ς𝑘: 𝑗,𝑘 ∈𝐸 1 − 𝑦𝑘

Achieving Our Objective

Objective

• For each 𝑗 ∈ 𝑛 and 𝑇 ⊆ 𝑛 ∖ 𝑗 , ℙ 𝐹𝑗

𝑑 ∩𝑘∈𝑇 𝐹 𝑘 𝑑 ≥ 1 − 𝑦𝑗

Recall

• ℙ 𝐹𝑗

𝑑 ∩𝑘∈𝑇 𝐹 𝑘 𝑑 = 1 − ℙ 𝐹𝑗 ∩ ∩ℓ∈𝑇1 𝐹ℓ 𝑑 ∩𝑘∈𝑇2 𝐹 𝑘 𝑑 ℙ ∩ℓ∈𝑇1 𝐹ℓ 𝑑 ∩𝑘∈𝑇2 𝐹 𝑘 𝑑

• ℙ 𝐹𝑗 ∩ ∩ℓ∈𝑇1 𝐹ℓ

𝑑

∩𝑘∈𝑇2 𝐹

𝑘 𝑑 ≤ ℙ 𝐹𝑗 ≤ 𝑦𝑗 ς𝑘: 𝑗,𝑘 ∈𝐸 1 − 𝑦𝑘

• ℙ ∩ℓ∈𝑇1 𝐹ℓ

𝑑 ∩𝑘∈𝑇2 𝐹 𝑘 𝑑 ≥ ς𝑘: 𝑗,𝑘 ∈𝐸 1 − 𝑦𝑘

• ⇒ ℙ 𝐹𝑗

𝑑 ∩𝑘∈𝑇 𝐹 𝑘 𝑑 ≥ 1 − 𝑦𝑗

∎

Circular logic

• We used the objective to lower bound the denominator

Induction to the Rescue

Objective

• For each 𝑗 ∈ 𝑛 and 𝑇 ⊆ 𝑛 ∖ 𝑗 , ℙ 𝐹𝑗

𝑑 ∩𝑘∈𝑇 𝐹 𝑘 𝑑 ≥ 1 − 𝑦𝑗

The issue

• Used the objective when bounding the denominator in the proof
• Parameters
• 𝑗 = ℓ ∈ 𝑇1
• 𝑇 = 𝑈

ℓ = 𝑠 ∈ 𝑇1: 𝑠 < ℓ ∪ 𝑇2

The fix

• Size of conditioned set
• We have 𝑈

ℓ ≤ 𝑇1 − 1 + 𝑇2 < 𝑇

• ⇒ when proving the objective for a set 𝑇, only require it for smaller subsets
• Apply induction on 𝑇
• Base case: 𝑇 = ∅ is trivial

Proof Recap

Ideas

• Chain rule: probability of intersection is product of conditional probabilities
• Prove ℙ 𝐹𝑗

𝑑 ∩𝑘∈𝑇 𝐹 𝑘 𝑑 ≥ 1 − 𝑦𝑗 by induction on 𝑇

• Separate conditioned events by dependence of 𝐹𝑗
• Simplify resulting expression by bounding the numerator
• Apply induction hypothesis to the denominator
• Substituting into chain rule gives result

Theorem 4.2.2 (Lovász Local Lemma; Erdős-Lovász, 1975) Let 𝐹1, 𝐹2, … , 𝐹𝑛 be events with a dependency digraph 𝐸. If there are 𝑦𝑗 ∈ 0,1 such that ℙ 𝐹𝑗 ≤ 𝑦𝑗 ς 𝑗,𝑘 ∈𝐸 1 − 𝑦𝑘 for all 𝑗 ∈ [𝑛], then ℙ ∩𝑗 𝐹𝑗

𝑑 ≥ ς𝑗 1 − 𝑦𝑗 .

Any questions?

§4 Latin Transversals

Chapter 4: The Lovász Local Lemma The Probabilistic Method

Latin Squares

Applications

• Experimental design
• Tournament scheduling
• Games and recreation
• Algebra
• Cayley table of a group is a Latin

square

Definition 4.4.1 (Latin square) A Latin square of order 𝑜 is an 𝑜 × 𝑜 array with entries from 𝑜 such that each symbol appears exactly once in each row and column.

1 2 3 4 5 2 3 4 5 1 3 4 5 1 2 4 5 1 2 3 5 1 2 3 4

Latin Transversals

Example

• Conducting a survey
• Public divided by metrics
• Age
• Height
• No. combinatorics courses taken
• Want a fair sample
• No group is overrepresented

Definition 4.4.2 (Latin transversal) Given an 𝑛 × 𝑜 array with entries in ℕ, a transversal is a selection of cells without any repeated row, column or symbol.

1 2 3 4 5 2 3 4 5 1 3 4 5 1 2 4 5 1 2 3 5 1 2 3 4

Latin Transversals

Example

• Conducting a survey
• Public divided by metrics
• Age
• Height
• No. combinatorics courses taken
• Want a fair sample
• No group is overrepresented

Definition 4.4.2 (Latin transversal) Given an 𝑛 × 𝑜 array with entries in ℕ, a transversal is a selection of cells without any repeated row, column or symbol.

1 2 3 4 5 2 3 4 5 1 3 4 5 1 2 4 5 1 2 3 5 1 2 3 4

Proposition 4.4.3 Any Latin square of order 𝑜 contains a transversal of size at least

𝑜 3.

An Extremal Problem

Large transversals?

• How large a transversal must a Latin square of order 𝑜 contain?

Proof

• Build a transversal greedily
• There are a total of 𝑜2 cells
• Each cell clashes with 3 𝑜 − 1 other cells
• Those in the same row, column or with the same symbol
• ⇒ we can select at least

𝑜2 3 𝑜−1 +1 cells before we run out

∎

An Upper Bound to the Extremal Problem

Proof

• Let 𝑜 = 2𝑙 and 𝑀 be the Cayley table of ℤ2𝑙
• 𝑀 𝑗, 𝑘 ≔ 𝑗 + 𝑘 (mod 2𝑙)
• Suppose we have a transversal
• Chosen cells:

𝑗, 𝜌 𝑗 : 𝑗 ∈ 2𝑙 for some permutation 𝜌 ∈ 𝑇2𝑙

• Rows, columns and symbols range over [2𝑙]
• ⇒ sum is 2𝑙

2

= 𝑙 2𝑙 − 1 ≡ 𝑙 (mod 2𝑙)

• But summing symbols = summing rows and columns:
• 𝑙 ≡ σ𝑗 𝑀 𝑗, 𝜌 𝑗

= σ𝑗 𝑗 + 𝜌 𝑗 = σ𝑗 𝑗 + σ𝑗 𝜌 𝑗 ≡ 𝑙 + 𝑙 ≡ 0 ≢ 𝑙 (mod 2𝑙) ∎

Proposition 4.4.4 For every even 𝑜 ∈ ℕ, there is a Latin square of order 𝑜 without a transversal of size 𝑜.

Conjecture 4.4.5 (Ryser-Brualdi-Stein, 1967+) Every Latin square of order 𝑜 admits a transversal of size 𝑜 − 1.

A Daring Conjecture

Odd orders

• Ryser conjectured that odd Latin squares have full transversals

Theorem 4.4.6 (Erdős-Spencer, 1991) Let 𝐵 be an 𝑜 × 𝑜 array with entries in ℕ. If no symbol appears more than

𝑜−1 4𝑓 times in 𝐵, then 𝐵 admits a transversal of size 𝑜.

Comparison to the conjecture

• Weak: Latin squares have each symbol appearing 𝑜 times
• Strong: In theorem, no restrictions on row or column repetitions!

Proof Framework

Goal

• Show that a random permutation can give a full transversal

Probability space

• Choose 𝜌 ∈ 𝑇𝑜 uniformly at random
• Potential transversal

𝑗, 𝜌 𝑗 : 𝑗 ∈ 𝑜

Bad events

• Chosen cells from distinct rows and columns by construction
• Only way to fail: repeat a symbol
• How do we define events to capture this?

Defining Failure

Events by symbol

• For each symbol 𝑗 ∈ ℕ appearing in 𝐵, define an event
• 𝐹𝑗 = two cells with the symbol 𝑗 are selected
• 𝜌 gives a transversal = ∩𝑗 𝐹𝑗

𝑑

Probabilities

• ℙ 𝐹𝑗 depends very much on structure of 𝐵
• If all 𝑗-entries are in the same row ⇒ ℙ 𝐹𝑗 = 0
• If Ω 𝑜 are on a diagonal ⇒ ℙ 𝐹𝑗 = Ω 1
• Expected number of events
• Hard to compute
• Could grow linearly

Many dependencies

• Different symbols that appear in the same row/column are dependent

Redefining Failure

Events by rows

• For 𝑗, 𝑘 ∈

𝑜 2 , define 𝐹𝑗,𝑘 = 𝐵𝑗,𝜌 𝑗 = 𝐵𝑘,𝜌 𝑘

• 𝜌 gives a transversal = ∩ 𝑗,𝑘 𝐹𝑗,𝑘

𝑑

Same issues as before

• Probabilities depend heavily on array 𝐵
• Expected number of events can be Ω 𝑜
• High dependence
• Knowing 𝐹𝑗,𝑘 occurs could tell us which elements are selected
• Affects distribution in other rows

Reredefining Failure

Events by cells

• Identify exactly where symbols are repeated
• For every pair of cells 𝑗, 𝑘 and 𝑗′, 𝑘′ with
• 𝐵 𝑗, 𝑘 = 𝐵 𝑗′, 𝑘′
• 𝑗 ≠ 𝑗′
• 𝑘 ≠ 𝑘′

define the event 𝐹𝑗,𝑘,𝑗′,𝑘′ = 𝜌 𝑗 = 𝑘 ∩ 𝜌 𝑗′ = 𝑘′

• 𝜌 gives a transversal =∩𝑗,𝑘,𝑗′,𝑘′ 𝐹𝑗,𝑘,𝑗′,𝑘′

Probabilities

• Each event occurs with probability

1 𝑜 𝑜−1

• Number of events depends on structure of 𝐵, but at most

1 2 ⋅ 𝑜2 ⋅ 𝑜−1 4𝑓

• ⇒ expected number of bad events can still be Ω 𝑜

Examining Independence

Neighbouring events

• Consider events 𝐹𝑗,𝑘,𝑗′,𝑘′ and 𝐹𝑞,𝑟,𝑞′,𝑟′
• Correspond to cells (𝑗, 𝑘), 𝑗′, 𝑘′ , 𝑞, 𝑟 and 𝑞′, 𝑟′
• Only one cell selected from each row/column
• ⇒ if 𝑗, 𝑗′ ∩ 𝑞, 𝑞′ ≠ ∅ or 𝑘, 𝑘′ ∩ 𝑟, 𝑟′ ≠ ∅, no independence

Non-neighbouring events

• Permutation restrictions are global in nature
• ⇒ information travels even when not sharing a row or column
• ⇒ cannot expect any independence

Who Needs Independence?

Proof of Lovász Local Lemma

• Used independence when bounding the numerator
• ℙ 𝐹𝑗 ∩𝑘∈𝑇2 𝐹

𝑘 𝑑 = ℙ 𝐹𝑗 ≤ 𝑦𝑗 ς𝑘: 𝑗,𝑘 ∈𝐸 1 − 𝑦𝑘

• (First) equality by independence
• (Second) inequality by assumption

Weakening condition

• We only use the inequality
• Could skip the intermediate equality
• ⇒ suffices to have ℙ 𝐹𝑗 ∩𝑘∈𝑇 𝐹

𝑘 𝑑 ≤ 𝑦𝑗 ς𝑘: 𝑗,𝑘 ∈𝐸 1 − 𝑦𝑘 for all 𝑗 ∈ [𝑛] and

𝑇 ⊆ 𝑛 ∖ 𝑘: 𝑗, 𝑘 ∈ 𝐸

Theorem 4.4.7 (Lopsided Lovász Local Lemma; Erdős-Spencer, 1991) Let 𝐹1, 𝐹2, … , 𝐹𝑛 be events in a probability space, let 𝑦1, 𝑦2, … , 𝑦𝑛 ∈ 0,1 , and let 𝐸 be a directed graph on the vertices 𝑛 . If, for every 𝑗 ∈ 𝑛 and 𝑇 ⊆ 𝑛 ∖ 𝑘: 𝑗, 𝑘 ∈ 𝐸 ∪ 𝑗 , we have ℙ 𝐹𝑗 ∩𝑘∈𝑇 𝐹

𝑘 𝑑 ≤ 𝑦𝑗

ෑ

𝑘: 𝑗,𝑘 ∈𝐸

1 − 𝑦𝑘 , then ℙ ∩𝑗 𝐹𝑗

𝑑 ≥ ς𝑗 1 − 𝑦𝑗 > 0.

The Lopsided Lovász Local Lemma

Strengthened result

• Using this observation, we can weaken the requirement in the Local Lemma
• Following version is useful in spaces with limited independence
• Most pairs of events should be positively correlated

Verifying the Condition

Proof idea

• Without loss of generality, may assume 𝑗 = 𝑘 = 1, 𝑗′ = 𝑘′ = 2
• Restrict to permutations 𝜌 satisfying ∩ 𝑞,𝑟,𝑞′,𝑟′ ∈𝑇 𝐹𝑞,𝑟,𝑞′,𝑟′

𝑑

• By modifying permutations, show that number of permutations with 𝜌 1 =

𝑠 and 𝜌 2 = 𝑡 is minimised (for 𝑠 ≠ 𝑡) when 𝑠 = 1 and 𝑡 = 2

Lemma 4.4.8 Let the events 𝐹𝑗,𝑘,𝑗′,𝑘′ be as previously defined, and let 𝑇 be a set of indices for events involving cells not sharing a row or column with 𝑗, 𝑘

• r 𝑗′, 𝑘′ . Then

ℙ 𝐹𝑗,𝑘,𝑗′,𝑘′ ∩ 𝑞,𝑟,𝑞′,𝑟′ ∈𝑇 𝐹𝑞,𝑟,𝑞′,𝑟′

𝑑

≤ 1 𝑜 𝑜 − 1 .

Verifying the Condition - Notation

Objective

• ℙ 𝐹1,1,2,2 ∩ 𝑞,𝑟,𝑞′,𝑟′ ∈𝑇 𝐹𝑞,𝑟,𝑞′,𝑟′

𝑑

≤

1 𝑜 𝑜−1

Notation

• Call 𝜌 “good” if 𝜌 ∈∩ 𝑞,𝑟,𝑞′,𝑟′ ∈𝑇 𝐹𝑞,𝑟,𝑞′,𝑟′

𝑑

• Let 𝑄

𝑠,𝑡 = {𝜌 good, 𝜌 1 = 𝑠, 𝜌 2 = 𝑡}

• Goal: 𝑄

1,2 ≤ 𝑄 𝑠,𝑡 for all 𝑠, 𝑡 ∈ 𝑜 2, 𝑠 ≠ 𝑡

Setting up the proof

• Goal: Construct an injection 𝑄

1,2 ↪ 𝑄 𝑠,𝑡

• Case: 𝑠, 𝑡 ∉ 1,2 (others similar)
• Let 𝜌 ∈ 𝑄

1,2 and let 𝑦 = 𝜌−1 𝑠 , 𝑧 = 𝜌−1(𝑡)

Verifying the Condition - Proof

Goal

• Injection 𝑄

1,2 ↪ 𝑄 𝑠,𝑡

• Given: 𝜌 ∈ 𝑄

1,2, 𝜌 𝑦 = 𝑠, 𝜌 𝑧 = 𝑡

Switching

• Define new permutation 𝜌∗ ∈ 𝑄

𝑠,𝑡

• 𝜌∗ 𝑨 =

𝑠 if 𝑨 = 1 𝑡 if 𝑨 = 2 1 if 𝑨 = 𝑦 2 if 𝑨 = 𝑧 𝜌 𝑨 otherwise

• 𝜌∗ is good: only change cells in the first two rows or columns, avoiding 𝑇
• ⇒ 𝜌∗ ∈ 𝑄

𝑠,𝑡

• The map 𝜌 ↦ 𝜌∗ is injective

∎

Finding Large Transversals

Proof

• We will apply the Lopsided Lovász Local Lemma
• 𝐸: edges between (𝑗, 𝑘, 𝑗′, 𝑘′) and (𝑞, 𝑟, 𝑞′, 𝑟′) if the corresponding cells share

a row or column

• Each event is adjacent to at most d ≔ 4 ⋅ 𝑜 ⋅

𝑜−1 4𝑓 − 1 = n n−1 e

− 1 other events

• We set 𝑦𝑗,𝑘,𝑗′,𝑘′ =

1 𝑒+1 for each event

• ⇒ inequality reduces to 𝑓𝑞 𝑒 + 1 ≤ 1, as in symmetric case
• ℙ 𝐹𝑗,𝑘,𝑗′,𝑘′ =

1 𝑜 𝑜−1 = 1 e d+1 .

∎

Theorem 4.4.6 (Erdős-Spencer, 1991) Let 𝐵 be an 𝑜 × 𝑜 array with entries in ℕ. If no symbol appears more than

𝑜−1 4𝑓 times in 𝐵, then 𝐵 admits a transversal of size 𝑜.

Theorem (Keevash-Pokrovskiy-Sudakov-Yepremyan, 2020+) Every Latin square of order 𝑜 admits a transversal of size 𝑜 − 𝑃 log 𝑜 log log 𝑜 .

Ryser’s Conjecture

State of the art

• More involved probabilistic proofs bring us much closer to Ryser’s Conjecture

Theorem (Kwan, 2016+) Almost all Latin squares of order 𝑜 have a transversal of size 𝑜.

Any questions?