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Randomness in Computing L ECTURE 21 Last time Probabilistic method - - PowerPoint PPT Presentation

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Randomness in Computing L ECTURE 21 Last time Probabilistic method Sample and Modify The Second Moment Method Today Probabilistic method The Second Moment Method Conditional expectation inequality Lovasz Local Lemma

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SLIDE 1

4/9/2020

Randomness in Computing

LECTURE 21

Last time

  • Probabilistic method
  • Sample and Modify
  • The Second Moment Method

Today

  • Probabilistic method
  • The Second Moment Method
  • Conditional expectation

inequality

  • Lovasz Local Lemma

Sofya Raskhodnikova;Randomness in Computing

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SLIDE 2

Threshold behavior in random graphs 𝑯 ∼ 𝑯(𝒐, 𝒒)

For many properties 𝓠, there exists function 𝑔 π‘œ s.t.

1. when π‘ž β‰ͺ 𝑔 π‘œ , probability that 𝐻 has 𝓠 β†’ 0 as π‘œ β†’ ∞ 2. when π‘ž ≫ 𝑔 π‘œ , probability that 𝐻 has 𝓠 β†’ 1 as π‘œ β†’ ∞

(It holds for all nontrivial monotone properties.)

4/9/2020

Sofya Raskhodnikova; Randomness in Computing

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SLIDE 3

The 2𝐨𝐞 moment method

4/9/2020

Sofya Raskhodnikova; Randomness in Computing

Theorem

If π‘Œ is a random variable with 𝔽 π‘Œ > 0, then

Pr π‘Œ = 0 ≀ Var π‘Œ 𝔽 π‘Œ

2

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SLIDE 4

Example: having a 4-clique

Proof: Let π‘Œ = number of 4-cliques in 𝐻. For every subset 𝐷 of 4 nodes, let π‘Œπ· be the indicator for 𝐷 being a 𝐿4. 𝔽 π‘Œ = ෍

𝐷

𝔽 π‘Œπ· = π‘œ 4 β‹… π‘ž6 1. π‘ž = 𝑝(π‘œβˆ’2/3) π‘žβˆ— = Pr π‘Œ β‰₯ 1 ≀ 𝔽[X]

1

= 𝔽 π‘Œ ≀ π‘œ4

4! β‹… π‘ž6 = π‘œ4 4! β‹… 𝑝 π‘œβˆ’(2/3)β‹…6 = π‘œ4 4! β‹… 𝑝 π‘œβˆ’4 = 𝑝(1)

4/9/2020

Sofya Raskhodnikova; Randomness in Computing

Theorem

Let 𝐻 ∼ π»π‘œ,π‘ž and π‘žβˆ— = Pr 𝐻 has a 𝐿4 . 1. If π‘ž β‰ͺ π‘œβˆ’2/3, then π‘žβˆ— β†’ 0 as π‘œ β†’ ∞ 2. If π‘ž ≫ π‘œβˆ’2/3, then π‘žβˆ— β†’ 1 as π‘œ β†’ ∞

= 𝒑 π‘œβˆ’2/3 = 𝝏 π‘œβˆ’2/3 Markov

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SLIDE 5

Review question

What is the expected number of copies of this graph in 𝐻 ∼ π»π‘œ,π‘ž? A. π‘œ 4 β‹… π‘ž6

  • B. 4 π‘œ

4 β‹… π‘ž6

  • C. 4 π‘œ

4 β‹… π‘ž5(1 βˆ’ π‘ž)

  • D. 6 π‘œ

4 β‹… π‘ž5(1 βˆ’ π‘ž)

  • E. None of the above

4/9/2020

Sofya Raskhodnikova; Randomness in Computing

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SLIDE 6

Example: having a 4-clique

Proof: Expected number of 4-cliques: 𝔽 π‘Œ = π‘œ 4 β‹… π‘ž6 2. π‘ž = πœ•(π‘œβˆ’2/3) 𝔽 π‘Œ β†’ ∞ as π‘œ β†’ ∞ Goal: Show Var π‘Œ β‰ͺ 𝔽 π‘Œ

2

Var π‘Œ = Var ෍

𝐷

π‘Œπ· = ෍

𝐷

Var π‘Œπ· + ෍

𝐷≠𝐸

Cov[π‘Œπ·, π‘ŒπΈ] Var π‘Œπ· = 𝔽 π‘Œπ·

2 βˆ’ (𝔽 π‘Œπ· )2= 𝔽 π‘Œπ· βˆ’ (𝔽 π‘Œπ· )2= π‘ž6 βˆ’ π‘ž12 ≀ π‘ž6

෍

𝐷

Var π‘Œπ· ≀ π‘œ 4 β‹… π‘ž6 = 𝑃(π‘œ4π‘ž6)

4/9/2020

Sofya Raskhodnikova; Randomness in Computing

Theorem

Let 𝐻 ∼ π»π‘œ,π‘ž and π‘žβˆ— = Pr 𝐻 has a 𝐿4 . 1. If π‘ž β‰ͺ π‘œβˆ’2/3, then π‘žβˆ— β†’ 0 as π‘œ β†’ ∞ 2. If π‘ž ≫ π‘œβˆ’2/3, then π‘žβˆ— β†’ 1 as π‘œ β†’ ∞

= 𝒑 π‘œβˆ’2/3 = 𝝏 π‘œβˆ’2/3 π‘«π’‘π’˜ 𝑍, π‘Ž = 𝔽 𝑍 βˆ’ πœˆπ‘ β‹… π‘Ž βˆ’ πœˆπ‘Ž = 𝔽 π‘π‘Ž βˆ’ πœˆπ‘πœˆπ‘Ž ≀ 𝔽 π‘π‘Ž

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SLIDE 7

Bounding the covariance

Case 1: |𝐷 ∩ 𝐸| is 0 or 1

4/9/2020

Sofya Raskhodnikova; Randomness in Computing

π‘«π’‘π’˜ 𝒀𝑫, 𝒀𝑬 ≀ 𝔽 𝒀𝑫 β‹… 𝒀𝑬

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SLIDE 8

Bounding the covariance

Case 2: 𝐷 ∩ 𝐸 = 2

4/9/2020

Sofya Raskhodnikova; Randomness in Computing

π‘«π’‘π’˜ 𝒀𝑫, 𝒀𝑬 ≀ 𝔽 𝒀𝑫 β‹… 𝒀𝑬

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SLIDE 9

Bounding the covariance

Case 3: 𝐷 ∩ 𝐸 = 3

4/9/2020

Sofya Raskhodnikova; Randomness in Computing

π‘«π’‘π’˜ 𝒀𝑫, 𝒀𝑬 ≀ 𝔽 𝒀𝑫 β‹… 𝒀𝑬

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SLIDE 10

Putting it all together

  • Var π‘Œ ≀ Var σ𝐷 π‘Œπ· = σ𝐷 Var π‘Œπ· + σ𝐷≠𝐸 Cov π‘Œπ·, π‘ŒπΈ

= O(π‘œ4π‘ž6 + π‘œ6π‘ž11 + π‘œ5π‘ž9)

  • Pr π‘Œ = 0 ≀ Var π‘Œ

𝔽 π‘Œ

2

4/9/2020

Sofya Raskhodnikova; Randomness in Computing

Theorem

Let 𝐻 ∼ π»π‘œ,π‘ž and π‘žβˆ— = Pr 𝐻 has a 𝐿4 . 2. If π‘ž ≫ π‘œβˆ’2/3, then π‘žβˆ— β†’ 1 as π‘œ β†’ ∞

= 𝝏 π‘œβˆ’2/3

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SLIDE 11

Conditional Expectation Inequality

  • Note that the indicators π‘Œπ‘— need not be independent.

Proof: Let Y = α‰Š1/X if π‘Œ > 0;

  • therwise.

Then XY = α‰Š1 if π‘Œ > 0; 0 otherwise. Pr π‘Œ > 0 = 𝔽 π‘ŒY

4/9/2020

Sofya Raskhodnikova; Randomness in Computing

Theorem

Let π‘Œ = Οƒπ‘—βˆˆ[π‘œ] π‘Œπ‘—, where each π‘Œπ‘— is an indicator R.V. Then

Pr π‘Œ > 0 β‰₯ ෍

π‘—βˆˆ[π‘œ]

Pr π‘Œπ‘— = 1 𝔽 π‘Œ | π‘Œπ‘— = 1

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SLIDE 12

Conditional Expectation Inequality

Proof: Pr π‘Œ > 0 = 𝔽 π‘ŒY = 𝔽 ෍

π‘—βˆˆ[π‘œ]

π‘Œπ‘— Y = ෍

π‘—βˆˆ[π‘œ]

𝔽[π‘Œπ‘—π‘] = ෍

π‘—βˆˆ[π‘œ]

𝔽[π‘Œπ‘—π‘ π‘Œπ‘— = 1 β‹… Pr π‘Œπ‘— = 1 + ෍

π‘—βˆˆ[π‘œ]

𝔽[π‘Œπ‘—π‘ π‘Œπ‘— = 0 β‹… Pr π‘Œπ‘— = 0 = ෍

π‘—βˆˆ[π‘œ]

𝔽[𝑍 π‘Œπ‘— = 1 β‹… Pr π‘Œπ‘— = 1 = ෍

π‘—βˆˆ[π‘œ]

𝔽[1/π‘Œ π‘Œπ‘— = 1 β‹… Pr π‘Œπ‘— = 1

β‰₯ ෍

π‘—βˆˆ[π‘œ]

Pr π‘Œπ‘— = 1 𝔽[π‘Œ π‘Œπ‘— = 1

4/9/2020

Sofya Raskhodnikova; Randomness in Computing

Theorem

Let π‘Œ = Οƒπ‘—βˆˆ[π‘œ] π‘Œπ‘—, where each π‘Œπ‘— is an indicator R.V. Then

Pr π‘Œ > 0 β‰₯ ෍

π‘—βˆˆ[π‘œ]

Pr π‘Œπ‘— = 1 𝔽 π‘Œ | π‘Œπ‘— = 1 Y = α‰Š1/X if π‘Œ > 0;

  • therwise.

𝒀 = σ𝒀𝒋 Linearity of expectation Law of Total Expectation = 𝟏 𝒀𝒋 = 𝟐

By Jensen’s inequality for convex function π’ˆ π’š = 𝟐/π’š, 𝔽 1 π‘Œ β‰₯ 1 𝔽[π‘Œ]

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SLIDE 13

π‘³πŸ“ thm, part 2: Alternative proof

Proof: Recall: π‘Œπ· = the indicator for 𝐷 being a 𝐿4.

Pr π‘Œ > 0 β‰₯ ෍

C

Pr π‘ŒC = 1 𝔽 π‘Œ | π‘Œπ· = 1 = π‘œ 4 π‘ž6 𝔽 π‘Œ | π‘Œπ· = 1 𝔽 π‘Œ | π‘Œπ· = 1 = 𝔽[෍

𝐷′

π‘ŒCβ€² | π‘Œπ· = 1] = ෍

𝐷′

𝔽[π‘ŒCβ€² | π‘Œπ· = 1] = ෍

𝐷′

Pr[π‘ŒCβ€² = 1 | π‘Œπ· = 1] = 1 + π‘œ βˆ’ 4 4 π‘ž6 + 4 π‘œ βˆ’ 4 3 π‘ž6 + 6 π‘œ βˆ’ 4 2 π‘ž5 + 4 π‘œ βˆ’ 4 1 π‘ž3

4/9/2020

Sofya Raskhodnikova; Randomness in Computing

Theorem

Let 𝐻 ∼ π»π‘œ,π‘ž and π‘žβˆ— = Pr 𝐻 has a 𝐿4 . 2. If π‘ž ≫ π‘œβˆ’2/3, then π‘žβˆ— β†’ 1 as π‘œ β†’ ∞

= 𝝏 π‘œβˆ’2/3 Conditional Expectation Inequality Symmetry 𝒀 = σ𝒀𝑫′ Linearity of expectation 𝒀𝑫′ is a 0-1 R.V. 𝑫′ = 𝑫 𝑫′ ∩ 𝑫 = βˆ… |𝑫′ ∩ 𝑫| = 𝟐 |𝑫′ ∩ 𝑫| = πŸ‘ |𝑫′ ∩ 𝑫| = πŸ’

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SLIDE 14

Avoiding bad events

  • Let 𝐢1 and 𝐢2 be (bad) events over a common probability space.
  • Q. If Pr 𝐢1 < 1 and Pr 𝐢2 < 1, does it imply Pr 𝐢1 ∩ 𝐢2 > 0?

(Is it possible to avoid both events)?

  • A. Not necessarily. E.g., for a single coin flip, let 𝐢1 = 𝐼, 𝐢2 = π‘ˆ

Then Pr 𝐢1 = Pr 𝐢2 = 1/2. But Pr 𝐢1 ∩ 𝐢2 = 0

  • Q. What if 𝐢1 and 𝐢2 are independent?
  • A. Yes. Pr 𝐢1 ∩ 𝐢2 = Pr 𝐢1 β‹… Pr 𝐢2 > 0
  • Q. What if Pr 𝐢1 <

1 2 and Pr 𝐢2 < 1 2 (but 𝐢1, 𝐢2 are dependent)?

  • A. Yes. By Union Bound, Pr 𝐢1 βˆͺ 𝐢2 ≀ Pr 𝐢1 + Pr 𝐢2 < 1. So,

Pr 𝐢1 ∩ 𝐢2 > 0.

4/9/2020

Sofya Raskhodnikova; Randomness in Computing

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SLIDE 15

Lovasz Local Lemma (LLL)

LLL states that as long as

  • 1. bad events 𝐢1, … , πΆπ‘œ have small probability,
  • 2. they are not ``too dependent’’,

there is a non-zero probability of avoiding all of them.

  • A dependency graph for events 𝐢1, … , πΆπ‘œ is a graph with vertex

set [π‘œ] and edge set 𝐹, s.t. βˆ€π‘— ∈ π‘œ , event 𝐢𝑗 is mutually independent of all events 𝐢

π‘˜

𝑗, π‘˜ βˆ‰ 𝐹}.

4/9/2020

Sofya Raskhodnikova; Randomness in Computing

Lovasz Local Lemma

Let 𝐢1, … , πΆπ‘œ be events over a common sample space s.t. 1. max degree of the dependency graph of 𝐢1, … , πΆπ‘œ is at most 𝑒 2. βˆ€π‘— ∈ π‘œ , Pr 𝐢𝑗 ≀

1 𝑓(𝑒+1)

Then PrΪπ‘—βˆˆ π‘œ ΰ΄₯

𝐢𝑗 > 0

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SLIDE 16

Example: Points on a circle

11π‘œ points are placed on a circle and colored with π‘œ different colors, so that each color is applied to exactly 11 points. Prove: There exists a set of π‘œ points, all colored differently, such that no two points in the set are adjacent.

4/9/2020

Sofya Raskhodnikova; Randomness in Computing