Adaptivity helps for testing juntas Rocco Servedio, Li-Yang Tan, - - PowerPoint PPT Presentation

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Adaptivity helps for testing juntas Rocco Servedio, Li-Yang Tan, - - PowerPoint PPT Presentation

Dec 02, 2022 359 likes •1.95k views

Adaptivity helps for testing juntas Rocco Servedio, Li-Yang Tan, John Wright Columbia TTIC CMU Adaptivity helps for testing juntas Rocco Servedio, Li-Yang Tan, John Wright Columbia TTIC CMU (work done while I was visiting Columbia)

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Adaptivity helps for testing juntas

Rocco Servedio, Li-Yang Tan, John Wright

Columbia CMU TTIC

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Adaptivity helps for testing juntas

Rocco Servedio, Li-Yang Tan, John Wright

Columbia CMU TTIC

(work done while I was visiting Columbia)

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Juntas

f : {0,1}n → {0,1}

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Juntas

f : → {0,1}

1 1 1 1 1 1

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Juntas

f : → {0,1} k-junta: f only depends on k bits

1 1 1 1 1 1

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Juntas

f : → {0,1} k-junta: f only depends on k bits

1 1 1 1 1 1

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Juntas

f : → {0,1} k-junta: f only depends on k bits

1 1 1 1 1 1 (a 3-junta)

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Juntas

f : → {0,1} k-junta: f only depends on k bits (k = 1: f is a dictator)

1 1 1 1 1 1 (a 3-junta)

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Juntas

f : → {0,1} k-junta: f only depends on k bits (k = 1: f is a dictator)

Key question: how to tell if f is a k-junta?

1 1 1 1 1 1 (a 3-junta)

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Queries

Given: ability to make queries x → f(x)

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Queries

Given: ability to make queries x → f(x) Nonadaptive: fix queries in advance

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Queries

Given: ability to make queries x → f(x) Nonadaptive: fix queries in advance

x1, x2, …
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Queries

Given: ability to make queries x → f(x) Nonadaptive: fix queries in advance x1, x2, … → f(x1), f(x2), ...

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Queries

Given: ability to make queries x → f(x) Nonadaptive: fix queries in advance x1, x2, … → f(x1), f(x2), ... Adaptive: choose queries based on answers

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Queries

Given: ability to make queries x → f(x) Nonadaptive: fix queries in advance x1, x2, … → f(x1), f(x2), ... Adaptive: choose queries based on answers x1

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Queries

Given: ability to make queries x → f(x) Nonadaptive: fix queries in advance x1, x2, … → f(x1), f(x2), ... Adaptive: choose queries based on answers x1 → f(x1)

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Queries

Given: ability to make queries x → f(x) Nonadaptive: fix queries in advance x1, x2, … → f(x1), f(x2), ... Adaptive: choose queries based on answers x1 → f(x1), x2

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Queries

Given: ability to make queries x → f(x) Nonadaptive: fix queries in advance x1, x2, … → f(x1), f(x2), ... Adaptive: choose queries based on answers x1 → f(x1), x2 → f(x2)

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Queries

Given: ability to make queries x → f(x) Nonadaptive: fix queries in advance x1, x2, … → f(x1), f(x2), ... Adaptive: choose queries based on answers x1 → f(x1), x2 → f(x2), x3

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Queries

Given: ability to make queries x → f(x) Nonadaptive: fix queries in advance x1, x2, … → f(x1), f(x2), ... Adaptive: choose queries based on answers x1 → f(x1), x2 → f(x2), x3 → f(x3)

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Queries

Given: ability to make queries x → f(x) Nonadaptive: fix queries in advance x1, x2, … → f(x1), f(x2), ... Adaptive: choose queries based on answers x1 → f(x1), x2 → f(x2), x3 → f(x3), ....

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Property testing

Goal: distinguish whether (unknown) f is

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Property testing

Goal: distinguish whether (unknown) f is

  • (Yes): a k-junta
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Property testing

Goal: distinguish whether (unknown) f is

  • (Yes): a k-junta
  • (No): not ɛ-close to a k-junta
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Property testing

Goal: distinguish whether (unknown) f is

  • (Yes): a k-junta
  • (No): not ɛ-close to a k-junta
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Property testing

Goal: distinguish whether (unknown) f is

  • (Yes): a k-junta
  • (No): not ɛ-close to a k-junta

f:

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Property testing

Goal: distinguish whether (unknown) f is

  • (Yes): a k-junta
  • (No): not ɛ-close to a k-junta

f: (ɛ-fraction)

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Property testing

Goal: distinguish whether (unknown) f is

  • (Yes): a k-junta
  • (No): not ɛ-close to a k-junta

f: (ɛ-fraction)

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Property testing

Goal: distinguish whether (unknown) f is

  • (Yes): a k-junta
  • (No): not ɛ-close to a k-junta

f: (ɛ-fraction) (k-junta)

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Property testing

Goal: distinguish whether (unknown) f is

  • (Yes): a k-junta
  • (No): not ɛ-close to a k-junta
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Property testing

Goal: distinguish whether (unknown) f is

  • (Yes): a k-junta
  • (No): not ɛ-close to a k-junta
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Property testing

Goal: distinguish whether (unknown) f is

  • (Yes): a k-junta
  • (No): ɛ-far from all k-juntas
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Property testing

Goal: distinguish whether (unknown) f is

  • (Yes): a k-junta
  • (No): ɛ-far from all k-juntas

Resources: Minimize query count q

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Property testing

Goal: distinguish whether (unknown) f is

  • (Yes): a k-junta
  • (No): ɛ-far from all k-juntas

Resources: Minimize query count q in terms of k and ɛ

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Property testing

Goal: distinguish whether (unknown) f is

  • (Yes): a k-junta
  • (No): ɛ-far from all k-juntas

Resources: Minimize query count q in terms of k and ɛ (no dependence on n!)

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Junta testing motivation

  • Boolean function version of finding a low

rank model for high dimensional data

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Junta testing motivation

  • Boolean function version of finding a low

rank model for high dimensional data

  • For k = 1, equivalent to dictatorship

testing, a basic topic in hardness of approximation

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Junta testing motivation

  • Boolean function version of finding a low

rank model for high dimensional data

  • For k = 1, equivalent to dictatorship

testing, a basic topic in hardness of approximation

  • One of the most basic Boolean function

properties.

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Prior work

adaptive nonadaptive

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Prior work

adaptive nonadaptive

[Bla08] O(k3/2log(k)3/ɛ)

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Prior work

adaptive nonadaptive

[Bla08] O(k3/2log(k)3/ɛ) [Bla08] (k/(ɛ log(k/ɛ))

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Prior work

adaptive nonadaptive

[Bla08] O(k3/2log(k)3/ɛ) [Bla08] (k/(ɛ log(k/ɛ)) [BGSMdW13] (k log(k))

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Prior work

adaptive nonadaptive

[Bla08] O(k3/2log(k)3/ɛ) O(k log(k) + k/ɛ) [Bla09] [Bla08] (k/(ɛ log(k/ɛ)) [BGSMdW13] (k log(k))

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Prior work

adaptive nonadaptive

[Bla08] O(k3/2log(k)3/ɛ) [CG04] (k) [Bla08] (k/(ɛ log(k/ɛ)) [BGSMdW13] (k log(k)) O(k log(k) + k/ɛ) [Bla09]

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Prior work

adaptive nonadaptive

[Bla08] O(k3/2log(k)3/ɛ) [CG04] (k) [Bla08] (k/(ɛ log(k/ɛ)) [BGSMdW13] (k log(k)) O(k log(k) + k/ɛ) [Bla09]

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[Bla08] (k/(ɛ log(k/ɛ)) [BGSMdW13] (k log(k)) O(k log(k) + k/ɛ) [Bla09]

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Annoyance: adaptive UB ≥ nonadaptive LB

[Bla08] (k/(ɛ log(k/ɛ)) [BGSMdW13] (k log(k)) O(k log(k) + k/ɛ) [Bla09]

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Annoyance: adaptive UB ≥ nonadaptive LB [Bla09]: does adaptivity even help?

[Bla08] (k/(ɛ log(k/ɛ)) [BGSMdW13] (k log(k)) O(k log(k) + k/ɛ) [Bla09]

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Annoyance: adaptive UB ≥ nonadaptive LB [Bla09]: does adaptivity even help?

[Bla08] (k/(ɛ log(k/ɛ)) [BGSMdW13] (k log(k)) O(k log(k) + k/ɛ) [Bla09]

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Annoyance: adaptive UB ≥ nonadaptive LB [Bla09]: does adaptivity even help?

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Annoyance: adaptive UB ≥ nonadaptive LB [Bla09]: does adaptivity even help? It should: [Bla09]’s O(k log(k) + k/ɛ) adaptive algorithm uses binary search.

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Annoyance: adaptive UB ≥ nonadaptive LB [Bla09]: does adaptivity even help? It should: [Bla09]’s O(k log(k) + k/ɛ) adaptive algorithm uses binary search. Adaptivity also helps for testing signed majority functions, read-once width-two OBDDs.

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Annoyance: adaptive UB ≥ nonadaptive LB [Bla09]: does adaptivity even help? It should: [Bla09]’s O(k log(k) + k/ɛ) adaptive algorithm uses binary search. Adaptivity also helps for testing signed majority functions, read-once width-two

  • OBDDs. Adaptive algos use binary search.
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Annoyance: adaptive UB ≥ nonadaptive LB [Bla09]: does adaptivity even help? It should: [Bla09]’s O(k log(k) + k/ɛ) adaptive algorithm uses binary search. Adaptivity also helps for testing signed majority functions, read-once width-two

  • OBDDs. Adaptive algos use binary search.
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Annoyance: adaptive UB ≥ nonadaptive LB [Bla09]: does adaptivity even help?

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Annoyance: adaptive UB ≥ nonadaptive LB [Bla09]: does adaptivity even help?

[Bla08] (k/(ɛ log(k/ɛ)) [BGSMdW13] (k log(k)) O(k log(k) + k/ɛ) [Bla09]

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Annoyance: adaptive UB ≥ nonadaptive LB [Bla09]: does adaptivity even help?

[Bla08] (k/(ɛ log(k/ɛ)) [BGSMdW13] (k log(k)) O(k log(k) + k/ɛ) [Bla09]

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Annoyance: adaptive UB ≥ nonadaptive LB [Bla09]: does adaptivity even help? Our work: yes it does

[Bla08] (k/(ɛ log(k/ɛ)) [BGSMdW13] (k log(k)) O(k log(k) + k/ɛ) [Bla09]

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Annoyance: adaptive UB ≥ nonadaptive LB [Bla09]: does adaptivity even help? Our work: yes it does. new nonadaptive LB

[Bla08] (k/(ɛ log(k/ɛ)) [BGSMdW13] (k log(k)) O(k log(k) + k/ɛ) [Bla09]

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Main result

Any nonadaptive algorithm requires

k log(k) ɛc log(log(k)/ɛc)

queries q ≥

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Main result

Any nonadaptive algorithm requires

k log(k) ɛc log(log(k)/ɛc)

queries (for any 0 < c < 1). q ≥

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Main result

Any nonadaptive algorithm requires

k log(k) ɛc log(log(k)/ɛc)

queries (for any 0 < c < 1). Set ɛ = 1/log(k). q ≥

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Main result

Any nonadaptive algorithm requires

k log(k) ɛc log(log(k)/ɛc)

queries (for any 0 < c < 1). Set ɛ = 1/log(k). Adaptive UB = O(k log(k) + k/ɛ) q ≥

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Main result

Any nonadaptive algorithm requires

k log(k) ɛc log(log(k)/ɛc)

queries (for any 0 < c < 1). q ≥ Set ɛ = 1/log(k). Adaptive UB = O(k log(k) + k/ɛ) = O(k log(k))

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Main result

Any nonadaptive algorithm requires

k log(k) ɛc log(log(k)/ɛc)

queries (for any 0 < c < 1). Set ɛ = 1/log(k). Adaptive UB = O(k log(k) + k/ɛ) = O(k log(k)) Our nonadapt LB = k log(k)1+c/log(log(k)) q ≥

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Our techniques

  • Basic ideas come from [CG04]’s (k)

adaptive lower bound

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Our techniques

  • Basic ideas come from [CG04]’s (k)

adaptive lower bound

  • [Bla08]’s (k/(ɛ log(k/ɛ)) nonadaptive lower

bound based on [CG04]’s lower bound

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Our techniques

  • Basic ideas come from [CG04]’s (k)

adaptive lower bound

  • [Bla08]’s (k/(ɛ log(k/ɛ)) nonadaptive lower

bound based on [CG04]’s lower bound

  • We give a new analysis of [Bla08]’s LB.
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Our techniques

  • Basic ideas come from [CG04]’s (k)

adaptive lower bound

  • [Bla08]’s (k/(ɛ log(k/ɛ)) nonadaptive lower

bound based on [CG04]’s lower bound

  • We give a new analysis of [Bla08]’s LB.
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[CG04] considers two distributions on n = (k+1)-variable functions:

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[CG04] considers two distributions on n = (k+1)-variable functions: Dyes: ● Pick i ~ {1,...,k+1} uar.

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[CG04] considers two distributions on n = (k+1)-variable functions: Dyes: ● Pick i ~ {1,...,k+1} uar.

  • Set fyes:{0,1}k+1→ {0,1} uar subject to

not depending on coordinate i.

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[CG04] considers two distributions on n = (k+1)-variable functions: Dyes: ● Pick i ~ {1,...,k+1} uar.

  • Set fyes:{0,1}k+1→ {0,1} uar subject to

not depending on coordinate i.

fyes(x1,...,0,…,xk+1) = fyes(x1,...,1,…,xk+1)

i i

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[CG04] considers two distributions on n = (k+1)-variable functions: Dyes: ● Pick i ~ {1,...,k+1} uar.

  • Set fyes:{0,1}k+1→ {0,1} uar subject to

not depending on coordinate i.

fyes(x1,...,0,…,xk+1) = fyes(x1,...,1,…,xk+1) = random {0,1}

i i

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[CG04] considers two distributions on n = (k+1)-variable functions: Dyes: ● Pick i ~ {1,...,k+1} uar.

  • Set fyes:{0,1}k+1→ {0,1} uar subject to

not depending on coordinate i.

fyes(x1,...,0,…,xk+1) = fyes(x1,...,1,…,xk+1) = random {0,1}

i i (for all x1,...,xk+1)

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[CG04] considers two distributions on n = (k+1)-variable functions: Dyes: ● Pick i ~ {1,...,k+1} uar.

  • Set fyes:{0,1}k+1→ {0,1} uar subject to

not depending on coordinate i.

fyes(x1,...,0,…,xk+1) = fyes(x1,...,1,…,xk+1) = random {0,1}

i i (for all x1,...,xk+1)

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[CG04] considers two distributions on n = (k+1)-variable functions: Dyes: ● Pick i ~ {1,...,k+1} uar.

  • Set fyes:{0,1}k+1→ {0,1} uar subject to

not depending on coordinate i.

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[CG04] considers two distributions on n = (k+1)-variable functions: Dyes: ● Pick i ~ {1,...,k+1} uar.

  • Set fyes:{0,1}k+1→ {0,1} uar subject to

not depending on coordinate i. Dno: ● Set fno:{0,1}k+1→ {0,1} uar.

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[CG04] considers two distributions on n = (k+1)-variable functions: Dyes: ● Pick i ~ {1,...,k+1} uar.

  • Set fyes:{0,1}k+1→ {0,1} uar subject to

not depending on coordinate i. (a k-junta) Dno: ● Set fno:{0,1}k+1→ {0,1} uar.

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  • Set fno:{0,1}k+1→ {0,1} uar.

[CG04] considers two distributions on n = (k+1)-variable functions: Dyes: ● Pick i ~ {1,...,k+1} uar.

  • Set fyes:{0,1}k+1→ {0,1} uar subject to

not depending on coordinate i. (a k-junta) (usually far from a k-junta) Dno:

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  • Set fno:{0,1}k+1→ {0,1} uar.

[CG04] considers two distributions on n = (k+1)-variable functions: Dyes: ● Pick i ~ {1,...,k+1} uar.

  • Set fyes:{0,1}k+1→ {0,1} uar subject to

not depending on coordinate i. (a k-junta) (usually far from a k-junta) Dno:

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[CG04] considers two distributions on n = (k+1)-variable functions: Dyes: ● Pick i ~ {1,...,k+1} uar.

  • Set fyes:{0,1}k+1→ {0,1} uar subject to

not depending on coordinate i. Dno: ● Set fno:{0,1}k+1→ {0,1} uar.

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[CG04] considers two distributions on n = (k+1)-variable functions: Dyes: ● Pick i ~ {1,...,k+1} uar.

  • Set fyes:{0,1}k+1→ {0,1} uar subject to

not depending on coordinate i. Dno: ● Set fno:{0,1}k+1→ {0,1} uar. [CG04 THM]: Need (k) queries to distinguish these distributions

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Given f, how to tell if from Dyes or Dno?

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Given f, how to tell if from Dyes or Dno? Idea: See if it has any irrelevant coords.

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Given f, how to tell if from Dyes or Dno? Idea: See if it has any irrelevant coords. For coord i:

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Given f, how to tell if from Dyes or Dno? Idea: See if it has any irrelevant coords. For coord i: ● Pick x uar.

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Given f, how to tell if from Dyes or Dno? Idea: See if it has any irrelevant coords. For coord i: ● Pick x uar.

  • Query f on x and x⊕i.
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Given f, how to tell if from Dyes or Dno? Idea: See if it has any irrelevant coords. For coord i: ● Pick x uar.

  • Query f on x and x⊕i.

Differ only on coord i.

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Given f, how to tell if from Dyes or Dno? Idea: See if it has any irrelevant coords. For coord i: ● Pick x uar.

  • Query f on x and x⊕i.

Differ only on coord i. Def: x and x⊕i form an i-twin.

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Given f, how to tell if from Dyes or Dno? Idea: See if it has any irrelevant coords. For coord i: ● Pick x uar.

  • Query f on x and x⊕i.

Differ only on coord i. Def: x and x⊕i form an i-twin.

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Given f, how to tell if from Dyes or Dno? Idea: See if it has any irrelevant coords. For coord i: ● Pick x uar.

  • Query f on x and x⊕i.
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Given f, how to tell if from Dyes or Dno? Idea: See if it has any irrelevant coords. For coord i: ● Pick x uar.

  • Query f on x and x⊕i.
  • If f(x) ≠ f(x⊕i), output relevant.
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Given f, how to tell if from Dyes or Dno? Idea: See if it has any irrelevant coords. For coord i: ● Pick x uar.

  • Query f on x and x⊕i.
  • If f(x) ≠ f(x⊕i), output relevant.
  • Repeat 10 log(k) times.
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Given f, how to tell if from Dyes or Dno? Idea: See if it has any irrelevant coords. For coord i: ● Pick x uar.

  • Query f on x and x⊕i.
  • If f(x) ≠ f(x⊕i), output relevant.
  • Repeat 10 log(k) times.
  • Output irrelevant.
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If i is relevant:

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If i is relevant: f(x) f(x⊕i)

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If i is relevant: f(x) f(x⊕i)

uar {0,1}

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If i is relevant: f(x) f(x⊕i)

uar {0,1} uar {0,1}

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If i is relevant: f(x) f(x⊕i)

uar {0,1} uar {0,1}

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If i is relevant: f(x) f(x⊕i)

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If i is relevant: f(x) f(x⊕i) = w/prob 1/2

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If i is relevant: f(x) f(x⊕i) = w/prob 1/2 f(x) f(x⊕i) ≠ w/prob 1/2

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If i is relevant: f(x) f(x⊕i) = w/prob 1/2 f(x) f(x⊕i) ≠ w/prob 1/2 ∴ will conclude relevant after O(1) i-twins

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If i is relevant: f(x) f(x⊕i) = w/prob 1/2 f(x) f(x⊕i) ≠ w/prob 1/2 ∴ will conclude relevant after O(1) i-twins If i is irrelevant:

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If i is relevant: f(x) f(x⊕i) = w/prob 1/2 f(x) f(x⊕i) ≠ w/prob 1/2 ∴ will conclude relevant after O(1) i-twins If i is irrelevant: f(x) f(x⊕i) = always

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If i is relevant: f(x) f(x⊕i) = w/prob 1/2 f(x) f(x⊕i) ≠ w/prob 1/2 ∴ will conclude relevant after O(1) i-twins If i is irrelevant: f(x) f(x⊕i) = always ∴ will query O(log(k)) i-twins

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If i is relevant: f(x) f(x⊕i) = w/prob 1/2 f(x) f(x⊕i) ≠ w/prob 1/2 ∴ will conclude relevant after O(1) i-twins If i is irrelevant: f(x) f(x⊕i) = always ∴ will query O(log(k)) i-twins Query cost:

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If i is relevant: f(x) f(x⊕i) = w/prob 1/2 f(x) f(x⊕i) ≠ w/prob 1/2 ∴ will conclude relevant after O(1) i-twins If i is irrelevant: f(x) f(x⊕i) = always ∴ will query O(log(k)) i-twins Query cost: (k+1) * O(1)

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If i is relevant: f(x) f(x⊕i) = w/prob 1/2 f(x) f(x⊕i) ≠ w/prob 1/2 ∴ will conclude relevant after O(1) i-twins If i is irrelevant: f(x) f(x⊕i) = always ∴ will query O(log(k)) i-twins Query cost: (k+1) * O(1) + O(log(k))

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If i is relevant: f(x) f(x⊕i) = w/prob 1/2 f(x) f(x⊕i) ≠ w/prob 1/2 ∴ will conclude relevant after O(1) i-twins If i is irrelevant: f(x) f(x⊕i) = always ∴ will query O(log(k)) i-twins Query cost: (k+1) * O(1) + O(log(k)) = O(k)

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[CG04]’s (k) lower bound

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[CG04]’s (k) lower bound

Key idea: ● Suppose you query f on x1,...,xq

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[CG04]’s (k) lower bound

Key idea: ● Suppose you query f on x1,...,xq

  • Want to test: is coord i relevant?
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[CG04]’s (k) lower bound

Key idea: ● Suppose you query f on x1,...,xq

  • Want to test: is coord i relevant?
  • Then x1,...,xq must have an

i-twin

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[CG04]’s (k) lower bound

Key idea: ● Suppose you query f on x1,...,xq

  • Want to test: is coord i relevant?
  • Then x1,...,xq must have an

i-twin LB: q points can have i-twins for ≤ q-1 coords.

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[CG04]’s (k) lower bound

Key idea: ● Suppose you query f on x1,...,xq

  • Want to test: is coord i relevant?
  • Then x1,...,xq must have an

i-twin LB: q points can have i-twins for ≤ q-1 coords. ∴ q = (k).

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Matching upper and lower bounds?

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Matching upper and lower bounds?

Algorithm was adaptive:

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Matching upper and lower bounds?

Algorithm was adaptive:

  • for relevant coords i, query O(1) i-twins
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Matching upper and lower bounds?

Algorithm was adaptive:

  • for relevant coords i, query O(1) i-twins
  • for irrelevant coords i, query O(log(k))
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Matching upper and lower bounds?

Algorithm was adaptive:

  • for relevant coords i, query O(1) i-twins
  • for irrelevant coords i, query O(log(k))

Can’t plan this in advance:

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Matching upper and lower bounds?

Algorithm was adaptive:

  • for relevant coords i, query O(1) i-twins
  • for irrelevant coords i, query O(log(k))

Can’t plan this in advance: x1,...,xq need O(log(k)) i-twins in all k+1 directions.

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Matching upper and lower bounds?

Algorithm was adaptive:

  • for relevant coords i, query O(1) i-twins
  • for irrelevant coords i, query O(log(k))

Can’t plan this in advance: x1,...,xq need O(log(k)) i-twins in all k+1 directions. ∴ q = (k log(k)) nonadaptive LB?

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Matching upper and lower bounds?

Algorithm was adaptive:

  • for relevant coords i, query O(1) i-twins
  • for irrelevant coords i, query O(log(k))

Can’t plan this in advance: x1,...,xq need O(log(k)) i-twins in all k+1 directions. ∴ q = (k log(k)) nonadaptive LB? (not quite)

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A nonadaptive algorithm.

[Fra83]: there are q = O(k log(k) / log log(k)) points x1,...,xq with log(k) i-twins for each i.

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A nonadaptive algorithm.

[Fra83]: there are q = O(k log(k) / log log(k)) points x1,...,xq with log(k) i-twins for each i.

k log(k) ɛc log(log(k)/ɛc)

q ≥

Recall our LB:

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A nonadaptive algorithm.

[Fra83]: there are q = O(k log(k) / log log(k)) points x1,...,xq with log(k) i-twins for each i.

k log(k) ɛc log(log(k)/ɛc)

q ≥

Recall our LB: Goal: ● show [Fra83] is optimal

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A nonadaptive algorithm.

[Fra83]: there are q = O(k log(k) / log log(k)) points x1,...,xq with log(k) i-twins for each i.

k log(k) ɛc log(log(k)/ɛc)

q ≥

Recall our LB: Goal: ● show [Fra83] is optimal

  • extend to general ɛ
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A nonadaptive algorithm.

[Fra83]: there are q = O(k log(k) / log log(k)) points x1,...,xq with log(k) i-twins for each i.

k log(k) ɛc log(log(k)/ɛc)

q ≥

Recall our LB: Goal: ● show [Fra83] is optimal

  • extend to general ɛ
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New distributions.

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Dno: ● Set fno:{0,1}k+1→ {0,1} random ɛ- biased.

New distributions.

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Dno: ● Set fno:{0,1}k+1→ {0,1} random ɛ- biased.

New distributions.

f(x) is independent from all other f(x’),

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Dno: ● Set fno:{0,1}k+1→ {0,1} random ɛ- biased.

New distributions.

f(x) is independent from all other f(x’), satisfies Pr[f(x) = 1] = ɛ

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Dno: ● Set fno:{0,1}k+1→ {0,1} random ɛ- biased.

New distributions.

f(x) is independent from all other f(x’), satisfies Pr[f(x) = 1] = ɛ

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Dno: ● Set fno:{0,1}k+1→ {0,1} random ɛ- biased.

New distributions.

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Dyes: ● Pick i ~ {1,...,k+1} uar. Dno: ● Set fno:{0,1}k+1→ {0,1} random ɛ- biased.

New distributions.

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Dyes: ● Pick i ~ {1,...,k+1} uar.

  • Set fyes:{0,1}k+1→ {0,1} random ɛ-

biased subject to not depending on coordinate i. Dno: ● Set fno:{0,1}k+1→ {0,1} random ɛ- biased.

New distributions.

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Dyes: ● Pick i ~ {1,...,k+1} uar.

  • Set fyes:{0,1}k+1→ {0,1} random ɛ-

biased subject to not depending on coordinate i. Dno: ● Set fno:{0,1}k+1→ {0,1} random ɛ- biased.

New distributions.

(a k-junta) Dno:

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Dyes: ● Pick i ~ {1,...,k+1} uar.

  • Set fyes:{0,1}k+1→ {0,1} random ɛ-

biased subject to not depending on coordinate i. Dno: ● Set fno:{0,1}k+1→ {0,1} random ɛ- biased.

New distributions.

(a k-junta) Dno: (usually ɛ-far from a k-junta)

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Distributions studied in [Bla08].

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Distributions studied in [Bla08]. His LB: (k/(ɛ log(k/ɛ))

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Distributions studied in [Bla08]. His LB: (k/(ɛ log(k/ɛ)) Main tool: Edge-isoperimetric inequality

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Distributions studied in [Bla08]. His LB: (k/(ɛ log(k/ɛ)) Main tool: Edge-isoperimetric inequality: points x1,...,xq can only have O(q log(q)) i-twins

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Distributions studied in [Bla08]. His LB: (k/(ɛ log(k/ɛ)) Main tool: Edge-isoperimetric inequality: points x1,...,xq can only have O(q log(q)) i-twins

  • Only about total # of i-twins.
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Distributions studied in [Bla08]. His LB: (k/(ɛ log(k/ɛ)) Main tool: Edge-isoperimetric inequality: points x1,...,xq can only have O(q log(q)) i-twins

  • Only about total # of i-twins.
  • Could be few directions have lots of i-twins.
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Distributions studied in [Bla08]. His LB: (k/(ɛ log(k/ɛ)) Main tool: Edge-isoperimetric inequality: points x1,...,xq can only have O(q log(q)) i-twins

  • Only about total # of i-twins.
  • Could be few directions have lots of i-twins.
  • Want edge-iso ineq. about most directions.
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Our main tool: New edge-iso inequality

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Our main tool: New edge-iso inequality Suppose x1,...,xq have m i-twins in d directions. Then q ≥ md / log(m).

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Our main tool: New edge-iso inequality Suppose x1,...,xq have m i-twins in d directions. Then q ≥ md / log(m).

  • m = log(k) and d = k gives
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Our main tool: New edge-iso inequality Suppose x1,...,xq have m i-twins in d directions. Then q ≥ md / log(m).

  • m = log(k) and d = k gives

q ≥ k log(k)/log(log(k))

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Our main tool: New edge-iso inequality Suppose x1,...,xq have m i-twins in d directions. Then q ≥ md / log(m).

  • m = log(k) and d = k gives
  • Generalization of [Fra83]

q ≥ k log(k)/log(log(k))

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Other ideas

  • Proofs are quite technical
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Other ideas

  • Proofs are quite technical
  • Answer you get for single query f(xj) is not

too important

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Other ideas

  • Proofs are quite technical
  • Answer you get for single query f(xj) is not

too important

  • Analyze a specific martingale w/r/t

f(x1), f(x2), …, f(xq)

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Other ideas

  • Proofs are quite technical
  • Answer you get for single query f(xj) is not

too important

  • Analyze a specific martingale w/r/t

f(x1), f(x2), …, f(xq)

  • Use McDiarmid’s inequality (with bad events)
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Open problem

Prove a separation between adapative and nonadaptive when ɛ = const.

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Thanks!