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3-Query Dictator TestingPowerPoint Presentation

3-Query Dictator Testing

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3-Query Dictator Testing

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Ryan O’Donnell

Yi Wu

joint work with

Carnegie Mellon University

Carnegie Mellon University

TexPoint fonts used in EMF.

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Input:

Output: Assignment: vi2 {0,1}

Desideratum: Satisfy as much as possible.

w1

w2

Definition: 0 · OPT · 1 is max. possible

w3

Definition: · k vbls per constraint: = “Max-kCSP”

w4

w5

w6

Fixing “type” of constraints special cases:

w7

w8

Max-3Sat

Max-3Lin

w9

+

¢ ¢ ¢

¢¢¢

¢ ¢ ¢

¢ ¢ ¢

= 1

Max-2SatMax-3SatMax-kSatMax-kLinMax-kCSPMax-CutMax-Directed-CutMin-BisectionSparsest-CutBalanced-SeparatorVertex-CoverIndependent-SetCliqueApproximate-Graph-ColoringMin-Multiway-CutMetric-Labeling0-ExtensionCut-Norm

Max-2SatMax-3SatMax-kSatMax-kLinMax-kCSPMax-CutMax-Directed-CutMin-BisectionSparsest-CutBalanced-SeparatorVertex-CoverIndependent-SetCliqueApproximate-Graph-ColoringMin-Multiway-CutMetric-Labeling0-ExtensionCut-Norm

Max-2SatMax-3SatMax-kSatMax-kLinMax-kCSPMax-CutMax-Directed-CutMin-BisectionSparsest-CutBalanced-SeparatorVertex-CoverIndependent-SetCliqueApproximate-Graph-ColoringMin-Multiway-CutMetric-Labeling0-ExtensionCut-Norm

Input:

Output: Assignment: vi2 {0,1}

Desideratum: Satisfy as much as possible.

w1

w2

Definition: 0 · OPT · 1 is max. possible

w3

Definition: · 3 vbls per constraint: = “Max-3CSP”

w4

w5

w6

w7

w8

w9

+

¢ ¢ ¢

¢¢¢

= 1

Max-Blah is c vs. s easy:satisfying ¸ s when OPT ¸ c is in poly time.

Max-Blah is c vs. s hard: satisfying ¸ s when OPT ¸ c is NP-hard.

1

[Cook71]

(.96)

[Johnson74]

[AS, ALMSS92]

[BGS95]

3/4

(.74)

[Trevisan96]

s

5/8

[TSSW96]

(.514)

[Håstad97]

1/2

[Trevisan97]

[Zwick98,02]

= in poly time

= NP-hard

(.367)

[KS06]

1/4

1/8

0

(OPT)

c

1

[Zwick98], on his 1 vs. 5/8 easiness result for Max-3CSP:

“We conjecture that this result is optimal.”

“… the hardest satisfiable instances of Max-3CSP [for the algorithm] turn out to be instances in which all clauses are NTWclauses.”

[Håstad97], p. 65, Concluding remarks:

The technique of using Fourier transforms to analyze [Dictator Tests] seems very strong. It does not, however, seem universal even limited to CSPs. In particular, an open question that remains is to decide whether the NTW predicate is non-approximable beyond the random assignment threshold [5/8] on satisfiable instances.

NTW(a,b,c) = 1

,

# 1’s among a,b,c is zero, one, or three – i.e., Not Two

“

”

Property Testing problem

Query access to unknown Boolean function f : {0,1}n {0,1}

Want to test if f is a Dictator:

f(x1, …, xn) = xifor some i.

Can only make a constant number of queries

- And by constant, I mean 3
- Or fewer
- And the queries must be non-adaptive

x, y, z

Tester

randomly chooses: i) 3 strings,x, y, z2 {0,1}n,

ii) a 3-bit predicate, φ:{0,1}3→ {acc, rej}

f : {0,1}n {0,1}

f(x), f(y), f(z)

“accepts” iff φ(f(x), f(y), f(z)) = acc

“Tester uses predicate set Φ”$ Φ = {possible φ’s tester may choose}

“Completeness” ¸ c$ all n Dictators accepted w. prob. ¸ c

“Soundness” · s$ “very non-Dictatorial f” accepted “w. prob. · s + o(1)”

Usually: “Every f which is ±-far from all Dictators is accepted w. prob. · s.”

[Håstad97]: Too hard! Relax.

Definition: f is quasirandom if

fixing any O(1) input bits changes bias by at most o(1).

Remark: Dictators are the epitome of not being quasirandom.

Formally: f is (²,±)-quasirandom if for all 0 < |S| · 1/±.

Definition: f is quasirandom if

fixing any O(1) input bits changes bias by at most o(1).

Not quasirandom:Dictators

“Juntas”

Epitome of quasirandom: Constants (f ´ 0, f ´ 1)

Majority

Large Parities: f(x) = where |S| > ω(1)

“Dictator-vs.-quasirandom” Tests:

Formally: Given a sequence of tests ( Tn),

Soundness · s $ every quasirandom f accepted w. prob. · s + o(1)

Soundness · s $ for all ´ > 0, exists ², ± > 0, for all suff. large n,

Tn accepts every (²,±)-quasirandom f w. prob. · s + ´

Meta-Theorem:

Suppose you build a Dictator-vs.-quasirandom test with:

completeness ¸ c, soundness · s,

tester uses predicate set Φ.

Then Max-Φ is c vs. s + ² hard.

(Max–Φ is the CSP where all constraints are from the set Φ.)

[Zwick98], on his 1 vs. 5/8 easiness result for Max-3CSP:

“We conjecture that this result is optimal.”

“… the hardest satisfiable instances of Max-3CSP [for the algorithm] turn out to be instances in which all clauses are NTWclauses.”

[Håstad97], p. 65, Concluding remarks:

The technique of using Fourier transforms to analyze [Dictator Tests] seems very strong. It does not, however, seem universal even limited to CSPs. In particular, an open question that remains is to decide whether the NTW predicate is non-approximable beyond the random assignment threshold [5/8] on satisfiable instances.

“

”

Theorem:

a. There is a 3-query Dictator-vs.-quasirandom test, using NTW predicate,with completeness c = 1 and soundness s = 5/8. [Pf: Fourier analysis.]

b. Every 3-query Dictator-vs.-quasirandom test, using any mix of predicates, with completeness c = 1 has soundness s ¸ 5/8. [Pf: Uses Zwick’s SDP alg.]

Not a Theorem: Max-NTW is 1 vs. 5/8 hard.

Why? Meta-Theorem problematic… maybe with Khot’s “2-to-1 Conjecture”…??

x

)

f (

NTW(

y

f (

)

z

f (

)

)

D =

=

p

q

r

s

t

w. prob.

Test: Choose triple (x, y, z) from Dn.

Problem: Constant functions

Solution: By “odd-izing” (“folding”) trick, may assume f(:x) = :f(x)

Issue: Reqs. uniform distr. on x, y, z

x

)

f (

NTW(

y

f (

)

z

f (

)

)

D =

=

p

w. prob.

Test: Choose triple (x, y, z) from Dn.

Problem: Majority

Corr[xi, yi] = Pr[xi = yi] – Pr[xiyi] = 2p

Solution: Make p very small

x

)

f (

NTW(

y

f (

)

z

f (

)

)

D =

=

w. prob.

Test: Choose triple (x, y, z) from Dn.

Problem? Large odd Parity

Solution: Don’t take ± = 0!

x

)

f (

NTW(

y

f (

)

z

f (

)

)

D =

=

±

w. prob.

Test: Choose triple (x, y, z) from D±n.

Fact:

D =

(1 – ±) D + ±D

±

EQU

XOR

Equivalent test:

1. Form “random restriction” fw with ¤-probability 1 – ±.

2. Do BLR test on fw, but also accept (0,0,0).

Pr[acc. odd f] ·

relatively standardFourier manips

Håstad’s term: ·± when f is (±2,±2)-quasirandom

Handle with careful use of the “hypercontractive inequality”

Long story short: last term always

Prove Max-3CSP is 1 vs. 5/8 + ² hard.

Prove Max-3CSP is 1 vs. 5/8 + ² hard assuming Khot’s 2-to-1 Conjecture.

Tackle Max-2Sat. [cf. Austrin07a, Austrin07b]

Max-4CSP?