Identifying an Honest EXP NP Oracle Among Many Shuichi Hirahara The - - PowerPoint PPT Presentation

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Identifying an Honest EXP NP Oracle Among Many Shuichi Hirahara The University of Tokyo CCC 18/6/2015 Dishonest oracle Honest oracle Overview Queries Which is honest? Selector Our Contributions 1. We formulate the notion of selector.

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SLIDE 1

Identifying an Honest EXPNP Oracle Among Many

Shuichi Hirahara

The University of Tokyo

CCC 18/6/2015

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SLIDE 2

Overview

Our Contributions

  • 1. We formulate the notion of selector.
  • βˆƒ Selector ⟺ Able to remove short advice
  • 2. We prove the existence of a selector for EXPNP-

complete languages.

Dishonest oracle Honest oracle Selector

Which is honest?

Queries

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SLIDE 3

Background: Instance checker

  • Introduced by Blum & Kannan (1989).
  • An instance checker for a function 𝑔 checks if a

given oracle correctly computes 𝑔 𝑦 on input 𝑦 in polynomial time.

Possibly buggy program

(modeled by black-box access to an oracle)

Is the program buggy

  • r correct on 𝑦?

Instance checker for 𝑔 Queries Answers

Given: input 𝑦

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SLIDE 4

Background: Instance checker

οƒΌThere exist instance checkers for P#P-, PSPACE-, EXP-complete languages. οƒΌAny languages with an instance checker must be in NEXP ∩ coNEXP. (Note: NEXP βŠ† EXPNP)

Possibly buggy program

(modeled by black-box access to an oracle)

Is the program buggy

  • r correct on 𝑦?

Instance checker for 𝑔 Queries Answers

Given: input 𝑦 [LFKN92, Sha92, BFL91]

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SLIDE 5

(Probabilistic) Selector for SAT

Given:

  • 1. An input πœ’, and
  • 2. access to two oracles
  • ne of which is honest.

Task:

compute SAT πœ’ with the help of the oracles

Selector for SAT

Is πœ’ satisfiable?

Given: input πœ’

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SLIDE 6

(Probabilistic) Selector for SAT

Given:

  • 1. An input πœ’, and
  • 2. access to two oracles
  • ne of which is honest.

Task:

compute SAT πœ’ with the help of the oracles

Selector for SAT

Queries: Is πœ” satisfiable?

Is πœ’ satisfiable?

πœ” is not satisfiable! πœ” is satisfiable!

Given: input πœ’

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SLIDE 7

(Probabilistic) Selector for SAT

Given:

  • 1. An input πœ’, and
  • 2. access to two oracles
  • ne of which is honest.

Task:

compute SAT πœ’ with the help of the oracles Dishonest oracle

Selector for SAT

Queries: Is πœ” satisfiable?

Is πœ’ satisfiable?

πœ” is not satisfiable!

Correct answers

Honest oracle

Arbitrary answers

πœ” is satisfiable!

Given: input πœ’

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SLIDE 8

Why is it called β€œselector”?

Given:

  • 1. An input πœ’, and
  • 2. access to two oracles
  • ne of which is honest.

Task:

compute SAT πœ’ with the help of the oracles Dishonest oracle

Selector for SAT

Is πœ’ satisfiable? YES

Correct answers

Honest oracle

Arbitrary answers

YES

Given: input πœ’

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SLIDE 9

Why is it called β€œselector”?

Given:

  • 1. An input πœ’, and
  • 2. access to two oracles
  • ne of which is honest.

Task:

compute SAT πœ’ with the help of the oracles Dishonest oracle

Selector for SAT

Is πœ’ satisfiable? YES

Correct answers

Honest oracle

Arbitrary answers

YES

Given: input πœ’

The answer must be YES!!

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SLIDE 10

Why is it called β€œselector”?

Given:

  • 1. An input πœ’, and
  • 2. access to two oracles
  • ne of which is honest.

Essential Task:

determine which is honest when they disagree on πœ’. Dishonest oracle

Selector for SAT

Is πœ’ satisfiable? NO

Correct answers

Honest oracle

Arbitrary answers

YES

Given: input πœ’

Which is honest?

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SLIDE 11

Definition of (Probabilistic) Selector

Selector 𝑇 for a language 𝑀

(polynomial-time probabilistic oracle TM)

Queries: 𝑀 π‘Ÿ = ?

Correct answers (YES/NO) Arbitrary answers

Definition (Selector) A selector 𝑇 for a language 𝑀 is a polynomial-time probabilistic oracle machine such that 𝐡0 = 𝑀 or 𝐡1 = 𝑀 ⟹ Pr 𝑇𝐡0,𝐡1 𝑦 = 𝑀 𝑦 β‰₯ 0.99

Oracle 𝐡0 Oracle 𝐡1

(for any 𝐡0, 𝐡1 βŠ† 0,1 βˆ—, 𝑦 ∈ 0,1 βˆ—)

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SLIDE 12

Definition of Deterministic Selector

Selector 𝑇 for a language 𝑀

(polynomial-time deterministic oracle TM)

Queries: 𝑀 π‘Ÿ = ?

Correct answers (YES/NO) Arbitrary answers

Definition (Deterministic Selector) A deterministic selector 𝑇 for a language 𝑀 is a polynomial- time deterministic oracle machine such that 𝐡0 = 𝑀 or 𝐡1 = 𝑀 ⟹ 𝑇𝐡0,𝐡1 𝑦 = 𝑀 𝑦

Oracle 𝐡0 Oracle 𝐡1

(for any 𝐡0, 𝐡1 βŠ† 0,1 βˆ—, 𝑦 ∈ 0,1 βˆ—)

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SLIDE 13

Selector for PNP-complete languages (1/2)

  • Def. (Lexicographically Maximum Satisfying Assignment)

Input: a Boolean formula πœ’: 0, 1 π‘œ β†’ 0, 1 and an index 𝑙 Output: the 𝑙th bit of the lexicographically maximum satisfying assignment of πœ’. Goal: to construct a deterministic selector for this language. Given: an input πœ’, 𝑙 and two oracles .

[Krentel 1988]

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SLIDE 14

Selector for PNP-complete languages (2/2)

  • Make queries πœ’, 1 , … , (πœ’, π‘œ) to the oracles:
  • If 𝑀0 = 𝑀1, then output the 𝑙th bit of 𝑀0(= 𝑀1).
  • Else, we have 𝑀0 β‰  𝑀1. We assume 𝑀0 < 𝑀1.

Evaluate πœ’ 𝑀1 .

  • We trust if πœ’ 𝑀1 = 1 and trust otherwise.

𝑀0 ∈ 0, 1 π‘œ 𝑀1 ∈ 0, 1 π‘œ The lexicographically maximum satisfying assignment of πœ’ is... (i.e. output the 𝑙th bit of 𝑀1) (i.e. output the 𝑙th bit of 𝑀0) Show us the 1st bit of the satisfying assignment!

Q.E.D.

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SLIDE 15

Identifying an Honest EXPNP oracle

Theorem (Main Result)

There exists a selector for EXPNP-complete languages.

  • Def. (An π…π˜ππŽπ-complete language)

Input: a succinctly described Boolean formula Ξ¦: 0, 1 2π‘œ β†’ 0, 1 and an index 𝑙 Output: the 𝑙th bit of the lexicographically maximum satisfying assignment of Ξ¦.

Proof sketch:

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SLIDE 16

Proof Sketch of the Main Theorem

We are given a (succinctly described) exponential- sized formula Ξ¦ and two oracles .

π‘Š

0 ∈ 0, 1 2π‘œ

π‘Š

1 ∈ 0, 1 2π‘œ

  • Proof strategy: the same with PNP-complete languages

Step 1: Which is the larger? (i.e. compute π‘Š

0 < π‘Š 1)

Step 2: Is π‘Š

1 a satisfying assignment of Ξ¦?

οƒΌ Can be done in the same way with MIP = NEXP.

[Babai, Fortnow, Lund (1991)].

οƒΌ Binary search & Polynomial identity testing

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SLIDE 17

Instance Checker vs. Selector

Selector Instance checker

Counterexample:

EXPNP-complete languages

οƒΌ The task of selectors is strictly easier than instance checking.

(unless EXPNP = NEXP)

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SLIDE 18

Motivation: Removing short advice

Q.

When can we remove short advice?

[Karp & Lipton (1980)]

SAT ∈ 𝐐/𝐦𝐩𝐑 ⟹ SAT ∈ 𝐐

  • A. When we have a selector.

[Trevisan & Vadhan (2002)]

EXP βŠ† 𝐂𝐐𝐐/𝐦𝐩𝐑 ⟹ EXP βŠ† 𝐂𝐐𝐐

οƒΌ This follows from the instance checkability of EXP-complete languages.

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SLIDE 19

βˆƒ Selector ⟹ Able to remove 1-bit advice

𝑁 π‘Ÿ, 0

Queries π‘Ÿ

𝑁 π‘Ÿ, 1

  • 1. Suppose 𝑀 is computable with 1-bit advice.

i.e. βˆƒ machine 𝑁 such that, given advice β€œ0” or β€œ1”, 𝑁 computes 𝑀 correctly.

Advice β€œ0” Advice β€œ1”

=

def

=

def

  • 2. Define two oracles as follows:

One of these oracles is honest! ⟹The selector can compute 𝑀 correctly (without any advice).

(We assume that 𝑀 is paddable and π‘Ÿβ€²s are

  • f the same length.)

(𝑁 π‘Ÿ, 0 = 𝑀 π‘Ÿ or 𝑁 π‘Ÿ, 1 = 𝑀 π‘Ÿ for any π‘Ÿ ∈ 0,1 π‘š)

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SLIDE 20

Key Lemma: β€œAmong Many”

(Honest)

=

(Honest) Identifying an honest oracle among polynomially many Identifying an honest oracle among two

  • utputs 𝑀(𝑦)
  • utputs 𝑀(𝑦)

οƒΌ Able to remove advice of 1 bit οƒΌ Able to remove advice of 𝑃 log π‘œ bits

On input 𝑦, On input 𝑦,

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SLIDE 21

Our Results

  • Thm. (βˆƒ Selector ⟺ Able to remove short advice)

For any paddable language 𝑀, the following are equivalent:

  • 1. There exists a deterministic selector for 𝑀.

2. 𝑀 ∈ 𝐐/𝐦𝐩𝐑 implies 𝑀 ∈ 𝐐 under any relativized world. In other words, 𝑀 ∈ 𝐐𝑆/𝐦𝐩𝐑 ⟹ 𝑀 ∈ 𝐐𝑆 βˆ€π‘†: oracle οƒΌ The converse direction (2 ⟹ 1) also holds in this sense!

  • The notion of selector provides a general framework to

remove short advice:

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SLIDE 22

Our Results

  • Thm. (βˆƒ Selector ⟺ Able to remove short advice)

For any paddable language 𝑀, the following are equivalent:

  • 1. There exists a probabilistic selector for 𝑀.

2. 𝑀 ∈ 𝐂𝐐𝐐/𝐦𝐩𝐑 implies 𝑀 ∈ 𝐂𝐐𝐐 under any relativized world.

In other words, 𝑀 ∈ 𝐂𝐐𝐐𝑆/𝐦𝐩𝐑 ⟹ 𝑀 ∈ 𝐂𝐐𝐐𝑆 βˆ€π‘†: oracle

  • The notion of selector provides a general framework to

remove short advice:

Technical Remark: This works for any type of advice for BPP, because we can remove the most powerful advice 𝑗. 𝑓. BPP//log .

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SLIDE 23

Key Lemma to remove 𝑃 log π‘œ advice

(Honest)

=

(Honest) Identifying an honest oracle among polynomially many Identifying an honest oracle among two

  • utputs 𝑀(𝑦)
  • utputs 𝑀(𝑦)

οƒΌ Able to remove advice of 1 bit οƒΌ Able to remove advice of 𝑃 log π‘œ bits

On input 𝑦, On input 𝑦,

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SLIDE 24

Proof of the Key Lemma (1/2)

  • We have a selector that identifies an honest
  • racle among two.
  • Given input 𝑦 and many oracles
  • Ask them about 𝑦: 𝑀 𝑦 is
  • Divide them into two teams:

(Honest)

1 1

We claim 𝑀 𝑦 = 0. We claim 𝑀 𝑦 = 1.

Which team should we trust?

(Honest)

Idea: β€œTournament”

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SLIDE 25

Proof of the Key Lemma (2/2)

We claim 𝑀 𝑦 = 0. We claim 𝑀 𝑦 = 1. (Honest)

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SLIDE 26

Proof of the Key Lemma (2/2)

We claim 𝑀 𝑦 = 0. We claim 𝑀 𝑦 = 1. (Honest)

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SLIDE 27

Proof of the Key Lemma (2/2)

We claim 𝑀 𝑦 = 0. We claim 𝑀 𝑦 = 1. (Honest)

I doubt !!

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SLIDE 28

Proof of the Key Lemma (2/2)

We claim 𝑀 𝑦 = 0. We claim 𝑀 𝑦 = 1. (Honest)

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SLIDE 29

Proof of the Key Lemma (2/2)

We claim 𝑀 𝑦 = 0. We claim 𝑀 𝑦 = 1. (Honest)

1

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SLIDE 30

Proof of the Key Lemma (2/2)

We claim 𝑀 𝑦 = 0. We claim 𝑀 𝑦 = 1. (Honest)

1

I doubt !!

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SLIDE 31

Proof of the Key Lemma (2/2)

We claim 𝑀 𝑦 = 0. We claim 𝑀 𝑦 = 1. (Honest)

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SLIDE 32

Proof of the Key Lemma (2/2)

We claim 𝑀 𝑦 = 0. We claim 𝑀 𝑦 = 1. (Honest)

1

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SLIDE 33

Proof of the Key Lemma (2/2)

We claim 𝑀 𝑦 = 0. We claim 𝑀 𝑦 = 1. (Honest)

1

I doubt !!

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SLIDE 34

Proof of the Key Lemma (2/2)

We claim 𝑀 𝑦 = 0. We claim 𝑀 𝑦 = 1. (Honest)

I trust , so the answer is 1! οƒΌThe honest oracle always wins!

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SLIDE 35

Proof of the Key Lemma (summary)

  • Pick arbitrary two oracles from each team.

I claim 𝑀 𝑦 = 0.

  • Run the selector on input 𝑦.
  • If outputs 0, then we doubt (the oracle loses!);
  • therwise we doubt .

I claim 𝑀 𝑦 = 1.

  • Continue it until one of the teams wins

and trust the team.

Q.E.D.

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SLIDE 36

Removing advice of size 𝑃 log π‘œ

𝑁 π‘Ÿ, 0 𝑁 π‘Ÿ, 1

π‘Ÿ π‘Ÿ

Advice β€œ0” Advice β€œ1”

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SLIDE 37

Removing advice of size 𝑃 log π‘œ

𝑁 π‘Ÿ, 00 𝑁 π‘Ÿ, 01 𝑁 π‘Ÿ, 10 𝑁 π‘Ÿ, 11

π‘Ÿ π‘Ÿ π‘Ÿ π‘Ÿ

Advice β€œ00” Advice β€œ01” Advice β€œ10” Advice β€œ11” οƒΌ Able to remove advice of 2 bits

Corollary (Selector β‡’ Removing short advice)

Suppose that there exists a selector for a paddable language 𝑀. Then, 𝑀 ∈ 𝐐/𝐦𝐩𝐑 implies 𝑀 ∈ 𝐐 (under any relativized world).

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SLIDE 38

βˆƒ selector ⟸ Removing short advice

If no selector exists, then for some relativized world, even 1-bit advice cannot be removed: βˆƒπ‘† 𝑑. 𝑒. 𝑀 ∈ 𝐐𝑆/1 & 𝑀 βˆ‰ 𝐐𝑆 Claim (βˆƒ Selector ⟸ Able to remove short advice)

Suppose 𝑀 ∈ 𝐐/𝐦𝐩𝐑 implies 𝑀 ∈ 𝐐 under any relativized world. Then, there exists a deterministic selector for 𝑀.

  • We prove the contraposition:
  • Constructing such an oracle 𝑆 by diagonalization.
  • 1-bit advice can tell which is honest. (i.e. 𝑀 ∈ 𝐐𝑆/1)
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SLIDE 39

Other Results: Deterministic Selector

Theorem (Deterministic Selector)

  • 1. There exists a deterministic selector for PSPACE-complete

languages.

  • 2. Any languages with a deterministic selector sit within

PSPACE.

PSPACE

PSPACE complete βˆƒ deterministic selector Deterministic selectors are in PSPACE

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SLIDE 40

Other Results: Probabilistic Selector

Theorem (Upper Bound for Probabilistic Selector)

Any languages with a selector sit within 𝑇2

π‘“π‘¦π‘ž.

EXPNP

EXPNP complete

𝑇2

π‘“π‘¦π‘ž

βˆƒ selector Selectors are in 𝑇2

π‘“π‘¦π‘ž

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SLIDE 41

Conclusions

οƒΌThe existence of a selector ⟺ A property of removing short advice under any relativized world. οƒΌThere exists a selector for EXPNP-complete languages.

  • We can efficiently identify an honest EXPNP-complete
  • racle among many.
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SLIDE 42

Future Work

  • Closing the gap between EXPNP and 𝑇2

π‘“π‘¦π‘ž

  • Does there exist a selector for promise-𝑇2

π‘“π‘¦π‘ž-complete

languages?

  • Does there exist a selector for NEXP-complete

languages?

  • Although NEXP βŠ† EXPNP, the set of languages with a

selector is not closed under downward reduction.

  • What about removing advice of polynomial size?
  • e.g. 𝑀

βˆ€π‘†, 𝑀 ∈ P𝑆/poly β‡’ 𝑀 ∈ MA𝑆