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

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.  ∃ Selector ⟺ Able to remove short advice 2. We prove the existence of a selector for EXP NP - complete languages.

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. Instance checker for 𝑔 Queries Answers Given: input 𝑦 Possibly buggy program (modeled by black-box access to an oracle) Is the program buggy or correct on 𝑦 ?

Background: Instance checker  There exist instance checkers for P #P -, PSPACE -, EXP -complete languages. [LFKN92, Sha92, BFL91]  Any languages with an instance checker must be in NEXP ∩ coNEXP . (Note: NEXP ⊆ EXP NP ) Instance checker for 𝑔 Queries Answers Given: input 𝑦 Possibly buggy program (modeled by black-box access to an oracle) Is the program buggy or correct on 𝑦 ?

(Probabilistic) Selector for SAT Given: 1. An input 𝜒 , and 2. access to two oracles one of which is honest. Task: compute SAT 𝜒 with the help of the oracles Given: input 𝜒 Is 𝜒 satisfiable? Selector for SAT

(Probabilistic) Selector for SAT 𝜔 is not 𝜔 is Given: satisfiable! satisfiable! 1. An input 𝜒 , and 2. access to two oracles one of which is honest. Task: compute SAT 𝜒 Queries: with the help of the oracles Is 𝜔 satisfiable? Given: input 𝜒 Is 𝜒 satisfiable? Selector for SAT

(Probabilistic) Selector for SAT 𝜔 is not 𝜔 is Given: satisfiable! satisfiable! 1. An input 𝜒 , and 2. access to two oracles Dishonest oracle Honest oracle one of which is honest. Task: compute SAT 𝜒 Queries: with the help of the oracles Is 𝜔 satisfiable? Arbitrary Correct answers answers Given: input 𝜒 Is 𝜒 satisfiable? Selector for SAT

Why is it called “selector”? Given: YES YES 1. An input 𝜒 , and 2. access to two oracles Dishonest oracle Honest oracle one of which is honest. Task: compute SAT 𝜒 Is 𝜒 satisfiable? with the help of the oracles Arbitrary Correct answers answers Given: input 𝜒 Selector for SAT

Why is it called “selector”? Given: YES YES 1. An input 𝜒 , and 2. access to two oracles Dishonest oracle Honest oracle one of which is honest. Task: compute SAT 𝜒 Is 𝜒 satisfiable? with the help of the oracles Arbitrary Correct answers answers Given: input 𝜒 The answer must be YES!! Selector for SAT

Why is it called “selector”? Given: YES 1. An input 𝜒 , and 2. access to two oracles Dishonest oracle Honest oracle one of which is honest. NO Essential Task: determine which is honest when they disagree on 𝜒 . Is 𝜒 satisfiable? Arbitrary Correct answers answers Given: input 𝜒 Which is honest? Selector for SAT

Definition of (Probabilistic) Selector Definition (Selector) A selector 𝑇 for a language 𝑀 is a polynomial-time probabilistic oracle machine such that 𝐵 0 = 𝑀 or 𝐵 1 = 𝑀 ⟹ Pr 𝑇 𝐵 0 ,𝐵 1 𝑦 = 𝑀 𝑦 ≥ 0.99 (for any 𝐵 0 , 𝐵 1 ⊆ 0,1 ∗ , 𝑦 ∈ 0,1 ∗ ) Oracle 𝐵 1 Oracle 𝐵 0 Queries: 𝑀 𝑟 = ? Arbitrary Correct answers (YES/NO) answers Selector 𝑇 for a language 𝑀 (polynomial-time probabilistic oracle TM)

Definition of Deterministic Selector Definition (Deterministic Selector) A deterministic selector 𝑇 for a language 𝑀 is a polynomial- time deterministic oracle machine such that 𝐵 0 = 𝑀 or 𝐵 1 = 𝑀 ⟹ 𝑇 𝐵 0 ,𝐵 1 𝑦 = 𝑀 𝑦 (for any 𝐵 0 , 𝐵 1 ⊆ 0,1 ∗ , 𝑦 ∈ 0,1 ∗ ) Oracle 𝐵 1 Oracle 𝐵 0 Queries: 𝑀 𝑟 = ? Arbitrary Correct answers (YES/NO) answers Selector 𝑇 for a language 𝑀 (polynomial-time deterministic oracle TM)

Selector for P NP -complete languages (1/2) [Krentel 1988] Def. (Lexicographically Maximum Satisfying Assignment) Input: a Boolean formula 𝜒: 0, 1 𝑜 → 0, 1 and an index 𝑙 Output: the 𝑙 th bit of the lexicographically maximum satisfying assignment of 𝜒 . Goal: to construct a deterministic selector for this language. Given: an input 𝜒, 𝑙 and two oracles .

Selector for P NP -complete languages (2/2) Show us the 1 st bit of the satisfying assignment! • Make queries 𝜒, 1 , … , (𝜒, 𝑜) to the oracles: The lexicographically maximum satisfying 𝑤 0 ∈ 0, 1 𝑜 𝑤 1 ∈ 0, 1 𝑜 assignment of 𝜒 is... • If 𝑤 0 = 𝑤 1 , then output the 𝑙 th bit of 𝑤 0 (= 𝑤 1 ) . • Else, we have 𝑤 0 ≠ 𝑤 1 . We assume 𝑤 0 < 𝑤 1 . Evaluate 𝜒 𝑤 1 . • We trust if 𝜒 𝑤 1 = 1 and trust otherwise. ( i.e. output the 𝑙 th bit of 𝑤 1 ) ( i.e. output the 𝑙 th bit of 𝑤 0 ) Q.E.D.

Identifying an Honest EXP NP oracle Theorem (Main Result) There exists a selector for EXP NP -complete languages. Proof sketch: Def. (An 𝐅𝐘𝐐 𝐎𝐐 -complete language) Input: a succinctly described Boolean formula Φ: 0, 1 2 𝑜 → 0, 1 and an index 𝑙 Output: the 𝑙 th bit of the lexicographically maximum satisfying assignment of Φ .

Proof Sketch of the Main Theorem  Proof strategy: the same with P NP -complete languages We are given a (succinctly described) exponential- sized formula Φ and two oracles . 1 ∈ 0, 1 2 𝑜 0 ∈ 0, 1 2 𝑜 𝑊 𝑊 Step 1: Which is the larger? ( i.e. compute 𝑊 0 < 𝑊 1 )  Binary search & Polynomial identity testing Step 2: Is 𝑊 1 a satisfying assignment of Φ ?  Can be done in the same way with MIP = NEXP . [Babai, Fortnow, Lund (1991)].

Instance Checker vs. Selector Counterexample: EXP NP -complete languages (unless EXP NP = NEXP ) Selector Instance checker  The task of selectors is strictly easier than instance checking.

Motivation: Removing short advice [Karp & Lipton (1980)] SAT ∈ 𝐐/𝐦𝐩𝐡 ⟹ SAT ∈ 𝐐 [Trevisan & Vadhan (2002)] EXP ⊆ 𝐂𝐐𝐐/𝐦𝐩𝐡 ⟹ EXP ⊆ 𝐂𝐐𝐐  This follows from the instance checkability of EXP-complete languages. Q. When can we remove short advice? A. When we have a selector.

∃ Selector ⟹ Able to remove 1-bit advice 1. Suppose 𝑀 is computable with 1-bit advice. i.e. ∃ machine 𝑁 such that, given advice “0” or “1”, 𝑁 computes 𝑀 correctly. ( 𝑁 𝑟, 0 = 𝑀 𝑟 or 𝑁 𝑟, 1 = 𝑀 𝑟 for any 𝑟 ∈ 0,1 𝑚 ) 2. Define two oracles as follows: def def = = Advice “0” Queries 𝑟 Advice “1” 𝑁 𝑟, 1 𝑁 𝑟, 0 (We assume that 𝑀 is paddable and 𝑟 ′ s are of the same length.) One of these oracles is honest! ⟹ The selector can compute 𝑀 correctly (without any advice) .

Key Lemma: “Among Many” Identifying an honest oracle among polynomially many Identifying an honest oracle among two (Honest) (Honest) = outputs 𝑀(𝑦) outputs 𝑀(𝑦) On input 𝑦 , On input 𝑦 ,  Able to remove advice of 𝑃 log 𝑜 bits  Able to remove advice of 1 bit

Our Results  The notion of selector provides a general framework to remove short advice: Thm. ( ∃ Selector ⟺ Able to remove short advice) For any paddable language 𝑀 , the following are equivalent: 1. There exists a deterministic selector for 𝑀 . 𝑀 ∈ 𝐐/𝐦𝐩𝐡 implies 𝑀 ∈ 𝐐 under any relativized world. 2. In other words, 𝑀 ∈ 𝐐 𝑆 /𝐦𝐩𝐡 ⟹ 𝑀 ∈ 𝐐 𝑆 ∀𝑆: oracle  The converse direction ( 2 ⟹ 1) also holds in this sense!

Our Results  The notion of selector provides a general framework to remove short advice: Thm. ( ∃ Selector ⟺ Able to remove short advice) For any paddable language 𝑀 , the following are equivalent: 1. There exists a probabilistic selector for 𝑀 . 𝑀 ∈ 𝐂𝐐𝐐/𝐦𝐩𝐡 implies 𝑀 ∈ 𝐂𝐐𝐐 under any relativized world. 2. In other words, 𝑀 ∈ 𝐂𝐐𝐐 𝑆 /𝐦𝐩𝐡 ⟹ 𝑀 ∈ 𝐂𝐐𝐐 𝑆 ∀𝑆: oracle Technical Remark: This works for any type of advice for BPP, because we can remove the most powerful advice 𝑗. 𝑓. BPP//log .

Key Lemma to remove 𝑃 log 𝑜 advice Identifying an honest oracle among polynomially many Identifying an honest oracle among two (Honest) (Honest) = outputs 𝑀(𝑦) outputs 𝑀(𝑦) On input 𝑦 , On input 𝑦 ,  Able to remove advice of 𝑃 log 𝑜 bits  Able to remove advice of 1 bit

Proof of the Key Lemma (1/2) • We have a selector that identifies an honest oracle among two. (Honest) • Given input 𝑦 and many oracles • Ask them about 𝑦 : 𝑀 𝑦 is 0 0 1 1 • Divide them into two teams: Idea: “Tournament” Which team We claim should we trust? We claim 𝑀 𝑦 = 0 . 𝑀 𝑦 = 1 . (Honest)

Proof of the Key Lemma (2/2) We claim We claim 𝑀 𝑦 = 1 . 𝑀 𝑦 = 0 . (Honest)

Proof of the Key Lemma (2/2) We claim We claim 𝑀 𝑦 = 1 . 𝑀 𝑦 = 0 . (Honest) 0

Proof of the Key Lemma (2/2) We claim We claim 𝑀 𝑦 = 1 . 𝑀 𝑦 = 0 . (Honest) 0 I doubt !!

Proof of the Key Lemma (2/2) We claim We claim 𝑀 𝑦 = 1 . 𝑀 𝑦 = 0 . (Honest)

Proof of the Key Lemma (2/2) We claim We claim 𝑀 𝑦 = 1 . 𝑀 𝑦 = 0 . (Honest) 1