Uniform Hardness vs. Randomness Tradeoffs for Arthur-Merlin Games. Danny Gutfreund, Hebrew U. Ronen...

Uniform Hardness vs. Randomness Tradeoffs for Arthur-Merlin Games.

Danny Gutfreund, Hebrew U.Ronen Shaltiel, Weizmann

Inst.Amnon Ta-Shma, Tel-Aviv U.

Arthur-Merlin Games [BM] Interactive games in which the all-

powerful prover Merlin attempts to prove some statement to a probabilistic poly-time verifier.

Merlin Arthur“xL”

toss coinsmessage

message

I accept

Arthur-Merlin Games [BM] Completeness: If the statement is

true then Arthur accepts. Soundness: If the statement is

false then Pr[Arthur accepts]<½.


toss coinsmessage

message

I accept

Arthur-Merlin Games [BM] Completeness: If the statement is

true then Arthur accepts. Soundness: If the statement is

false then Pr[Arthur accepts]<½.

The class AM: All languages L which have an Arthur-Merlin protocol.

Contains many interesting problems not known to be in NP.

Example: Co-isomorphism of Graphs. L={G1,G2: the labeled graphs G1,G2 are

not isomorphic}. L in coNP and is not known to be in NP.

Merlin Arthur(G1,G2 ) L

Randonly chooses:

b {1,2} random permutation of

Gb

“The graph Gc was permuted”

Decides which of the two graphs

was permuted.

Verifies that c=b.

The big question:

Does AM=NP?

In other words: Can every Arthur-Merlin protocol be replaced with one in which Arthur is deterministic?

Note that such a protocol is an NP proof.

Derandomization: a brief overview A paradigm that attempts to transform:

Probabilistic algorithms => deterministic algorithms. (P BPP EXP NEXP).

Probabilistic protocols => deterministic protocols. (NP AM EXP NEXP).

We don’t know how to separate BPP and NEXP.

Can derandomize BPP and AM under natural complexity theoretic assumptions.

Hardness versus Randomness Initiated by [BM,Yao,Shamir].

Assumption: hard functions exist.

Conclusion: Derandomization.

A lot of works: [BM82,Y82,HILL,NW88,BFNW93, I95,IW97,IW98,KvM99,STV99,ISW99,MV99, ISW00,SU01,U02,TV02]

A quick surveyAssumption: There exists a function in

DTIME(2O(n)) which is hard for “small” circuits.

ClassBPPAM

A hard function for:

Deterministic circuits

Nondeterministic circuits

High-endBPP=PAM=NP

Low-endBPPSUBEXPAM NSUBEXP

Hardness versus Randomness





Exists pseudo-random generator


Pseudo-random generators A pseudo-random generator (PRG) is an algorithm

that stretches a short string of truly random bits into a long string of pseudo-random bits.

pseudo-random bits

PRG seed

Pseudo-random bits are indistinguishable from truly random bits for feasible algorithms.

For derandomizing AM: Feasible algorithms = nondeterministic circuits.

??????????????

Pseudo-random generators for nondeterministic circuits Nondeterministic circuits can identify pseudo-

random strings. Given a long string, guess a short seed and check

that PRG(seed)=long string. Can distinguish between random strings and

pseudo-random strings. Assuming the circuit can run the PRG!! The Nisan-Wigderson setup: The circuit cannot run

the PRG!! For example: The PRG runs in time n5 and fools

(nondeterministic) circuits of size n3. Sufficient for derandomization!!

The Nisan-Wigderson setting We’re given a function f which is:

Hard for small circuits. Computable by uniform machines with “slightly”

larger time. Basic idea:

G(x)=x,f(x) “f(x) looks random to a small circuit that sees x”.

Warning: no composition theorems. Correctness proof of PRG can’t use it’s efficiency.

The PRG runs in time “slightly” larger than the size of the circuit.



Exists pseudo-random generator


PRG’s for nondeterministic circuits derandomize AM We can model the AM protocol as a

nondeterministic circuit which gets the random coins as input.


random message

message

I accept

Hardwire input




random input

Nondeterministic guess

I accept

inputNondeterministic guessHardwire input



We can use pseudo-random bits instead of truly random bits.


pseudo-random input


I accept

Nondeterministic guess inputHardwire input

PRG’s for nondeterministic circuits derandomize AM We have an AM protocol in which Arthur

acts deterministically. (Arthur sends all pseudo-random strings

and Merlin replies on each one.) Deterministic protocol => NP proof.


pseudo-random input


I accept

A quick surveyAssumption: There exists a function in

DTIME(2O(n)) which is hard for “small” circuits.

ClassBPPAM

A hard function for:

Deterministic circuits

Nondeterministic circuits

High-endBPP=PAM=NP

Low-endBPPSUBEXPAMNSUBEXP

Uniform Hardness versus Randomness The conclusion in the results above involve

only uniform classes (BPP,AM,P,NP). The assumptions involve nonuniform classes. All the results above assume hardness for

circuits (nonuniform machines). Can we get derandomization from uniform

assumptions? Follow from uniform assumptions such as

EXP≠PH [KL79]. A stronger notion of uniformity was considered

in [IW98,TV02].

A closer look at nonuniform tradeoffs for BPP [BFNW93]Assumption: Hard function for:

circuits. EXP≠P/poly

Conclusion: Derandomization of: probabilistic

algorithms. BPP SUBEXP

Impagliazzo-Wigderson 98: A uniform tradeoff for BPP Assumption: Hard function for:

probabilistic algorithms. EXP≠BPP

Conclusion: Derandomization of: probabilistic

algorithms. BPP * SUBEXP*Pseudo-

containment

Impagliazzo-Wigderson 98: A uniform tradeoff for BPP

Assumption: Hard function for

probabilistic algorithms.

Conclusion: Derandomization* of

probabilistic algorithms.

Either the assumption isn’t true:

probabilistic algorithms are very

strong.

Or the assumption is true: Derandomization*

of probabilistic algorithms.

Our result: A uniform tradeoff for AM

Assumption: Hard function for Arthur-Merlin protocols.

Conclusion: Derandomization* of

Arthur-Merlin protocols.

Either the assumption isn’t true:

Arthur-Merlin protocols are very strong.

Or the assumption is true: Derandomization* of

Arthur-Merlin protocols.

[IW98 :]low-end. )Weak assumption and conclusion(. Our result: high-end. )Strong assumption and conclusion(.

Motivation: weak unconditional derandomization We believe that AM=NP (= Σ1). We only know that AM is in Σ3. Goal: Unconditional proof that AMΣ2 (or even

AMΣ2-SUBEXP). Conditional => Unconditional ?? Basic idea: AM is either weak or very strong.

If AM can be derandomized (AM=NP) then AMΣ2.

If AM is very strong (AM=EXP) then AMΣ2.

Main problem: replace ‘*’ with ‘’.

Pseudo-containmnets [Kab99]: * Intuitively, Containment only on feasibly

generated inputs. L =* L’ if it is infeasible to generate

counterexamples to the statement L=L’. No feasible algorithm R can output inputs

which are in one language but not in the other (for a specified input length).

C * D if for every L in C there exists L’ in D such that L =* L’.

Formally, =* and * are relative to some complexity class of feasible R’s.

Formal statement of our result If E=DTIME(2O(n)) is not in

AMTIME(2an), for some constant a>0 AM * NP. AM coAM = NP coNP.

The class AM coAM contains: co-isomorphism of graphs. SZK (Statistical Zero Knowledge).

The proof

We want to show that

Hard function for AM (EXP≠AM)

Derandomization of AM

No derandomization of AM

No Hard function for AM (EXP=AM)

Basic idea: Use nonuniform tradeoff

No Hard function for nondeter. Circuits (EXP NP/poly)



Nonuniform tradeoff

[MV99,SU01]

Goal

Want to prove

Can’t prove it in general. Can prove it for the circuits

constructed in phase 1.

Attempt: Prove that EXPNP/poly => EXPAMLet f be an EXP complete function.

Merlin Arthurf(x)=b

The circuit Cf has a small nondeterminist

ic circuit C

Verifies that C(x)=b

Problems:

1. Arthur cannot “run” C. It is a nondeterministic circuit.

2. How can Arthur be sure that C(x)=f(x)?

Thm: [BFL91] EXPP/poly => EXPAMLet f be an EXP complete function.

Merlin Arthurf(x)=b

The circuit Cf has a small deterministic

circuit C


Instance Checker [BK95]: A probabilistic poly-time T which gets oracle access to a function g.

• g=f => Pr[Tg(x)=f(x)]=1.

• g≠f => Pr[Tg (x) =fail]>½.

Thm: [BFL91] EXPP/poly => EXPAMLet f be an EXP complete function.

Merlin Arthurf(x)=b

The circuit Cf has a small deterministic

circuit C


by running TC(x)

Instance Checker [BK95]: A probabilistic poly-time T which gets oracle access to a function g.

• g=f => Pr[Tg(x)=f(x)]=1.

• g≠f => Pr[Tg (x) {fail,f(x)}]>½.

By sending C ,Merlin commits

to some function g!

Nondeterministic Circuits A nondeterministic circuit for f is a

deterministic circuit C(x,y) such that: f(x)=1 => exists y, C(x,y)=1. f(x)=0 => for all y, C(x,y)=0.

Arthur cannot use C to evaluate f. Merlin can help Arthur to evaluate f:

Arthur sends an input x. If f(x)=1, Merlin can send y s.t. C(x,y)=1.

If f(x)=0 ??

Pairs of Nondeterministic Circuits By our assumption EXPNP/poly.

fEXP => f has a nondeterministic circuit. => neg(f) has a nondeterministic circuit!

Arthur can ask Merlin to send both circuits C,C’ for f,neg(f). If f(x)=1, Merlin sends y s.t. C(x,y)=1. If f(x)=0, Merlin sends y s.t. C’(x,y)=1.

There are appropriate witnesses for both cases.

Attempt 2: Prove that EXP in NP/poly => EXP in AMLet f be an EXP complete function.

Merlin Arthurf(x)=b

The circuits C,C’f and neg(f) have small

nondeterministic circuits C,C’

Computes queries x1,..,xt

for the instance checker .I want to evaluate f at

x1,..,xt

Appropriate witnesses for x1,..,xt

Verifies that f(x)=b using the

instance checker .Is it true that by sending C,C’

Merlin commits himself to some function g?

Single Valued pairs of Nondeterministic Circuits If Merlin sends C,C’ which accept all

inputs, he is not at all commited: For every x he can “open” x as both 0 and 1.

A pair (C,C’) defines a function g only if L(C’)=L(C)c . Such a pair is called “single valued”.

Can Arthur verify that C,C’ is a single valued pair?

The big picture

Nondeterministic circuits for EXP (EXPNP/poly)



Nonuniform tradeoff

[MV99,SU01]

Goal

Want to prove

Can’t prove it in general. Can prove it for the circuits

constructed in phase 1.

The argument

EXP is computable by pairs of nondeterministic

circuits which can be certified (probabilistically)

as single valued.



Goal

The protocol I just showed

Nonuniform hardness vs. randomness tradeoff with a

resilient reconstruction .

The final protocol: Using cerified circuits Let f be an EXP complete function.

Merlin Arthurf(x)=b

The certified circuits C,C’f and neg(f) have small

nondeterministic circuits C,C’

Computes queries x1,..,xt

for the instance checker .I want to evaluate f at

x1,..,xt

Appropriate witnesses for x1,..,xt

Verifies that f(x)=b using the

instance checker .As C,C’ are certified!

Merlin commits himself to some function g!

Resilient reconstruction algorithms

EXP is computable by pairs of nondeterministic

single-valued circuits


Nonuniform tradeoff

[MV99,SU01]

The proofs give efficient (prob) “reconstruction algorithms” R(x,a):

If the derandomization fails on x, then there exists an a such that R(x,a) outputs a single-valued pair C,C’ for f.

What does R do when x and a are incorrect?

We cannot expect R to output circuits for f.

We can hope that R outputs a single-valued pair for some function g! We call such an R resilient.

Resilient reconstruction gives certified pairs When Merlin sends the circuits C,C’ he will also

send x and a. Arthur verifies that R(x,a)=(C,C’). This guarantees that (C,C’) is a single-valued

pair of nondeterministic circuits. Open problem: Does there exist a resilient

reconstruction algorithm? We show that the reconstruction algorithm of

[MV99] is “somewhat resilient”. It is resilient to errors in a, but vulnerable to

errors in x. (This is why we get * ).

Partial resiliency We show: the (probabilistic) reconstruction

algorithm of [MV99] is resilient to errors in a. If the derandomization fails on x then for

every a w.h.p. R(x,a) outputs a single-valued pair C,C’ for some function g.

We only get ‘*’ containments because of this weak resiliency.

We cannot trust Merlin to send x, so when the derandomization fails we need a feasible way to come up with x’s on which it failed.

Stronger partial resiliency Actually, we can handle some errors in x. Previous slide: If the derandomization of the

AM language L fails on x then resiliency… Stronger resiliency: If x is not in L then

resiliency… We can trust Merlin to send x if he can give

an AM proof that xL. We can trust Merlin when L is in AM

intersect coAM. No ‘*’ for AM intersect coAM.

ConclusionsMain result: Either Arthur-Merlin protocols are very strong. Or Arthur-Merlin protocols can be

derandomized on feasibly generated inputs.

The technique: Uses nonuniform hardness vs. randomness. Resiliet reconstruction algorithms. Enables using a modified [BFL] protocol.

Open problems: 1. A low-end result. We show that the [MV99] generator has

a (partially) resilient reconstruction algorithm.

The [MV99] result only works for the high-end.

A low-end result by [SU01] which is not even partially resilient!

Open problem: Prove a low-end version of our result.

Open problems: Remove pseudo-containments We show that the [MV99] generator has a

partially resilient reconstruction algorithm. Construct a generator with a fully resilient

reconstruction algorithm. This will remove the * (pseudo-

containment). Solving both open problems will give an

unconditional proof that AMΣ2-SUBEXP!

That’s it…

Date post:	14-Dec-2015
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Uniform Hardness vs. Randomness Tradeoffs for Arthur-Merlin Games. Danny Gutfreund, Hebrew U. Ronen...

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