Algebrization: A New Barrier in Complexity Theory

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Algebrization A New Barrier in Complexity Theory

• Scott Aaronson (MIT)
• Avi Wigderson (IAS)

NEXP?P/poly
NP?SIZE(n)
4xyw-12yz17xyzw-2x-2y-2z-2w
IPPSPACE
MIPNEXP
NEXP?P/poly?NEXPMA
-15xyz43xy-5x
13xw-44xzx-7y
PP?SIZE(n)
PP?P/poly?PPMA
PromiseMA?SIZE(n)
MAEXP?P/poly
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What To Call It?

• Algebraic Relativization?
• Algevitization?
• Algevization?
• Algebraicization?
• Algebraization?
• Algebrization?

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Any proof of P?NP will have to defeat two
terrifying monsters
P?NP
Furthermore, even our best weapons seem to work
against one monster but not the other
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Yet within the last decade, weve seen circuit
lower bounds that overcome both barriers
Buhrman-Fortnow-Thierauf 1998 MAEXP ?
P/polyFurthermore, this separation doesnt
relativize Vinodchandran 2004 PP ? SIZE(nk)
for every fixed kA. 2006 Vinodchandrans
result is non-relativizing
Santhanam 2007 PromiseMA ? SIZE(nk) for fixed k
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Bottom Line Relativization and natural proofs,
even taken together, are no longer insuperable
barriers to circuit lower bounds
Obvious Question Santhanam 2007 Is there a
third barrier?
This Talk Unfortunately, yes.
Algebrization A generalization of
relativization where the simulating machine gets
access not only to an oracle A, but also a
low-degree extension à of A over a finite field
or ring

• We show
• Almost all known techniques in complexity theory
  algebrize
• Any proof of P?NPor even PRP or
  NEXP?P/polywill require non-algebrizing
  techniques

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Your result here
BFT, Vinodchandran, Santhanam,
GMW?
Relativizing
Naturalizing
Furst-Saxe-Sipser, Razborov-Smolensky, Raz,
dozens more
Toda, Impagliazzo-Wigderson,
Valiant-Vazirani, Kannan, hundreds more
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Definitions
The inclusion C?D relativizes if CA?DA for all
oracles A CApoly Polynomial-size queries to A
onlyCAexp Exponential-size queries also
allowed
Given an oracle AAn with An0,1n?0,1, an
extension à of A is a collection of polynomials
ÃnZn?Z satisfying (i) Ãn(x)An(x) for all
Boolean x?0,1n,(ii) deg(Ãn)O(n),(iii)
size(Ãn(x)) ? p(size(x)) for some polynomial p,
where
Note Can also consider extensions over finite
fields instead of the integers. Will tell you
when this distinction matters.
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A complexity class inclusion C?D algebrizes if
CA?Dà for all oracles A and all extensions à of
A Proving C?D requires non-algebrizing techniques
if there exist A,Ã such that CA?DÃ A separation
C?D algebrizes if CÃ?DA for all A,Ã Proving C?D
requires non-algebrizing techniques if there
exist A,Ã such that CÃ?DA
Notice weve defined things so that every
relativizing result is also algebrizing.
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Related Work
Low-degree oracles have been studied before for
various reasons (recently by JKRS07) Fortnow94
defined a class O of oracles such that
IPAPSPACEA for all A?O Since he wanted the same
oracle A on both sides, he had to define A
recursively(take a low-degree extension, then
reinterpret as a Boolean function, then take
another low-degree extension, etc.) Proving
separations in his model seems extremely hard
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Why coNP?IP Algebrizes
Recall the usual coNP?IP proof of LFKN
Bullshit!
The only time Arthur ever has to evaluate the
polynomial p directly is in the very last
roundwhen he checks that p(r1,,rn) equals what
Merlin said it does, for some r1,,rn chosen
randomly in the previous rounds.
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How was the polynomial p produced? By starting
from a Boolean circuit, then multiplying together
terms that enforce correct propagation at each
gate
A
Ã(x,y)g (1-Ã(x,y))(1-g)
A(x,y)g (1-A(x,y))(1-g)
Arthur and Merlin then reinterpret p not as a
Boolean function, but as a polynomial over some
larger field.
But what if the circuit contained oracle gates?
Then how could Arthur evaluate p over the larger
field?
Hed almost need oracle access to a low-degree
extension à of A.
Hey, wait
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Other Results That Algebrize
PSPACEApoly ? IPÃ Shamir NEXPApoly ?
MIPÃ BFL PPÃ ? PÃ/poly ? PPA ?
MAÃ LFKN NEXPÃpoly ? PÃ/poly ? NEXPApoly
? MAÃ IKW MAEXPÃexp ? PA/poly BFT PPÃ ?
SIZEA(n) Vinodchandran PromiseMAÃ ?
SIZEA(n) Santhanam ? OWF secure against PÃ ?
NPA ? CZKÃ GMW
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Proving P?NP Will Require Non-Algebrizing
Techniques
Theorem There exists an oracle A, and an
extension Ã, such that NPÃ?PA.
Proof Let A be a PSPACE-complete language, and
let à be the unique multilinear extension of
A. Then à is also PSPACE-complete BFL. Hence
NPÃ PA PSPACE.
By the same argument, P?NP cant even be proved
with double-algebrizing or triple-algebrizing
techniques
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Harder Example Proving PRP Will Require
Non-Algebrizing Techniques(hence PNP as well)
Theorem There exist A,Ã such that RPA?PÃ.
Whats the difficulty here, compared to
standard oracle separation theorems?
Since à is a low-degree polynomial, we dont have
the freedom to toggle each Ã(x) independently.
I.e. the algorithm were fighting is no longer
looking for a needle in a haystackit can also
look in the haystacks low-degree extension!
We will defeat it anyway.
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Theorem Let F be a field, and let Y?Fn be the
set of points queried by the algorithm. Then
there exists a polynomial pFn?F, of degree at
most 2n, such that (i) p(y)0 for all y?Y.(ii)
p(z)1 for at least 2n-Y Boolean points
z.(iii) p(z)0 for the remaining Boolean points.
Y
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Proof Given a Boolean point z, let ?z be the
unique multilinear polynomial thats 1 at z and 0
at all other Boolean points. Then we can express
any multilinear polynomial r as
A standard diagonalization argument now yields
the separation between P and RP we wantedat
least in the case of finite fields.
Requiring r(y)0 for all y?Y yields Y linear
equations in 2n unknowns. Hence there exists a
solution r such that r(z)?0 for at least 2n-Y
Boolean points z. We now set
In the integers case, we can no longer use
Gaussian elimination to construct r. However, we
(i.e. Avi) found a clever way around this problem
using Chinese remaindering and Hensel lifting,
provided every query y satisfies
size(y)O(poly(n)).
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Other Oracle Results We Can Prove By Building
Designer Polynomials
?A,Ã NPA ? coNPÃ ?A,Ã NPA ? BPPÃ (only for
finite fields, not integers) ?A,Ã NEXPÃexp ?
PA/poly ?A,Ã NPÃ ? SIZEA(n)
By contrast, MAEXP ? P/poly and PromiseMA ?
SIZE(n) algebrize!
We seem to get a precise explanation for why
progress on non-relativizing circuit lower bounds
stopped where it did
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From Algebraic Query Algorithms to Communication
Protocols
A(000)1A(001)0A(010)0A(011)1
A(100)0A(101)0A(110)1A(111)1
A0
A1
Truth table of a Boolean function A
Alice and Bobs Goal Compute some property of
the function A0,1n?0,1, using minimal
communication
Let ÃFn?F be the unique multilinear extension of
A over a finite field F
Theorem If a problem can be solved using T
queries to Ã, then it can also be solved using
O(TnlogF) bits of communication between Alice
and Bob
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This argument works just as well in the
randomized world, the nondeterministic world, the
quantum world
Proof Given any point y?Fn, we can write
Also works with integer extensions (we didnt
have to use a finite field).
The protocol is now as follows
Ã(y1)Ã0(y1)Ã1(y1)
y1 (O(nlogF) bits)
Ã1(y1) (O(logF) bits)
Theorem If a problem can be solved using T
queries to Ã, then it can also be solved using
O(TnlogF) bits of communication between Alice
and Bob
y2 (O(nlogF) bits)
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The Harvest Separations in Communication
Complexity Imply Algebraic Oracle Separations
?(2n) randomized lower bound for Disjointness KS 1987 Razborov 1990 ? A,Ã NPA ? BPPÃ
?(2n/2) quantum lower bound for Disjointness Razborov 2002 ? A,Ã NPA ? BQPÃ
?(2n/2) lower bound on MA-protocols for Disjointness Klauck 2003 ? A,Ã coNPA ? MAÃ
Exponential separation between classical and quantum communication complexities Raz 1999 ? A,Ã BQPA ? BPPÃ
Exponential separation between MA and QMA communication complexities Raz-Shpilka 2004 ? A,Ã QMAA ? MAÃ
Advantages of this approach à is just the
multilinear extension of A! Works automatically
with integer extensions
Disadvantage The functions achieving the
separations are more contrived (e.g. Disjointness
instead of OR).
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Can also go the other way algebrization-inspired
communication protocols
Klauck 2003 Disjointness requires ?(?N)
communication, even if theres a Merlin to prove
Alice and Bobs sets are disjoint
Obvious Conjecture Klaucks lower bound can be
improved to ?(N)
This conjecture is false! We give an MA-protocol
for Disjointness (and indeed Inner Product) with
total communication cost O(?N log N)
Hardest communication predicate?
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O(?N log N) MA-protocol for Inner Product
A?N??N?0,1
B?N??N?0,1
Claimed value S for S
r?RF
Alice and Bobs Goal Compute
First step Let F be a finite field with
F?N,2N. Extend A and B to degree-(?N-1)
polynomials
Now let
If Merlin is honest, then
But how to check SS?
If S?S, then
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Conclusions
Arithmetization had a great run. It led to
IPPSPACE, the PCP Theorem, non-relativizing
circuit lower bounds Yet we showed its
fundamentally unable to resolve barrier problems
like P vs. NP, or even P vs. BPP or NEXP vs.
P/poly. Why? It doesnt pry open the black-box
wide enough. I.e. it uses a polynomial-size
Boolean circuit to produce a low-degree
polynomial, which it then evaluates as a black
box. It doesnt exploit the small size of the
circuit in any deeper way. To reach this
conclusion, we introduced a new model of
algebraic query complexity, which has independent
applications (e.g. to communication complexity)
and lots of nooks and crannies to explore in its
own right.
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Open Problems
Develop non-algebrizing techniques! Do there
exist A,Ã such that coNPA ? AMÃ? Improve
PSPACEApoly ? IPÃ to PSPACEÃpoly IPÃ The
power of double algebrization Integer queries
of unbounded size Algebraic query lower bounds ?
communication lower bounds? Generalize to
arbitrary error-correcting codes (not just
low-degree extensions)? Test if a low-degree
extension came from a small circuit?