SSATSSATA new characterization of NPA new characterization of NP

and the hardness of approximating CVP.

joint work with G. joint work with G. Kindler, R. Raz, and S. , R. Raz, and S. SafraSafra

Lattice ProblemsLattice Problems Definition: Given v1,..,vkRn,

The lattice L=L(v1,..,vk) = {aivi | integers ai}

SVP: Find the shortest non-zero vector in L.

CVP: Given a vector yRn, find a vL closest to y.

shortestyclosest

Lattice Approximation ProblemsLattice Approximation Problems gg-Approximation version: Find a vector whose

distance is at most gg times the optimal distance.

gg-Gap version: Given a lattice LL, a vector yy, and a number dd, distinguish between– The ‘yes’ instances (dist(y,L)<d)(dist(y,L)<d)– The ‘no’ instances (dist(y,L)>gd)(dist(y,L)>gd)

If gg-Gap problem is NP-hard, then having a gg-approximation polynomial algorithm --> P=NP.

Lattice Problems - Brief HistoryLattice Problems - Brief History [Dirichlet, Minkowsky] no CVP algorithms… [LLL] Approximation algorithm for SVP, factor 2factor 2n/2n/2 [Babai] Extension to CVP [Schnorr] Improved factor, (1+(1+))nn for both CVP and SVP

[vEB]: CVP is NP-hard [ABSS]: Approximating CVP is

– NP hard to within any constant– Quasi NP hard to within an almost polynomial factor.

Lattice Problems - Recent HistoryLattice Problems - Recent History [Ajtai96]: average-case/worst-case equiv. for SVP. [Ajtai-Dwork96]: Cryptosystem [Ajtai97]: SVP is NP-hard (for randomized reductions). [Micc98]: SVP is NP-hard to approximate to within some constant

factor.

[LLS]: Approximating CVP to within n1.5 is in coNP. [GG]: Approximating SVP and CVP to within n is in coAMNP.

Lattice ProblemsLattice Problems Definition: Given v1,..,vkRn,

The lattice L=L(v1,..,vk) = {aivi | integers ai}

SVP: Find the shortest non-zero vector in L.

CVP: Given a vector yRn, find a vL closest to y.

shortestyclosest

Reducing g-SVP to g-CVP Reducing g-SVP to g-CVP [GMSS98][GMSS98]

shortest: b1-b2

b1

b2

The lattice LThe lattice L

The lattice L’ The lattice L’ L L

Reducing g-SVP to g-CVP Reducing g-SVP to g-CVP [GMSS98][GMSS98]

b1

2b2

shortest vector in L = shortest vector in L = cciibbii

Note: at least one coef. ci of the shortest vector must be odd

CVP oracle:apx. minimize ||c1b1+2c2b2-b2||

The ReductionThe Reduction

Where B(j) = (b1,..,bj-1,2bj,bj+1,..,bn)

Input:Input: A pair (B,d), B=(b A pair (B,d), B=(b11,..,b,..,bnn) and d) and dRR

for j=1 to n: for j=1 to n: invoke the CVP oracle on(Binvoke the CVP oracle on(B(j)(j),b,bjj,d),d)

Output:Output: The OR of all oracle replies. The OR of all oracle replies.

SSATSSATA new Characterization of NPA new Characterization of NP

and the hardness of approximating CVP

Hardness of approx. CVP Hardness of approx. CVP [DKRS][DKRS]

g-CVP is NP-hard for g=n1/loglog n

n - lattice dimension

Improving – Hardness (NP-hardness instead of quasi-

NP-hardness)– Non-approximation factor (from

2(logn)1-)

[ABSS] reduction: uses PCP to show – NP-hard for g=O(1)– Quasi-NP-hard g=2(logn)1- by repeated blow-up.

Barrier - 2(logn)1- const >0

SSAT: a new non-PCP characterization of NP. NP-hard to approximate to within g=n1/loglogn .

SATSAT

Input: =f1,..,fn Boolean functions ‘tests’x1,..,xn’ variables with range {0,1}

Problem: Is satisfiable?

Thm (Cook-Levin): SAT is NP-complete (even when depend()=3)

SAT as a consistency problemSAT as a consistency problemInput=f1,..,fn Boolean functions - ‘tests’x1,..,xn’ variables with range Rfor each test: a list of satisfying assignments

ProblemIs there an assignment to the tests that is consistent?

g(w,x,z) h(y,w,x)(1,0,7)(1,3,1)(3,2,2)

f(x,y,z)(0,2,7)(2,3,7)(3,1,1)

(0,1,0)(2,1,0)(2,1,5)

Super-AssignmentsSuper-Assignments

||SA(f)|| = |1|+|-2|+|+2| = 5 Norm SA - Averagef||A(f)||

A natural assignment for f(x,y,z)

(1,1,2) (3,1,1) (3,2,5) (3,3,1) (5,1,2)

1

0

A(f) = (3,1,1)

f(x,y,z)’s super-assignment

SA(f) = +3(5,1,2)-2(3,1,1)2(3,2,5)

3210

-1-2

(1,1,2) (3,1,1) (3,2,5) (3,3,1) (5,1,2)

ConsistencyConsistency

A(f) = (3,2,5)A(f)|x := (3)

x f,g that depend on x: A(f)|x = A(g)|x

In the SAT case:

ConsistencyConsistency

SA(f) = +3(11,1,2) -2(33,2,5) 2(33,3,1)

Consistency:Consistency: x f,g that depend on x: SA(f)|x = SA(g)|x

SA(f)|x := +3(1) 0(3)

-2+2=0

3210

-1-2

(3,2,5)

(3,3,1)

(1) (2) (3)

(1,1,2)

g-g-SSAT - DefinitionSSAT - DefinitionInput:=f1,..,fn tests over variables x1,..,xn’ with range Rfor each test fi - a list of sat. assign.

Problem: Distinguish between[Yes] There is a natural assignment for [No] Any non-trivial consistent super-assignment is of norm > g

Theorem: SSAT is NP-hard for g=n1/loglog n.

(conjecture: g=n , = some constant)

Take a PCP test-system = {f1,...,fn }

Attempt at reducing PCP to SSATAttempt at reducing PCP to SSAT

Satisfying assignment for

Assignment (to vars.) satisfies only

fraction of

No instances

Is there a super-assignment for a ‘no’ instance,consistentsmall-norm (less than g=n1/loglog n)

Yes instances

the the GAPGAP

g(x,z) h(y,z)f(x,y)(1,2)(2,2)(2,1)

A PCP A PCP no-instance no-instance

(1,3)(3,3)(3,1)

(1,5)(5,5)(5,1)

Best assignment satisfies 2/3 of = {f,g,h}

x <--- 1y <--- 2z <--- 3

g(x,z) h(y,z)f(x,y)(1,2)(2,2)(2,1)

An SSAT An SSAT ‘almost-yes’-instance‘almost-yes’-instance

(1,3)(3,3)(3,1)

(1,5)(5,5)(5,1)

f(x,y) <-- +1(1,2) -1(2,2) +1(2,1) g(x,z) <-- +1(1,3) -1(3,3) +1(3,1) h(y,z) <-- +1(1,5) -1(5,5) +1(5,1)

+1-1+1

x0 x1

f( x0 x1 )

+1(1)

+1 (1 2)-1 (2 2)+1 (2 1)

+1(1)

x2 x3 x4 x5 x6

f( x0 x1 x2 x3 x4 x5 x6 )

f( x0 x1 x2 x3 x4 x5 x6 )

+1(1)

+1 (1 2 3 4 5 6 0 )-1 (2 2 2 2 2 2 2 )+1 (2 1 0 6 5 4 3 )

+1(1) +1(3)-1(2)+1(0)

+1(4)-1(2)+1(6)

+2(5)-1(2)

+1(6)-1(2)+1(4)

+1(0)-1(2)+1(3)

Original variables

Low Degree ExtensionLow Degree Extension embed variables in a domain {1..h}d

extend the domain {1..p}d (ph3, prime)

Extensionvariables

Original variables

Replace each testtest with several new testsnew tests depending on the original variables original variables and some new extension variablesextension variables..

satisfying assignment = a Low-Degree-Extension Low-Degree-Extension

Consistently Reading an LDFConsistently Reading an LDF

Consistency Lemma:Consistency Lemma:low-norm super-assignment for tests --> global super-LDF

that agrees with the tests.

Deduce a satisfying assignment for almost all of ‘s tests.

Suppose we hadSuppose we had......

A Consistent-Reader for LA Consistent-Reader for LDFsDFs

using composition-recursionusing composition-recursion

Short representation.Short representation. Negligible error.Negligible error.

in one piece, by writing its coefficients:

there are too manytoo many degree-h polynomials: there are ph such polynomials

(where h = n1/loglogn, p h3).

in many smaller pieces:

Representing a Representing a degree-h degree-h LDFLDF

test test test testtesttest

A Consistency LemmaA Consistency Lemma

Consistency:Consistency: For every pair of cubes with mutual points --their super-LDFs agree.

Global super-LDF:Global super-LDF: Agreeing with the cubes’ super-LDFs

almost

for almost all cubes.

‘cube’ = constant-dimensional affine subspace

Embedding ExtensionEmbedding Extension

(x,y) (x, x2, x4, y, y2, y4)

x

y

X1 X2 X3

y1

y2

y3

f(.)=x5y2 fe(.)=x1x3y2

A Tree A Tree of Consistent Readers of Consistent Readers

lower degree

lower degree

The low-degree-extensiondomain

lower dimension

lower dimension

SSAT is NP-hard to approximateSSAT is NP-hard to approximateto within to within g = ng = n1/loglogn1/loglogn

f(w,x)f’(z,x)

00000000

Reducing SSAT to CVPReducing SSAT to CVPf,(1,2) f’,(3,2)

f,f’,x

wwwwwwww

I

ww0w

00w0

*123

Yes --> Yes: dist(L,target) = n

No --> No: dist(L,target) > gn

Choose w = gn + 1

00w0

A consistency gadgetA consistency gadget

*123

wwww

ww0w

w0ww

w0ww

w0ww

w0ww

w0ww

w0ww

w0ww

w0ww

w0ww

w0ww

w0ww

w0ww

00w0

A consistency gadgetA consistency gadget

*123

wwww

ww0w

w0ww

000w

0w00

www0

+ b3 a1 + a2 = 1

+ b2 a1 + + a3 = 1

+ b1 a2 + a3 = 1

a1 a2 a3 b1 b2 b3

a1 + a2 + a3 = 1

ConclusionConclusion SSATSSAT is NP hard to approx. to within

g=g=nn1/loglog n1/loglog n

CVPCVP is NP-hard to approximate to within the samethe same gg

Future Work:– Increase to g=nncc,, c c constant.constant.

– Extend CVP to SVP reduction