1 within an almost polynomial factor is np-hard approximating closest vector irit dinur joint work...

1

Within an Almost Polynomial Factor

is NP-hard

Approximating Closest Vector

Irit DinurJoint work with G. Kindler and S. Safra

2

Lattice Problems

Definition: Let v1,..,vk be vectors in Rn.The lattice L=L(v1,..,vk) is the set

{aivi | integers ai}. SVP: Find the shortest non-

zero vector in L.

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

shortest

y

closest

3

Lattice Approximation Problems

g-Approximation version: Find a vector whose distance is at most g times the optimal distance.

g-Gap version: Distinguish between two sets of instances: The ‘yes’ instances (dist(y,L)<d) The ‘no’ instances (dist(y,L)>gd)

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

4

Lattice Problems - Brief History

[Dirichlet, Minkowsky] no CVP algorithms…

[LLL] Approximation algorithm for SVP, factor 2n/2

[Babai] Extension for CVP [Schnorr] Improved factor, (1+)n

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.

5

Lattice Problems - Recent History

[Ajtai96]: average-case/worst-case equivalence for SVP.

[Ajtai-Dwork96]: Cryptosystem [Ajtai97]: SVP is NP-hard in l2. [Micc98]: SVP is hard to

approximate within some constant.

[LLS]: Approximating CVP to within n1.5 is in coNP.

[GG]: Approximating SVP and CVP to within n is in coAMNP.

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Our Results

g-CVP is NP-hard for g=2(logn)1-

o(1)

n - lattice dimension o(1) - 1/loglogcn for any c<0.5

SSAT is NP-hard with gap g

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SAT

Input=f1,..,fn Boolean functions

x1,..,xn’ variables with range {0,1}

ProblemIs satisfiable?

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

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SAT as a consistency problem

Input=f1,..,fn Boolean functions

x1,..,xn’ variables with range {0,1}

ProblemIs there an assignment to the functions that is consistent and satisfying?f(x,y,z) f(x,y’,z’)

(1,0,0) (1,1,0)

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SAT to SIS

Given a SAT instance:f(x,y), g(x,z), h(z,w)

SATSIS

1000001,),,(

0111000,),,(

0001011,),,(

0010100,),,(

0,1,0,0,1,1,1,0,1,0,0,1,

zhg

zhg

xgf

xgf

hhggff

1,00,1

0,11,1

0,01,0

110000

001100

000011

h

g

f

1

1

1

1

1

1

1

10

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Shortest Integer Solution (SIS)

SIS Input: vectors v1,..,vk,t

Problem: Find the shortest integer linear combination of the vi’s that reaches t.

Translating SIS to CVP: Multiply by a large number w Add a distinct ‘counting

coordinate’ per vector

SIS CVP

100

0

,,10

00121

kwvwvwv

L

0

0

wt

y

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Reducing SAT to SIS

Satisfying assignment for

NO satisfying assignment

IS with size=||

IS is with size>||

Yes instances No instances

g

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PCP:fractional gap SAT

Input=f1,..,fn Boolean functions

x1,..,xn variables with range R

ProblemDistinguish between [yes] is satisfiable [no] is no more than 1/R satisfiable

Thm: [RS,DFKRS] PCP is NP-complete for any R<2(logn)1-

even when depend()=O(1)

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Reducing PCP to CVP

Satisfying assignment for

Assignment satisfies only 1/g of

CVP solution with dist<d

CVP solution is of dist >gd

Yes instances No instances

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Super-Assignments

Assign a linear combination of values to each function

f(x,y,z)’s super-assignment

A(f) = c1(1,1,2)+c2(3,2,5)+c3(3,3,1)+...

Natural Assignment:

A(f) = 1·(1,1,2) ||A(f)|| = |c1| + |c2| + |c3| +... Norm A - Averagef||A(f)||

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Consistency

Project the super-assignment to a variable

f(x,y,z)’s super-assignmentA(f) = c1(1,1,2)+c2(3,2,5)+c3(3,3,1)+...

x(A(f)) = c1(1)+(c2+c3)(3)+...

x’s super-assignment

Consistency: x f,g that depend on x

x(A(f)) = x(A(g))

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SSAT - Super-SAT

Input=f1,..,fn Boolean functions

x1,..,xn variables with range R Problem

Distinguish between [Yes] There is a natural assignment [No] Any consistent super-assignment

is of norm > g

Thm

SSAT is NP-hard for g=2(logn)1-o(1)

(perhaps g=nc...)

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SSAT to SIS Reduction;

Given an SSAT instance:f(x,y), g(x,z), h(z,w)

SSATSIS

0110002,),,(

1001001,),,(

1100000,),,(

0011102,),,(

0001011,),,(

0010000,),,(

0,2,1,1,2,1,1,0,1,2,2,1,

zhg

zhg

zhg

xgf

xgf

xgf

hhggff

1,22,1

0,11,2

1,12,0

110000

001100

000011

h

g

f

1

1

1

1

1

1

1

1

1

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1

1

1

1

1

1

The Reduction

Focus on f,g and x

SSATSIS

1

1

1

non-triviality

consistency of f,g on x

f f g g

f g x

f g x

f g x

, , , , , , , ,

( , ), ,

( , ), ,

( , ), ,

12 21 01 12

0 0 0 0 1

1 1 0 1 0

2 0 1 1 1

1 1 0 0

0 0 1 1

0 0 0 0

f

g

h

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SSAT to SIS Reduction;

solution <--> super-assignment

SSATSIS

1

2

2

3

4

5

f <-- -1(1,2) + 2(2,1)g <-- 2(0,1) + -3(1,2)h <-- -4(1,2) + 5(2,1)

solutionf f g g h h, , , , , , , , , , , ,12 21 01 12 11 2 0

0 0 0 1 0 0

1 0 1 0 0 0

0 1 1 1 0 0

0 0 0 0 1 1

0 0 1 0 0 1

0 0 0 1 1 0

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Canellations 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)

x <-- +1·(1)

Y <-- +1·(1)

z <-- +1·(1)

Consistency: All variables areassigned (1) with coefficient +1

Norm: 3 = |+1| + |-1| + |+1|

SSAT

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Low Degree Functions

SSAT

x1 x2 x3

x4 x5 x6

x7 x8 x9

y1 y2 y3 ...

Add auxiliary variables y1 y2

y3 … representing the Low Degree Extension of x1 x2 x3 …

Fd

Hd

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Low Degree Functions

SSAT

f(x1, x9) ---> f’(x1 , x9 , y1 , y3 , y9 ,y2,... )

f’ accepts only low-degree functions whose restrictions to x1 , x9 satisfy f.

x1 x9

y1 y3

y9

y2

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The New Function System

SSAT

Variables - for points in Fd. Functions - Every f in is replaced by planes that

contain its variables.

Super-assignments - ‘super-polynomials’ on a plane that are legal.

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Few Cancellations SSAT

Two distinct LDFs agree on very few points

(hd/|F|) A super-assignment of LDFs, of reasonable size (<g), can cancel

very few variables

(g2hd/|F|)

Choose large |F|=2(logn)1-o(1)

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A Consistency Lemma

SSAT

Assign each plane a super-polynomial (<g) consistently

global super-polynomial G that agrees with ‘most’ planes.

f

gx

A(f) = 1·p1 + -3·p2

A(g) = 1·p3 + -3·p4

p1(x) = p3(x)

p2(x) = p4(x)

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OK but too large...

SSAT

An Assignment to gives an assignment to the planesAny consistent super-assignment of norm <g satisfies most of

Size: The range of the functions is TOO LARGE(there are over FH possible LDFs, F=2(logn)1-o(1) ,H=2c(logn)1-o(1) )

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Recursion

SSAT

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Recursion

SSAT

t11+1

t0t1t3 + 1

Manifold Equations

t1=t02

t2=t12

t3=t22

1/ times

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Conclusion

SSAT is NP hard with g=2(logn)1-o(1)

CVP is NP-hard to approximate to within

the same g

Future Work: Increase g to nc

Extend CVP to SVP reduction