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
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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.
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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.
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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
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