1 Within an Almost Polynomial Factor is NP-hard Approximating Closest Vector Irit Dinur Joint work with G. Kindler and S. Safra

2 Lattice Problems uDefinition: Let v 1,..,v k be vectors in R n.The lattice L=L(v 1,..,v k ) is the set {  a i v i | integers a i }. uSVP: Find the shortest non- zero vector in L. uCVP: Given a vector y  R n, find a v  L closest to y. shortest y closest

3 Lattice Approximation Problems ug-Approximation version: Find a vector whose distance is at most g times the optimal distance. ug-Gap version: Distinguish between two sets of instances:  The ‘yes’ instances (dist(y,L)<d)  The ‘no’ instances (dist(y,L)>gd) uIf g-Gap problem is NP-hard, then having a g-approximation polynomial algorithm --> P=NP.

4 Lattice Problems - Brief History u[Dirichlet, Minkowsky] no CVP algorithms… u[LLL] Approximation algorithm for SVP, factor 2 n/2 u[Babai] Extension for CVP u[Schnorr] Improved factor, (1+  ) n for both CVP and SVP u[vEB]: CVP is NP-hard u[ABSS]: Approximating CVP is  NP hard to within any constant  Quasi NP hard to within an almost polynomial factor.

5 Lattice Problems - Recent History u[Ajtai96]: average-case/worst-case equivalence for SVP. u[Ajtai-Dwork96]: Cryptosystem u[Ajtai97]: SVP is NP-hard in l 2. u[Micc98]: SVP is hard to approximate within some constant. u[LLS]: Approximating CVP to within n 1.5 is in coNP. u[GG]: Approximating SVP and CVP to within  n is in coAM  NP.

6 Our Results ug-CVP is NP-hard for g=2 (logn) 1- o(1)  n - lattice dimension  o(1) - 1/loglog c n for any c<0.5 uSSAT is NP-hard with gap g

7 SAT uInput  =f 1,..,f n Boolean functions x 1,..,x n’ variables with range {0,1} uProblem Is  satisfiable? uThm: (Cook) SAT is NP-complete (even when depend(  )=3)

8 SAT as a consistency problem uInput  =f 1,..,f n Boolean functions x 1,..,x n’ variables with range {0,1} uProblem Is 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)

9 SAT to SIS uGiven a SAT instance: f(x,y), g(x,z), h(z,w) SAT  SIS 1,0 0,1 1,1 0,0 1,0

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11 Shortest Integer Solution (SIS) uSIS Input: vectors v 1,..,v k,t  Problem: Find the shortest integer linear combination of the v i ’ s that reaches t. uTranslating SIS to CVP:  Multiply by a large number w  Add a distinct ‘ counting coordinate ’ per vector SIS  CVP

12 Reducing SAT to SIS  Satisfying assignment for  NO satisfying assignment  IS with size=|  |  IS is with size>|  | Yes instancesNo instances g

13 PCP: fractional gap SAT uInput  =f 1,..,f n Boolean functions x 1,..,x n variables with range R uProblem Distinguish between  [yes]  is satisfiable  [no]  is no more than 1/R satisfiable uThm: [RS,DFKRS] PCP is NP- complete for any R<2 (logn) 1-  even when depend(  )=O(1)

14 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 instancesNo instances

15 Super-Assignments uAssign a linear combination of values to each function f(x,y,z)’s super-assignment A(f) = c 1 (1,1,2)+c 2 (3,2,5)+c 3 (3,3,1)+... uNatural Assignment: A(f) = 1·(1,1,2) u||A(f)|| = | c 1 | + | c 2 | + | c 3 | +... uNorm A - Average f ||A(f)||

16 Consistency uProject the super-assignment to a variable f(x,y,z)’s super-assignment A(f) = c 1 (1,1,2)+c 2 (3,2,5)+c 3 (3,3,1)+...  x (A(f)) = c 1 (1)+(c 2 +c 3 )(3)+... x’s super-assignment uConsistency:  x  f,g that depend on x  x (A(f)) =  x (A(g))

17 SSAT - Super-SAT uInput  =f 1,..,f n Boolean functions x 1,..,x n variables with range R uProblem Distinguish between  [Yes] There is a natural assignment  [No] Any consistent super- assignment is of norm > g uThm SSAT is NP-hard for g=2 (logn) 1-o(1) (perhaps g=n c...)

18 SSAT to SIS Reduction; uGiven an SSAT instance: f(x,y), g(x,z), h(z,w) SSAT  SIS 1,2 2,1 0,1 1,2 1,1 2,0

19 The Reduction uFocus on f,g and x SSAT  SIS non-triviality consistency of f,g on x

20 SSAT to SIS Reduction; solution super-assignment SSAT  SIS f <-- -1(1,2) + 2(2,1) g <-- 2(0,1) + -3(1,2) h <-- -4(1,2) + 5(2,1) solution

21 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 are assigned (1) with coefficient +1 Norm: 3 = |+1| + |-1| + |+1| SSAT

22 Low Degree Functions SSAT x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 y 1 y 2 y 3... zAdd auxiliary variables y 1 y 2 y 3 … representing the Low Degree Extension of x 1 x 2 x 3 … FdFd HdHd

23 Low Degree Functions SSAT f( x 1, x 9 ) ---> f’(x 1, x 9, y 1, y 3, y 9,y 2,... ) f’ accepts only low-degree functions whose restrictions to x 1, x 9 satisfy f. x 1 x 9 y 1 y 3 y 9 y 2

24 The New Function System SSAT zVariables - for points in F d. zFunctions - Every f in  is replaced by planes that contain its variables. zSuper-assignments - ‘super-polynomials’ on a plane that are legal.

25 Few Cancellations SSAT uTwo distinct LDFs agree on very few points (hd/|F|) uA super-assignment of LDFs, of reasonable size (<g), can cancel very few variables (g 2 hd/|F|) uChoose large |F|= 2 (logn) 1-o(1)

26 A Consistency Lemma SSAT uAssign each plane a super- polynomial (<g) consistently  global super-polynomial G that agrees with ‘most’ planes. f g x A(f) = 1·p ·p 2 A(g) = 1·p ·p 4 p 1 (x) = p 3 (x) p 2 (x) = p 4 (x)

27 OK but too large... SSAT An Assignment to  gives an assignment to the planes Any consistent super- assignment of norm <g satisfies most of  Size: The range of the functions is TOO LARGE (there are over F H possible LDFs, F= 2 (logn) 1-o(1),H= 2 c(logn) 1-o(1) )

28 Recursion SSAT

29 Recursion SSAT t t 0 t 1 t Manifold Equations t 1 =t 0 2 t 2 =t 1 2 t 3 =t 2 2 1/  times

30 Conclusion uSSAT is NP hard with g= 2 (logn) 1-o(1) uCVP is NP-hard to approximate to within the same g uFuture Work:  Increase g to n c  Extend CVP to SVP reduction