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SSAT A new characterization of NP

SSAT A new characterization of NP. and the hardness of approximating CVP. joint work with G. Kindler , R. Raz, and S. Safra. Lattice Problems. Definition: Given v 1 ,..,v k  R n , The lattice L=L(v 1 ,..,v k ) = {  a i v i | integers a i }

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SSAT A new characterization of NP

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  1. SSAT A new characterization of NP and the hardness of approximating CVP. joint work with G. Kindler, R. Raz, and S. Safra

  2. Lattice Problems • Definition: Given v1,..,vkRn, The lattice L=L(v1,..,vk)={aivi | integers ai} • SVP: Find the shortest non-zero vector in L. • CVP: Given a vector yRn, find a vLclosest to y. y shortest closest

  3. Lattice Approximation Problems • g-Approximation version: Find a vector whose distance is at mostg times the optimal distance. • g-Gap version:Given a lattice L, a vector y, and a number d, distinguish between • 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 to CVP • [Schnorr] Improved factor,(1+)nfor 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 equiv. for SVP. • [Ajtai-Dwork96]: Cryptosystem • [Ajtai97]:SVP isNP-hard (for randomized reductions). • [Micc98]:SVP is NP-hardto approximate to within some constant factor. • [LLS]: Approximating CVP to within n1.5 is in coNP. • [GG]: ApproximatingSVP and CVP to within n is in coAMNP.

  6. Lattice Problems • Definition: Given v1,..,vkRn, The lattice L=L(v1,..,vk)={aivi | integers ai} • SVP: Find the shortest non-zero vector in L. • CVP: Given a vector yRn, find a vLclosest to y. y shortest closest

  7. b2 shortest: b2-2b1 b1 Reducing g-SVP to g-CVP [GMSS98] The lattice L

  8. The lattice L’’ L The lattice L’ L shortest vector in L = cibi Reducing g-SVP to g-CVP [GMSS98] CVP oracle: apx. minimize ||c1b1+2c2b2-b2|| L’’=span (2b1,b2) L’=span (b1,2b2) Note: at least one coef. ci of the shortest vector must be odd

  9. The Reduction Input: A pair (B,d), B=(b1,..,bn) and dR for j=1 to n: invoke the CVP oracle on(B(j),bj,d) Output: The OR of all oracle replies. Where B(j) = (b1,..,bj-1,2bj,bj+1,..,bn)

  10. SSATA new Characterization of NP and the hardness of approximating CVP

  11. Hardness of approx. CVP [DKRS] g-CVPis 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-)

  12. [ABSS] reduction:uses PCP to show • NP-hard for g=O(1) • Quasi-NP-hardg=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 .

  13. SAT 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)

  14. SAT as a consistency problem Input =f1,..,fn Boolean functions - ‘tests’ x1,..,xn’ variables with range R for each test: a list of satisfying assignments Problem Is there an assignment to theteststhat is consistent? f(x,y,z) g(w,x,z) h(y,w,x) (0,2,7) (2,3,7) (3,1,1) (1,0,7) (1,3,1) (3,2,2) (0,1,0) (2,1,0) (2,1,5)

  15. f(x,y,z)’s super-assignment SA(f)=-2(3,1,1)+2(3,2,5)+3(5,1,2) 3 2 1 0 -1 -2 (1,1,2) (3,1,1) (3,2,5) (3,3,1) (5,1,2) ||SA(f)|| =|-2|+|2|+|3|= 7Norm SA - Averagef||A(f)|| Super-Assignments A natural assignment for f(x,y,z) A(f) = (3,1,1) 1 0 (1,1,2) (3,1,1) (3,2,5) (3,3,1) (5,1,2)

  16. Consistency In the SAT case: A(f) =(3,2,5) A(f)|x :=(3) x  f,gthat depend on x:A(f)|x= A(g)|x

  17. -2+2=0 SA(f)|x :=+3(1) 0(3) (1,1,2) 3 2 1 0 -1 -2 (3,3,1) (1) (2) (3) (3,2,5) Consistency SA(f) =+3(1,1,2)  -2(3,2,5)  2(3,3,1) Consistency:x  f,g that depend on x:SA(f)|x= SA(g)|x

  18. g-SSAT - Definition Input: =f1,..,fn tests over variables x1,..,xn’ with range R for 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)

  19. Attempt at reducing PCP to SSAT Take a PCP test-system  = {f1,...,fn } Yes instances No instances  Assignment (to vars.) satisfies only  fraction of  the GAP Satisfying assignment for Is there a super-assignment for a ‘no’ instance, consistent small-norm (less than g=n1/loglog n)

  20. A PCP no-instance g(x,z) h(y,z) f(x,y) (1,2) (2,2) (2,1) (1,3) (3,3) (3,1) (1,5) (5,5) (5,1) Best assignment satisfies 2/3 of = {f,g,h} x <--- 1 y <--- 2 z <--- 3

  21. +1 -1 +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) An SSAT ‘almost-yes’-instance g(x,z) h(y,z) f(x,y) (1,2) (2,2) (2,1) (1,3) (3,3) (3,1) (1,5) (5,5) (5,1)

  22. f(x0x1x2 x3 x4 x5 x6) x0 x1 x2 x3 x4 x5 x6 f(x0x1) +1(1 2) -1(2 2) +1(2 1) +1(1) +1(1)

  23. +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) f(x0x1x2 x3 x4 x5 x6) +1(1 23 4 5 6 0) -1(2 22 2 2 2 2) +1(2 10 6 5 4 3) +1(1) +1(1) mod 7 linear extension

  24. Original variables Original variables Extension variables Low Degree Extension • embed variables in a domain {1..h}d • extend the domain {1..p}d (ph3, prime)

  25. Consistently Reading an LDF • Replace each testwith several new testsdepending on the original variables and some newextension variables. • satisfying assignment = a Low-Degree-Extension

  26. Suppose we had... • Consistency Lemma: low-norm super-assignment fortests --> globalsuper-LDF that agrees with the tests. • Deduce a satisfying assignment for almost all of ‘s tests.

  27. A Consistent-Reader for LDFs using composition-recursion • Short representation. • Negligible error.

  28. Representing a degree-h LDF • in one piece, by writing its coefficients: there are too manydegree-h polynomials: there are phsuch polynomials (whereh = n1/loglogn, p  h3). • in many smaller pieces:

  29. almost for almost all cubes. A Consistency Lemma ‘cube’ = constant-dimensional affine subspace test test test test test test Consistency:For every pair of cubes with mutual points -- their super-LDFs agree. Global super-LDF:Agreeing with the cubes’ super-LDFs

  30. fe(.)=x1x3y2 Embedding Extension X1 X2 X3 x f(.)=x5y2 y1 y2 y3 y (x, x2, x4, y, y2, y4) (x,y)

  31. lower dimension lower degree lower dimension lower degree A Tree of Consistent Readers The low-degree-extension domain

  32. SSAT is NP-hard to approximateto within g = n1/loglogn

  33. Yes --> Yes: dist(L,target) = n w w 0 w 0 0 w 0 No --> No: dist(L,target) > gn I Choose w = gn + 1 Reducing SSAT to CVP f,(1,2) f’,(3,2) w w w w w w w w * 1 2 3 f,f’,x 0 0 0 0 0 0 0 0 f(w,x) f’(z,x)

  34. A consistency gadget w w 0 w 0 0 w 0 w w w w * 1 2 3

  35. a1 a2 a3 b1 b2 b3 w 0 w w w 0 w w w 0 w w w 0 w w w 0 w w w 0 w w w 0 w w w 0 w w w 0 w w w 0 w w w 0 w w w 0 w w a1 + a2 +a3 = 1 a2 +a3 +b1 = 1 a1 + +a3 +b2 = 1 a1 + a2 +b3 = 1 A consistency gadget w 0 w w w w 0 w w w w 0 0 w 0 0 0 0 w 0 0 0 0 w w w w w * 1 2 3

  36. Conclusion • SSAT is NP-hard to approx. to within g=n1/loglog n • CVPis NP-hard to approximate to within the sameg • Future Work: • Increase to g=nc, c constant. • Extend CVP to SVP reduction

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