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

2. 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 } SVP: Find the shortest non-zero vector in L. CVP: Given a vector y  R n, find a v  L closest to y. shortest y closest

3. Lattice Approximation Problems g g g-Approximation version: Find a vector whose distance is at most g times the optimal distance. gLy d g-Gap version: Given a lattice L, a vector y, and a number d, distinguish between (dist(y,L)<d) –The ‘yes’ instances (dist(y,L)<d) (dist(y,L)>gd) –The ‘no’ instances (dist(y,L)>gd) gg 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… factor 2 n/2 [LLL] Approximation algorithm for SVP, factor 2 n/2 [Babai] Extension to CVP (1+  ) n [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 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 n 1.5 is in coNP. [GG]: Approximating SVP and CVP to within  n is in coAM  NP.

6. 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 } SVP: Find the shortest non-zero vector in L. CVP: Given a vector y  R n, find a v  L closest to y. shortest y closest

7. Reducing g-SVP to g-CVP [GMSS98] shortest: b 1 -b 2 b1b1 b2b2 The lattice L

8. The lattice L’  L Reducing g-SVP to g-CVP [GMSS98] b1b1 2b 2 shortest vector in L =  c i b i Note: at least one coef. c i of the shortest vector must be odd CVP oracle: apx. minimize || c 1 b 1 +2c 2 b 2 -b 2 ||

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

10. SSAT A new Characterization of NP and the hardness of approximating CVP

11. Hardness of approx. CVP [DKRS] g-CVP is NP-hard for g=n 1/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-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=n 1/loglogn.

13. SAT Input:  =f 1,..,f n Boolean functions ‘tests’ x 1,..,x n’ 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  =f 1,..,f n Boolean functions - ‘tests’ x 1,..,x n’ variables with range R for each test: a list of satisfying assignments Problem Is 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)(1,0,7)(1,3,1)(3,2,2) f(x,y,z) (0,2,7)(2,3,7)(3,1,1)(0,2,7)(2,3,7)(3,1,1) (0,1,0)(2,1,0)(2,1,5)(0,1,0)(2,1,0)(2,1,5)

15. Super-Assignments ||SA(f)|| = |1|+|-2|+|+2| = 5 Norm SA - Average f ||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) 1010 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) 3 2 1 0 -2 (1,1,2) (3,1,1) (3,2,5) (3,3,1) (5,1,2)

16. Consistency 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:

17. Consistency 133 SA(f) = +3(1,1,2)  -2(3,2,5)  2(3,3,1) Consistency: Consistency:  x  f,g that depend on x: S A(f)| x = S A(g)| x SA(f)| x := +3(1)  0(3) -2+2=0 3 2 1 0 -2 (3,2,5) (3,3,1) (1) (2) (3) (1,1,2)

18. g-SSAT - Definition Input:  =f 1,..,f n tests over variables x 1,..,x n’ with range R for each test f i - 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=n 1/loglog n. (conjecture: g=n ,  = some constant)

19. Take a PCP test-system  = {f 1,...,f n } Attempt 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, consistent small-norm (less than g=n 1/loglog n ) Yes instances the GAP

20. g(x,z)h(y,z) f(x,y) (1,2) (2,2) (2,1) A PCP no-instance (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. g(x,z)h(y,z) f(x,y) (1,2) (2,2) (2,1) An SSAT ‘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

22. x0x0 x1x1 f( x 0 x 1 ) +1 (1) +1 (1 2) -1 (2 2) +1 (2 1) +1 (1) x2x2 x3x3 x4x4 x5x5 x6x6 f( x 0 x 1 x 2 x 3 x 4 x 5 x 6 )

23. +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)

24. Original variables Low Degree Extension embed variables in a domain {1..h} d extend the domain {1..p} d (p  h 3, prime) Extension variables Original variables

25. test new tests original variables extension variables. Replace each test with several new tests depending on the original variables and some new extension variables. Low-Degree-Extension satisfying assignment = a Low-Degree-Extension Consistently Reading an LDF

26. 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 had...

27. A Consistent-Reader for L DFs using composition-recursion Short representation. Short representation. Negligible error. Negligible error.

28.  in one piece, by writing its coefficients: too many there are too many degree-h polynomials: there are  p h such polynomials (where h = n 1/loglogn, p  h 3 ).  in many smaller pieces: Representing a degree-h LDF

29. test A 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

30. Embedding Extension (x,y) (x, x 2, x 4, y, y 2, y 4 ) x y X 1 X 2 X 3 y1y2y3y1y2y3 f(. )=x 5 y 2 f e (. )=x 1 x 3 y 2

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

32. SSAT is NP-hard to approximate to within g = n 1/loglogn

33. f(w,x) f’(z,x) 0000000000000000 Reducing SSAT to CVP f,(1,2) f’,(3,2) f,f’,x wwwwwwwwwwwwwwww I ww0www0w 00w000w0 *123*123 Yes --> Yes: dist(L,target) = n No --> No: dist(L,target) > gn Choose w = gn + 1

34. 00w000w0 A consistency gadget *123*123 wwwwwwww ww0www0w

35. w0www0ww w0www0ww w0www0ww w0www0ww w0www0ww w0www0ww w0www0ww w0www0ww w0www0ww w0www0ww w0www0ww w0www0ww 00w000w0 *123*123 wwwwwwww ww0www0w w0www0ww 000w000w 0w000w00 www0www0 + b3+ b3 a 1 + a 2 = 1 + b2+ b2 a 1 + + a 3 = 1 + b1+ b1 a 2 + a 3 = 1 a 1 a 2 a 3 b 1 b 2 b 3 a 1 + a 2 + a 3 = 1

36. Conclusion SSAT SSAT is NP hard to approx. to within g= n 1/loglog n g= n 1/loglog n CVP CVP is NP-hard to approximate to within the same g Future Work: n c, c constant. –Increase to g=n c, c constant. –Extend CVP to SVP reduction