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 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L closest to y.
y
closest
shortest
Lattice Approximation Problems

g-Approximation version: Find a vector whose
distance is at most g 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 gapproximation polynomial algorithm --> P=NP.
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+)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.
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 n1.5 is in coNP.
 [GG]: Approximating SVP and CVP to within n is in
coAMNP.

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L closest to y.
y
closest
shortest
Reducing g-SVP to g-CVP
[GMSS98]
shortest: b1-b2
b1
b2
The lattice L
Reducing g-SVP to g-CVP
[GMSS98]
CVP oracle:
apx. minimize
||c1b1+2c2b2-b2||
b1
2b2
The lattice L’  L
shortest vector in L = cibi
Note: at least one coef. ci of the shortest vector must be odd
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)
SSAT
A new Characterization of NP
and the hardness of approximating CVP
Hardness of approx. CVP
[DKRS]
g-CVP is NP-hard for g=n1/loglog n
n - lattice dimension
Improving
– Hardness (NP-hardness instead of quasi-NPhardness)
– Non-approximation factor (from 2(logn)1-)

[ABSS] reduction: uses PCP to show
– NP-hard for g=O(1)
1-
– Quasi-NP-hard g=2(logn) by repeated blow-up.

Barrier -

SSAT: a new non-PCP characterization of NP.
NP-hard to approximate to within g=n1/loglogn .
2(logn)
1-
const >0
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)
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 the tests that 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)
Super-Assignments
A natural assignment for f(x,y,z)
1
0
A(f) = (3,1,1)
(1,1,2) (3,1,1) (3,2,5) (3,3,1) (5,1,2)
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
-1
(1,1,2) (3,1,1) (3,2,5) (3,3,1) (5,1,2)
-2
||SA(f)|| = |1|+|-2|+|+2| = 5
Norm SA - Averagef||A(f)||
Consistency
In the SAT case:
A(f)
= (3,2,5)
A(f)|x := (3)
x  f,g that depend on x: A(f)|x = A(g)|x
Consistency
SA(f) = +3(1,1,2)

-2(3,2,5)

2(3,3,1)
-2+2=0
SA(f)|x := +3(1)

0(3)
(1,1,2)
3
(3,3,1)
2
1
0
-1
(1)
(2)
(3)
-2
(3,2,5)
Consistency: x  f,g that depend on x: SA(f)|x = SA(g)|x
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=n
(conjecture: g=n ,  = some constant)
1/loglog n
.
Attempt at reducing PCP to SSAT
Take a PCP test-system  = {f1,...,fn }
Yes instances
 Satisfying
assignment for 
No instances
the
GAP
 Assignment (to vars.)
satisfies only
 fraction of 
Is there a super-assignment for a ‘no’ instance,
consistent
small-norm (less than g=n1/loglog n)
A PCP no-instance
f(x,y)
g(x,z)
h(y,z)
(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
An SSAT ‘almost-yes’-instance
+1
-1
+1
f(x,y)
g(x,z)
h(y,z)
(1,2)
(2,2)
(2,1)
(1,3)
(3,3)
(3,1)
(1,5)
(5,5)
(5,1)
f(x,y) <-- +1(1,2)
g(x,z) <-- +1(1,3)
h(y,z) <-- +1(1,5)



-1(2,2)
-1(3,3)
-1(5,5)



+1(2,1)
+1(3,1)
+1(5,1)
f( x0
+1 (1
-1 (2
+1 (2
x1 ) x2
x4
x6 )
x5
2)
2)
1)
x0
+1(1)
x3
+1(1)
x1
x2
x3
x4
x5
x6
f( x0
x1
x2
x3
x4
x5
x6 )
+1 (1
-1 (2
2
2
1
3
2
0
4
2
6
5
2
5
6
2
4
0 )
2 )
3 )
+1 (2
+1(1)
+1(1)
+1(3)
+1(4)
+2(5)
+1(6)
+1(0)
-1(2)
-1(2)
-1(2)
-1(2)
-1(2)
+1(0)
+1(6)
+1(4)
+1(3)
Low Degree Extension


embed variables in a domain {1..h}d
extend the domain {1..p}d (ph3, prime)
Original
variables
Extension
variables
Consistently Reading an LDF


Replace each test with several new tests depending on the
original variables and some new extension variables.
satisfying assignment = a Low-Degree-Extension
Suppose we had...

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.
A Consistent-Reader for LDFs
using composition-recursion
Short representation.
 Negligible error.

Representing a degree-h LDF

in one piece, by writing its coefficients:
too many degree-h
there are
polynomials:
h
there are  p such polynomials
(where h = n1/loglogn, p  h3).

in many smaller pieces:
A Consistency Lemma
‘cube’ = constant-dimensional
affine subspace
test
test
test
test
test
test
almost
Consistency: For every pair of cubes with mutual points -their super-LDFs agree.
Global super-LDF: Agreeing with the cubes’ super-LDFs
for almost all cubes.
Embedding Extension
X1 X2 X3
x
f(.)=x5y2
y
(x,y)
fe(.)=x1x3y2
y1
y2
y3
(x, x2, x4, y, y2, y4)
A Tree
of Consistent Readers
The low-degree-extension
domain
lower dimension
lower degree
lower dimension
lower degree
SSAT is NP-hard to approximate
to within g = n1/loglogn
Reducing SSAT to CVP
f,(1,2)
f,f’,x
*
1
2
3
w
w
0
w
0
0
w
0
I
f(w,x)
f’(z,x)
f’,(3,2)
w
w
w
w
w
w
w
w
0
0
0
0
0
0
0
0
Yes --> Yes:
dist(L,target) = n
No --> No:
dist(L,target) > gn
Choose w = gn + 1
A consistency gadget
*
1
2
3
w
w
0
w
0
0
w
0
w
w
w
w
A consistency gadget
*
1
2
3
a1
a2
a3
b1
b2
b3
www
000
www
www
www
00w
ww0
www
www
00w
www
ww0
ww0
00w
ww0
ww0
ww0
000
www
ww0
ww0
000
ww0
www
a1 + a2 + a3
a2 + a3
a1 +
a1 + a2
+ a3
w
w
w
w
= 1
+ b1
= 1
+ b2
= 1
+ b3
= 1
Conclusion
 SSAT is NP hard to approx. to within
g=n1/loglog n

CVP is NP-hard to approximate to within
the same g
 Future Work:
– Increase to g=nc, c constant.
– Extend CVP to SVP reduction