SSATSSATA new characterization of NPA new characterization of NP
and the hardness of approximating CVP.
joint work with G. joint work with G. Kindler, R. Raz, and S. , R. Raz, and S. SafraSafra
Lattice ProblemsLattice Problems
Definition: Given v1,..,vkRn,
The lattice L=L(v1,..,vk) = {aivi | integers ai}
SVP: Find the shortest non-zero vector in L.
CVP: Given a vector yRn, find a vL closest to y.
shortesty
closest
Lattice Approximation ProblemsLattice Approximation Problems
gg-Approximation version: Find a vector whose distance is at most gg times the optimal distance.
gg-Gap version: Given a lattice LL, a vector yy, and a number dd, distinguish between– The ‘yes’ instances (dist(y,L)<d)(dist(y,L)<d)– The ‘no’ instances (dist(y,L)>gd)(dist(y,L)>gd)
If gg-Gap problem is NP-hard, then having a gg-approximation polynomial algorithm --> P=NP.
Lattice Problems - Brief HistoryLattice Problems - Brief History
[Dirichlet, Minkowsky] no CVP algorithms… [LLL] Approximation algorithm for SVP, factor 2factor 2n/2n/2 [Babai] Extension to CVP [Schnorr] Improved factor, (1+(1+))nn 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 HistoryLattice 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 coAMNP.
Lattice ProblemsLattice Problems
Definition: Given v1,..,vkRn,
The lattice L=L(v1,..,vk) = {aivi | integers ai}
SVP: Find the shortest non-zero vector in L.
CVP: Given a vector yRn, find a vL closest to y.
shortesty
closest
Reducing g-SVP to g-CVP Reducing g-SVP to g-CVP [GMSS98][GMSS98]
The lattice LThe lattice L
shortest: b2-2b1
b1
b2
Reducing g-SVP to g-CVP Reducing g-SVP to g-CVP [GMSS98][GMSS98]
shortest vector in L = shortest vector in L = cciibbii
Note: at least one coef. ci of the shortest vector must be odd
The lattice L’’The lattice L’’ L L The lattice L’The lattice L’ L L
CVP oracle:apx. minimize ||c1b1+2c2b2-b2||
L’=span (bL’=span (b11,2b,2b22))L’’=span (2bL’’=span (2b11,b,b22))
The ReductionThe Reduction
Where B(j) = (b1,..,bj-1,2bj,bj+1,..,bn)
Input:Input: A pair (B,d), B=(b A pair (B,d), B=(b11,..,b,..,bnn) and d) and dRR
for j=1 to n: for j=1 to n: invoke the CVP oracle on(Binvoke the CVP oracle on(B(j)(j),b,bjj,d),d)
Output:Output: The OR of all oracle replies. The OR of all oracle replies.
SSATSSATA new Characterization of NPA new Characterization of NP
and the hardness of approximating CVP
Hardness of approx. CVP Hardness of approx. CVP [DKRS][DKRS]
g-CVP is 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-)
[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=n1/loglogn .
SATSAT
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 problemSAT as a consistency problemInput=f1,..,fn Boolean functions - ‘tests’
x1,..,xn’ variables with range R
for each test: a list of satisfying assignments
ProblemIs 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)
f(x,y,z)
(0,2,7)(2,3,7)(3,1,1)
(0,1,0)(2,1,0)(2,1,5)
Super-AssignmentsSuper-Assignments
||SA(f)|| = |-2|+|2|+|3| = 7 Norm SA - Averagef||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)
1
0
A(f) = (3,1,1)
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)
ConsistencyConsistency
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:
ConsistencyConsistency
SA(f) = +3(11,1,2) -2(33,2,5) 2(33,3,1)
Consistency:Consistency: x f,g that depend on x: SA(f)|x = SA(g)|x
SA(f)|x := +3(1) 0(3)
-2+2=0
3
2
1
0
-1
-2
(3,2,5)
(3,3,1)
(1) (2) (3)
(1,1,2)
g-g-SSAT - DefinitionSSAT - 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)
Take a PCP test-system = {f1,...,fn }
Attempt at reducing PCP to SSATAttempt 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,consistentsmall-norm (less than g=n1/loglog n)
Yes instances
the the GAPGAP
g(x,z) h(y,z)f(x,y)
(1,2)(2,2)(2,1)
A PCP A PCP no-instance no-instance
(1,3)(3,3)(3,1)
(1,5)(5,5)(5,1)
Best assignment satisfies 2/3 of = {f,g,h}
x <--- 1y <--- 2z <--- 3
g(x,z) h(y,z)f(x,y)
(1,2)(2,2)(2,1)
An SSAT An SSAT ‘almost-yes’-instance‘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+1
x0 x1
f( x0 x1 )
+1(1)
+1 (1 2)-1 (2 2)+1 (2 1)
+1(1)
x2 x3 x4 x5 x6
f( x0 x1 x2 x3 x4 x5 x6 )
f( x0 x1 x2 x3 x4 x5 x6 )
+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)
Original
variables
Low Degree ExtensionLow Degree Extension
embed variables in a domain {1..h}d
extend the domain {1..p}d (ph3, prime)
Extensionvariables
Original
variables
Replace each testtest with several new testsnew tests depending on the original variables original variables and some new extension variablesextension variables..
satisfying assignment = a Low-Degree-Extension Low-Degree-Extension
Consistently Reading an LDFConsistently Reading an LDF
Consistency Lemma: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 hadSuppose we had......
A Consistent-Reader for LA Consistent-Reader for LDFsDFs
using composition-recursionusing composition-recursion
Short representation.Short representation. Negligible error.Negligible error.
in one piece, by writing its coefficients:
there are too manytoo many degree-h polynomials:
there are ph such polynomials
(where h = n1/loglogn, p h3).
in many smaller pieces:
Representing a Representing a degree-h degree-h LDFLDF
test test test testtesttest
A Consistency LemmaA 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
Embedding ExtensionEmbedding Extension
(x,y) (x, x2, x4, y, y2, y4)
x
y
X1 X2 X3
y1
y2
y3
f(.)=x5y2 fe(.)=x1x3y2
A Tree A Tree of Consistent Readers of Consistent Readers
lower degree
lower degree
The low-degree-extensiondomain
lower dimension
lower dimension
SSAT is NP-hard to approximateSSAT is NP-hard to approximateto within to within g = ng = n1/loglogn1/loglogn
f(w,x)f’(z,x)
00000000
Reducing SSAT to CVPReducing SSAT to CVPf,(1,2) f’,(3,2)
f,f’,x
wwwwwwww
I
ww0w
00w0
*123
Yes --> Yes: dist(L,target) = n
No --> No: dist(L,target) > gn
Choose w = gn + 1
00w0
A consistency gadgetA consistency gadget
*123
wwww
ww0w
w0ww
w0ww
w0ww
w0ww
w0ww
w0ww
w0ww
w0ww
w0ww
w0ww
w0ww
w0ww
00w0
A consistency gadgetA consistency gadget
*123
wwww
ww0w
w0ww
000w
0w00
www0
+ b3 a1 + a2 = 1
+ b2 a1 + + a3 = 1
+ b1 a2 + a3 = 1
a1 a2 a3 b1 b2 b3
a1 + a2 + a3 = 1
ConclusionConclusion
SSATSSAT is NP-hard to approx. to within
g=g=nn1/loglog n1/loglog n
CVPCVP is NP-hard to approximate to within
the samethe same gg
Future Work:– Increase to g=nncc,, c c constant.constant.
– Extend CVP to SVP reduction
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