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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
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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.
shortestyclosest
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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.
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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.
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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.
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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.
shortestyclosest
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Reducing g-SVP to g-CVP Reducing g-SVP to g-CVP [GMSS98][GMSS98]
shortest: b1-b2
b1
b2
The lattice LThe lattice L
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The lattice L’ The lattice L’ L L
Reducing g-SVP to g-CVP Reducing g-SVP to g-CVP [GMSS98][GMSS98]
b1
2b2
shortest vector in L = shortest vector in L = cciibbii
Note: at least one coef. ci of the shortest vector must be odd
CVP oracle:apx. minimize ||c1b1+2c2b2-b2||
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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.
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SSATSSATA new Characterization of NPA new Characterization of NP
and the hardness of approximating CVP
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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-)
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[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 .
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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)
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SAT as a consistency problemSAT as a consistency problemInput=f1,..,fn Boolean functions - ‘tests’x1,..,xn’ variables with range Rfor 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)
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Super-AssignmentsSuper-Assignments
||SA(f)|| = |1|+|-2|+|+2| = 5 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) = +3(5,1,2)-2(3,1,1)2(3,2,5)
3210
-1-2
(1,1,2) (3,1,1) (3,2,5) (3,3,1) (5,1,2)
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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:
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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
3210
-1-2
(3,2,5)
(3,3,1)
(1) (2) (3)
(1,1,2)
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g-g-SSAT - DefinitionSSAT - DefinitionInput:=f1,..,fn tests over variables x1,..,xn’ with range Rfor 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)
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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
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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
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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
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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 )
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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)
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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
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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
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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......
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A Consistent-Reader for LA Consistent-Reader for LDFsDFs
using composition-recursionusing composition-recursion
Short representation.Short representation. Negligible error.Negligible error.
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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
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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
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Embedding ExtensionEmbedding Extension
(x,y) (x, x2, x4, y, y2, y4)
x
y
X1 X2 X3
y1
y2
y3
f(.)=x5y2 fe(.)=x1x3y2
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A Tree A Tree of Consistent Readers of Consistent Readers
lower degree
lower degree
The low-degree-extensiondomain
lower dimension
lower dimension
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SSAT is NP-hard to approximateSSAT is NP-hard to approximateto within to within g = ng = n1/loglogn1/loglogn
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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
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00w0
A consistency gadgetA consistency gadget
*123
wwww
ww0w
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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
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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