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Overview of the Previous Overview of the Previous LectureLecture
Gap-QS[O(n),,2||-1]
Gap-QS[O(1),,2||-1]
QS[O(1),]
Solvability[O(1),] 3-SAT
This will imply a strong PCP
characterization of NP
??
3
AimAim
To reduce Gap-QS[O(n),,2||-1] to
Gap-QS[O(1),,2||-1].
p1(x1,...,xn) = 0
p2(x1,...,xn) = 0
...
pn(x1,...,xn) = 0
q1(x16,x21,x32) = 0
q2(x13,x26) = 0
q3(x1,xn,x5n) = 0
...
qm(x3,x4,x5,xn+1) = 0
each equation may depend on many variables
each equation depends only on a constant number of variables
4
New VersionNew Version
Definition (Gap-QS*[D, ,]): Instance: a set of n conjunctions of constant
number of quadratic equations (polynomials) over . Each equation depends on at most D variables.
Problem: to distinguish between:
There is an assignment satisfying all the conjunctions.
No more than an fraction of the conjunctions can be satisfied simultaneously.
5
Conjunctions Rather Than Just Conjunctions Rather Than Just EquationsEquationsAn example for a NO instance of Gap-QS*[1,Z2,½]
x = 0 y = 0
x = 0 y = 1
x = 1 y = 0
Nevertheless, we can satisfy more than a half of the
equations!!
Henceforth, we’ll assume the number of equations in all the
conjunctions is the same. Is this a restriction?
6
RelaxationRelaxation
Claim: Gap-QS*[O(1),,2||-1] reduces to Gap-QS[O(1),,3||-1] (When || is at most polynomial in the size of the input).
Proof: Given an instance of Gap-QS*[O(1),,2||-1], replace each conjunction with all linear combinations of its polynomials.
Make sure that:1) The number of linear combinations over is polynomial.2) The dependency of each linear combination is constant.
7
Correctness of the ReductionCorrectness of the Reduction
If the original system had a common solution, so does the new system.
Otherwise, fix an assignment to the variables of the system and observe the two instances:
8
AnalysisAnalysis
2||-1
fraction of
unsatisfied
conjunctions
fraction of satisfied
conjunctions
polynomials originating
from the blue set
polynomials originating from
the pink set
all satisfied
2||-1
fraction of satisfied polynomials originating
from unsatifiable conjunctions ||-1
9
RelaxationRelaxation
Yes instance of Gap-QS*[O(1),,2||-1] are transformed into Yes instances of Gap-QS[O(1),,3||-1].
No instance of Gap-QS*[O(1),,2||-1] are transformed into No instances of Gap-QS[O(1),,3||-1].
The construction is efficient when || is at most polynomial in the size of the input.
What proves the claim.
10
AmplificationAmplification
Claim: Gap-QS[O(1),,3||-1] reduces to Gap-QS[O(1),,2||-1] (When ||>3 is at most polynomial in the size of the input).
Proof: Given an instance of Gap-QS[O(1),,3||-1], generate the set of all linear combinations of N polynomials.
Make sure that:1) The number of linear combinations over is polynomial.2) The dependency of each linear combination is constant.
A constant to be determined
later.
11
Correctness of the ReductionCorrectness of the Reduction
Again if the original system had a common solution, so does the new system.
Otherwise, fix an assignment and observe a linear combination. – the probability all the N polynomials are
satisfied (3||-1)N
– if not all the polynomials are satisfied, the probability the combination is satisfied ||-1
– When ||>3 for N=4 we get the desired error probability.
12
What Have We Done So Far?What Have We Done So Far?
Gap-QS[O(n),,2||-1]
Gap-QS[O(1),,2||-1]
Gap-QS*[O(1),,2||-1]
Gap-QS[O(1),,3||-1]
QS[O(1),]
Solvability[O(1),] 3-SAT
This will imply a strong PCP
characterization of NP ??
13
New AimNew Aim
To reduce Gap-QS[O(n),,2||-1] to Gap-QS*[O(1),,2||-1].
p1(x1,...,xn) = 0
p2(x1,...,xn) = 0
...
pn(x1,...,xn) = 0
q1(xn,x2n)=0 q2(x1,x2)=0
q3(x13)=0 q2(x12,x234)=0
q4(xn)=0 q5(x1)=0 q4(x1)=0
...
qm(x3,x4,x5,xn+1)=0 qn(x1)=0
each equation may depend on many variables
each conjunction is composed of O(1) equations, which depend only on O(1) variables
14
Rewriting the EquationsRewriting the Equations
Every quadratic polynomial P(x1,...,xD) can be written as
for some series of coefficients {cij}, where
(1) aij=xixj for any 1ijD
(2) ai0=xi for any 1iD
(3) a00=1
Dji ,0
ijij acdepends on the assignment
depends on the polynomial
15
Representing Polynomials As Representing Polynomials As SumsSums For any given point xd we can associate
each aij with a point in Hd (H is some finite field, d=O(log|H|D)).
Now the evaluation of P in this point can be written as
A(h)(h)cdHh
P
16
Representing Polynomials As Representing Polynomials As SumsSumsThe evaluation of every quadratic polynomial at
a point can be written as
where fP(h)=LDEcP(h)·LDEA(h).
dHh
P(h)f
Its total degree is at most 2·d·(|H|-1)
17
Partial SumsPartial Sums
Define:
Hh
d1jj1Hh
d1fd1j
)h,...,h,a,...,f(a...)a,...,a(j,Sum
Sumf can be thought of as a polynomial of total degree at most
2·d2·(|H|-1)
18
Verifying the Polynomial ZeroesVerifying the Polynomial Zeroes