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Probabilistically Checkable Proofs and Hardness of Approximation. S.Safra some slides borrowed from Dana Moshkovits. The Crazy Tea Party. Problem To seat all guests at a round table, so people who sit an adjacent seats like each other. John. Jane. Mary. Alice. Bob. - PowerPoint PPT Presentation
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1
S.Safra
some slides borrowed from Dana Moshkovits
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The Crazy Tea PartyProblem To seat all guests at a round table, so
people who sit an adjacent seats like each other.
John Mary Bob Jane Alice
John
Mary
Bob
Jane
Alice
3
Solution for the Example
Alice
Bob
Jane
John
Problem To seat all guests at a round table, so people who sit an adjacent seats like each other.
Mary
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Naive Algorithm
• For each ordering of the guests around the table– Verify each guest likes the guest
sitting in the next seat.
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How Much Time Should This Take? (worse case)guests steps
n (n-1)!
5 24
15 87178291200
100 9·10155
say our computer is capable of 1010
instructions per second, this will still take 3·10138 years!
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ToursProblem Plan a trip that visits every site exactly
once.
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Solution for the ExampleProblem Plan a trip that visits every site exactly
once.
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Is a Problem Tractable?
• YES! And here’s an efficient algorithm for it
• NO! and I can prove it
and what if neither is the case?
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Growth Rate: Sketch
10n
n2
2n
n! =2O(n lg
n)
input length
time
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The World According to Complexity
reasonable unreasonable
polynomial nO(1)
exponential 2nO(1)
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Could one be Fundamentally Harder
than the Other?
Tour
Seating
?
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Relations Between Problems
Assuming an efficient procedure for problem A, there is an efficient procedure for
problem B
B cannot be radically harder than A
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Reductions
B p A
B cannot be radically harder than A
In other words: A is at least as hard as B
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Which One is Harder?
Tour
Seating
?
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Reduce Tour to SeatingFirst Observation: The problems aren’t so
different
site guest
“directly reachable from…”
“liked by…”
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Reduce Tour to SeatingSecond Observation: Completing the circle
• Let’s invite to our party a very popular guest,• i.e one who can sit next to everybody else.
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Reduce Tour to Seating
• If there is a tour, there is also a way to seat all the imagined guests around the table.
. . . . . .
popular guest
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Reduce Tour to Seating
• If there is a seating, we can easily find a tour path (no tour, no seating).
. . .
popular guest
. . .
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Bottom Line
The seating problem is at least as hard as the tour problem
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What have we shown?
• Although we couldn’t come up with an efficient algorithm for the problems
• Nor to prove they don’t have one,• We managed to show a very
powerful claim regarding the relation between their hardness
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Furthermore
• Interestingly, we can also reduce the seating problem to the tour problem.
• Moreover, there is a whole class of problems, which can be pair-wise efficiently reduced to each other.
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NPC
NPC
Contains thousands of distinct problem
exponential algorithms
efficient algorithms
?
each reducible to all others
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How can Studying P vs NP Make You a Millionaire?
• This is the most fundamental open question of computer science.
• Resolving it would grant the solver a great honor
• … as well as substantial fortune…www.claymath.org/prizeproblems/pvsnp.htm
• Huge philosophical implications:– No need for human ingenuity!– No need for mathematicians!!!
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Constraints Satisfaction
DefDef Constraints Satisfaction Problem (CSP):– InstanceInstance:
• Constraints: A set of constraints = { 1, …, l } over two sets of variables, X of range RX and Y of range RY
• Determinate: each constraint determines the value of a variable yY according to the value of some xX
xy : RX RY , satisfied if xy(x)=y
• Uniform: each xX appears in dX of , and each yY appears in dY of , for some global dX and dy
– OptimizeOptimize:• Define () = maximum, over all assignments to X and Y
A: X RX; Y RY
of the fraction of satisfied
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Cook’s Characterization of NP
ThmThm: It is NP-hard to distinguish between () = 1 () < 1
For any language L in NP
testing membership in L
can be reduced to...
CSP
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Showing hardness
From now on, to show a problem NP-hard, we merely need to reduce CSP to it.
any NP problem
can be reduced to...
CSP
new, hardproblem
can be reduced to...
Cook’s Thm
will imply the new problem is NP-hard
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Max Independent-Set
Instance: A graph G=(V,E) and a threshold k.Problem: To decide if there is a set of
vertices I={v1,...,vk}V, s.t. for any u,vI: (u,v)E.
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Max I.S. is NP-hard
Proof: We’ll show CSPp Max I.S.
≤p1 12 78346 43 416x y x y x y, ,...,
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The reduction: Co-Partite Graph
• G comprise k=|X| cliques of size |RX| - a vertex for each plausible assignment to x:
kk
An edge: two assignments
that determine a
different value to same
yE {(<i,j1>, <i,j2>) | iM, j1≠j2 RX}
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Proof of CorrectnessAn I.S. of size k must contain exactly one
vertex in every clique.
kk
A satisfying assignment
implies an I.S. of size k
An I.S. of size k corresponds to a
consistent, satisfying assignment
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Generalized Tour Problem
• Add prices to the roads of the tour problem• Ask for the least costly tour
$8
$10
$13$12
$19
$3$17
$13
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Approximation
• How about approximating the optimal tour? • I.e – finding a tour which costs, say, no more
than twice as much as the least costly.
$8
$10
$13$12
$19
$3$17
$13
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Hardness of Approximation
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Promise Problems
• Sometimes you can promise something about the input
• It doesn’t matter what you say for unfeasible inputs
I know my graph has clique of size n/4! Does it have a clique of size
n/2?
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Promise Problems & Approximation
• We’ll see promise problems of a certain type, called gap problems, can be utilized to prove hardness of approximation.
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Gap Problems (Max Version)
• Instance: …
• Problem: to distinguish between the following two cases:
The maximal solution B
The maximal solution ≤ A
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Idea
• We’ve shown “standard” problems are NP-hard by reductions from CSP.
• We want to prove gap-problems are NP-hard
• Why won’t we prove some canonical gap-problem is NP-hard and reduce from it?
• If a reduction reduces one gap-problem to another we refer to it as gap-preserving
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Gap-CSP[]Instance: Same as CSPProblem: to distinguish between the
following two cases:There exists an assignment that satisfies all constraints.No assignment can satisfy more than of the constraints.
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PCP (Without Proof)
Theorem [FGLSS, AS, ALMSS]: For any >0,
Gap-CSP[] is NP-hard,as long as |RX|,|RY| ≥ -O(1)
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Why Is It Called PCP? (Probabilistically Checkable Proofs)
CSP has a polynomial membership proof checkable in polynomial time.
1 12 78346 43 416x y x y x y, ,...,
x1
x2
x3
x4
x5
x6
x7
x8
yn-3
yn-2
yn-1
yn
. . .
My formula is satisfiable!
Prove it!
This assignment satisfies it!
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Why Is It Called PCP? (Probabilistically Checkable Proofs)
…Now our verifier has to check the assignment satisfies all constraints…
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Why Is It Called PCP? (Probabilistically Checkable Proofs)
While for gap-CSP the verifier would be right with high probability, even by:
(1)pick at random a constant number of constraints and
(2)check only those
In a NO instance of gap-CSP, 1-
of the constraints are not satisfied!
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Why Is It Called PCP? (Probabilistically Checkable Proofs)
• Since gap-CSP is NP-hard, All NP problems have probabilistically checkable proofs.
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Hardness of Approximation
• Do the reductions we’ve seen also work for the gap versions (i.e approximation preserving)?
• We’ll revisit the Max I.S. example.
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The same Max I.S. ReductionAn I.S. of size k must contain exactly one vertex in every part.
kk
A satisfying assignment
implies an I.S. of size k
An I.S. of size k corresponds to a
consistent assignment
satisfying of
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Corollary
Theorem: for any >0,Independent-set is hard to
approximate to within any constant factor
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Chromatic Number
• Instance: a graph G=(V,E).• Problem: To minimize k, so that
there exists a function f:V{1,…,k}, for which
(u,v)E f(u)f(v)
skip
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Chromatic NumberObservation:
Each color class is an
independent set
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Clique Cover Number (CCN)
• Instance: a graph G=(V,E).• Problem: To minimize k, so that
there exists a function f:V{1,…,k}, for which
(u,v)E f(u)=f(v)
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Clique Cover Number (CCN)
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Observation
Claim: The CCN problem on graph G is the CHROMATIC-NUMBER problem on the complement graph Gc.
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Reduction Idea
.
.
.
CLIQUE CCN
.
.
.
q
same under cyclic shift
clique preserving
m G G’
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Correctness
• Given such transformation:– MAX-CLIQUE(G) = m CCN(G’) = q– MAX-CLIQUE(G) < m CCN(G’) > q/
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Transformation
T:V[q]
for any v1,v2,v3,v4,v5,v6,
T(v1)+T(v2)+T(v3) T(v4)+T(v5)+T(v6) (mod q)
{v1,v2,v3}={v4,v5,v6}T is unique for triplets
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Observations
• Such T is unique for pairs and for single vertices as well:
• If T(x)+T(u)=T(v)+T(w) (mod q), then {x,u}={v,w}
• If T(x)=T(y) (mod q), then x=y
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Using the Transformation
0 1 2 3 4 … (q-1)
vi
vj
T(vi)=1
T(vj)=4
CLIQUE
CCN
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Completing the CCN Graph Construction
T(s)
T(t)
(s,t)ECLIQUE
(T(s),T(t))ECCN
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Completing the CCN Graph Construction
T(s)
T(t)
Close the set of edges under shift:
For every (x,y)E,
if x’-y’=x-y (mod q), then (x’,y’)E
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Edge Origin Unique
T(s)
T(t)
First Observation: This edge comes
only from (s,t)
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Triangle Consistency
Second Observation: A
triangle only comes from a triangle
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Clique Preservation
Corollary: {c1,…,ck} is a clique in the CCN graph
iff {T(c1),…,T(ck)} is a clique in the CLIQUE graph.
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What Remains?
• It remains to show how to construct the transformation T in polynomial time.
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Corollaries
Theorem: CCN is NP-hard to approximate within any constant factor.
Theorem: CHROMATIC-NUMBER is NP-hard to approximate within any constant factor.
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Max-E3-Lin-2
DefDef: Max-E3-Lin-2– Instance: a system of linear equations
L = { E1, …, En } over Z2
each equation of exactly 3 variables(whose sum is required to equal either 0 or 1)
– Problem: Compute (L)
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Main Theorem
Thm [Hastad]: gap-Max-E3-Lin-2(1-, ½+) is NP-hard.
That is, for every constant >0 it is NP-hard to distinguish between the case 1- of the equations are satisfiable and the case ½+ are.
[ It is therefore NP-Hard to approximateMax-E3-Lin-2 to within 2- constant >0]
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This bound is Tight!
• A random assignment satisfies half of the equations.
• Deciding whether a set of linear equations have a common solution is in P (Gaussian elimination).
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Proof OutlineThe proof proceeds with a reduction from gap-
CSP[], known to be NP-hard for any constant >0
Given such an instance , the proof shows a poly-time construction, of an instance L of Max-E3-Lin-2 s.t. () = 1 (L) ≥ 1 - L
() < (L) ≤ ½ + L
Main Idea:Replace every x and every y with a set of variables
representing a binary-code of their assigned values.Then, test consistency within encoding and any xy
using linear equations over 3 bits
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Long-Code of R
• One bit for every subset of R
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Long-Code of R
• One bit for every subset of R
to encode an element eR
00 00 11 11 11
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The Variables of L
Consider an instance of CSP[], for small constant (to be fixed later)
L has 2 types of variables:
1.a variable z[y,F]z[y,F] for every variable yY and a subset F F P[R P[Ryy]]
2.a variable z[x,F]z[x,F] for every variable xX and a subset F F P[R P[RXX]]
In fact use a “folded” long-code, s.t. f(F)=1-f([n]\F)
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Linearity of a Legal-Encoding
An Boolean function f: P[R] f: P[R] Z Z22,, if legal long-code-word , is a linear-function, that is, for every F, G F, G P[R] P[R]::
f(F) + f(G) F) + f(G) f(F f(FG)G)
where FFG G P[R] P[R] is the symmetric difference of F and G
Unfortunately, any linear function (a sum of a subset of variables) will pass this test
89
The Distribution DefDef: denote by the biased,
product distribution over P[RX], which assigns probability to a subset H as follows:Independently, for each aRX, let– aH with probability 1-– aH with probability One should think of as a multiset of subsets in
which every subset HH appears with the appropriate probability
90
The Linear Equations
L‘s linear-equations are the union, over all ,, of the following set of equations:
FF P[R P[RYY]],, GG P[R P[RXX]] and HH
denote denote FF**== xxyy-1-1(F) (F)
z[y,F] + z[x, G] z[y,F] + z[x, G] z[x, F z[x, F** G G H] H]
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Correctness of ReductionPropProp: if (() = 1) = 1 then (L(L) = 1-) = 1-
ProofProof: let AA be a satisfying assignment to ..Assign all LL ‘s variables according to the legal-
encoding of A’s values.A linear equation of LL, corresponding to xxyy,F,G,H,F,G,H, would be unsatisfied exactly if A(x)HH, which occurs with probability over the choice of H.
LLC-LemmaLLC-Lemma: (L(L) = ½+) = ½+/2/2 (() > 4) > 422
= 2= 2(L) -1(L) -1
Note: independent of ! (Later we use that fact to set small enough for our needs).
92
Denoting an Assignment to L
Given an assignment AL to L’s variables:
For any xX, denote by fx : P[RX] {-1, 1} the
function comprising the values AL assigns to
z[x,*] (corresponding to the long-code of the
value assigned to x)
For any yY, denote by fy : P[RY] {-1, 1} the
function comprising the values AL assigns to
z[y,*] (corresponding to the long-code of the
value assigned to y)Replacing 1 by -1 and 0 by 1
93
Distributional Assignments
Consider a CSP instance Let (R)(R) be the set of all distributions over R
DefDef: A distributional-assignment to isA: X A: X (R(RXX); Y ); Y (R(RXX))
Denote by (()) the maximummaximum over distributional-assignments A of the averageaverage probability for to be satisfied, if variables’ values are chosen according to A
Clearly (() ) (()). . Moreover
PropProp: : (() ) (())
94
The Distributional-Assignment A
DefDef:: Let A be a distributional-assignment to according to the following random processes:
• For any variable xxXX
– Choose a subset SSRRXX with probability
– Uniformly choose a random aS.• For any variable yYY
– Choose a subset SRY with probability
– Uniformly choose a random bS.
2
xf S
2
xf S
For such functions, the squares of the coefficients constitute a distribution
95
What’s to do:
Show that AALL‘s expected success on xxyy is > 4422 in two steps:
First show that AALL‘s success probability, for any xxyy
Then show that value to be 4422
X
12 2
y x y xS Rf odd S f S S
odd(xy(S)) = {b| #{aS| xy(a) = b} is odd}
96
Claim 1
Claim 1Claim 1: : AALL‘s success probability, for any xxyy
ProofProof::That success probability is
Now, taking the sum for only the cases in which Sy=odd(xy(Sx)), results in the claimed inequality.
X
12 2
y x y xS Rf odd S f S S
xy Y x X
2 2
y y x x a S x y yS R ,S Rf S f S Pr a S
100
High Success Probability
'y x x'
y Y x X x X
' ' 'y x x x x'
y Y x X x X
x
x
*F,G,H y x x
' *y y x x x x S S SF,G,H
S R ,S R ,S R
'y y x x x x S S S S SF G H
S R ,S R ,S R
2 Sy x y y x x
S R
E f F f G f F G H
f S f S f S E U F U G U F G H
f S f S f S E U F E U G E U H
f odd S f S 1 2
X
102
Related work• Thm (Friedgut): a Boolean function f with small average-
sensitivity is an [,j]-junta
• Thm (Bourgain): a Boolean function f with small high-frequency weight is an [,j]-junta
• Thm (Kindler&Safra): a Boolean function f with small high-frequency weight in a p-biased measure is an [,j]-junta
• Corollary: a Boolean function f with small noise-sensitivity is an [,j]-junta
• [Dinur, S] Showing Vertex-Cover hard to approximate to within 10 5 – 21
• Parameters: average-sensitivity [BL,KKL,F]; high-frequency weight [H,B], noise-sensitivity [BKS]
103
Boolean Functions and Juntas
A Boolean function
Def: f is a j-Junta if there exists J[n]where |J|≤ j, and s.t. for every x
f(x) = f(x J)
• f is (, j)-Junta if j-Junta f’ s.t.
n
f : P n T,F
f : 1,1 1,1
n
f : P n T,F
f : 1,1 1,1
x
f x f ' xPr x
f x f ' xPr
104
Motivation – Testing Long-code
• Def (a long-code test): given a code-word w, probe it in a constant number of entries, and– accept w.h.p if w is a monotone
dictatorship– reject w.h.p if w is not close to any
monotone dictatorship
105
Motivation – Testing Long-code
• Def(a long-code list-test): given a code-word w, probe it in a constant number of entries, and– accept w.h.p if w is a monotone
dictatorship,– reject w.h.p if a Junta J[n] s.t. f is close
to f’ and f’(F)=f’(FJ) for all F
• Note: a long-code list-test, distinguishes between the case w is a dictatorship, to the case w is far from a junta.
106
Motivation – Testing Long-code
• The long-code test, and the long-code list-test are essential tools in proving hardness results.
Examples …
• Hence finding simple sufficient-conditions for a function to be a junta is important.
107
Noise-Sensitivity• Idea: check how the value of f changes
when the input is changed not on one, but on several coordinates.
[n]x
Iz
108
Noise-Sensitivity
• Def(,p,x[n] ): Let 0<<1, and xP([n]).
Then y~,p,x, if y = (x\I) z where– I~
[n] is a noise subset, and– z~ p
I is a replacement.
Def(-noise-sensitivity): let 0<<1, then
• Note: deletes a coordinate in x w.p. (1-p),adds a coordinate to x w.p. p.
Hence, when p=1/2: equivalent to flipping each coordinate in x w.p. /2.
[n]x
Iz
[n] [n]p ,p,xx~ ,y~
ns f = Pr f x f y
[n] [n]p ,p,xx~ ,y~
ns f = Pr f x f y
109
Noise-Sensitivity – Cont.
• Advantage: very efficiently testable (using only two queries) by a perturbation-test.
• Def (perturbation-test): choose x~p, and y~,p,x, check whether f(x)=f(y). The success is proportional to the noise-sensitivity of f.
• Prop: the -noise-sensitivity is given by
2S
S
2 ns f =1 1 f S 2S
S
2 ns f =1 1 f S
110
Related Work
• [Dinur, S] Showing Vertex-Cover hard to approximate to within 10 5 – 21
• [Bourgain] Showing a Boolean function with weight <1/k on characters of size larger than k, is close to a junta of size exponential in k ([Kindler, S] similar for biased, product distribution)