11

Tight Hardness Results forTight Hardness Results forSome Approximation Some Approximation

ProblemsProblems

[Raz,Håstad,...][Raz,Håstad,...]Adi Akavia Adi Akavia

Dana Moshkovitz & Ricky Rosen Dana Moshkovitz & Ricky Rosen S. SafraS. Safra

22

“ “Road-Map” for Chapter IRoad-Map” for Chapter I

Parallel repetition lemma

XY is NP-hard

expander

par[, k]

Gap-3-SAT-7

Gap-3-SAT

33

Maximum SatisfactionMaximum Satisfaction

DefDef: : Max-SATMax-SAT InstanceInstance::

A set of variables A set of variables Y = { YY = { Y11, …, Y, …, Ym m }} A set of Boolean-functions (A set of Boolean-functions (local-testslocal-tests) over ) over YY

= { = { 11, …, , …, ll } }

MaximizationMaximization:: We define We define (() =) = maximum, over all assignments to maximum, over all assignments to

YY, of the fraction of , of the fraction of ii satisfiedsatisfied

StructureStructure:: Various versions of Various versions of SATSAT would impose structure would impose structure

properties on properties on YY, , YY’s range and ’s range and

44

Max-E3-Lin-2Max-E3-Lin-2

DefDef: : Max-E3-Lin-2Max-E3-Lin-2 InstanceInstance: a system of linear : a system of linear

equationsequationsL = { EL = { E11, …, E, …, Enn } } over over ZZ22

each equation of exactly each equation of exactly 33 variablesvariables[whose sum is required to equal either [whose sum is required to equal either 00 or or 11]]

ProblemProblem: Compute : Compute (L)(L)

55

ExampleExample

Assigning Assigning xx1-61-6=1, x=1, x7-97-9=0=0 satisfy all but the third satisfy all but the third equation.equation.

No assignment can satisfy No assignment can satisfy all equation, as the sum all equation, as the sum of all leftwing of of all leftwing of equations equals zero equations equals zero (every variable appears (every variable appears twice) while the rightwing twice) while the rightwing sums to 1.sums to 1.

Therefore, Therefore, (L)=5/6(L)=5/6..

x1+x2+x3=1 (mod 2)x4+x5+x6=1

x7+x8+x9=1

x1+x4+x7=0

x2+x5+x8=0

x3+x6+x9=0

66

2-Variables Functional SAT

DefDef[ [ XY ]: ]: over over variables variables X,YX,Y of range of range RRxx,R,Ryy respectively respectively each each is of the form is of the form xxyy: R: RxxRRyy

an assignment an assignment A(XA(XRRxx, Y, YRRyy)) satisfies satisfies xxyy iff iffxxyy(A(x))=A(y)(A(x))=A(y)

[ Namely, every value to [ Namely, every value to xx determines exactly 1 satisfying value for determines exactly 1 satisfying value for yy]]

ThmThm:: distinguishing between distinguishing between AA satisfies all satisfies all AA satisfies satisfies << fraction of fraction of

isis NP-hard NP-hard as long as as long as |R|Rxx|,|R|,|Ryy|> |> -0(1)-0(1)

77

Proof OutlineProof OutlineDefDef: : 3SAT3SAT is is SATSAT where every where every ii is is

a disjunction of a disjunction of 33 literals. literals.

DefDef: : gap-3SAT-7 gap-3SAT-7 is is gap-3SATgap-3SAT with the with theadditional restriction, that everyadditional restriction, that everyvariable appearsvariable appears in in exactly exactly 77 local- local-teststests

TheoremTheorem: : gap-3SAT-7gap-3SAT-7 is NP-hard is NP-hard

XY is NP-hard

par[, k]

Gap-3-SAT-7

Gap-3-SAT

?

88

ExpandersExpanders

DefDef: a graph : a graph G(V,E)G(V,E) is a is a c-expander c-expander if for if for everyevery SSV, V, |S||S| ½|V| ½|V|: : |N(S)\S| |N(S)\S| c·|S| c·|S|

[where [where N(S)N(S) denotes the set of neighbors of denotes the set of neighbors of SS]]

LemmaLemma: For every : For every mm, one can construct in , one can construct in poly-time a poly-time a 33-regular, -regular, mm-vertices, -vertices, cc-expander, for some constant -expander, for some constant c>0c>0

CorollaryCorollary: a cut between: a cut between S S andand V\S V\S, for, for |S| |S| ½|V| ½|V| must containmust contain > c·|S| > c·|S| edgesedges

XY is NP-hard

par[, k]

Gap-3-SAT-7

Gap-3-SAT

Expanders?

99

Reduction Using ExpandersReduction Using Expanders

Assume Assume ’’ for which for which ((’) ’) is either is either 11 or or 1-1-2020/c/c. . is is ’’ with the following changes: with the following changes:• an occurrence of an occurrence of yy in in ii is replaced by a is replaced by a

variable variable xxy,iy,i

• Let Let GGyy, for every , for every yy, be a , be a 33-regular, -regular, cc--expander over all occurrences expander over all occurrences xxy,iy,i of of yy

• For every edge connecting For every edge connecting xxy,i y,i to to xxy,jy,j in in GGyy, add , add to to the clauses the clauses ((xxy,iy,i x xy,jy,j) ) and and (x(xy,iy,i x xy,jy,j))

It is easy to see that:It is easy to see that:1.1. ||| | 10 | 10 |’|’|2.2. Each variable Each variable xxy,iy,i of of appears in exactly appears in exactly 77 i i

ensuring equality

constructible by the Lemma

1010

Correctness of the Correctness of the ReductionReduction

1.1. is completely satisfiable iff is completely satisfiable iff ’’ is is2.2. In case In case ’’ is unsatisfiable: is unsatisfiable: ((’) <’) < 1-201-20/c/c

Let Let AA be an optimal assignment to be an optimal assignment to Let Let AAmajmaj assign assign xxy,iy,i the value assigned by the value assigned by AA to the to the majority, over majority, over jj, of variables , of variables xxy,jy,j

Let Let FFAA and and FFAAmajmaj be the sets of be the sets of

ununsatisfied by satisfied by AA and and AAmajmaj respectively: respectively:

|||·(1-|·(1-(()) = |F)) = |FAA| = |F| = |FAAFFAAmajmaj|+|F|+|FAA\F\FAAmajmaj

| |

|F|FAAFFAAmajmaj|+½c|F|+½c|FAAmajmaj

\F\FAA| | ½c|F ½c|FAAmajmaj| |

andand since since AAmajmaj is in fact an assignment to is in fact an assignment to ’’

(() ) 1- ½c(1- 1- ½c(1- ((’))/10 < 1- ’))/10 < 1- ½c(20½c(20/c)/10= 1-/c)/10= 1-

1111

NotationsNotations

DefDef: For a : For a 3SAT3SAT formula formula over Boolean variables over Boolean variables ZZ, , Let Let ZZkk be the set of all be the set of all kk-sequences of -sequences of ’s ’s

variablesvariables Let Let kk be the set of all be the set of all kk-sequences of -sequences of ’s ’s

clausesclauses

DefDef: Let : Let Y=ZY=Zk k and and X=X=kk, and, and RRYY be the set of all assignments to be the set of all assignments to k k variablesvariables RRXX be the set of all sat. assignments to be the set of all sat. assignments to k k

clausesclauses

DefDef: For any set of : For any set of kk variables variables yyZZkk, and a set of , and a set of k k clauses clauses xxkk, , denote denote y xy x yy is a choice of one variable of is a choice of one variable of each clause in each clause in xx..

1212

Parallel SATParallel SAT

DefDef: for a : for a 3SAT3SAT formula formula over Boolean over Boolean variables variables ZZ, let , let par[par[, k], k] be the be the following 2-var functional SATfollowing 2-var functional SATpar[par[, k], k]’s ’s variablesvariables:: Y=ZY=Zkk (one var. for each (one var. for each k zk z’s) each ’s) each yy‘s ‘s

range is range is RRYY

XXkk (one var. for each (one var. for each k k ‘s) each ‘s) each xx‘s ‘s range is range is RRXX

par[par[, k], k] local-testslocal-tests:: xxyy for every for every y xy x which which acceptsaccepts if if xx’s value’s value restricted torestricted to y y isis y y’s value’s value[namely, if the assignments are [namely, if the assignments are consistentconsistent] ]

|RY|=2k

|RX|=7k

1313

Gap Increases with kGap Increases with k

Note that if Note that if (() = 1) = 1 then then

(par[(par[, k]) = 1, k]) = 1

On the other hand, if On the other hand, if is not satisfiable: is not satisfiable:

LemmaLemma: : (par[(par[, k]) , k]) (())c·kc·k for some for some c>0c>0

ProofProof: :

first note thatfirst note that1-1-(par[(par[, 1]) , 1]) (1- (1-(())/3))/3

now, to prove the lemma, apply the Parallel-now, to prove the lemma, apply the Parallel-Repetition lemma [Raz] to Repetition lemma [Raz] to par[par[, 1], 1]

In any assignment to s variables, any unsatisfied clause in ”induces“ at least 1 (out of corresponding 3) unsatisfied par[par[, 1], 1]

XY is NP-hard

par[, k]

Gap-3-SAT-7

Gap-3-SAT

Parallel repetition

lemma

1414

ConclusionConclusion: : XY is NP-hard

distinguishdistinguish between: between: AA satisfies all satisfies all AA satisfies satisfies <<pp fraction of fraction of

isis NP-hard NP-hard as long as as long as |R|Rxx|,|R|,|Ryy|| pp-0(1)-0(1)

1515

“ “Road-Map” for Chapter IIRoad-Map” for Chapter II

() = ()

LLC-LemmaLLC-Lemma:: (L) = ½+/2 (par[,k]) > 42

Long code

L

XY

is NP-hard

1616

Main TheoremMain Theorem

ThmThm: : gap-gap-Max-E3-Lin-2Max-E3-Lin-2(1-(1-, ½+, ½+)) is is NPNP--hardhard..

That is, for every constant That is, for every constant 0<0<<¼<¼ it is it is NP-hardNP-hard to distinguish between the to distinguish between the case where case where 1-1- of the equations are of the equations are satisfiable and the case where satisfiable and the case where ½+½+ are.are.

[ It is therefore [ It is therefore NP-HardNP-Hard to approximate to approximateMax-E3-Lin-2Max-E3-Lin-2 to within factor to within factor 2-2- for any for any constant constant 0<0<<<¼¼]]

1717

This bound is tightThis bound is tight

A random assignment satisfies half of A random assignment satisfies half of the equations.the equations.

Deciding whether a set of linear Deciding whether a set of linear equations have a common solution is equations have a common solution is in in PP (Gaussian elimination). (Gaussian elimination).

1818

Distributional Distributional AssignmentsAssignments

Let Let be a be a SATSAT instance over variables instance over variablesZZ of range of range RR..

Let Let (R)(R) be all distributions over be all distributions over RR

DefDef: a : a distributional-assignmentdistributional-assignment to to is is A: Z A: Z (R)(R)

Denote byDenote by (() ) thethemaximummaximum over distributional-assignments over distributional-assignments AA of the of theaverageaverage probability for probability for to be satisfied, to be satisfied,if variables` values are chosen according to if variables` values are chosen according to AA

ClearlyClearly (() ) (()). Moreover. Moreover

PropProp: : (() ) (() )

LLC-LemmaLLC-Lemma:: (L) = ½+/2 (par[,k]) > 42

Long code

L

XY is

NP-hard

1919

Distributional-assignment to Distributional-assignment to

1 1

1

1

0

00

0

x1

x2

x3

xn

2020

Restriction and ExtensionRestriction and Extension

DefDef: :

For any For any yyYY over over RRYY and and xxXX over over RRXX s.t s.t xxyy

The The natural restrictionnatural restriction of an of an aaRRXX to to RRYY is denoted is denoted aa|y|y

The The elevationelevation of a subset of a subset FFP[RP[RYY]] to to RRXX is the subset is the subset FF**P[RP[RXX]] of all of all members members aa of of RRXX for which for which xxyy(a)(a) F F

F* = { a | aF* = { a | a|y|y F } F }

2121

Long-CodeLong-CodeIn the long-code the set of legal-words consists In the long-code the set of legal-words consists of all monotone dictatorshipsof all monotone dictatorshipsThis is the most extensive binary code,This is the most extensive binary code,as its bits represent all possible binary values overas its bits represent all possible binary values overnn elements elements

LLC-LemmaLLC-Lemma:: (L) = ½+/2 (par[,k]) > 42

Long code

L

XY is

NP-hard

2222

Long-CodeLong-Code

Encoding an element Encoding an element ee[n][n] :: EEee legally-encodeslegally-encodes an element an element ee if if EEee = f = fee

FF FF TT TT TT

2323

Long-Code over Range RLong-Code over Range RBP[R] BP[R] the set of all the set of all

subsets of subsets of RR of size of size ≤½|R|≤½|R|

Our long-codeOur long-code: in our context there’re two : in our context there’re two types of domains “types of domains “RR”:”: RRxx and and RRyy . .

DefDef: an : an RR-long-code-long-code has 1 bit for each has 1 bit for each FF BP[R] BP[R]

DefDef: a : a legal-long-code-wordlegal-long-code-word encoding an encoding an element element eeRR, is , is EERR

ee: BP[R] : BP[R] {-1, 1} {-1, 1} that that assigns the Boolean value assigns the Boolean value eeFF to every to every subset subset F F BP[R] BP[R]

|BP[R]| = 2|R|-1-1

2424

Linearity of a Legal-Linearity of a Legal-EncodingEncoding

An assignment An assignment A : BP[R] A : BP[R] {-1,1} {-1,1}, if , if legal, is a legal, is a linear-functionlinear-function, i.e., , i.e., F, G F, G BP[R] BP[R]::

A(F) A(F) A(G) A(G) A(F A(FG) G)

Unfortunately, any character is linear Unfortunately, any character is linear as wellas well

2525

The Variables of LThe Variables of L

Consider Consider ( (xxyy) ) withwithsmall constant small constant pp (to be (to befixed later)fixed later)

LL has 2 types of has 2 types of variablesvariables::

1.1. a variable a variable z[y,F]z[y,F] for every variable for every variable yy of of and a subset and a subset F F BP[R BP[RYY]]

2.2. a variable a variable z[x,F]z[x,F] for every variable for every variable xx of of and a subset and a subset F F BP[R BP[RXX]]

LLC-LemmaLLC-Lemma:: (L) = ½+/2 (par[,k]) > 42

Long code

L

XY is

NP-hard

2626

The Distribution The Distribution

DefDef: denote by : denote by the distribution the distribution over all subset of over all subset of RRxx, which assigns , which assigns probability to a subset probability to a subset HH as follows: as follows:

Independently, for each Independently, for each a a R Rxx, let, let aaHH with probability with probability aaHH with probability with probability 1-1-

One should think of as a multiset of subsets in which every subset HH appears with the appropriate

probability

2727

Linear equationLinear equation

LL‘s ‘s linear-equationslinear-equations are the are the union, over all union, over all xxyy par[ par[,k],k], , of:of:

xxyy , F, G, H , F, G, H there’s a testthere’s a test

z[y, F] + z[x, G] = z[x, Fz[y, F] + z[x, G] = z[x, F* * G G H ] (mod H ] (mod 2)2)

‘‘F*F*GGHH’’ is the symmetric symmetric differencedifference of the

extension of FF to SSCC, GG and HH

2828

Revised RepresentationRevised Representation

Multiplicative RepresentationMultiplicative Representation:: True True -1 -1 False False 1 1 L:L:

z[X,*], z[Y,*] z[X,*], z[Y,*] {-1, 1} {-1, 1} z[X, F] • z[Y, G] • z[X, F* • G • H ] = 1z[X, F] • z[Y, G] • z[X, F* • G • H ] = 1

Representation by Fourier

Basis

Claim 2

(par[(par[,k]) > 4,k]) > 422

Claim 1

Claim 3:The expected success of the distributional assignment on [C,V]par[,k] is at least 4 4 22

General

Fourier Analysis facts

Multiplicative

representation

2929

PropProp: if : if (() = 1) = 1 then then (L(L) = 1-) = 1-

ProofProof::Let Let AA be a satisfying assignment be a satisfying assignment to to . Assign all variables of . Assign all variables of LL

according to the legal encoding of according to the legal encoding of AA’s values.’s values.A linear equation of A linear equation of LL, corresponding to , corresponding to

X,Y,F,G,HX,Y,F,G,H, would be unsatisfied exactly if , would be unsatisfied exactly if A(x)A(x)HH, which occurs with probability , which occurs with probability over over the choice of the choice of HH..

LLC-LemmaLLC-Lemma: : (L(L) = ½+) = ½+/2/2 (() > 4) > 422

= 2= 2(L) -1(L) -1

Note: independent of pp! (Later we define pp small enough).

LLC-LemmaLLC-Lemma:: (L) = ½+/2 (par[,k]) > 42

L

XY is

NP-hard

3030

Hardness of approximating Max-E3-Hardness of approximating Max-E3-Lin-2Lin-2

Main TheoremMain Theorem::

For any constant For any constant >0>0::gap-Max-E3-Lin-2(1-gap-Max-E3-Lin-2(1-,½+,½+) ) is is NP-NP-hardhard..

ProofProof::By By propositionproposition

((’) = 1 ’) = 1 (L(L’’) ) 1- 1-

3131

Lemma Lemma Main Theorem Main Theorem

PropProp::Let Let be a constant be a constant >0>0 s.t.: s.t.:

(1-(1-)/(½+)/(½+/2) /2) 2- 2-SetSet pp < 4 < 433 Then Then (() < 1 ) < 1 (L(L’’) ) ½+ ½+/2 /2 ½+ ½+

ProofProof: : Assume, by way of contradiction, Assume, by way of contradiction, that that (L) (L) ½+ ½+/2 /2 then:then:4433 > > ((’)’)c·k c·k (() > 4) > 422,,

which implies that which implies that > > .. Contradiction!Contradiction!

of the parallel repetition

lemma

3333

An Assignment An Assignment to L to L

For any variable For any variable xx of of

The set The set z[x,*]z[x,*] of variables of of variables of L L represents the represents the

long-code of long-code of xx

Let be the Fourier-Coefficient Let be the Fourier-Coefficient <<|z[x,*]|z[x,*],,ss>>

For any variable For any variable YY of of

The set The set z[Y,*]z[Y,*] of variables of of variables of L L represent the represent the

long-code of long-code of yy

LetLet be the Fourier-Coefficient be the Fourier-Coefficient <<|z[y,*]|z[y,*],,ss>>

xA (S)xA (S)

yA (S)yA (S)

3434

The Distributional AssignmentThe Distributional Assignment

DefDef: :

Let Let be a be a distributional-assignmentdistributional-assignment to to as follows:as follows:

For any variable For any variable xx Choose a set Choose a set SSRRxx with probability , with probability , Uniformly choose a random assignment Uniformly choose a random assignment aa S S..

For any variable For any variable yy Choose a set Choose a set S S RRyy with probability , with probability , Uniformly choose a random assignment Uniformly choose a random assignment bb S S..

2

xA (S) 2

xA (S)

2

SyA 2

SyA

3535

Home AssignmentHome Assignment

Given an assignment to a Longcode Given an assignment to a Longcode A:BP[R] A:BP[R] {-1, 1} {-1, 1}, show that for any , show that for any (constant) (constant) > 0 > 0::

| {e | {e R | R | (E(Eee, A) > ½ + ½, A) > ½ + ½ } | } | -2-2

where where (A(A11, A, A22)) is the fraction of bits is the fraction of bits AA1 1 andand A A22 differ on. differ on.

3636

What’s Ahead:What’s Ahead:

We would show that We would show that ‘s expected success ‘s expected success on on xxyy is is > 4> 422 in two steps: in two steps:

First we show (First we show (claim 1claim 1) that ) that ‘s success ‘s success probability, for any probability, for any xxyy is is

Then show (Then show (claim 3claim 3) that value to be ) that value to be 4422

x

y xS R

2 2SA S| y A S

x

y xS R

2 2SA S| y A S

3737

Claim 1Claim 1

Claim 1Claim 1: : The The success probabilitysuccess probability of of on on

xxy y is is

ProofProof::That success probability is at leastThat success probability is at least

and there is at least one and there is at least one bbSS s.t. s.t. bb|y|y S’ S’

Multiplicative representatio

n

Representation by Fourier

Basis

Claim 2

(par[(par[,k]) > 4,k]) > 422

Claim 3:The expected success of the distributional assignment on [C,V]par[,k] is at least 4 4 22

General Fourier Analysis

facts

Claim 1

go to claim3

Ry x

Rx

2 2

b SS' R ,S R

2 2

|yb SS R

b| y S'y x

b| y Sxy |y

A S' A S Pr

A SA S Pr

Ry x

Rx

2 2

b SS' R ,S R

2 2

|yb SS R

b| y S'y x

b| y Sxy |y

A S' A S Pr

A SA S Pr

x

|yS R

22 -1A S A S Sy x

x

|yS R

22 -1A S A S Sy x

3838

Lemma’s Proof - Lemma’s Proof - Claim 2 Claim 2 (1)(1)

Claim 2Claim 2::

ProofProof::The test accepts iffThe test accepts iff z[y, F]•z[x, G]•z[x,F*•G•H] = 1z[y, F]•z[x, G]•z[x,F*•G•H] = 1

By our assumption, this happens with probabilityBy our assumption, this happens with probability

/2+½/2+½..

Now, according to the definition of the expectation: Now, according to the definition of the expectation:

EExxyy, f, g, h , f, g, h [z[y, F]•z[x, G]•z[x, F*•G•H]] = [z[y, F]•z[x, G]•z[x, F*•G•H]] =

1•(1•(½½++/2) + (-1)•(1 -(/2) + (-1)•(1 -(½½++/2)) = /2)) =

go to claim3

y x|y

2 SΕ A S A S 1 2ε δ

x y S Rx

y x|y

2 SΕ A S A S 1 2ε δ

x y S Rx

3939

Lemma’s Proof - Lemma’s Proof - PropositionProposition

X

|S|2F,G,H y x|y

S R

E z y, F z x, G z x, F* G H A S A S 1 2ε

X

|S|2F,G,H y x|y

S R

E z y, F z x, G z x, F* G H A S A S 1 2ε

1 2

Y1 X X2

1 2

Y1 X X2

1 2 22

Y1

F,G,H

y x 1 S x 2 STT R ,S R ,S R

y x 1 x 2 F,G,H S STT R ,S R ,S R

y x 1 x 2 F,G,H S S ST S |yT R ,S S

E z y, F z x, G z x, F* G H

A T F A S G A S F * G H

A T A S A S E F G F * G H

A T A S A S E F F * G G H

C C2

1 2 22

Y1 X X2

x x

,S S

y x 1 x 2 F G S S H ST S |yT R ,S R ,S R

S S2y x x y x|y |y

S R S R

A T A S A S E [ F ] E [ G ] E [χ H ]

A S A S A S 1 1 1 2 A S A S 1 2

1 2

Y1 X X2

1 2

Y1 X X2

1 2 22

Y1

F,G,H

y x 1 S x 2 STT R ,S R ,S R

y x 1 x 2 F,G,H S STT R ,S R ,S R

y x 1 x 2 F,G,H S S ST S |yT R ,S S

E z y, F z x, G z x, F* G H

A T F A S G A S F * G H

A T A S A S E F G F * G H

A T A S A S E F F * G G H

C C2

1 2 22

Y1 X X2

x x

,S S

y x 1 x 2 F G S S H ST S |yT R ,S R ,S R

S S2y x x y x|y |y

S R S R

A T A S A S E [ F ] E [ G ] E [χ H ]

A S A S A S 1 1 1 2 A S A S 1 2

4040

Lemma’s Proof - Lemma’s Proof - Claim 3Claim 3

Claim 3Claim 3: The expected success of the : The expected success of the

distributional assignment distributional assignment

on on xxy y is at least is at least 4422

Proof:Proof: Claim 1Claim 1 gives us the initial lower gives us the initial lower

bound for the expected success:bound for the expected success:

x y

x

2 2 1y x

S R

E (A S | y ) (A S ) | S |

x y

x

2 2 1y x

S R

E (A S | y ) (A S ) | S |

4141

Lemma’s Proof - Lemma’s Proof - Claim 3Claim 3

As we’ve already As we’ve already seenseen,, . Hence, . Hence, our lower-bound takes the form ofour lower-bound takes the form of

Or alternatively,Or alternatively,

Which allows us to use the known Which allows us to use the known inequality inequality E[xE[x22]]E[x]E[x]2 2 and getand get

x y

x x

22 2 1y x x

S R S R

E [( (A S | y ) (A S ) | S | ) ( A S )]

x y

x x

22 2 1y x x

S R S R

E [( (A S | y ) (A S ) | S | ) ( A S )]

12 22

y xS

[( A S | (A S ) | S | ) ]

x yRx

E y 12 22

y xS

[( A S | (A S ) | S | ) ]

x yRx

E y

12 22

y xS

( [ A S | (A S ) | S | ])

x y

xR

E y 12 22

y xS

( [ A S | (A S ) | S | ])

x y

xR

E y

2xA 1

xs R

S 2xA 1

xs R

S

4242

Lemma’s Proof - Lemma’s Proof - Claim 3Claim 3

ByBy auxiliary lemmasauxiliary lemmas (4(4|S|)|S|)-1/2-1/2 e e-2-2|S| |S|

(1-2 (1-2))|S||S|, , i.e.i.e. |S||S|-1/2-1/2 (4(4))1/21/2 ·(1-2 ·(1-2))|S||S|, , which yields the following boundwhich yields the following bound

That is, That is,

Now applying claim 2 results the Now applying claim 2 results the desired lower bounddesired lower bound

x y

x

1 |S|2 22y x

S R

(E [ A S | y (A S ) (4 ) 1 2 ])

x y

x

1 |S|2 22y x

S R

(E [ A S | y (A S ) (4 ) 1 2 ])

x y

|S|2 2y x

S X

4 (E [ A S | y (A S ) 1 2 ])

x y

|S|2 2y x

S X

4 (E [ A S | y (A S ) 1 2 ])

x

|S| 22 2y x

S R

4 (E[ A S | y (A S ) 1 2 ]) 4

x

|S| 22 2y x

S R

4 (E[ A S | y (A S ) 1 2 ]) 4

4343

Lemma’s Proof -Lemma’s Proof -ConclusionConclusion

We showed that there is an We showed that there is an

assignment scheme with expected assignment scheme with expected

success of at least success of at least 4422 , ,

There There existsexists an assignment that satisfies an assignment that satisfies at least at least 4422 of the of the tests in tests in

(() > 4) > 422

Q.E.D.Q.E.D.

4444

Home AssignmentHome Assignment

Show it is Show it is NPNP-hard, for any -hard, for any > 0 > 0, given a , given a 3SAT3SAT instance instance , to distinguish between the case , to distinguish between the case where where (() = 1) = 1, and the case in which , and the case in which (() < ) < 7/8+7/8+

HintHint: Let : Let ’s variables be as in ’s variables be as in LL, and , and ’s clauses ’s clauses to take the formto take the form

f OR g OR ‘f* + g + h’f OR g OR ‘f* + g + h’

for for ff and and gg chosen in the same way as in chosen in the same way as in LL,,while while hh is chosen as follows: is chosen as follows: h(b) = 1h(b) = 1 for for bb such that such that f(bf(b|V|V)) and and g(b)g(b) are both are both FALSEFALSE For all other For all other bb’s, independently for each ’s, independently for each bb, , h(b)=1h(b)=1 with with

probability probability , and , and 00 with probability with probability 1-1-